CALCULATIONS

IN
LABORATORY SCIENCE
UPDATED FOR 2023

ALLAN DEACON
CALCULATIONS IN LABORATORY
SCIENCE

Allan Deacon BSc, PhD, FRCPath

Former Consultant Clinical Scientist, King’s College and Bedford
Hospitals NHS Foundation Trusts

Editors:
Roy Sherwood BSc, MSc, DPhil
Consultant Clinical Scientist, King’s College Hospital, London

James Hooper MD, MSc, FRCPathConsultant Chemical Pathologist, Royal

Brompton Hospital, London

ACB VENTURE PUBLICATIONS

DIRECTOR OF PUBLICATIONS AND COMMUNICATIONS
Chairman - Kamaljit Chatha

British Library Cataloguing in Publication Data

A catalogue record for the book is available from the British Library

ISBN 978-0-902429-43-7, EAN 9780902429437, ACB Venture Publications

Printed by Piggott Black Bear, Cambridge

Cover by Alan Sherwood, Aspire Design Studios, London

© Copyright 2009. Association for Clinical Biochemistry, 130-132 Tooley St, London
SE1 2TU

All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system or transmitted in any form or by any means electronic, mechanical, photocopying,
recording or otherwise, without written permission from the publisher.
Preface
“What one fool can do, another can” (Ancient Simian Proverb)

Clinical biochemistry is predominantly a quantitative science in which both the

acquisition and utilisation of laboratory data requires some understanding of the
underlying mathematical principles involved. The Royal College of Pathologists
acknowledges the importance of this by including calculation questions in the
practical section of part 1 of its fellowship examination in chemical pathology
(FRCPath). Examiners report that these questions are often done badly and
candidates display not only poor numeracy skills, but a lack of understanding of
the basic underlying physical chemistry and physiology. Sadly, calculations
receive only scant attention in most textbooks and on most undergraduate and
postgraduate courses.

I have previously tried to address this problem in two ways. Firstly, by holding
tutorials: initially for local trainees in the Clinical Biochemistry Department at
King’s College Hospital and in later years on regional and national training courses
organised by the Association for Clinical Biochemistry. Secondly, by publishing
worked answers to past FRCPath examination calculation questions in the
ACB News. However, trainees often express the need for a comprehensive textbook
which not only presents worked examples, but brings together the relevant
mathematics and basic science. This book is an attempt to meet that need.

This book was originally intended for trainees in clinical biochemistry, particularly
those preparing for the FRCPath examination. However, I hope it will also prove
useful to trainees in other clinical sciences and to undergraduate and postgraduate
students in any of the life sciences. Each chapter takes a topic and explains the
relevant physical chemistry and/or physiology. I have tried to derive the various
formulae and mathematical procedures from first principles whenever possible in
the belief that an understanding of their basis leads not only to their correct
application, but helps the reader develop these methods to new problems which
he/she may encounter in the future. Questions are included in the text, which I
hope the reader will attempt before looking at the worked answer. At the end of
each chapter there is a set of additional questions, the answers to which are
provided in the Appendix.

Allan Deacon
March 2009
Contents

Chapter Page

1. Units and their manipulation 1

2. Laboratory manipulations 13

3. Acid-base, pH and buffers 27

4. Spectrophotometry 51

5. Renal function 77

6. Osmolality 105

7. Basic pharmacokinetics 113

8. Body fluids and electrolytes 139

9. Enzymology 157

10. The basis of statistics 193

11. Analysis of means and variance 219

12. Correlation and regression 239

13. Clinical utility of laboratory tests 265

14. Statistical power 287

15. Miscellaneous topics 297

Appendix: Answers to additional questions 333

Worked answers to further questions 339

Index 492
Important notice
Although ACB Venture Publications has made every effort to ensure the accuracy
of the information contained in this book, the responsibility for the patient is
ultimately that of the medical practitioner ordering or performing/supervising the
investigations. All drugs and intravenous fluids must be prescribed by a registered
medical practitioner and administered by an individual authorised to do so. The
publishers, authors and editors do not assume any liability for any injury and/or
damage to persons or property arising from this publication.
 UNITS AND THEIR MANIPULATION

Chapter 1

Units and their manipulation

“A numerical result without units is meaningless”

Failure to pay attention to units is probably the commonest cause of error when
performing even the simplest of calculations. In clinical biochemistry we often use
several units for the same property. For example concentration may be expressed as
mmol/L, mg/100 mL or g/L and volume may be expressed as L, mL or μL. Therefore
ability to manipulate units is an essential prerequisite for the successful completion of
most calculations.

Units used in clinical biochemistry

International standardization of units has obvious advantages. In the UK and Europe

clinical laboratories have attempted to adopt Systeme International d’Unites (SI units),
whereas in the USA mass units are still used.

The SI units for the three basic dimensions are:

Length - metre (m)

Mass - kilogram (kg)
Time - second (s)

Under this system, volume is defined as a cubic quantity of length, for example m3, cm3.
However, most laboratories (and scientific journals) have retained the old metric unit of
volume, the litre (L). 1L is equivalent to the SI volume of 1 dm3.

Where the molecular weight (MW) of the substance being measured is known, the SI unit
of quantity is the mole or multiple of a mole but the metric unit of volume, the litre, is
still used. For example, glucose concentration is expressed as mmol/L. One mole is the
formula weight of a substance measured in grams. Often the term molar is used instead of
mol/L and is abbreviated to M. A term in common use for the symbol “ / ” is “per” and
means the value obtained when one quantity is divided by another. Alternatively, the
divisor may be written to the power of “ -1”. For example 1 mol/L may also be written
1 mol L-1.

1
CHAPTER 1

Question: Q1(1)

180 g of glucose (formula C6H12O6) is dissolved in 1 L of distilled water. What is the

concentration of glucose in (a) g/L, (b) mg/100 mL and (c) mol/L?

Answer: Q1(1)

a) Since 180 g of glucose was dissolved in 1L of water the concentration is 180 g/L.

b) “Milli” means a thousandth of. Therefore 1 g of glucose contains 1000 mg.

Therefore 180 g contains 180 x 1000 = 180,000 mg.
Similarly, 1L of water contains 1000 mL which is 10 multiples of 100 mL.
Therefore the concentration is 180,000/10 = 18000 mg/100 mL.
NB. mg/100 mL is sometimes written mg% or mg/dL.

c) The molecular weight of glucose is (6 x 12) + (12 x 1) + (6 x 16) = 180.

Therefore 180 g of glucose contains 180/180 = 1 mole. Since it is dissolved in
1 litre of water the concentration is 1 mol/L or 1 mol L-1 or 1 M.

When the molecular weight is not known, or the analyte being measured is a mixture of
substances of differing molecular weights (such as plasma total protein) then mass units
may be used. e.g. plasma total protein is expressed as g/L.

When concentration is measured indirectly by some property which is not easily related
to a given mass of analyte (e.g. enzyme activity), or using a method calibrated against a
standard of unknown purity (e.g. some hormones) then arbitrary units are used. e.g.
creatine kinase activity is expressed as U/L. When an internationally accepted standard
preparation is used to calibrate the assay, results may be expressed as IU/L.

Dealing with both large and small numbers

Units are often adapted by adding a prefix to simplify dealing with very small and very
large numbers. For example, volume may be expressed in litres, mL or μL and
concentration may be expressed as mmol/L or μmol/L, which may cause problems, for
example, when calculating glomerular filtration rate from creatinine concentrations
expressed as μmol/L in blood and mmol/L in urine, particularly when by convention the
result is required in mL/min. This practice has arisen in order to avoid numbers which
are clumsy. For example, 25 μL is far more convenient to handle than 0.000025 L. A list
of common prefixes is given in Figure 1.1.

2
 UNITS AND THEIR MANIPULATION

Name Symbol Meaning Example

kilo- k 103 x kg body weight

deci- d 10-1 x dL (0.1L or 100 mL)

centi- c 10-2 x cm
(1/100th of a metre)

milli- m 10-3 x mmol

(1/1000th of a mole)

micro- μ 10-6 x μL
(1/1,000,000 th of a litre)

nano- n 10-9 x nm
(1/1,000,000,000th of a metre)

pico- p 10-12 x pg
(1/1,000,000,000,000th
of a gram)

femto- f 10-15 x fmol

(1/1,000,000,000,000,000
of a mole)

Figure 1.1 Prefixes for powers of 10 with common applications

3
CHAPTER 1

An alternative approach is to express a numerical value as a number in the range 1-10

multiplied by an appropriate power of 10. For example, the weight 0.005 g could be
written as 5.0 x 10-3 g as well as 5 mg. As will be discussed later, this practice enables a
numerical result to be expressed to a given number of significant figures.

The general term 10x means 10 multiplied by itself x number of times, so that the
expression y x 10x means y multiplied by 10, x times. Since our number system is based
on 10 all this means is moving the decimal place to the right x number of times. For
example 3.125 x 102 simply means that the decimal point is moved 2 places to the right
to become 312.5. The number 14 x 103 does not have a decimal point (or to be correct it
is implied that the decimal point is placed after the last digit, i.e. 14) so instead of moving
the decimal point 3 digits to the right 3 noughts are added to give 14000. Similarly,
for 6.25 x 104 the decimal point is moved two places to the right then two noughts added
to give 62500.

On the other hand, 10-x means 1 divided by 10x i.e. 1/10x (in a similar way that 1 mol L-1
means 1 mol/L). Therefore, instead of moving the decimal point one place to the right or
adding a nought for each increment in x, the opposite applies and the decimal point is
moved to the left; when the number of digits is exhausted, noughts are placed in front of
it. For example 6834 x 10-3 means 6.834, 24.52 x 10-2 means 0.2452 and 6.35 x 10-4
means 0.000635.

Significant figures
The way in which a numerical value is written makes a statement about the reliability or
accuracy of that value. For example, the concentration of an analyte determined by a
colourimetric assay written as 0.103562 mol/L implies a greater degree of accuracy (and
hence precision of its method of measurement) than if it were written 0.104 mol/L. This
result must have been calculated from an absorbance reading with only three (or possibly
four) digits, so it is misleading to quote the result to 6 significant figures. The reliability
of the measurement process permits the reporting of the result to only 3 significant
figures at the most.

As a general rule one should express a result with no more (or only a little more)
precision than the accuracy of the data from which it was calculated. When the result is
calculated from more than one piece of data then the accuracy of the least precise piece
of data should be used.

4
 UNITS AND THEIR MANIPULATION

For example, consider a creatinine clearance calculated from plasma and urine creatinine
measurements and a 24 h urine volume of 1836 mL. Creatinine measurements are made
to only 3 figures and the reliability of the third must be doubtful. Although the volume is
known to 4 figures, the accuracy of timed urine collections is notoriously poor and the
volume is only likely to be accurate to 2 figures at the most.

The accuracy of the final result of a calculation can never be greater than that of the least
accurate measurement used in its calculation.

When the final result has been calculated and clearly has a greater number of significant
figures than is warranted, the unwanted digits can be removed either by rounding or
truncation. In rounding, the value of the retained digit is increased by 1 only if the
discarded digit(s) begin with 5, 6, 7, 8 or 9. In truncation the unwanted digits are simply
removed. For example the value 2.3478 can be rounded to 2.35 or truncated to 2.34.
Rounding is slightly more accurate than truncation and is preferred.

During the intermediate stages of a calculation, errors can accumulate if intermediate

values are rounded off to the desired final number of significant figures. A fairly reliable
rule is:

Use one more significant figure in the calculations than one expects to retain in the final
result.

The exception to this rule is calculations which involve small differences between nearly
equal numbers. For example, if a beaker weighs 20.4675 g empty but 20.5796 g after
addition of a sample then it would be wrong to subtract 20.5 from 20.6 since this would
give only one significant figure in the answer, whereas the data would clearly support
more.

Sometimes it is difficult to convey the number of significant figures in an answer. For

example a result of 100 g contains only one significant figure (zeros do not count!) One
solution is to place a decimal point after the result so that it becomes 100. which indicates
three significant figures. It is more difficult to convey two significant figures.
One solution it to change the units and express the result as 0.1 kg (1 significant figure),
0.10 kg (2 significant figures) or 0.100 kg (3 significant figures). Alternatively the result
could be expressed as a number multiplied by 10 to the power of another number e.g.
1 x 102 g. It is then easy to convey the number of significant figures e.g. 1 x 102
(one significant figure), 1.0 x 102 (two significant figures) or 1.00 x 102 (3 significant
figures) etc.

5
CHAPTER 1

What happens to units during a calculation?

There are two general rules which must be remembered:

• For the operations of addition and subtraction, the dimensions and the units must
be the same and remain unchanged after the calculation.

• For the operations of multiplication and division, the dimensions are multiplied
and divided just as are the numbers, the result being the product or quotient of the
dimensions.

For example, to add together the two weights 0.952 g and 0.23 mg it is necessary to either
convert the first weight to mg then add it to the second, yielding a result in mg, or, to
convert the second weight to grams then add it to the first, in which case the answer will
be in g.

The calculation of molar absorptivity involves combining several units. The expression
for calculating molar absorptivity (ε) is:

ε = Absorbance
Concentration x path length

Where as absorbance does not have any units, concentration is in mol/L and the optical
path length is expressed in cm. Substituting these units into this expression gives:

ε = 1 = L
mol/L x cm mol x cm

so that the units for ε are L/mol/cm or L. mol-1 cm-1.

Care needs to be taken when calculating ratios of concentrations. If the two analytes and
their units are identical, then the result does not have any units. For example, to calculate
the ratio of plasma to urinary calcium, when both are expressed in the same units, then
the units cancel:

Plasma calcium (mmol/L) = mmol x L

Urine calcium (mmol/L) = L x mmol

6
 UNITS AND THEIR MANIPULATION

If both concentrations are in mass units, then both concentrations can be converted to SI
units by multiplying the mass concentration by the atomic weight (40). Since both
operations are identical, the ratio will be the same and will not have any units:

Plasma calcium (mmol/L) = Plasma calcium (mg/L) x 40

Urine calcium (mmol/L) Urine calcium (mg/L) x 40

If the concentrations are for different analytes then this is no longer true. The ratio of
their concentrations calculated from mass units will be different to that calculated from SI
units. For example, calculation of the calcium:creatinine ratio in a random urine in which
the concentration of calcium is 2.5 mmol/L and that of creatinine is 5.0 mmol/L gives:

Urine calcium = 2.5 mmol/L = 0.5

Urine creatinine 5.0 mmol/L

If the concentrations are converted to mass units by multiplication of the calcium

concentration by its atomic weight (40) and the creatinine concentration by its molecular
weight (113) then a different ratio is obtained:

Urine calcium (mg/L) = 2.5 (mmol/L) x 40 = 0.18

Urine creatinine (mg/L) 5.0 (mmol/L) x 113

Thus whilst SI concentrations of different analytes are comparable, mass concentrations

are not. 1 mg of calcium is not equivalent to 1 mg of creatinine. To avoid confusion it is
best to give the units of the components of the ratio in parentheses after the ratio i.e. 0.5
(mol/mol) or 0.18 (g/g).

Question Q1(2)

a) 45 mL of solution A and 2.65 L of solution B are mixed. What is the total

volume?

b) How many grams of a substance are contained in 350 mL of a solution containing

50 g/L?

7
CHAPTER 1

Answer Q1(2)

a) Total volume = Volume A + Volume B

The units of the two volumes are different (i.e. 45 mL and 2.65 L)
Before they can be added together one must be converted to the other so
that the units are the same. There are 1000 mL in a L, therefore if 45 mL
is divided by 1000 then it becomes 0.045L. (Alternatively 2.65 L could be
multiplied by 1000 to convert it to 2650 mL).

Total volume = 0.045 L + 2.65 L = 2.695 L (or 2695 mL)

b) Amount of substance = volume x concentration

Both volume and concentration contain a volume term, but their units are
different i.e. mL and L. These need to be converted to the same units. If
the volume is converted from mL to L by dividing by 1000 then it
becomes 0.35 L So that:

Amount of substance = 0.35 x 50 = 17.5

To find out what units to use, carry out the calculation with units rather
than numbers:

Amount of substance = L x g/L which can be written L x g

L

Since the litres (the Ls) cancel, then the final units are g
so the amount of substance contained in 350 mL is 17.5 g.

Interconversion of mass and SI units

Since SI units of concentration used in clinical biochemistry are mol/L or multiples of
mol/L i.e. mmol/L, μmol/L etc, then conversion to and from mass units requires
knowledge of the molecular or atomic weight. The two sets of units are related by the
expression:

Concentration (mol/L) = Concentration (g/L)

Molecular or atomic weight

8
 UNITS AND THEIR MANIPULATION

This relationship still applies to units with different prefixes as long as the same prefix is
used on both sides of the equation. For example if the mass concentration is in mg/L then
in SI units the concentration will be in mmol/L and not mol/L; and if the mass
concentration is in mg/100 mL then the SI units will be mmol/100 mL. The above
relationship can be manipulated to carry out the reverse calculation, e.g. to convert
mmol/L to mg/L.

Use of equivalent weights

Chemists have, in the past, attempted to simplify calculations by introducing the concept
of equivalent weight. The equivalent weight of one substance reacts exactly with the
equivalent weight of another. For example in titrimetric analysis one mole of
hydrochloric acid neutralises one mole of sodium hydroxide, whereas one mole of
sulphuric acid (which yields two titratable hydrogens) will neutralize two moles of
sodium hydroxide. Therefore one mole of sulphuric acid is equivalent to two moles of
hydrochloric acid. These principles also apply to redox reactions involving metal ions.
Monovalent ions, such as sodium and potassium, are equivalent to one hydrogen ion so
that their equivalent weights are equal to their atomic weights. However, divalent ions ,
such as calcium and magnesium are equivalent to two hydrogen ions and their equivalent
weight is one half their atomic weights. In general:

Equivalent weight = molecular weight

Valency

number of equivalents = weight in g

Equivalent weight

number of equivalents = weight in g x valency

molecular weight

In the case of univalent ions the units will be numerically the same e.g. 140 mmol/L of
sodium is the same as 140 mEq/L. For divalent ions 1 mol contains 2 Eq, e.g. 1 mmol/L
of magnesium is the same as 2 mEq/L. Results are no longer reported as mEq/L in the
UK but may be found in the literature. The term “normal,” often abbreviated as N, may
also be encountered which is the concentration in equivalents per litre e.g. 1 molar
(or M) saline (1 mol/L) is equivalent to 1 normal (or N) saline (1 Eq/L). This should not
be confused with “normal” or “physiological saline (9 g/L).

Question: Q1(3)

a) Express 360 mg% glucose as mmol/L

b) Express 140 mmol/L saline as g/100 mL
c) Convert 5.64 mEq/L calcium to mmol/L.

9
CHAPTER 1

Answer Q1(3)

a) mmol/L = mg/L
MW

First calculate the molecular weight of glucose – formula C6H12O6

C6 = Atomic weight of glucose x 6 = 12 x 6 = 72

H12 = Atomic weight of hydrogen x 12 = 1 x 12 = 12
O6 = Atomic weight of oxygen x 6 = 16 x 6 = 96
SUM: 180

Glucose concentration is 360 mg/% i.e. 360 mg/100 mL

Concn (mg/L) = concn (mg/100mL) x 10 = 360 x 10 = 3600 mg/L

And glucose concn (mmol/L) = 3600 = 20 mmol/L

180

b) g/L = mol/L x MW

Sodium chloride concn = 140 mmol/L = 140 = 0.14 mol/L

1000
(Since there are 1000 mmol in a mol)

Molecular weight of sodium chloride (NaCl) = 23 + 35.5 = 58.5

NaCl (g/L) = 0.14 x 58.5 = 8.19 g/L

And NaCl (g/100 mL) = g/L = 8.19 = 0.82 g/100 mL (2 sig figs)
10 10

d) Calcium is a divalent cation (Ca++) and therefore equivalent to two hydrogens

Therefore mEq/L = mmol/L x 2

And calcium concentration (mmol/L) = mEq/L = 5.64 = 2.82 mmol/L

2 2

10
 UNITS AND THEIR MANIPULATION

Blood urea nitrogen (BUN)

It is common practice in the USA to express plasma urea concentration in terms of its
contribution to the nitrogen content of plasma, usually using mg% (i.e. mg/100 mL or
mg/dL) as units. The formula of urea is CO(NH2)2 so that each molecule contains
2 nitrogen atoms. Therefore one mole of urea contains one mole of molecular nitrogen
(N2) and the molecular weight of nitrogen (N2) is twice the atomic weight (14) i.e. 28.
Since:

Concentration (g/L) = Concentration (mol/L) x Molecular weight

we can write for urea:

BUN (mg/L) = Urea (mmol/L) x 28

and if BUN is to be expressed as mg/100 mL (mg%), then both sides are divided by 10:

BUN (mg/100mL) = Urea (mmol/L) x 28 = Urea (mmol/L) x 2.8

10
Alternatively, working in atoms of nitrogen rather than moles of molecular nitrogen, the
atomic weight of 14 is used but the molar concentration of urea is still multiplied by 2
since one mole of urea contains 2 atoms of nitrogen (as compared to one mole of
nitrogen):

BUN (mg/L) = 2 x urea concentration (mmol/L) x atomic weight (14)

The overall result is the same. The decision of whether to work in atoms or moles of
nitrogen often causes confusion. The two expressions for the inter-conversion of BUN
and urea concentrations are:

BUN (mg/100mL) = Urea (mmol/L) x 2.8

Urea (mmol/L) = BUN (mg/100mL)

2.8

Question Q1 (4)

Express BUN 14 mg% as urea concentration in mmol/L.

11
CHAPTER 1

Answer Q1 (4)

Urea (mmol/L) = BUN (mg%)

2.8

Substituting BUN = 14 mg%

Urea (mmol/L) = 14 = 5 mmol/L

2.8

Further questions

Atomic weights: C = 12; H = 1; O = 16; Ca = 40; N = 14

1. Convert the following: a) 125 mg% to g/L; b) 0.25 mol/L to mmol/L; c) 0.236
nmol/L to μmol/L; d) 1.6 mg/L to ng/mL.

2. Convert the following concentrations to “SI” units: a) plasma glucose 120 mg%;
b) serum calcium 4.0 mEq/L; c) BUN 21 mg%; d) Serum creatinine 0.66 mg%.

3. Convert the following: a) plasma glucose from 20 mmol/L to mg/100 mL; b)

serum calcium from 3.2 mmol/L to mEq/L; c) serum urea from 30.6 mmol/L to
mg% BUN; d) serum creatinine from 250 μmol/L to mg%.

4. Convert the following: a) 1.5 x 10-3 M to mmol/L; b) 1.25 x 10-5 M to μmol/L;

c) 2.5 x 102 mg/100 mL to g/L; d) 3.25 x 10-6 mmol/L to μmol/L.

5. After incubation of an enzyme with substrate for 30 min the concentration of

product in the reaction mixture was 3.00 x 10-3 M. a) How many mmol of
product would be present in 100 mL of the reaction mixture; and b) what is the
rate of formation of product in 250 mL of reaction mixture expressed as
μmol/min?

6. If an acid dissociates in solution to give its conjugate base and hydrogen ions,
what are the units of its dissociation constant if urine contains 0.1 M of
undissociated acid, 25 x 10-5 mol/L of its conjugate base and 120 nmol/L of
hydrogen ions? NB the dissociation constant is the product of the concentrations
of conjugate base and hydrogen ions divided by the concentration of
undissociated acid.

12
 LABORATORY MANIPULATIONS

Chapter 2

Laboratory manipulations

In this chapter calculations involved in common laboratory manipulations such as

preparing solutions of desired concentrations, preparing dilutions from a stock solution,
calculating concentrations obtained on mixing solutions etc are described.

Preparing solutions from a solid material

If the required concentration is in mass units then calculation of the amount to be

weighed out is a relatively simple matter provided attention is paid to the units involved
For example to prepare 1 L of a solution containing 10 g/L, 10 g of the substance will
need to be weighed out. The volume of solution required needs to be taken into
consideration as well as the final concentration. To prepare 500 mL of the same solution
then half the amount would be required i.e. 5 g. In general:

Weight required = Required concentration x volume

It is important that the units should be compatible. If the concentration is in g/L then the
volume should be expressed as litres and the calculated amount to be weighed out will be
in grams.

Question Q2(1)

Calculate the number of grams of glucose needed to prepare 2 L of a solution with a

concentration of 150 mg/dL (150 mg%).

13
 CHAPTER 2

Answer Q2(1)

Since the weight is required in grams and the final volume in litres then the final
concentration is first converted from mg/dL to g/L.

Since there are 1000 mg in a gram, 150 mg is equivalent to 150 / 1000 = 0.15 g

Since there are 10 dL in 1 L then 1 dL is equivalent to one tenth of a litre = 0.1 L

Therefore the final concentration is 0.15 g / 0.1 L. Multiplication by 10 converts to g/L

i.e. 0.15 x 10 = 1.5 g/L

Therefore weight required (g) = Concentration (1.5 g/L) x final vol (2 L) = 3.0 g

If the target concentration is given in SI units then the weight of the substance to be
weighed out must be calculated using both the molecular weight (MW) of the substance
and the final volume required. It is usually simplest to first convert the target
concentration from SI to mass units:

Concentration (mass units) = Concentration (SI units) x Molecular weight

Again the prefix to the concentration terms (i.e. m, n or μ) must be the same for both the
mass and SI units. For example, if the SI concentration is in mmol/L then the mass
concentration will be in mg/L.

Suppose we needed to prepare 500 mL of a solution of sodium chloride with a

concentration of 140 mmol/L. The first thing would be to calculate the concentration of
sodium chloride in mass units (i.e. mg/L):

The atomic weights of sodium and chlorine are 23 and 35.5 respectively. Therefore the
molecular weight of sodium chloride (NaCl) is 23 + 35.5 = 58.5.

NaCl concentration (mg/L) = 140 (mmol/L) x MW (58.5) = 8190 mg/L

The amount of NaCl required to prepare 500 mL of solution will be half of this:
8190 / 2 = 4095 mg. Since most balances have scales in g rather than mg, division by
1000 (since there are 1000 mg in a g) gives the weight in g (4095/1000 = 4.095 g).

Sometimes the chemical required to prepare a solution may not be in exactly the same
form as that described in a method. For example, a method sheet for a manual glucose

14
 LABORATORY MANIPULATIONS

method dictates that 1 L of a 500 mg% solution of glucose is prepared by dissolving

5.00 g of glucose in water and making the final volume up to 1 L. However, if only
glucose monohydrate is available then a glucose solution containing the same
concentration of glucose can still be prepared if the difference in molecular weights is
taken into account. First convert the glucose concentration to SI units. The molecular
weight of glucose (formula C6H12O6) is 180. Since 5.00 g of glucose is normally
weighed out and made up to 1L the final concentration is 5.00 g/L.

Therefore glucose (mol/L) = glucose (g/L) = 5.00 = 0.0278 mol/L

Molecular weight 180

Since one mole of glucose monohydrate (formula C6H12O6.H2O) contains one mole of
glucose, then the concentration of glucose monohydrate will also be 0.0278 mol/L.
Conversion of the glucose monohydrate concentration to mass units will give the weight
of glucose monohydrate to be weighed out.

MW of glucose monohydrate = MW glucose + MW water = 180 + 18 = 198

Glucose monohydrate (g/L) = Glucose monohydrate (mol/L) x MW

= 0.0278 x 198 = 5.50 g/L

Therefore 5.50 g of glucose monohydrate will need to be weighed out instead of 5.00 g of
glucose. Note that the same result can be obtained by multiplying the weight of glucose
by the ratio of the molecular weights of the hydrated to the anhydrous form:

Wt glucose monohydrate (g) = wt glucose (5.00g) x 198 = 5.50 g

180

Question 2(2)

a) How many grams of anhydrous disodium hydrogen phosphate will be needed to

prepare 2 litres with a concentration of 50 mmol/L.

b) Instructions for preparing 1L of a phosphate buffer state that 12.00 g of anhydrous

sodium dihydrogen phosphate are required. If this material is unavailable how
many grams of sodium dihydrogen phosphate dihydrate would be required?

(MWs: Na = 23, P = 31)

15
 CHAPTER 2

Answer Q2(2)

a) First calculate MW of anhydrous disodium hydrogen phosphate (Na2HPO4):

Na2 = 2 x 23 = 46
H = 1 = 1
P = 31 = 31
O4 = 4 x 16 = 64
142

Next convert the concentration from mmol/L to mol/L then to g/L:

Since there are 1000 mmol in 1 mol,

50 mmol/L is the same as 50 = 0.05 mol/L

1000

Concentration (g/L) = Concentration (mol/L) x MW

= 0.05 x 142 = 7.1 g/L

To prepare 2 L, twice this amount will be needed, i.e. 2 x 7.1 = 14.2 g

b) MW NaH2PO4 = 23 + (2 x 1) + 31 + (4 x 16) = 120

MW NaH2PO4.2H2O = 120 + 2 (2 + 16) = 156

12.0 g/L of NaH2PO4 is equivalent to 12.0 = 0.1 mol/L

120

0.1 mol/L NaH2PO4.2H2O contains 12.0 x 156 = 15.6 g/L

120

Correcting for purity

For many chemicals used in the laboratory the percentage purity is significantly less than
100%. If a compound has a purity of x %, this means that each 100 g of the material
contains x g of the compound. It follows that each g will contain only x/100 g.
Therefore, to prepare a solution containing W g/L then the weight required is W x 100 /x
g/L. In general:

16
 LABORATORY MANIPULATIONS

Weight of impure material = Weight of pure material x 100

% purity

Again the weight units must be the same on both sides of the expression.

Preparing solutions from liquids

Some chemicals used to prepare reagents are themselves liquids. If the liquid is weighed
then the procedure is the same as for solids, providing allowance is made for any
departure from a purity of 100%. However, if the liquid is measured in volume then
allowance must be made for the fact that most liquid chemicals do not have a density of
one.

The units of density are weight/volume. If a liquid has a density of x g/mL, then this
means that each mL contains x g of the compound. In general:

Density = weight
volume

The term specific gravity (SG) is often used. This is the density expressed as a ratio to the
density of water (which is 1 g/mL). For most practical purposes SG and density can be
considered as being the same thing (although SG, being a ratio, does not have units). By
re-arranging the above equation, it is a simple matter to calculate the volume which
contains the target weight of a compound. Suppose we wished to prepare 1 L of an
ethanol standard solution, containing 800 mg/L of ethanol, from ethanol which has an
SG of 0.79. The units for density must be the same as for concentration, the density
(=SG) is in g/mL, the weight of ethanol per litre must also be in g (i.e. 800 mg / 1000 =
0.8 g /L).

Volume (mL) = weight (g) = 0.8 = 1.01 mL

Density (g/mL) 0.79

Question Q2(3)

How many mL of hydrochloric acid (SG 1.16) are required to prepare 500 mL of 2.0
molar hydrochloric acid. The purity of the acid is 32 % w/w.

17
 CHAPTER 2

Answer Q2(3)

MW hydrochloric acid (HCl) = 1 + 35.5 = 36.5

Weight (g) of pure acid required to make 1 L 2.0 M HCl = 2.0 x 36.5 = 73 g

Weight required to make 500 mL 2.0 M HCl = 73 / 2 = 36.5 g

Since HCl has a purity of 32 %w/w, the weight of HCl (SG 1.16) required is

36.5 x 100 = 36.5 x 100 = 114.0 g

% purity 32

Using the density of 1.16, the volume can be calculated:

Volume (mL) = weight (g) = 114.0 = 98.3 mL

Density (g/mL) 1.16

Dealing with dilutions

In the laboratory we often need to calculate the final concentration of a substance after a
given dilution, the volume of a stock solution which has to be diluted to give a target final
concentration or how much liquid to add to a set volume of a stock solution to give a
desired concentration. All of these problems are variations on a single theme and are
approached in the same way. First it is important to realise that the total amount of a
substance (whether expressed in mass or SI units) in a solution is the product of
concentration and volume (the volume in the concentration term must be in the same
units as the volume of solution).

Total amount = concentration x volume

For example:

2 L of 0.1 M sodium hydroxide (0.1 mol/L) will contain 2 x 0.1 = 0.2 mols

500 mL of 100 mM glucose (100 mmol/L) will contain 0.5 x 100 = 50 mmol

1.5 L of 20% sodium chloride (20% = 20 g/100 mL = 200 g/L) contains 1.5 x 200 =
300 g

18
 LABORATORY MANIPULATIONS

If a finite amount of solution is diluted with solvent then the total amount of the solute in
the final solution will be the same. It is only the volume (which has become larger) and
the concentration (which has become lower) that have changed:

Initial amount of solute = Final amount of solute

Since the amount of a solute is the product of concentration and volume, then the
following expression can be written:

initial concentration x initial volume = final concentration x final volume

All dilution problems can be solved using this equation. If any three of the terms are
known then this formula can be rearranged to obtain the remaining term. In the
laboratory there are four applications which are commonly used:

1. The concentration of an analyte in a biological sample may exceed the working

range of the assay. In this situation it is common practice to dilute the sample
before carrying out the analysis and then to calculate the concentration in the
undiluted sample. For example, 0.1 mL of urine is diluted to 2.0 mL with water;
the creatinine concentration measured on the diluted urine is 150 μmol/L.

Initial concentration = ? Final concentration = 150 μmol/L

Initial volume = 0.1 mL Final volume = 2.0 mL

initial concentration x initial volume = final concentration x final volume

This equation can be rearranged if both sides are divided by the initial volume
(the initial volume terms on the left hand side cancel each other).

initial concentration = final concentration x final volume

initial volume

= 150 x 2.0 = 3000 μmol/L = 3.0 mmol/L

0.1

19
 CHAPTER 2

In other words the result for the diluted sample is simply multiplied by the
dilution (2.0/0.1 = 20). If the dilution is prepared manually, it is often simpler to
add 0.1 mL of urine to 2.0 mL of water, giving a final volume of 2.1 mL. The
method of calculation is exactly the same, and the final dilution will be 21
(2.1/0.1) instead of 20.

2. Calculation of the volume of a stock solution of a chemical which will need to be

diluted to produce a given volume of solution with a target concentration. In this
situation the initial volume is unknown but the initial concentration, final volume
and final concentration are all known. For example, we may wish to calculate
how much 1.0 M hydrochloric acid will need to be diluted to 100 mL to give a
concentration of 0.025 M.

Initial concentration = 1.0 M Final concentration = 0.025 M

Initial volume = ? Final volume = 100 mL

initial volume = final concentration x final volume

Initial concentration

= 0.025 x 100 = 2.5 mL

1.0

3. Calculation of the volume of diluent to be added to a given volume of stock

solution to achieve a target concentration. This is often necessary if we wish to
make best use of all of a stock solution remaining in a near empty bottle. For
example, if 325 mL of a stock diluent containing 0.5 M phosphate buffer remains,
how much water will need to be added if we wish to prepare the maximum
volume of buffer with a phosphate concentration of 0.05 M.

Initial concentration = 0.5 M Final concentration = 0.05 M

Initial volume = 360 mL Final volume = ?

Final volume = Initial concentration x initial volume

Final concentration

Final volume = 0.5 x 360 = 3600 mL

0.05

20
 LABORATORY MANIPULATIONS

The volume to be added can be calculated from the initial and final volumes:

Final volume = initial volume + volume added

Volume added = final volume - initial volume

= 3600 - 360 = 3240 mL

4. Calculation of the final concentration achieved when a known dilution is prepared

from a solution of known initial concentration. For example, if 0.1 mL of a
standard solution containing 50 mmol/L glucose is added to 2.0 mL of diluent,
what is the final concentration?

Initial concentration = 50 mmol/L Final concentration = ?

Initial volume = 0.1 mL Final volume = 0.1 + 2.0 = 2.1 mL

Final concentration = Initial concentration x initial volume

Final volume

= 50 x 0.1 = 2.38 mmol/L

2.1

Question Q2(4)

A working reagent for a phosphate assay is prepared by mixing 100 mL of stock reagent
with 900 mL of diluent. If only 360 mL of diluent is available, how much stock reagent
must be added to obtain the maximum volume of working reagent?

21
 CHAPTER 2

Answer Q2(4)

In this situation, the initial concentration of the stock reagent can be regarded as 100%.
The final concentration in the diluted stock will be 10% (since 1 part of stock reagent is
mixed with 9 parts of diluent giving a working dilution of 1 in 10).

Stock reagent Working reagent

Initial concentration = 100% Final concentration = 10%

Initial volume = Volume stock mL Final volume = (360 + Volume stock) mL

Volume stock (mL) x 100% = (360 + Volume stock) mL x 10%

Expanding brackets on the right hand side of the equation:

Volume stock x 100 = 3600 + (Volume stock x 10)

Moving (volume stock x 10) to the left hand side:

(Volume stock x 100) - (Volume stock x 10) = 3600

Taking the “volume stock” term outside of the brackets, re-arranging and solving:

Volume stock (100 - 10) = 3600

Volume stock = 3600 = 3600 = 40 mL

(100 – 10) 90

Preparing a series of dilutions

Often a set of concentrations of a substance will need to be prepared, either to construct a

standard curve, investigate the effect of varying a constituent in an assay or to determine
the Km or Ki of an enzyme. The golden rule is to first prepare a suitable amount of the
solution with the highest concentration, then to set about preparing dilutions from this.
Two approaches can be used, depending on whether or not the steps in concentration
need to be equal.

22
 LABORATORY MANIPULATIONS

For example, to set up a manual glucose method then a likely set of standards to cover the
range of clinical interest would be solutions containing 5, 10, 15, 20 and 25 mmol/L of
glucose. The first step is to prepare a stock glucose solution containing the highest
concentration of glucose required i.e. 25 mmol/L. The lowest concentration required is
5 mmol/L which is 1/5th the concentration of the stock standard. As the steps are equal
(5 mmol/L difference between each adjacent standard) then the second lowest
concentration would be 2/5th of the stock, the next 3/5th etc. If 1 mL of each standard is
required, then the lowest standard will contain 1/5 of 1 which is 0.2 mL of stock, the next
standard will require twice this (0.4 mL of stock) etc. To ensure that the total volume is
the same (1 mL) the volume of diluent (water) will decrease by 0.2 mL each time the
volume of stock increases by 0.2 mL:

Required concentration (mmol/L): 5 10 15 20

Volume of stock glucose (25 mmol/L) 0.2 0.4 0.6 0.8

Volume diluent (water) 0.8 0.6 0.4 0.2

The scale can be altered to either increase or decrease the volumes of standards prepared
– as long as the ratios of stock standard to diluent remain the same. Intermediate
concentrations can be introduced by extrapolation. For example to prepare a glucose
standard containing 7.5 mmol/L, 0.3 mL of stock would be mixed with 0.7 mL of diluent,
and to prepare a standard containing 2.5 mmol/L, 0.1 mL of stock would be mixed with
0.9 mL of diluent.

An alternative approach is to prepare doubling dilutions. This results in a series in which

each increment doubles the concentration of the solute i.e. the increments are not equal.
Doubling dilutions are extremely easy to prepare and afford a means of quickly spanning
a wide concentration range if there is uncertainty of the working concentration range
required. As before, a stock solution is prepared containing the highest concentration
required. A series of tubes is prepared. Each tube contains either stock solution or
diluent, the volume of which is equal to the final volume required for each dilution.
The first tube contains stock solution only, all of the remaining tubes contain diluent
only. An equal volume of diluent is added to the first tube and mixed. The concentration
of the contents of the first tube is therefore ½ of the stock concentration. The same
volume of the contents of the first tube are transferred to the second tube and mixed.
The concentration in the second tube is therefore ½ of the concentration in the first tube
(½ x ½ = ¼). The same volume of the contents of the second tube are then transferred
to the third tube and mixed which gives a concentration ½ of the concentration in the
second tube (½ x ¼ = 1/8). This process can be repeated ad infinitum.

23
 CHAPTER 2

For example, if a stock solution has a concentration of 200 mmol/L, then preparing serial
dilutions would give the following concentrations:

Tube number 1 2 3 4 5 6 7

Vol stock solution (mL) 1.0 - - - - - -

Vol diluent (mL) 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Final concentration (mmol/L) 100 50 25 12.5 6.25 3.125 1.563

1 / final concentration (L/mmol) 0.01 0.02 0.04 0.08 0.16 0.32 0.64

The main drawback (apart from the concentrations not being evenly spaced) is that
awkward numbers are soon encountered. This need not be a problem if the data is
subsequently processed and plotted by computer. However, if the reciprocal of
concentration is to be plotted (e.g. to determine the Km of an enzyme) then the numbers
produced are much easier to handle. If the material is in short supply then preparing
doubling dilutions affords an economical and easy way to produce a wide range of
concentrations provided minimal volumes are used. Serological titrations always employ
doubling dilutions.

Further questions

(Atomic weights: H = 1; C = 12; O = 16; P = 31; Na = 23; K = 39; Ca = 40; S = 32)

1. How many grams of albumin are required to prepare 100 mL of a solution

containing 70 g/L?

2. Calculate the concentration of sodium ions (in mmol/L) in a solution prepared by

dissolving 85 g of sodium chloride in 1 litre of water.

3. What weight of calcium carbonate must be dissolved in 500 mL of dilute acid to

provide a calcium standard containing 5.0 mmol/L ?

4. A solution contains 5 % sucrose. How much of this solution would you dilute to
prepare 500 mL of 1 % sucrose?

5. 50 μL of urine is added to 5 mL of water. What is the resulting dilution of the

urine?

24
 LABORATORY MANIPULATIONS

6. Concentrated sulphuric acid (SG 1.84) is 96% by weight H2SO4. Calculate the
volume of concentrated acid required to prepare 1 L of 0.1M H2SO4.

7. The following solutions were mixed together:

50 mL potassium chloride (5.0 g/L)

100 mL sodium chloride (50 g/L)

Calculate the molar concentrations of potassium, sodium, and chloride ions.

8. If you have available 650 mL of 95 % ethanol, how much water would you add to
obtain the maximum volume of 65 % ethanol?

9. In order to prepare 1 L of a stock standard solution containing 0.2 mol/L, the

appropriate amount of sodium dihydrogen orthophosphate dihydrate should be
weighed out. Due to an error, the same weight of anhydrous sodium dihydrogen
orthophosphate used. Working standard was prepared by taking 5 mL of this stock
standard and diluting it to 250 mL. What is the phosphate concentration (in
mmol/L) of the working standard?

10. Solution A contains 12.0 g of anhydrous sodium dihydrogen phosphate per litre.
What is the phosphate concentration expressed as mmol/L? What volume of
solution A needs to be diluted to 1 L to give a phosphate concentration of
4 mmol/L.

25
CHAPTER 2

26
 ACID-BASE, pH AND BUFFERS

Chapter 3

Acid-base, pH and buffers

What are acids and bases?

An acid is defined as a substance with a tendency to lose a proton (hydrogen ion) and
a base as a substance with a tendency to gain a proton. It follows that there must be
a relationship between an acid and a base. Whenever an acid loses a proton the anion
formed will have a tendency to regain the proton, and hence it will be a base.
Therefore, in general:

AH H+ + B-
Acid Base

The acid and base (which differ only by a proton) are said to form a conjugate pair;
every acid must have its conjugate base, and every base its conjugate acid e.g.

CH3COOH H+ + CH3COO-
acetic acid acetate ion

OH- + H+ H2O
hydroxide ion water

Strictly speaking, an alkali is any substance which can produce hydroxyl ions in
aqueous solution but the terms base and alkali are often interchanged.

Some species, such as the bicarbonate ion can act as both proton donors and acceptors
and therefore function as both an acid and a base:

HCO3- H+ + CO32-
bicarbonate ion carbonate ion

H+ + HCO3- H2CO3
carbonic acid

Water can dissociate to produce a proton and its conjugate base (hydroxyl ion):

H2O H+ + OH-

27
CHAPTER 3

When as acid is added to water this equilibrium is driven to the left and the
concentration of hydroxide ions decrease i.e. H+ >> OH-. When a base
(e.g. hydroxide) is added to water then this equilibrium is again driven to the left but
this time protons are removed i.e. OH- >> H+.

The fact that in aqueous solutions hydrogen ions are hydrated to form hydroxonium
ions (H3O+) and free hydrogen ions do not exist is usually ignored.

How is the degree of acidity/alkalinity expressed?

Since most solutions we encounter are aqueous, water is taken as a reference point,
and any solution in which the hydrogen ion concentration is greater than pure water is
said to be acidic; any solution in which it is less is said to be alkaline.
The dissociation constant (K) for water can be defined as:

K = [H+][OH-] ………………………………………….Eq.3.1
[ H2O ]

where the brackets denote the concentrations of each species in mol/L. The
concentration of water can be regarded as constant (since its concentration is very
high and the molecule is only weakly ionized) and therefore the water concentration
term is incorporated into the ionization constant, which is then referred to as the ionic
product of water (Kw):

Kw = [H+ ] [ OH- ] = 10-14 mol/L ……………………….. Eq.3.2

Since [H+] = [OH-] then the concentration of each of these ions must be 10-7 mol/L.
Therefore any solution in which [H+] >10-7 mol/L will be acidic with the degree of
acidity related to [H+]. Conversely, any solution with [H+] < 10-7 mol/L will be
alkaline.

The range of [H+] usually encountered by the chemist is very wide and quite often the
[H+] is very low e.g. the [H+] of blood is typically 0.000000040 mol/L. In order to
compress the scale and simplify the expression of low concentrations (and hopefully
make calculations simpler) Sorensen, in 1909, devised the logarithmic pH scale.

What are logarithms?

Numbers can be written as some other number (called the base) raised to the power of
another number (called the logarithm) i.e. baselogarithm. The power or logarithm defines
the number of times the base value must be multiplied by itself to give the number.
A few examples are given in Fig 3.1.

28
 ACID-BASE, pH AND BUFFERS

Mathematicians have found that working with logarithms has advantages in some
situations. In clinical biochemistry we only need to use logarithms to two different
bases: 10 (known as common logarithms) and 2.718 (known as natural or Napierian
logarithms). Values for both types of logarithms are readily available on most pocket
calculators or can be obtained from tables of logarithms. For example the logarithm of
31.62 is 1.5, a result which is difficult to obtain by manual calculation. In this chapter
we will only use common logarithms (natural logarithms may be used in later
chapters).

Number Base Logarithm Evaluation

4 2 2 4 = 22 = 2 x 2

8 2 3 8 = 23 = 2 x 2 x 2

16 2 4 16 = 24 = 2 x 2 x 2 x 2

9 3 2 9 = 32 = 3 x 3

16 4 2 16 = 42 = 4 x 4

25 5 2 25 = 52 = 5 x 5

125 5 3 125 = 53 = 5 x 5 x 5

100 10 2 100 = 102 = 10 x 10

1000 10 3 1000 = 103 = 10 x 10 x 10

Figure 3.1 Examples of logarithms

The reverse of a logarithm is the antilogarithm which is the number which gave rise
to the logarithm in the first place. The practical importance of antilogarithms is that at
the end of a calculation we often end up with a result which is a logarithm; we then
need to determine the number which would give rise to this antilogarithm.
Fortunately antilogarithms are also available on most pocket calculators or can be
obtained from tables of antilogarithms (or by using tables of logarithms backwards).
The abbreviations “log” and “antilog” are respectively used for logarithm and
antilogarithm.

29
CHAPTER 3

A general notation relating a number (N) to its logarithm (x) to a base (b) can be used:

If , N = bx e.g. 100 = 102

then
logbN = x e.g. log10100 = 2
and
antilog b x = N e.g antilog102 = 100

Number (N): 0.001 0.01 0.1 1 10 100 1000

Log10N (x): -3 -2 -1 0 1 2 3

3
Log N (x)

0
0.0001 0.001 0.01 0.1 1 10 100 1000
-1
Number (N)

-2

-3

• The logarithm of 1 is zero

• Numbers between 1 and zero have negative logarithms

• Logarithms do not have units

• Negative numbers do not have logarithms

Figure 3.2 Relationship between numbers and their common logarithms

30
 ACID-BASE, pH AND BUFFERS

As illustrated in fig 3.2, log10100 = 2 and log101000 = 3. This process can be

continued indefinitely. For example, log1010000 = 4 , i.e. 10000 = 10 x 10 x 10 x
10 = 104. Therefore, for common logarithms (where the base 10 is used), as a number
increases by a factor of 10, it’s log increases by 1. For example, increasing 100 by a factor of
10 gives 1000 whereas its log increases by 1, from 2 to 3. The converse is true, as a number
decreases by a factor of 10 its log decreases by 1. What would happen if 100 was decreased
by a factor of 10 to become 10? The log would have to decrease by 1 i.e. change from 2 to 1
so that log1010 = 1. This is not really surprising since the logarithm is the number of base
terms which must be multiplied together to give the number; in this case the log is 1 which
means that only one term is used so that the number is unchanged i.e. 101 = 10. If 10
is decreased by a factor of 10 to give 1, then its log must also decrease by 1 and
therefore becomes zero. This gives the surprising result that log101 = 0. If this
process is taken one stage further and 1 is decreased by a factor of 10 to 0.1 then its
log decreases by 1 from 0 to -1, so that log100.1 = -1. For numbers below 1 the
logarithm becomes increasingly negative. As a consequence negative numbers cannot
have logs and the log of zero has no meaning. Fig 3.2 illustrates the relationship
between a number and its common logarithm.

What is pH? Why is it used?

pH is a logarithmic scale devised by Sorensen to simplify expression of the wide

ranges in hydrogen ion concentration encountered by chemists. Since in most
biological systems the hydrogen ion concentration is very low (much less than 1
mol/L) this inevitably means that the logarithm of the molar hydrogen ion
concentration will be a negative number. For example, at a hydrogen ion
concentration of 0.0000001 mol/L the logarithm would be -7. Negative numbers are
inconvenient so the sign is changed to a positive value. Therefore, a value can be
defined called the pH which is the negative logarithm to the base 10 of the molar
hydrogen ion concentration.

pH = - log10 [H+] OR pH = log10 1 ………….. Eq 3.3.

[H+]

Properties of pH are shown in Fig 3.3.

Question: Q3(1)

a) Calculate the hydrogen ion concentration of blood with a pH of 6.95.

b) Treatment with bicarbonate halves the hydrogen ion concentration. What is

the new pH?

31
CHAPTER 3

Answer to Q3(1)

a) Reverse the expression for pH so that the concentration term is on the left:

- log 10 [H+] = pH

Multiply both sides by minus 1 in order to change the signs:

log 10 [H+] = -pH , then take antilogarithms of both sides:

[H+] = antilog 10 (- pH)

substitute pH = 6.95 and evaluate:

[H+] = antilog 10 (- 6.95) = 0.000000112 = 1.12 x 10-7 mol/L

Multiply by 109 to convert to the more manageable nanomolar units:

[H+ = 1.12 x 10-7 x 109 = 112 nmol/L

N.B. 10-7 means “move the decimal point seven places to the left” whereas
109 means “move the decimal point nine places to the right”. The net result is
to move the decimal point 2 places (9-7 = 2) to the right so that 1.12 becomes
112.

b) If the hydrogen ion concentration is halved by the bicarbonate treatment then

the new concentration will be 56 nmol/L or 0.56 x 10-7 mol/L (= 0.000000056
mol/L). Substitute this new value into the expression for pH:

pH = - log 10 [H+] = - log 10 (0.000000056) = - (-7.25) = 7.25

Some calculators are cannot cope a large number of digits. Use can be made of the
following property of logarithms:

log ( A x B ) = log A + log B

In Q3(1) to evaluate the log of 0.000000056 first write the number in exponential
form

log10 0.000000056 = log10 (5.6 x 10-8)

evaluate each component separately then add to give the final result:

log10 (5.6 x 10-8) = log10 5.6 + log10 10-8 = 0.75 + (-8) = - 7.25

32
 ACID-BASE, pH AND BUFFERS

Physiological hydrogen ion concentrations are always in the nanomolar range so a

useful trick to simplify calculations is to keep the nonmolar term separate throughout
the calculation.

Thus, in question Q3(1) above, an alternative expression for pH can be used:

pH 6.95 = 10-6.95

Since 1 mol = 1,000,000,000 nmol, then 1 mol/L can be written as 109 mol/L. If
concentration of hydrogen ions is expressed as nmol/L, then pH 6.95 can be written:

pH 6.95 = 10 9-6.95 = 10 2.05

The antilog of 2.05 is 112 and so the hydrogen ion concentration is 112 nmol/L.

Similarly for part (b) where the concentration of hydrogen ions is halved to
56 nmol/L:

pH = - log10 [H+] = -log10 (56 x 10-9) = - (log10 56 + log10 10-9)

Since log1010-9 is - 9, then pH = - (log1056 - 9) = - (1.75 - 9) = - (-7.25) = 7.25

As

pH = - log 10 [H+] ……………………… Eq 3.3

.
• Since the pH scale is logarithmic (to the base 10) a change in 1 pH unit
corresponds to a 10-fold change in hydrogen ion concentration.

• Since the hydrogen ion concentration of pure water is 10-7 mol/L it follows that
neutral pH is 7.

• pH changes in the opposite direction to hydrogen ion concentration. Solutions

with pH greater than 7 are alkaline and solutions with pH less than 7 are acidic.

• The small ‘p’ means ‘minus the logarithm to the base 10’ and should not be
confused with capital ‘P’ which denotes partial pressure e.g. Pco2.

The expression for pH can be rearranged to calculate [H+ ] from pH:

[H+] = antilog 10 (- pH) ……………………… Eq. 3.4

Figure 3.3 Characteristics of pH

33
CHAPTER 3

WHAT ARE BUFFERS? HOW DO THEY WORK?

A buffer is a solution which resists change in pH when an acid or alkali is added.

Many buffers consist of a mixture of a weak acid (HA) and its salt (MA). The salt
can be regarded as completely ionised, but the weak acid is only partially ionised:

Weak acid: HA H+ + A-

Salt: MA M+ + A-

The dissociation constant of the weak acid ( Ka ) is given by:

Ka = [H+] [A-] ……………………………………..Eq.3.5

[HA]

The high concentration of the common anion (A-) from the salt component suppresses
ionisation of the weak acid (HA). The pH of the buffer solution will therefore depend
on the relative amounts of weak acid and salt present. If a small amount of strong
acid (HX) is added then the concentration of hydrogen ion will increase. To maintain
constant Ka these hydrogen ions are “removed” by combination with the common
anion (A-) to form undissociated weak acid:

H+ + X- + A- HA + X-

Hence the effect of the addition of the strong acid has been buffered. Consequently
the strong acid anion (X-) replaces the common anion (A-) with only a small change in
the [A-]/[HA] ratio so that pH does not change significantly.

If some alkali (XOH) is added then the hydroxyl ions (OH-) react with, and therefore
lower, the hydrogen ion concentration (with the formation of water). In order to
maintain constant [H+][A-]/[HA] ratio more weak acid dissociates so that the H+
consumed by reaction with hydroxyl ion is replaced. This process can be considered
as the titration of hydroxyl ions with weak acid (HA) so that the common anion (A-)
replaces the added hydroxyl ions:

OH- + HA H2O + A-

Buffering results in a relatively constant pH but at the expense of altering the

concentrations of the buffer components i.e. [HA] and [A-].

34
 ACID-BASE, pH AND BUFFERS

BUFFER CALCULATIONS

The expression for the dissociation constant of a weak acid (Eq.3.5) can be written
slightly differently:

Ka = [H+] x [A-]
[HA]

taking logarithms gives:

log10 Ka = log10 [H+] + log10 [A-]

[HA]

Note that the logarithms of two numbers multiplied together is the same as the sum of
their individual logarithms e.g. log (A x B) = log A + log B.

This equation can be rearranged by transferring log10Ka to the right hand side (which
then becomes negative) and log10 [H+] to the left hand side (which also becomes
negative):

-log10 [H+] = -log10 Ka + log10 [A-]

[HA]

-log10 [H+] is the same as pH and a term pKa can be defined as –log10 Ka, so that this
equation becomes:

pH = pKa + log10 [salt] ……………….…. Eq3.6

[ acid]

This is known as the Henderson Hasselbalch equation and can be used to calculate
the amounts of salt and acid which must be used to prepare a buffer solution of a
desired pH. Important quantitative properties of buffers are summarised in fig 3.4.

Question: Q3(2)

Calculate the amount in grams of lactic acid which must be added to 3 g of sodium
hydroxide to give 1 litre of a solution with a pH of 4.5 (pKa of lactic acid is 3.86;
atomic weight of sodium is 23).

35
CHAPTER 3

Answer to Q3(2)

The following reaction occurs between lactic acid (LactH) and sodium hydroxide:

LactH + NaOH LactNa + H2O

Both sodium lactate and lactic acid can dissociate to give lactate ions (Lact-):

LactNa Lact- + Na+

LactH Lact- + H+

The relationship between the concentrations of lactate (a salt) and lactic acid (an acid)
is governed by the Henderson Hasselbalch equation:

pH = pKa + log10 [Lact-]

[LactH]

In order to make use of this equation it is necessary to assume that the concentration
of Lact- is equal to that of sodium, i.e.

1. That sodium lactate is completely dissociated, and

2. That the proportion of Lact- derived from lactic acid is negligible compared to
that derived from sodium lactate. This proportion will be the same as the
hydrogen ion concentration which can be calculated from the pH and is
0.0001 mol/L – clearly insignificant.

Next calculate the molar concentration of sodium hydroxide and use it in place of
[Lact-]:

Molecular weight (MW) of NaOH = 23 + 16 + 1 = 40

Molar concentration of NaOH = Concentration (g/L) = 3.0 = 0.075 mol/L

MW 40

Reaction of sodium hydroxide with lactic acid (LactH) yields approximately an equal
amount of lactate (Lact-) ions:

NaOH + LactH H2O + Lact - + Na+

36
 ACID-BASE, pH AND BUFFERS

Substitute Lact- = 0.075, pH = 4.5 and pKa = 3.86 into the Henderson
Hassebalch equation and solve for [LactH]:

4.5 = 3.86 + log10 0.075

[LactH]

Rearranging gives: 4.5 - 3.86 = log10 0.075

[LactH]

0.64 = log10 0.075

[LactH]

taking antilogs of both sides: antilog 0.64 = 0.075

[LactH]

4.365 = 0.075
[LactH]

Rearranging: [LactH] = 0.075 = 0.0172 mol/L

4.365

The total lactate needed is the sum of the ionised and unionised acid:

Total [Lact] = [LactH] + [Lact -] = 0.0172 + 0.075 = 0.0922 mol/L

Since all of this lactate must originate from the lactic acid added to the sodium
hydroxide, calculate the weight of lactic acid required to give a concentration of
0.0922 mol/L:

Wt lactic acid (g) = Concentration (mol/L) x MW

MW of lactic acid (formula C3H6O3) = (3 x 12) + (6 x 1) + (3 x 16) = 90

Wt of lactic acid required = 0.0922 x 90 = 8.30 g (2 sig figs)

Question: Q3(3)

A buffer is required for a chromatographic procedure which has a pH of 7.0 and a

total phosphate concentration of 0.1 mol/L. Calculate the amounts of anhydrous
sodium dihydrogen phosphate and disodium hydrogen phosphate which need to be
weighed to produce 1 litre of buffer. The pKa of the dissociation is 6.82. (Atomic
weights: Na = 23, P = 31).

37
CHAPTER 3

Buffers are a mixture of a weak acid and its salt, the pH of which is defined by
the Henderson Hasselbalch equation:

pH = pKa + log10 [salt] …………………… Eq.3.6

[acid]

• pH depends on the ratio [salt]

[acid]

• when [salt] = [acid], [salt] = 1, log10 1 = 0 so pH = pKa

[acid]

when [salt] = 0.1, log10 0.1 = -1 so pH = pKa - 1

[acid]

when [salt] = 10, log10 10 = 1 so pH = pKa + 1

[acid]

Buffering capacity (i.e. the ability to resist a change in pH) is maximal when
pH = pKa. Outside the pH range pKa - 1 to pKa + 1 buffering action is
minimal since the further the [salt]/[acid] ratio is from 1 the greater the resulting
change in the logarithm of this ratio and hence pH when the same amount of
acid or alkali is added.

• Outside 2 pH units either side of the pKa the solution can be considered, for
practical purposes, to consist entirely of acid or salt.

• For a weak base (B) the pKa describes the dissociation of its conjugate acid:

BH+ B + H+

Therefore the Henderson Hasselbalch equation can also be applied to a buffer

consisting of a weak base and its salt since [salt] = [B] and [acid] = [BH+].

Figure 3.4 Quantitative properties of buffers

38
 ACID-BASE, pH AND BUFFERS

Answer to Q3(3)

The dissociation being considered is:

H2PO4- HPO42- + H+
dihydrogen hydrogen
phosphate ion phosphate ion

Phosphoric acid is a tribasic acid (with 3 pKa values) but at near neutral pH the
contributions from the other species (H3PO4 and PO43-) are negligible.

Substituting [salt] = [HPO42-], [acid] = [H2PO4-], pKa = 6.82 and pH = 7.0 into the
Henderson Hasselbalch equation gives:

7.0 = 6.82 + log10 [HPO42-]

[H2PO4-]

Rearranging gives:

log10 [HPO42-] = 7.0 - 6.82 = 0.18

[H2PO4-]

Taking antilogs:

[HPO42-] = antilog10 0.18 = 1.51 ……………………. Eq.3.7

[H2PO4-]

This equation now contains two unknowns and so at first sight appears impossible to
solve. However, a further piece of information is given, namely, that the total
phosphate concentration must be 0.1 mol/L. There are only two forms of phosphate
to consider, therefore:

[HPO42-] + [H2PO4-] = [total phosphate] = 0.1 mol/L

Rearranging to obtain an expression for one phosphate species in terms of the other (it
doesn’t matter which one) gives

[HPO42-] = 0.1 - [H2PO4-] …………………… Eq.3.8

which can be substituted into equation Eq. 3.7 to give an expression containing one
variable which can then be solved:

0.1 - [H2PO4-] = 1.51

[H2PO4- ]

0.1 - [H2PO4-] = 1.51 [H2PO4-]

39
CHAPTER 3

0.1 = 1.51 [H2PO4-] + [H2PO4-] = 2.51 [H2PO4-]

[H2PO4-] = 0.1 = 0.040 mol/L

2.51

The other unknown, [HPO42-], can be obtained by substituting [H2PO4-] = 0.040 into
equation Eq.38:

0.1 = [HPO42-] + 0.040

[HPO42-] = 0.1 - 0.040 = 0.060 mol/L

Next calculate the weight required for each phosphate salt:

concn (g/L) = concn (mol/L) x MW

For anhydrous sodium dihydrogen phosphate (NaH2PO4):

MW = 23 + (2 x 1) + 31 + (4 x 16) = 120

Wt required per litre = 0.040 x 120 = 4.80 g

For anhydrous disodium hydrogen phosphate (Na2HPO4):

MW = (2 x 23) + 1 + 31 + (4 x 16) = 142

Wt required per litre = 0.060 x 142 = 8.52 g

THE BICARBONATE BUFFER SYSTEM

Carbonic acid (H2CO3) dissociates into hydrogen and bicarbonate ions:

H2CO3 H+ + HCO3-

Since carbonic acid is a weak acid, equilibrium is in favour of undissociated acid.

Carbonic acid arises by the reaction of dissolved carbon dioxide with water, a reaction
catalysed by the enzyme carbonate dehydratase (CD):

H2O + CO2 H2CO3

40
 ACID-BASE, pH AND BUFFERS

These two reactions can be linked together (each with their own equilibrium
constants, K1 and K2):

CO2 + H2O H2CO3 H+ + HCO3-

The bicarbonate buffer system is central to acid-base homeostasis for two reasons:

• It affords a means of transporting the very insoluble CO2 generated in tissues

to the lungs where it can be eliminated. This is facilitated by the high
concentration of carbonate dehydratase in erythrocytes which ensures rapid
conversion of carbon dioxide to bicarbonate and vice versa. The hydrogen
ions released when bicarbonate is generated are buffered by the high
concentrations of haemoglobin also present in erythrocytes.

• It is an open system. The two components feeding into the system (carbon
dioxide and bicarbonate ions) can be generated by the lungs and kidney
respectively to ensure a constant CO2/HCO3- ratio, and hence pH.

The equilibrium constants for these two reactions involving carbonic acid are given
by:
K1 = [H2CO3] and K2 = [H+] [HCO3-]
[H2O] [CO2] [H2CO3]

In practice it is not possible to measure the very low concentrations of carbonic acid
present in blood directly. Instead carbon dioxide is measured and so the above two
equations are combined so that the carbonic acid term is eliminated and the buffer
system can be described in terms of carbon dioxide and bicarbonate ion terms only:

Rearranging the first term gives an expression for carbonic acid concentration:

[H2CO3] = K1[H2O] [CO2]

This expression can then be substituted for [H2CO3] in the expression for K2:

K2 = [H+] [HCO3-]
K1 [H2O] [CO2]

Which can be rearranged to give an expression for [H+]:

[H+] = K1K2 [H2O [CO2]

[HCO3-]

The concentration of water can be considered constant and combined with K1 and K2
to give a new constant, K1'

[H+] = K1' [CO2] ………………………….. Eq.3.9

[HCO3-]

41
CHAPTER 3

In routine practice the CO2 content of blood is expressed as its partial pressure, Pco2.
Pco2 is the partial pressure of a gaseous phase which is in equilibrium with the
sample. This practice has arisen because calibrants are prepared by equilibrating
blood with gaseous mixtures with known CO2 content. Henry’s Law states that the
amount of gas physically dissolved in a solution is proportional to the partial
pressure of that gas. The constant of proportionality is the Bunsen solubility
coefficient, α:
[CO2] = α Pco2

If Pco2 is expressed in kiloPascals (kPa) this constant is 0.225, if it expressed in mm

of Hg then it is 0.03. One Pascal (the SI unit of pressure) is the pressure exerted by
1 Newton acting on an area of 1 square metre. The Newton is the SI unit of force.
One Newton is the force required to give a mass of 1 kg an acceleration of 1 metre per
second per second.

By substituting αPco2 for [CO2] in equation Eq.3.9 the following expression is

obtained:
[H+] = K1′ α Pco2 ………………………Eq.3.10
[HCO3-]

The constant K1' can be combined with the solubility coefficient, α, to give a new
constant. If [H+] is expressed in nmol/L, [HCO3-] in mmol/L and Pco2 in kPa then
this constant is approximately 180:

[H+] = 180 Pco2 ………….. Eq.3.11

[HCO3-]

if Pco2 is measured in mm Hg, then the value of the constant is 24.

Alternatively, if equation 3.10 is inverted and logarithms of both sides taken, then it
becomes:

log10 1 = log10 1 + log10 [HCO3-]

[H+] K1' α Pco2

By definition log10 1/ [H+] is the pH and log10 1/ K1' is the pK1' and if these values are
substituted into the above equation then the result is the familiar Henderson
Hasselbalch equation. If 6.1 is substituted for pK1' , and Pco2 is expressed in kPa,
then it becomes:

pH = 6.1 + log10 [HCO3-]………….. Eq.3.12

0.225 Pco2

42
 ACID-BASE, pH AND BUFFERS

Both equations Eq.3.11 and Eq.3.12 contain three variables. Change in one variable
must be accompanied by a change in at least one other. This is of importance for two
reasons:

• Blood gas instruments only measure two variables (hydrogen ion

concentration/pH and Pco2). The third variable, the actual bicarbonate
concentration [HCO3-], is calculated by substituting pH and Pco2 in to the
Henderson Hasselbalch equation.

• It is impossible to carry out experiments in which only one component is

varied in order, for example, to investigate the susceptibility of
chemoreceptors to single acid-base parameters.

It is important to remember that constants such as pK and α depend on other variables

e.g. temperature.

In recent years there has been a move towards expressing the acid-base status of blood
in terms of concentration using nanomolar units. The normal pH (7.4) then becomes
40 nmol/L (antilog10 (-7.4) = 4.0 x 10-8 mol/L = 40 nmol/L) – quite an easy number
to manage. The advantages of using hydrogen ion concentration instead of pH are:

• It allows a more intuitive approach i.e. acidosis is associated with an increase

in [H+] rather than a fall in pH

• The changes are linear. If pH is used then [H+] has to increase 10-fold before
the pH increases by a value of 1

• Unlike pH, [H+] is linearly related to both Pco2 and [HCO3-]. This makes it
easier to calculate an expected compensatory change and helps interpretation
of patients’ blood gas results.

There is little doubt that using concentration is simpler than using pH (otherwise the
three variables would have three very different types of units). Adoption of [CO2]
instead of Pco2 would further simplify matters since all components would then be
expressed as concentrations!

Question Q3(4)

The SHO in ITU carried out a blood gas analysis but failed to record all of the results
in the patient’s notes. The only available results are:

H+ concentration = 93 nmol/L
Actual bicarbonate = 21 mmol/L

Calculate the pH, Pco2 (in kPa) and carbon dioxide concentration (in mmol/L).
Assume that the solubility coefficient of CO2 (in kPa) is 0.225.

43
CHAPTER 3

Answer Q3(4)

Since there are 1,000,000,000 (i.e. 109) nmol in a mol, then 93 nmol/L can be written
as 93 x 10-9 mol/L.

Since pH = - log10 [H+] and [H+] = 93 x 10-9, then

pH = -log10 (93 x 10-9), which can be written pH = - [log1093 + log10 10-9]

Since log1093 = 1.97 and log10 10-9 = -9

pH = - [1.97 + (-9)] = 9 – 1.97 = 7.03

The Henderson-Hasselbalch equation for the HCO3-/CO2 pair is:

pH = pKa + log10 [HCO3-]

α Pco2

substitute: pH = 7.03, pKa = 6.1, [HCO3-] = 21 mmol/L , α = 0.225

then solve for Pco2

7.03 = 6.1 + log10 21

0.225 Pco2

7.03 - 6.1 = log10 21

0.225 Pco2

0.93 = log 10 21
0.225 Pco2

antilog10 0.93 = 21
0.225 Pco2

Pco2 = 21 = 21 = 21 = 10.9 kPa

0.225 antilog10 0.93 0.225 x 8.51 1.92

[CO2] is calculated from the Pco2 and α:

[CO2] = α Pco2 = 0.225 x 10.9 = 2.45 mmol/L

44
 ACID-BASE, pH AND BUFFERS

URINARY BUFFERS

The maximum hydrogen ion gradient from tubular lumen to blood which can be
generated by the kidney tubule is approximately 600:1. Since the hydrogen ion
concentration of normal blood is about 40 nmol/L (pH = 7.4) this means that the
maximum hydrogen ion concentration in urine will be 40 x 600 = 24000 nmol/L,
which corresponds to a pH of 4.62.

The human body normally produces approximately 70 mmol of hydrogen ions per day
of fixed acid (i.e. from sources other than carbon dioxide) which is excreted almost
exclusively in the urine. The vast majority of these hydrogen ions are buffered by
base in the urine, principally phosphate and ammonia:

H+ + HPO42- H2PO4- pKa = 6.8

H+ + NH3 NH4+ pKa = 9.8

The high pKa of ammonia means that at physiological pH and below the majority
exists as ammonium ions. For this reason it has been debated whether or not
excretion of urinary ammonium ions truly reflects excretion of acid. However, unlike
phosphate, where its availability in the tubular fluid is relatively independent of
hydrogen ion status, ammoniogenesis increases during acidosis and is therefore
closely linked to acid-base status.

A quantity called the titratable acidity can be obtained by titrating urine with alkali
back to the pH of blood (7.4). It therefore approximates to:

Titratable acidity (mol/L) = [H+] buffered by phosphate + free unbuffered [H+]

Whereas the total acid excretion includes hydrogen ions buffered with ammonia:

[Total acid excretion] = [Titratable acidity] + [NH4+]

Question: Q3(5)

Over a 24 h period a patient excretes 1 litre of urine with a pH of 5.5, containing

50 mmol of inorganic phosphate and 10 mmol of ammonium. Assuming the blood pH
is 7.40 and that the pKa2 for phosphate is 6.82, calculate:

a) The amount of free hydrogen ion excreted

b) The amount of hydrogen ion buffered by phosphate and ammonium ions.

45
CHAPTER 3

Answer to Q3(5)

a) pH meters measure free hydrogen ions. Therefore convert the urinary pH to

hydrogen ion concentration:

pH = - log10 [H+]

rearranging gives: [H+] = antilog (-pH)

substituting pH = 5.5: [H+] = antilog (- 5.5)

= - (- 3.16 x 10-6)

= 3.16 x 10-6 mol/L (approx 0.003 mmol/L)

Since the urine volume is 1 L, then in 24 h approx 0.003 mmol of acid is excreted
as free hydrogen ions (H+).

b) At the initial pH of the glomerular filtrate (7.40) a significant proportion of

phosphate exists in the “acid “ form (i.e. H2PO4-). As the pH of the tubular
fluid falls to that of urine (pH = 5.5) the proportion of this “acid” form increases
as hydrogen ions are buffered. Therefore the amount of hydrogen ions buffered
will be the difference:

[secreted H+ buffered by phosphate] = [H2PO4-] pH = 5.5 - [H2PO4-] pH = 7.40

The concentration of H2PO4- both before and after buffering the secreted
hydrogen ions can be calculated using the Henderson Hasselbalch equation since
the total phosphate concentration (50 mmol/L) is known.

At pH 7.40:

7.4 = 6.82 + log10 [HPO42-]

[H2PO4-]

7.4 - 6.82 = log10 [HPO42-]

[H2PO4-]

Since [total phosphate] = [HPO42- ] + [H2PO4-] = 50 mmol/L

[HPO42-] = 50 - [H2PO4-]

Substituting for [HPO42-] in the Henderson Hasselbalch equation:

7.4 - 6.82 = log10 (50 - [H2PO4-])

[H2PO4-]

46
 ACID-BASE, pH AND BUFFERS

antilog10 (7.4 - 6.82) = 50 - [H2PO4-]

[H2PO4-]

3.80 = 50 - [H2PO4-]
[H2PO4-]

3.80 [H2PO4-] = 50 - [H2PO4-]

3.80 [H2PO4-] + [H2PO4-] = 50

4.80 [H2PO4-] = 50

[H2PO4-] = 50 = 10.4 mmol/L

4.80

At pH 5.5:

5.5 = 6.82 + log10 [HPO42-]

[H2PO4-]

The above calculation procedure used at pH 7.40 is repeated to give:

[H2PO4-] = 33.3 mmol/L

By subtracting the concentration of H2PO4- at pH 7.40 from this value the

concentration of secreted hydrogen ion buffered by phosphate can be calculated:

[H+ buffered by phosphate] = 33.3 - 10.4 = 22.9 mmol/L

Since ammonia is almost entirely present as ammonium ions at pH 7.4, lowering

the pH to 5.5 will not alter the amount of hydrogen ions buffered. Therefore the
concentration excreted as NH+ is 10 mmol/L. Since the 24 urine volume is 1 L:

Total buffered H+ in urine =

H+ buffered by phosphate + H+ buffered by ammonia

= 22.9 + 10

= 32.9 mmol (33 to 2 sig figs)

By comparison the amount excreted as free hydrogen ions (0.003 mmol) is

insignificant.

47
CHAPTER 3

FURTHER QUESTIONS

1. What is the pH of 0.5 per cent (w/v) hydrochloric acid (assume complete
dissociation, atomic weight Cl = 35.5)?

2. The reference range for blood pH is often quoted as 7.35-7.45. Express this
range in terms of nannomoles of hydrogen ion per litre.

3. If the pH of urine is 6.0 and of blood 7.40, what is the gradient of hydrogen
ion concentrations across the tubular cell walls?

4. Determine the secondary dissociation constant of phosphoric acid if blood of

pH 7.00 contains 12.85 mg disodium hydrogen orthophosphate and 6.88 mg
sodium dihydrogen orthophosphate per 100 mL of plasma.

5. What weight of anhydrous sodium carbonate and sodium bicarbonate would

be required to prepare 500 mL of 0.2 M buffer pH 10.7 (pKa HCO3- =
10.3)?

6. Isotonic sodium lactate, pH 7.4, is commonly administered intravenously to

combat metabolic acidosis. How many ml of concentrated lactic acid (85%
w/w, density 1.2) and how many grams of anhydrous sodium lactate would be
used to prepare 2.5 L of this solution (pKa lactic acid = 3.86)?

7. A 24 h urine collection has a pH of 5.5 and total phosphate content of

65 mmol. If the arterial pH is 7.40 and the pKa for phosphate is 6.80, how
many millimoles of hydrogen ion are excreted as titratable acidity using
HPO42- as buffer?

8. A buffer solution (pH 4.74) contains acetic acid (0.1 mol/L) and sodium
acetate (0.1 mol/L) i.e. it is a 0.2M acetate buffer. Calculate the pH after
addition of 4 mL of 0.025 M hydrochloric acid to 10 mL of the buffer.

48
 ACID-BASE, pH AND BUFFERS

49
CHAPTER 3

50
 SPECTROPHOTOMETRY

Chapter 4

Spectrophotometry

Basic principles

A photometer is a device for measuring the amount of light transmitted through (or
absorbed by) a solution. The wavelength of light absorbed will depend on the chemical
structure of the compound present in the solution. A photometer therefore consists of a
light source which generates a beam of light which passes through a cell (cuvette)
containing the solution under analysis; the light which is transmitted through the solution
falls on photodetector and generates an electric current proportional to the intensity of the
light, which is then translated into a reading. Maximum sensitivity and specificity is
achieved if the beam of light reaching the sample cell is parallel and of a constant
wavelength (i.e. is monochromatic) and is of the wavelength which gives maximum
absorption (minimum transmission) of the light. To achieve this, instruments isolate a
portion of spectrum of white light generated from the light source (usually a bulb) by
placing a monochromator in the light path before reaching the sample cell. In simple
filter photometers a glass filter is used, whereas spectrophotometers use a prism or
diffraction grating. Further details can be found in the standard textbooks of analytical
chemistry.

Consider an incident light beam with intensity Io passing through a square cell containing
a solution of a compound which absorbs light at the wavelength being used. The
intensity of the light reaching the detector, I, will be less than Io. However it is the
fraction of light absorbed (or transmitted) which is related to the concentration of the
compound of interest. The fraction of incident light reaching the detector, I/Io, is known
as the transmittance (T). If expressed as a percentage then the term percentage
transmittance (%T) is used:

T = I and %T = I x 100 ……….. Eq.4.1

Io Io

Therefore measurement of the transmitted intensity, I, by itself is useless, we also need to

know the intensity of the incident light, Io. A measure of Io can easily be made if the
sample cell is removed from the light path. In practice, instead of taking a reading of Io,
the instrument is set to a transmittance of 100% or an absorbance of zero (see later for

51
CHAPTER 4

definition of these terms). Other factors affect the value of Io when a reading of the
sample is made, including a small amount of incident light reflected by the surface of the
cell, absorbed by the material of the cell and by the solvent and/or components of the
reagent. Therefore instead of setting the instrument to 100 %T or zero absorbance with
the cell compartment empty (i.e. against air) it is customary to use a “blank” consisting of
a cell containing either solvent alone or reagent without the analytical sample added. In a
single beam spectrophotometer the blank is inserted into the cell compartment and the
instrument blank value set, then the sample inserted etc. This has the disadvantage that
error will be introduced if the instrument “drifts”. This difficulty is overcome in double
beam spectrophotometers in which the light source is split into two equal beams, one
passing through the blank or reference and the other through the sample position,
enabling the blank to be monitored continuously. Note that T and %T are ratios and so do
not have units.

The absorption laws

Light is absorbed only when a photon collides with a molecule. It is not surprising
therefore that the chance of a photon of light colliding with a molecule in solution, and
hence the amount of light absorbed, will depend on the concentration of the compound in
solution and the path length or thickness of the cell. This simple concept gives rise to the
two absorption laws:

Bouger’s Law or Lambert’s Law: The fraction of light absorbed is proportional to the
thickness of the absorber.

Beer’s Law: The fraction of light absorbed by a compound in solution is proportional to

its concentration.

Taking as an example a solution of a compound at a concentration of 1 g/L which absorbs

half of the light passing through it, then:

T = I = 0.5 and %T = I x 100 = 50%

Io Io

If the concentration is increased by 1 g/L to 2 g/L (i.e. doubled) then a half of the light
which would have been transmitted by the solution containing 1 g/L, will be absorbed
with a result that only a quarter of Io will reach the detector Therefore of the incident light
intensity I0, one half is transmitted after passing through a cell containing 1 g/L of the
compound and a quarter after passing through a solution containing 2 g/L. If the
concentration is increased by a further 1 g/L to 3 g/L then only an eighth is transmitted

52
 SPECTROPHOTOMETRY

Io Io/2 Io/4 Io/8 Io/16

Light

Concentration (g/L) 0 1 2 3 4

T = I / Io 1 0.5 0.25 0.125 0.0625

%T = I x 100 100 50 25 12.5 6.25

Io

Log10 %T 2 1.699 1.398 1.097 0.796

A = 2 - log10 %T 0 0.301 0.602 0.903 1.204

Figure 4.1 Relationship between % transmittance (%T), log10% transmittance

(log %T) and absorbance (A) illustrated by increasing the concentration of
absorbing species in steps of 1 g/L

and if it is increased to 4 g/L, one sixteenth is transmitted. Therefore, of the incident light
1/2 is absorbed by 1 g/L, 3/4 by 2 g/L, 7/8 by 3 g/L and 15/16 by 4 g/L. It is clear that
the relationship between concentration and transmittance is non-linear (Fig 4.1) and is in
fact a geometric progression in which subsequent increase in concentration by 1 g/L
decreases the transmitted light by a factor of 2. Taking logs of % transmittance (it
doesn’t matter to which base, but 10 is usually used) converts to a linear relationship with

53
CHAPTER 4

concentration (logarithms are explained in chapter 3). However, at a concentration of

zero, %T is 100%, log10%T is 2.0 and the value of log10%T decreases as the concentration
of the absorbing species increases. It would be far more convenient if a value of zero was
obtained when the concentration was zero, and the value then increased with increasing
concentration. To achieve this an entity called absorbance (A) was created, which is the
logarithm of the reciprocal of T:

A = log10 1 or A = log10 Io ……………………… Eq.4.2

T I

Note that absorbance is a logarithmic function and so does not have units. Absorbance is
some times called “optical density” and abbreviated “OD”.

If however %T is used then %T can be converted to T by dividing by 100 to give %T/100

so that the above expression becomes:

A = log10 100
%T

Since log10 100 can also be written as : log10 100 - log10 T and since log10 100 = 2
T
it follows that: A = 2 - log10 %T ……… Eq.4.3

This is a very convenient way to inter-convert A and %T. For example, when the
concentration of the absorbing species is zero the %T is 100, the logarithm of 100 is 2 so
that the absorbance must be 2 – 2 which is zero. When %T is 50, the logarithm of 50 is
1.699 so that the absorbance is 2 – 1.699 which is equal to 0.301. A plot of absorbance
versus concentration is linear and passes through the origin (Fig 4.1). Similar reasoning
shows that absorbance is also linearly related to cell path length. Therefore both Beer’s
and Lambert’s Laws can be redefined as follows:

Beer’s Law: Absorbance(A) is directly proportional to concentration(c).

Lambert’s Law: Absorbance(A) is directly proportional to cell path length(b).

Question Q4(1)

A particular sample of a solution of a coloured substance, which is known to obey the

Beer-Lambert Law shows 70% transmittance when measured in a 1 cm cell. Calculate
the percent transmittance and absorbance of this solution if measured in a 0.5 cm cell and
of a solution of twice the original concentration.

54
 SPECTROPHOTOMETRY

Answer Q4(1)

First calculate the absorbance of the original solution:

A = 2 - log10 %T = 2 - log10 70 = 2 - 1.845 = 0.155

According to Lambert’s law, if the cell path length is halved by making the reading in a
0.5 cm cell, then the absorbance will be halved:

A in 0.5 cm cell = 0.155 = 0.078 (3 sig figs)

2

Next calculate %T by substituting A = 0.078 into:

A = 2 - log10 %T

0.078 = 2 - log10 %T

Rearrange and evaluate:

log10 %T = 2 - 0.078 = 1.922

%T = antilog10 1.922 = 84% (2 sig figs)

If the initial concentration is doubled, then, according to Beer’s law, the absorbance will
also double:

A = 2 x 0.155 = 0.310

Convert to %T as above:

0.310 = 2 - log10 %T

log10 %T = 2 - 0.310 = 1.69

%T = antilog10 1.69 = 49 % (3 sig figs)

Combining these two laws gives: A is proportional to bxc

And introducing a proportionality constant (a) gives the Beer-Lambert equation:

A = abc …….. Eq.4.4

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CHAPTER 4

Where a is a proportionality constant known as the “absorptivity”. The units for a will
be the reciprocal of the units of b and c, and can be evaluated by substituting the units for
A, b and c into equation Eq 4.4 then rearranging it. For example, if b is in cm and c is in
mol/L (A of course, has no units), then the units of a will be L.mol-1,cm-1:

A = a x (mol/L) x (cm)

a = A = L = L/mol/cm OR L.mol-1.cm-1
(mol/L) x (cm) mol x cm

When concentration is in molar units, then a is termed the “molar absorptivity,” and the
symbol ε is used. Terms and units used in spectrophotometry are defined in Fig 4.2.

NAME SYMBOL DEFINITION UNITS

Transmittance T I/Io None

% Transmittance %T I x 100 /Io %

Absorbance A log10 (Io/I) None

Optical density OD “ “

Molar absorptivity ε A when c = 1 mol/L, b =1cm L.mol-1cm-1

Path length b Width of cell cm

Wavelength λ Distance between light waves nm

Absorption maximum λmax λ at peak maximum nm

Figure 4.2 Terms and units used in spectrophotometry. Io = intensity of incident

light (or reference), I = intensity of transmitted light, c = concentration

Question Q4(2)

The absorbance of a solution of pure bilirubin in chloroform when measured in a cuvette

with a 0.5 cm path length (using a 0.5 cm cuvette containing chloroform to zero the
instrument) is 0.268. If the concentration of bilirubin in the solution is 4 mg/L, calculate
the molar absorptivity of chloroform. The molecular weight of bilirubin is 584.

56
 SPECTROPHOTOMETRY

Answer Q4(2)

A = a bc

Where: A = absorbance = 0.268

a = molar absorptivity = ?

b = cuvette path length = 0.5 cm

c = molar concentration =

= concentration (mg/L)
1000 x Molecular weight

= 4 = 0.000 00685 mol/L (8.83 x 10-6 mol/L)

1000 x 584

Substituting these values:

0.268 = a x 0.5 x 0.000 00685 (units: cm x mol)

L
Rearranging and solving for a:

a = 0.268 (units: L x 1)
0.5 x 0.000 00685 mol cm

= 78,800 L/mol/cm (or L.mol-1cm-1)

Note that the answer is rounded to 3 significant figures because the absorbance
measurement is only given to 3 decimal places.

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CHAPTER 4

How are absorbance measurements used to calculate the

concentration of an analyte in a biological specimen?

Various approaches can be used, but all are based on the simple relationship between
absorbance and concentration (Eq. 4.4):

A = abc

Since cuvettes with path lengths of 1 cm are almost universally used in

spectrophotometric analysis, then the above expression can be simplified to:

A = ac

which can be rearranged to give:

c = A i.e. concentration = absorbance …… Eq.4.5

a absorptivity

so that concentration can be calculated from an absorbance reading if the absorptivity is

known. Note that the units of absorptivity should be appropriate for the units of
concentration being used. For example, if concentration is in μmol/L then the
micromolar absorptivity should be used (units: L.μmol-1cm-1). The assumption is also
made that at zero concentration the absorbance is zero (i.e. the instrument is either zeroed
on, or the measurement made with reference to, a cuvette of the same path length
containing either the same solvent or the reagent being used.

1. How to use a documented value for absorptivity (a)

If the absorptivity of the absorbing species (that is, the analyte or a chromogen formed by
the reaction of the analyte with a reagent) is known, then the concentration of the
unknown can be calculated from its absorbance reading. It is vital to allow for any
dilution involved and differences in concentration units.

Question 4(3):

0.1 mL of serum is mixed with 3.0 mL of a reagent which forms a coloured product with
glucose. After the reaction has reached equilibrium the absorbance (versus a reagent
blank) in a 1 cm cuvette was found to be 0.250. If the absorpivity of the chromogen is
933 L.mol-1cm-1 what is the serum glucose concentration expressed as mmol/L?

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 SPECTROPHOTOMETRY

Answer Q4(3):

The expression relating absorbance to absorptivity (Eq. 4.5) is used:

c = A
a

Where c = glucose concentration in mol/L = ?

A = absorbance reading = 0.250
a = absorptivity of the chromogen = 933 L.mol-1cm-1

Substituting these values:

c = 0.250 mol/L
933

Note that as the units of absorptivity of the chromogen are L.mol-1cm-1, the calculated
concentration of glucose is in mol/L NOT mmol/L. If the above result is multiplied by
1000 (since there are 1000 mmol in a mol) then the result will be in mmol/L. Since the
cuvette path length is also 1 cm then no correction need be made for differences in path
length.

c = 0.250 x 1000 mmol/L

933

The absorptivity relates to the concentration of glucose derived chromogen in the solution
of which the absorbance is being measured. However, 0.1 mL of serum was mixed with
3.0 mL reagent (i.e. diluted to 3.1 mL) before the absorbance reading was made.
Therefore to convert the calculated concentration of chromogen (c) to glucose
concentration in the undiluted sample, it is multiplied by 3.1 and divided by 0.1:

Serum glucose = 0.250 x 1000 x 3.1 = 8.3 mmol/L

933 0.1

Note that the absorbance reading is multiplied by the reciprocal of absorptivity. Therefore
if a large number of analyses are to be carried out then it is relatively easy to program this
factor into the memory of a pocket calculator so that concentrations may be read off
directly upon entering absorbance readings. Alternatively, using the spreadsheet facility
of a PC, a table can be generated with absorbance values and their corresponding
concentrations. It is important to allow for any differences in units and any dilution of
the biological sample. If question 4(3) is taken as an example:

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CHAPTER 4

c = A x 1 x 1000 x 3.1 mmol/L

933 0.1

c = A x 33.2 mmol/L

so that multiplication of the absorbance reading by 33.2 gives the serum glucose
concentration in mmol/L.

2. Using a standard solution

Suppose we have two samples: sample 1 with concentration c1 gives rise to an

absorbance of A1, sample 2 with a concentration c2 gives rise to absorbance A2. Then we
can write two relationships for absorptivity:

a = A1 and a = A2
c1 c2

Since absorptivity (a) is constant then A1 = A2

c1 c2

which can be rearranged to: c2 = A2 x c1

A1

If solution 1 is a standard solution in which the concentration of the analyte being

measured is accurately known, and solution 2 is the biological specimen in which the
concentration of the analyte is unknown, then:

Concentrationunknown = Absorbance unknown x Concentration standard ………. Eq.4.6

Absorbancestandard

It is important that the absorbances are measured against an appropriate blank.

Question Q4(4):

A method for the measurement of serum glucose involves adding 0.1 mL of sample
(serum, water or standard) to 3 mL of reagent, then after 10 min incubation at room
temperature, measuring the absorbance at 500 nm in a cuvette with a 1 cm path length
using an identical cuvette containing distilled water as reference. The readings using
serum, standard or water as sample were 0.302, 0.353 and 0.052 respectively. If the
concentration of glucose in the standard is 10 mmol/L, calculate the glucose
concentration in the serum.

60
 SPECTROPHOTOMETRY

Answer Q4(4):

Since all the absorbances are measured against a cuvette containing water as reference,
the first step is to subtract the absorbance of the blank (in which 0.1 mL of distilled water
is used as the sample) so that in the absence of analyte the absorbance reading will be
zero:

Sample A (versus water) A (corrected for blank)

Blank 0.052 0.000

Standard 0.353 0.301

Serum 0.302 0.250

Since the glucose concentration in the standard is known to be 10 mmol/L

Serum glucose (mmol/L) = Aserum (corrected) x 10

Astandard (corrected)

= 0.250 x 10 = 8.3 mmol/L

0.301

Failure to correct each reading for the absorbance reading of the blank (0.052) would
result in an erroneous answer of 8.6 mmol/L (0.302 x 10 / 0.353).

Since the dilutions of the sample (serum or standard) with reagent are identical, the
dilutions cancel and do not need to be included in the calculation. However, any dilution
of the sample which is different to that of the standard must be taken into account.

3. Using a calibration curve

If Beer’s Law is obeyed and cuvettes with a constant path length are used then the
equation relating absorbance to concentration (Eq. 4.4) is a linear function:

A = ac

In other words if, for a series of solutions, absorbance (A) is plotted on the vertical axis
and concentration (c) on the horizontal axis then a straight line is obtained which passes

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CHAPTER 4

through the origin and has a slope (i.e. gradient) of a (Fig 4.3). Such a plot is called a
calibration curve. There are various ways in which a calibration curve can be used to
calculate the concentrations of analyte in an unknown sample:

c A A-A0 A4- A0

c0 A0 0
A3- A0
c1 A1 A1-A0
Absorbance (A)
A2- A0
c2 A2 A2-A0
A1 - A0
c3 A3 A3-A0

c4 A4 A4-A0 A0
c3 c4
-0.5 0.5 1.5
c2
2.5 3.5

0 c1

Figure 4.3 Construction of a calibration curve from a set of absorbance

readings obtained for a series of standard solutions. c0 is a
reagent blank (standard with zero concentration) with an
absorbance of A0

Question Q4(5)

A manual method for the determination of plasma glucose involves mixing 0.1 mL of
sample (water, glucose standard solution or plasma sample for the reagent blank, standard
and unknown sample respectively) with 3 mL of reagent, then after 10 min incubation at
room temperature, measuring the absorbance of the chromogen at 500 nm. The following
results were obtained:
Sample Absorbance

Water (zero standard) 0.085

Standard, 3 mmol/L 0.175
“ , 6 “ 0.265
“ , 9 “ 0.355
“ , 12 “ 0.445
Plasma A 0.220
“ B 0.800
“ “ (1 in 4 dilution) 0.246

Before the absorbance measurements were made, the instrument was set to zero using an
identical 1 cm cuvette containing distilled water. Determine the concentrations of glucose
in plasma samples A and B.

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 SPECTROPHOTOMETRY

Answer Q4(5):

There are several approaches which can be used. The simplest is to plot the data with
absorbance (A) on the vertical axis glucose concentration (c) on the horizontal axis:

0.5

0.4
A (absorbance vs water)

Plasma B (diluted 1 in 4)

0.3

Plasma A

0.2

0.1

0
0 1 2 3 4 5 6 7 8 9 10 11 12
c (glucose concentration, mmol/L)

A straight line is then drawn through the points in such a way as to produce the best
visual fit to the data. It is then an easy matter to read off the values for the plasma
samples from this curve.

For plasma A (absorbance = 0.220) a horizontal line is drawn from the absorbance value
on the vertical scale until it intersects the calibration curve. From this point a vertical line
is drawn downwards until it intersects the horizontal axis. The reading at this point gives
the glucose concentration in the sample as 4.5 mmol/L.

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CHAPTER 4

This procedure cannot be followed for the neat (undiluted) absorbance value for plasma B
(0.80) since this value is beyond the range of the calibration curve i.e. the absorbance
value of the highest standard is only 0.445. Using a ruler the calibration curve could be
extrapolated to this value but this would assume that the calibration is linear to this point.
This is unlikely to be true if either the linear range of the instrument is exceeded or the
capacity of the reagent is exhausted. The absorbance for plasma B which had been
previously diluted 1 in 4 is well within the working range of the assay (0.310) and
reading the plasma glucose concentration from the calibration curve using an identical
procedure as for plasma A gives a value of 7.5 mmol/L. Since the 0.1 mL of plasma B
sample had been diluted 1 in 4 prior to assay, this result is multiplied by 4 to give a final
glucose result of 30 mmol/L.

Another approach is to determine the equation which describes the line through the
standards (the calibration curve) then use it to arithmetically calculate the concentration
of glucose in the plasma sample. The first step is to draw a right angled triangle (ABC)
for which the calibration curve is the hypotenuse:

0.5 B

0.4
A (absorbance vs water)

0.3

0.360

0.2

0.1

A
0.085 C
0
0 1 2 3 4 5 6 7 8 9 10 11 12
c (glucose concentration, mmol/L)

The linear equation describing the calibration curve takes the form:

A = Intercept + ( c x slope)

64
 SPECTROPHOTOMETRY

The “intercept” is the point at which the calibration curve crosses the vertical axis i.e. the
value of the absorbance (A) when glucose concentration (c) equals zero, and is 0.085.
The slope is the change in absorbance which is observed for a 1mmol/L increment in
glucose concentration i.e. BC divided by AC:

Slope = (0.445 – 0.085) = 0.360 = 0.030

12 12

So that the equation for the calibration curve is:

A = 0.085 + 0.03 c

Which can be re-arranged to:

c = (A - 0.085)
0.03

By substituting absorbances obtained for the plasma samples, their glucose concentration
is easily calculated:

Plasma A glucose = (0.220 - 0.085) = 4.5 mmol/L

0.03

For plasma B the absorbance obtained from the diluted sample should again be used:

Plasma B glucose (1 in 4 dilution) = (0.310 – 0.085) = 7.5 mmol/L

0.03

Plasma B glucose (neat) = 4 x 7.5 = 30 mmol/L

Some modern spectrophotometers can be calibrated directly as the blank and standard
absorbances are read so that as unknown samples are read the result is displayed directly
in concentration units rather than absorbance values. Again it is important to emphasize
that this equation should only be used if the calibration curve is linear and for the
concentration range covered by the standards. The glucose concentration can be
calculated for the undiluted plasma B sample (A = 0.80) from the above equation:

Plasma B glucose = (0.800 - 0.085) = 23.8 mmol/L

0.03

Clearly this value is erroneously low and illustrates the danger of extrapolating the
standard curve beyond the highest standard used in its construction.

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CHAPTER 4

Sometimes the scatter of the points of the calibration curve make it difficult to decide
where to draw the line of best fit. Under these circumstances the statistical line of best fit
can be calculated using statistical methods. Modern instruments often use very complex
mathematical techniques to calculate the best equation which describes the calibration
curve.

An alternative approach is to subtract the reagent blank absorbance (the absorbance

obtained when the glucose concentration of the standard is zero) form all readings (or
alternatively make all readings using the reagent blank as reference): :

Sample Absorbance Absorbance –Blank absorbance

Water (zero standard) 0.085 0.000

Standard, 3 mmol/L 0.175 0.090
“ , 6 “ 0.265 0.180
“ , 9 “ 0.355 0.270
“ , 12 “ 0.445 0.360
Plasma A 0.220 0.135
“ B 0.800 0.715
“ “ (1 in 4 dilution) 0.246 0.161

This has the effect of forcing the calibration curve through the origin:

0.4
A (absorbance corrected for blank)

0.3

Plasma B diluted 1in 4

0.2

0.1 Plasma A

0
0 1 2 3 4 5 6 7 8 9 10 11 12
c (glucose, mmol/L)

66
 SPECTROPHOTOMETRY

Note that the slope is the same (0.03) although the intercept on the absorbance axis
(previously 0.085) is now zero. Thus the equation describing the calibration curve is now
simpler:

A (blank corrected) = 0.03 c

Which can easily be rearranged to

c = A (blank corrected) mmol/L

0.03

and used to calculate the glucose concentration for plasma samples. It is vital that the
reagent blank is also subtracted from the absorbance readings for the plasma samples
before their concentrations are either calculated from the above formula or read directly
from the calibration curve.

An alternative approach when the absorbance of a plasma sample is beyond the working
range of the calibration curve is to dilute the final reaction mixture, either with distilled
water or reagent until the absorbance falls within range. This practice overcomes non-
linearity due to the spectrophotometer but not if non-linearity is due to exhaustion of the
reagent or one of its components. If the reaction mixture was diluted with more reagent
then the correction to be applied is the same as if the assay had been repeated with diluted
plasma sample. However, if the reaction mixture has been diluted with water then the
reagent blank is no longer appropriate since the contribution of the reagent to the final
absorbance reading has been reduced. The reagent blank to be subtracted should first be
divided by the dilution factor used.

If the relationship between absorbance and concentration is linear (i.e. Beer’s Law is
obeyed) then the minimum number of data points required to determine the linear
equation relating the two quantities is 2. The term single point calibration is often used
which is, strictly speaking, incorrect. A second point is always required to define a
straight line (the shortest distance between two points in the same plane). It is always
inferred that this second point is the value when concentration is zero i.e. the blank.
There are other approaches to using a set of standard absorbances. In general if there are
n standards with their corresponding absorbances then the ratios of their absorbances to
their concentrations are equal:

A1 = A2 = A3 = A4 = …… An
c1 c2 c3 c4 cn

The only time this relationship does not hold is when the concentration (c0) is zero, since
any number divided by zero becomes infinite. A further requirement is that all
measurements should be made using an appropriate blank as reference.

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CHAPTER 4

The individual ratios of A/c can be averaged to produce a mean value for A/c which can
then be used to calculate the concentrations in unknown samples:

Concentration of unknown = mean (c/A)standards x Aunknown

Although this procedure is simple and takes account of the imprecision of absorbance
measurements of the standards it is not recommended. The absorbances of the individual
standards provide a further piece of information: confirmation that the relationship
between absorbance and concentration is in fact linear. Therefore, a calibration curve
should always be plotted or accessed mathematically.

Dealing with mixtures

Often we encounter solutions which contain more than one light-absorbing species. If
their absorption spectra overlap then the presence of one of these species may interfere
with the determination of the other. Fig 4.4 shows the absorption spectra of two
compounds, A and B. At the wavelength of maximum absorption (λmax) of either
compound there is significant absorption of light by the other. The spectrum of a mixture
of A and B at the same concentrations present in their separate solutions has an
absorption maximum somewhere in between that of the individual compounds. Provided
there is no interaction between the two compounds, their individual spectra are additive.

Beer’s law states that absorbance (A) is equal to absorptivity (a) multiplied by
concentration (c):

A = ac

Similar equations can be written for the two species A and B:

AA = aA x cA

AB = aB x cB

Provided there is no interaction between A and B, their absorbances are additive, so that
the total absorbance (Atotal) is the sum of their individual absorbances:

Atotal = AA + AB

Which can be written in terms of their individual concentrations and absorptivities:

Atotal = (aA x cA) + (aB x cB) …………… Eq.4.7

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 SPECTROPHOTOMETRY

A+B

A
A B

λA λB

Wavelength (λ)

Figure 4.4 Over-lapping spectra of two compounds (A and B) and

a mixture of both (A + B)

At first sight it may appear impossible to use the measured absorbances to calculate the
individual concentrations of A and B in a mixture of both, even if the individual
absorptivites of A and B are known i.e. the equation contains two unknowns. However,
if the absorbance is also measured at a second wavelength (usually the λmax of the other
species) then another equation for the measured absorbance can be set up similar to
Eq. 4.7.

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CHAPTER 4

In general, if there are two species, A and B

A has absorptivities of aAλ1 at wavelength 1 and aAλ2 at wavelength 2

B “ “ “ aBλ1 “ “ “ “ aBλ2 “ “ “

A1 is the measured absorption of the mixture at wavelength 1

A2 “ “ “ “ “ “ “ “ “ 2

cA is the concentration of A in the mixture

cB “ “ “ “ B “ “ “

Then two equations can be written for the measured absorption, one for each wavelength:

A1 = aAλ1 cA + aBλ1 cB
……….. Eq.4.8
A2 = aAλ2 cA + aBλ2 cB

These form a set of 2 simultaneous equations (each containing the same two unknowns)
which can then be solved in the usual way (see Fig 4.5).

The same principle can be applied to mixtures containing 3, 4 or even more components.
Absorbance measurements must be made at the same number of wavelengths as the
components in the mixture. However, the simultaneous equations become increasing
complex and difficult to solve.

Question Q4(6)

A chromatographic method for the separation of 4 different drugs fails to completely

resolve two of them (drugs A and B). Fortunately drugs A and B have overlapping
spectra with the following absorptivities (L.mol-1cm-1):

260 nm 280 nm
Drug A 100 500
Drug B 1000 200

Fractions from peaks A and B were pooled and the absorbance of the mixture measured
in a cuvette with a 1 cm light path using solvent as reference. The absobance reading at
260 nm was 0.4 and at 280 nm 0.8. Calculate the individual concentrations of A and B in
the pooled fractions.

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 SPECTROPHOTOMETRY

Simultaneous equations are a set of two or more equations in two or more unknowns that
are simultaneously true. For example, consider two equations in which x and y are
unknown but in which the values of x and y are identical:

2x + 3y = 16 ……………… (i)

3x + 2y = 14 …………..….. (ii)

Either or both equations are multiplied by factors in order to render one of the terms
equal. Subtraction of one equation from the other then eliminates one of the variables.
The resulting equation is then solved for the other variable. Simple inspection often
suggests a suitable factor to use but an approach which always works is to multiply the
first equation by the constant in front of one of the variables in the second equation and
the second equation by the constant in front of the same variable in the first equation. In
the above example if the whole of equation (i) is multiplied by 3 (the constant in front of
x in equation (ii)) which then becomes equation (iii), and the whole of equation (ii) is
multiplied by 2 (the constant in front of x in equation (i)), which then becomes equation
(iv) then both equations contain 6x. Subtraction of equation (iv) from (iii) eliminates
variable x, and the result (equation (v)) can be solved for the other variable (y):

6x + 9y = 48 ……………… (iii)

6x + 4y = 28 ………………. (iv)

5y = 20 ………………. (v)

y = 20 = 4
5

This value for y can then be substituted into either equation (i) or (ii) which is then solved
for the other variable (x). Using equation (i):

2x + (3 x 4) = 16

2x = 16 - (3 x 4x) = 16 - 12 = 4

x = 4/2 = 2

Figure 4.5 A method for the solution of a pair of simultaneous equations

containing two unknowns

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CHAPTER 4

Answer Q4(6):

At each wavelength the measured absorbance is equal to the sum of the absorbances due
to dugs A and B. The absobance due to either drug is given by its molar concentration
multiplied by its molar absorptivity at that particular wavelength. Therefore an eauation
for the measured absorbance at each wavelength can be set up:

At 260 nm: A260 = aA260 cA + aB260 cB …………………………..(i)

At 280 nm: A280 = aA280 cA + aB280.cB …………………………(ii)

Substituting values for the absorbances, molar concentrations and molar absorptivities
into equations (i) and (ii):

At 260 nm: 0.4 = 100 cA + 1000 cB ………………………….(i)

At 280 nm: 0.8 = 500 cA + 200 cB ………………………….(ii)

Multiply equation (i) by 5 throughout (to become equation (iii)), then subtract equation
(ii) from (iii) so as to eliminate the cA term:

2.0 = 500 cA + 5000 cB ………………………...(iii)

0.8 = 500 cA + 200 cB …………………………(ii)

1.2 = 4800 cB

Solve for cB:

cB = 1.2 = 0.00025 mol/L = 0.25 mmol/L

4800

Substitute 0.00025 for cB in equation (i), then solve for cA:

0.4 = 100 cA + (1000 x 0.00025)

0.4 = 100 cA + 0.25

100 cA = 0.4 - 0.25 = 0.15

cA = 0.15 = 0.0015 mol/L = 1.5 mmol/L

100 x

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 SPECTROPHOTOMETRY

FURTHER QUESTIONS

1. An aqueous solution in a 1 cm cell has an absorbance of 0.23 when read against a

water blank at 500 nm. Assuming Beer’s Law is obeyed, what volume of this
solution would need to be added to 100 mL of water to give a solution which
absorbs 30% of the light entering it under the same measurement conditions?

2. Calculate the absorbances corresponding to the following percentage

transmittance readings:

a) 95 b) 75 c) 50 d) 25 e) 10 f) 1

3. Calculate the % of incident light transmitted by solutions with the following

absorbances:

a) 0.1 b) 0.25 c) 0.50 d) 0.75 e) 1.00 f) 2.00

4. A solution of a compound (concentration 100 mmol/L) was placed in a cuvette

with a 1 cm light path and the percentage of incident light transmitted was 18.4.
Calculate the molar absorptivity of the compound.

5. The transmittance of a solution of NADH at 340 nm is 45%. What is the

absorbance at 340 nm of a 1 in 5 dilution of this solution?

6. 75 mg of faeces were homogenised in 1 mL of concentrated hydrochloric acid,

then 3 mL diethylether added, mixed, 3 mL of water added and mixed again.
After centrifugation the aqueous phase (volume 4.5 mL) was scanned in a
spectrophotometer using a cell with a 1 cm pathlength and the peak height at 405
nm due to porphyrin, after applying a background correction, was 0.35
absorbance units. A separate 0.250 g portion of faeces was dried in a 100oC oven
for 3 hours after which it’s weight was 0.125 g. Given that the molar absorption
coefficient of porphyrin is 2.75 x 105 L/mol/cm calculate the porphyrin
concentration in nmol/g dry weight of faeces.

7. A solution containing a substance of molecular weight 400 at a concentration of

3 g/L transmitted 75% of incident light of a particular wavelength in a 1 cm
cuvette. Calculate the % of incident light of the same wavelength that would be
transmitted by a solution of the same substance at a concentration of 4 g/L and
calculate the molar absorption coefficient for that substance at this wavelength.

73
CHAPTER 4

8. The absorbances of a solution containing NAD and NADH in a 1cm light path
cuvette were 0.337 at 340 nm and 1.23 at 260 nm. The molar extinction
coefficients are:

NAD: 1.8 x 104 at 260 nm, 1.0 x 10-3 at 340 nm

NADH: 1.5 x 104 at 260 nm, 6.3 x 103 at 340 nm

Calculate the concentrations of NAD and NADH in the solution.

9. 25 mg of bilirubin (C33H36O6N4) were dissolved in 4 mL of dimethyl sulphoxide;

200 µL of this solution was diluted to 250 mL with chloroform. This solution
gave an absorbance of 0.502 when measured in a 1 cm cell against a chloroform
blank.

Given that the molar absorptivity of bilirubin under these conditions is 6.07 x 104,
calculate the percentage purity of the bilirubin.

10. A method for creatinine determination based on the Jaffe reaction involved
mixing 0.1 mL of sample with 2.5 mL alkaline picrate reagent, incubating for
10 min at room temperature, then measuring the absorbance at 530 nm in a 1-cm
cuvette in a spectrophotometer set to read zero using a cuvette containing distilled
water. The following readings were obtained:

Blank (water as ample) 0.050

Creatinine standard (200 μmol/L) 0.250
Serum sample 0.125
Urine sample (prediluted 1 in 50 with water) 0.200

Calculate the creatinine concentration in the serum (in μmol/L) and urine (in
mmol/L).

11. A standard curve for a plasma glucose method was set up by preparing a series of
dilutions of a stock glucose standard (containing 50 mmol glucose/L) and
measuring the absorbance at 500 nm in a 1 cm cuvette using a blank with zero
glucose concentration to zero the instrument. The following readings were
obtained:

Glucose (mmol.L): 5 10 15 20 25 30
Absorbance: 0.102 0.203 0.305 0.375 0.410 0.432

Does the method obey Beer’s Law? What glucose concentration corresponds to
an absorbance reading of 0.250?

74
 SPECTROPHOTOMETRY

75
CHAPTER 4

76
 RENAL FUNCTION

Chapter 5

Renal function

The kidney has multiple functions but in routine clinical practice very few of these are
formally assessed. A proportion of the blood supplied to each kidney is filtered at the
glomerulus to produce a cell-free ultrafiltrate which is virtually protein-free but
otherwise has the same composition as plasma. The tubules then modify this filtrate by
reclaiming components (reabsorption) or by adding further components to it (secretion)
before it is transported via the ureters to the bladder then excreted. In other words the
tubules convert the glomerular filtrate into urine. These two processes, filtration or
glomerular function and tubular function can be quantitatively assessed in the laboratory.

The Glomerular Filtration Rate (GFR)

Glomerular filtration is usually quantified as the rate of formation of filtrate. For a

normal adult the glomerular filtration rate (GFR) is in the order of 100 mL/min (or 0.10
L/min). In other words, each minute the body produces approximately 100 mL of
glomerular filtrate i.e. filters about 100 mL of plasma. Using this information it is easy to
calculate the total amount of filtrate produced in any given time period e.g. 24 h. If the
concentration of any component of plasma freely filtered at the glomerulus is known,
then, since the concentration of that component is the same in the filtrate, it is also
possible to calculate the total amount of that component filtered in any time period.

Question Q5(1)

A normal subject has a GFR of 120 mL/min and a plasma creatinine concentration of
100 μmol/L. Calculate total volume (in litres) of filtrate produced over a 24 h period and
the 24 h excretion of creatinine (in mmol) assuming that none of the filtered creatinine is
reabsorbed by the tubules.

77
CHAPTER 5

Answer Q5(1)

The longer the time period the more filtrate is produced. Since 120 mL is produced per
minute then twice this amount is produced in 2 minutes (2 x 120 mL = 240 mL), three
times this amount in 3 minutes (3 x 120 mL = 360 mL) etc. Therefore if the rate of
filtration (the GFR) is multiplied by the time period (using the same units of time) then
the result is the total volume of filtrate produced over that time period:

Vol filtrate (mL) = GFR (mL/min) x time period (min)

In this instance the GFR is 120 mL/min and the time period is 24 h. Obviously the same
units for time must be used so 24 h is multiplied by number of minutes in an hour (60) to
give 24 x 60 = 1440 min. From this the volume of filtrate formed in 24 h can be
calculated:

Vol filtrate (mL) = 120 x 1440 = 172800 mL/24 h

To convert from mL to L divide by 1000 (since there are 1000 mL in a L):

Vol filtrate (L) = 172800/1000 = 173 L (to 3 significant figures)

This volume is considerably greater than the total amount of plasma in the human body
(approx 3.5 L) or of the total water content (approx 42 L) and emphasises the importance
of the renal tubules in reclaiming the vast majority of filtered water to reduce the volume
of filtrate to the daily output of urine (in the order of 1-2 L) and thus avoid dehydration.

Calculation of the total amount of a solute in this volume of filtrate is analogous to

calculating the amount of a compound needed to prepare a given amount of solution of a
given concentration (see chapter 2). All that is need is to multiply the volume of the
solution (in this case glomerular filtrate formed in 24 h) by the concentration of the solute
(in this case the concentration of creatinine in the glomerular filtrate will be the same as
its plasma concentration):

Creatinine filtered (μmol/24 h) = Volume filtrate (L/24 h) x Plasma creatinine

(μmol/L)

= 173 x 100 = 17300 μmol/24 h

Division by 1000 converts to mmol/L (since there are 1000 μmol in 1 mmol):

Creatinine filtered = 17300/1000 = 17.3 mmol/24 h

If none of this filtered creatinine is reabsorbed by the tubules then this represents the
24 urinary excretion of creatinine.

78
 RENAL FUNCTION

The manipulations involved in calculating amounts of substances filtered can be

summarised:

Volume of filtrate = GFR x Time ………………………………..… Eq. 5.1

Amount filtered = Volume of filtrate x Plasma concentration .…. Eq. 5.2

Amount filtered = GFR x Time x Plasma concentration …….... Eq. 5.3

Rate of filtration of substance = GFR x Plasma concentration …….... Eq. 5.4

It is vital to always ensure that the units are compatible. These calculations always
assume that the plasma concentration of the substance remains constant over the time
period being considered and that it is freely filtered at the glomerulus.

Rate of urinary excretion

This is usually assessed by making a timed collection of urine usually, but not always,
over a 24 h period. The total volume of urine is measured and the concentration of the
analyte of interest measured in an aliquot of the urine collection. Multiplication of the
urine volume by the analyte’s concentration in the urine aliquot gives its total urinary
excretion over the collection period:

Total excreted = Urine volume x Urine concentration ………. Eq. 5.5

Again, care must be taken that units are compatible. Note that if rate of urine production
is used instead of urine volume then this calculation gives the rate of urinary excretion of
the compound.

Question Q5.2

A patient was asked to collect urine over a 24 h period. The volume was found to be
1.5 L and the concentration of creatinine in an aliquot of this urine 8.0 mmol/L.
Calculate the 24 h urinary excretion of creatinine in mmol.

79
CHAPTER 5

Answer Q 5(2)

Creatinine excretion(mmol) = Urine volume (L) x Creatinine concentration (mmol/L)

= 1.5 x 8.0 = 12.0 mmol

Clearance and GFR

Consider a hypothetical compound which is only excreted by the kidney (and is not
further metabolised by the body) and is in a steady state i.e. the rate of excretion via the
kidneys is equal to the rate of formation by the body (or rate of infusion into the body) so
that its plasma concentration remains constant. In practice this can be achieved in two
ways:

• Administration of an exogenous compound (e.g. inulin – a fructose polymer)

intravenously. By definition when a steady state is reached the rate of excretion is
equal to the rate of infusion and urinary collections are unnecessary.

• Using a compound which is produced endogenously by the body at a constant rate

(e.g. creatinine).

Provided this compound is freely filtered at the glomerulus then its rate of filtration is
given by Eq.5.4:

Rate of filtration = GFR x Plasma concentration

And its rate of excretion in the urine can be described by Eq.5.5:

Rate of excretion = Rate of urine formation x Urine concentration

If all of the compound that is filtered at the glomerulus is excreted in the urine i.e. it is
neither reabsorbed from nor further amounts secreted into the filtrate, then:

Rate of filtration = Rate of excretion

If we substitute for rates of filtration and excretion the following expression is obtained:

GFR x Plasma concentration = Rate of urine formation x Urine concentration

80
 RENAL FUNCTION

which can be rearranged to give:

GFR = Rate of urine formation x Urine concentration …….. Eq. 5.6

Plasma concentration

Therefore, provided the above conditions are met, if a timed urine is collected and the
concentration of the analyte is measured in both plasma and urine, then the GFR is easily
calculated. Again it is vital that all the units are compatible.

Another way of looking at the GFR is that it is the clearance of the substance being
considered i.e. the volume of plasma from which the substance is completely removed or
cleared in unit time. This volume can easily be determined if urinary excretion is divided
by plasma concentration:

Urinary clearance (L) = Urinary excretion

Plasma concentration

If the substance is reabsorbed or secreted by the tubules then of course the measured
urinary excretion of the compound will not be equal to the rate of its filtration at the
glomerulus. In other words the GFR will not be equal to the measured clearance. If a
compound is reabsorbed by the tubules (for example urea) then only a proportion of the
filtered compound will be excreted in the urine and the measured clearance will be much
lower than the GFR. In the case of urea, the amount reabsorbed is very dependent upon
the state of hydration (and hence the urine flow rate) and attempts have been made to
correct for this using the square root of the urine volume in the calculation. If a substance
is actually secreted into the tubules then the urinary excretion will be greater than the
amount filtered and the measured clearance will be greater than the GFR.

The clearance of any substance can be measured, but the value obtained will only give a
measure of GFR if it is excreted by glomerular filtration alone.

Question Q 5(3)

A 24 h urine (volume 2.4 L) was collected from a patient in ITU and the creatinine
concentration of an aliquot was found to be 6.0 mmol/L. The creatinine concentration of
a plasma sample collected during this 24 h period is 500 μmol/L. Calculate this patient’s
creatinine clearance in mL/min.

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CHAPTER 5

Answer Q 5(3)

Creatinine clearance = Rate of urinary creatinine excretion

Plasma creatinine concentration

Urine flow rate (L/min) = Urine volume (L/24 h) = 2.4 = 0.00167 L/min
24 x 60 1440

(Division by 24 converts the urine flow to L/h, then division by 60 converts to L/min).

Rate of excretion (μmol/min) = Urine flow rate (L/min) x Urine concentration (μmol/L)

Since the urine creatinine concentration is given in mmol/L it is first multiplied by 1000
to covert to μmol/L, then:

Rate of excretion (μmol/min) = 0.00167 x 6.0 x 1000 = 10.0 μmol/min

Division of this rate of urinary excretion of creatinine by its plasma concentration gives
the volume of plasma completely cleared of creatinine in each minute . i.e. the clearance:

Clearance (L/min) = Rate of excretion (μmol/min)

Plasma concentration (μmol/L)

Clearance = 10.0 = 0.020 L/min

500

Multiplication by 1000 converts the clearance from L/min to mL/min:

Clearance = 0.020 x 1000 = 20 mL/min

Alternatively the equation described in Fig 5.1 could be used.

It should be noted that the calculation of clearance involves three measurements: urine
volume, urine concentration and plasma concentration. The combination of the errors
involved in these measurements is considerable. It cannot be emphasised too strongly
that the largest source of error in a clearance measurement is the accuracy of the timed
urine collection. Although clearance measurements are conventionally expressed as
mL/min, the result implies an unrealistic degree of accuracy and it would be better if
results were expressed as L/min using only 2 significant figures. For example a clearance
of 123 mL/min would become 0.12 L/min.

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 RENAL FUNCTION

The clearance of a substance is given by the equation:

Clearance = U x V ………………………… Eq. 5.7

P

Where: U = concentration of the substance in urine

V = rate of urine formation
P = concentration of the same substance in plasma

The units must always be compatible. In the case of creatinine clearance, the clearance is
usually expressed as mL/min so the units of the individual measurements are first
adjusted as follows:

U: urine creatinine is usually reported in mmol/L. Since plasma creatinine is

reported in μmol/L, the urine creatinine is multiplied by 1000 to convert it to the
same units.

V: urine is usually collected over a 24 h period and its volume expressed in litres.
Since the clearance is required in mL/min this volume is multiplied by 1000 (to
convert from litres to mL) and divided by 24 (to convert from 24 h to 1 h) then 60
(to convert from hours to minutes).

P: plasma creatinine is usually expressed as μmol/L and is unchanged.

Introducing these adjustments leads to the equation:

Creatinine clearance (mL/min) = U x 1000 x V x 1000

24 x 60 x P

Which (to 2 significant figures) simplifies to:

Creatinine clearance (mL/min) = U x V x 700 ……………. Eq. 5.8

P

If different units or urine collection times are used then the factor 700 must be adjusted.

Figure 5.1 Calculation of creatinine clearance

83
CHAPTER 5

Relationship between clearance, plasma concentration and

urinary excretion

Equation 5.7, clearance = UV , contains 3 variables:

P
Clearance, which may or may not be equal to the GFR
Plasma concentration (P)
Urinary excretion of the substance (UV)

Alteration of one variable must result in the alteration of at least one other. Consider a
steady state in which clearance, P and UV are constant (Fig 5.2).

P
200

150 Change in
clearance

UV
% of 100
initial
value

Clearance

50

0 Time

Figure 5.2 Effect of a change in clearance (shown by arrow) on plasma

concentration (P) and urinary excretion (UV) of a substance

If at some point (shown by the arrow) the clearance is halved (for example by removal of
one kidney) then the immediate effect will be for the urinary excretion (UV) to halve. As
time progresses the plasma concentration (P) will rise (since less of the substance is being

84
 RENAL FUNCTION

removed from the plasma by filtration) and as a result the amount filtered per unit time
(given by: clearance x P) will also rise. This will be reflected in the urinary excretion
(UV) which will also increase. Eventually a new steady state will be achieved in which
the urinary excretion is unchanged but the plasma concentration is doubled (Fig 5.2).
The relationship between clearance, P and UV is summarised in Fig 5.3:

3000

2500 0.004

2000

1500
1/P
P
0.002

1000

500

0 0
0 50 100 150
0 50 100 150
Clearance Clearance

Figure 5.3 Relationship between plasma concentration (P), reciprocal of plasma

concentration (1/P) and clearance

Question Q 5(4)

A patient with a creatinine clearance of 120 mL/min has a plasma creatinine

concentration of 100 μmol/L. Assuming no tubular action on filtered creatinine, what
concentration of creatinine would you expect to find in a 6h urine collection which has a
total volume of 500 mL? A healthy kidney is removed by surgery for transplantation to a
relative. What concentration of creatinine would you expect to find in a 6 h urine
(volume 400 mL) collected immediately following surgery. At a follow-up clinic six
weeks later his plasma creatinine was found to be 200 μmol/L. Estimate the likely
creatinine content of a repeat 6 h urine collected the day before his clinic appointment
assuming his creatinine clearance is unchanged.

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CHAPTER 5

Answer Q 5(4)

Clearance (mL/min) = (U x 1000) x (V x 1000) ……Eq. 5.8

(h x 60) x P

Rearranging: U = Clearance x (h x 60) x P

1000 x (V x 1000)

Clearance = U x V …………………. Eq. 5.7

P
Pre-operatively:

Clearance = 120 mL/min

h = urine collection period = 6 h
V = urine volume = 500 mL = 0.5 L
P = plasma creatinine = 100 μmol/L

U = 120 x 5 x 60 x 100 = 8.6 mmol/L

1000 x 0.5 x 1000

Immediately following removal of one kidney:

Assuming both kidneys function equally and there is no pre-renal impairment post-
operatively then the clearance will be a half of the previous value.

Creatinine clearance = 120/2 = 60 mL/min

P is initially unchanged, V = 400 mL and h is again 6

Substitution of these values into the rearranged Eq. 5.8 allows calculation of the new
value of U:

U = 60 x 6 x 60 x 100 = 5.4 mmol/L

1000 x 0.4 x 1000

Follow-up at six weeks:

GFR = unchanged = 60 mL/min

P = 200 µmol/L
Both V and U are unknown

Eq. 5.8 can be rearranged in a slightly different way to give:

U x V = Clearance x (h x 60) x P
1000 x 1000

86
 RENAL FUNCTION

Therefore U x V = 60 x 6 x 60 x 200 = 4.3 mmol

1000 x 1000

Correction of GFR for Body Surface Area

Kidney size and hence GFR increases with increasing body size. This is not surprising
since as body size increases so does body water and hence plasma volume. Furthermore,
a larger body produces a greater amount of waste products to be excreted. For example,
the GFR of a 2 year-old child is about 45 mL/min, a reflection of lower body size than an
adult.

To compensate for variations in body size, GFR or clearance is often related to body
surface area. Therefore results are either expressed as mL/min/m2 or relative to an
“average” body surface area of 1.73 m2 i.e. as mL/min/1.73 m2. To do this it is first
necessary to calculate the body surface area. Body surface area can be estimated from
both height (in cm) and body weight (in Kg) using the following formula:

log10 A = (0.425 x log10 W) + (0.725 x log10 H) - 2.144

where A is body surface area in m2 , W is body weight (in Kg) and H is height (in cm).

If the measured GFR or clearance is divided by this surface area then the result becomes
mL/min/m2. If this result is then multiplied by 1.73 then the GFR or clearance is corrected
to the average or standard surface area of 1.73 m2. These calculations are summarised in
Fig 5.6.

Question Q 5(5)

A patients measured creatinine clearance is 60 mL/min. Correct this value to a standard

body surface area of 1.73 m2 given that the body weight is 38.5 kg and height 102 cm.

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CHAPTER 5

Correction of GFR (or clearance) for body surface area

A = antilog10 [ (0.425 x log10 W) + (0.725 x log10 H) - 2.144] …..… Eq. 5.10

Where: A = body surface area in m2

W = body weight in Kg
H = body height in cm

GFR expressed per m2 body surface area:

Corrected GFR (mL/min/m2) = Measured GFR (mL/min) ………...…. Eq. 5.11

A (m2)

GFR expressed as a ratio to “standard” surface area of 1.73 m2:

Corrected GFR (mL/min/1.73m2) = Measured GFR (mL/min) x 1.73 …. Eq. 5.12

A (m2)

Figure 5.6 Correction of GFR (or clearance) for body surface area

Again it should be emphasized that this correction is NOT the same as estimating
clearance from plasma creatinine measurement. In order to calculate clearance form
plasma creatinine then the body weight (W) will be used twice: once to estimate clearance
(uncorrected for body size) from plasma creatinine (using Eq. 5.9) then again to calculate
body surface area (using Eq. 5.11) for correction of this value to standard body surface
area.

Answer Q 5(5)

First calculate the surface area using Eq 5.11:

A = antilog10 [(0.425 x log10 W) + (0.725 x log10 H) - 2.144]

Substitute W = 38.5 kg and H = 102 cm then evaluate A:

A = antilog10 [(0.425 x log10 38.5) + (0.725 x log10 102) - 2.144]

88
 RENAL FUNCTION

= antilog10 [(0.425 x 1.59) + (0.725 x 2.00) - 2.144]

= antilog10 [0.676 + 1.45 - 2.144]

= antilog10 - 0.018 = 0.96 m2

Next correct the clearance for a body surface area of 1.73m2 using Eq. 5.12:

Corrected clearance (mL/min/1.73 m2) = Measured clearance x 1.73

A

Substitute: measured clearance = 60 mL/min and A = 0.96 m2

Corrected clearance = 60 x 1.73 = 108 mL/min/1.73 m2 (2 sig figs)

0.96

In this example correction of a low clearance for the small body size resulted in a normal
value.

Calculation of creatinine clearance directly from plasma

creatinine concentration

Since the largest source of error in a clearance measurement is the accuracy of the timed
urine collection and continuous urine collections are very inconvenient, attempts have
been made to derive the urinary excretion of creatinine from sources other than urinary
analysis. Creatinine originates from two sources:

• From creatine (and creatine phosphate) which is released continuously from

muscle. Under normal circumstances the rate of creatinine production is relatively
constant and proportional to the total body muscle mass.

• From dietary sources (including dietary creatine).

Therefore methods of calculation have been devised which attempt to relate muscle mass
to the rate of creatinine added to the body pool (and hence excreted in urine). Dietary
sources of creatinine are usually ignored.

a) Method of Cockroft and Gault

Cockroft and Gault measured urinary creatinine output and derived its relationship with
both body weight and age which they then used to develop a formula which estimates

89
CHAPTER 5

creatinine clearance from plasma creatinine, age and body weight (Fig 5.5).
This equation assumes that the relationship between body weight and muscle mass is
constant and may be unreliable in obese and oedematous subjects. Note that body weight
and age are ONLY used to estimate urinary creatinine output which is then incorporated
into the standard equation for creatinine clearance (Eq. 5.8). It does NOT correct the
clearance to body surface area – if this is required then an additional calculation is
required.

Question Q 5(5)

A 40-year old lady has a plasma creatinine concentration of 153 μmol/L. Estimate her
creatinine clearance (in mL/min) given that her body weight is 60 kg.

90
 RENAL FUNCTION

The Cockcroft-Gault equation

Slope = -0.0018
Rate of urinary
creatinine
excretion
mmol/24 h/kg
Intercept = 0.248

Age (years)

Rate of urinary excretion (mmol/24 h/kg) = (-0.0018 x age) + 0.248

The rate of urinary excretion is the same as U x V.

U x V (mmol/24 h) = Body wt (kg) [ 0.248 - (0.0018 x age) ]

Simplifying:

U x V = 0.0018 Body wt (0.248 - 0.0018 x age)

0.0018 0.0018

U x V = 0.0018 Body wt (kg) (138 - age)

U x V is then substituted into Eqn 5.9:

Creatinine clearance = 0.0018 Body wt (kg) (138 - age) x 700

Plasma creatinine (μmol/L)

Which simplifies to the following expression (to 2 sig figs):

Creatinine clearance (mL/min) = (140 - age (y)) x Body wt (kg) x 1.3 …. Eq. 5.13
Plasma creatinine (μmol/L)

The above data was validated for men. It is common practice to multiply this value
by 0.85 for women to correct for a lower muscle mass.

Figure 5.5. The Cockroft and Gault equation for predicting creatinine clearance
from age, body weight and plasma creatinine

91
CHAPTER 5

Answer Q 5(6)

Creatinine clearance (mL/min) = (140 - age in yrs) x Body Wt (kg) x 1.2

Plasma creatinine (μmol/L)

Substitute Age = 40 years

Body weight = 60 kg
Plasma creatinine = 153 μmol/L

Creatinine clearance (mL/min) = (140 - 40) x 60 x 1.2

153

= 100 x 60 x 1.2 = = 47 mL/min (2 sig figs)

153

Since the patient is female this result is multiplied by 0.85 to correct for a lower muscle
mass than males (for which the formula was derived).

Creatinine clearance = 47 x 0.85 = 40 mL/min (2 sig figs)

b) Formulas derived from comparative studies of GFR and serum creatinine

A purely mathematical approach is employed in order to derive equations which convert

serum creatinine to estimated GFR (eGFR), taking into account variables such as age, sex
and ethnicity which are likely to affect creatinine production. GFR was always obtained
by a reliable clearance method (e.g. iothalamate) and serum creatinine measurements
aligned to a reference method. GFR, and hence eGFR, is expressed for an average body
surface area of 1.73 m2.

The earliest equation to be widely recommended in most official guidelines was derived
by the Modification of Diet in Renal Disease (MDRD) study group. Later the Chronic
Kidney Disease Epidemiology Collaboration (CKD-EPI) published a new equation using
pooled data from multiple studies for both its development and validation1. Their
approach was a little different in that they divided the data into two groups – those with
serum creatinine below a certain value (called a “knot”) and those above it – resulting in
two equations (although they can be combined into one provided appropriate constants
are selected according to plasma creatinine and sex). The CKD-EPI equation performed
better than the MDRD equation, especially at higher GFR. At the time of writing the
CKD-EPI equation is recommended by the National Institute of Clinical Excellence
(NICE). An easy to use version of this equation is included in Fig 5.6.
1
Levey AS, Stevens LA, Schmid CH et al. A new equation to estimate glomerular
filtration rate. Ann Intern Med 2009; 150(9): 604-612.

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 RENAL FUNCTION

These formulae are still undergoing development and it is likely that guidelines may
change in the near future.

Chronic Kidney Disease Epidemiology Collaboration (CKD-EPI) formula

GFR (mL/min/1.73 m2) = 141 x [serum creatinine (µmol/L) x 0.011312]α

k

x 0.993Age x [1.018 if female] x [1.159 if black]

Appropriate values for α and k are:

Male Female
Serum creatinine (µmol/L) ≤80 >80 ≤62 >62
k 0.9 0.9 0.7 0.7
α -0.411 -1.209 -0.329 -1.209

Fig 5.6 CKD-EPI formula for estimation of GFR (eGFR) from serum creatinine
concentration

Question Q 5(7)

Use the CKD-EPI formula to estimate the GFR for a 55-year old Caucasian woman
whose serum creatinine is 125 μmol/L.

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CHAPTER 5

Answer Q 5(6)

Substitute into the CKD-EPI equation:

Serum creatinine = 125 µL Age = 55 years

k = 0.7 (since patient is female)
α = -1.209 (since patient is female with creatinine > 62)
Include the factor of 1.018 since the patient is female but not the factor of 1.159
since she is Caucasian.

GFR = 141 x [125 x 0.011312]-1.209 x 0.99355 x 1.018

0.7

= 141 x 0.4274 x 0.6795 x 1.018 = 42 µmol/L/1.73 m2

Measures of Tubular Function – Fractional Excretion (FE)

and Tubular Maximum (Tm)
The fractional excretion (FE) expresses the excretion of a solute as a fraction of the
amount filtered at the glomeruli:

Fractional excretion (FE) = Amount excreted in urine

Amount filtered

or Rate of urinary excretion

Rate of filtration

The rate the substance is excreted in urine can be calculated from the concentration of the
substance in urine (Usubstance) and the urine flow rate (V):

Rate of excretion = Usubstance x V

The rate of filtration is the product of GFR and plasma concentration (P):

94
 RENAL FUNCTION

Rate of filtration = Psubstance x GFR

These two expressions can be combined to give a formula to calculate FEsubstance:

FEsubstance = Usubstance x V………………………Eq. 5.14

Psubstance x GFR

If GFR is estimated at the same time and using the same urine collection by measuring
the concentrations of creatinine in the same plasma and urine, then:

GFR = (Ucreatinine x V)
Pcreatinine

This value for GFR can then be substituted into Eq. 5.14 to give:

FEsubstance = Usubstance x V x Pcreatinine

Psubstance x Ucreatinine x V

The V terms cancel so that the expression simplifies to:

FEsubstance = Usubstance x Pcreatinine ………....Eq. 5.15

Ucreatinine x Psubstance

Question Q 5.8

The following results were obtained in a 20 year-old male admitted after a car crash and
found to be oliguric:

Plasma: Sodium = 125 mmol/L Creatinine = 200 μmol/L

Urine: Sodium = 60 mmol/L Creatinine = 1.2 mmol/L

Calculate the fractional excretion of sodium.

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CHAPTER 5

Answer Q5(8)

FENa = UNa x Pcreat

Ucreat x PNa

Substituting the relevant values after making sure that units are appropriate:

PNa = 125 mmol/L

UNa = 60 mmol/L
Pcreat = 200 μmol/L
Ucreat = 1.2 mmol/L = 1.2 x 1000 = 1200 μmol/L

FENa = 60 x 200 = 0.08

1200 x 125

Many substances filtered at the glomerulus are reabsorbed by the tubules. If the rate of
filtration is less than the rate of tubular re-absorption for the substance then the urinary
excretion is zero. The proportion of the filtered substance that is reabsorbed is called the
tubular reabsorption (TR). Since the rate of filtration of a substance is given by the
product of its concentration and GFR, the rate of filtration increases with plasma
concentration. Most tubular transport mechanisms are concentration dependent so as the
concentration of the substance in the glomerular filtrate increases the rate of reabsorption
also increases but eventually a threshold is reached as the tubular transport mechanism
becomes saturated and no further substance can be reabsorbed and any excess appears in
the urine. This maximal rate of tubular transport is called the Tm value of the substance.
This gives rise to the concept of the renal threshold for a substance and is illustrated in
Fig 5.7. Since by definition the proportion filtered is 1, the numerical relationship
between FE and TR is:

FE = 1 - TR ……………………………. Eq. 5.16

If the rate of filtration (GFR x P) is less then the value for TR then none of the substance
appears in the urine and the value for FE is negative. However, if the rate of filtration
exceeds the TR value then the renal threshold is exceeded and the substance appears in
urine and is possible to calculate TR by rearranging Eq. 5.17:

TR = 1 - FE ………………………... Eq. 5.17

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 RENAL FUNCTION

In other words the fraction that is not excreted must have been reabsorbed. To convert
this fraction to an absolute concentration (i.e. the amount of substance reabsorbed per
given volume of filtrate) then the value for TR is multiplied by the plasma concentration.
This is the same thing as the ratio of the maximal rate of reabsorption (Tm) to the GFR –
i.e. the Tm/GFR:

Tm/GFR = TR x P ……………. Eq. 5.18

This method of calculation assumes that the plasma concentration (P) is well above the
renal threshold. Units are concentration e.g. mmol/L.

Question Q 5(9)

The following results were obtained for urine and plasma from a fasting adult:

Plasma phosphate 0.65 mmol/L Plasma creatinine 105 μmol/L

Urine phosphate 11.5 mmol/L Urine creatinine 4.64 mmol/L

Calculate: a) The fractional excretion of phosphate (FEP)

b) The fractional tubular reabsorption of phosphate (TRP)
c) The renal tubular reabsorption of phosphate (TmP/GFR)

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The renal threshold

Rate of Filtered
Reabsorbed
filtration
or
reabsorption
or
excretion Tm Excreted

Renal
Threshold P
Tm/GFR

Figure 5.7 The renal threshold for a substance filtered at the glomerulus then
reabsorbed by the tubules

Answer Q5(9)

a) To calculate the fractional excretion of phosphate (FEP) use Eq. 5.15:

FEP = Uphosphate x Pcreatinine

Ucreatinine x Pphosphate

Where: Uphosphate = 11.5 mmol/L

Pcreatinine = 105 μmol/L
Ucreatinine = 4.64 mmol/L = 4.64 x 1000 = 4640 μmol/L
Pphosphate = 0.65 mol/L

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 RENAL FUNCTION

Substituting these values:

FEP = 11.5 x 105 = 0.40

4640 x 0.65

b) The proportion of filtered phosphate that is reabsorbed (TRP) is the difference

between the fraction filtered (by definition 1) and the fraction excreted (FEP):

TRP = 1 - FEP

TRP = 1 - 0.40 = 0.60

c) Since the TRP is the fraction of filtered phosphate that is reabsorbed, multiplication
by the plasma phosphate concentration gives the maximum rate of reabsorption of
phosphate (in mmol) per litre of glomerular filtrate (TmP/GFR):

TmP/GFR = TRP x Pphosphate

TmP/GFR = 0.60 x 0.65 = 0.39 mmol/L glomerular filtrate

Reference: Payne RB. Renal tubular reabsorption of phosphate (TMP/GFR):

indications and interpretation. Ann Clin Biochem 1998; 35: 201-206.

Osmolar Clearance and Free Water Clearance

If osmolality is substituted for creatinine in the calculation of creatinine clearance then

the result (Fig 5.8) is the osmolar clearance (Cosm). Osmolality is a measure of the sum
of the individual osmotically active species present in plasma or urine (see chapter 6).
Just as the creatinine clearance can be thought of as the volume of plasma from which all
creatinine is removed per unit of time, osmolar clearance can be regarded as the volume
of plasma from which all filterable solutes are removed in any given time period. It can
also be thought of as the minimum volume of urine required to excrete a solute load in
isosmolar (the same osmolality as plasma) form.

The calculated difference between the actual urine volume and the osmolar clearance is
known as free water clearance. In a hyposmolar urine the free water clearance is
positive; in a hyperosmolar urine it is negative. The free water clearance can also be
viewed as the volume of pure water subtracted from (positive free water clearance) or
added to (negative free water clearance) the plasma per unit time.

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The renal excretion of a large water load is limited by the diluting capacity of the tubules
and the amount of solute available for excretion. The minimum osmolality of urine is
about 50 mmol/kg. On a medium protein diet 1200 mmol is excreted per day (mainly
urea) and the maximum urine volume is 1200/50 = 24 L. In starvation however, the
source of solute is tissue breakdown (approx. 100-200 mmol/day. There is little ability to
excrete free water and hyponatraemia myoccur if water intake is greater than 2 to 4 L/day
in starvation.

Calculation of osmolar clearance and free water clearance

If: V = urine flow rate

Posm = plasma osmolality
Uosm = urine osmolality

Then the osmolar clearance (Cosm) is given by:

Cosm = Uosm x V ………………………… Eq. 5.19

Posm

And the free water clearance (Cwater) is given by:

Cwater = V - Cosm ……………………… Eq. 5.20

Figure 5.8 Calculation of osmolar clearance and free water clearance

Question Q 5(10)

A 24 h urine collection (volume 1.20 L) has an osmolality of 750 mOsm/kg. If the

plasma osmolality is 300 mOsm/kg calculate:

a) The osmolar clearance in mL/min

b) The free water clearance in mL/min.

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 RENAL FUNCTION

Answer Q 5(10)

a) Using Eq. 5.19: Cosm = Uosm x V

Posm

Posm = 300 mOsm/Kg

Uosm = 750 mOsm/Kg
V = urine flow rate = 1.20 L/24 h

Since Cosm is required in mL/min, the flow rate must first be converted to
mL/min by multiplying by 1000 (1000 mL in 1L) and dividing by 24 (to
convert to h) then 60 (to covert to min):

V = 1.20 x 1000 = 0.833 mL/min

24 x 60

Substituting into Eq. 5.19:

Cosm = 750 x 0.833 = 2.08 mL/min

300

b) Calculate the free water clearance using Eq 5.20

Cwater = V - Cosm

Ensuring that the units are the same:

V = 0.833 mL/min
Cosm = 2.08 mL/min

Cwater = 0.833 - 2.08 = -1.25 mL/min (2 sig figs)

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FURTHER QUESTIONS

1. An aliquot of a 24 h urine (volume 1850 mL) has a creatinine concentration of

8500 μmol/L. Calculate the 24 h urinary creatinine excretion expressing the result
as mmol/24 h.

2. A patient has a GFR of 110 mL/min. If the plasma creatinine concentration is 180
μmol/L how many mmol of creatinine are filtered in 12 h?

3. A urine collection (volume 3.2 L) was handed in by a patient which he said he had
collected over the previous day. Calculate the creatinine clearance given that the
urine was found to have a creatinine concentration of 7.2 mmol/L. The plasma
creatinine concentration taken during the collection was 94 µmol/L. Give the
most likely cause for this result.

4. The concentration of a compound in the plasma of a normal adult is 10 mg/L.

The GFR is 110 mL/min and 316.8 mg of the compound are excreted over 24 h in
a urine volume of 1584 mL. Comment on these findings.

5. A subject with a GFR of 100 mL/min was infused with a 'drug' X at a rate of
100 µmol/min and the plasma concentration reaches a steady state value of
200 µmol/L. It is known that this drug is not metabolized or excreted by organs
other than the kidney. What is the clearance of this drug? Comment on the result.

6. A patient who is severely water depleted and excreted only 100 mL of urine in the
last 6 hours was a short time before, found to have a creatinine clearance of
100 mL/min with a plasma creatinine concentration of 100 µmol/L. If renal
function has remained unchanged what concentration of creatinine would you
expect to find in the latest 100 mL (6 h collection) specimen of urine?

7. Estimate the effect on urinary sodium excretion in a person with a GFR of

95 mL/min and plasma sodium of 140 mmol/L, of a 1% decrease in the overall
reabsorption of sodium.

8. The following data were obtained for a hypertensive patient on a low sodium diet:

Plasma: creatinine = 200 μmol/L sodium = 155 mmol/L

24 H Urine: creatinine = 12.5 mmol/L volume = 1250 mL

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 RENAL FUNCTION

If the renal tubules reabsorb 90% of filtered sodium, how many grams of
sodium are excreted in the same 24 h period?

9. The following results were obtained in a 20 year old male admitted after a car
crash and found to be oliguric:

Plasma Na 125 mmol/L

K 5.0 mmol/L
Urea 25.0 mmol/L
Creatinine 200 µmol/l

Urine Na 60 mmol/L
Creatinine 1.2 mmol/L
Osmolality 200 mOsm/Kg

Calculate the fractional excretion of sodium.

10. A 45 year old lady has a body weight of 56 kg and a height of 155 cm. If her
plasma creatinine is 150 μmol/L estimate her GFR expressing the result as
mL/min/1.73 m2.

11. Calculate the tubular maximum reabsorptive capacity (Tm/GFR) for glucose from
the following data:

Plasma glucose 10 mmol/L Plasma creatinine 120 μmol/L

Urine glucose 50 mmol/L Urine creatinine 6.0 mmol/L

The urine (volume 30 mL) was collected over a 15 minute period.

12 A 6 h urine collection (volume 800 mL) has an osmolality of 200 mOsm/kg. If

the plasma osmolality is 260 mOsm/kg calculate the free water clearance in
mL/min.

13. An estimation of glomerular filtration rate can be calculated using the abbreviated
MDRD (Modified Diet in Renal Disease) formula:

GFR (mL/min/1.73 m2) = 186 x [serum creatinine x 0.011312]-1.154

x [age in years]-0.203 x 0.742 if female and/or x 1.21 if Afro American origin
(where serum creatinine is in μmol/L)

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Calculate the GFR for a 55 year old Caucasian women whose serum creatinine is
125 μmol/L, and her creatinine clearance, given that a 24 h urine collection with a
volume of 1.1 L had a creatinine concentration of 4.7 mmol/L.

Comment critically on the two values.

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 OSMOLALITY

Chapter 6

Osmolality

Determination of plasma and urine osmolality can be useful in the assessment of

electrolyte disorders. Comparison of plasma and urine osmolalities is invaluable in
investigating renal water regulation in the setting of severe electrolyte disturbances as
may occur in diabetes insipidus and the syndrome of inappropriate antidiuresis.

What is osmolality?
Osmosis is the process by which solvent moves from an area of low solute concentration,
through a semi-permeable membrane, to an area of high solute concentration. A semi-
permeable membrane is one which is permeable to solvent but not solute. The pressure
of the solvent movement will depend on the gradient in solute concentrations separated
by the membrane and is known as osmotic pressure. In the human body there is no active
mechanism for the transport of water into and out of cells; water always follows an
osmotic gradient. Most cell membranes are freely permeable to water. The main
exception to this are the membranes of cells lining the collecting tubules in the kidney,
which only become permeable to water in the presence of antidiuretic hormone (ADH).
The cell membranes are however, selectively permeable to solutes such as sodium and
glucose a property which is modulated by hormonal action e.g. aldosterone and insulin.

The osmotic pressure of a solution depends upon the concentration of particles in

solution, which is termed its osmolality. The osmolality of a solution is the total
concentration (in mols) of all osmotically active species present. A molecule such as
sodium chloride which can dissociate into two ionic particles (sodium and chloride ions)
will have an osmolality of approximately twice that of an equimolar solution of a species
such as glucose which does not dissociate into constituent ions. However, since
electrolyte dissociation is always incomplete and there are always associations between

105
CHAPTER 6

solute and solvent, electrolytes do not behave in an ideal manner. In these circumstances
osmolality can be calculated in the following manner:

Osmolality = osmol/kg water = Ǿ n C ……………… Eq. 6.1

Where: Ǿ = osmotic coefficient

n = number of particles into which each molecule in the
solution potentially dissociates
C = molality in mol/kg water

In routine clinical biochemistry it is usual to assume that osmotic coefficients are always
equal to 1.

Confusion often arises between the terms osmolality and osmolarity. These terms may
be defined as follows:

• Osmolality expresses concentrations relative to the mass of solvent

e.g. mols per kg of water.

• Osmolarity expresses concentrations relative to volume of solution

e.g, mols per litre of solution.

Osmometers always measure osmolality. Osmotic concentration calculated from molar

concentrations gives osmolarity. In routine clinical biochemistry the difference is nearly
always ignored and the two terms used interchangeably.

Properties other than osmotic pressure are also dependent upon the concentration of
solute particles in solution. These include increasing vapour pressure, raising boiling
point and decreasing freezing point. These are all known as colligative properties and
each may be used to measure to obtain a measure of the active solute concentration of a
solution. In routine clinical practice osmometers based on depression of freezing point
are usually used to obtain a measure of osmolality.

Question Q 6(1)

Estimate the osmolality (in mOsmol/Kg) of (a) 20% glucose, (b) Physiological saline,
(c) a mixture containing equal volumes of 20% glucose and physiological saline, and (d)
50 mmol/L calcium chloride. Assume ideal behaviour.

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 OSMOLALITY

Answer Q6(1)

(a) Glucose does not dissociate into ions in solution so its osmolality will be
approximately equal to its molar concentration.

Concentration (mmol/L) = Concentration (mg/L)

MW

Concentration (mg/L) = Concentration (g/100 mL) x 10 x 1000

Substitute the concentration of 20%, which is the same as 20 g/100 mL:

Concentration (mg/L) = 20 x 10 x 1000 = 200,000 mg/L

Glucose formula: C6H12O6 C6 = 6 x 12 = 72

H12 = 12 x 1 = 12
O6 = 6 x 16 = 96
MW = 180

Concentration (mmol/L) = 200,000 = 1100 mmol/L (2 sig figs)

180

Therefore the osmolality of 20% glucose is approximately 1100 mOsm/kg

(b) Physiological saline contains 0.9% sodium chloride.

Concentration (mmol/L) = Concentration (mg/L)

MW

Concentration (mg/L) = Concentration (g/100mL) x 10 x 1000

Substitute the sodium chloride concentration of 0.9%, which is the same as

0.9g/100 mL

Concentration (mg/L) = 0.9 x 10 x 1000 = 9000 mg/L

Formula sodium chloride : NaCl Na = 23

Cl = 35.5
MW = 58.5

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CHAPTER 6

Concentration (mmol/L) = 9000 = 154 mmol/L

58.5

Since each molecule potentially dissociates into two ionic species:

NaCl Na+ + Cl-

Then assuming ideal behaviour the osmolality will be approximately twice the
molar concentration of sodium chloride:

Osmolaity = 2 x 154 = 308 mOsm/kg

(c) If equal volumes of 20% glucose and physiological saline are mixed then the
resulting osmolality will be one half the sum of the individual osmolalities:

Osmolality = Osmolality20%glucose + OsmolalityPhysiological saline

2

Osmolality = 1100 + 308 = 1408 = 700 mOsm/kg (2 sig figs)

2 2

(d) Calcium chloride (CaCl2) potentially dissociates into 3 ionic species:

CaCl2 Ca++ + 2 Cl-

Therefore the osmolality will be approximately three times the concentration:

Osmolality = 3 x 50 = 150 mOsm/kg

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 OSMOLALITY

Plasma osmolality and the osmolal gap

The major osmotically active species present in normal plasma are Na+, Cl-, glucose and
urea. The simplest formula for calculating osmolality from the concentrations of these
species is:

Osmolality = 2 [Na+] + glucose + urea ………… Eq. 6.2

mOsm/kg mmol/L mmol/L mmol/L

Another version of this formula includes a term for potassium concentration i.e. 2[K+].
The concentration of sodium is multiplied by 2 to allow for the associated anions (mainly
chloride and bicarbonate). However, this simple formula does not give very good
agreement with measured osmolality since the following assumptions are made:

• That all important osmotically active species are accounted for.

• That all potential dissociations are complete.

• That the anions associated with Na+ and K+ are free to contribute to omolality and
are not part of a macromolecule (e.g. protein).

• That the activity of each species is the same as concentration i.e. the ions exhibit
ideal behaviour.

• That the millimolal concentration of each ion (mmol/kg water) is the same as its
millimolar concentration (mmol/L plasma). This is not true since plasma is
approximately 95% water.

In an attempt improve the agreement between measured and calculated osmolality

various modifications of the above formula have been proposed. One of the most popular
is:

Osmolality = 1.86 [Na+] + [glucose] + [urea] + 9 ……….. Eq. 6.3

mOsm/kg mmol/L mmol/L mmol/L

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CHAPTER 6

It is important to remember that osmolality is calculated from three individual

measurements, each with its own inherent imprecision so that the combined imprecision
of the final result may be greater than the imprecision of measured osmolality. Within
these limitations agreement between calculated and measured osmolality is reasonable for
plasma but NOT for urine.

In routine clinical practice the commonest reason for comparing measured with
calculated osmolality is to obtain evidence for the presence of, or to quantify, an
unmeasured osmotically active species. If no such species is present then, within the
limitations discussed above, there should be good agreement between measured and
calculated osmolality. In other words the difference between the two values (known as
the osmotic gap) should be approximately zero. If the osmolal gap has a significant
numerical value then this is good evidence that an unmeasured osmotically active species
is present and the size of the gap is proportional to its concentration.

Osmolal gap = Measured osmolality - Calculated osmolality ….. Eq. 6.4

mOsm/kg mOsm/kg mOsm/kg

The osmolal gap is often calculated if a patient is suspected of having ingested large
amounts of a volatile compound such as ethanol, methanol or ethylene glycol. Provided
only a single compound has been ingested and its identity is known then it is possible to
derive an approximate value for its concentration which is adequate to act as a guide for
treatment. Apart from the limitations inherent in calculating osmolality, errors may occur
because volatile solvents such has ethanol do not behave entirely as expected with some
osmometers.

Question Q 6(2)

The following data were obtained on the plasma from an unconscious man:

sodium = 140 mmol/L

urea = 7.0 mmol/L
glucose = 7.0 mmol/L
osmolality = 353 mosmol/kg

Assuming that the osmolar gap is due solely to ethanol, calculate the plasma ethanol
concentration in mg/dL.

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 OSMOLALITY

Answer Q 6(2)

First calculate the osmolality from the concentrations of sodium, urea and glucose:

Calculated osmolality = 1.86 [Na+] + [urea] + [glucose] + 9

= (1.86 x 140) + 7.0 + 7.0 + 9

= 283 mOsmol/kg

The osmolal gap is the difference between measured and calculated osmolality:

Osmolal gap = Measured osmolality - Calculated osmolality

= 353 - 283

= 70 mOsmol/kg

If this osmolal gap is due entirely to ethanol, then the ethanol concentration is
70 mmol/L.

To convert to mg/dL, multiply by the MW and divide by 10.

MW ethanol (C2H5OH) = (2 x 12) + 16 + (6 x 1) = 46

Therefore, ethanol = 70 x 46 = 322 mg/dL

10

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CHAPTER 6

ADDITIONAL QUESTIONS

1. Calculate the approximate osmolality of a glucose/saline infusion containing

equal proportions of 5% glucose and 0.9% sodium chloride.

2. Calculate the approximate osmolality of a solution containing 10% mannitol and

0.9% saline (MW mannitol = 182).

3. A patient was mistakenly given 500 mL 20% mannitol (C6H14O6 ) intended for the
patient in the next bed instead of the same volume of normal (0.9%) saline.
Calculate the extra osmolal load given over that which would have resulted from
isotonic saline.

4. What increase in plasma osmoality would result from a plasma ethanol

concentration of 92 mg/dL?

5. A 45 year old man is brought to casualty following a fit. He had been working
alone late in a garage, when he was found by the security guard who called an
ambulance. On admission, he has a large bruise on the left temple and is semi-
comatose, he smells of alcohol. The admitting team request urea and electrolytes,
glucose and an alcohol and blood gas estimation and arrange an urgent CT scan.
The results are as follows:

Sodium 141 mmol/L Urea 3.5 mmol/l

Ethanol 270 mg/dL Glucose 3.2 mmol/L

The CT scan does not show any bony injury or evidence of intracranial bleed.
The neurological registrar is called and asks for an osmolal gap to help provide a
quick estimation of whether there is a possibility that other toxic substances
present in the garage, such as antifreeze, have been taken in any quantity.

The measured osmolality is 330 mOsm/kg

a) Calculate the osmolal gap

b) Show whether the alcohol concentration explains the observed osmolal
gap, explaining any assumptions you make in the process.

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 BASIC PHARMACOKINETICS

Chapter 7

Basic pharmacokinetics

Pharmacokinetics may be defined as what the body does with drugs and includes such
processes as absorption, distribution, metabolism and elimination. In practice
phamacokinetic factors determine not only the plasma concentration of the “active” drug
but the amount of active drug reaching its site of action. Therapeutic drug monitoring
(TDM) may be defined as the use of drug measurements in body fluids as an aid to the
use of drugs in the management (i.e. cure, alleviation or prevention) of disease.

Bio-availability (F)
The proportion of administered drug absorbed is termed its bio-availability (F) and
depends on the drug, the individual and the dosage form of the drug. Bio-availability is
defined as the proportion of administered drug which reaches the circulation:

Bio-availability (F) = Dose reaching circulation …………….. Eq. 7.1

Dose administered

This expression can be rearranged to give the dose absorbed for any administered dose:

Dose reaching circulation = F x Dose administered …………... Eq. 7.2

Salt-conversion factor (S)

Often a drug are administered in several chemical forms. For example phenytoin may be
given as the free acid (MW 252) or as its sodium salt (MW 274). Clearly the amount of
drug administered if the same weight of each compound is given, will be different.
A quantity termed the salt-conversion factor or salt fraction (S) may be defined:

Salt conversion factor (S) = Molecular weight of free drug .. Eq. 7.3
Molecular weight of compound administered

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CHAPTER 7

Therefore for sodium phenytoin S would be 252/274 = 0.92 equation 7.2 can be modified
to incorporate the salt conversion factor:

Dose reaching circulation = F x S x Dose administered …………... Eq. 7.4

Volume of distribution (Vd)

Once a drug has been absorbed by the body its distribution depends on such factors as its
relative solubility in water and fat, binding by plasma protein, availability of active
transport mechanisms and regional blood flow. In the laboratory the relationship
between the volume of a solution, the weight of chemical used in its preparation and the
concentration of the substance is familiar to us all:

Concentration = Weight and Volume = Weight

Volume Concentration

Therefore for a given weight of a substance its concentration is inversely proportional to

volume The same concept can be applied to a drug, the concentration of which is
measured in plasma:

Volume of distribution (Vd) = Amount of drug in body ………… Eq. 7.5

Plasma concentration

The volume of distribution is a theoretical concept. If the drug is distributed throughout

the ECF only then Vd will approximate to the ECF volume. If the drug is lipid soluble
then Vd may be far in excess of the volume of total body water.

Question Q7(1)

Calculate the theoretical maximum plasma concentration if 500 mg of the sodium salt of
a drug is administered to a 70 kg male. Assume the drug (MW of the parent drug 345
Daltons) is only distributed throughout the extra-cellular fluid (the volume of which is
20% of the body weight) and its bio-availability is 0.85.

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 BASIC PHARMACOKINETICS

Answer Q7(1)

First calculate the salt-conversion factor (S):

S = MW parent drug
MW sodium salt of drug

MW parent drug (free acid) = 345

For its sodium salt, Na (atomic weight 23) replaces a hydrogen atom (atomic weight 1).

Therefore MW sodium salt of drug = 345 - 1 + 23 = 367

S = 345 = 0.94
367

Dose absorbed = F x S x Dose administered

Since F = 0.85 and 500 mg was administered:

Dose absorbed = 0.85 x 0.94 x 500 = 400 mg

The volume of distribution (Vd) is the ECF volume which is 20% of the body weight
(70 kg) – assuming density of ECF is approximately 1:

Vd = 70 x 20 = 14 litres
100

Vd = Dose absorbed
Plasma concentration

Therefore: 14 (litres) = 400 (mg)

Plasma concentration (mg/L)

Re-arranging: Plasma concentration = 400 = 29 mg/L (2 sig figs)

14

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CHAPTER 7

CLEARANCE OF A DRUG

Drugs can be cleared from the body by two principal routes:

• Renal excretion (Clr)

• Metabolism by the liver (Clm)

The total clearance of a drug is the sum of both individual clearances:

Total clearance = Clr + Clm

In practice hepatic function is difficult to quantify and most dosage calculations take into
account impairment of renal function only.

Following a single dose of a drug a plot of plasma drug concentration versus time is
usually non-linear (Fig 7.1). As the drug is cleared the amount of drug removed from
plasma in unit time (i.e. the rate of fall in concentration) decreases as plasma
concentration falls. In fact the rate of fall with time can be expressed as:

d Cpt is proportional to Cpt

dt

where Cpt is the concentration of the drug at time t.

The symbols d Cpt does NOT mean d multiplied by Cpt but a very small (infinitesimally
so ) change in Cpt. Similarly d t is a minute change in t. Mathematicians call d Cpt / d t
the first differential of concentration with respect to time and it can be regarded as the
slope of a tangent drawn to the curve at any defined point ( i.e. any value of t ). When the
differential d Cpt / d t is proportional to a single concentration term then the elimination
process is said to follow first-order kinetics. If d Cpt / d t is independent of concentration
(i.e. constant) then the elimination is said to follow zero-order kinetics and, if
proportional to two concentration terms, second-order kinetics etc.

If a proportionality constant (kd) is introduced then the following equation is obtained:

d Cpt = kd. Cpt

dt

where kd is known as the elimination rate constant which has units time.-1 This equation
can be rearranged to give:

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 BASIC PHARMACOKINETICS

d Cpt = kd . d t
Cpt

In practice it is easier to measure absolute concentrations (and this is what we require in

therapeutic calculations) than it is to measure the rates of change in concentration, so
mathematicians use a technique known as integration to convert the rate equation into an
expression relating absolute concentration to time. Integration is always carried out
between defined limits – in this case between zero time (when the initial concentration is
given the symbol Cp0) and time t (when the concentration is given the symbol Cpt).
Integration of the above expression between these limits gives the equation:

Cpt = Cp0 x e –kd.t ………………. Eq. 7.6

Where e is a universal constant which is encountered in nature whenever we are dealing

with exponential growth or decay and has a value of approximately 2.718. Eq 7.6 is still
non-linear but can be converted into a linear form by taking logarithms. However, since
e appears in the equation it is more convenient to take logarithms to the base e rather than
to the base 10. Logarithms to base e (loge is conventionally denoted ln) are also known as
natural logarithms or Napierian logarithms. Logarithms to the base e behave exactly the
same as logarithms to the base 10 (see Fig 3.2). For example ln a.b = ln a + ln b, ln
a/b = ln a - ln b and ln ab = b. ln a. Therefore ln e –kd.t = -kdt. ln e which, since ln
e = 1, becomes -kd.t. Therefore taking natural logarithms of Eq 7.8 gives the
expression:

ln Cpt = ln Cp0 - kd.t ……………. Eq. 7.7

Therefore if ln Cpt is plotted against t then a straight line is obtained with slope -kd and
intercept on the vertical axis equal to ln Cp0 (see Fig 7.1).

If log10Cpt is plotted instead of ln Cpt then a straight line is still obtained, but this time the
slope is equal to - kd/2.303. This value can be derived using the relationship between
natural and common logarithms: ln x = ln 10 x log10 x = 2.303 log10 x. Eq. 7.7
can then be written:

2.303 log10 Cpt = 2.303 log10 Cp0 - kd.t

and dividing both sides by 2.303 gives:

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log10 Cpt = log10 Cp0 - kd . t

2.303

Most modern calculators readily produce natural logarithms so it is rarely necessary to

work in common logarithms. The elimination rate constant, kd, is easily calculated if we
have two plasma drug concentrations at different times:

If Cp1 = plasma drug concentration at time t1

and Cp2 = plasma drug concentration at time t2

then each pair of values can be substituted into Eq 7.7 to give the following two
equations:

ln Cp1 = ln Cp0 - kd . t 1
ln Cp2 = ln Cp0 - kd . t 2

Subtraction of one equation from the other eliminates the Cp0 term to give:

(ln Cp1 - ln Cp2) = - kd . t 1 + kd. t2

which can be rearranged to give an expression for kd:

kd = (ln Cp1 - ln Cp2) …………….. Eq. 7.8

( t2 - t1 )

The units of kd are time.-1

Question Q 7(2)

A 15 year old boy presents to casualty following a convulsion. It turns out that he had
swallowed 30 of his mother’s lithium tablets about 10 hours previously. On admission
his lithium concentration is 4.1 mmol/L. A decision needs to be made whether to
haemodialyse him to reduce the lithium concentration. As this is not going to be
available quickly, the physicians want to know how long he will have toxic levels just
with endogenous clearance. How long it will be before his lithium concentration drops
to the relatively safe level of 1.5 mmol/L below which toxicity is unlikely, given an
elimination rate constant of 0.05 h.-1

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Answer Q 7(2)

Let Cp0 = initial drug concentration = 4.1 mmol/L

Cpt = safe drug level at time t = 1.5 mmol/L
t = time taken to achieve the safe level of 1.5 mmol/L = ??
kd = elimination rate constant = 0.05 h-1

Substitute these values into the first order rate equation (Eq. 7.7) and solve for t:

ln Cpt = ln Cp0 - kd . t

ln 1.5 = ln 4.1 - 0.05. t

0.405 = 1.411 - 0.05. t

0.05. t = 1.411 - 0.405 = 1.006

t = 1.006 = 20.12 h (20 h to 2 sig figs)

0.05

A quantity known as the elimination half-life (t 1/2) can be defined as the time taken for
the plasma drug concentration or total body content of the drug to fall by 50%. Fig 7.1
shows that after one half-life the plasma concentration falls to 50%, after two half-lives to
25%, after three half-lives to 12.5% etc.

The elimination rate constant is related to the elimination half-life. If the initial drug
concentration is Cp0, then after one half-life (when t = t ½), the concentration Cpt will be
one half of the initial value i.e. Cp0/2. If these values are substituted into the logarithmic
form of the first-order rate equation (Eq. 7.7) then the following expression is obtained:

ln Cp0 / 2 = ln Cp0 - kd. t 1/2

which can be rearranged to give:

kd . t ½ = ln Cp0 - ln Cp0 / 2

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Since the difference between the logarithms of two individual numbers is the same as the
logarithm of one number divided by the other, this expression can also be written:

kd . t ½ = ln. Cp0 x 2
Cp0

The Cp0 terms cancel so this expression becomes:

kd . t ½ = ln 2

ln 2 is 0.693, so rearrangement gives the following expression for the half-life:

t½ = 0.693 ……………………. Eq. 7.9

kd

Rearranging Eq 7.9 gives kd = 0.693 / t ½ which can then be substituted into the
Eq. 7.7 to give an alternative formula:

ln Cpt = ln Cp0 - 0.693. t

t 1/2

which can be rearranged to give the following alternative expression for half-life:

t½ = 0.693. t ………………….. Eq. 7.10

ln Cpt - ln Cp0

Question Q 7(3)

A bolus of 6 g drug is given IV and plasma concentration of the drug determined at

intervals giving the following data:

Time since dose (h) Plasma concentration (mg/L)

2.5 32
5 10

What is the half-life of the drug?

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Time (t) Concentration (Cp) ln. Cp

0 100 4.61
1 50 3.91
2 25 3.22
3 12.5 2.53
4 6.25 1.83
5 3.125 1.14

100

Cp
50

0
0 1 2 3 4 5 6
t

5
Slope = - k d
4
3
ln Cp
2 Intercept = ln
Cp 0
1
0
0 1 2 3 4 5 6

Figure 7.1 Clearance of a drug (concentrations expressed as percentage of initial

value) with time (expressed as number of half-lives)

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Answer Q 7(3)

The 2.5 h blood sample can be considered as the initial sample (Cp0) and the 5 h sample
as the sample when t = 2.5 h (i.e. 5.0-2.5) with concentration Cpt. Therefore:

Cp0 = 32 mg/L
Cpt = 10 mg/L
t = 2.5 h

These values can then be substituted into Eq. 7.10 and solved for t ½ :

t½ = 0.693. t
ln Cpt - ln Cp0

t½ = 0.693 x 2.5
ln 32 - ln 10

t½ = 1.73
3 .47 - 2.30

t½ = 1.73 = 1.5 h (2 sig figs)

1.17

Note that ln Cp means the natural logarithm of concentration and not ln multiplied by Cp.
Therefore ln Cpt - ln Cp0 is NOT the same as ln (Cpt - Cp0).

An alternative approach to this problem would be to plot the natural logarithm of the two
drug concentrations (32 and 10 mg/L) against the times (2.5 and 5 h), measure the slope
to derive kd, then divide 0.693 by kd to obtain the half-life.

So far we have been dealing with the kinetics of elimination following administration of a
single dose of a drug. In many situations patients receive maintenance therapy: that is the
drug is taken at regular intervals. The dose is repeated well before the previous dose has
been eliminated from the body. Eventually, after multiple dosing, the situation is
reached where the rate of administration of a drug (RA) is equal to the rate of elimination
(RE) so that a constant steady state plasma concentration (Cpss) is achieved. The
clearance of a drug (Cl) can be defined as the volume of plasma from which the drug is
completely cleared per unit of time. Therefore the rate of elimination is the product of
clearance and plasma steady state concentration:

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 BASIC PHARMACOKINETICS

RE = Cl x Cpss

The rate of administration is the amount of drug administered per unit time and depends
on the dose administered, bioavailability (F), salt factor (S) and interval between doses
(τ):

RA = Dose x F x S
τ

When a steady state is reached, RA = RE, and substitution of the expressions for RA
and RE gives:

Dose x F x S = Cl x Cpss
τ

which can be rearranged to give an expression for the calculation of clearance:

Cl = Dose x F x S ……………….. Eq. 7.11

τ x Cpss

For a drug which is eliminated by glomerular filtration alone, its clearance is equal to the
patient’s GFR (se chapter 5).

For drugs which exhibit first order kinetics, the elimination rate constant (kd) can be
defined as the fraction of the drug which is cleared in unit time:

kd = Amount of drug cleared in unit time = RE

Total amount of drug Ab

Where Ab is the amount of drug in the body and is given by:

Ab = Vd x Cpss

The rate of excretion (RE) = Cl x CPss

Substituting these values into the expression for kd gives the following:

kd = Cl x Cpss
Vd x Cpss

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Canceling the Cpss terms gives a relationship between Vd, clearance and kd:

kd = Cl ……….. Eq. 7.12

Vd

Practical applications of pharmacokinetics

The equations developed so far can be applied to answer many practical problems in drug
therapy.

1. Calculation of plasma concentration any time after a loading dose or bolus of a

drug is given.

First the theoretical initial plasma concentration (Cp0) is calculated from the dose and
volume of distribution (Vd) by substituting the expression for total amount absorbed
(Eq. 7.4) into the rearranged expression for Vd (Eq. 7.5):

Amount of drug in body = Dose x S x F

Plasma concentration (Cp0) = Amount of drug in body

Total volume ( Vd )

Cp0 = Dose x S x F ………..…… Eq. 7.13

Vd

The values for Cp0, t and kd (if kd is not known it can be calculated from clearance and Vd
using Eq. 7.12) can be substituted into Eq. 7.7 and solved for Cpt.

Question Q 7(4)

A 70 kg lady is given an oral dose of carbamazepine of 400 mg. What is the plasma
carbamazepine concentration 24 h later assuming a volume of distribution of 1.0 L/kg,
a clearance of 0.05 L/h/kg, salt factor of 1 and a bioavailability of 0.75.

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Answer Q 7(4)

First calculate the theoretical initial drug concentration (Cp0) by substituting values for
dose, S, F and Vd into Eq. 7.13:

Cp0 = Dose x S x F
Vd

Where: Dose = 400 mg

S = 1
F = 0.75
Vd = 1.0 L/kg = 1.0 x 70 = 70 L for patient’s weight

Cp0 = 400 x 1 x 0.75 = 4.29 mg/L

70

Next substitute values for Cp0, t and kd (which must first be calculated from Vd and the
clearance) into Eq. 7.7 and solve for Cpt:

ln Cpt = ln Cp0 - kd . t

Where: Cpt = drug concentration at 24 h post dose = ?

Cp0 = theoretical initial drug concentration = 4.29 mg/L
t = time since dose = 24 h
kd = elimination rate constant calculated using Eq. 7.14:

kd = Cl = 0.05 = 0.05 h-1

Vd 1.0

NB since the ratio of Cl to Vd is being calculated it is not

necessary to correct these values for body weight.
The units must however be compatible.

ln Cpt = ln 4.29 - (0.05 x 24)

ln Cpt = 1 .46 - 1.20 = 0.26

Cpt = antilge 0.26 = 1.3 mg/L (2 sig figs)

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2. Calculation of maintenance dose

Sometimes it is desirable to know which dose to give a patient (to be repeated at a time
interval, τ) to achieve a target steady state plasma concentration (Cps). This is easily
calculated by rearranging Eq. 7.11 to give:

Maintenance dose = CPss x Cl x τ …………. Eq. 7.14

S x F

It is important to appreciate that a new a steady state will not be achieved until at least
5 half lives have elapsed.

3. Effect of a change in maintenance dose

This is simply calculated by inserting the new dose into Eq. 7.14. Alternatively the old
steady state concentration can be multiplied by the fractional increase in dose.

New Cpss = Old Cpss x fractional change in dose

Alternatively to calculate the dose necessary to change the steady state concentration by a
set amount all that is necessary is to calculate the maintenance dose (using Eq. 7.14)
required to achieve the difference in steady state concentration between the original value
and the target level then add this to the original maintenance dose. Again it is necessary
to wait at least 5 half-lives before the new steady state is achieved.

Question Q 7(5)

A 60 kg patient requires phenobarbitone to be given at 12 hourly intervals. Calculate:

a) The dose required to achieve an average steady state plasma concentration of

25 mg/L

b) The new average steady state plasma phenobarbitone concentration if the dose
was increased to 120 mg.

Assume a clearance of 5 mL/h/kg and that both bioavailability and salt conversion factors
are 1.

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 BASIC PHARMACOKINETICS

Answer Q 7(5)

a) The maintenance dose is calculated by substituting values for Cpss, t and clearance
into Eq. 7.14:

Maintenance dose = Cpss x Cl x τ

S x F

Where Cpss = target average steady state plasma concentration = 25 mg/L

τ = dosing interval = 12 h
S = salt conversion factor = 1
F = bioavailability = 1
Cl = clearance = 5 mL/h/kg
= 5 x 60 = 300 mL/h/60 kg
= 300 = 0.3 L/h/60 kg
1000

Maintenance dose = 25 x 0.3 x 12

1 x 1

= 90 mg

a) If the dose is increased from 90 mg to 120 mg, then the fractional increase is
given by:

New dose - old dose = 120 - 90 = 1.33

Old dose 90

The new average steady state concentration can be calculated by multiplying the
old average steady state plasma concentration by this fractional increase in dose:

New Cpss = Old Cpss x Fractional increase in dose

= 25 x 1.33

= 33 mg/L (2 sig figs)

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4. Calculation of a loading dose (LD)

Equation Eq. 7.5 can be re-arranged to give an expression for the amount of drug in the
body:

Amount of drug in body = Volume of distribution (Vd) x Plasma concentration (Cp)

The amount of drug in the body is the dose reaching the circulation so that Eq 7.4 can be
substituted for the left hand side of this equation:

F x S x Dose administered = Vd x Cp

The dose administered is then the loading dose (LD) required to achieve the desired
plasma concentration of the drug (Cp) which can be rearranged to give the following
expression:

Loading dose (LD) = Vd x Cp …………….. Eq. 7.15

F x S

If the patient is already receiving the drug in question then this formula is modified to
take account of the pre-existing drug concentration. The loading dose is then the amount
of drug which needs to be administered to raise the plasma concentration from the initial
concentration (Cpinitial) to the desired concentration (Cpdesired):

Loading dose (LD) = Vd x (Cpdesired - Cpinitial ) …… Eq. 7.16

F x S

Question Q7(6)

Calculate the loading dose of digoxin (bioavailability = 0.75, salt factor = 1) required to
achieve an initial plasma concentration of 1.5 μg/L in a 60 kg man (assume volume of
distribution = 7 L/kg):

a) If the patient has never taken digoxin

b) If the patient is currently on digoxin with a plasma concentration of 0.5 μg/L.

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 BASIC PHARMACOKINETICS

Answer Q 7(6)

First calculate the volume of distribution (Vd) for the patient:

Vd = Vd (L/kg) x Body weight (kg)

= 7 x 60

= 420 L

Formula used to calculate loading dose:

LD = Vd x (Cpdesired - Cpinitial)
F x S

a) If patient is not already on digoxin then Cpinitial is zero:

LD = 420 x 1.5 = 840 μg

0.75 x 1

b) If the patient is already on digoxin and Cpinitial is 0.5 μg/L:

LD = 420 x (1.5 - 0.5) = 560 μg

0.75 x 1

5. Calculation of time to reach a steady state

Figure 7.1 shows that following a single dose of a drug its concentration falls to 50% of
the original value after one half-life, 25% after two halve lives, 12.5% after three half
lives etc. After five half lives the amount remaining is negligible at 3.25%.

Consider the situation when the same dose of a drug is administered using a dosing
interval of one half-life. If the theoretical maximum plasma concentration achieved (Cp0)
is 100%, then after one half-life the concentration of the original dose will be 50%.
Administration of a second dose at the end of one half-life contributes 50% to the plasma

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concentration when two half lives have passed since the first dose, whereas the
contribution at this time from the first dose will be one half of 50% which is 25%.
Therefore after two halve lives the plasma concentration will be 50% + 25% = 75% of
the theoretical maximum (i.e 75% of the steady state concentration). If this process is
repeated for 5 half lives then the following pattern emerges:

Number of half lives since initial dose

Dose 1 2 3 4 5

1 50 25 12.5 6.25 3.125

2 50 25 12.5 6.25
3 50 25 12.5
4 50 25
5 50

Total 50 75 87.5 93.8 96.9

Figure 7.2 Cumulative drug concentration (expressed as percentage of theoretical

maximum) for a drug administered at a dosage interval equal to its
half-life

The concentration when a steady state is reached is given by the expression:

Cpss = Cp0 + Cp0 + Cp0 + Cp0 + Cp0 + Cp0 + ……

2 4 8 16 32 64

For practical purposes the plasma concentration after five half lives have elapsed
approximates to the average steady state concentration. A similar argument, although
more complex can be applied if the dosage interval is shorter than the half-life. A dosage
interval of less than the half-life minimises oscillations around the average steady state
concentration. If the dosage interval is considerably longer than the half-life then a
steady state is never achieved and virtually all of the drug is removed from the body
between each dose.

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6. Calculation of pharmacokinetic parameters from a patient’s data

Sometimes reliable pharmacokinetic data is not available or there may be considerable

patient to patient variation in the handling of a drug. In this situation it is often possible
to estimate kinetic parameters using the patient’s own data. Sometimes a single test dose
is employed to do this. It is vital however that exact details of the dose and timings of
blood samples is accurately recorded. Provided clearance of the drug follows first order
kinetics then it is possible to derive all kinetic parameters of interest from measurements
made on two blood samples drawn at appropriate times.

Suppose that the dose is given at time t0 and that samples are drawn at times t1 and t2 and
that the measured plasma concentrations of these latter two samples are Cp1 and Cp2
respectively. The elimination rate constant can then be calculated by substituting these
values into Eq. 7.8:

kd = (ln Cp1 - ln Cp2)

t2 - t1

The half-life can then be calculated by substituting the value for kd into Eq. 7.9:

t½ = 0.693
kd

The theoretical initial plasma concentration (Cp0) can then be calculated by substituting
either pair of concentration and times (it doesn’t matter which) into the logarithmic
integrated first-order rate equation (Eq. 7.7), together with the value for kd then solving
for Cp0:

ln Cpt = ln Cp0 - kd . t

Cp0 = antiloge (ln Cp0 - kd. t)

The volume of distribution (Vd) is obtained by substituting the value of Cp0 and the dose
into Eq. 7.13:

Cp0 = Dose x F x S
Vd

Which can be rearranged to give a value for Vd / (F x S):

Vd = Dose
F x S Cp0

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Note that F and S are unknown but are incorporated into the apparent value for Vd.
Dosage adjustments can be made using the apparent Vd without knowledge of the actual
values for F and S.

Substitution of kd and Vd/ (F x S) into Eq. 7.12 allows calculation of clearance (or more
precisely Cl / (F x S):

Cl = kd x Vd
F x S F x S

The maintenance dose at dosage interval τ to achieve an average steady state drug
concentration Cpss can then be obtained from Eq. 7.14.

Maintenance dose = Cpss x Cl x τ

F x S

Again values for F and S do not need to be determined since they are already
incorporated into the apparent clearance Cl / (F x S).

Question Q 7(7)

A 60 kg patient is given 4 g of a drug and blood samples drawn at timed intervals with
the following results:

Time (h) Plasma drug concentration (mg/L)

2 30
3 7

a) What is the half-life of the drug?

b) Calculate the apparent volume of distribution in L/kg.
c) Calculate the apparent clearance in L/h/kg.
d) What maintenance dose (to be administered once daily) would be needed to
achieve an average plasma steady state concentration of 15 mg/L?

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 BASIC PHARMACOKINETICS

Answer Q 7(7)

a) To calculate the half-life first calculate the elimination rate constant using the
logarithmic form of the integrated first-order rate equation (Eq. 7.7):

ln Cpt = ln Cp0 - kd . t

Use the 1st (2 h) sample as the initial sample so that Cp0 = 30 mg/L
Use the 2nd (4 h) sample as Cpt (7 mg/L )
Therefore t = 4 - 2 = 2 h

ln 7 = ln 30 - kd . 2

1.95 = 3.40 - 2 kd

2 kd = 3.40 - 1.95 = 1.45

kd = 1.45 = 0.725 h-1

2

The half-life (t1/2) is then calculated from kd using Eq. 7.9:

t1/2 = 0.693 = 0.693 = 0.96 h

kd 0.725

b) First calculate the theoretical value for Cp0 by substituting values for kd, Cpt and t
(either set of t and Cpt values can be used) into Eq. 7.7 but this time use the actual
sample times.

Using the 2h results, t = 2h, Cpt = 30 mg/L

ln 30 = ln Cp0 - 2 kd

3.40 = ln Cp0 - 2 x 0.725

ln Cp0 = 3.40 + 1.45 = 4.85

Cp0 = antiloge 4.85 = 128 mg/L

Then calculate the apparent volume of distribution using Eq. 7.13.

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CHAPTER 7

Cp0 = Dose x F x S
Vd

Which can be rearranged to give a value for the apparent volume of distribution:

Apparent volume of distribution i.e Vd = Dose

F x S Cp0

Substitute Cp0 = 7.03 mg/L and dose = 4 g = 4000 mg

Vd = 4000 = 31.3 L
F x S 128

Divide by body weight (60 kg) to give Vd in trerms of L/kg:

Vd = 31.3 = 0.52 L/kg

F x S 60

c) The apparent clearance can be calculated by substituting kd and Vd into Eq. 7.12:

Apparent clearance i.e. Cl = kd x Vd

F x S F x S

= 0.725 x 0.52

= 0.38 L/h/kg

d) The maintenance dose required to achieve an average steady state plasma

concentration (Cpss) of 15 mg/L can be obtained by substituting values into
Eq. 7.14:

Maintenance dose = Cpss x Cl x τ

F x S

Cpss = 15 mg/L
Cl / (F x S) = 0.38 L/h/kg = 0.38 x 60 = 23 L/h/kg
τ = dosing interval = 24 h

Maintenance dose = 15 x 23 x 1 = 345 mg

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 BASIC PHARMACOKINETICS

7. More complex situations

Although many drugs follow the principle outlined above (i.e. single compartment
models obeying first-order kinetics) there are some notable exceptions. These drugs are
often cleared by saturable mechanisms so that a concentration is reached at which the
rate of elimination becomes independent of concentration. This is very important
clinically since only small increases in dose above a threshold value can easily lead to
toxic levels of the drug. A good example of this is phenytoin the clearance of which
displays a mixture of zero and first-order kinetics. The best model to use under these
circumstances is the equation of Michaelis and Menton which describes the rate of an
enzyme reaction (v) in terms of substrate concentration (s) in terms of two constants the
Km (which can be shown to be the substrate concentration at half-maximal velocity) and
the maximal velocity (Vmax):

v = Vmax x s
Km + s

Substituting the administration rate (which in a steady state is equal to the rate of
elimination) for v:

v = Dose x S x F
τ

and the average steady state drug concentration (Cpss) for s, gives the expression:

Dose x S x F = Vmax x Cpss

τ Km + Cpss

which can be rearranged to give an equation to calculate the dose required to give the
desired average steady state plasma drug concentration for the dosing interval:

Dose = VMax x Cpss x τ ……… Eq. 7.17

( Km + Cpss ) x S x F

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CHAPTER 7

ADDITIONAL QUESTIONS

1. An antidepressant drug has a biological half-life of 30 hours. How long will it

take a plasma concentration of 50 mg/L to fall to 20 mg/L?

2. A 15 year old boy presents to casualty following a convulsion. It turns out that he
had swallowed 30 of his mother’s lithium tablets about 10 hours previously.
On admission his lithium concentration is 4.1 mmol/L. A decision needs to be
made whether to haemodialyse him to reduce the lithium concentration. As this is
not going to be available quickly, the physicians want to know how long he will
have toxic levels just with endogenous clearance. Estimate the following,
indicating clearly any assumptions you have made:

a) The likely volume of distribution of the lithium at this stage in the

situation, given a body weight of 65 kg.

b) How long it will be before his lithium concentration drops to the relatively
safe level of 1.5 mmol/L below which toxicity is unlikely, given a
clearance of 0.03 L/h/kg.

3. A 60 mg dose of a drug is given to a male experimental subject who weighs

80 kg. Assuming that the drug is completely absorbed and distributed evenly
throughout the total body water, estimate the potential peak plasma level. If the
drug were distributed only within the extracellular compartment, what would the
plasma level be?

4. A bolus of 6 g drug is given IV and 3 blood samples collected at intervals.

Time mg/L
2.5h 32
5h 10
7.5h 3

a) What is the half-life of the drug?

b) What is the volume of its distribution?

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 BASIC PHARMACOKINETICS

5. The plasma concentration of a drug is found to be 200 nmol/L at 9.00 am. It’s
elimination follows first order kinetics with a rate constant is 0.34/h. Calculate the
times at which the plasma concentrations will reach 100 nmol/L and 75 nmol/L.

6. A patient in casualty with a suspected adrenal crisis is given an IV dose of

hydrocortisone at 18.00. The medical team on take wish to carry out a short
synacthen test to confirm the diagnosis but there will be a significant contribution
form the administered drug until its concentration has fallen to 10% of the peak
value. If the half-life of hydrocortisone is 2 h, what is the earliest time at which the
test can be carried out?

7. The SHO decides to treat a patient (weight 80 kg) with intravenous theophylline (salt
factor = 0.8). What loading dose would you recommend in order to achieve a
theophylline level of 12 mg/L given a volume of distribution of 0.5 L/kg and an
initial plasma theophylline level of 4 mg/L?

8. A patient, unable to take oral medication, had been receiving intravenous valproate
for a number of days and an average steady state level of 75 mg/L. After regaining
consciousness the doctors wished two change to an oral twice daily regimen.
In order to maintain the same average steady state concentration what dose would
you recommend. Assume a clearance of 10 mL/h/kg, a bioavailability of 0.7 and a
salt factor of 0.85.

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138
 BODY FLUIDS AND ELECTROLYTES

Chapter 8

Body fluids and electrolytes

Distribution of body water

The human body contains approximately sixty per cent of its weight of water. For an
average adult male weighing 70 kg this amounts to 42 L. Of this about a third (14 L) is
contained in the extracellular fluid (ECF) and two thirds (28 L) in the intracellular fluid
(ICF). The ECF and ICF are separated from each other by cell membranes which are
semi-permeable but allow ready diffusion of water into and out of cells. Both of these
compartments are in osmotic equilibrium with each other. Blood plasma constitutes
approximately one quarter of the ECF (3.5 L), most of the remainder is present as
interstitial fluid (10.5 L) which surrounds cells in the various tissues of the body. Plasma
and interstitial fluid are separated by capillary walls which are freely permeable to water
and electrolytes but minimally permeable to proteins such as albumin. The distribution
of ECF between plasma and the interstitial fluid is controlled mainly by haemodynamic
factors, integrity of capillary walls and the plasma albumin concentration. A small
proportion of the ECF is present in the various body cavities (e.g. peritoneal, pleural,
pericardial, synovial, spinal cavities). These relationships are illustrated in Fig 8.1.

Composition of body fluids

The fluid compartments differ in their solute composition. Sodium is the main cation in
the ECF whereas potassium is the main cation within the ICF (see Fig 8.1). This status is
maintained by the action of the sodium pump. The cell membrane of most tissues is
impermeable to glucose in the absence of insulin. Since the interstitial fluid is created by
filtration of plasma it is an ultrafiltrate of plasma with essentially the same solute
composition but a much lower protein content. For practical purposes the concentration
of glucose and electrolytes in the plasma and interstitial fluid can be regarded as identical.

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a) Major fluid compartments of the body

% composition of body

0 5 20 60

Total body water (42 L)

ECF (14 L)
ICF (28 L)
Interstitial fluid
(10.5 L)

Plasma
(3.5L)

b) Typical electrolyte compositions of the ECF and ICF

Typical concentration (mmol/L)

Electrolyte ECF ICF
Sodium 140 12
Potassium 4 145
Chloride 100 4
Bicarbonate 26 8

Figure 8.1 a) Distribution of body water between fluid compartments (ECF =

extracellular fluid, ICF = intracellular fluid) – values shown for a
typical adult male.
b) Typical electrolyte concentrations in the ECF and ICF.

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 BODY FLUIDS AND ELECTROLYTES

Estimation of fluid losses

This can be extremely difficult since the mechanisms and consequences are complex and
vary according to the nature of the fluid lost. In general however, several approaches can
be used:

• Clinical assessment – such as presence of thirst, dry mouth, decreased sweating,

tachycardia, decreased skin turgor, decreased urine output

• Loss of body weight – short term changes in body weight are due to changes in
hydration. Difficult to assess in immobilised patients e.g. in ITU

• Fluid balance charts recording fluid input versus output – negative fluid balance
consistent with fluid losses (this is not helpful if dehydration is already
established). Accurate fluid balance charts are difficult to maintain and rely on
estimating fluid loss via lungs, skin etc

• Laboratory measurements – haemoconcentration (raised haematocrit and serum

proteins), increased serum creatinine and urea (with the increase in urea being
disproportionate to creatinine), hypernatraemia, low volume concentrated urine
(unless losses are due to renal losses).

A full discussion is beyond the scope of this book, but here are two areas where
calculations are involved:

1. Calculation of fluid balance

When a patient is normally hydrated:

Net fluid intake = Net fluid losses

The difficulty is identifying and accurately quantifying all the fluid losses and gains.
Small errors in the calculation of daily balance can accumulate over time with
catastrophic consequences. If the net fluid loss exceeds the net fluid gain then the patient
becomes water depleted i.e. is in negative fluid balance:

Fluid balance = Net fluid intake - Net fluid loss ……….. Eq. 8.1

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Fluid is gained by drinking. As food has a water content fluid is also gained by feeding,
either orally, by naso-gastric tube feeding or by intravenous infusion. Another source of
water gain which must be considered is water produced metabolically when fats and
carbohydrates are oxidised. Fluid can be lost in urine, as water vapour via the lungs,
sweating and evaporation through the skin and in faeces. In the hospitalised patient
losses via the lungs may be increased if the patient is on a ventilator, losses via the skin
may increased in burns patients or if the patient is pyrexial, and further losses may occur
by vomiting, nasogastric suction and through drains and fistulae. Typical values for a
normal adult in fluid balance are shown in figure 8.2.

Fluid gains Fluid losses

Oral (water) 1200 mL Urine 1600 mL

Oral (food) 800 mL Lungs 300 mL
Metabolism 500 mL Skin 400 mL
Faeces 200 mL
Total 2500 mL Total 2500 mL

Figure 8.2. Typical gains and losses in a normal adult in perfect fluid balance

In practice it is difficult to measure the so-called insensible losses i.e. losses via the lungs,
skin and faeces, and it is customary to assume a value of approximately 900 mL per day.
Water production via metabolism, the insensible gain, is also difficult to determine, so an
average value of about 500 mL per day is assumed. Therefore, the net insensible loss for
an adult is normally in the order of 900 - 500 = 400 mL per day.

Question Q 8(1)

During a 24 h period a patient recovering from intestinal surgery receives 1.5 L of fluid
by intravenous infusion. The total urine output during this period is 1200 mL and a
further 450 mL of fluid was removed by nasogastric suction. Estimate the patients fluid
balance.

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Answer Q8(1)

Fluid gained Fluid lost

iv fluids = 1500 mL Urine = 1200 mL

Nasogastric = 450 mL
Insensible losses = 400 mL

Total = 1500 mL Total = 2050 mL

Fluid balance = Fluid gained - Fluid lost

= 1500 - 2050

= - 550 mL

Therefore the patient is a negative fluid balance of about 550 mL/24 h.

2. Calculation of water loss from plasma sodium

The consequences of fluid loss depend upon the nature of the fluid lost. If the loss is
isotonic, i.e. water and sodium are lost in the same proportion, due for example to
haemorrhage, then the osmolality of the plasma (and ECF) remains unchanged and there
is no stimulus for the osmotic shift of fluid from the ICF to the ECF. In other word acute
loss of isotonic fluid is confined to the ECF. On the other hand if pure water loss occurs
from the plasma (and ECF) then the osmolality (and hence sodium concentration) of the
plasma (and ECF) rises and provides an osmotic stimulus for the shift of water from the
ICF to the ECF until a point is reached when the two compartments are again in osmotic
equilibrium. Note that at equilibrium the osmolality in both compartments will be higher
than normal, but not as high as it would be if there had been no fluid shift from the ICF to
ECF. In other word the loss in fluid volume is shared between the two compartments and
so helps protect the circulating plasma volume (see Fig 8.3). In reality pure water loss
rarely occurs, so calculation of the approximate fluid loss from changes in plasma sodium
often serve as a (rough) guide for fluid replacement. The small difference between
osmolality and osmolarity (see Chapter 6) will be ignored and it will be assumed that
each mmol/L of solute contributes an osmolality of 1 mOsm/kg.

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a) Isotonic fluid loss

100%

Isotonic fluid loss

ICF ECF Ik
I

b) Pure water loss

100%

ICF ECF
Pure Water
water loss shift

Figure 8.3 Comparison of the effects on ECF and ICF volumes of a) isotonic
fluid loss, and b) pure water loss

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 BODY FLUIDS AND ELECTROLYTES

If only pure water loss has occurred then the total amount of osmotically active species
present in the body (both in the ICF and ECF) remains constant. The total amount of
osmotically active species is the product of the volume of total body water and the plasma
osmolality (the osmolality of all compartments must be the same since they are in
osmotic equilibrium). Before the fluid loss occurred:

Total solutes (mOsm) = Initial osmolality (mOsm/kg) x Initial vol body water (L)

After loss of fluid and osmotic equilibrium has been reached:

Total solutes (mOsm) = Final osmolality (mOsm/kg) x Final vol body water (L)

Assuming no solutes have been lost then these two quantities are equal and we can write:

Final osmolality (mOS/kg) x Final vol (L)

= Initial osmolality (mOsm/kg) x Initial vol (L)

which can be rearranged to give an expression for the final body water volume:

Final vol (L) = Initial osmolality (mOsm/kg) x Initial vol (L)

Final osmolality (mOsm/kg)

The volume of fluid lost is the difference between the initial volume and the final
volume:

Fluid loss (L) = Initial volume (L) - Final volume (L)

Substituting for final vol gives:

Fluid loss (L) =

Initial vol (L) - [Initial osmolality (mOsm/kg) x Initial vol (L)] ……….. Eq. 8.2
Final osmolality (mOsm/kg)

This formula can be simplified by taking the value for initial body water of 42 L (based
on the average 70 kg male) and an average normal initial osmolality of 285 mOsm/kg:

Fluid loss (L) = 42 - { 285 x 42 }

Final osmolality (mOsm/kg)

Which becomes:

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CHAPTER 8

Fluid loss (L) = 42 - { 12000 } …… Eq. 8.3

Osmolality (mOsm/kg)

Since sodium and its associated anions normally account for most of the osmolality of
plasma then this expression can be further simplified by using the plasma sodium
concentration assuming that it was initially normal (140 mmol/L), so that 140 x 42 =
5880 mmol:

Fluid loss (L) = 42 - { 5880 } ………………… Eq. 8.4

Sodium (mmol/L)

It cannot be emphasised too strongly that this formula can only give a crude estimate of
fluid loss and is based on the following assumptions:

• That pure water loss has occurred

• That the plasma sodium was initially normal at 140 mmol/L (if the previous
sodium concentration was known then it could be taken into account).

• There are no other abnormal amounts of osmotically active species present.

• There is no significant solute loss, gain or transfer between compartments.

• The initial body weight of the patient is 70 kg and contains 60 per cent water. If
the body weight is known then a correction could be made, but this is more
difficult for obese patients.

Question Q 8(2)

An adult dehydrated male has a plasma sodium of 165 mmol/L due to water depletion.
Estimate the fluid deficit.

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 BODY FLUIDS AND ELECTROLYTES

Answer Q 8(2)

Initial vol (L) x Intial Na (mmol/L) = Final vol (L) x Final Na (mmol/L)

Assume initial vol = 42 L; initial Na = 140 mmol/L

Final vol (L) = ?, Final Na = 165 mmol/L.

Substituting these values gives:

42 x 140 = Final vol (L) x 165

Final vol (L) = 42 x 140 = 36 L (2 sig figs)

165

Fluid loss (L) = Initial vol (L) - Final vol (L)

= 42 - 36 = 6L

2. Effect of hyperglycaemia on plasma sodium concentration

A rise in plasma glucose (for example due to insulin deficiency) is immediately

accompanied by an increase in plasma (and hence ECF) osmolality which will stimulate
the hypothalamic osmoreceptors which initiates an increase thirst and the release of
antidiuretic hormone (ADH) from the posterior pituitary. What happens next depends
upon whether the patient is able to drink an adequate amount of fluid (see Fig 8.4), but
the end result is that there is a fall in plasma sodium concentration. It is often useful to
predict this fall in sodium since if it does not account for the observed hyponatraemia
then there must be another cause.

a) Assuming free access to fluid

The initial rise in osmolality due to the hyperglycaemia stimulates thirst via the
hypothalamic osmoreceptors. The patient will respond to thirst by taking in fluids until a
point is reached at which the plasma (and hence ECF) osmolality is returned to normal
but at the expense of diluting sodium and other solutes i.e. to produce a dilutional
hyponatraemia. Since the plasma osmolality is unchanged it follows that the rise in

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plasma glucose concentration must be equal to the fall in sodium and its associated
anions. Therefore the fall in sodium concentration must be equal to one half the rise in
plasma glucose concentration:

Fall in plasma sodium (mmol/L) = Rise in plasma glucose (mmol/L) ….. Eq. 8.5
2

b) Assuming no intake of water

If there is no intake of water then the plasma osmolality will remain elevated (because the
cell membranes are essentially impermeable to sodium and, when insulin deficient,
impermeable to glucose). The osmotic stimulus will result in a shift of water from the
ICF to the ECF (and hence plasma). An equilibrium is established in which the
osmolality of both compartments is again equal but higher than normal. In other words
the osmotic load is shared between both the ECF and ICF compartments. Therefore the
increase in osmolality must be equal to the increase in plasma glucose concentration:

↑ plasma glucose = ↑plasma osmolality = ↑ECF osmolality = ↑ICF osmolality

The osmotic load in the ECF due to this accumulated glucose can be calculated from the
rise in plasma glucose concentration (Δ glucose) and the ECF volume:

Osmotic load (mOsm) = Δ glucose (mmol/L) x ECF vol (L)

At equilibrium the rise in osmolality (Δ osmolality) (which must be the same for both
compartments) is given by the expression:

Δ osmolality (mOsm/kg) = Osmotic load (mOsm)

Total body fluid volume (L)

Substituting Δ glucose x ECF vol for the osmotic load and (ECF + ICF) for the total
body fluid volume gives:

Δ osmolality (mOsm/kg) = Δ glucose (mmol/L) x ECF vol (L)

[ECF vol (L) + ICF vol (L)]

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 BODY FLUIDS AND ELECTROLYTES

Hyperosmolar

Iso-osmolar
ECF

Rise in plasma glucose

ICF ECF ICF

No intake of fluids
Free access to Water shifts from
oral fluids ICF to ECF
ECF osmolality Both compartments
is restored become hyper-osmolar
Hyperosmolar

Iso-osmolar

ICF ECF ICF ECF

Figure 8.4 Effect of a rise in plasma (and hence ECF) glucose on the osmolality
of both the ECF and ECF compartments when there is either free
intake of water or no intake at all. In both instances addition of water
to the ECF compartment dilutes sodium and other electrolytes whilst
the plasma glucose remains elevated

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Although some fluid has shifted between compartments, the ratio of (ECF + ICF) to ECF
is still approximately 3:1 (since one third of body water is in the ECF). Therefore this
expression becomes:

Δ osmolality (mOsm/Kg) = Δ glucose (mmol/L) ..…… Eq. 8.6

3

Since the rise in osmolality is less than the rise in plasma glucose, the concentration of
other solutes (predominantly sodium and its associated anions) must have fallen by an
amount equal to the difference between the two:

Δ (Na + Cl) (mmol/L) = Δ osmolality (mmol/L) - Δ glucose (mmol/L)

Δ (Na + Cl) (mmol/l) = {Δ glucose (mmol/L) / 3} - Δ glucose (mmol/L)

Which is the same as:

Δ (Na + Cl) (mmol/L) = - 2 x Δ glucose (mmol/L)

3
Since the concentration of sodium and its associated anions (mainly Cl-) are assumed
equal, then division by 2 gives the change in sodium concentration, which then cancels
with 2 in the numerator, to give the expression:

Δ Na (mmol/L) = - Δ glucose (mmol/L) ………. Eq. 8.6

3

This calculation assumes that there has been no net gain or water loss by the body and no
transfer of solutes between the ECF and ICF.

Question Q 8(3)

A male adult insulin dependent diabetic forgot to take his insulin. His blood glucose rose
from 5 mmol/L to 20 mmol/L in 2 hours. During this period he did not pas any urine.
Calculate the likely effect on his plasma sodium concentration assuming:

a) Free access to oral fluids

b) No intake of fluids occurs.

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 BODY FLUIDS AND ELECTROLYTES

Answer Q 8(3)

a) Initially the increase in plasma glucose results in an increase in plasma osmolality.

Hyperosmolality stimulates thirst and the patient drinks fluids which dilutes
plasma until the stimulus is removed i.e. plasma osmolality has returned to
normal. Since the plasma osmolality is unchanged but plasma glucose has risen it
follows that the concentration of other solutes in plasma (principally sodium and
its associated anions) must have fallen by an amount equal to the rise in glucose.

Fall in concentration of non-glucose solutes = rise in glucose concentration

= 20 - 5

= 15 mmol/L

Approximately one half of these solutes will be present as sodium, therefore:

Fall in sodium 15 = 7.5 mmol/L

2

b) Initially the increase in plasma glucose results in an increase in plasma osmolality.

If there is no intake of fluids, then water moves by osmosis from the ICF to the
ECF (which includes plasma) until equilibrium is established i.e. osmolalities of
ICF and ECF are equal but elevated. The osmotic load is therefore shared
between the two compartments.

Rise in amount of glucose in body (mmol)

= Rise (∆) in plasma glucose concentration (mmol//L) x ECF vol (mmol/L)

This rise in the amount of glucose present in the body is responsible for the rise in
overall osmolality. Therefore the increase in osmolality can be calculated by
dividing this amount by the volume of total body water (i.e. ICF + ECF):

∆ osmolality (mOsm/kg) = ∆ glucose (mmol/L x ECF Vol (L)

ECF vol (L) + ICF vol (L)

Although there is small shift in fluid between compartments, approximately one

third of body water is located in the ECF, the ratio ECF /(ECF + ICF) is roughly
3. Since the plasma glucose rose from 5 mmol/L to 20 mmol/L, the rise in
osmolality is:

∆ osmolality = 20 - 5 = 5 mOsm/kg
3

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As the plasma osmolality has risen by 5 mOsm/kg and the plasma glucose has risen by 15
mmol/L, it follows that the concentrations of the other solutes in plasma (principally
sodium and its associated anions) must have fallen by an amount equal to the difference
between the two:

Fall in concentration of non-glucose solutes

= ∆ osmolality - ∆ glucose

= 5 - 15 = - 10 mmol/L

Approximately one half of these solutes will be present as sodium, therefore:

Fall in sodium 10 = 5 mmol/L

2

Direct versus indirect reading ion-specific electrodes

The advent of ion-specific electrodes for the determination of plasma/serum sodium soon
led to discrepancies when compared with values determined by traditional flame
photometry. The difficulty arose because these electrodes measure sodium activity in
plasma water. Plasma contains appreciable protein (normally about 70 g/L). This
protein, together with its hydration shell, occupies a significant proportion of plasma
volume and reduces the amount of plasma water available to dissolve sodium and other
ions. The “normal” plasma sodium determined by flame photometry is approximately
140 mmol/L of plasma. If this plasma contains 70 g/L of protein and we assume that its
volume is approximately 70 mL/L (or 0.07 L/L), then this means that a litre of plasma
contains only 1-0.07 = 0.93 L of water. Therefore the concentration of sodium in
plasma water can be calculated as follows:

Na (mmol/L water) = 140 = 151 mmol/L

0.93

This is the value which will be obtained for plasma sodium containing 140 mmol/L
plasma when measured on whole undiluted plasma in an ion-specific electrode system.
Since the measurement is made directly on whole plasma rather than diluted plasma it is
said to be a “direct reading electrode”. In general:

[Na+] (plasma water) = [Na+] (plasma) ..……… Eq. 8.7

1 - Plasma protein (Kg/L)

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 BODY FLUIDS AND ELECTROLYTES

If, on the other hand, the plasma sample is first diluted with an aqueous diluent before the
electrode measurement is taken then the discrepancy almost disappears. These
instruments are known as “indirect reading electrodes” Consider the same sample,
containing 140 mmol of sodium per L plasma, first diluted 1 in 20 before the sodium
measurement is made. This is equivalent to diluting 0.05 L of plasma to 1 L. The
sodium present in 1 L of plasma diluted 1 in 20 will be 140 x 0.05 = 7 mmol. The
amount of water this sodium is dissolved in will be equal to the amount of water in the
0.05 L plasma sample plus the water added to it.

Water in 0.05 L plasma = 0.05 L - Volume due to protein

If the protein is 70 g/L (approx 0.07 L/L), then 0.05 L of plasma is occupied by 0.05 x
0.07 = 0.0035 L of protein. The plasma sample therefore contains 0.05 - 0.0035 =
0.0465 L of water. To dilute the plasma 1 in 20, the amount of diluent added is 1 - 0.05
= 0.95 L. Therefore:

Total amount of water in 1 L of 1 in 20 dilution of plasma = 0.95 + 0.0465 = 0.9965 L.

Concentration of sodium = 7 = 7.025 mmol/L plasma water

0.9965
Allowing for the 1 in 20 dilution:

Plasma sodium = 7.025 x 20 = 140.5 mmol/L

The reason for this discrepancy is that dilution results in a decrease in the proportion of
water displaced by protein. Plasma sodium measured with a flame photometer gives
similar readings to indirect reading electrodes.

It has become common practice for instrument manufacturers to “adjust” the calibration
of their direct reading ISE instruments so that the discrepancy with flame photometers
disappears, but only at a “normal” plasma protein concentration (usually 70 g/L):
[Na+] (plasma) = [Na+] (water) x 0.93

If the plasma protein differs markedly from “normal” then the discrepancy with flame
photometer readings reappears. In the above example, if the instrument is adjusted so
that the plasma water sodium of 151 mmol/L reads 140 mmol/L then a sample with the
same plasma sodium concentration but containing 50 g/L (0.05 kg/L) protein is
measured, then the concentration of sodium in plasma water will be:

[Na+] (plasma water) = 140 = 140 = 147 mmol/L

1 - 0.05 0.95
If the instrument makes the same adjustment (by assuming that the plasma protein is
70 g/L), then the reading given will be:
Reported [Na+] = 147 x 0.93 = 137 mmol/L.

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CHAPTER 8

Question Q 8(4)

A plasma contains 140 mmol/L of sodium and 95% water by volume. Neglecting sodium
binding by plasma proteins, calculate the apparent plasma sodium concentration
determined from measurements with an electrode system which responds to water sodium
(a) in undiluted plasma, and (b) in plasma diluted 1 in 20 with water.

Answer Q 8(4)

a) Undiluted plasma – contains 95% water

Water concentration of plasma = 1 x 95 = 0.95 L/L

100

Therefore concentration of sodium in plasma water is:

[Na+] (plasma water) = 140 = 147 mmol/L

0.95

b) Plasma diluted 1 in 20 with water:

Working on a volume of 1 L i.e. 50 mL (0.05 L) plasma diluted to 1 L with water:

Amount of sodium in 1 L diluted plasma = 140 = 7 mmol

20

Amount of water in diluted plasma = Water from plasma + water from diluent

= (0.05 x 95) + 0.95 = 0.9975 L

100

[Na+] (plasma water) = 7 = 7.02 mmol/L diluted plasma

0.9975

Multiplication by the dilution factor of 20 gives the sodium concentration in

undiluted plasma:

[Na+] in undiluted plasma = 7.02 x 20 = 140 mmol/L (3 sig figs)

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 BODY FLUIDS AND ELECTROLYTES

The anion gap

To maintain electrical neutrality the sum of concentrations of cations (sodium, potassium,
calcium, magnesium etc) must equal that of all the anions (chloride, bicarbonate,
phosphate, sulphate and proteins). Some, but not all of these are frequently measured in
clinical practice. If only sodium chloride and bicarbonate are measured then the following
relationship can be written:

[Na+] + [unmeasured cations] = [Cl-] + [HCO3-] + [unmeasured anions]

Which can be rearranged to:

[Na+] = [Cl-] + [HCO3-] + [unmeasured anions] - [unmeasured cations]

Since the concentrations of unmeasured anions always exceeds that of unmeasured

cations, their difference is defined as the anion gap:

Anion gap = [unmeasured anions] - [unmeasured cations]

Substituting anion gap into the previous expression gives:

[Na+] = [Cl-] + [HC03-] + Anion gap

which can rearranged to give the following expression for the anion gap:

Anion gap = [Na+] - [Cl-] - [HCO3-] ………………….. Eq. 8.8

The anion gap was originally developed as a quality control tool when it was noted that in
most patients the difference between the sodium concentration and the sum of the
chloride and bicarbonate concentrations was always approximately 12 mmol/L.
Sometimes the potassium concentration is included in the calculation. Small deviations
from the reference range for the anion gap (7-16 mmol/L) are usually due to marked
changes in plasma calcium, potassium, phosphate and negatively charged proteins.
However, the principal use of the anion gap is as an aid in the differential diagnosis of
non-respiratory acidosis. A markedly raised anion gap indicates the presence of excess
unmeasured anions of metabolic acids e.g. ketoacids, lactic, salicylic, oxalic (from
metabolism of ethylene glycol) and formic acids (from metabolism of methanol).

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Question Q 8(5)

The following results were obtained on a young adult in the Accident and Emergency
Department:
Plasma sodium = 140 mmol/L
Plasma chloride = 97 mmol/L
Plasma bicarbonate = 8 mmol/L

Calculate the anion gap.

Answer Q 8(5)

Anion gap = [Na+] - {[Cl-] + [HCO3-]}

= 140 - {97 + 8}

= 140 - 105

= 35 mmol/L

ADDITIONAL QUESTIONS
1. Over a 24 h period a patient recovering from intestinal resection receives 2 L of
fluids intravenously and 750 mL orally but does not eat any solids over this
period. The urine output over the same period is 1.25 L and 600 mL of fluid is
lost via a fistula. Is the patient in positive or negative fluid balance and by how
much?

2. A patient known to have diabetes insipidus is admitted in coma. His plasma

osmolality is 324 mosm/kg. If his weight is 85 kg, estimate his body water deficit.

3. A male adult insulin dependent diabetic forgot to take his insulin. His blood
glucose concentration, which was 5 mmol/L, rose to 15 mmol/|L in two hours.
Estimate the effect on his plasma sodium concentration, assuming that no other
water intake nor loss of water from the body takes place during this time,
indicating what assumptions you make.

4. A plasma sample with a total protein content of 70 g/L gave identical sodium
results of 140 mmol/L when measured using either a direct-reading ion-selective
electrode or a flame photometer. What plasma sodium result would you expect
the ion-selective electrode to give with the same plasma sample if its total protein
concentration had been 90 g/L?

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 ENZYMOLOGY

Chapter 9

Enzymology

Why measure enzyme activity?

Enzymes are proteins which catalyse chemical reactions. Enzymes are of interest to the
clinical biochemist for a number of reasons:

• Enzymes are often released from tissues into the circulation as a result of disease
e.g. the release of aspartate aminotransferase from the liver affected by hepatitis.

• Inherited diseases are often due to a deficiency in a particular enzyme e.g.

glucose-6-phosphatase in type-I glycogen storage disease.

• Enzymes may be used as analytical tools e.g. hexokinase and glucose-6-phosphate

dehydrogenase in the assay of glucose.

Several different approaches can be used to quantify enzymes:

• Measurement of the amount of enzyme protein following isolation from the

analytical sample. This approach is seldom used since enzyme purification is a
lengthy procedure during which some losses are inevitable.

• Since enzymes are proteins, immunoassay can be used (i.e. mass measurements).
This approach has been used for the assay of the MB isoenzyme of creatine
kinase.

• Measurement of the rate of the enzyme reaction (i.e. catalytic activity).

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Catalytic activity

In routine clinical practice enzymes are usually quantified by measuring their catalytic
activity. The rate of an enzyme reaction is dependent upon many factors (see Fig 9.1)
but, with a few exceptions and provided the conditions of the assay are carefully chosen,
the rate of the reaction is always proportional to the concentration of the enzyme in the
reaction mixture. The rate (or activity) will vary depending upon the analytical
conditions used and over the past few decades considerable effort has been expended by
biochemists to standardise assay conditions so that activity measurements obtained in
individual laboratories are comparable. Reaction conditions have been optimised so that
the highest rate (maximal sensitivity) is obtained and small variations in conditions
(substrate and cofactor concentrations, pH etc) have minimal effect. Whichever assay is
used the rate is proportional to enzyme concentration but the actual rate depends on the
reaction conditions employed.

 Enzyme concentration

 Nature of the substrate and any cofactor(s)

 Concentration of substrate and any cofactors

 Buffer and its concentration

 pH

 Ionic strength

 Inhibitor/activator concentration

 Temperature

Figure 9.1 Factors affecting enzyme activity

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 ENZYMOLOGY

How is activity measured?

An enzyme catalyses the conversion of its substrate into a product. Therefore the course
of the reaction can be followed by either measuring the disappearance of substrate or
formation of product (Fig 9.2).

Rate of consumption of substrate(s) = Rate of formation of product(s)

Product(s)

Concentration

Substrate(s)

Time

Figure 9.2 Progress curves of an enzyme catalysed reaction

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CHAPTER 9

If the product of the reaction does not have a physical property (such as absorbance) by
which its appearance can be monitored then either a reagent is added to form a suitable
derivative or a second enzyme is added to convert the product into another compound
which is easily measurable. When a second (or third) enzyme is added in this way the
assay is said to be coupled. It is vital that the concentration of the second enzyme is
present in excess so that product is removed as soon as it is formed i.e. the first enzyme
reaction (the one we wish to measure) is always rate limiting.

There are two approaches which can be used to obtain a rate measurement:

• Take measurements (of product or substrate concentration) at two points in time,

calculate the difference then divide by the time period to give the rate (i.e. change
of concentration per unit time). These are usually referred to as fixed-time
methods. The term end-point assay is often used but is incorrect because enzyme
activity is always a rate measurement and therefore needs at least two
measurements. A second reading is always required and sometimes it is assumed
(rightly or wrongly) that a reagent blank gives a reliable measure of concentration
at time zero.

• Monitor the reaction continuously and take a rate measurement over a suitable
time period. These continuous monitoring methods are preferred since it is
possible to evaluate the reaction progress and ensure that a true initial rate
measurement is taken and is constant, avoiding errors due to any lag-phase.

Whichever approach is used, the timing of the measurements is critical. Initially (or
possibly after any lag-phase) the rate of reaction at any given enzyme concentration is
constant. However, as the reaction progresses substrate is consumed and its
concentration falls and eventually a point is reached at which substrate availability
become rate limiting and the rate of the reaction falls i.e. the progress curve becomes
non-linear. This is illustrated in Fig 9.3. Unless a lag-phase is observed, measurement
over the segment 0-A gives a true initial rate whereas measurement over the segments 0-B
or 0-C gives an artificially low result. At point C the substrate is completely exhausted
(or the reaction is at equilibrium) and the reading is actually a measure of substrate rather
than enzyme concentration. Sometimes a lag phase is observed so that the initial rate is
less than optimal (see inset to Fig 9.3). In this situation the optimal rate is given over the
segment A-B not 0-A.

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 ENZYMOLOGY

Concn

0 AB
Time

Concentration

0 A B C

Time

Figure 9.3 Effect of measurement period on rate of an enzyme-catalysed

reaction. Inset shows an enzyme catalysed reaction with a lag-phase

Units for expressing enzyme activity

In order to make use of enzyme activity measurements for diagnostic purposes it is
obviously essential to express the result in a way which makes comparison with a
reference range or a patient’s previous result easy. In the early days of diagnostic
enzymology units were often named after the originator of the method used. For example

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King-Armstrong (KA) units were used for alkaline phosphatase (the amount of enzyme
present in 100 mL of serum that will split 1mg of phenol from phenylphosphate in 1
hour). Very soon a plethora of units were in use and attempts were made to standardise
enzyme units. This may seem a rather pointless exercise since although units may be
identical the enzyme activity will not be identical unless the reaction conditions are held
constant. Nevertheless, the Commission on Enzymes of the International Union of
Biochemistry propose that enzyme activity should be expressed in terms of international
units. One international unit (U) is the quantity of enzyme that catalyses the reaction of
1 μmol of substrate per minute. Catalytic concentration is to be expressed in terms of
U/L or mU/L, whichever gives the more convenient numerical value.

The international unit itself may eventually be replaced by a new unit termed the katal
and concentrations expressed as katals per litre (kat/L). One katal is the amount of
enzyme which catalyses the reaction of 1 mol of substrate per second.

When there is some uncertainty about the exact nature of the substrate (e.g. where the
substrate is a macromolecule such as starch or a protein) then units are still expressed as
the amount of a group or reside released per unit time (e.g. glucose units or amino acids
formed per minute).

Calculating enzyme activity

This involves converting physical measurements (e.g. absorbances) made over timed
interval(s) into substrate concentration units which are then used to derive a rate for the
enzyme catalysed reaction. This rate then needs to be converted to the concentration of
enzyme units in the clinical sample. The following example illustrates the process:

Question Q 9(1)

50 µL serum is added to 2 mL NADH solution (0.17 mmol/L) in Tris buffer (5.6

mmol/L) and incubated at 37°C for 10 min. 0.2 mL sodium pyruvate solution (13.5
mmol/L) is added and the rate of absorbance change monitored at 340 nm in a cuvette
with a path-length of 0.5cm. The absorbance readings at 30 s and 60 s are 0.183 and
0.148 respectively. Calculate the LDH activity in the serum (molar absorption coefficient
of NADH at 340 nm = 6.30 x 103 L/mol/cm).

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Answer Q 9(1)

The reaction catalysed by lactate dehydrogenase (LDH) is:

Pyruvate + NADH + H+ → Lactate + NAD+

This is the reverse of the normal reaction. The cofactor NADH absorbs at 340 nm,
whereas absorbance due to NAD+ is negligible. Therefore as the reaction progresses the
absorbance at 340 nm falls at a rate which is equal to the rate of consumption of the
substrate (pyruvate).

The concentration of NADH (c) at any point in time can be calculated from the
absorbance reading (A), the cuvette path length (b) and the molar absorptivity of NADH
(a) using equation Eq 4.4:

A = abc rearranged to give c = A

ab

Therefore at the first time the absorbance reading is taken (t1 = 30 seconds), A = 0.183,
b = 0.5 cm and a = 6.3 x 103 L/mol/cm so the concentration of NADH (c1) in mol/L
can be calculated as follows:

c1 = 0.183 = 5.8 x 10-5 mol/L

6.3 x 103 x 0.5

The same calculation could be performed for the second absorbance reading (when t2 =
60 seconds and A2 = 0.148) to give the concentration (c2) at time t2. The two equations
for the calculation of the concentrations at t1 and t2 are therefore:

At t1: c1 = A1
ab

At t2: c2 = A2
ab

These two expressions can be combined to calculate the change in concentration (∆c)
over the time period t2 – t1 (∆t):

∆c = c1 – c2 = A1 - A2 = (A1 – A2) = ∆A
ab ab ab ab

(Mathematicians use the symbol Δ to denote a difference or change between two values.)

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This gives the decrease in concentration of substrate over the time period (∆t) in units of
mol/L. Since International Units use concentration expressed as μmol/L, this value must
be multiplied by 1,000,000 (since there are 1,000,000 μmol in a mol):

∆c = ∆A x 1,000,000 μmol/L reaction mixture

a x b

Division by the time interval (t2 - t1 = ∆t) gives the rate of change in concentration as
μmol/sec/L reaction mixture. Multiplication by 60 converts this rate to μmol/min/L
reaction mixture:

LDH activity = ∆c = ∆A x 1,000,000 x 60 μmol/min/L reaction mixture

∆t ∆t x a x b

It is usual to express enzyme activity in serum as U/L of serum, not U/L of reaction
mixture. Therefore the dilution of the serum in the reaction mixture needs to be taken
into account. Multiplying by the total reaction volume and dividing by the sample volume
gives:

LDH activity = ΔA x 1,000,000 x 60 x Total reaction volume μmol/min/L serum

Δt x a x b x Sample volume

Substituting:

ΔA = A1 - A2 = 0.183 - 0.148 = 0.035

Δt = t2 - t1 = 60 - 30 = 30 seconds
a = molar absorptivity of NADH = 6.30 x 103 L/mol/cm
b = cuvette path length = 0.5 cm
Total reaction volume = 0.05(serum) + 2(NADH/buffer) + 0.2(substrate) = 2.25 mL
Sample volume (serum) = 0.05 mL

LDH activity = 0.035 x 1,000,000 x 60 x 2.25 = 10000 U/L

30 x 6.30 x 103 x 0.5 x 0.05

If a large number of calculations are to be performed for the same assay then it would be
simpler to combine all the terms (except ΔA) to produce a factor which could then be
used to obtain enzyme activity directly i.e. ΔA x Factor (28600) = LDH activity
(U/L).

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 ENZYMOLOGY

1. Divide change in absorbance by the time period (in min) over which the
measurements were taken to give rate i.e. ΔA/min

2. Divide by molar absorptivity (units: L/mol/cm) and cuvette path length (cm)

3. Multiply by 1,000,000 to convert from mol to μmol

4. Multiply by total reaction volume (mL) and divide by sample volume (mL)

For an enzyme assay utilizing NADH/NAD as cofactor (monitored at 340 nm) the
formula used is:

Enzyme activity (U/L) = ΔA/min x 1,000,000 x Total vol (mL)

6.30 x 103 x Path length (cm) x Sample vol (mL)

which simplifies to:

Enzyme activity (U/L) = ΔA/min x 160 x Total vol (mL)

Path length (cm) x Sample vol (mL)

Figure 9.4 Steps in the calculation of enzyme activity

Conversion of enzyme units

The older literature is full of enzyme data expressed in units other than U/L. It is
sometimes useful to convert these values to the corresponding activity in U/L. For
example, King-Armstrong (KA) units were used for many years to report alkaline
phosphatase activity. One KA unit is the amount of enzyme in 100 mL of serum that will
split 1 mg of phenol from phenylphosphate in 1 hour and can be written:

1 KA unit = 1 mg phenol/h/100 mL serum

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To convert to activity expressed as international units (U/L) it is necessary to determine

the number of μmol of phenol formed in 1 min in 1 L of serum. The following steps are
involved:

1. Multiply by 1,000 to convert mg to μg

2. Divide by the molecular weight of phenol (94) to convert from μg to μmol

3. Divide by 60 to convert reaction period from h to min

4. Multiply by 10 to convert from 100 mL serum to 1 L serum

The final result is: 1 KA unit = 1,000 x 10 = 1.77 U/L

94 x 60

Therefore:

Alk phos (U/L) = Alk phos (KA units) x 1.77

Alk phos (KA units) = Alk phos (U/L)

1.77

It is important to remember that even after converting enzyme activity from one unit to
another, the numerical result will still depend on the reaction conditions used.

Question Q 9(2)

A transaminase result is quoted in the literature as 207 Karmen units. One Karmen unit is
the amount of transaminase that will produce an absorbance change of 0.001 in a 1cm
cuvette at 340 nm (a coupled reaction) in 1 min per 1 mL serum (in a total volume of 3
mL). Assuming the molar absorptivity of NADH is 6.30 x 103 L/mol/cm, express the
transaminase activity as a) international units per L of serum, and b) katals/L.

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Answer Q 9(2)

a) 1 Karmen Unit = 0.001 A/min/mL serum (in a total volume of 3 mL)

1 U/L = 1 μmol/min/L serum

Therefore:

1. Divide by molar absorptivity of NADH to covert from absorbance (A) to mol/L

2. Multiply by 1,000,000 to convert from mol/L to μmol/L

3. Multiply by 3 to allow for dilution of 1 ml serum to 3 mL in the assay

Therefore:

1 Karmen unit = 0.001 x 1,000,000 x 3 U/L

6.30 x 103

1 Karmen unit = 0.476 U/L

Therefore 207 Karmen units = 0.476 x 207 = 99 U/L (2 sig figs)

b) 1 Katal/L = 1 mol/sec/L

To convert to U/L:

1. Multiply by 1,000,000 to convert from mol to μmol

2. Multiply by 60 to convert from seconds to minutes

Therefore:

1 katal/L = 1,000,000 x 60 = 60 x 106 U/L

or 1 U/L = 1 = 16.7 x 10-9 katal/L

60 x 106

Therefore, 99 U/L = 99 x 16.7 x 10-9 = 1.65 x 10-6 katal/L = 1.65 μkatal/L

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Effect of substrate concentration on the rate of an enzyme

catalysed reaction – the Michaelis-Menten equation

Figure 9.1 listed factors which influence the rate of an enzyme catalysed reaction i.e.
enzyme activity. These factors are best studied by keeping them all, except the one under
investigation, constant. So far we have only considered the effect of variation of the
amount of enzyme on activity since this is the variable of most interest in clinical
practice. In the absence of complicating factors the rate of an enzyme reaction is directly
proportional to enzyme concentration. However, the relationship between reaction rate
and substrate concentration is a little more complex. If we take the simplest possible
enzyme reaction in which a single substrate (S) binds to enzyme (E) to form an essential
intermediate the enzyme-substrate complex (ES) which decomposes to release free
enzyme and the reaction product (P) this may be represented schematically as:

k+1 k+2
E + S ES E + P

k-1 k-2

The rate constants for the various reactions are denoted k+1, k-1, k+2 and k-2. Note that
rate constants of reverse reactions are given a minus sign. The rate of each reaction is the
product of the molar concentrations of the reactants and the respective rate constant.
Therefore we can write the following equations (in which square brackets denote molar
concentrations) for the rates of formation and decomposition of the enzyme-substrate
complex ES:

Rate of formation of ES = k+1[E][S] + k-2[E][P]

Rate of decomposition of ES = k-1[ES] + k+2[ES]

A few milliseconds after the enzyme and substrate are mixed [ES] builds up and does not
change provided [S] is in large excess and k+1>>k+2. This condition is called a steady
state in which the rate of formation of ES is balanced by its rate of decomposition so that
[ES] is constant. Therefore the following steady state equation can be written :

Rate of formation of ES = Rate of decomposition of ES

k+1[E][S] + k-2[E][P] = k-1[ES] + k+2[ES]

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However, in enzyme kinetics we make measurements at an early stage of the enzyme-

catalyzed reaction when [P] will be very small and rate of formation of ES from E and P
is very low and can be ignored. Thus if we measure initial rates only then the above
equation can be simplified to:

k+1[E][S] = k-1[ES] + k-2[ES]

then rearranged to give an expression for [E]/[ES]:

k+1[E][S] = [ES](k-1 + k-2)

[E] = (k-1 + k-2)

[ES] k+1 [S]

A constant, called the Michaelis-Menten constant (Km) can be defined as:

Km = k-1 + k-2
k+1

So substituting Km for (k-1 + k-2)/k+1 gives:

[E] = Km ………………………. Eq. 9.1

[ES] [S]

The total enzyme concentration [E]Total is the sum of the free enzyme and the enzyme-
substrate complex. Therefore we can write the following enzyme conservation equation:

[E]Total = [E] + [ES]

Re-arranging gives [E] = [E]Total - [ES], so that an expression for [E]/[ES] can be
written in terms of concentrations of enzyme components:

[E] = [E]Total - [ES]

[ES] [ES]

which can also be written:

[E] = [E]Total - 1 …………………… Eq. 9.2

[ES] [ES]

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Combining the two expressions for [E]/[ES] gives:

[E]Total - 1 = Km
[ES] [S]

Multiplication throughout by [S] gives:

[E]Total[S] - [S] = Km
[ES]

[E]Total[S] = Km + [S]
[ES]

which can be re-arranged to give an expression for [ES]:

[ES] = [E]Total[S]
Km + [S]

The rate of formation of product (v) can be expressed as a function of the rate constant
k+2 and the concentration of the enzyme-substrate complex:

v = k+2[ES] ……………………………. Eq. 9.3

Substitution of the expression for [ES] gives:

v = k+2[E]Total[S]
Km + [S]

K+2 and [E]Total are constant and can be replaced by a single constant called maximal
velocity (Vmax) to give the Michaelis-Menten equation:

v = Vmax [S] …………………………………. Eq. 9.4

Km + [S]

Consider the behaviour of this equation at the two extremes of substrate concentration.
When [S] is very low, for example much lower than the Km (i.e. Km >>> [S]), then the
Michaelis-Menten equation approximates to v = Vmax[S]/Km which is of the general form
v = constant x [S]. This is a linear expression so under these conditions the rate is
directly proportional to substrate concentration as illustrated in Fig 9.5. When a reaction

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rate is proportional to a single concentration term it is said to follow first-order kinetics.

At very high substrate concentrations most of the enzyme exists as the enzyme-substrate
complex i.e. the enzyme becomes saturated with substrate. Under these conditions [S]
>>> Km so that the Michaelis-Menten equation approximates to v = Vmax[S]/[S] = Vmax
and the rate becomes essentially independent of substrate concentration and constant (i.e.
v = Vmax). Note that during derivation of the Michaelis-Menten equation Vmax contained
the term [E]Total so that although the rate is independent of substrate concentration it is
still dependent on the amount of enzyme present. When a reaction rate is independent of
concentration it is said to follow zero-order kinetics. This is also illustrated in Fig 9.5.
At intermediate concentrations the reaction follows a mixture of first and zero-order
kinetics and the plot of rate versus substrate concentration is curved. In fact it is
hyperbolic with the rate approaching Vmax asymptotically as [S] approaches infinity.

One special situation worth considering is the significance of the substrate concentration
at half-maximal velocity (i.e. when v = Vmax/2). We can then write:

Vmax = Vmax [S]

2 Km + [S]

which can be re-arranged to give an expression for [S]:

[S] = Vmax (Km + [S])

2 Vmax

Cancelling the Vmax terms and re-arranging gives:

2 [S] = Km + [S]

and subtracting [S] from both sides gives:

[S] = Km

Therefore the Km is the substrate concentration at half-maximal velocity (it therefore

follows that the units for Km are concentration units) This is probably the most useful
definition of Km since it makes no assumption about the relative magnitude of the
individual rate constants. If k+1>>k+2 then k+2 can be ignored, the ES complex is in
equilibrium with free enzyme and substrate and Km approximates to the dissociation
constant of the enzyme-substrate complex:

Km ≈ k+1 = [E][S]
k-1 [ES]

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v Zero-order
v = Vmax

First-order
v = Vmax x [S] Mixture of zero and first-order
Km
v = Vmax[S]
v = Vmax/2 Km + [S]

[S] = Km

[S]

Figure 9.5 Effect of increasing substrate concentration [S] upon the initial
velocity (v) of an enzyme-catalysed reaction. At low substrate
concentrations the reaction follows first order kinetics; at high
substrate concentrations first order kinetcs. At all concentrations the
rate is described by the Michaelis-Menten equation. At half maximal
velocity the substrate concentration is equal to the Michelis-Menten
constant (Km)

Question Q 9(3)

Enzymologists recommend that whenever possible the substrate concentration in an

enzyme assay should be at least ten times the Michaelis constant (Km). What is the rate
of reaction achieved (expressed as multiples of the maximal velocity), for an enzyme
reaction which obeys simple Michaelis-Menten kinetics, when the substrate concentration
is exactly ten times the Km value?

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Answer Q 9(3)

The Michaelis-Menten equation relating initial velocity to substrate concentration is:

v = Vmax [s]
Km + [s]

v = initial velocity
Vmax = maximal velocity (at infinite substrate concentration)
Km = Michaelis-Menten constant = substrate concentration at half-maximal velocity
[s] = initial molar substrate concentration

Substituting 10 Km for [s]:

v = Vmax 10 Km
Km + 10 Km

Substituting (Km + 10 Km) = 11 Km, then cancelling Km gives the value of v:

v = Vmax 10 Km = 10 Vmax = 0.91 Vmax (2 sig figs)

11 Km 11

Therefore the initial rate is approximately 90 per cent of the maximal rate (Vmax).

Although the Michaelis-Menten equation has been derived for the simplest case of a
single substrate reaction its application is by no means limited to this. The equation can
be written in a more general form:

v = Vmaxapp [S] …………………… Eq. 9.5

Kmapp + [S]

in which Kmapp and Vmaxapp are not true constants but apparent values for Km and Vmax
which depend on the concentrations of activators, inhibitors, second substrates etc which
are held constant whilst [S] is varied. Variation of these apparent constants with other
parameters depends on the kinetic mechanisms involved. Therefore simple modification
of the Michaelis-Menten equation can be used to study enzyme inhibition, activation, pH
effects and multiple-substrate reactions.

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Graphical solutions of the Michaelis-Menten equation

In theory it possible to obtain estimates of Km and Vmax from plots of v versus [S] but
deciding when the rate is maximal is difficult since the rate approaches Vmax
asymptotically. To overcome this practical difficulty various graphical solutions have
been proposed.

The double-reciprocal plot of Lineweaver-Burk. Inversion of the Michaelis-Menten

equation (Eq 9.4) gives the expression:

1 = Km + [S]
v Vmax[S]

Separating the left hand side into two components gives:

1 = Km + [S]
v Vmax[S] Vmax[S]

If the [S] terms are cancelled in the second component of this expression on the right
hand side then this equation can be rewritten in a more useful form:

1 = Km x 1 + 1 …………….. Eq. 9.6

v Vmax [S] Vmax

Therefore a plot of 1/v versus 1/[S] is linear, with slope Km/Vmax and intercept on the 1/v
axis of 1/Vmax (see Fig 9.6a). It can easily be shown that the intercept on the 1/[S] axis is
-1/Km. Double reciprocal plots are easy to interpret and computation of Km and Vmax is
straightforward.

The [S]/v versus [S} plot of Hanes. If the double-reciprocal equation of Lineweaver and
Burk (Eq 9.6) is multiplied throughout by [S] then another linear from is obtained:

[S] = Km + [S] x 1 ………… Eq. 9.7

v Vmax Vmax

Thus a plot of [S]/v versus [S] is linear with slope 1/Vmax and intercepts on the [S]/v and
[S] axes of Km/Vmax and –Km respectively (Fig 9.6b).

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a) Double reciprocal plot (Lineweaver & Burk )

0.12
1/v
0.1

0.08

0.06 Slope = K m /V max

Intercept
0.04
= -1/K m

0.02 Intercept = 1/V max

0
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1
1/[S]

0.7 b) Plot of [S]/v versus [S] (Hanes )

0.6
[S]/v
0.5

0.4
Slope = 1/V max
0.3
Intercept
= -K m 0.2

0.1 Intercept = K m /V max

0
-10 -5 0 5 10 15 20 25 30
[S]

Figure 9.6 Double reciprocal plot (a) and [S]/v versus [S] plot (b) for the same set
of data

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CHAPTER 9

60 a) Plot of v versus v /[S] (Eadie-Hofstee )

50 Intercept = V max

40

v 30

20 Intercept
Slope = -K m = V max /K m
10

0
0 2 4 6 8 10 12

v /[S]

b) Direct linear plot

(Eisenthal & Cornish Bowden ) Km
60

V max
50 V max

40

30

20

10

0
-25 -20 -15 -10 -5 0 5 10 15 20 25
Km

Figure 9.7 Plot of v versus v/[S] (a) and the direct linear plot (b) for the same
data depicted in Fig 9.6

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 ENZYMOLOGY

The v versus v/[S] plot of Eadie-Hofstee. Division of both sides of the Michaelis-Menten
equation by [S] gives:

v = Vmax
[S] Km + [S]

Multiplication of both sides by (Km + [S]) yields the following expression:

v(Km + [S]) = Vmax

[S]

which can be simplified to:

vKm + v = VMax
[S]

then rearranged to give a linear expression for v and v/[S]

v = Vmax - Km v ………. Eq. 9.8

[S]

Therefore a plot of v versus v/[S] is linear with slope –Km and v and v/[S] intercepts of
Vmax and Vmax/Km respectively (Fig 9.7a).

The direct linear plot of Eisenthal and Cornish-Bowden. These authors use a rather
unique approach in which the constants Vmax and Km are treated as variables and plotted as
points in observational space whereas values for v and [S] are treated as constants and
plotted as lines in parameter space. This strange concept is probably easier to understand
if the Eadie-Hofstee form of the Michaelis-Menten equation (Eq. 9.5) is re-arranged
slightly to:

Vmax = v Km + v ………….. Eq. 9.9

[S]

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This equation is now that of a straight line of the form y = ax + b, in which the variables x
and y are now Km and Vmax respectively with a lope of v/[S] and y intercept of v. We are
used to thinking of Km and Vmax being constant with an infinite number of values of v and
[S] which can satisfy the Michaelis-Menten equation. This concept is now reversed with
v and [S] being constant but with an infinite number of values for Km and Vmax which can
satisfy the equation. As before we can calculate the values for the intercepts on the x and
y axes when Vmax is plotted against Km:

When Km = 0, (v/[S])Km is also zero so that Eq 9.6 becomes: Vmax = v. In other

words the y intercept is v.

Similarly when Vmax = 0, Eq 9.6 becomes: 0 = v Km + v

[S]

which can be re-arranged to v = - Km v

[S]

Multiplying both sides by [S]/v and gives: Km = - [S]. Therefore the intercept on
the x axis is –[S].

This means that a line joining a pair of values for v and [S] plotted on the y and x axes
respectively if extrapolated must pass through the point in the Vma x- Km space with
coordinates Vmax and Km for that enzyme. A similar argument applies for other pairs of
values for v and [S] for the same enzyme. They must all pass through the point with
coordinates Vmax and Km. In other words they must all intercept at the same point and this
point has the coordinates Vmax and Km.

The procedure is illustrated in Fig 9.7b for the same data as that used in Figs 9.6a and b
and 9.7a. The values for v and [S] are marked on the Vmax and Km axes respectively,
then each pair of values joined by a straight line which is extrapolated into the positive
Vmax-Km quadrant. All the lines intercept at a common point with coordinates Vmax and
Km.

Difficulties often occur in deciding upon the exact intersection point. Frequently several
such points occur as a result of experimental variation (this is analogous to deciding
where to draw a straight line through a series of points on a conventional plot). In the
worse-case scenario if there are n pairs of observations then there is a maximum of 0.5n
(n - 1) possible intersections. The authors recommend that the median of all possible
intersections are used (if three lines intersect at one point the this is treated as three
intersections rather than one in finding the median, if there are four lines intersecting then
this counts as six intersections etc).

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Although the Lineweaver-Burk plot is widely used for kinetic analysis of enzyme
reactions, the use of reciprocals means that small experimental errors can result in large
errors in graphically determined values for Km and Vmax. It has been argued that the
Eadie-Hofstee plot (i.e. v versus v/[S]) results in less error. It is apparent by inspection of
the plots in Figs 9.6 and 9.7, which are all based on the same v and [S] data, that each
method of plotting results in a different “spread” of results e.g. the double reciprocal plot
compresses the points at high substrate concentrations. This pitfall can be overcome by
careful selection of concentration values employed in the experiment. The direct linear
plot overcomes some of these limitations. Nowadays computer statistics packages are
frequently used to fit data directly to the Michaelis-Menten equation.

Enzyme inhibition

Some enzyme inhibitors act irreversibly by forming a covalent bond with an amino acid
residue in the enzyme thereby rendering it inactive. However, most inhibitors bind
reversibly to the enzyme so that an equilibrium is established between the free enzyme
(E) and inhibitor enzyme (EI) complex, the position of which is determined by the
inhibitor constant (Ki) which reflects the affinity of the inhibitor (I) for the free enzyme:

E + I EI Ki = [E][I]
[EI]

Clearly the inhibitor can lower the reaction velocity (v) by either decreasing the
numerator of the Michaelis-Menten equation (i.e. value of the Vmaxapp) or increasing the
denominator (i.e. the value for Kmapp). There are three main types of inhibitors which
exert their effects on v in different ways:

Competitive inhibitors, often structurally related to the natural substrate bind reversibly
with the enzyme at or near the active site. The inhibitor and substrate therefore compete
for the enzyme according to the scheme:

E + S ES E + P
+
I

EI

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Therefore the inhibitor effectively removes a portion of the enzyme from the reaction.
The enzyme conservation equation therefore becomes:

[E]Total = [E] + [ES] + [EI]

By substituting [E][I]/Ki for [EI] and grouping the [E] terms this becomes:

[E]Total = [ES] + [E](1 + [I]/Ki)

which can the be incorporated into the steady state equation for [ES] (Eq 9.1) to derive
the following variation of the Michaelis-Menten equation for competitive inhibition:

v = Vmax[S] …………………… Eq. 9.10

Km(1 + [I]/Ki) + [S]

Therefore the Km has been increased since it is now multiplied by a number which is
always greater then one, which in turn reduces the value of v. Note that it is possible to
reduce the relative contribution of the Km(1 + [I]/Ki) term by increasing the value of [S].
In other words it is theoretically possible to overcome competitive inhibition at high
substrate concentrations.

Non-competitive inhibitors bind reversibly at areas other than the active site. Binding of
inhibitor and substrate is independent so it is possible not only to form complexes
between enzyme and inhibitor (i.e. EI) but between inhibitor, substrate and enzyme
(i.e. EIS) according to the scheme:

E + S ES E + P
+ +
I I

EI + S EIS

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 ENZYMOLOGY

The EIS complex cannot breakdown to reaction products so again a proportion of the
enzyme becomes unavailable to take part in the reaction. Inhibition cannot be overcome
by increasing substrate concentration since binding at the two sites in independent.
In fact the dissociation constants (Kis) of the EIS and EI complexes are identical:

Ki = [E][I] = [ES][I]
[EI] [EIS]

and the enzyme conservation equation becomes:

[E]Total = [E] + [ES] + [EI] + [EIS]

By substituting [E][I]/Ki for [EI], [ES][I]/Ki for [EIS] and grouping the [E] and [ES]
terms this becomes:

[E]Total = [E](1 + [I]/Ki) + [ES](1 + [I]/Ki)

which can the be incorporated into the steady state equation for [ES](Eq 9.1) then used to
derive the following variation of the Michaelis-Menten equation for non-competitive
inhibition:

v = Vmax[S] …………………… Eq. 9.11

(1 + [I]/Ki) (Km + [S])

Therefore the value of Km is unchanged but the Vmax is divided by a number greater than
one so its value and hence the rate of the reaction is reduced.

An uncompetitive inhibitor combines only with the ES complex, not the free enzyme
according to the scheme:

E + S ES E + P
+
I

Ki = [ES][I]
EIS [EIS]

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0.25

Non-competitive
0.2

1/v Competitive

0.15

Uncompetitive
0.1

Uninhibited
0.05

0
-0.6 -0.4 -0.2 0 0.2 0.4 0.6
1/[S]

Type of inhibition Kmapp Vmaxapp

Competitive Km(1 + [I]/Ki) Vmax

Non-competitive Km Vmax/(1 + [I]/Ki)

Uncompetitive Km/(1 + [I]/Ki) Vmax/(1 + [I]/Ki)

Figure 9.8 Double reciprocal plots for an enzyme illustrating the effects of
competitive, non-competitive and uncompetitive inhibition. The inset
shows the effects on the apparent Km and Vmax. For each plot Km =
5 mmol/L, Vmax = 50 μmol/min, Ki = 10 mmol/L and the inhibitor
concentration is 20 mmol/L

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The enzyme conservation equation is therefore:

[E]Total = [E] + [ES] + [EIS]

Substituting [EIS] = [ES][I]/Ki and grouping the [ES] terms this becomes:

[E]Total = [E] + [ES] (1 + [I]/Ki)

which can be incorporated into the steady state equation for [ES] (Eq 9.1) then used to
drive the following variation of the Michaelis-menten equation for uncompetitive
inhibition:

v = Vmax[S] ………… Eq. 9.9

Km + [S](1 + [I]/Ki)

As for non-competitive inhibition the Vmax is divided by a number greater than one so its
value and hence the rate of the reaction is reduced. It is interesting to note that the
apparent Km is actually reduced suggesting that the substrate is more avidly bound to the
enzyme in the presence of inhibitor. However this is insufficient to overcome the
reduction in apparent Vmax.

Inhibition data can be transformed to linear equations in the same way as for uninhibited
reactions. Figure 9.8 shows a double reciprocal plot for an enzyme inhibited in three
different ways by the same inhibitor concentration. Simple inspection of the curves
allows identification of the mode of inhibition. Values for Kmapp and Vmaxapp can be
obtained by analogous procedures used for Km and Vmax. Substitution of values for Km
and Vmax into Kmapp and Vmaxapp respectively, together with inhibitor concentration permits
calculation of Ki. Another approach attributed to Dixon is to plot 1/v versus [I] using two
or more substrate concentrations; the best fit lines at each substrate concentration should
intercept at a value of [I] equal to Ki. The preferred approach is to determine values for
Kmapp and Vmaxapp over a wide range of inhibitor concentrations then to use secondary
plots to determine Ki.

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Question Q 9(4)

Drug A is routinely used in the treatment of rheumatoid arthritis. It is metabolized

in vivo to its active metabolite B by an enzyme PP. The possibility of introducing drug C
into the treatment regimen is being investigated and there are some concerns that drug C
may inhibit the metabolism of drug A.

8 cuvettes were set up (numbered 1 to 8) each with an optical path length of 1 cm. 0.6
mL of a stock solution of drug A (5 mmol/L) was diluted to 25 mL with buffer, then
used to prepare a series of dilutions from this diluted substrate as follows:

0.5 mL of diluted A was pipetted into cuvettes 1 and 2.

0.5 mL of buffer was added to cuvette 2, mixed, then 0.5 mL transferred to cuvette 3.
0.5 mL of buffer was added to cuvette 3, mixed, then 0.5 mL transferred to cuvette 4.
0.5 mL of buffer was added to cuvette 4, mixed, then 0.5 mL removed and discarded.

0.5 mL of buffer was then added to each cuvette.

An identical set of dilutions of diluted A was prepared in cuvettes 5 to 8, except that 0.5
mL of a solution of drug C (50 mmol/L) was added at the final stage instead of buffer.

1 mL of enzyme PP solution and 1 mL of a second enzyme (which was not rate limiting
but converts B into a coloured product with a molar absorptivity at 505 nm of 5500
L/mol/cm) was added to each cuvette, the contents mixed then incubated for exactly 5
minutes. The absorbance of each cuvette at 505 nm was measured versus a cuvette
containing distilled water (there was no significant reagent blank).

The following absorbance readings were obtained:

Cuvette No 1 2 3 4 5 6 7 8
Absorbance 0.400 0.330 0.250 0.167 0.330 0.250 0.167 0.100

a) Calculate the initial concentration of substrate (A) in each cuvette

b) Calculate the rate of formation of B (expressed as umol /L/min)
c) Determine the Km of enzyme PP for A
d) Determine the type of inhibition present
e) Determine the inhibitor constant of C
f) Comment on the likely consequences of introducing drug C into the
regimen for patients already receiving drug A.

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Answer Q 9(4)

a) Concentration of stock A = 5 mmol/L = 0.005 mol/L

Concentration of diluted A (mol/L)

= Concentration of stock A (mol/L) x Vol of stock A (mL)

Volume of diluted A (mL)

= 0.005 x 0.6 = 0.00012 mol/L = 1.2 x 10-4 mol/L

25

Each cuvette contains: Substrate 0.5 mL

Buffer or inhibitor 0.5 mL
Enzyme solution (PP) 1.0 mL
Second enzyme 1.0 mL
Total volume 3.0 mL

Therefore the concentration of substrate in cuvettes 1 (and 5) is given by:

Initial substrate concentration (mol/L)

= Concentration of added substrate (mol/L) x Vol substrate added (mL)

Final volume in cuvette (mL)

= 1.2 x 10-4 x 0.5 = 0.20 x 10-4 = 20.0 x 10-6 mol/L

3.0

Since doubling dilutions were prepared the concentration in each subsequent

cuvette is reduced by one half of the concentration in the previous cuvette.
Therefore the concentrations are:

Cuvette No: 1 2 3 4 5 6 7 8
Substrate (x 106 mol/L) 20 10 5 2.5 20 10 5 2.5

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b) Assuming the initial rate of the conversion of of A to B is rate limiting and

that the absorbance of the coloured product of the second enzyme obeys the
Beer-Lambert Law then the absorbance is proportional to the concentration of
substrate A consumed in the reaction catalyzed by enzyme PP. Therefore:

A = a x b x c

Where A = absorbance at 505 nm

a = molar absorptivity = 5,500 L/mol/cm
b = cuvette path length = 1 cm
c = concentration (mol/L)

Re-arrange to give the concentration of B in the cuvette:

c = A = A mol/L
a x b 5,500 x 1

Divide by 5 to obtain the concentration of B produced per minute (since a 5

minute reaction time was used) then multiply by 1,000,000 to convert from
mol to μmol:

v = A x 1,000,000 = A x 36.4 μmol/L/min

5,500 x 1 x 5

Cuvette No: 1 2 3 4 5 6 7 8
Absorbance 0.400 0.330 0.250 0.167 0.330 0.250 0.167 0.100
v (μmol/min) 14.6 12.0 9.1 6.1 12.0 9.1 6.1 3.6

c) To determine the Km and answer the rest of the questions some graphical
presentation of the data is required. Although not ideal, the double reciprocal
plot is simplest. Cuvettes 1 to 4 are without inhibitor, cuvettes 5 to 8 have the
same substrate concentrations but with inhibitor present.

Concentration of inhibitor solution C = 50 mmol/L = 50 x 10-3 mol/L

Final concentration of inhibitor C (mol/L)

= Initial inhibitor concentration (mol/L) x Vol inhibitor added (mL)

Final volume in cuvette (mL)

= 50 x 10-3 x 0.5 = 8.33 x 10-3 mol/L

3.0

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 ENZYMOLOGY

Cuvette No Inhibitor [S] 1/[S] v 1/v

(x 10 M) (x 106mol/L)
3

1 0 20 0.05 14.6 0.0685

2 0 10 0.10 12.0 0.0833
3 0 5 0.20 9.1 0.110
4 0 2.5 0.40 6.1 0.164
5 8.33 20 0.05 12.0 0.0833
6 8.33 10 0.10 9.1 0.110
7 8.33 5 0.20 6.1 0.164
8 8.33 2.5 0.40 3.6 0.278

0.3

0.25

+ inhibitor
0.2

1/v
0.15

No inhibitor
0.1

0.05

0
-0.25 -0.15 -0.05 0.05 0.15 0.25 0.35 0.45
6
1/[S] ( x 10 L/mol )

From the graph the intercept on the 1/[S] axis for the uninhibited reaction is
approximately - (0.20 x 106) L/mol. This corresponds to -1/Km:

-1 = - (0.20 x 106) L/mol

Km

Therefore Km = -1 = 5.0 x 10-6 mol/L

- (0.20 x 106)

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d) Drug C lowers the activity of the enzyme by increasing the value of the Km
without altering the Vmax i.e. the double reciprocal plots for the inhibited and
uninhibited reaction intersect on the 1/v axis. Therefore drug C is a competitive
inhibitor of enzyme PP.

e) From the graph the intercept on the 1/[S] axis for the inhibited reaction is
approximately - (0.11 x 106 ) L/mol. This corresponds to -1/Kmapp:

-1 = - (0.11 x 106) L/mol

Km app

Therefore Kmapp = -1 = 9.09 x 10-6 mol/L

- (0.11 x 106)

Since we are dealing with a competitive inhibitor:

Kmapp = Km (1 + [I]/Ki) (see Fig 9.8)

Where Km = Km of uninhibited reaction = 5.0 x 10-6 mol/L

[I] = inhibitor concentration = 8.33 x 10-3 mol/L
Ki = inhibitor constant

Substituting these values and solving for Ki:

9.09 x 10-6 = 5.0 x 10-6 {1 + (8.33 x 10-3)/Ki}

1 + (8.33 x 10-3) = 9.09 x 10-6

Ki 5.0 x 10-6

(8.33 x 10-3) = 1.818 - 1 = 0.818

Ki

Ki = 8.33 x 10-3 = 1.0 x 10-2 mol/L (2 sig figs)

0.818

f) The inhibitor constant (Ki) of drug C is considerably higher than the Km for the
substrate A (1.0 x 10-2 mol/L compared to 5.0 x 10-6 mol/L). Both of these
constants are inversely proportional to the affinity of the enzyme PP for the
substrate and inhibitor. Thus the affinity of the enzyme for the substrate, A, is
considerably greater than its affinity for the inhibitor, C. Therefore the effect of

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 ENZYMOLOGY

introducing drug C into the regimen for patients receiving drug A will depend on
the relative concentrations of the two drugs. If their therapeutic concentrations
are similar, or the concentration of A is greater than C, then C will have little
effect on the metabolism of drug A or the optimum dose required to achieve
therapeutic levels of its active metabolite B. If, however, the plasma level of drug
C is considerably higher than that of drug A, then inhibition of the metabolism of
drug A will occur and higher plasma levels of drug A will be achieved, with the
consequence of decreased levels of the active metabolite B. As a result higher
doses of dose A will be required to achieve the same therapeutic result.

If drug C is not only an inhibitor of PP but a substrate for this enzyme then the
metabolism of drug C will also be affected by drug A which will have
consequences on the levels of drug C (and its metabolites) achieved.

ADDITIONAL QUESTIONS

1. An assay mixture for the measurement of lactate dehydrogenase constituted

2.7 mL of buffered NADH and 100 µL of serum. The reaction was started by
adding 100 µL of sodium pyruvate. The absorbance change over 5 minutes was
0.150 when measured in a 0.5 cm light path at 340 nm. Assuming the molar
absorbtivity of NADH at 340 nm is 6.30 x 103 L.mol-1cm-1, calculate the enzyme
activity in international units per litre of serum..

2. An assay for alkaline phosphatase activity involved mixing 0.5 mL of serum with
2.7 mL buffer, allowing temperature to reach equilibrium then starting the
reaction by adding 0.2 mL of substrate (4-nitrophenyl phosphate). The increase in
absorbance in a 1cm cuvette due to the liberation of product (4-nitrophenol) was
0.180 over a 5-minute period. Calculate the alkaline phosphtase activity
expressing the result as a) international units per litre of serum, and b) katals per
litre of serum. Assume that the molar absorptivity of 4-nitrophenol is 1.88 x 104
L/mol/cm.

3. The Somogyi saccharogenic method for the assay of amylase involves measuring
the rate of release of glucose from substrate. One Somgyi unit is the amount of
enzyme catalysing the release of 1 mg of glucose in 30 min per 100 mL serum.
Derive a factor to convert Somogyi units to international units per litre of serum.

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4. One Wroblewski-laDue unit is the amount of lactate dehydrogenase which results

in an absorbance change (due to NADH) at 340 nm of 0.001 per minute per mL
serum in a reaction mixture with a total volume of 3 mL. Derive a factor to
convert Wroblewski-LaDue units to International units per litre of serum. Assume
the molar absorptivity of NADH is 6.3x 103 L/mol/cm.

5. If the Km of an enzyme which obeys simple Michaelis-Menten kinetics is 2.5

mmol/L, what velocity (expressed as a multiple of Vmax) would be obtained at a
substrate concentration of 10 mmol/L?

6. What information can be obtained from the double-reciprocal plot for an enzyme
under the following conditions: a) 1/v = 0 when 1/[S] = -12.5 x 106 L/mol, b)
1/[S] = 0 when 1/v = 5.2 x 106 min.L/mol, c) 1/[S] = 0 when 1/v = 6.5 x 106
min/mol and the slope of the line is 100 min/L?

7. You carry out an enzyme experiment in which the substrate concentration is

expressed as mmol/L and the reaction velocity in μmol/L/min. What would be
the units for the axes of the three following plots: a) 1/[S] versus 1/v, b) [S]/v
versus [S], c) v versus v/[S]?

8. Mucic acid is an inhibitor of β-glucuronidase. The following data were obtained

using phenolphthalein glucuronide as substrate, in the presence and absence of
mucic acid (concentration in the assay = 1.0 x 10-4 mol/L).

Substrate Reaction velocity

Concentration
(mmol/L) No inhibitor Mucic acid

0.5 33 9
1.0 50 17
2.0 67 29
4.0 80 44
10 91 67

Determine the type of inhibition and the enzyme-inhibitor dissociation constant.

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 ENZYMOLOGY

9. An experiment was conducted to study the effect of pH on the activity of lactate

dehydrogenase. Using a histidine buffer at pH 5.5 and 7.4 the reaction was
monitored by following the increase in absorbance at 340 nm due to the reduction
of NAD. The following data were obtained:

Lactate Reaction velocity

concentration
mmol/L pH 7.4 pH 5.5

1 12 33
2 21 50
4 35 67
10 57 83
20 73 91

Stating any assumptions that you make determine the pH at which the enzyme has
greatest affinity for the substrate.

10. The apparent Km and Vmax of an enzyme were measured over a range of inhibitor
concentrations and the following data obtained:

Inhibitor Apparent value

concentration Km Vmax
(mmol/L) (mmol/L) (μmol/min)

5 10 7.5
10 7 5
15 5 4
20 4 3

Determine the mode of inhibition and the inhibitor constant (Ki).

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 THE BASIS OF STATISTICS

Chapter 10

The basis of statistics

“If your experiment needs staticism then you ought to have done a better experiment”

In the perfect world if we were to measure the creatinine concentration in a serum sample
a large number of times then we would obtain exactly the same result every time. In
practice, however, the result is not always the same due to analytical imprecision. The
results obtained would vary but only by a small amount, often the same result would be
obtained more than once so that the results would tend to cluster around a particular
value. The value around which results would cluster is not necessarily the true value due
to inherent inaccuracy of the method. Similarly if serum samples were collected from the
same “normal” individual on a number of occasions and the creatinine concentration
measured in each sample then a similar cluster of results would be obtained but the
spread would be much wider due to intra-individual variation being added to the
analytical imprecision. On the other hand if samples were collected from a number of
“normal” individuals then the spread of results would be even wider due to a contribution
from inter-individual variation in addition to intra-individual variation and analytical
imprecision. An important consequence is that if two different results are obtained we
cannot be sure that the change is real since it may be explained by the expected analytical
impression and/or intra-individual variation.

The science of statistics gives us the tools to deal with this random variation due to
analytical imprecision and biological variation in order that we can extract maximum
information from data that we obtain in the clinical laboratory. Nowadays anyone can
use computers (and some pocket calculators) to do statistical calculations. However,
correct interpretation of the statistical parameters produced requires some understanding
of the underlying principles.

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The presentation and description of laboratory data

Figure 10.1a gives the results obtained when creatinine concentration was measured in
sera from 60 “normal” individuals. Simple inspection of the data shows that:

• The results are not all the same

• All the results fall within the range 35 to 122 μmol/L

• The same result often appears more than once

In most areas of statistics an extremely useful first step is to express the results in the
form of a diagram. Depending upon the number of results the data is first grouped into
class intervals of equal size. The interval used should be chosen to make sure that most
intervals contain more than one result. For the creatinine data in Fig 10.1 a class interval
spanning a concentration range of 10 μmol/L ensures that each group contains at least one
result with one of the groups containing as many as sixteen values (Fig 10.1b). If a graph
is plotted with the concentration intervals as the x axis and frequency as the y axis then
the result (Fig 10.1cc) is a frequency distribution or histogram.

Visual inspection of the distribution reveals that the class interval with the highest
number of results is the 70-79 μmol/L group and that there are approximately equal
numbers of results below this group as above it i.e. the overall shape is symmetrical. As
we move further away from this group on either side of the diagram the number of results
in each group diminishes. If we were to join up the peaks of each class interval then the
result would be a continuous bell-shaped curve which mathematicians refer to as a
normal or Gaussian distribution.

It is often useful to find a mathematical way to describe this distribution or curve.

We need to convey two things:

1. The peak value. Mathematicians call this the measure of central location.

2. The spread of results (i.e. a measure of the variability of the results) or a measure
of the width of the bell-shaped curve. Mathematicians call this the measure of
dispersion.

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 THE BASIS OF STATISTICS

a) Individual creatinine results (μmol/L)

80 62 91 81 111 75 78 45 91 72 65 50 57 85 103 75 75
63 47 93 77 122 60 76 97 35 54 106 90 80 80 74 41 78
76 115 76 52 87 75 82 64 93 71 79 100 67 68 74 59 69
63 68 65 89 72 82 68 80 83

b) Frequency of results grouped in intervals spanning 10 μmol/L

Interval: 30-39 40-49 50-59 60-69 70-79 80-89 90-99 100-109 110-119 102-129
Frequency: 1 3 5 12 16 11 6 3 2 1

c) Histogram of frequency versus concentration interval

16

14

12

10

0
30 40 50 60 70 80 90 100 110 120 130
Creatinine (μmol/L)

Figure 10.1 A set of 60 serum creatinine results obtained on healthy individuals

(a), grouped into concentration intervals (b) and plotted as a
histogram (c)

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Measures of central location

The arithmetic mean or average: This simply is the sum of all the individual results in
the series (if the same result is encountered more than once then it is counted more than
once), divided by the number of results. If we denote each individual result in the series
by the symbol x (so that the first is x1, the second x2, etc), the symbol Σ to mean the sum
of all values of the series and n as the number of results, then we can write the following
expression for the mean (m):

Mean (m) = Σ x ………………….. Eq. 10.1

n

The median: This is simply the value such that half of the data points fall above it and a
half below it. In other words if we have 100 results and arrange then in ascending order,
then the value of the fiftieth result is the median.

The mode: The mode is the most frequently occurring result or the class interval
containing the most results.

Measures of dispersion
The standard deviation (SD or s): This is a measure of the average difference of all the
values from the mean. If the mean result (m) is subtracted from an individual result (x)
then the result is the difference or deviation of that result from the mean i.e. (x – m). If
this is done for each data point (i.e. each individual result) and these deviations are added
together, then the result can be expressed mathematically as Σ (x – m). If this value is
divided by the number of results, n, then the result would be expected to be a measure of
the average difference of all the results from the mean. However, this is not the case, the
result comes out at zero. The reason for this is that the normal distribution is symmetrical
with an approximately equal number of results both below and above the mean. The
deviation of a result below the mean is negative, the deviation of a result above the mean
is positive. Therefore the positive deviations cancel the negative deviations so that their
sum is zero. To overcome this problem mathematicians square each deviation so as to
always give a positive result. A positive deviation multiplied by a positive deviation
gives a positive result, as does a negative deviation multiplied by a negative deviation.
If these are then added together then the result is the sum of squares of the deviations,
denoted by the expression:

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 THE BASIS OF STATISTICS

Σ(x – m)2. Mathematicians use this trick in many areas of statistics. If the sum of
squares is divided by the number of data points, n, then the result is a measure of
dispersion known as the variance, which is denoted by the symbol s2. If the square root of
the variance is taken (to allow for taking squares of the deviations in the first place), then
the result is the standard deviation, denoted by the symbol s or SD, a value which is more
easily related to the dispersion of results in the distribution. This simple concept is
complicated by the fact that instead of dividing by n it is customary to divide by n-1. n-1
is known as the degrees of freedom. The reason for this that when the sum of deviations
(or their squares) is calculated, use of the value for the mean restricts the freedom of the
individual values. Suppose we had six results, 1, 2, 3, 4, 5 and 6 (the numbers on a dice).
Their sum is 21 and their mean (21/6) is 3.5. The deviations (x – m) for the first five
values are –2.5, -1.5, -0.5, +0.5, +1.5 and their sum, Σ (x – m), is –2.5. Since the
deviations must add up to zero, it follows that the deviation for the sixth value must be
+2.5. Therefore the sixth value must be the sum of the mean and its deviation i.e.
3.5 + 2.5 = 6. In other words the final value in the series is fixed, cannot vary and so
does not add any useful information Therefore for practical purposes there are only 5
results contributing to the sum of squares. Another way of looking at this is if a dice is
lying with the six side face down then we do not need to turn the dice over to know what
the value on the hidden face is! In practice the larger the number of data points the less
important the difference between the number of values (n) and their degrees of freedom
(n-1); above n=30 this difference is usually ignored. The expressions for variance and
standard deviation are:

Variance ( s2 ) = Σ ( x – m )2 ………………. Eq. 10.2

n–1

Standard deviation ( s or SD ) = √s2 = Σ ( x – m )2 ….… Eq. 10.3

n-1

The units of standard deviation are the same as the units of the data used in its
calculation. In an attempt to standardise the expression of s, clinical biochemists often
use the term coefficient of variation, denoted cv. This is the standard deviation divided
by the mean; the result is usually expressed as a percentage:

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Coefficient of variation (%cv ) = s x 100 ……………Eq 10.4

m

The hope was that by expressing imprecision in this way, the same numerical result
would be obtained over the entire concentration range of the assay. However, this is
rarely the case.

The range: This is simply the difference between the highest and the lowest value in the
set of data. This is the least reliable measure of dispersion.

The interquartile range: The data are arranged in ascending order and grouped into
four equal sets (known as quartiles). The middle two sets (comprising the middle fifty
per cent of the data points) form the interquartile range.

Is the data “normally distributed”?

Often the histogram of a set of data is not a typical bell-shaped Guassian distribution.
Fig 10.2 shows two ways in which the curve may deviate from normality. In Fig 10.2a,
the two curves are not symmetrical but skewed. Statistical packages often calculate a
parameter called the skew. A skew of 0 indicates no skew, a positive value indicates
skew to the right (curve A) and a negative value skew to the left (curve B). A skewed
distribution can often be converted to a reasonably “normal” distribution by taking
logarithms of the data e.g. the distribution of serum bilirubin concentrations in normal
adults is normally skewed to the right whereas the distribution of the logarithm of
concentration becomes relatively normal.

In Fig 10.2b the curves differ in how “peaked” or “flat” they are; this is known as
kurtosis. Again statistical packages often calculate a value for kurtosis. A truly Guassian
curve has a kurtosis of 3 (some computer programs convert this value to zero).

Formal tests of normality include:

• The Anderson Darling test

• The Kolmogorov Smirnov test
• The Shapiro Wilks W test
• The Shapiro Francia W test

A discussion of these tests is beyond the scope of this book but may be found in standard
statistical texts.

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 THE BASIS OF STATISTICS

a) Skewness b) Kurtosis

A B
B

A
C

Figure 10.2 Deviations from true Guassian distributions a) Curve A

is skewed to the right (positive skewness), curve B is skewed to
the left (negative skewness); b) Curve B is “normal” or
mesokurtic, curve A is more “peaked” or leptokurtic, curve C is
flatter than normal or platykurtic

Question Q 10(1)

A laboratory had just changed its method for serum creatinine. To check that there had
been no change in their reference range they analysed serum samples collected from 12
members of staff and obtained the following results (arranged in ascending order): 44,
58, 60, 68, 70, 75, 76, 78, 80, 90, 95, 106 μmol/L. Calculate the mean, variance,
standard deviation and coefficient of variation.

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Answer Q 10(1)

Construct a table with the individual creatinine concentrations (x) in the first column:

x (x – m) (x – m)2 x2

44 -31 961 1936

58 -17 289 3364
60 -15 225 3600
68 -7 49 4624
70 -5 25 4900
75 0 0 5625
76 1 1 5776
78 3 9 6084
80 5 25 6400
90 15 225 8100
95 20 400 9025
106 31 961 11236

Total: 900 0 3170 70670

Add these together to give their sum (Σx). The number of results (n) is 12.
From Σx and n, the mean (m) can be calculated:

Mean (m) = Σx = 900 = 75 μmol/L

n 12

Next subtract the mean (m) from each individual value of x so as to give a column for
individual deviations (x – m). Note that the sum of all these, Σ(x –m), is zero and cannot
be used in the calculation of variance. Instead calculate the square of these deviations
i.e. (x – m)2 and enter in the third column. These are then added together to give the
sum of squares of the individual deviations, Σ(x – m)2, referred to by mathematicians as
simply the sum of squares. This value can then be used to calculate the variance:

Variance ( s2 ) = Σ ( x – m )2 = 3170 = 3170 = 288 μmol/L

n–1 (12 – 1 ) 11

The standard deviation is simply the square root of the variance:

Standard deviation (s) = √ s2 = √ 288 = 17 μmol/L

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The coefficient of variation is the standard deviation expressed as a percentage of the

mean:

Coefficient of variation (cv) = s x 100 = 17 x 100 = 23% (2 sig figs)

m 75

Before the advent of computers or sophisticated pocket calculators it was often easier to
calculate the sum of the squares of individual values (Σx2), then calculate the sum of
squares of the deviations using the identity:

Σ(x – m)2 = Σx2 - (Σx)2 or Σx2 - n m2

n

i.e. Σ(x – m)2 = 70670 - 9002 = 70670 - 67500 = 3170

12

or Σ(x – m)2 = 70670 - (12 x 752) = 70670 - 67500 = 3170

Does a single result belong to the population?

This a question we often try to answer in clinical biochemistry. The creatinine data
quoted in Fig 10.1 and Question Q 10(1) were obtained from normal individuals. We
may want to know if the creatinine result obtained from a patient is “abnormal” i.e. is it
“different” from the reference population? There is no perfect way to answer this
question. Statisticians try and deal with this problem by reformulating the question as
“suppose this result does belong to this normal population what is the likelihood that this
result could have been obtained by pure chance?” If it is improbable that it arose by
chance then it is probably “significantly “ different (in this case abnormal). This begs
the question as to how unlikely the event has to be for the result to be considered different
or belonging to a different population? By convention a probability of less than 1 in 20
(i.e. 0.05 or 5%) is taken as a cut-off indicating that the likelihood of the result not being
abnormal as so low as to be negligible. This value was suggested by the eminent
mathematician Fischer and is not based on any theoretical consideration. Statistics can
only answer the question “how likely it is that an event has occurred by chance”, whether
the difference matters is a subjective one!

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But how is this probability obtained from the individual result and the data in the
reference population? The reference population when plotted (see Fig 10.1) shows a bell-
shaped curve and its characteristics are partially determined by the peak value (the mean,
m) and the width (standard deviation, s). Mathematicians have shown that the curve can
be described mathematically by the exponential equation:

– (x – m)2
2 s2
y = 1 x e ……….……………… Eq. 10.5
s√2 π

In other words the value of y is a complicated function of both m and s (some forms of
this equation use μ and σ instead of m and s but the subtle difference need not concern us
here). Rest assured that you will never need to manipulate this equation. However, what
we need to know is not the value of y but the probability of obtaining any particular value
of x. This probability is given by the area under the curve if a perpendicular line is drawn
at point x. Since all results for all the population must fall somewhere within the curve
(probability = 1) the total area must equal 1. This area can be calculated by a complex
mathematical function obtained by integrating equation Eq. 10.5. Thus from the values
of x, m and s it is possible to calculate the probability of obtaining a value x; if it is less
than 0.05 then the result is significantly different. It is obviously inconvenient to have to
perform such a complex calculation every time. To get around this difficulty
mathematicians always reduce their data to a “normalized population” in which the mean
is always zero and the standard deviation one. As a result the calculation need only be
done once and is used to generate a statistical table in which the probability of obtaining
any value of x can be easily obtained. The mean is subtracted from the value then
divided by the standard deviation to give a standard score, z:

z = x - m ………………………………. Eq. 10.6

s

z (sometimes called the normal deviate, d, or the standard deviation index, SDI) is
therefore the number of standard deviations the value of x is away from the mean. z is
always normally distributed with a mean of zero and standard deviation of one. Such a
normalized Guassian distribution (in which z is the horizontal axis) is shown in Fig 10.3.
The curve has a mean of zero and a standard deviation of one. Since all results of the
population must fall somewhere in the area between the curve and the z axis, the total
area under the curve is the probability (P) of obtaining any value and is one.

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67%

95%

-2s -s m s 2s
z -2 -1 0 1 2
P 0.025 0.34 0.50 0.84 0.975

Figure 10.3 Normalised Guassian distribution in which the mean is zero and
standard deviation 1. Values of P show the probability of obtaining a
value to the left of the z value i.e. area under the curve to the left of a
perpendicular line drawn at the value for z

The probability of obtaining a result between any two given values is the area under the
curve between the two values. For example, the probability of obtaining a value between
the mean minus one s and the mean plus one s is two-thirds or 0.67 (equal to 67%). The
probability of obtaining a value between the mean minus 2s and the mean plus 2s is 19/20
or 0.95 (equal to 95%). Strictly speaking 1.96s should be used rather than 2s. It follows
that the probability of obtaining a result outside the mean ± 2s (or more correctly mean
± 1.96s) is 1 – 0.95 = 0.05 (or 5%). This is exactly the level of probability which
statisticians regard as significant when deciding whether a particular value belongs to a
given population.

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Since 5% of values fall within the mean ±1.96s and the normal curve is symmetrical it
follows that 2.5% or results will be below the mean – 1.96s and a further 2.5% will be
above the mean + 1.96s. This is quite an important point since in some situations we
only wish to know if a result is significantly greater than the mean + 1.96s or is
significantly less than the mean – 1.96s. In this case we use the value of z which
excludes 10% of results (P=0.1) since only a half of these (5%) will be greater (or less
than) the range encompassing 90% of the values. The value of z which gives rise to
range which excludes the lowest and highest 5% of results is 1.645. A table of z values
with their corresponding probabilities is given in Fig 10.4.

P 0.33 0.10 0.05 0.02 0.01 0.002 0.001

z 1.0 1.645 1.96 2.326 2.576 3.090 3.291

Figure 10.4 Percentage points of the normal distribution. This table gives the
percentage points most frequently required for significance tests for a
normal variable having zero mean and unit standard deviation. Thus,
the probability of obtaining a departure from the mean of more than
1.96 standard deviations in either direction is 0.05 or 5%

The range of values obtained at any probability level is known as the confidence limits.
The 95% confidence limits of a set of results can be calculated from the corresponding z,
the mean (m) and standard deviation (s) as follows:

Upper limit = m + (z x s)

Lower limit = m - (z x s)

In general the 95% confidence limits (where z = 1.96) are:

Mean - ( 1.96 x s ) to mean + (1.96 x s) ……….. Eq. 10.7

It is common practice in clinical biochemistry to use the 95% confidence limits obtained
for the concentration of an analyte from normal subjects as the reference range. Any
value obtained for a patient outside this range is usually regarded as abnormal. The truth

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is that the probability of obtaining a result outside of this range is 0.05 or 1 in 20 – we are
simply making a subjective judgment that it is abnormal. If we were to measure this
analyte in twenty healthy individuals then we would expect one of them to have a result
outside of the reference range. Similarly, if we were to measure twenty different analytes
in the same patient then again one of the results would most likely fall outside of the
reference range.

Calculation from the creatinine data for 60 normal individuals depicted in Fig 10.1 gives
a mean of 76 μmol/L and standard deviation of 17 μmol/L. From these figures it is
possible to calculate the 95 % confidence limits (when z = 1.96) as follows:

95% confidence limits = mean ± (1.96 x s)

= 76 ± (1.96 x 17)
= 76 ± 33 (to 2 sig figs)
= (76 – 33) to (76 + 33)
= 43 to 109 μmol/L

The chance of obtaining a result outside of these limits is 100 - 95 = 5% (1 in 20) and
if it occurs is probably abnormal. The chance of obtaining a result less than 43 μmol/L is
2.5% (1 in 40), and of obtaining a result greater than 109 μmol/L is also 2.5% (1 in 40). If
we wish to know the chance of obtaining a result greater than 115 μmol/L then the first
thing to do it to calculate the z score:

z = x - m = 115 - 60 = 55 = 3.24
s 17 17

Inspection of the table in Fig 10.4 shows that there is no value of z corresponding exactly
to z = 3.24, but that when z = 3.09 the probability (P value) is 0.002. Therefore the
chance of obtaining a value outside the mean ± 3.24s is slightly less than 0.002.
The chance of obtaining a result greater then mean + 3.24s is one half of this i.e. 0.001
(or 1 in 1000).

Question Q 10(2)

The imprecision of a certain assay for Troponin I yields a coefficient of variation of 13%
between 0.3 and 0.5 µg/L, around the decision point for myocardial infarction of
0.4 µg/L. A result of 0.46 µg/L is obtained on a sample. Assuming that is the true level
of Troponin I, give an estimate of the probability that analysis of that same sample would
give a result below the decision point.

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Answer Q10(2)

The true result (0.46 μg/L) can be considered as the mean with a coefficient of variation
(cv) of 13%. The first step it to calculate the standard deviation (s). cv, mean (m) and s
are related as shown in Eq 10.4:

cv (%) = s x 100
m

Which can be rearranged to:

s = cv (%) x m
100

Substitute for cv and m and solve for s:

s = 13 x 0.46 = 0.060 μg/L

100

Therefore the analyses of the sample are distributed with a mean of 0.46 µg/L and s
of 0.06 µg/L. We want find out what proportion of results will be below the decision
point of 0.4 µg/L. To do this we need to 'normalize' the data so that the mean is zero and
the SD =1. i.e. calculate the standard deviate -'z':

z = decision point - m = 0.4 - 0.46 = -0.06 = -1

s 0.06 0.06

Therefore the decision point is –1s from the mean. From the table in Fig 10.4 we can see
that the probability of obtaining a result which differs from the mean by more than 1s in
either direction (i.e. when z = 1 or –1) is 0.33.

Therefore the probability of obtaining a result below the decision point is one half of 0.33
i.e. 0.17 (2 sig figs).

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Utilization of the Normal Distribution in Quality Control

Another application of the normal distribution is the analysis of quality control data. It is
common practice to include quality control samples into an analytical run to check that
the method is performing to specification. The characteristics of the quality control (QC)
material are first determined by replicate analyses of the material (usually twenty) then
calculating the mean and standard deviation. The same material is then analyzed in each
batch of samples (sometimes more than once). If the method is performing to
specification then the results for the QC material should fall within the 95% confidence
limits (mean ± 2s) most of the time. In fact the distribution of results should belong to
the same normal distribution as when the characteristics were originally determined i.e.
the results should cluster around the mean.

A useful way of plotting the data is to turn the histogram on it’s side and construct a y
axis with horizontal lines representing the mean, 1s, -1s, 2s etc. The results are plotted
along the x axis and should fall on either side of the mean with equal frequency, most
results will fall within the mean ±s, fewer results between the s and 2s limits, very few
between 2s and 3s with only very occasional results outside the 3s limits. This is known
as a Levy-Jennings Chart and an example is shown in Fig 10.4.

The only limitation of using 95% confidence limits (i.e. mean ± 2s) as the only criterion
is that by definition 1 in 20 analytical runs will be rejected. For a multi-channel analyzer
measuring 20 analytes this means that on average one channel will be rejected every run.
In other words this criterion is too sensitive. Westgard has devised a set of criteria, the
Westgard Rules, to improve the power of QC data to detect “real” errors without an
unacceptably high lever of “false rejections”. These rules are based upon the fact that the
probability of obtaining two consecutive results outside the 95% confidence limits is the
product of the individual probabilities that one result is outside these limits and is
considerably lower and so increases the likelihood that the method is out of control. This
idea is extended to four results being between the s and 2s limits and ten results being one
side of the mean etc.

Question Q 10(3)

Calculate the probability of obtaining:

a) a QC result outside the mean ± 3s range; b) two QC results between the mean 2s and
mean 3s limits, and c) four results between the mean – s and mean – 2s limits.

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3s
2s x x
s x x x x x
m x x x x x x xx x x
x x x x x
-s
x
-2s
-3s x

Figure 10.4 Levy-Jennings Quality Control Chart showing 24 sequential results

for the same control sample

Answer Q 10(3)

a) The probability of obtaining a result outside the mean ± 3s range is the P value
corresponding to a z score of 3. An exact value for z = 3 is not given in Fig 10.4
but the nearest value (z = 3.09) can be used as an approximation and corresponds
to a P value of 0.002. The chance of obtaining a result outside the mean ± 3s
range is therefore 1 in 1/0.002 which is 1 in 500. This value will occur so
infrequently that the method is almost certainly out of control, and is known as the
Westgard 13s rule.

a) The probability of obtaining a result outside the mean ± 2s range can again be
obtained by looking up the P value corresponding to z = 2 in Fig 10.4. Again the
exact value for z = 2 is not given but the nearest (z = 1.96) is a good
approximation and corresponds to a P value of 0.05. There is a slight
complication here in that we really wish to know the P value when z is between
±2 and ±3, not simply when it is greater than ±2. To allow for this all we need to
do is to subtract the P value for z = 3 (P3) from the P value for z = 2 (P2) to give
the probability of obtaining a value between 2 and 3 standard deviations from the
mean (P2-3):

P2-3 = P2 - P3 = 0.05 - 0.002 = 0.048

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Furthermore, we need the probability that the result is between the mean +2s and
mean +3s, not simply between the mean ± 2s and mean ± 3s limits on either side
of the mean. Therefore the probability of obtaining a result between the mean
+ 2s and mean +3s limits is one half of 0.048 i.e. 0.024. The likelihood of
obtaining two results between the mean + 2s and mean + 3s limits is the product
of the probability of obtaining each result between these limits:

0.024 x 0.024 = 0.000576

which is the same as 1 in 1/0.000576 = 1 in 1736

This is similar to flipping a coin twice. The chance of heads the first time is 0.5
the second time 0.5. Therefore the chance of obtaining heads on both occasions is
0.5 x 0.5 = 0.25. Another way of looking at this is that there are four possible
results, heads and tails, heads and heads, tails and heads and tails and tails: there
are 4 equally likely results but only one of these is heads on both tosses, so that
the probability is 1 in 4 or 0.25).

b) As shown above, the probability (P2s) of obtaining a result outside the mean ± 2s
limits is 0.05. From Fig 10.4 it can be seen that the chance (Ps) of obtaining a
result outside the mean ± s range is 0.33. Therefore the probability (P2s-s) of
obtaining a result between the ± s and ± 2s limits is given by:

P2s-s = Ps - P2s = 0.33 - 0.05 = 0.28

The probability of obtaining a result between the mean –s and mean –2s is one
half of this i.e. 0.14. The likelihood of obtaining four results within this range is
obtained by multiplying this probability by itself four times:

0.14 x 0.14 x 0.14 x 0.14 = 0.00038 (2 sig figs)

This calculation can be simplified as follows:

0.14 x 0.14 x 0.14 x 0.14 = 0.144 = antilog10 (4 x log10 0.14)

= antilog10 ( 4 x (-0.854)) = antilog10 (-3.42) = 0.00038

A probability of 0.00038 is a chance of 1 in 1/0.00038 = 1 in 2632

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Variances are additive

One of the reasons why statisticians prefer to work with squares of deviations rather than
directly with deviations is that the former are additive. This is particularly true of the
variance i.e. the standard deviation squared (s2) which is calculated directly from the sum
of squares of the deviations about the mean (Σ(x – m)2). If there are two independent
sources of variation (with variances s12 and s22) contributing to a measurement, then the
total variation (with variance stotal2) is described by:

stotal2 = s1 2 + s22 ………………………… Eq. 10.8

It is important to note that it is the variances only (s2) which are additive, not the standard
deviations (s). If we wish to calculate the combined standard deviation from the
individual standard deviations then these are first squared, added together then the square
root taken of the product:

stotal = √ (s12 + s22) …………………… Eq. 10.9

Similarly coefficients of variation (cvs) are not additive unless they are first squared:

cvtotal = √ (cv12 + cv22) ………………. Eq. 10.10

One of the most useful applications of this concept in clinical biochemistry is in

examining the effects of population variation and analytical impression on laboratory
data. If we were to measure the concentration of an analyte, say creatinine, in sera
collected from a large number of individuals then it would be possible to calculate the
mean, variance etc for the population (stotal2). However, if we were to take one of these
samples and analyze it a large number of times, then for that sample we derive both a
mean and a variance due to the analytical imprecision of the method (sanalytical2).

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Clearly the analytical imprecision is contributing to the overall variation in the patient
results so that the true population variation, that is the biological variation (sbiological2) is
much lower. Since the component variances are additive, and assuming that the
analytical variance is the same at all analyte concentrations encountered in the
population, we can write:

stotal = √ (sbiological2 + sanalytical2) ………….. Eq. 10.11

The biological variation can be calculated by rearranging this expression:

sbiological = √ (stotal2 - sanalytical2) …………….. Eq. 10.12

The biological variation can be further subdivided into its components e.g.
intra-individual and inter-individual variations and equation Eq. 10.11 re-written:

stotal = √ (sintra-individual2 + sinter-individual + sanalytical2) ……… Eq. 10.13

Another application of this principle is the analysis of the impression of the steps in an
analytical process on the total performance. For example, if a method involved pipetting
7 mL of reagent, then this could be achieved by either pipetting 5 mL and 2 mL from
separate bulb pipettes or by pipetting 7 mL from a graduated pipette. The combined error
from using separate bulb pipettes could be calculated from stotal = √ (s5mL2 + s2mL2) and
compared with the variance obtained from using the 7 mL graduated pipette.

Question Q 10(4)

A laboratory using a method with an analytical coefficient of variation of 5% at a

concentration of 100 mmol/L for a serum constituent examined samples from a healthy
population and found a Gaussian distribution with 95% reference range of
74-126 mmol/L. If the method coefficient of variation had been 22%, what reference
range would the laboratory have found?

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Answer Q10(4)

The first step is to calculate the overall standard deviation (sTotal) and mean (m) from the
reference range obtained with the original method (with analytical cv of 5%).

The 95% reference limits incorporates the mean ± 2s, i.e. spans 4s units

Therefore sTotal = Range of 95% limits = 126 - 74 = 52 = 13 mmol/L

4 4 4

The mean is given by the lower reference limit + 2s.

Therefore, mean (m) = 74 + (2 x 13) = 74 + 26 = 100 mmol/L

Convert the original analytical cv (5%) to its standard deviation (sAnalytical):

Coefficient of variation (cv%) = s x 100

m

Rearranging, s = cv (%) x m
100

Therefore analytical s (sAnalytical) = 5 x 100 = 5 mmol/L

100

The measured variation will reflect both the biological and analytical variations. Since it
is variances and not standard deviations which are additive, then the square of the total
standard deviation (sTotal) is equal to the sum of the squares of both the biological
(sBiological) and analytical (sAnalytical) standard deviations.:

(sTotal) 2 = (sBiological) 2 + (sAnalytical) 2

Substitute sTotal = 13 mmol/L and sAnalytical = 5 and solve for sBiological

132 = (sBiological) 2 + 52

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 THE BASIS OF STATISTICS

(sBiological) 2 = 132 - 52

= 169 - 25 = 144

sBiological = √ 144 = 12 mmol/L

If new analytical cv is 22%, then sAnalytical = 22 x 100 = 22 mmol/L

100

and the new sTotal can be calculated (assuming biological variation remains unchanged):

sTotal 2 = 122 + 222 = 144 + 484 = 628

sTotal = √ 628 = 25 mmol/L (to 2 sig figs)

Lower limit of reference range = m - 2sTotal = 100 - (2 x 25)

= 100 - 50 = 50 mmol/L

Upper limit of reference range = m + 2sTotal = 100 + (2 x 25)

= 100 + 50 = 150 mmol/L

New reference range = 50 - 150 mmol/L

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When is a change in a test result significant?

When monitoring patients it is helpful to know by how much the concentration of any
given analyte has to change before the change is clinically significant. To answer this
question allowance has to made for the effects of both analytical and within-individual
variation. One way to address this problem would be to analyse each sample a number of
times and compare their means using a suitable statistical test. In day-to-day practice we
do not have this luxury, only single measurements. However, all laboratories should
have some idea of the total variability for each of their analytes (which includes both
analytical and within-individual variation).

Suppose the result for an analysis is x1 on the first occasion and x2 on the second
occasion, then we could calculate the difference (x1 – x2). We could treat the value for (x1
– x2) as a variable which forms a normal Guassian distribution. In other words, if the
analysis of the two samples was repeated a large number of times then a histogram could
be constructed with frequency plotted against (x1 – x2). If the two results (x1 and x2 are
not significantly different) then their difference (x1 – x2) should be within the 95%
confidence limits of a distribution with mean of zero and their combined standard
deviation (s1,2). In other words the difference in results (x1 – x2) can be normalized to give
a value for z if it is divided by s12:

z = (x1 – x2) - 0
s1,2

The distribution would have a mean of zero with a standard deviation of 1. If there was
no significant difference between the results obtained on the two occasions then the peak
of the histogram (i.e. the mean value for (x1 – x2) would be zero). For a value to be
significantly different from the mean (in this case zero) at a probability level of 5% (i.e. P
= 0.05) the value for z would need to be 1.96. Therefore for the difference (x1 – x2) to be
significantly different from zero we substitute z = 1.96 into the above expression:

1.96 = (x1 – x2) …………………………… Eq. 10.14

s1,2

The value for s is actually the combined standard deviations for the two measurements (s1
and s2). As seen in the previous section, the combined value for s when two results are
combined (i.e. added or subtracted) is the square root of the sum of their squares:

s1,2 = √ (s12 + s22)

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However, if the value for s is the same at both concentrations then this expression
simplifies to:

s1,2 = √ (2s2) = √2 x s = 1.414 s

Substituting s1,2 = 1.414 s into Eq 10.14 gives:

1.96 = (x1 – x2)

1.414 s

Which can be rearranged and simplified to:

(x1 – x2) = 2.8 s …………………………… Eq. 10.15

Therefore for a change in a result to be significant at the 5% level of probability the two
results must differ by at least 2.8s. It is important that the value for s is the same at both
concentrations.

Question Q 10(5)

While trying to follow the National Service Framework guidelines for coronary heart
disease a doctor prescribed a statin to lower the cholesterol of a patient with coronary
heart disease. The patient's original cholesterol level was 5.8 mmol/L and at the next
visit the doctor was delighted to find that it was just below the target level of 5.0 mmol/L
at 4.9 mmol/L and discharged the patient. The patient, a statistician, was less sure the
treatment had been responsible. Given that the physiological coefficient of variation for
cholesterol is 6% and the analytical coefficient of variation is 3%, calculate the least
significant change (at p<0.05) in cholesterol as a percentage at his original level, and
determine whether the second measurement was significantly different from the first.

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Answer Q 10(5)

The total CV is the square root of the sum of the squares of the physiological and
analytical CVs:

Total CV(%) = √ Analytical %CV 2 + Physiological %CV 2

= √ (32 + 62 ) = √ (9 + 36 ) = √ 45 = 6.7%

Next calculate the standard deviation (s):

CV(%) = s x 100 therefore s = CV(%) x mean

Mean 100

Substitute CV = 6.7% and the original level (5.8 mmol/L) as the mean:

s = 6.7 x 5.8 = 0.389 mmol/L

100

For two results to be significantly different (at p <0.05) they have to be at least
2.8 standard deviations apart (2.8s).

Therefore the least significant change is 2.8 x 0.389 = 1.09 mmol/L

Which expressed as a percentage of the original measurement is 1.09 x 100 = 18.8%

5.8

Next calculate the difference between the first and second measurement as a percentage
of the first measurement:

(5.8 - 4.9) x 100 = 15.5%

5.8

which is less than 18.8% so that the change in cholesterol is not statistically significant.

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ADDITIONAL QUESTIONS

1. The following results were obtained for a QC sample:

Total protein (g/L): 70, 68, 71, 65, 68, 70, 73, 69, 75, 74, 69, 71

Calculate the mean, variance, standard deviation, coefficient of variation and

95 per cent confidence limits.

2. Serum thyroxine was measured in 10,000 healthy male adults. Assuming a

Gaussian distribution the normal range was calculated to be 50-150 nmol/L.
How many results are expected to be above 165 nmol/L?

3. Calculate the least significant difference for a change in cholesterol if the intra-
individual coefficient of variation for cholesterol is 4.7% and the analytical
coefficient of variation, 2.4%. A patient was changed from Atorvastatin 80 mg to
Rosuvastatin 40 mg and the total cholesterol fell from 6.9 to 5.9 mmol/L.
Calculate the percentage change in cholesterol and state whether this is
significant.

4. Your on-call laboratory service uses 30 different methods, each of which has a
1% probability of failing QC criteria during the course of a night. Assuming that
QC of any method is independent of that of the other methods, what is the
probability that on any one night all methods will pass the QC criteria?

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5. You attempt to derive a reference range for TSH for an ethnic minority
population. The first 10 samples give the following results:

Result n
Between 0.5 and 1.49 5
Between 1.5 and 2.49 3
Between 2.5 and 3.49 0
Between 3.5 and 4.49 1
Between 4.5 and 5.49 1

On the basis of these results, what range of TSH values would encompass 95% of
the ethnic minority population?

6. You are required to pipette a 9ml volume and have available a 10 ml graduated
pipette which has a 2% CV associated with it’s delivery volume and 5 and 2 ml
volumetric pipettes each of which has a 1%CV associated with their delivery
volumes. What is the error of pipetting a 9 mL volume, expressed as plus/minus
mL volume?

a) using the graduated pipette

b) using the volumetric pipettes

7. It has been suggested that a proposed analytical goal for an analyte is that the
between batch analytical coefficient of variation should not exceed one half of
the “true biological” inter-individual coefficient of variation. Calculate the
percentage “expansion” of the measured reference range over the true biological
reference range when this analytical goal is exactly met.

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Chapter 11

Analysis of means and variances

Comparison of means (the z- and t-test)

Sometimes we do not want to ask the question “is a single result significantly different
from a given population” but “is the mean of a set of results significantly different from
the mean of another set”. The two problems are approached in a similar manner but
difficulties arise because the value of the mean is influenced by the number of results
used in its calculation.

a) The standard error

Consider the data in Fig 10.1 for creatinine results obtained with sera from sixty normal
individuals (mean = 76 μmol/L; standard deviation = 17 μmol/L). Suppose we were to
take two of these results (n=2) at random and calculate their mean, then repeat this
process a large number of times. The results for the means could then be plotted in the
form of a histogram similar to the individual results in Fig 10.1. Such a histogram is
called the sampling distribution of the mean. The peak value (mean) would be the same
but there would be one important difference: the distribution of results would be much
narrower and the value of its standard deviation would be lower. The standard deviation
of the sampling distribution of the mean is called the standard error of the mean (SEm) to
distinguish it from the standard deviation of individual results. This process could be
repeated by taking three results (n=3) at random, repeating this process a large number of
times and plotting the sampling distribution of the mean in a similar manner. The mean
of this new distribution would be unchanged but the distribution of results would be less
and their standard deviation (i.e. their standard error) would be lower then when the
means of two results were calculated. This process could be repeated with increasing
sample size (n) and we would find that as n increases the standard error is reduced.
An example is shown in Fig 11.1.

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n = 10

n=5

n=2

n=1

Figure 11.1 Effect of sample size (n) on the sampling distribution of the mean

Mathematicians have calculated that the standard error (SEm) is in fact equal to the
standard deviation (s) divided by the square root of the number of results (n) used in its
calculation:

SEm = s or s2 ……………………Eq. 11.1

√n n

Provided the data follow a Guassian distribution the values for the mean (values of m) of
samples of size n will be distributed with the peak of the bell shaped curve having the
overall true mean (which we shall call μ ) and a standard deviation equal to the standard
error of the mean (SEm).

Just as with single measurements, the data can be “normalized” to produce an overall
mean of zero and a standard error of one by calculating the z value:

z (or t) = m - μ ………….. Eq. 11.2

SEm

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 ANALYSIS OF MEANS AND VARIANCES

b) Comparing the mean of a sample with a known standard

In order to test whether the mean (m) of a sample size n is significantly different from a
hypothetical mean μ of a Guassian distribution first calculate the z (or t) value:

z (or t) = m - μ = m - μ …………..Eq 11.3

SEm s/√n

then look up the corresponding probability (P value) in tables of z (see Fig 10.4). It is
important to note that this approach is only valid if the standard error calculated from the
sample (i.e. s/√n) is a reasonable estimate of the true standard error of the mean. This in
unlikely to be the case unless the value for n is relatively large (greater than 30). William
Gossett, who published under the pen-name “Student” noted that in small samples, the
sample s underestimates the population s. To get around this problem the “t-distribution”
was introduced, which is similar to a normal z-distribution in being symmetrical about a
mean of zero and is bell-shaped, but differs in that it is flatter (more dispersed) and its
dispersion varies according to the size of the sample. The larger the value of n, the more
closely a t-distribution resembles a z-distribution. Fortunately statisticians have
calculated tables of t for us (portion shown in Fig 11.2) and all that is required is to read
off the P-value for the corresponding value of both t and the degrees of freedom (equal to
n – 1).

Degrees Value of P
of
freedom 0.10 0.05 0.02 0.01 0.002 0.001
…. ……. …… …… ……. …….. …….
…. …… …… ……. ……. …….. ……..
6 1.943 2.447 3.143 3.707 5.208 5.959
7 1.895 2.365 2.998 3.499 4.785 5.408
8 1.860 2.306 2.896 3.355 4.501 5.041
9 1.833 2.262 2.821 3.250 4.297 4.781
10 1.812 2.228 2.764 3.169 4.144 4.587
….. …… ……. …… ……. ……. ……
30 1.697 2.042 2.457 2.750 3.385 3.646

Figure 11.2 Portion of a table of “Student’s” t-distribution. e.g. to find the

probability (P) of obtaining a value for t greater than 2.9 for a sample
of size 10 (n=10), look across the row for 9 degrees of freedom (n-1) to
find a value for P. In this case the next lowest value for t (2.821) occurs
when P is 0.02

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b) Comparing the means of two samples

More often we wish to compare two groups of samples (samples 1 and 2), each of which
has its own mean (m1 and m2), standard deviation (s1 and s2) and sample size (n1 and n2).
Thus the mean of each group of samples has its own standard error (s1/√n1 and s2/√n2). In
order to calculate a z (or t) value we need to know the combined standard error (SEm1,2)
i.e. the standard error of the difference between estimates of the two means (m1-m2). This
combined standard error is calculated in the same was as we calculate the combined
standard deviation of biological and analytical variation. Like standard deviations,
standard errors are not additive but their squares are (and since we are dealing with
squares the signs are always positive):

(SEm1,2)2 = SEm12 + SEm22 = (s1/√n1)2 + (s2/√n2)2

= s1 2 + S22 = s1 2 + s2 2
(√n1)2 (√ n2)2 n1 n2

Taking square roots gives:

SEm1,2 = s1 2 + s2 2 …………………… Eq. 11.4

n1 n2

Therefore dividing the difference between the means by their combined standard error
gives the corresponding z (or t value):

z (or t) = m1 - m2 …………………………….. Eq .11.5

s1 2 + s22
n1 n2

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 ANALYSIS OF MEANS AND VARIANCES

If the values for n1 and n2 are greater than 30, then the value for P is obtained from tables
of z. If however, the value of n1 and/or n2 is less than 30 then the value for P is obtained
from tables of t and the degrees of freedom (DF) calculated from the expression:

DF = (s12/n1 + s22/n2)2 ………. Eq. 11.6

[(s12/n1)2/(n1 – 1)] + [(s22/n2)2/(n2 – 1)]

For the special case where s1 and s2 are equal, simplified versions of these formulae can
be used. However, it is first necessary to carry out a variance ratio test to see if this is
indeed the case so it is probably simplest to stick with only one formula that can be used
whatever the relative magnitudes of the individual variances.

Question Q 11(1)

In January a laboratory analysed a quality control sample for sodium 10 times and
obtained a mean result of 150 mmol/L with a standard deviation of 4 mmol/L.
In February the same sample was analysed 10 times and gave a mean of 154 mmol/L
with a standard deviation of 2 mmol/L. Has there been a significant change in
performance between January and February?

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Answer Q 11(1)

t is calculated using Eq 11.5:

t = m1 - m2

s1 2 + s22
n1 n2

where: m1 = 150 mmol/L m2 = 154 mmol/L

s1 = 4 mmol/L s2 = 2 mmol/L
n1 = 10 n2 = 10

Therefore t = 150 - 154

2
√ [ ( 4 )/10 + ( 22 )/10 ]

= -4 = -4 = -4 = -2.84
√ ( 1.6 + 0.4 ) √2 1.41

Since n1 and n2 are small (less than 30), the degrees of freedom (DF) is calculated using
Eq. 10.21:

DF = (s12/n1 + s22/n2)2
[(s12/n1)2/(n1 – 1)] + [(s22/n2)2/(n2 – 1)]

Therefore DF = ( 42/10 + 22/10 ) = 13 (2 sig figs)

[(42/10)2/(10-1) + (22/10)2/(10-1)]

From tables of t it can be seen that for 13 degrees of freedom, at the 5% level of
probability t should be outside the limits -2.16 to +2.16 to be statistically significant. In
fact a t value of -2.84 is also significant at the 2% level of probability. Therefore we can
conclude that there has been a change in performance between January and February.

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 ANALYSIS OF MEANS AND VARIANCES

Dealing with paired data (the paired t-test)

Sometimes individual measurements are made on a series of samples in order to

determine if the two measurements differ e.g. when comparing two analytical methods a
series of specimens with different values may be analysed by two different methods.
Although the two sets of data could be compared using a standard t-test this would be a
very inefficient way of determining if the two methods of measurement yield different
results. This is because the variation in results due to the specimens having different
analyte concentrations far outweighs the smaller difference in the two sets of results due
to the methods used. A better way to answer this question would be to carry out a
paired t-test.

For each pair of results, one value is subtracted from the other to give the difference (d).
For n pairs of results there will therefore be n values for d. We can calculate the mean
(md) and standard deviation (sd) for these values of d. If there is no difference between
the two sets of results then the average value of d would be zero. To test whether there is
a statistically significant difference of md from zero we calculate t in which we are
comparing the value for md with zero assuming that values for md are normally
distributed with a standard error of sd/√n:

t = md ………………………………. Eq 11.7
sd /√n

Where, md = ∑ d /n and, sd = √ [ ∑ (d - md)2/ (n– ) ]

Question Q 11(2)

It is suspected that the glucose results obtained with near patient testing (NPT) device on
the ward are positively biased. One of the investigations into the problem involves
analyzing a series of blood specimens on both the NPT device (A) and an analyzer in the
laboratory which measures whole blood glucose (B), with the following results:

Specimen No 1 2 3 4 5 6 7 8 9 10
Glucose (mmol/L) NPT (A) 4.5 6.8 3.2 5.8 8.9 9.5 4.8 7.3 5.1 7.8
Glucose (mmol/L) lab (B) 4.2 7.0 2.8 5.6 8.7 9.7 4.9 6.8 4.6 7.7

Do these results support the suspicion of bias?

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Answer Q 11(2)

The variabilities of the results in groups A and B are due to differing glucose
concentrations in the specimens and to the analytical variation between the instruments.
Therefore a standard t-test comparing the means of both sets of results would be
inappropriate for comparing the analytical performance of method B with method A. As
the data are paired, i.e. the same samples were assayed by both instruments, a paired t-test
can be used.

Calculate the difference (d) between each pair of results: d = A - B

A B d d2 d - md (d - md)2

4.5 4.2 0.3 0.09 0.13 0.017

6.8 7.0 -0.2 0.04 -0.37 0.137
3.2 2.8 0.4 0.16 0.23 0.053
5.8 5.6` ` 0.2 0.04 0.03 0.001
8.9 8.7 0.2 0.04 0.03 0.001
9.5 9.7 -0.2 0.04 -0.37 0.137
4.8 4.9 -0.1 0.01 -0.27 0.073
7.3 6.8 0.5 0.25 0.33 0.109
5.1 4.6 0.5 0.25 0.33 0.109
7.8 7.7 0.1 0.01 -0.07 0.005

∑ d = 1.70 ∑ d2 = 0.93 ∑(d - md)2 = 0.642

If there is no bias then the differences between each pair of results (d) would be very
small and the average would be very close to zero. A paired t-test is used to compare the
mean difference (i.e. the mean of d) with a hypothetical value of zero taking into account
the standard error of the values of d. The mean and standard error of the difference
(between values of d) is calculated in the usual way:

md = ∑d = 1.70 = 0.17 mmol/L

n 10

sd = √ [∑(d – md)2 / (n-1)]

= √ (0.642 / (10 – 1) = √ 0.0713

= 0.27 mmol/L (2 sig figs)

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 ANALYSIS OF MEANS AND VARIANCES

Alternatively, sd = √ [(∑ d2 - (∑d)2/n ) / (n-1)]

= √ [( 0.93 - 1.702/10) / (10 – 1)]

= √ [( 0.93 - 0.289 ) / 9]

= √0.0712 = 0.27 mmol/L (2 sig figs)

Next calculate t:

t = md
sd/√n

t = 0.17 = 0.17 x √10 = 0.17 x 3.16 = 1.99

0.27/√10 0.27 0.27

From tables of t, for 9 (i.e. n-1) degrees of freedom the probability of obtaining a t value
of 1.99 is greater than 0.05. Therefore, the mean difference (0.17) is NOT significantly
different to zero at the 5 per cent level of probability so the data does NOT demonstrate
any bias between the two methods.

Variance ratio (the F-test)

Sometimes it is not the means that wish to compare but the variation in results. For
example, in question Q 11(1) a quality control serum was analysed 10 times in January
giving a standard deviation (s) of 4 mmol/L, and when analysed 10 times again in
February the standard deviation was 2 mmol/L. The variance ratio test (F-test) can be
used to determine whether or not there has been a significant change in precision.
The variance ratio is simply the highest variance (s1) divided by the lowest variance (s2):

F = s1 2 …………………………….. Eq. 11.8

s2 2

where s1 is greater than s2.

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Therefore F = 42 = 16 = 4.0
22 4

The next step is to look up the probability of obtaining this value for F from tables of F.
Unlike tables of t, the columns refer to the degrees of freedom of s1 and the rows to the
degrees of freedom of s2. Therefore, there is a separate table for each level of probability
(typically 5 and 1%). The degrees of freedom are n –1 for each variance.

DF1 …. 7 8 9 10 12 15 ….
DF2

…. …. …. …. …. …. …. …. ….
7 …. 5.59 4.74 4.35 4.12 3.97 3.87 ….
8 …. 5.32 4.46 4.07 3.84 3.69 3.58 ….
9 …. 5.12 4.26 3.86 3.63 3.48 3.37 ….
10 …. 4.96 4.10 3.71 3.48 3.33 3.22 ….
11 …. 4.84 3.98 3.59 3.36 3.20 3.09 ….
12 …. 4.75 3.89 3.49 6.26 3.11 3.00 ….
…. …. …. …. …. …. …. …. ….

Figure 11.3 Portion of a table of variance ratio (F = s12/s22) values, for which
degrees of freedom are DF1 and DF2 for s1 and s2 respectively (where
s1 > s2), corresponding to P = 0.025 (2.5%)

Therefore, when DF1 = 9 and DF2 = 9 (since DF = n – 1 = 10 – 1 = 9),

the probability that the observed value for F is 3.86 is exactly 0.05 (or 5%). Since the F
value of 4 is greater than this the probability that it occurred by pure chance is less than
0.05 so that there has been a significant change in precision.

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 ANALYSIS OF MEANS AND VARIANCES

Analysis of variance (ANOVA)

An alternative to comparing means of two sets of data is to analyse the variability that
exists within the data. Consider the following two sets of data obtained by analyzing the
same QC sample five times in each of two analytical runs (A and B):

A B

6.5 6.0
6.2 5.8
6.8 5.4
5.8 5.6
6.3 5.9

Mean: 6.34 5.74 Overall mean = 6.04

Normally the data would be analysed by calculating the mean and variances of data sets
A and B separately then applying a t-test. ANOVA involves calculating the variance of
the combined data, but in three separate ways:

Between groups variance The variation between results within each group is eliminated
by substituting each group mean for the individual results. In this example group A
would consist of 5 results each of 6.34 and group B of 5 results each of 5.74 with an
overall mean of 6.04.

Between groups variance = [ (6.34 - 6.04)2 + (6.34 - 6.04)2 + (6.34 - 6.04)2 +

(6.34 - 6.04)2 + (6.34 - 6.04)2 + (5.74 - 6.04)2 +

(5.74 - 6.04)2 + (5.74 - 6.04)2 + (5.74 - 6.04)2

+ (5.74 - 6.04)2 ] / (n -1 )

which can be simplified to:

Between groups variance = [5(6.34 - 6.04)2 + 5(5.74 - 6.04)2] / (n-1)

Because we have substituted means for individual results, n is only 2 so that n-1 =1.

Therefore between groups variance = 0.45 + 0.45 = 0.90

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Within groups variance This is the variance for the individual results calculated using
the appropriate group (A or B) mean, but combining both sets of data:

Within groups variance = [ (6.5 - 6.34)2 + (6.2 - 6.34)2 + (6.8 - 6.34)2

+ (5.8 - 6.34)2 + (6.4 - 6.34)2 + (6.0 - 5.74)2

+ (5.8 - 5.74)2 + (5.4 - 5.74)2 + (5.6 - 5.74)2

+ (5.9 - 5.74)2 ] / (n - 2)

Although there were 10 results initially, two means were used in the calculation, so that
there are n - 2 = 10 - 2 = 8 degrees of freedom. Evaluation of the above expression
gives a within groups variance of 0.10.

Sometimes the within groups variance is referred to as the residual variance.

Total variance This is the combined variance of the two groups using the overall mean
value (in this case 6.04):

Total variance = [(6.5 - 6.04)2 + (6.2 - 6.04)2 + (6.8 - 6.04)2 +

(5.8 - 6.04)2 + (6.4 - 6.04)2 + (6.0 - 6.04)2 +

(5.8 - 6.04)2 + (5.4 - 6.04)2 + (5.6 - 6.04)2 +

(5.9 - 6.04)2] / (n - 1)

There are 10 results initially, one mean was used in the calculation, so that there are n - 1
= 10 - 1 = 9 degrees of freedom. Evaluation of the above expression gives a total
variance of 0.19.

The null hypothesis which is used is that if the two sets of data are from the same
population then the between groups variance will not be significantly different from the
within groups (residual) variance. This can be tested by calculation of the variance ratio:

F = Between groups variance ………….. Eq. 11.9

Within groups (residual) variance

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 ANALYSIS OF MEANS AND VARIANCES

In this example, F is 0.90/0.10 = 9.0 with 1 and 8 degrees of freedom. From tables of F
at the 5% level of probability the F value would be 5.32. Since the value obtained is
higher than this then the two sets of data are significantly different at the 5% level of
probability (i.e. P <0.05). ANOVA for two sets of data is rarely used since it is much
easier to compare the means directly with a t-test. In fact the P value obtained by
ANOVA is exactly the same as that obtained with a t-test.

The value of ANOVA comes into its own when comparing more than two sets of data.
For example, if we had four sets of data we wished to compare, A, B, C and D then one
option would be to carry out t-test between each possible combination of data. The
various combinations are: A-B, A-C, A-D, B-C, B, D and C-D making six t-tests in all.
Clearly it would be simpler to first carry out an ANOVA to se if a difference exists
between any of the groups of data. There is another reason for using ANOVA in
preference to multiple t-tests. If we are looking for a difference which is significant at the
5% level and no significant difference really exists between two sets of data then a
significant value for t will be obtained on five occasions out of a hundred by chance alone
i.e. a false positive rate of 5% will be obtained. If numerous t-tests are carried out then
the incidence of false positives is even higher.

An underlying assumption when carrying out ANOVA is that the variances of the
individual groups are homogeneous i.e. they are not significantly different from each
other. This can be confirmed by first carrying out a variance ratio test on the variances of
the two groups which have the highest and lowest variance. If the variances are not
homogeneous then the problem can often be overcome by first transforming the data.

A difficulty which often arises is that the size of the groups are not equal i.e. some of the
data is missing. Computer packages, which are usually used to perform these
calculations nowadays, have the facility to correct for missing data.

If ANOVA does not reveal a significant difference between the groups then further
statistical analysis is not usually necessary. However, if a difference is demonstrated then
ANOVA cannot tell us which group(s) is/are significantly different from the rest. This
can only be done by carrying out appropriate t-tests – although simple inspection of the
data usually suggest which group(s) is/are likely to be different.

It is common practice to first calculate sums of squares rather then variances. This is
because sums of squares are additive e.g. the between groups and within groups sums of
squares should add up to the total – a useful check on the calculations (a similar check
can be applied to degrees of freedom). In fact it is only necessary to calculate two of the
sums of squares e.g. the total and within groups, then obtain the between groups sum of
squares by subtraction.

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CHAPTER 11

1. Let u = number of groups of data and v = the number of data points in each group.
Arrange the data (values of x) into a table with u columns and v rows.

2. At the foot of each column enter values for ∑x, n, mean, ∑x2, (∑x)2/n and sum of
squares [ i.e. ∑x2 – (∑x)2/n ] for each group.

Group No
1 2 3 4 … u SUM

1 x1,1 x2,1 x3,1 x4,1 …. xu,1

Sample No 2 x1,2 x2,2 x3,2 x4,2 …. xu,2
3 x1,3 x2,3 x3,3 x4,3 …. xu,3
4 x1,4 x2,4 x3,4 x4,4 …. xu,4
… …. …. …. …. …. ….
v x1,v x2,v x3,v x4,v …. xu,v
∑x … … … … … … … ∑. ∑x
n … … … … … … … u x v
Mean … … … … … … …
∑x2 … … … … … … … ∑. ∑x2
(∑x)2/n … … … … … … … ∑. (∑x)2/n
∑x – (∑x)2/n …
2
… … … … … …

3. Calculate the totals, for each group, of ∑x, ∑x2 and (∑x2)/n and enter in a new
column to the right. Complete the following table using these values:

Source of Sum of squares Degrees of Mean square

Variance (SS) freedom (DF) (Variance,s2)

Between groups ∑. (∑x)2/n - (∑. ∑x)2/uv u-1 SS/DF

Within groups ∑. ∑x2 - ∑. (∑x)2/n u(v -1) SS/DF

Total ∑. ∑x2 - (∑. ∑x)2/uv (uv) - 1 SS/DF

4. Divide the between groups mean square by the within groups mean square to
obtain F and look up its P value in tables of F with u - 1 and u(v - 1) degrees of
freedom.

Figure 11.4 Simplified procedure for performing one-way ANOVA

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 ANALYSIS OF MEANS AND VARIANCES

The example discussed above in which the data was divided into several sets is known as
one-way analysis of variance. This simple concept can be extended to two-way,
three-way etc analysis of variance. For example if we were conducting a study to
compare various drug treatments with a group of patients receiving each treatment then
we would need to carry out a one-way ANOVA. If however, the patients were divided
into males and females then the creation of these extra groups would require a two-way
ANOVA. Details of these techniques can be found in statistics textbooks, but nowadays
computer packages are usually used to perform these calculations.

Question Q11(3)

A laboratory was attempting to optimize a new alkaline phosphatase assay. Four

different buffers (A, B, C and D) were each used to assay the same serum sample ten
times with the following results (expressed as IU/L):

Buffer
A B C D

175 170 175 168

160 162 168 158
162 180 198 183
189 165 174 174
177 165 178 178
165 158 182 162
171 164 184 176
190 191 178 193
162 176 201 168
170 168 194 175

Is there a significant difference in the measured activity between any of the four buffers?

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Answer Q11(3)

There are two methods available for calculation of the between groups, within groups and
total sums of squares:

Method 1 Direct calculation from the means

First calculate the mean of each column and the overall mean:

MeanA = (175 + 160 + 162 + ………..162 + 170)/10 = 172.1

MeanB = (170 + 162 + 180 + ………..176 + 168)/10 = 169.9

MeanC = (175 + 168 + 198 + ………..201 + 194)/10 = 183.2

MeanC = (168 + 158 + 183 + ………..168 + 175)/10 = 173.5

Mean (A..D) = (172.1 + 169.9 + 183.2 + 173.5)/4 = 174.7

The between groups sum of squares is calculated from the group means and overall
means:

Between groups sum of squares = 10 [(172.1 - 174.7) + (169.9 - 174.7) +

(183.2 - 174.7) + (173.5 - 174.7)]

= 1035 [degrees of freedom = 4 - 1 = 3]

The within groups sum of squares is calculated from the individual results and the group
means:

Within-groups sum of squares = (175 - 172.1) + (160 - 172.1) + ……..(170 - 172.1)

+ (170 - 169.9) + (162 - 169.9) + ………(168 - 169.9)

+ (175 - 183.2) + (168 - 183.2) + ………(194 - 183.2)

+ (168 - 173.5) + (158 - 173.5) + ………(175 - 173.5)

= 3944 [degrees of freedom = 4 (10 - 1) = 36]

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 ANALYSIS OF MEANS AND VARIANCES

The total sum of squares is calculated from the individual results and the overall mean:

Total sum of squares = (175 - 174.7) + (160 - 174.7) + ……. (170 - 174.7)

+ (170 - 174.7) + (162 - 174.7) + …… (168 - 174.7)

+ (175 - 174.7) + (168 - 174.7) + …… (194 - 174.7)

+ (168 - 174.7) + (158 - 174.7) + ……. (175 - 174.7)

= 4978 [degrees of freedom = (4 x 10) - 1 = 39]

Method 2. Via individual sums of squares

This is the procedure shown in Fig 11.4:

Buffer
A B C D

175 170 175 168

160 162 168 158
162 180 198 183
189 165 174 174
177 165 178 178
165 158 182 162
171 164 184 176
190 191 178 193
162 176 201 168
170 168 194 175
TOTALS
∑x 1721 1699 1832 1735 6987
n 10 10 10 10 40
Mean 172.1 169.9 183.2 173.5 698.7
∑x2 297229 289535 336714 301955 1225433
(∑x)2/n 296184 288660 335622 301023 1221489
∑x - (∑x)2/n
2
1045 875 1092 933 3944

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Number of groups (u) = 4, number in each group (v) = 10.

Between groups sum of squares = ∑.(∑x)2/n - (∑.∑x)2/uv

= 1221489 - (69872/40) = 1035

Within groups sum of squares = ∑.∑x2 - ∑.(∑x)2/n

= 1225433 - 1221489 = 3944

Total sum of squares = ∑.∑x2 - (∑.∑x)2/uv

= 1225433 - (69872/40) = 4979

Whichever method was used to calculate the sum of squares, the procedure to calculate
the F value is the same:

Source Sum of squares DF Variance Variance ratio (F)

Between groups 1035 3 345 3.15

Within groups 3944 36 110
Total 4979 39 128

From tables the probability of obtaining an F value of greater than 2.84 (for 3 and 36
degrees of freedom) is 0.05 (5 per cent). Therefore the data are not homogeneous i.e. at
least one of the groups of data are significantly different to the rest. ANOVA cannot tell
us which group(s) is/are different, t-tests must be performed on paired groups of data.

236
 ANALYSIS OF MEANS AND VARIANCES

FURTHER QUESTIONS

1. The following analytical results were obtained on the same QC sample: 109, 91,
105, 112, 90, 115, 89, 113, 93, 94. Calculate the mean, standard deviation and
standard error of the mean.

2. Two laboratories measured sodium in the same plasma sample ten times. One
laboratory obtained a mean of 145 mmol/L with an SD of 3 mmol; the other
obtained a mean of 147 mmol/L with an SD of 2 mmol/L. Do the laboratories
differ in their bias or imprecision?

3. Serum thyroxine was measured in 500 healthy adults. Assuming a Gaussian

distribution, the normal range was calculated to be 50-150 nmol/L. What is the
probability that the mean of a set of 9 results taken at random from this population
is greater than 125 nmol/L?

4. It is suspected that an instrument used for near patient measurement of cholesterol

is showing positive bias. The following data are the results of paired analyses of
samples from ten patients measured on the standard laboratory analyser (A) and
the instrument under investigation (B). Assuming that the results from the main
analyser are correct, is there any evidence of bias?

A B

6.8 7.2
4.2 4.5
5.0 4.8
5.6 5.9
8.5 8.7
2.9 2.8
4.8 4.9
7.6 8.1
6.5 6.4
5.0 5.2

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CHAPTER 11

5. Four laboratories in a managed network compared the performance of their serum

cholesterol assays by measuring the same sample 10 times with the following
results:

Lab
A B C D

7.6 7.5 7.3 7.7

7.3 7.6 7.4 7.8
7.5 7.2 7.7 7.4
7.7 7.5 7.8 7.5
7.5 7.7 7.4 7.2
7.6 7.4 7.2 7.5
7.4 7.8 7.5 7.3
7.8 7.5 7.6 7.6
7.2 7.3 7.5 7.6
7.5 7.4 7.6 7.4

Is there any significant difference in bias for serum cholesterol at this

concentration between the four laboratories?

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 CORRELATION AND REGRESSION

Chapter 12

Correlation and regression

During graphical analysis of laboratory data the question often arises as to whether or not
there is a valid relationship between variables and, if there is, where the line of best fit
should be drawn? The statistical techniques of correlation and regression seek to answer
these questions. Provided there is a linear relationship between two variables then the
best fit will be a straight line. Often the data is best described by a curve in which case
more advanced techniques are needed which are beyond the scope of this book.
Fortunately it is often possible to transform non-linear data (e.g. by taking logarithms or
reciprocals) to a reasonably straight line which can be fitted by linear techniques.

The graphical presentation of data

It is always a good idea to first plot the data before proceeding to statistical analysis.
This is to check that the data is likely to fit a linear equation, to detect outliers and
confirm that there is a reasonable spread of data points throughout the range for both sets
of values. It is customary to use the horizontal axis for the independent variable which is
often denoted x. The vertical axis is usually used for the dependent variable which is
often denoted y. The independent variable is the one whose value is accurately known
and the dependent variable is some function of the dependent variable. Sometimes it is
not obvious which are the dependent and independent variables.

It is customary to write a linear equation in the form:

y = bx + a ……………………………………………. Eq. 12.1

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CHAPTER 12

30
a) Plot of: y = 2x + 5
25

20

y Slope = 2
15

10
x intercept
= -2.5
5 y intercept = 5

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11
x

30 b) Plot of: y = -2x + 25

25 y intercept = 25

20
y
15 Slope = - 2

10
x intercept = 12.5

0
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Figure 12.1 Plots of two equations showing the significance of the slope and
intercept

240
 CORRELATION AND REGRESSION

Where x and y are the independent and dependent variables respectively and a and b are
constants. b is the slope of the line i.e. the rate of change of y with x. b can be
determined by drawing a right-angled triangle with the line through the data as the
hypoteneuse and measuring the vertical and horizontal sides and dividing the former by
the latter. Alternatively the angle between the line through the data with the base of the
triangle can be measured and its tangent calculated. It is important to take into account
the scales for x and y. a is a constant and represents the intercept on the y axis if the line
through the data is extrapolated. If a has a value of zero then the line passes through the
origin i.e. where the x and y axes intercept.

Figure 12.1 shows a plot of two sets of data of y against x. In both cases the x values are
1, 2, 3,…..10. In (a) the values of y are calculated according to the equation y = 2x + 5.
Therefore the slope of the line is 2 i.e. when x increases by a value of 1, y increases by a
value of 2. The value of the y intercept is 5. This is the value of y when x = 0 in which
case the expression is y = (2 x 0) + 5 which simplifies to y = 5. When y = 0 however,
the expression becomes 0 = 2x + 5 which can be rearranged to 2x = -5 so that x = -5/2
= -2.5 which is the intercept on the x axis.

In (b) the values for y are calculated according to the equation y = -2x + 25. Therefore
the slope of the line is -2 i.e. as x increases by 1, y decreases by a value of 2. The value
of the y intercept is 25. This is the value of y when x = 0 in which case the expression is y
= (-2 x 0) + 25 which simplifies to y = 25. When y = 0 however, the expression becomes
0 = -2x + 25 which can be rearranged to 2x = 25 so that x = 25/2 = 12.5 which is the
intercept on the x axis.

The correlation coefficient (r)

The Pearson correlation coefficient (r) is a number between -1 and +1 whose sign is the
same as the slope of the best fit line to the data and the magnitude of which is related to
the degree of linear association between the two variables. Figure 12.3 shows a plot in
which all points fall exactly on a line with a positive slope so that r = +1. Figure 12.4
shows a similar plot in which all the points fall exactly on a line with negative slope so
that r = -1. However, the data in Fig 12.5 are scattered widely so that there is no linear
relationship between x and y and as a result r = 0. In practice r values are usually
encountered between 0 and 1 or 0 and -1: the nearer r is to 1 (or -1) the more significant
the linear relationship between the two variables.

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CHAPTER 12

35 mx

30 (x - m x )
A B

25 (x - m x )(y - m y ) (x - m x )(y - m y )
(y - m y )
Negative Positive
20
y
15 my

C D
10
(x - m x )(y - m y ) (x - m x )(y - m y )
5
Positive Negative

0
0 5 10 15 20 25 30
x

Figure 12.2 Computation of (x – mx) and (y – my) for a single point (part of a set of
data). The intersecting axes representing the means (mx and my)
divide the graph area into 4 quadrants in which the product (x – mx)
(y – my) is either positive or negative

The idea behind the correlation coefficient is that it should measure the degree of
association between x and y. The measure used is the sum of the products of the
individual deviations of x from its mean and of y from its mean. Figure 12.2 shows the
mode of calculation of both deviations for a single point (which is only one of a series of
values for x and y). A vertical axis is drawn to represent the mean of the values of x (mx)
and a horizontal axis to represent the mean of all the values of y (my). The intersection of
these two axes always falls on the best straight-line fit to the data. The distance of the
observed value of x (10) from the mean of all values of x (20) i.e. (x – mx), is 10 – 20 =
-10. The distance of the observed value of y (30) from the mean of all values of y (15)
i.e. (y – my) is 30 – 15 = 15. Their product, (x – mx)(y – my) is therefore –10 x 15 = -150.

242
 CORRELATION AND REGRESSION

Note that the intersecting axes of mx and my divide the area of the plot into four quadrants
– A, B, C and D. In quadrants A and D the product of the deviations is always negative,
whereas in quadrants B and C it is always positive. The product of each pair of
deviations may be thought of as a “turning moment” about the intersection point of the mx
and my axes. In fact r is often called Pearson’s product-moment coefficient. If the sum of
all the values of (x – mx)(y – my) is divided by the number of data points (or more
specifically the number of data points minus one) then the resulting parameter is known
as the covariance of x and y (cxy):

cxy = ∑(x – mx)(y – my)

n-1

Division by the square root of the products of the individual variances of x and y:

sx2 = ∑(x – mx)2 and sy2 = ∑(y – my)2

n–1 n-1

corrects for the total variability in the data and yields an expression for Pearson’s
correlation coefficient (r):

r = cxy
√ sx2.sy2

Substitution of the expressions for cxy, sx2 and sy2 gives:

r = ∑(x – mx)(y – my)/(n – 1)

√ {[∑(x – mx)2/(n – 1)][∑(y – my)2/(n – 1)]}

Cancelling the (n – 1) values gives the following simpler expression for r:

r = ∑(x – mx)(y – my) ………………….. Eq. 12.2

√ [ ∑(x – mx)2.∑(y – my)2]

243
CHAPTER 12

The following version of this expression is often easier to evaluate:

r = ∑xy - ∑x∑y/n …..…….. Eq. 12.3

√ { [∑x – (∑x) /n] [∑y – (∑y) /n] }
2 2 2 2

The way in which the products of the two deviations affect r is illustrated in Figs 12.3 to
12.6. These figures all use the same values for x (i.e. 5, 10, 15 and 20) and y (10, 20, 30
and 40) but differ in the way each value of y is matched with a value of x. Since all the
values for x and y are the same it follows that ∑(x – mx)2 and ∑(y – my)2, and therefore
√ [∑(x – mx)2∑(y – my)2] are also identical. However, the value of ∑(x – mx)(y- my) varies
according to the pairing of x and y. This is not surprising since the product (x – mx)
(y – my) will not only depend upon the value of y but the value of x with which it is
associated.

In Fig 12.3 the values of x and y are matched so that all the data points fall on a straight
line described by the relationship y = 2x. All data points fall into quadrants B and C so
that each value of (x – mx)(y – my) and hence r is positive. In fact the values for
∑(x – mx)(y – my) and √ [∑(x – mx)2∑(y – my)2] are identical (= 250) so that r = +1.
However, in Fig 12.4 the values for x and y are matched in such a way that the data points
all fall on a straight line described by y = 50 – 2x. All the points fall into quadrants A and
D so that each value of (x – mx)(y – my) and hence r is negative. The value for
∑(x – mx)(y – my) is –250 whereas the value for √[∑(x - mx)2∑(y – my)2] is +250 so that
their ratio (r) is -1.

In fig 12.5 the values of x and y are matched in such a way that there is no linear
relationship between x and y. All four data points fall into different quadrants giving two
negative and two positive results for (x – mx)(y – my) which cancel each other exactly.
Therefore the value for ∑(x – mx)(y – my), and hence r, is zero.

Figure 12.6 shows the situation more commonly encountered where x and y are matched
in such a way that the data can be described by a linear expression (in this case y = 2x)
but with none of the points falling on the fitted line. All of the points still fall into
quadrants B and C but the matching of the values of x and y still results in lower values
for (x – mx)(y – my) giving a total of 150. Hence r is 150/250 = 0.6.

244
 CORRELATION AND REGRESSION

mx
45

40

35

30

25 my
y
20

15

10

0
0 5 10 15 20 25
x

x y (x – mx) (x – mx)2 (y – my) (y – my)2 (x – mx)(y – my)

5 10 -7.5 56.25 -15 225 112.5

10 20 -2.5 6.25 -5 25 12.5
15 30 2.5 6.25 5 25 12.5
20 40 7.5 56.25 15 225 112.5

∑= 50 100 0 125 0 500 250

mx = ∑x = 50 = 12.5 ; my = ∑y = 100 = 25
n 4 n 4

r = ∑(x – mx)(y – my) = 250 = 250 = 1

√ [∑(x – mx)2∑(y – my)2] √ [ 125 x 500] 250

Figure 12.3 A set of four data points which fall exactly on a straight line described
by the relationship y = 2x yielding a correlation coefficient (r) of one,
showing the method for calculation of deviations and sums of squares
and products

245
CHAPTER 12

45 mx
40

35

30

25 my
y
20

15

10

0
0 5 10 15 20 25
x

x y (x – mx) (x – mx)2 (y – my) (y – my)2 (x – mx)(y – my)

5 40 -7.5 56.25 15 225 -112.5

10 30 -2.5 6.25 5 25 -12.5
15 20 2.5 6.25 -5 25 -12.5
20 10 7.5 56.25 -15 225 -112.5

∑ = 50 100 0 125 0 500 -250

mx = ∑x = 50 = 12.5; my = ∑y = 100 = 25
n 4 n 4

r = ∑(x – mx)(y – my) = -250 = -250 = -1

√ [∑(x – mx)2∑(y – my)2] √ [125 x 500] 250

Figure 12.4 A set of four data points which fall exactly on a straight line described
by the relationship y = 50 - 2x yielding a correlation coefficient (r) of
minus one, showing the method for calculation of deviations and sums
of squares and products

246
 CORRELATION AND REGRESSION

45
mx
40

35

30

25 my
y
20

15

10

0
0 5 10 15 20 25
x

x y (x – mx) (x – mx)2 (y – my) (y – my)2 (x – mx)(y – my)

5 20 -7.5 56.25 -5 25 37.5

10 40 -2.5 6.25 15 225 -37.5
15 10 2.5 6.25 -15 225 -37.5
20 30 7.5 56.25 5 25 37.5

∑ = 50 100 0 125 0 500 0

mx = ∑x = 50 = 12.5; my = ∑y = 100 = 25
n 4 n 4

r = ∑(x – mx)(y – my) = 0 = 0 = 0

√[∑(x – mx) ∑(y – my) ]
2 2
√ [125 x 500) 250

Figure 12.5 The same set of data used in Figs 12.3 and 12.4 but paired in such a
way as to yield no correlation (r = 0), showing the method for
calculation of deviations and sums of squares and products. Note that
each data point falls into a different quadrant

247
CHAPTER 12

45
mx
40

35

30

25 my
y
20

15

10

0
0 5 10 15 20 25
x

x y (x – mx) (x – mx)2 (y – my) (y – my)2 (x – mx)(y – my)

5 20 -7.5 56.25 -5 25 37.5

10 10 -2.5 6.25 -15 225 37.5
15 40 2.5 6.25 15 225 37.5
20 30 7.5 56.25 5 25 37.5

∑ = 50 100 0 125 0 500 150

mx = ∑x = 50 = 12.5; my = ∑y = 100 = 25
n 4 n 4

r = ∑(x – mx)(y – my) = 150 = 150 = 0.6

√ [∑(x – mx)2∑(y – my)2] √ [ 125 x 500] 250

Figure 12.6 The same set of four data points used in Fig 12.3 to but with the
y values interchanged to give a poor correlation (r = 0.6) yet still
fitting the function y = 2x, showing the method for calculation of
deviations and sums of squares and products. Note that all four data
points still fall into the positive quadrants

248
 CORRELATION AND REGRESSION

The standard error of r (SEr) is given by the surprisingly simple expression:

SEr = 1 – r2 …………………………. Eq. 12.4

n-2

Division of r by SEr yields an expression for t:

t = r = r √ (n – 2) ………………… Eq. 12.5

SEr √ (1 – r2)

The probability (P) of obtaining this value for t can be obtained from standard tables of t
where there are n – 2 degrees of freedom. Alternatively tables of r are available from
which the value of P can be read directly for any number of degrees of freedom.

A significant correlation only means that there is an association between x and y, it does
NOT necessarily mean that a change in x causes a change in y. In other words
correlation does not equal causation.

Nowadays these calculations are usually performed on a pocket calculator or with a

computer statistics package.

Another way of analysing comparison data is to use the analysis of variance approach
described in chapter 11. For each pair of values of x and y there is a value for y (which
we shall call yfit) which falls exactly on the line of best fit (of course if the x,y data point
happens to fall on the line of best fit then y = yfit). The sum of squares (SS) for the
regression is the sum of the differences of yfit from the horizontal axis described by my:

SSregression = ∑(yfit – my)2

249
CHAPTER 12

This tells us how far the predicted values differ from the overall mean (analogous to the
between sum of squares used in chapter 11).

The residual sum of squares reflects the difference between the original data and the
fitted line:

SSresidual = ∑(y – yfit)2

These sum of squares can be used to calculated mean squares and hence a value for the
variance ratio (F) which can in turn be used to test the hypothesis that the line of best fit
is significantly different from the horizontal axis. Alternatively the sum of squares can be
used to calculate the coefficient of determination (R2) which is simply the proportion of
the total variance described by regression:

R2 = SSregression
SSregression + SSresidual

Fortunately R2 turns out mathematically to be the square of the correlation coefficient:

R2 = r2 = [∑ (x – mx) (y – my) ]2 ……….…… Eq. 12.6

[∑ (x – mx)] [∑(y – my)]

R2 expresses the proportion of variance of the dependent variable explained by the

independent variable. For example, if R2 = 0.75 then 75% of the variation in y is
accounted for by the variation in x. As for r a perfect correlation would have a value of 1
(since the square of one is one) and if there is no relationship then R2 equals zero. Note
that as R2 is the square of the correlation coefficient its value is always positive regardless
of whether the slope of the line is positive or negative.

Linear regression

Whilst correlation seeks to establish the degree of relationship between variables,

regression analysis attempts to determine the expression which best describes the
relationship i.e. the line of best fit. One approach would be to manually draw the line of
best fit by eye then measure the slope and intercept of the line to determine the constants
in the linear equation. However, this would be subject to operator error and would give
no measure of the reliability of the constants obtained.

250
 CORRELATION AND REGRESSION

Since x is the independent variable it is assumed to be without error. Most values for y
will not fall on the line of best fit due to inherent imprecision of y. However, the linear
relationship between x and y predicts the expected value of y (which we shall call yfit) for
any given value of x:

yfit = bx + a

The deviation of the observed value for y from the regression line is (y – yfit), which is
also known as the residual (e), and can be either positive or negative depending upon
whether the observed value falls above or below the line (see Fig 12.7):

e = y – yfit = y - bx - a

y – yfit = 0
y = yfit x
y x Regression line
y - yfit
yfit = b.x + a

yfit = b.x + a
y - yfit
y x

x x x

Figure 12.7 Regression line determined for three values of the dependent variable
(x,x and x) and the corresponding values for the independent variable
(y,y and y). The regression line is drawn (from calculated values for
slope b and intercept a) such that the sum of the squares of the
residuals ( Σ(y – yfit)2 = (y – yfit)2 + (y – yfit)2 ) + (y – yfit)2 is a minimum

251
CHAPTER 12

As in other situations these positive and negative residuals cancel but the sum of their
squares always gives a positive value which is a measure of the overall residuals of the
observations from the line. The problem is to derive values for a and b such that
∑(y – yfit)2 is a minimum (hence this calculation is also known as the method of least
squares). Mathematicians deal with this problem by equating the two derivatives of
∑(y – yfit)2 with respect to both a and b, to zero, then solving the resulting simultaneous
equations for a and b. The solution for the slope of the line (b) is given by the
expression:

b = ∑(x – mx) (y – my) …………………… Eq. 12.7

∑(x – mx)2

The slope of the regression line (b) is also known as the regression coefficient. A similar
expression can be derived to determine a. It is simpler, however, to use the fact that the
line must pass through the intersection point of the means of x and y, so that equation
12.1 becomes my = b.mx + a. Simple rearrangement, with substitution of the value
for b enables determination of the value of a:

a = my - b.mx ……………………. Eq. 12.8

The regression process assumes that the distribution of residuals (y – yfit) about the
regression line is Guassian i.e. that there are approximately equal numbers of
observations each side of the line, that the residuals are independent of the value of x and
that most observations are close to the line with relatively few a large distance form it.
The standard deviation of the residuals (sres, sometimes known as syx) is a valuable
indicator (the lower the better) of the goodness of fit of the data to a straight line:

sres = Σ (y – yfit)2 ……………………….. Eq.12.9

(n – 2)

which can be shown to be algebraically equal to:

sres = (sy)2 (1 – r2) (n – 1) ……………… Eq. 12.10

(n – 2)

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 CORRELATION AND REGRESSION

Having obtained a value for the slope (b) the question often arises as to whether or not it
is significant i.e. whether it is really different from zero (in which there would be no
relationship between x and y). This question can be answered by dividing the difference
between b and 0 (actually b) by the standard error of the estimate of b so as to obtain a
corresponding value for t:

SEb = sres ………………… Eq 12.11

√ Σ(x – mx)2

t = b √ Σ(x – mx)2 ………………………….. Eq 12.12

sres

where t has n-2 degrees of freedom. If n is greater than 30 then the probability can be
obtained from tables of z. Ninety-five percent confidence limits for the slope are given
by b ± 1.96 t.

Question Q 12(1)

The following results for total calcium and albumin were obtained for a series of serum
samples:

Sample Albumin (g/L) Calcium (mmol/L)

1 23 1.95
2 26 2.20
3 30 2.10
4 33 2.25
5 36 2.22
6 40 2.35
7 44 2.32
8 48 2.40
9 52 2.52

Is there a significant linear relationship between serum total calcium and albumin?
Derive an expression to “correct” serum calcium to a “normal” albumin concentration of
40 g/L.

253
CHAPTER 12

Answer Q 12(1)

First plot and inspect the data:

2.6

2.5

2.4
Calcium (mmol/L)

2.3

2.2

2.1

1.9

1.8

1.7
20 25 30 35 40 45 50 55
Album in (g/L)

Inspection of the data suggests that it could be reasonably described by a linear

expression (although it could be coming non-linear at the extremes). Construct a table
with columns for x (albumin g/L), x2, y (calcium mmol/L), y2 and xy. Calculate the sum
of each column i.e. Σx, Σx2, Σy, Σy2 and Σxy:

x x2 y y2 xy

23 529 1.95 3.803 44.85

26 676 2.20 4.840 57.20
30 900 2.10 4.410 63.00
33 1089 2.25 5.063 74.25
36 1296 2.22 4.928 79.92
40 1600 2.35 5.523 94.00
44 1936 2.32 5.382 102.08
48 2304 2.40 5.760 115.20
52 704 2.52 6.350 131.04

Σx = 332 Σx2 = 13034 Σy = 20.31 Σy2 = 46.059 Σxy = 761.54

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 CORRELATION AND REGRESSION

Calculate the correlation coefficient (r) using Eq 12.3:

r = Σxy – (ΣxΣy/n)
√ {[Σx2 – (Σx)2/n] [Σy2 – (Σy)2/n]}

r = 761.54 - (332 x 20.31/9)

√ {[13034 – (332 x 332/9)] [46.059 – (20.31 x 20.31/9)]}

= 761.54 - 749.21
√ { [13034 – 12247] [46.059 – 45.833]}

= 12.33 = 12.33
√ {787 x 0.226} √177.86

= 12.33 = 0.925
13.336

From tables of r, the probability of obtaining a value of 0.925 for n-2 degrees of freedom
(7) is approximately 0.001 (i.e. 0.1%). Therefore the correlation is highly significant.
(Alternatively t can be calculated using Eq 12.5 and its P value obtained from tables of
t for 7 degrees of freedom.) Furthermore:

R2 = 0.9252 = 0.856

So that 85.6% of the variability of y (total calcium) can be explained by variation in x

(albumin).

The slope of the regression line of y upon x can be obtained using Eq. 12.7:

b = ∑(x – mx) (y – my)

∑(x – mx)2

which can also be written:

b = Σxy - (ΣxΣy/n)
Σx2 - (Σx)2/n

Therefore b = 761.54 - (332 x 20.31/9) = 12.33 = 0.0157 mmol/g

13034 – (332 x 332/9) 787

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CHAPTER 12

The standard deviation of the residual (sres) is calculated from Eq 12.10:

sres = (sy)2 (1 – r2) (n – 1)

(n – 2)

(sy)2 = Σ(y – my)2 = Σy2 - (Σy)2/n

n-1 n-1

Therefore sres = [ Σy2 – (Σy)2/n ] (1 – r2) (n – 1)

(n – 1)(n – 2)

Substituting for Σy2, Σy, r and n:

sres = [46.059 – (20.312/9)] (1 – 0.9252) = 0.068

(9 – 2)

The standard error of the slope (SEb) is calculated from Eq 12.11:

SEb = sres = 0.068 = 0.0024

√ Σ(x – mx)2 √ [13034 – (3322/9) =]

and the corresponding t value from Eq. 12.12

t = b √ Σ(x – mx)2 = 0.0157 = 6.5

sres 0.0024

From tables when t = 6.5 for 7 degrees of freedom, P<0.001. Therefore the regression
coefficient is significantly different from zero.

The value for the intercept (a) can be obtained by substituting for b, mx and my in
Eq. 12.8:

a = my - b.mx = Σy/n - b.Σx/n

= 20.31/9 - 0.0157 x 332/9 = 2.26 - 0.579 = 1.68 mmol/L

Therefore the equation for the regression line of y upon x is:

y = 0.0157 x + 1.68

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 CORRELATION AND REGRESSION

To draw the regression line calculate the value of y for two carefully chosen values of x
(say 20 and 55 g/L) using the regression equation, plot the points on a graph of y versus x
and join them up:

when x = 20 g/L, y = (0.0157 x 20) + 1.68 = 1.99 mmol/L

when x = 55 g/L, y = (0.0157 x 55) + 1.68 = 2.54 mmol/L

2.6
2.5
Measured Ca (mmol/L)

2.4
2.3
2.2
Regression line
2.1
of y upon x
2
1.9
1.8
1.7
20 25 30 35 40 45 50 55
Albumin (g/L)

The regression equation shows that for each increase in albumin by 1 g/L, the measured
calcium also increases by 0.0157 mmol/L. To “correct” a measured calcium
concentration to the value expected if the albumin was “normal” (40 g/L) the difference
between the measured albumin and 40 g/L is multiplied by 0.0157 then added to the
measured calcium:

Corrected Ca (mmol/L) = Measured calcium (mmol/L)

+ 0.0157 (40 – measured albumin, g/L)

If the measured albumin is greater than 40 g/L, then the expression 0.0157 (40 –
measured albumin) becomes negative so that it is subtracted from the measured calcium.

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Other modes of regression

Regression of x upon y. If it is y that is the independent variable then a regression of x

upon y is performed. This involves minimizing the sum of the residuals between x and x
fit (Fig 12.3(b) rather than between y and yfit (Fig 12.8(a)). The expression for the slope
becomes:

b = ∑(x – mx) (y – my)

∑(y – my)2

Sometimes it is not clear which is the independent variable. In this case a regression of
both y upon x and of x upon y can be performed. Two different regression lines are
obtained which intercept at the intersection of the two means (mx and my).

Deming regression Ideally if neither x nor y can be identified as the independent variable
then the best solution is to carry out a regression using the method of Deming. This
involves calculating a regression line such that the sum of squares of residuals between
the data points and lines drawn perpendicular to the regression line is minimal (see Fig
12.8 (c)). The process is rather complicated but computer programmes are available to
perform the calculations.

Regression through the origin If the nature of the problem dictates that the regression
line must pass through the origin then there is a simple technique to make sure this is so.
Two points are needed to draw a straight line. The origin will be one and the other is at
the intersection of the two means (see Fig 12.8 (d)).

b = my = Σy/n = Σy
mx Σx/n Σx

Weighted regression The underlying assumption with linear regression is that the
standard deviation of y is constant throughout the range of values. This is often not true
and techniques exist for weighting each value to allow for variations in imprecision.

Multiple regression Sometimes we are dealing with more that one independent
variable. In this situation the partial regression coefficients between pairs of variables are
calculated. These techniques are beyond the scope of this book.

Non-linear regression Although non-linear data can often be transformed then analysed
by linear regression this is not always the case. Non-linear methods are available but
again are beyond the scope of this book.

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 CORRELATION AND REGRESSION

a) Regression of y upon x b) Regression of x upon y

y y

00
0 x x

a) Regression of y upon x
c) “Deming” regression d) Regression through the origin

yMean x
y x
y

x xMean

0 0
x x

Figure 12.8 Modes of regression. In each case the sum of the squares of the
distance from the line ( ) is minimized in the direction
shown

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Method comparison – a special case?

Method comparisons are usually carried out to determine if two methods yield the same
results with patient samples (o rare sufficiently similar for routine purposes). The
correlation coefficient is unhelpful since it only indicates if there is a relationship
between the two sets of results, not whether the results are comparable; furthermore it
would be surprising if the correlation was not significant between two methods designed
to measure the same analyte. Simple regression methods rely on the assumption that the
independent variable is determined without error – an assumption which is rarely true.
Regression methods do however, provide the relationship between the two methods.

Since two methods are measuring the same analyte, the goal is determine whether or not
the two sets of results are significantly different. One way to do this is by the paired t-test
(see Chapter 11). However, if the two sets of results are identical then the relationship
between them should be described by a straight line which passes through the origin and
has a slope of one (i.e. the data is best described by the expression y = x). Visual
assessment of the data can be made if the two sets of results (x and y) are plotted against
each other and the line y = x drawn (Fig 12.9). Altman and Bland took this approach one
step further and plotted the absolute difference d ( equal to y – x) against the mean of
each pair of data, (x + y)/2. If there is no significant difference between the two methods
then the data should scatter evenly about a horizontal line with a value of zero (Fig 12.9);
in other words the mean difference (md) should be close to zero.

md = Σ d …………………………… Eq. 12.13

n

Calculation of the standard deviation of the differences gives a measure of the scatter
about the line.

sd = √ [ Σ (d – md)2/(n – 1) ] ……………………. Eq. 21.14

The 95% confidence limits of the mean difference are known as the “95% limits of
agreement”:

95% limits of agreement = md – 1.96sd to md + 1.96sd …….. Eq. 21.15

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 CORRELATION AND REGRESSION

25
a)
20
Method B

15

10
Line of equivalence
Method B = Method A
5

0
0 5 10 15 20 25
Method A

1.2
1 b)
d = Method B - Method A

0.8
0.6
0.4
0.2
0
-0.2 0 5 10 15 20 25
-0.4
-0.6
-0.8
-1 Mean = (Method B + Method A)/2

Figure 12.9 Data from a comparison of two methods – A and B. a) plot of

Method B versus Method A results, b) difference plot (Altman &
Bland) of same data with 95 % limits of agreement shown

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CHAPTER 12

A t-test can be applied to determine if the mean difference is significantly different from
zero:

t = md …………………………………. Eq. 12.16

sd/√n

These calculations are identical to those of the paired t-test (Chapter 11).

It is important to carefully inspect the data to ensure that the differences are normally
distributed about their mean throughout the range of results. For example, if there is a
significant slope in the y versus x plot of the data then the difference plot will show an
even more exaggerated change in d with concentration, which can be removed if the
percentage difference is plotted instead of the absolute difference.

The advantage of the Altman-Bland approach is that it focuses on the differences of

individual results throughout the range. The same information can be obtained from
plots of y versus x, but small deviations are not obvious.

ADDITIONAL QUESTIONS

1. Regression analysis of results using new standards (y) against old standards (x)
showed a linear relationship. The regression coefficient (slope) was 1.10 and the
intercept on the y axis 1.0 mmol/L. Calculate the results which would be expected
using new standards for the analysis of old standards containing (a) 15 mmol/L
and (b) 150 mmol/L.

2. A laboratory changed its method for the assay of serum alkaline phosphatase
activity. Assay of a selection of patient’s samples by both methods yielded the
following data:

ALP (Old method), IU/L: 50 350 700 100 1500 2000 420 1200
ALP (New method), IU/L: 40 190 350 90 750 1500 280 600

A gastroenterologist has been using ALP to monitor patients on treatment. Use

these data to derive an expression to convert the new ALP results to the results
expected by the old method.

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 CORRELATION AND REGRESSION

3. An endocrinologist has been using serum prolactin measurements to assess the

response of patients with prolactinoma to treatment with a new drug. The
following data were obtained for a series of patients:

Drug dosage (mg/kg body wt): 50 100 150 200 250 300 350 400
Prolactin (IU/L) 750 1500 350 400 2000 1250 500 1800

Do these data show a linear relationship between drug dosage and serum prolactin
concentration?

4. A research paper contains the following statement:

“A good correlation was obtained when 45 patient samples were analysed by

methods A and B (r = 0.90, B = 1.05A – 10)….” Comment on this statement.

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 CLINICAL UTILITY OF LABORATORY TESTS

Chapter 13

Clinical utility of laboratory tests

A laboratory test for a specific disease may yield either a positive or a negative result.
This may be achieved either by a qualitative test (in which the result is either positive or
negative) or a quantitative result which is either above (positive) or below (negative) a
selected “cut-off” value. In the ideal world all patients with the disease in question, and
only these patients, would give a positive result. Furthermore, all patients without the
disease, and only these patients, would produce negative results. In other words there
would be no false positives or false negatives. However, in reality a proportion of both
false positives (positive results in patients without disease) and false negatives (negative
results in patients with disease) are always obtained. The classification of positive and
negative results is shown in Fig 13.1. Correct interpretation of laboratory results demands
an appreciation of the likelihood of the result identifying the presence (or absence) of
disease.

Positive result Negative result Total

Patients with disease TP FN TP + FN

Patients without disease FP TN TN + FP

Total TP + FP TN + FN TP + FN + TN +FP

Figure 13.1 Classification of positive and negative test results.

TP = true positive; FP = false positive
TN = true negative; FN = false negative

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Sensitivity and specificity

The sensitivity of a test is the proportion of patients with disease that are identified by the
test. The number with disease giving a positive result is the true positives (TP) and the
total number with disease is the sum of the true positives and false negatives (TP + FN):

Sensitivity = TP …………………….. Eq. 13.1

TP + FN

The specificity of a test is the proportion of patents without disease that are identified by
the test. The number without disease giving a negative result is the true negatives (TN)
and the total number without disease is the sum of the true negatives and false positives
(TN + FP):

Specificity = TN ………………………… Eq. 13.2

TN + FP

Sometimes these values are multiplied by 100 to give results for sensitivity and
specificity as percentages.

The efficiency of a test is the proportion of all results which are true results:

Efficiency = TP + FN = sensitivity + specificity …… Eq. 13.3

total tested 2

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 CLINICAL UTILITY OF LABORATORY TESTS

Determinations of sensitivity and specificity, like any other biological measurement, are
subject to error. Consider a test with a true sensitivity of 0.50 (50%) determined on a
group of 50 patients with a particular disease. If the sensitivity was repeatedly
determined on groups of 50 patients with the disease then the sensitivity would not come
out at exactly 0.5 every time but would cluster around this value in a similar way to the
mean determined for a continuous variable. Situations where there are only two possible
outcomes (positive or negative) belong to a special distribution called the binomial
distribution. Fortunately if the sample size (n) is large enough (>30) then the binomial
approximates to a normal distribution, and this can be extremely useful.

If p is the proportion of n individuals with the disease giving a positive result

(i.e. estimated sensitivity), then the variance, standard deviation and 95% confidence
limits of p are calculated as follows.

s2 = p (1- p) n ……………………….. Eq. 13.4

s = √ [ p (1 – p) n] ……………….. Eq. 13.5

95% confidence limits = p + 1.96 s to p – 1.96 s ……. Eq. 13.6

Question Q 13(1)

A test for a certain disease gave a 5% false positive rate and a 2% false negative rate.
What is the sensitivity and specificity of the test?

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Answer Q13(1)

If the false negative rate is 2% then this means that for every 100 patients with disease,
2 false negatives will be obtained, the remainder will give positive results.

Therefore FN = 2 and TP = 100 – 2 = 98 and (TP + FN) = 2 + 98 = 100

Sensitivity = TP = 98 = 0.98 (or 98%)

TP + FN 100

If the false positive rate is 5% then this means that for every 100 patients without disease
5 false positive results will be obtained, the remainder will give negative results.

Therefore FP = 5 and TN = 100 – 5 = 95 and (TN + FP) = 5 + 95 = 100

Specificity = TN = 95 = 0.95 (or 95%)

TN + FP 100

Predictive values

Sensitivity and specificity define the performance of a test when applied to populations of
individuals who either have (sensitivity) or do not have (specificity) the disease in
question. In practice tests are applied to populations made up of a mixture of subjects
with or without disease. The proportion of the two populations (i.e. the prevalence of
disease in the population being tested) can have a profound effect on the predictive value
of a test. The predictive value of a test is the probability that a subject with a positive
result has the disease (positive predictive value, denoted PV+) or the probability that a
subject with a negative result does not have the disease (negative predictive value
denoted PV-).

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 CLINICAL UTILITY OF LABORATORY TESTS

Positive predictive value (PV+) = TP ……………… Eq. 13.7

(TP + FP)

Negative predictive value (PV-) = TN ……………….. Eq. 13.8

(TN + FN)

Prevalence of disease = (TP + FN) ……….. Eq. 13.9

(TP + FN + TN + FP)

Consider the example in question Q 13.1 where there is a false positive rate of 5%. In a
population in which a half of all individuals have disease (i.e. the prevalence of disease is
0.5 or 50%) it is possible to calculate the number of each possible outcome of the test
given that the sensitivity is 0.98 and specificity 0.95:

If prevalence is 0.5, then proportion of patients with disease (TP + FN) = 0.5.

Since sensitivity = TP , TP = sensitivity (TP + FN) = 0.98 x 0.5 = 0.49

TP + FN

In other words:

TP = Sensitivity x Prevalence …………………….. Eq. 13.10

If sensitivity is expressed as a percentage then the prevalence is multiplied by sensitivity

and divided by hundred i.e. in the above example the sensitivity is 85% so that the
prevalence is multiplied by 85/100.

Similarly if prevalence is 0.5, then proportion without disease (TN + FP) = 0.5

If the prevalence is expressed as a proportion of one rather than absolute numbers or a

percentage then (TN + FP), also expressed as proportion of one, is obtained by
subtracting the prevalence from one:

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CHAPTER 13

(TN + FP) = 1 - prevalence …………………… Eq. 13.11

Since specificity = TN , TN = specificity (TN + FP) = 0.95 x 0.5 = 0.475

TN + FP

In general:

TN = Specificity x (1 – prevalence) …………… Eq. 13.12

Since (1 - prevalence ) = (TN + FP) = 0.5 and TN = 0.45, FP can be calculated:

FP = (1 - prevalence ) - TN = 0.5 - 0.475 = 0.025

Since TP = 0.49 and FP = 0.025 the proportion of patients with a positive result who have
disease (i.e. the positive predictive value) is

0.49 = 0.95 (or 95%)

(0.49 + 0.025)

Therefore 0.95 (95%) of positive results are due to disease.

Consider a population consisting of 10 patients with and 90 patients without, disease (i.e.
prevalence = 0.1 = TP + FN).

As above, TP = sensitivity (TP + FN) = 0.98 x 0.1 = 0.098

Since we are dealing with proportions all groups must add up to one

i.e. TP + FN + TN + FP = 1, and (TP + FN) = 0.1

then TN + FP = 1 - (TP + FN) = 1 - 0.1 = 0.9

Therefore TN = specificity (TN + FP) = 0.95 x 0.9 = 0.855

and FP = (TN + FP) - TN = 0.90. - 0.855 = 0.045

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 CLINICAL UTILITY OF LABORATORY TESTS

The proportion of true positives has fallen from 0.49 to 0.098 and false positives risen
from 0.025 to 0.045. Therefore, the proportion of positive results which are due to
disease (i.e. the positive predictive value) is:

PV+ = 0.098 = 0.098 = 0.685

(0.098 + 0.025) 0.143

Therefore only 0.685 (or 68.5%) of positive results are due to the presence of disease.
Continuation of this process for other prevalences yields the data in Fig 13.2.

Prevalence TP FP PV+

1 in 2 4,900 250 0.95

1 in 4 2,450 375 0.87
1 in 10 980 450 0.69
1 in 40 245 488 0.33
1 in 100 98 495 0.17
1 in 1,000 10 499 0.02
1 in 10,000 1 500 0.002

Figure 13.2 Effect of disease prevalence upon the number of true positives (TP),
false positives (FP) and positive predictive value (PV+) for a test with
a sensitivity of 98% and specificity of 95% applied to a population of
10,000 subjects

Therefore when the prevalence is very low the number of false positives exceed the true
positives.

Question Q13(2)

In a cancer clinic where the prevalence of ovarian malignancy is 40%, a tumour marker
has a specificity of 88% and a sensitivity of 92%. Calculate the predictive value of a
positive test result.

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CHAPTER 13

Question Q13(2)

To solve this type. of problem the calculations can be based on actual numbers of results,
percentages or proportions of one. It is often simplest to work with proportions.

Construct a 2 x 2 contingency table (as Fig 13.1):

Positive result Negative result Total

Patients with disease TP FN Prevalence

Patients without disease FP TN 1 - prevalence

The prevalence of disease is 40 % (0.4 as a proportion) so that (1 – prevalence) is

(1 – 0.4) = 0.6. The above table then becomes:

Positive result Negative result Total

Patients with disease TP FN 0.4

Patients without disease FP TN 0.6

The next task is to determine values for TP, FN, FP and TN. We are given values for
sensitivity and specificity so can write:

Sensitivity = TP = 0.92 and specificity = TN = 0.88

TP + FN TN + FP

Since the prevalence is 0.4 and is equal to (TP + FN), and (1 – prevalence) is 0.6 and is
equal to (TN + FP), both of these expressions can be re-written:

Sensitivity = TP = 0.92 and specificity = TN = 0.88

0.4 0.6

These expressions are then rearranged and solved for TP and TN:

TP = 0.92 x 0.4 = 0.368 and TN = 0.88 x 0.6 = 0.528

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 CLINICAL UTILITY OF LABORATORY TESTS

Both of these values are now added to the contingency table:

Positive result Negative result Total

Patients with disease 0.368 FN 0.4

Patients without disease FP 0.528 0.6

Since prevalence = (TP + FN) and (1 – prevalence) = (TN + FP), values for FN and
FP can be obtained by subtraction of TP and TN from the corresponding totals in the last
column.

Positive result Negative result Total

Patients with disease 0.368 0.032 0.4

Patients without disease 0.072 0.528 0.6

The predictive value of a positive test (PV+) is then obtained by substitution of TP and
FP into equation Eq 13.7:

PV+ = TP = 0.368 = 0.368 = 0.84 (84%)

TP + FP 0.368 + 0.072 0.440

Question Q13(3)

A man has a PSA of 5 μg/L. 22% of patients with benign prostatic hypertrophy and 38%
of patients with prostatic cancer have concentrations of PSA between 4.1 and 10 μg/L.
What is the positive predictive value for a diagnosis of cancer of the result for this man in
this range, if the prevalence of cancer in his age group is 5% and benign prostatic
hypertrophy is 20%? Assume 2% of patients without any prostatic pathology have a PSA
>4.1 μg/L.

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Answer Q13(3)

This question differs from Q13(2) in that there are three instead of two groups of patients.
However, there are only two groups as far as the disease in question (prostatic cancer) is
concerned – those with cancer and those without. The only difference is that the group
without cancer is made up of two populations:

• those with benign prostatic hypertrophy (BPH)

• those without BPH and without prostatic cancer (CAP)

In order to calculate the predictive values first calculate the individual positive and
negative results for each of the three groups. This it is easier if everything is converted
to proportions of one rather than working with percentages.

CAP group, prevalence = 5% = 5/100 = 0.05

BPH group, prevalence = 20% = 20/100 = 0.20

The normal group (i.e. those with neither BPH or CAP) consists of the remaining patients

Therefore, prevalence of normals = 1.0 – (0.05 + 0.20) = 0.75

SensitivityCAP = 38% = 38/100 = 0.38

SensitivityBPH = 22% = 22/100 = 0.22

Set up a contingency table and enter the totals:

Positive result Negative result Total

CAP TPCAP FNCAP 0.05

BPH TPBPH FNBPH 0.20

Normals FP TN 0.75

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 CLINICAL UTILITY OF LABORATORY TESTS

For the group with CAP:

TPCAP = PrevalenceCAP x SensitivityCAP

= 0.05 x 0.38 = 0.019

Since TPCAP and FNCAP must add up to prevalenceCAP

FNCAP = 0.05 - 0.019 = 0.031

Similarly for the group with BPH:

TPBPH = 0.20 x 0.22 = 0.044

and FNCAP = 0.20 - 0.044 = 0.156

For the group without either disease we are told that 2% of patients have raised PSA
(i.e. false positives). Therefore FP = 2% = 2/100 = 0.02

Therefore, TN = 0.75 - 0.02 = 0.73

Enter these values to complete the contingency table:

Positive result Negative result Total

CAP 0.019 0.031 0.05

BPH 0.044 0.156 0.20

Normals 0.020 0.730 0.75

Total 0.083 0.917 1.00

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The positive predictive value (PV+) is obtained by substituting the proportion of true
positives and false positives into Eq 13.7, remembering that the false positives are made
up of two groups – those with BPH and those with neither BPH nor CAP:

PV+ = TPCAP
(TPCAP + TPBPH + FP)

= 0.019 = 0.23 (2 sig figs) or 23%

(0.019 + 0.044 + 0.02)

Therefore approximately only 1 in 4 positive results will be due to prostatic cancer.

The positive predictive value (the likelihood of disease when a test result is positive) is
markedly influenced by the prevalence of disease in the population being tested. The
prevalence of any disease is very low in the general population, becomes higher in a
population with symptoms of the disease in question and highest in patients referred by a
general practitioner to a hospital specialist.

This concept is of greatest significance when screening a well population for the presence
of a rare disease such as bowel cancer. False positives may be obtained due to
interference in the screening test used, dietary factors and bleeding from other sources.
As a result the predictive value of a positive screening test is low so that only a small
portion of patients with positive results will have bowel cancer. Detection of bowel
cancer then relies on a secondary test (such as colonoscopy). However, the screening test
is still valuable in that it enables selection of a sub-population of individuals with a higher
prevalence of disease which warrants the expensive and unpleasant secondary test.
On the other hand a false negative will be obtained if the tumour is not bleeding when the
screening test is carried out. When designing a screening program the incidence of false
positives and false negatives must be carefully balanced so as to achieve optimal
detection of disease with minimum cost and morbidity.

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 CLINICAL UTILITY OF LABORATORY TESTS

Receiver operator characteristic (ROC) curves

Although calculations of sensitivity, specificity etc are simple with a qualitative test
which yields either a positive or negative result, the majority of diagnostic tests are of a
quantitative nature yielding a result which is a continuous variable. The quantitative
result is converted into a qualitative one by using a decision level (or cut-off point).
Values above this level are classified as “positive” and below it as “negative”. This
provides an opportunity to manipulate the performance of a test by adjustment of the
decision level.

Figure 13.3a shows the result for an analyte which is able to distinguish patients with a
particular disease from healthy individuals. The distribution of results for each set of
patients is roughly Guassian, but the two distributions overlap. In this overlap area it is
impossible to use the test to distinguish patients with the disease from those who have
not. If a decision level at point “A” is used then there are no false negatives but as many
as a half of those without disease would yield false positive results (i.e. the sensitivity
would be virtually one but with a specificity of only 0.5). A decision level of “B”
produces fewer false positives but at the expense of missing a small number with disease.
Using “C” or “D” even fewer false positives are obtained but false negatives become
more frequent. At level “E” there are no false positives but at the expense of missing
about one half of patients with the disease.

A useful way of looking at the effect of changing decision levels is to plot the sensitivity
(y-axis) versus 1 – specificity (x-axis) at each decision level (see Fig 13.3b). The x-axis
therefore represents the false positive rate and the y-axis the true positive rate. For a
perfect test, the resulting receiver operator characteristic (ROC) curve would extend
from the lower left to the upper left then to the upper right. It is generally accepted that
for a test of no diagnostic value the curve would be a diagonal line from the origin with a
slope of one. When comparing the performance of different tests it is helpful to calculate
the area under the ROC curve. Other things being equal, the test with the highest area is
superior.

Although examination of the ROC curve is often helpful, usually other considerations
(the importance of avoiding false positives versus false negatives, the nature and cost of
any follow-up tests etc) determine which decision point to use.

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CHAPTER 13

(a) Healthy Diseased

No

Analyte value

A B C D E

(b)
1
0.9 D E
C
0.8
0.7 B
sensitivity

0.6
0.5 A
0.4
0.3
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
1 - specificity

Figure 13.3 Effect of variation of decision level (A,B, C, D and E) upon the
performance of a diagnostic test. a) Distribution of results from
healthy and diseased individuals; b) Same data plotted as a
receiver operator characteristic (ROC) curve

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 CLINICAL UTILITY OF LABORATORY TESTS

Odds and likelihood ratios

What the clinician really needs to know is the probability that a patient with a given test
result has the disease in question. To answer this question two parameters are required:

• The prevalence of the disease in the population to which the patient belongs

• The likelihood ratio of a positive test

The likelihood ratio positive (LR+) is defined as the ratio between the probability of
finding a positive test in the presence of disease and the probability of a positive test in
the absence of disease. The probability of a positive test in the presence of disease is
simply the sensitivity of the test. The probability of finding a negative test in the absence
of disease is the specificity so that the probability of a positive test when disease is absent
is simply (1 – specificity). Therefore LR+ can be calculated directly from the sensitivity
and specificity of the test:

LR+ = probability of +ve test with disease = sensitivity … Eq. 13.13

probability of +ve test without disease (1 – specificity)

A similar expression can be derived for likelihood ratio of a negative test (LR-):

LR- = probability of a –ve test with disease = (1 – sensitivity) … Eq. 13.4

probability of a -ve test without disease specificity

When comparing several different tests it is often helful to calculate the diagnostic odds
ratio, which is simply the ratio of LR+ to LR-, which can be simplified to:

Diagnostic odds ratio = sensitivity x specificity ………. Eq. 13.15

(1 – sensitivity) x (1 – specificity)

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The likelihood ratio positive is useful in converting the pre-test odds (the likelihood that
disease is present before the test is carried out) to the post-test odds. In the absence of
any data about the patient’s individual risk factors, the pre-test odds is the prevalence of
the particular disease in the population to which he/she belongs. For example if the
prevalence of disease is 0.2 (20%) then a group of 100 patients will contain 20 with
disease and 80 without disease. Therefore the odds of disease being present are given by
the ratio of the number of patients with disease to the number without i.e. 20/80 = 0.25.
In other words for each patient without disease there will 0.25 patients with disease or
rather for every 4 patients without disease there will be one with the disease or the odds
against disease being present are 4:1. In general:

Pre-test odds = prevalence …………………………. Eq. 13.16

(1 – prevalence)

provided the prevalence is expressed as an absolute proportion of one

Multiplication of the pre-test odds by the likelihood ratio positive gives the post-test
odds:

Post-test odds = pre-test odds x likelihood ratio positive…… Eq. 13.17

Post-test odds can be converted back to the probability of the patient having disease using
the expression:

Post-test probability = post-test odds ………………….. Eq. 13.18

(1 + post-test odds)

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 CLINICAL UTILITY OF LABORATORY TESTS

Provided reliable data is available on the prevalence of the disease and the diagnostic
performance of the laboratory test then calculation of post-test probability is the best way
of presenting laboratory data to the clinician. However, one problem that is not addressed
is the effect of the magnitude of the increase in the analyte above the decision point. A
result which is several times the decision value often is more likely to be associated with
disease than a result which is only just above this value.

Two further refinements of this technique are often used:

• Multiple tests. Provided the likelihood ratios of several different tests are known
then it sometimes possible to combine them to give a more reliable post-test
probability of disease than for either test on its own. An example of this is the
triple test to screen for Down’s syndrome.

• Sequential testing. A preliminary laboratory test may be applied to select a

population in which the prevalence of disease is enhanced, then apply a secondary
test to accurately identify those with the disease. In is vital that the two tests are
independent i.e. are not different ways of assessing the same thing.

Question Q 13(4)

A certain disease has a prevalence of 10 percent. A diagnostic test was applied to a

random sample of 500 individuals from this population and yielded 45 true positive and
40 false positive results. Calculate a) the likelihood ratio positive and, b) the post-test
probability of disease being present for a positive test result.

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Answer Q 13(4)

If the prevalence is 10% then for a sample of 500 individuals 50 will have the disease and
450 will be healthy. Use this data to set up a 2x2 contingency table:

Positive test Negative test Total

Diseased 45 FN 50

Healthy 40 TN 450

Subtract the positive test from the corresponding total to give to give the numbers of
negative results:

Positive test Negative test Total

Diseased 45 5 50

Healthy 40 410 450

Use these values to calculate the sensitivity and specificity:

Sensitivity = TP = 45 = 0.90
(TP + FN) 50

Specificity = TN = 410 = 0.911

(TN + FP) 450

a) Calculate the likelihood ratio positive (LR+) from the sensitivity and specificity
(Eq13.13):

LR+ = sensitivity = 0.90 = 0.90 = 10 (2 sig figs)

(1 – specificity) (1 – 0.911) 0.089

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b) First calculate the pre-test odds from the disease prevalence (Eq 13.16):

Pre-test odds = prevalence

(1 – prevalence)

The prevalence is converted to a proportion of one by dividing the number with

disease by the total number tested.

Prevalence = 50 = 0.1
500

Pre-test odds = 0.1 = 0.1 = 0.111 (i.e. 0.111 to 1)

(1 – 0.1) 0.9

The post-test odds is calculated by multiplication of the pre-test odds by the

likelihood ratio positive (Eq. 13.17):

Post-test odds = Pre-test odds x LR+

= 0.111 x 10 = 1.11

The post-test odds can be converted to a probability using Eq 13.18:

Post test probability = post test odds

(1 + post test odds)

= 1.11 = 1.11 = 0.53 (2 sig figs)

(1 + 1.11) 2.11

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ADDITIONAL QUESTIONS

1. A test for a particular disease has a sensitivity of 95% and a specificity of 95%.
Calculate the predictive value of both a positive and a negative test result in a
population in which the prevalence of the disease is:

a) 1 in 2
b) 1 in 5000

2. The table shows data from two urinary screening tests for the detection of
phaeochromocytoma.

Test Sensitivity Specificity

VMA 96.7% 99.1%

Total metanephrines 100% 98%

Both tests were used to screen a population of 100,000 hypertensive patients in

which the incidence of phaeochromocytoma is known to be 0.5%.

a) How many patients with phaeochromocytoma were missed by the VMA

test?

b) How many patients were incorrectly diagnosed as having

phaeochromocytoma using the metanephrine test?

c) Which test would you use to screen a hypertensive population for

phaeochromocytoma? Give reasons for your choice.

3. A new laboratory test has a sensitivity of 85% and a specificity of 90%.

The incidence of disease in a population considered at risk is 0.10. What is the
predictive value of

a) a positive result?
b) a negative result?

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4. A proposed diagnostic serological test for coeliac disease was evaluated in 200
consecutive patients referred to a paediatric gastroenterology service in whom the
condition was suspected clinically. The test result was compared with the
diagnosis as established by biopsy, withdrawal of gluten and response to
re-challenge. On this basis 76 children had the condition of whom only 64 gave a
positive test result: 10 positive test results occurred in children who were shown
not to have coeliac disease. Calculate the sensitivity and specificity of the test and
the predictive value of a positive result.

5. In a cancer clinic where the prevalence of ovarian malignancy is 40%, a tumour

marker has a specificity of 88% and a sensitivity of 92%. Calculate the predictive
value of a positive test result. If this test was used as a screening tool in all
patients attending a general gynaecological clinic with a cancer prevalence of
0.4%, what would be the predictive value of a positive test in this population?

6. A certain disease has a prevalence of 5 percent. A diagnostic test was applied to a

random sample of 400 individuals from this population and yielded 15 true
positive and 30 false positive results. Calculate: a) the positive predictive value of
the test applied to this population, b) the pre-test odds of disease, c) the likelihood
ratio positive; d) the post test odds of disease for a positive result, and e) the
post-test probability of disease for a positive result.

7. A two-stage sequential test strategy is used to screen for a rare inherited disease.
The prevalence of the disease is 0.0005. The initial test has a sensitivity of 98%
and specificity of 95%, the follow-up test a sensitivity of 95% and specificity of
99%. What is the probability of a patient with a positive result for the follow-up
test having the disease?

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Chapter 14

Statistical power

Power is the ability of a statistical test to detect a specified effect with a given probability.
For example, comparison of the means of two sets of data involves calculation of the
number of standard errors by which they differ (i.e. the z or t value) then looking up the
P value of this statistic in appropriate tables. Fig 14.1 shows the sampling distribution of
the mean for two sets of data. In Fig 14.1a it is apparent that the two distributions
overlap considerably. However, increasing the sample size from 10 to 50 lowers the
standard deviation of these distributions (i.e. the standard error of the means) so that the
distributions are less widely spread but without any shift of the overall mean values. This
is exactly what would be predicted from Fig 11.1. As a consequence the overlap between
the distributions has been eliminated. Since the standard error of each mean is given by
s/√n it is not surprising that increasing the value of n reduces the standard error
(i.e. spread) of each distribution so reducing the degree of overlap.

The likelihood of demonstrating a significant difference between the means of two sets of
data increases as the degree of overlap is reduced. The degree of overlap and hence the
chance of detecting a difference (i.e. power) is in turn determined by four factors:

• The sample size of each group.

• The magnitude of the difference between the two groups

• The standard deviation of the data

• The required level of statistical significance.

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a) n = 10 A B

b) n = 50

Figure 14.1 Sampling distributions of the means for two sets of data (A and B)
showing the effect of sample size (n) on the overlap between the two
distributions: a) small sample size (n = 10) resulting in large standard
errors of the means and overlap between both distributions, and b)
larger sample size (n = 50) resulting in smaller standard errors of the
means and virtually no overlap between the distributions. Note that
the overall means of A and B remain unchanged

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 STATISTICAL POWER

Figure 14.2 shows the frequency distribution of the same data but using the difference
between observed values for the means (i.e. mean B minus mean A) expressed as the
number of standard deviations from the overall mean (i.e. the z or t value). The blue
curve shows the distribution which would be obtained if there was no true difference
between the overall means of A and B; this is identical to the normalized Guassian curve
(with a mean of zero and standard deviation of one) used to generate values of P for
given values of z (or t). For any individual set of data the observed value of the mean is
compared with a decision level (DL, which is usually z = 1.96 and equivalent to a
probability of 0.05 – or 0.025 if a single sided z-test is performed) to decide whether or
not to accept the null hypothesis (Ho) that no real difference exists and the observed data
could have arisen by chance. The probability used for the decision level (the probability
that the value for z could have arisen by chance) is denoted as alpha (α) – or α/2 for a
single sided test.

There is however a risk in relying on the null hypothesisalone. If we set alpha at 0.05
then this means that a value for z of greater than 1.96 will be used to reject the null
hypothesis and we then accept that the means of the two sets of data are probably
different (i.e. belong to a different frequency distribution). However, by chance, z will be
greater than 1.96 (and P less than 0.05) on approximately one occasion in twenty even if
no real difference exists. Therefore the chance of a false positive (i.e. accepting that there
is a difference between the two means even when in fact none exists) is 1 in 20 (i.e. 0.05).
Rejection of the null hypothesis when it is true is known as a type I error, the chance of it
occurring is alpha (α). If a lower cut off point for P (i.e. α) is used then the risk of a type
I error is reduced. The chance of a type I error not occurring is (1 – α). It is important to
realize that rejection of the null hypothesis does not prove that the samples came from
different populations – it simply means that we have failed to prove that they do not come
from the same population.

The alternative hypothesis (H1) is that the two sets of data belong to different populations
i.e. the true difference between their means is not zero. Figure14.2 shows the distribution
for the difference between the means plotted as z values in red. Therefore if the data
belong to different populations then their values cluster around a value other than zero so
that the curve shifts to the right (or to the left if there is a negative difference). The
decision level cuts the H1 curve at a different point and divides the curve into two
portions. The segment to the left, denoted as beta (β), represents the chance of rejecting
the alternative hypothesis when it is true (or not rejecting the H0 hypothesis when it is
false) and is known as a type II error. The difference (1 – β) is the probability of
rejecting the null hypothesis when it is false, and is known as the power of the study.

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Null hypothesis (H0)

Decision level (DL)

Area = (1 – α)

Area = α/2

Alternative hypothesis (H1)

Area = β
Area = (1 – β) = power

Figure 14.2 Frequency distributions for the null (H0) and alternative (H1)
hypotheses showing the significance of alpha (α) and beta (β)

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 STATISTICAL POWER

The relationship between type I and type II errors is shown in Fig 14.3.

No true difference True difference

H0 not rejected True result (1 - α) Type II error (β)

H0 rejected Type I error (α) True result (1 –β)

Figure 14.3 The consequences of hypothesis testing

Ideally the power should be one (or 100%) so that the null hypothesis will always be
rejected whenever it is false. In practice such a high power is never achieved as there is
always a small risk of a type II error. Fortunately, the power of a test can be improved by
manipulating several key factors. First it is necessary to derive a relationship between
these factors.

Consider two sets of overlapping data, A and B, with means mA and mB and a common
standard deviation (s) and sample number (n). mB is greater than mA. The distance of the
decision level (DL) from mA is (DL - mA) and defines the value of α. If this distance is
divided by the standard error (s/√ n) then the corresponding value for α expressed as
standard errors from the mean (zα) is:

Zα = DL - mA = (DL - mA) x √ n
s/√ n s

Similarly the distance of DL from mB is (mB - DL) and defines the value of β. If this
distance is also divided by the standard error (s/√ n) then the corresponding value for β
expressed as standard errors from the mean (zβ) is:

zβ = mB - DL = (mB - DL) x √ n
s/√ n s

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Adding the values for zα and zβ gives:

zα + zβ = (DL - mA) x √ n + (mB - DL) x √ n

s s

Which can be simplified to:

zα + zβ = (DL - mA + mB - DL) √ n
s

The DL’s cancel giving:

zα + zβ = (mB - mA) √ n
s

Substituting ∆ for (mB - mA) gives:

zα + zβ = ∆ √ n ……………….. Eq. 14.1

s

A more useful form of this expression can be obtained by rearrangement to give a value
for √ n:

√n = s (zα + zβ)
∆

then squaring everything up:

n = [s (zα + zβ) / ∆] 2 …………………. Eq. 14.2

These expressions have two principal uses:

• Evaluation of published data. Quite often authors publish a P value without any
discussion of the power of their study. This is particularly important when no
significant difference is found since the power of the study may be too low to
stand a reasonable chance of detecting any true difference which may exist.
Evidence based reviews often demand some estimate of the power of published
studies cited. Values for n and s and the individual means (and hence ∆) are

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• usually given so it is possible to insert these into Eq 14.1 to obtain the value for
(zα + zβ). The P value can be converted to zα (by dividing it by s/√ n) then zβ
obtained by subtraction. zβ can be converted to β (by multiplying by √ n/s) then
subtracted from one to give the power. It is generally accepted that a “good”
power should be at least 80%.

• Experimental design. Designing an experiment which has a good (at least

70-80%) power ensures a good chance of successfully detecting any effect which
exists. Clearly it would only be worthwhile embarking on a study if there is a
good chance of detecting a scientifically or clinically significant effect. Clinical
trials, are expensive in terms of time and resources, may expose participants to
some degree of risk and so clearly are irresponsible and possibly clinically
unethical unless there is a reasonable chance of success. Grant awarding bodies
demand evidence of power for proposed studies. At the outset levels of statistical
significance (i.e. values for α and β) are set (often at 0.05 and 0.2) and it is usually
possible to decide on the magnitude of clinical effect or change which would be
worth detecting (i.e. a value for ∆). An estimate of the standard deviation (s) is
usually available. These values can be substituted into Eq 14.2 and solved to give
the sample size (n) required. It is important to remember that β always refers to a
single tailed test.

The above principles are also applicable to other study designs but unfortunately the
mathematics can become quite complex. Fortunately tables (and computer packages) are
available for the common study designs so direct calculation is rarely required. Note that
to simplify matters these tables often use the effect size (ES), which is the ratio of the
difference between the groups to s (i.e. ES = ∆/s).

Question Q 14.1

A study into the efficacy of a new cholesterol lowering drug involved measuring serum
cholesterol in 30 subjects both before and after administration of the drug. Using a
decision level of 5% the authors concluded that there was no effect of the drug upon
serum cholesterol concentrations (z = 0.97, P >0.1). The mean initial serum cholesterol
concentration was 7.0 mmol/L (SD = 2.0 mmol/L). After 4 weeks of treatment with the
drug the mean serum cholesterol was 6.5 mmol/L (SD= 2.0 mmol/L). Calculate a) the
power of their study, and b) the sample size needed to achieve a power of 90%.

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Answer Q 14(1)

a) The power of the study can be calculated using Eq. 14.1:

zα + zβ = ∆ √n
s

∆ = difference between the means for the two groups = 6.5-7.0 = - 0.5 mmol/L
n = number of subjects in the study = 30
s = standard deviation = 2.0 mmol/L

Substitution of these values for ∆, n and s into this equation allows evaluation of
the sum of the z values for α and β:

zα + zβ = - 0.5 √ 30 = - 0.5 x 5.48 = - 1.37

2.0 2.0

Since the probability (P) used as a decision level in this study is 0.05, the
corresponding z value (obtainable from tables) is 1.96. Therefore, α = 0.05 and
zα = - 1.96 (since we are using the negative part of the distribution) allowing
calculation of zβ:

zβ = -1.37 - (-1.96) = 0.59

From tables, the value for β (i.e. proportion of total area under the curve)
corresponding to zβ = 0.59 is 0.28.

Therefore power = (1 - β) = 1 - 0.28 = 0.72 (or 72%)

Therefore the study has a 72% chance of detected a change of 0.5 mmol/L at a
decision level of 5% probability.

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b) The sample size required to achieve the desired power can be calculated from
Eq. 14.2:

n = [ s ( zα + zβ ) / ∆ ] 2

s = standard deviation = 2.0 mmol/L

∆ = difference between the means of the two groups = 6.5 – 7.0 = - 0.5 mmol/L

The decision level used is a probability of 0.05 (corresponding to α). From

tables the corresponding z score (zα) is 1.96 standard errors.

The required power is 90% (or 0.9).

Therefore (1 – β) = 0.9

and β =I (1.0 - 0.9) = 0.1.

From tables the corresponding z value (remembering that we are dealing with a
one-tailed z-test) is 1.28.

Therefore (zα + zβ) = 1.96 + 1.28 = 3.24 standard errors

Substitute (zα+ zβ), s and ∆ into the above equation and solve for n:

n = [ (2.0 x 3.24)/-0.5]2

= 12.962 = 168

Therefore at least 168 subjects will need to be studied in order to achieve a 90%
chance of detecting a change in serum cholesterol of 0.5 mmol/L at the 5% level
of probability.

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FURTHER QUESTIONS

1. A study into the effect of nutritional supplements on patients with Crohn’s disease
involved measuring serum albumin both before and after supplementation for a
four-week period. During this period the mean serum albumin level increased
from 25 g/L to 30 g/L. The study involved 40 patients with a standard deviation
for albumin concentration of 10 g/L. What is the power of this study to detect a
5 g/L change in serum albumin at the 5% level of probability?

2. It is proposed to set up a study to determine the effect of dietary modification on

serum cholesterol. The population to be studied has a mean serum cholesterol of
7.5 mmol/L with standard deviation of 2.5 mmol/L. What number of participants
need to be recruited in order to demonstrate a lowering of serum cholesterol by
10% (using alpha = 0.05 as a critical value) with a power of 90%?

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Chapter 15

Miscellaneous topics

Recovery experiments

Studies on the performance of an analytical method often include recovery experiment in

which a known amount of the analyte (the “spike”) is added to a base material to produce
a “spiked” sample. The ideal outcome is that all of the added material will be recovered
when the sample is analysed (i.e. the recovery is 100 percent):

Recovery (%) = Amount recovered x 100%

Amount added

In most instances the base material will already contain some of the analyte in question
but this amount is not usually known with certainty. Therefore both the base material and
the spiked material are analysed. The amount recovered is calculated by subtraction of
the “base result” from the “spiked result” so that the above expression becomes:

Recovery (%) = (spiked result - base result) x 100 ………….. Eq. 15.1
spike added

Allowance must be made for dilution of the spike by the base material and vice versa.

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Sometimes recovery is calculated in a different way which assumes that the expected
result for the spiked sample is equal to the sum of the base result and the spike added:

Recovery (%) = Spiked result x 100

Base result + spike added

This approach is fundamentally flawed and should NOT be used. The expected
contribution of analyte in the base sample is not the same as its measured value (unless
the recovery of the method is exactly 100%). The expected result for the analyte in the
base sample is not known and therefore the expected result for the spiked base cannot be
calculated and the above equation does not give a valid recovery value. Only the value
for the “spike” is accurately known.

Most assays in routine use have recoveries in the range 90-100% with imprecisions of the
order of 5%. Imprecision of the assay can easily lead to unreliable estimates of recovery
and can be minimized by:

• Spiking the base material with concentrations similar to the endogenous analyte
concentration.

• Performing several replicates.

Sometimes it is useful to carry out recovery experiments with a range of spiked values
and using several different base materials.

Question Q 15(1)

A plasma glucose assay involved adding 0.1 mL of sample to 2.0 mL of reagent,

incubating at 37oC for 30 minutes then measuring the absorbance at 500 nm. Using
0.1 mL of water as sample the absorbance reading was 0.080, using 0.1 mL of a 10
mmol/L glucose standard it was 0.320 and using a patient’s plasma sample 0.200.

0.1 mL of a 50 mmol/L glucose solution was mixed with 0.9 mL of the same plasma
sample, then 0.1 mL of the mixture taken through the assay. The absorbance was 0.300.
Calculate the recovery of the method.

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Answer Q 15(1)

First calculate the concentration of glucose in the un-spiked plasma sample:

Assuming that absorbance is directly proportional to glucose concentration:

Plasma glucose (mmol/L) = Plasma absorbance

Glucose standard (mmol/L) Absorbance of standard

Rearranging:

Plasma glucose (mmol/L) = Plasma absorbance x glucose standard (mmol/L)

Absorbance of standard

Subtract the blank (water used as sample) absorbance from both standard and plasma
readings and substitute into the above equation:

Plasma glucose = (0.200 - 0.080) x 10 = 0.12 x 10 = 5 mmol/L

(0.320 - 0.080) 0.24

It is not necessary to calculate absolute amounts of glucose in the samples in order to

calculate recovery; concentrations can be used:

Recovery (%) = Recovered concentration x 100

Concentration added

However, by mixing the diluted glucose solution with the plasma sample the base
concentration and the spiked concentration has been diluted and must first be calculated:

Glucose from plasma = Plasma glucose (mmol/L) x Plasma volume (mL)

[Volume of plasma (mL) + volume of glucose solution (mL)]

= 5 x 0.9 = 4.5 mmol/L

(0.9 + 0.1)

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Glucose added = Concn of glucose solution (mmol/L) x Volume of glucose added (mL)
[Vol of glucose added (mL) + Volume of plasma (mL)]

= 50 x 0.1 = 5 mmol/L
(0.9 + 0.1)

Next calculate the measured glucose in the spiked plasma sample – remembering to
subtract the reagent blank:

Spiked sample glucose concentration = Spiked sample absorbance x Standard concn

Absorbance of standard

= (0.300 – 0.080) x 10
(0.320 – 0.080)

= 0.22 x 10 = 9.17 mmol/L

0.24

Calculate the recovery from the measured glucose concentration in the spiked plasma
(9.17 mmol/L), the base plasma sample (4.5 mmol/L) and the glucose added (5 mmol/L):

Recovery (%) =

[Glucose in spiked plasma (mmol/L) - glucose in base plasma (mmol/L)] x 100

Glucose added to base plasma (mmol/L)

= (9.17 – 4.5) x 100 = 4.67 x 100 = 93% (2 sig figs)

5 5

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Elimination of a tumour marker

Frequently tumour markers are measured following surgical resection of the tumour to
provide evidence that removal of the tumour is complete. If the tumour has been
completely removed then the concentration of the tumour marker should reflect the value
expected from the natural decay of the compound present at the time of surgery. If the
concentration has not fallen to this level then it is likely that the marker is still being
produced by residual tumour tissue (or from secondaries).

Tumour markers are usually cleared exponentially as are most drugs (see Chapter 7).
If

Cpt = concentration of tumour marker at time t

Cp0 = initial concentration of tumour marker
kd = elimination rate concentration

then as was shown in Chapter 7, a linear expression can be derived relating these
variables:

ln Cpt = ln Cp0 - kd.t ………………………….. Eq. 7.7

Provided the rate constant kd is known then the concentration at any given time can be
calculated or the time taken to reach a specified concentration estimated. The latter may
be particularly useful in predicting the time when the concentration of tumour marker
should be below the upper reference limit.

There are ways of manipulating this equation to give an expression that is simpler to
apply in common situations. Subtraction of one logarithm from another is the same thing
as calculating the logarithm of their ratios:

Therefore ln Cpt - ln Cp0 = ln (Cpt/Cp0) = ln CR

Where CR is the ratio of the concentration at time t (CPt) to the initial concentration
(Cp0). Eq. 7.7 can then be re-written as:

ln CR = - kd.t …………………….. Eq. 15.2

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Further simplification is possible. In Chapter 7 it was shown that the elimination rate
constant is related to the half-life (t1/2):

t 1/2 = 0.693 ……………. Eq. 7.9

kd

Substitution of kd = 0.693/t1/2 into expression Eq 15.2 gives:

ln CR = - 0.693. t …………. Eq. 15.3

t1/2

If the time (t) is expressed as the number of half lives (N) so that N = t/t½ then an even
simpler expression is produced:

ln CR = - 0.693 N ………… Eq. 15.4

If logarithms to the base 10 are used then 0.693 is divided by 2.303 (since ln CR = 2.303
log10 CR) and the expression becomes:

log10 CR = - 0.30 N ………… Eq. 15.65

Using this expression it is quite simple to determine the number of half-lives required for
a given change in concentration ratio or if the absolute time period is known then the
half-life of the tumour marker can be determined.

Question Q 15(2)

A tumour marker X is used to guide a decision on chemotherapy after the resection of the
main tumour mass. The concentration decays exponentially. If the half-life of the
tumour marker is less than 75 hours, then this is indicative of tumour clearance and
chemotherapy is withheld. If the half-life is greater than this, it indicates that residual
disease is present and chemotherapy is indicated. The precision of the assay is such that
measurements can be safely made at a precisely timed interval of more than 36 hours
from two or more days after surgery.

The level of X at 50 hours post surgery is 1756 ng/L and at 94 hours it is 1050 ng/L.
Calculate the half-life and indicate whether you can say with confidence whether
chemotherapy needs to be given.

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Answer Q 15(2)

There are two ways of solving this problem:

1. Using Eq 7.7:

ln CPt = ln Cp0 - kd.t

The 1st sample (collected at 50 h) can be regarded as the initial sample. Therefore:

CPt = tumour marker concentration at 94 h = 1050 ng/L

Cp0 = tumour marker concentration at 50 h = 1756 ng/L
t = time interval between the two samples = 94 - 50 = 44 h

Substituting these values into Eq. 7.7 allows calculation of kd:

ln 1050 = ln 1756 - 44 kd

6.96 = 7.47 - 44 kd

44 kd = 7.47 - 6.96 = 0.51

kd = 0.51 = 0.0116 h-1

44

Use Eq. 7.9 to covert the elimination rate constant (kd) to the half life (t½):

t½ = 0.693 = 0.693 = 60 h (2 sig figs)

kd 0.0116

Since the half-life is less than 75 h, the time interval is 44 h and the 1st sample was
taken at least 48 h after tumour removal we can conclude that chemotherapy can
be withheld.

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2. Using Eq 15.5

log10 CR = - 0.30 N

CR = concentration ratio = Cpt = 1050 = 0.598

Cpo 1756

Substituting CR = 0.598 gives: log10 0.598 = - 0.30 N

Rearranging and solving for N:

N = - log10 0.598 = - (- 0.223) = 0.744 half lives

0.30 0.30

Thus the concentration fell from 1756 ng/L to 1050 ng/L in 0.744 half-lives OR
44 h. This information can be used to calculate the half-life:

N = t/t½ so that t½ = t/N.

Substitute t = 44 h and N = 0.744 and solve for t½:

t½ = 44 = 59 days (2 sig figs)

0.744

Radioactive decay

Radioactive decay follows first order kinetics and therefore obeys the same mathematical
laws as the clearance of the drug (or elimination of a tumour marker discussed in the
preceding section). Equations Eq. 7.7 and 15.5 can be used if units of radioactivity are
substituted for concentration.

Question Q 15(3)

If the half life of a radionucleotide is 20 hours at the end of how many complete days will
the activity have fallen to less than 2% of the initial value?

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Answer Q 15(3)

Using equation Eq. 15.5:

log10 CR = - 0.30 N

Rearrangement gives:

N = - log10 CR
0.30

Since the activity falls from 100% to 2% of the initial value, then these figures can be
treated as concentrations Cp0 and Cpt respectively:

CR = Cpt = 2 = 0.02
Cp0 100

Substituting CR = 0.02 permits calculation of N:

N = - log10 0.02 = - (-1.70) = 1.70 = 5.67

0.30 0.30 0.30

Therefore the activity reached 2% of the initial value after 5.67 half lives have elapsed.
Multiplication of 5.64 by the half-life (20 h) gives the total time period (t) in hours.

t = 5.67 x 20 = 113.4

Division by 24 gives the number of days:

t = 113.4 = 4.7 days

24

Therefore 5 complete days must elapse before the activity falls below 2% of the initial
value.

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Exponential growth

The mathematics of exponential growth are the same as exponential decay or elimination
except that concentration increases with time. The elimination constant (kd) is replaced
by the specific growth rate (k) and the term -kd.t becomes positive (+k.t). As a
consequence Eq. 7.7 and Eq. 15.5 become:

ln Cpt = ln Cp0 + k.t ……… Eq. 15.6

log10 CR = 0.30 N ……………. Eq. 15.7

The half-life (t½) is replaced by doubling time (td) so that Eq. 7.9 can be re-written:

td = 0.693 ………………. Eq. 15.8

k

Question Q 15(4)

A woman had a beta hCG concentration measured at 265 IU/L and 11 days later,
following some abdominal pain, it was 820 IU/L. Assuming hCG rises exponentially in
early pregnancy, what has been the doubling time over this period? What is the
significance of the result you obtain?

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Answer Q 15(4)

This problem can be solved by direct application of Eq. 15.7:

log10 CR = 0.30 N

Rearranging gives: N = log10 CR

0.30

CR can be calculated taking the initial concentration (Cp0) as 265 IU/L, and the Cpt as
820 IU/L:

CR = Cpt = 820 = 3.095

Cp0 265

Substituting CR = 3.095 and solving for N:

N = log10 CR = log10 3.095 = 0.491 = 1.64

0.30 0.30 0.30

The doubling time (td) can be calculated from N (1.63) and the time period between
measurements (t = 11 days) using Eq. 15.8:

td = t = 11 = 6.7 days
N 1.63

The normal doubling time for hCG during early pregnancy is approximately 2 days.
Therefore this result is consistent with ectopic pregnancy.

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Urinary nitrogen excretion and nitrogen balance

Measurements of urinary nitrogen excretion are often used to provide an estimate of

nitrogen balance in patients on pareneteral nutrition. Nitrogen intake can be calculated
from the amount of feeding solutions given but measurements of total urinary nitrogen
are cumbersome and not widely available. Instead nitrogen excretion is estimated from
the 24 urinary urea excretion. The underlying assumption is that all amino acid nitrogen
is converted to urea, is not excreted by any other route (i.e. the gut, sweat or fistulae) and
that there is no other significant nitrogen loss in the urine (i.e. amino acids, ammonia etc).

Urea has the formula CO(NH2)2 so that each molecule contains two atoms of nitrogen.
Therefore each mmol of urea contains 2 mmol of nitrogen and since the atomic weight of
nitrogen is 14, this constitutes 2 x 14 = 28 mg of nitrogen. Division by 1000 converts
this figure to g of nitrogen:

Nitrogen excretion (g/24 h) = Urea excretion (mmol/24 h) x 28 ……. Eq. 15.9

1000

It has been advocated that a figure of 20% should be added to nitrogen excretion to allow
for other urinary losses and a further 2 g/day added to allow for losses by other routes.

Subtraction of this figure from the nitrogen intake gives the nitrogen balance:

Nitrogen balance = Nitrogen intake - Nitrogen excretion …….. Eq. 15.10

(g/24 h) (g/24 h) (g/24 h)

Question Q 15(5)

A patient receiving total parenteral nutrition is receiving 12 g nitrogen/24 h as amino

acids. Urinary urea excretion is 600 mmol/24 h. Indicating what assumptions you make,
calculate whether she is in positive or negative nitrogen balance.

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Answer Q15(5)

Use Eq. 15.9 to calculate nitrogen excretion from urea excretion:

Urinary nitrogen excretion (g/24 h) = Urinary urea excretion (mmol/24 h) x 28

1000

= 600 x 28 = 16.8 g/24 h

1000

Use Eq. 15.10 to calculate nitrogen balance:

Nitrogen balance (g/24 h) = Nitrogen intake (g/24 h) - Nitrogen excretion (g/24 h)

= 12 - 16.8 = - 4.8 g/24 h

Correcting for other urinary losses (+20%) and other routes of nitrogen excretion (+2 g)
gives a revised value for nitrogen excretion:

Corrected nitrogen excretion = (16.8 x 120) + 2 = 22.2 g/24 h

100

So that the corrected nitrogen balance becomes:

Corrected nitrogen balance = 12 - 22.2 = - 10.2 g/24 h

Tritratable gastric acidicity

Acid secretion studies are less frequently performed nowadays, the main remaining
application is in the follow-up of a raised plasma gastrin result. The documentation of a
raised basal acid output (BAO) in gastric juice provides strong evidence that a high
plasma gastrin concentration is caused by Zollinger-Ellison syndrome. After an overnight
fast gastric juice is collected over a timed period (usually 30 min), its volume measured
and an aliquot titrated with standardised sodium hydroxide:

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HCl + NaOH → NaCl + H20

In gastric Known
fluid concentration

There has been considerable debate over the years as to which pH to use as the endpoint
since secreted hydrochloric acid may be buffered by gastric proteins.

At the equivalence point the total amount of acid in the gastric fluid aliquot is equal to the
amount of sodium hydroxide added. The total amount of acid or alkali is equal to the
volume used multiplied by its concentration. Therefore, we can write:

M1 x V1 = M2 x V2 …………….. Eq. 15.11

where M1 = molar concentration of hydrochloric acid in the gastric fluid

V1 = volume of aliquot of gastric fluid used in the titration
M2 = molar concentration of sodium hydroxide
V2 = titre of sodium hydroxide solution

Eq. 15.11 can be rearranged to determine M1 which can be converted to the total acid
secreted and finally the rate of acid secretion.

Question Q 15(6)

During a test of gastric acid secretion, 28mL of gastric juice was aspirated over a single
30 min period from a fasting patient before the administration of pentagastrin.
The volume of 0.1 M sodium hydroxide solution required to titrate a 5 mL aliquot of the
gastric juice to pH 7.4 was 14 mL. Calculate the basal acid secretion rate in mmol/h.

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Answer Q 15(6)

Since the answer is required in mmol/h it is easiest to work in mmol from the outset.
Use Eq. 15.11:

M1 x V1 = M2 x V2

M1 = millimolar concentration of acid in gastric fluid = ? mmol/L

V1 = volume of gastric fluid aliquot used in titration = 5 mL
M2 = millimolar concentration of sodium hydroxide = 0.1 x 1000 = 100 mmol/L
V2 = titre of sodium hydroxide solution = 14 mL

Substitute for V1, M2 and V2 in Eq. 15.11 and solve for M1:

M1 x 5 = 100 x 14

M1 = 100 x 14 = 280 mmol/L

5

Multiply the acid concentration (280 mmol/L) by the gastric fluid volume in litres (28 mL
divided by 1000) to give the total acid output in millimols:

Total acid output = 280 x 28 = 7.84 mmol

1000

Multiply by 2 (since gastric fluid was collected over 30 min) to obtain the secretion rate
in mmol/h:

Secretion rate = 7.84 x 2 = 16 mmol/h (2 sig figs)

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Internal standardisation

Chromatographic techniques are often subject to variability (e.g. due to instability of the
detector) or to unpredictable losses (due to the preliminary sample preparation which may
involve extraction and/or derivatization). Addition of an internal standard, which is a
compound chemically related to the analyte with similar properties but resolvable by
chromatography, can be used to minimize these problems. Instead of plotting the peak
height or area of the standard against standard concentration, the ratio of peak height (or
area) of the standard to that of the internal standard is used. An example is shown in
Figure 15.1 in which both methods of plotting the standard curve are used. The same
amount of internal standard is also added to each of the unknown samples, the peak
height or (area) ratio similarly calculated and the results obtained by interpolation with
the standard curve.

The underlying assumption of this approach is that the internal standard behaves exactly
as the analyte being measured i.e. the ratio of analyte to internal standard remains
constant throughout the various stages of the analytical process.

Question Q 15(7)

An HPLC method for estimation of plasma phenylalanine uses N-methyl 1-phenylalanine

(NMP) as internal standard. 200 μL of NMP has been added to 200 μL aliquots of
standard or sample prior to analysis. The following peak areas were obtained:

Sample Peak area

NMP Phenylalanine

Standard (500 μmol/L) 20,000 81,000

Patient 18,000 120,000

Calculate the phenylalanine concentration in the patient’s sample.

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(a) 2 mg/L
2

3 mg/L

S
i 1 mg/L
g 1 1mg/L
n
a 1 mg/L
l
1 mg/L

0 2 4 0 2 4 0 2 4
Retention time (min)

(b)
Area x
or
peak area
ratio x x
x

xx

0 1 2 3

Standard concentration (mg/L)

Figure 15.1 a) Chromatographs of standards at three levels, spiked with equal

amounts of internal standard (IS), b) Standard curves of standard
peak area and standard:internal standard peak area ratios

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Answer Q 15(7)

First calculate the peak area ratio (PAR) of the phenylalanine peak to that of the internal
standard (NMP) for the standard and patient:

PAR (Standard) = 81, 000 = 4.05

20,000

PAR (Patient) = 120,000 = 6.67

18,000

Assuming that PAR is proportional to concentration

PAR (standard) = PAR (unknown)

Concn (standard) Concn (unknown)

Which can be rearranged to give:

Concn (unknown) (μmol/L) = PAR (unknown) x Standard concn (500 μmol/L)

PAR (Standard)

Patient phenylalanine concn = 6.67 x 500 = 823 μmol/L

4.05

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Population genetics – the Hardy-Weinberg equilibrium

Consider an ideal population in which there is a gene locus with two alleles A and a with
gene frequencies p and q i.e:

p = frequency of the dominant allele (A)

q = frequency of the recessive allele (a)

Mating results in the random combination of a male gamete (A or a) with a female

gamete (A or a). The possible results, with their frequencies, can be obtained by
constructing a Punnett’s square (Fig 15.2):

Male gametes

A a
(p) (q)

A (p) AA Aa
Female (p2) (pq)
gametes
a (q) Aa aa
(pq) (q2)

Figure 15.2 Punnetts square showing the allele frequencies and resulting genotype
frequencies for a two allele system (A and a with frequencies p and q )
in the first generation

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Therefore the recombination of A and a gametes from both parents results in three
genotypes – AA, Aa and aa with frequencies p2, 2pq and q2. Since the frequencies of all
genotypes must add up to 1 the following useful expressions can be written:

For a pair of alleles: p + q = 1 …………… Eq. 15.12

For genotypes: p2 + 2pq + q2 = 1 …………… Eq. 15.13

Mathematically, Eq. 15.13 is the expansion of (p + q)2.

Subsequent mating of this population to produce a second generation can only produce
offspring with genotypes AA, Aa or aa. The frequencies can be calculated from the
frequencies of the possible mating combinations:

For example AA can arise from the combinations AA x AA, AA x Aa and Aa x Aa. The
frequencies can be derived from the frequencies derived after the first cross as follows:

AA x AA has only one possible outcome (AA) , therefore frequency of AA p2 x

p2 = p4

AA x Aa can give rise to AA in two ways: A from the male or A from the female.
The total initial frequency of Aa is 2pq (one pq from males and one pq from
females). Therefore the frequency of AA in this cross is p2 x 2pq = 2p3q

Aa x Aa. The initial frequency of each of these is 2pq, so that the frequency of the
A gamete from each parent is pq . Therefore frequency of AA in the second cross
= pq x pq = p2q2

Adding the frequency of Aa offspring gives p4 + 2p3q + p2q2, which can also be
written as factors with p2 outside the bracket: p2 (p2 + 2pq + q2). Since we know
that (p2 + 2pq + q2) = 1 (Eq. 15.14), the frequency of AA remains at p2.

Repeating this process for all the possible matings in the second generation gives the data
in Fig 15.3. The frequency of the various genotypes after this second mating remains
unchanged. This calculation could be repeated for further generations but the frequencies
of genotypes AA, Aa and aa would always remain at p2, 2pq and q2.

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Frequency of offspring
Mating type Frequency
AA Aa aa

AA x AA p4 p4 - -
AA x Aa 4p3q 2p3q 2p3q -
Aa x Aa 4p2q2 p2q2 2p2q2 p2q2
AA x aa 2p2q2 - 2p2q2 -
Aa x aa 4pq3 - 2pq3 2pq3
aa x aa q4 - - q4

Total p2(p2+2pq+q2) 2pq(p2+2pq+q2) q2(p2+2pq+q2)

Relative frequency p2 2pq q2

Figure 15.3 Frequency of various types of offspring due to different matings in the
second generation

This principle was discovered by Godfrey Hardy and Wilhelm Weinverg and is known
as the Hardy-Weinberg equilibrium:

“Gene frequencies and genotype ratios in a randomly breeding population

remain constant from generation to generation”.

For this equilibrium to hold true seven conditions must be met:

• mutation is not occurring

• natural selection is not occurring
• the population is infinitely large
• all members of the population breed
• all mating is totally random
• everyone produces the same number of offspring
• there is no migration in or out of the population

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Identity of p and q for a population depends on the phenotypic expression of the alleles:

1. In-complete dominance i.e. homozygosity for the dominant gene is required in

order for the phenotype to be expressed. In this instance the proportion of the
population expressing the genotype (AA) is the same as p2. p is then the square
root of this value. Subtraction of p from 1 gives q (Eq. 15.12).

2. Complete dominance i.e. both homozygotes (AA) and heterozygotes (Aa) express
the phenotype. Therefore, the proportion of individuals without this phenotype
gives the value of q2. q is the square root of this value. Subtraction of q from 1
gives p.

3. Autosomal recessive i.e. only those individuals homozygous for the recessive
gene (aa) express the phenotype. Therefore the proportion of individuals with this
phenotype represent the q2 frequency. q is the square root of this value.
Subtraction of q from 1 gives p.

Question Q 15(8)

Phenylketonuria is inherited in an automal recessive manner. If the prevalence of the

disorder is 10 in 10,000 estimate the percentage of heterozygous carriers in the
population.

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Answer Q 15(8)

For an autosomal recessive mode of inheritance the frequency of the phenylketonuria

phenotype represents the autosmal recessive genotype i.e. q2

q2 = 1 in 10,000 = 1 = 0.0001
10,000

q = √ 0.0001 = 0.01

Using Eq. 15.12: p + q = 1

p = 1 - q = 1 - 0.01 = 0.99

The carriers are heterozygous with a frequency of 2pq. Therefore:

Frequency of carriers = 2 x 0.99 x 0.01 = 0.0198

Multiply by 100 to convert to percentage = 0.0198 x 100 = 1.98 (approx 2%)

When data can be broken down into a set of compartments (i.e. grouped frequencies), for
each of which there is an observed number (O) and an expected (E) number of
individuals, the goodness-of fit- (Χ2) is given by:

Χ2 =
Σ ( O - E )2
E
…………… Eq. 15.14

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Care must be taken to use the correct number of degrees of freedom. For a contingency
table of r rows and c columns there are (r - 1)(c - 1) degrees of freedom. If the data is
only grouped into k classes then there are (k – 1) degrees of freedom. The probability of
obtaining any value of Χ2 can be obtained by looking up its P value in tables of Χ2.
In general, every additional restraint removes one degree of freedom.

The Χ2 test is extremely useful in genetics to assess whether the distribution of genotypes
fits the predicted pattern. For example if the observed and expected (i.e. predicted)
frequencies were:

Genotype Observed Expected

AA 80 100
Aa 240 200
Aa 80 100

Χ2 = Σ (O - E )2
E
= (80 - 100)2 + (240 - 200)2
100 200
+ (80 - 100)2
100

= 16.0 ( 2 degrees of freedom)

From tables of Χ2, the P value for a Χ2 value of 16 with 2 degrees of freedom is less than
0.001. Therefore the observed frequencies do not fit with the predicted model.

It is very important to appreciate that the Χ2 test should ONLY be used for grouped
frequencies. This is a frequently abused test. In spite of the fact that Eq. 15.14 uses the
terms “observed” and “expected” results, it should NOT be used evaluate data from
method comparison studies.

Question Q 15(9)

The prevalence of a metabolic disease which is inherited in an autosomal recessive

manner due to a single allele is 1 in 10,000. A survey identified 1% of the population as
asymptomatic carriers. Are these data consistent with the population in a
Hardy-Weinberg equilibrium?

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 MISCELLANEOUS TOPICS

Answer Q 15(9)

Let the dominant gene for the disorder be A and the recessive gene a. As the inheritance
of the disease is autosomal recessive only the homozygous recessive genotype (aa)
expresses the disease.

The incidence of the receive disorder (aa) = 1 in 10,000 = 1 = 0.0001

10,000

The incidence of carriers (Aa) is 1 % = 1 = 0.01

100

Since the total must equal 1, the incidence of the homozygous dominant genotype
(which does not express disease nor have carrier status) can be calculated:

Incidence of homozygous dominant (AA) = 1 - (0.0001 + 0.01)

= 1 - 0.0101 = 0.9899

To summarise the observed frequencies of the three genotypes:

Genotype AA Aa aa
Observed frequency 0.9899 0.01 0.0001

To calculate the expected frequency if Hardy-Weinberg equilibrium is present, determine

values for p and q then calculate the expected frequencies for the three genotypes.

Frequency of affected individuals (aa) = 0.0001 = q2

Therefore q = √ q2 = √ 0.0001 = 0.01

Since p + q = 1, p = 1 - q

Therefore p = 1 - 0.01 = 0.99

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Using these values of p and q the expected frequencies can be calculated as follows:

Frequency of AA = p2 = 0.992 = 0.9801

Frequency of Aa = 2pq = 2 x 0.99 x 0.01 = 0.0198

Frequency of aa = q2 = 0.012 = 0.0001

If all the data is tabulated then Χ2 can be calculated at the same time:

Genotype Frequency (O - E) (O - E)2 (O - E)2/E

Observed Expected

AA 0.9899 0.9801 0.0098 0.00009604 0.00009799

Aa 0.01 0.0198 - 0.0098 0.00009604 0.00485051

Aa 0.0001 0.0001 0.0000 0.00000000 0.00000000

Total 1.0000 1.0000 0.0000 0.00019208 0.00497509

Χ2 is the sum of all the values in the final column = 0.0050 (2 sig figs)

Normally the degrees of freedom would be 3 - 1 = 2. However, since one of the

observations (frequency of the disease) was used to estimate the expected values,
a further degree of freedom is lost leaving only one.

From tables, the value of P for Χ2 = 0.0050 is somewhere between 0.05 and 0.95.
Therefore there is no significant difference between the observed and expected values so
that the data fit with the Hardy-Weinberg equilibrium.

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Competitive binding assays

Many assays in laboratory medicine utilise the property of the analyte in question to bind
specifically and reversibly to a protein. More often than not this protein is an antibody
and so these techniques are known as immunoassays. The antibody (Ab) binds to the
analyte in question, the antigen (Ag) to form the antibody-antigen complex (AbAg)
according to the equilibrium:

k1
Ab + Ag AbAg
k-1

K = k1 = [AbAg]
k-1 [Ab] [Ag]

where
k1 = the rate constant for the forward reaction
k-1 = the rate constant of the reverse reaction
K = the binding constant for the overall reaction

If antigen is labelled in some way (Ag*) and if the label does not interfere with binding
then both the labelled and unlabelled antigen compete for the binding sites according to
the scheme:

Ab + Ag* AbAg*
+
Ab

AbAg

This is analogous to the scheme for competitive inhibition of an enzyme except that the
antibody-antigen complexes do not break down to produce reaction products.

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Since the labelled and unlabelled antigen compete for the same binding sites it follows
that if the biological sample containing unlabelled antigen and a reagent with labelled
antigen are both added to a solution containing the antibody then the proportion of
AbAg* and AbAg formed will reflect the proportion of Ag* and Ag in the reaction
mixture. In fact the concentration of AbAg* will be inversely proportional to the
concentration of Ag in the sample. This is the basic principle of all competitive binding
assays. A plethora of techniques have been developed which differ in the nature of the
label used, whether separation of bound from free label is necessary (and the technique
used to achieve this), the nature of the binding protein etc. Additionally non-competitive
assays have been developed.

The simplest type of immunoassay is probably the radioimmunoassay in which the

analyte competes for antibody binding sites with an antigen to which a radioisotope label
has been attached. At equilibrium we are left with a reaction mixture which contains free
analyte (Ag), bound analyte (AbAg), free label (Ag*), bound label (AbAg*) and free
antibody (Ab). In order to be able to measure the radioactivity due to bound label alone it
is necessary to separate bound from free label i.e. AbAg* from Ag*. This is usually
achieved by precipitation (e.g. using a second antibody or polyethylene glycol). The
radioactivity in the precipitate is a measure of AbAg*. The higher the concentration of
antigen in the sample (Ag) the lower the radioactivity in the precipitate. Unfortunately
there is no simple mathematical relationship between the analyte concentration in the
sample (Ag) and the radioactivity in the precipitate. Numerous complex mathematical
procedures have been devised to enable calculation of analyte concentration from the
count rate which are beyond the scope of this book. However, in many cases it is
possible to obtain a reasonable linear relationship over a limited working range for the
assay by plotting the proportion of label bound (usually expressed as a percentage)
against the logarithm of analyte concentration (Fig 15.4). It is usual to carry out all
measurements in duplicate to minimize imprecision.

In order to express the count rate (B) as a percentage of the maximum binding (B0) the
total binding (TB) is determined using a standard containing zero concentration of analyte
so that there is no competition for binding sites. In practice there is always a small
amount of non-specific binding (NSB) of label to the walls of the reaction tube. NSB is
assessed by setting up a tube containing label but no sample or antibody which is then
carried through the entire assay. The count rate for the NSB tube is subtracted from that
of the TB tube in order to obtain a total binding value:

B0 = TB - NSB ……………… Eq. 15.15

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 MISCELLANEOUS TOPICS

%
Bound
A

Log concentration

Figure 15.4 Schematic diagram for the dose response curve of a typical
radioimmunoassay. The area between A and B is the analytically
useful range of the assay

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The NSB counts are also subtracted from the count rate for each standard and sample
before expression as a percentage of B0:

B/B0 (%) = (Standard/sample cpm - NSB) x 100 …….. Eq. 15.16

B0

A standard curve is then plotted of B/B0(%) versus the logarithm of concentration.

Values for unknowns are then obtained from the standard curve in the usual way
(remembering to take the antilogarithm of the result). Nowadays, computer programmes
are usually used to carry out the entire calculation including fitting the best curve to the
standard data.

It is customary to set up a tube containing label alone which is not taken through the
entire assay procedure but used to obtain a measure of the total count rate (TC). This,
together with the B0 result is a useful quality control tool to ensure that the antibody is
able to bind a reasonable amount of label.

Question Q 15(10)

The following data were obtained for a plasma cortisol radioimmunoassay using second
antibody separation. Calculate the cortisol concentration in the serum sample.

Sample Primary Label 2nd antibody Average cpm

Antibody
None - + - 12,500
Buffer - + + 200
Buffer + + + 8,050
50 nmol/L standard + + + 4,250
100 “ “ + + + 3,720
200 “ “ + + + 3,190
400 “ “ + + + 2,675
800 “ “ + + + 2,120
1600 “ “ + + + 1,600
3200 “ “ + + + 1,065
Serum + + + 2,490

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Answer Q 15(10)

First identify tubes to be used for total counts (TC), non-specific binding (NSB) and total
binding (TB).

TC is the tube which contains label and nothing else (it is not used in the calculation of
results). This is the first tube, therefore TC = 12,500 cpm.

NSB is the tube which contains label, buffer and second antibody and is taken through the
entire assay procedure. This is the second tube so NSB = 200 cpm.

TB is the tube which contains everything except the analyte in question (i.e. buffer
instead of sample). This is the third tube, therefore TB = 8,050 cpm.

Maximum binding (B0) when analyte is absent is obtained by subtraction of NSB from
TB:

B0 = TB - NSB = 8,050 - 200 = 7,850 cpm.

The NSB is also subtracted from each standard or sample count rate before it is expressed
as a ratio to B0 (See Eq. 15.16). This is best laid out in tabular form:

Sample Log10 conc cpm cpm – NSB B/B0(%)

TC 12,500
NSB 200
TB 8,050 7,850 100
Standard 50 nmol/L 1.70 4,250 4,050 51.6
“ 100 “ 2.00 3,720 3,520 44.8
“ 200 “ 2.30 3,190 2,990 38.1
“ 400 “ 2.60 2,675 2,475 31.5
“ 800 “ 2.90 2,120 1,920 24.5
“ 1600 “ 3.20 1,600 1,400 17.8
“ 3200 “ 3.51 1,065 865 11.0
Serum 2,490 2,290 29.2

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60

50 Cortisol standard curve

40
B /B 0 (%)

30

20

10

0
1.5 2 2.5 3 3.5 4
Log cortisol conentration

Note that the value for B0 (100%) cannot be plotted since the log of zero (concentration
for TB) has no meaning.

Serum sample B/B0 (%) = 29.2.

From standard curve corresponding log concentration = 2.7

Therefore cortisol = antilog 2.7 = 500 nmol/L

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 MISCELLANEOUS TOPICS

ADDITIONAL QUESTIONS

1. A 0.5 mL sample of urine is extracted into dichloromethane. An aliquot of the

extract is analysed by HPLC and found to give an apparent original concentration
of 320 nmol/L of analyte Y. 100 µL of Y standard with a concentration of 880
nmol/L is added to a further 0.5 mL sample of the same urine and the sample
mixed. 0.5 mL of the mixed sample is then processed as before, giving a
measured concentration of 405 nm/L. Calculate the recovery of analyte Y.

2. A new method for HCG in urine is being evaluated. The concentration in a

sample from a pregnant woman is measured at 8240 IU/L. A 50 µL aliquot of an
international standard containing 50,000 IU/L is added to 450 µL of the same
urine sample and the sample mixed. On measuring the mixed sample, the new
concentration is found to be 12100 IU/L. What is the recovery of HCG by this
method?

3. Measurement of plasma AFP is used to monitor a patient with a teratoma. If the

initial concentration was 10,200 U/L what plasma level would you expect to find
21 days after successful surgery? Assume the half-life of AFP is 5.5 days.

4. A radioisotope has a half-life of 21 days. How long will it take for the activity to
fall to 10% of the initial value?

5. In normal pregnancy serum beta hCG has a doubling time of approximately

2 days. How long will it take for the serum level to increase ten-fold?

6. A patient receiving parenteral nutrition is receiving 11.8 g nitrogen/24 h as amino

acids. Urinary urea excretion is 580 mmol/24 h. Indicating what assumptions
you make, calculate whether she is in positive or negative nitrogen balance.

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7. A 30 min basal gastric secretion sample (total volume 27 mL) required 2.5 mL of
0.1 M NaOH to titrate 5 mL of the material to pH 7.4. Calculate the basal acid
secretion rate in mmol/h.

8. A five day faecal fat collection was homogenised and diluted to 1500 mL.
A 10 mL aliquot of the homogenate was subjected to hydrolysis and the
fatty acids were extracted. The volume of 0.05 M sodium hydroxide required to
effect neutralisation was 48 mL. Calculate the fat excretion in mmol/24 h.

9. Gas chromatography for a drug involves adding equal amounts of internal

standard to standard or sample prior to analysis. The following peak areas were
obtained:

Sample Peak area

Internal standard Drug

Standard (200 nmol/L) 50,000 200,000

Patient 40,000 150,000

Calculate the drug concentration in the sample.

10. Genotyping of a group of 100 unrelated individuals for a two-allele polymorphism

showed that the allele frequencies were:

A 0.65
B 0.35

Calculate the expected percentages of heterozygotes (AB) and homozygotes

(AA and BB) in the group.

11. The prevalence of an inherited metabolic disease (inherited in an autosomal

recessive manner due to a single allele) is 1 in 2,500. A survey identified 1 in 50
of the population as asymptomatic carriers. Is this finding consistent with a
population in a Hardy-Weinberg equilibrium?

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 MISCELLANEOUS TOPICS

12. The following data were obtained for a digoxin radioimmunoassay employing
PEG precipitation of the primary antibody. The assay was performed in duplicate.
Calculate the digoxin concentration in the serum sample.

Sample Duplicate cpm

1 2
TC 15,100 15,900
NSB 320 380
TB 11,350 11,650
0.2 nmol/L standard 10,320 10,980
0.4 “ “ 9,250 8,340
0.8 “ “ 6,782 6,630
1.2 “ “ 5,104 5,890
2.4 “ “ 3,700 3,430
4.8 “ “ 1,350 1,650
Patient serum 4,350 5,000

331
CHAPTER 15

332
 ANSWERS TO FURTHER QUESTIONS

Appendix I

Answers to further questions

Chapter 1 Chapter 2

1. a) 1.25 g/L 1. 7g
b) 250 mmol/L 2 1453 mmol/L
c) 0.000236 μmol/L 3. 0.25 g
d) 1600 ng/mL 4. 100 mL
2. a) 6.7 mmol/L 5. 101
b) 2.0 mmol/L 6. 8.8 mL
c) 7.5 mmol/L 7. Potassium = 0.022 mol/L
d) 58 μmol/L Sodium = 0.57 mol/L
3. a) 360 mg/100 mL Chloride = 0.59 mol/L
b) 6.4 mEq/L 8. 950 mL
c) 86 mg% 9. 5.2 mmol/L
d) 2.8 mg% 10. Phosphate = 100 mmol/L
4. a) 1.5 mmol/L 40 mL needed
b) 12.5 μmol/L
c) 2.5 g/L
d) 3.25 x 10-3 μmol/L
5. a) 0.30 mmol
b) 25 μmol/min/250 mL
6. 3.0 x 10-10 mol/L

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APPENDIX I

Chapter 3 Chapter 4

1. 0.86 1. 207 mL
2. 36 to 45 nmol/L 2. a) 0.022
3. 25:1 b) 0.125
4. 6.80 c) 0.301
5. Na2CO3 = 7.58 g d) 0.602
NaHCO3 = 2.39 g e) 1
6. 39.8 g sodium lactate f) 2
0.22 mL lactic acid 3. a) 79 %
7. 49 mmol b) 56 %
8. 4.65 c) 32 %
d) 18 %
e) 10 %
f) 1 %
4. 7.35 L.mol-1.cm-1
5. 0.069
6. 153 nmol/g dry wt
7. 68 %
16.7 L.mol-1.cm-1
8. NADH = 53.5 μmol/L
NAD = 23.7 μmol/L
9. 97 %
10. Serum creat = 75 μmol/L
Urine creat = 7.5 mmol/L
11. Linear to 15 mmol/L
12.5 mmol/L

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 ANSWERS TO FURTHER QUESTIONS

Chapter 5 Chapter 7

1. 15.7 mmol/24 h 1. 40 h
2. 14.3 mmol/12 h 2. a) 39 L
3. Clearance = 170 mL/min b) 20 h
? >24h collection 3. 1.25 mg/L (total body water)
4. Filtration = 1584 mg/24 h 3.75 mg/L (ECF only)
? tubular reabsorption 4. a) 1.5 h
5. 500 mL/min? b) 59 L
? tubular secretion 5. 2.0 h (for 100 nmol/L)
6. 36 mmol/L 2.9 h (for 75 nmol/L)
7. Increased Na excretion 6. 6.6 h
by 192 mmol/24 h 7. 400 mg
8. 28 g Na (or 71 g NaCl)
9. 0.069 (= 6.9%)
10. 40 mL/min/1.73 m2
11. 9 mmol/L glomerular filtrate
12. 0.51 mL/min
13. GFR = 39 mL/min/ 1.73 m2
Clearance = 28 mL/min
? incomplete urine collection
Failure to correct clearance to
body surface area
Tubular secretion of creatinine

Chapter 6 Chapter 8

1. 293 mOsm/kg 1. Positive by 500 mL

2. 857 mOsm/kg 2. 6.6 L (or 5L)
3. 396 mOsm 3. Na+ decrease by 3 mmol/L
4. 20 mOsm/kg 4. 143 mmol/L
5. a) Gap = 52 mOsm/kg
b) Ethanol = 59 mOsm/kg
Ethanol explains gap.

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APPENDIX I

Chapter 9 Chapter 11

1. 276 IU/L 1. m = 101.1

2. a) 113 IU/L s = 10.6
b) 1.88 x 10-6 kat/L SEm = 3.4
3. 1.85 2. t = -1.75 (Not sig)
4. 476 F = 2.25 (Not sig)
5. 0.8 Vmax 3. 0.01
6. a) Km = 8.0 x 10-8 mol/L 4. Paired t = -1.98
b) Vmax = 1.9 x 10-7 mol/min Not significant
c) Vmax = 1.5 x 10-7 mol/min 5. F = 2.34
Km = 1.5 x 10-5 mol/L 6. Not significant
7. a) 1/[S] = 103 L/mol, 1/v = 106 min/mol
b) [S]/v = 103 min/L, [S] = 10-3 mol/L
c) v = 10-6 mol/min, v/[S] = 10-3 L/min
8. Competitive, Ki = 2.6 x 10-5 mol/L
9. Km @ pH 7.4 = 6.2 x 10-3 mol/L
Km @ pH 5.5 = 1.8 x 10-3 mol/L
Assuming equilibrium conditions
highest affinity at pH 5.5
10. Uncompetitive, Ki approx 4 x 10-3 mol/L

Chapter 10 Chapter 12

1. Mean = 70.25 g/L 1. a) 17.5 mmol/L

Variance = 7.82 g/L b) 166 mmol/L
SD = 2.80 g/L 2. OLD =
CV = 4.0% (NEW + 41)/0.65
95% limits = 64.8 - 75.7 g/L 3. r = 0.30 (Not sig)
2. 10 4. ? relationship linear
3. Least sig change = 14.8% ? sresidual (syx)
Actual change = 14.5%
Not significant
4. 0.74
5. Less than 6 mU/L
(0.50-5.51 mU/L)
6. a) ± 0.36 mL
b) ± 0.11 mL
7. 11.8%

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 ANSWERS TO FURTHER QUESTIONS

Chapter 13 Chapter 15

1. a) PV(+) = 0.95 or 95% 1. 94%

PV(-) = 0.95 or 95% 2. 94%
b) PV(+) = 0.004 or 4% 3. 723 U/L
PV(-) = 1.00 or 100% 4. 70 days
2. a) 16.5 (2 sig figs) 5. 6.6 days
b) 1990 6. Negative
c) Metanephrines since (-4.4 g/24 h or
no false negatives -9.7 g/24 h)
3. a) 0.49 or 49% 7. 2.7 mmol/h
b) 0.98 or 98% 8. 24 mmol fat/24 h
4. SENS = 0.84 or 84% as triglyceride
SPEC = 0.92 or 92% 9. 188 nmol/L
PV+ = 0.86 or 86% 10. Heterozygotes 45.5%
5. Prev = 40 % PV(+) = 0.84 or 84% Homozygotes 54.5%
Prev = 0.4 % PV(+) = 0.030 or 3.0% 11. Consistent
6. a) PV(+) = 0.33 or 33% X2 = 0.010 (not sig)
b) Pre-test odds = 0.053 12. Digoxin = 1.6 nmol/L
c) LR+ = 9.4
d) Post-test odds = 0.50
e) Post-test prob = 0.33 or 33%
7. 0.48 or 48%

Chapter 14

1. 0.88 or 88% (2 sig figs)

2. 95

337
APPENDIX I

338
 WORKED ANSWERS TO FURTHER QUESTIONS

Worked Answers to Further Questions

Chapter 1

(Atomic weights: C = 12; H = 1; O = 16; Ca = 40; N = 14)

1. Convert the following: a) 125 mg% to g/L; b) 0.25 mol/L to mmol/L; c) 0.236
nmol/L to μmol/L; d) 1.6 mg/L to ng/mL.

a) Convert 125 mg% to g/L

125 mg% is the same as 125 mg/100 mL

Multiply by 10 to convert volume from 100 mL to 1000 mL (i.e. 1 L)

Divide by 1000 to convert from mg to g (there are 1000 mg in one g)

125 mg% = 125 x 10 = 1.25 g/L

1000

b) Convert 0.25 mol/L to mmol/L

There are 1000 mmol in one mol. Therefore multiply by 1000.

0.25 mol/L = 0.25 x 1000 = 250 mmol/L

c) Convert 0.236 nmol/l to μmol/L

One nmol = 1.0 x 10-9 mol, one μmol = 1.0 x 10-6 mol.
Therefore one μmol = 1.0 x 103 nmol = 1000 nmol
Division of 1 nmol/L by 100 converts to μmol/L

0.236 nmol/L = 0.236 = 0.000236 μmol/L

1000

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

d) Convert 1.6 mg/L to ng/mL

There are 1,000,000 (or 1.0 x 106) ng in 1 mg

There are 1000 mL in one L

Therefore multiplication by 1,000,000 and division by 1000 converts from

mg/L to ng/mL:

1.6 mg/L = 1.6 x 1,000,000 = 1600 ng/mL

1,000

2. Convert the following concentrations to “SI” units: a) plasma glucose 120 mg%;
b) serum calcium 4.0 mEq/L; c) BUN 21 mg%; d) Serum creatinine 0.66 mg%.

a) Convert plasma glucose = 120 mg% to SI units

‘SI’ units for glucose are mmol/L

120 mg% can also be written 120 mg/100mL

Concentration (mmol/L) = Concentration (mg/L)

Molecular weight

Multiplication of concentration in mg/100 mL by 10 converts to mg/L

Formula of glucose = C6H12O6

Atomic wt carbon = 12, therefore C6 = 6 x 12 = 72

Atomic wt hydrogen = 1, therefore H12 = 12 x 1 = 12
Atomic wt oxygen = 16, therefore O6 = 6 x 16 = 96

Molecular weight of glucose = 180

Therefore 120 mg% glucose = 120 x 10 = 6.7 mmol/L (2 sig figs)

180

340
 WORKED ANSWERS TO FURTHER QUESTIONS

b) Convert serum calcium from 4.0 mEq/L to ‘SI’ units

The SI units for calcium are mmol/L

Concentration (mEq/L) = Concentration (mmol/L) x valency

Calcium ions are divalent (i.e. Ca++) so that the valency is 2

Therefore, serum calcium (mmol/L) = 4.0 = 2.0 mmol/L

2

c) Convert BUN 21 mg% to urea ‘SI’ units

SI units for serum urea = mmol/L

mg% can also be written mg/100 mL

Multiplication of BUN mg% by 10 converts to mg/L (since 1 L = 1000 mL)

Division of blood urea nitrogen (BUN) in mg% by the molecular weight of

nitrogen gives the blood urea nitrogen in mmol/L.

The formula of molecular nitrogen is N2. The atomic weight of nitrogen is

14.

Molecular weight of nitrogen (N2) = 2 x 14 = 28

The formula for urea is CO(NH2)2. Therefore each mol of urea contains one
mol of nitrogen (N2).

Therefore, serum urea (mmol/L) = BUN (mg%) x 10

28

Serum urea (mmol/L) = 21 x 10 = 7.5 mmol/L

28

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

d) Convert serum creatinine 0.66 mg% to ‘SI’ units

The SI units for creatinine are μmol/L

There are 1000 μg in one mg so that multiplication by 1000 converts from

mg% to μg%

μg% can also be written μg/100 mL

Multiplication by 10 converts from μg/100 mL to μg/L (1 L = 1000 mL)

Division by the molecular weight of creatinine converts from μg/L to

μmol/L. Formula of creatinine is: C4H7ON3

Carbon atomic wt = 12, C4 = 4 x 12 = 48

Hydrogen atomic wt = 1, H7 = 7 x 1 = 7
Oxygen atomic weight = 16, O = 1 x 16 = 16
Nitrogen atomic wt = 14, N3 = 3 x 14 = 42

Creatinine molecular weight = 113

Creatinine (μmol/L) = Creatinine (mg%) x 10 x 1000

Molecular weight

Creatinine (mmol/L) = 0.66 x 10 x 1000 = 58 μmol/L (2 sig figs)

113

3. Convert the following: a) plasma glucose from 20 mmol/L to mg/100 mL;

b) serum calcium from 3.2 mmol/L to mEq/L; c) serum urea from 30.6 mmol/L to
mg% BUN; d) serum creatinine from 250 μmol/L to mg%.

a) Convert plasma glucose 20 mmol/L to mg/100 mL

Division by 10 converts from mmol/L to mmol/100 mL (since I L = 1000 mL)

Multiplication by the molecular weight of glucose converts from mmol/100

mL to mg/100 mL. Formula of glucose is C6H12O6.

342
 WORKED ANSWERS TO FURTHER QUESTIONS

Atomic wt carbon = 12, therefore C6 = 6 x 12 = 72

Atomic wt hydrogen = 1, therefore H12 = 12 x 1 = 12
Atomic wt oxygen = 16, therefore O6 = 6 x 16 = 96

Molecular weight of glucose = 180

Therefore, glucose (mg/100mL) = Glucose (mmol/L) x 180

10

Glucose (mg/100 mL) = 20 x 180 = 360 mg/100mL

10

b) Convert serum calcium from 3.2 mmol/L to mEq/L

Concentration (mEq/L) = Concentration (mmol/L) x valency

Calcium ions are divalent (i.e. Ca++) so that the valency is 2

Therefore, calcium (mEq/L) = calcium (mmol/L) x 2

Calcium (mEq/L) = 3.2 x 2 = 6.4 mEq/L

c) Convert serum urea from 30.6 mmol/L to mg% BUN

Division by 10 converts from mmol/L to mmol/100 mL (since 1 L = 1000

mL), which can also be written mmol%.

Since the formula of urea is CO(NH2)2, each mol contains 1 mol of molecular
nitrogen (N2). The atomic weight of nitrogen is 14, so its molecular weight
(N2) is 2 x 14 = 28. Therefore, multiplication of urea in mmol% by 28 gives
the BUN in mg%.

Therefore, BUN (mg%) = Urea (mmol/L) x 28

10

BUN (mg%) = 30.6 x 28 = 86 mg% (2 sig figs)

10

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

d) Convert serum creatinine from 250 μmol/L to mg%

Division by 10 converts from μmol/L to μmol/100 mL (since 1 L = 1000

mL), which can also be written as μmol%.

Division by 1000 converts from μmol% to mmol% (since 1000 μmol =

1 mmol).

Multiplication by the molecular weight of creatinine converts from mmol% to

mg%. Creatinine has the formula C4H7ON3.

Carbon atomic wt = 12, C4 = 4 x 12 = 48

Hydrogen atomic wt = 1, H7 = 7 x 1 = 7
Oxygen atomic weight = 16, O = 1 x 16 = 16
Nitrogen atomic wt = 14, N3 = 3 x 14 = 42

Creatinine molecular weight = 113

Therefore, creatinine (mg%) = creatinine (μmol/L) x 113

10 x 1000

Creatinine (mg%) = 250 x 113 = 2.8 mg% (2 sig figs)

10 x 1000

4. Convert the following: a) 1.5 x 10-3 M to mmol/L; b) 1.25 x 10-5 M to μmol/L;

c) 2.5 x 102 mg/100 mL to g/L; d) 3.25 x 10-6 mmol/L to μmol/L.

a) Convert 1.5 x 10-3M to mmol/L

‘M’ is an abbreviation for mol/L. There are 1000 mmol in one mol.

Therefore, multiplication of a concentration in mol/L by 1000 converts it to

mmol/L.

1.5 x 10-3 is the same as 1.05

103

103 means 10 multiplied by itself 3 times i.e. 10 x 10 x 10 = 1000

344
 WORKED ANSWERS TO FURTHER QUESTIONS

Therefore, 1.5 x 10-3 = 1.5 = 0.0015

1000

Another way of looking at it is that the power minus 3 means that we must
move the decimal point 3 places to the left (as opposed to a positive power
which would have meant moving it 3 places to the right). Moving the decimal
point one pace to the left gives 0.15, 2 places gives 0.015 and 3 places gives
0.0015.

Combining these moves:

1.5 x 10-3 M = 0.0015 x 1000 = 1.5 mmol/L

In other words whenever we see a molar concentration with the term ‘ x 10-3,
it is really the same as the concentration in mmol/L.

Another way to approach this problem is that multiplying by 1000 to convert

from mol to mmol is the same as multiplying by 103 in which case the
calculation becomes:

1.0 x 10-3 x 103 = 1.0

since the 10-3 and 103 cancel (i.e. we move the decimal point 3 places to the
left then back 3 places to the right to the original position).

b) Convert 1.25 x 10-5 M to μmol/L

The symbol ‘M’ stands for mol/L.

Multiplication by 1,000,000 converts from mol to μmol (since there are

1,000,000 μmol in one mol). 1,000,000 can also be written as 106

Therefore 1.25 x 10-5 M = 1.25 x 10-5 x 106 μmol/L

Since 10-5 x 106 = 101 (which is simply 10) the calculation becomes:

1.25 x 10-5 M = 1.25 x 10 = 12.5 μmol/L

This is the same as moving the decimal point 5 places to the left, then 6 places

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

to the right i.e. a net movement of one place to the right.

c) Convert 2.5 x 102 mg/100 mL to g/L

Multiplication by 10 converts from mg/100 mL to mg/L (since 1 L = 1000

mL).

Division by 1000 converts from mg/L to g/L (since there are 1000 mg in one
g)

2.5 x 102 mg/100 mL means 2.5 multiplied by 10 squared (i.e. 100) =

250 mg/100 mL

Therefore, 2.5 x 102 mg/100 mL = 250 x 10 = 2.5 g/L

1000

Another way of doing this is that we first move the decimal point 2 places to
the right (the same as multiplying by 102), a further one place to the right to
convert from 100 mL to 100 mL (making 3 moves to the right altogether).
A further 3 moves to the left (the same as dividing by 1000 to convert from
mg to g) takes us back to the starting position and an answer of 2.5 g/L.

d) Convert 3.25 x 10-6 mmol/L to μmol/L

Multiplication of mmol/L by 1000 converts to μmol/L (since 1 mmol = 1000

μmol). Multiplication by 1000 is the same as multiplication by 103

Therefore, 3.25 x 10-6 mmol/L = 3.25 x 10-6 x 103 = 3.25 x 10-3 μmol/L

3.25 x 10-3 μmol/L can also be written 0.00325 μmol/L.

(The same result is obtained by moving the decimal point 6 places to the left
then back 3 places to the right).

346
 WORKED ANSWERS TO FURTHER QUESTIONS

5. After incubation of an enzyme with substrate for 30 min the concentration of product
in the reaction mixture was 3.00 x 10-3 M. a) How many mmol of product would be
present in 100 mL of the reaction mixture; and b) what is the rate of formation of
product in 250 mL of reaction mixture expressed as μmol/min?

a) How many mmol of product is present in 100 mL of reaction mixture?

If the concentration in the reaction mixture is 3.00 x 10-3 M then each litre
contains 3.0 x 10-3 mol of product.

Division by 10 gives the number of mol in 100 mL (since 1 L = 1000 mL)

Multiplication by 1000 converts from mol to mmol (since 1 mol =

1000 mmol)

3.00 x 10-3 can also be written as 3.00 or 3.00

103 1000

Therefore concentration (mmol/100mL) = concentration (mol/L) x 1000

10

Amount of product in 100 mL = 3.00 x 1000 = 0.30 mmol

10 x 1000

b) What is the rate of formation of product in 250 mL of reaction mixture

expressed as μmol/min?

3.00 x 10-3 M means 3.00 x 10-3 mol/L

Multiplication by 1,000,000 converts from mol/L to μmol/L (since 1 mol =

1,000,000 μmol). 1,000,000 can also be written 106.

Division by 4 gives the amount of product present in 250 mL (since 1000/4

= 250).

Since this amount of product was formed over 30 min, division by 30 gives
the amount which would be formed in one minute.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Therefore number of μmol formed in 250 mL in one minute

= Molar concentration x 106

4 x 30

= 3.00 x 10-3 x 106 = 25 μmol/min/250 mL

4 x 30

NB 10-3 x 106 = 103 i.e. move decimal point 3 places to the left,
then 6 places to the right making 3 places to the right overall, which is the
same as multiplying by 1000.

6. An acid dissociates in solution to give its conjugate base and hydrogen ions.
Calculate the dissociation constant if urine contains 0.1 M of undissociated acid,
25 x 10-5 mol/L of its conjugate base and 120 nmol/L of hydrogen ions? NB the
dissociation constant is the product of the concentrations of conjugate base and
hydrogen ions divided by the concentration of undissociated acid.

The dissociation of the acid can be written:

Acid ↔ Conjugate base- + H+

0.1M 25 x 10-5 mol/L 120 nmol/L

Before calculating the dissociation constant (K) convert all concentrations to the same
units. It doesn’t really matter which units but conventionally molar concentrations
(mol/L) are used.

0.1 M undissociated acid is the same as 0.1 mol/L of acid, which can also be written
as 1.0 x 10-1 mol/L.

There are 1,000,000,000 (i.e. 109) nmol in one mol so that 120 nmol/L hydrogen ions
can be written 120 x 10-9 mol/L or more conveniently 1.20 x 10-7 mol/L (moving the
decimal point 2 places to the left and decreasing the power of 10 by 2 from -9 to -7).

The conjugate base concentration is already in mol/L, but would be more

conventionally written as 2.5 x 10-4 (reduction from 25 to 2.5 involves moving the
decimal point one place to the left so that the power of 10 must be increased by one
i.e. changed from -5 to -4).

348
 WORKED ANSWERS TO FURTHER QUESTIONS

The expression for the dissociation constant, with molar concentrations in square
brackets [] can be written:

K = [Conjugate base] (mol/L) x [Hydrogen ions] (mol/L)

[Acid] (mol/L)

An “expression” for the units of K can be written:

(mol/L) x (mol/l)
(mol/L)

One set of (mol/) above the line cancels with (mol/L) below the line leaving mol/L as
the units.

Calculation of K from these data gives:

K = 2.5 x 10-4 x 1.20 x 10-7 = 2.5 x 1.20 x 10-10 = 3.0 x 10-10 mol/L
1.0 x 10-1

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Chapter 2

(Atomic weights: H = 1; C = 12; O = 16; P = 31; Na = 23; K = 39; Ca = 40; S = 32;

Cl = 35.5)

1. How many grams of albumin are required to prepare 100 mL of a solution

containing 70 g/L.

70 g/L is the same as 70 g/1000 mL (since 1L = 1,000 mL)

If 1000 mL contain 70 g of albumin then each mL contains 70 g

1,000

And 100 mL must conation 100 times this amount, so that the weight required to
make 100mL of 70 g/L albumin

= 70 x 100 = 7g
1,000

Another way of looking at this is that 100 mL is one tenth of 1 L (i.e. 1,000 mL)
so that one tenths of the amount present in l L is required.

2. Calculate the concentration of sodium ions (in mmol/L) in a solution prepared by

dissolving 85 g of sodium chloride in 1 litre of water.

Concentration of sodium chloride = 85 g/L

Division by the molecular weight of sodium chloride gives the concentration in

mol/L:

Concentration (mol/L) = Concentration (g/L)

Molecular weight

Multiplication by 1,000 converts from mol/L to mmol/L (since 1 mol = 1,000

mmol)

350
 WORKED ANSWERS TO FURTHER QUESTIONS

Formula for sodium chloride = NaCl

Atomic weight of Na = 23; atomic weight of Cl = 35.5

Therefore molecular weight of NaCl = 23 + 35.5 = 58.5

Therefore, NaCl (mmol/L) = NaCl (g/L) x 1,000

58.5

Since each molecule of NaCl dissociates to give one ion of Na+, this is also the
concentration of sodium ions in mmol/L.

Concentration of Na+ (mmol/L) = 85 x 1,000 = 1453 mmol/L

58.5

3. What weight of calcium carbonate must be dissolved in 500 mL of dilute acid to

provide a calcium standard containing 5.0 mmol/L ?

Calcium carbonate has the formula CaCO3 so that each mol contains 1 mol of
calcium. Therefore, the standard solution will need to contain 5.0 mmol/L of
CaCO3.

Atomic wt calcium = 40, therefore Ca = 1 x 40 = 40

Atomic wt carbon = 12, therefore C = 1 x 12 = 12
Atomic wt of oxygen = 16, therefore O3 = 3 x 16 = 48

Molecular weight of CaCO3 = 100

1 L of 1 mol/L will contain 100 g of CaCO3

1 L of 1mmol/L will contain 100 g of CaCO3 (since 1 mol/L = 1,000 mmol/L)

1,000

1 L of 5.0 mmol/L will contain 100 x 5.0 g CaCO3

1,000

500 mL of 5.0 mmol/L will contain 100 x 5.0 g CaCO3 (1L = 1,000mL)
1,000 x 2

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

= 0.25 g (= 250 mg)

4. A solution contains 5% sucrose. How much of this solution would you dilute to
prepare 500 mL of % sucrose?

The total amount of sucrose (as opposed to concentration) remains the same after
dilution. The amount of sucrose in a given volume of solution is equal to the
volume multiplied by concentration. Therefore the following expression can be
written:

Initial volume x Initial concentration = Final volume x Final concentration

The units for volume and concentration must be the same on both sides of the
equation.

Initial volume = unknown

Initial concentration = 5%
Final volume = 500 mL
Final concentration = 1%

Substituting these values the initial volume can be calculated:

Initial volume (mL) x 5 = 500 x 1

Initial volume (mL) = 500 x 1 = 100 mL

5

Another way to do this is that the final concentration (1%) is one fifth of the
initial value (5%) so that the 5% sucrose solution has to be diluted 5-fold.
The volume required is 500 mL so that one fifth of this, 100 mL has to be diluted.

5. 50 μL of urine is added to 5 mL of water. What is the resulting dilution of the

urine?

Both volumes must be expressed in the same units. Multiplication of the volume
of water (5 mL) by 1000 gives its volume in μL (5,000 μL) since 1 mL =
1,000 μL.

The total volume of diluted urine is the sum of the volumes of urine and water:

Final volume (μL) = 50 + 5,000 = 5,050 μL

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 WORKED ANSWERS TO FURTHER QUESTIONS

The dilution is the number of times the urine was diluted which is the final
volume divided by the initial volume:

Dilution = Final volume = 5,050 = 101

Initial volume 50

N.B. Concentration is the reciprocal of dilution. In this case 1/101 = 0.0099.

6. Concentrated sulphuric acid (SG 1.84) is 96% by weight H2SO4. Calculate the
volume of concentrated acid required to prepare 1 L of 0.1M H2SO4.

First calculate the molecular weight of sulphuric acid:

Atomic wt of hydrogen = 1, 2H = 2x1 = 2

Atomic wt of sulphur = 32, S = 1 x 32 = 32
Atomic wt of oxygen = 16, 4O = 4 x 16 = 64

Molecular weight H2SO4 = 98

Therefore 1 L of 1 mol/L requires 98 g H2SO4

1 L of 0.1 mol/L requires 98 g H2SO4 (since it is 1/10th the strength of 1 mol/L)

10

Since the sulphuric acid is only 96% pure this weight must be multiplied by
100/96:

Weight H2SO4 required = 98 x 100 = 10.21 g (4 sig figs)

10 x 96

The volume required can be calculated from the specific gravity:

Specific gravity (SG) = Weight

Volume

Volume = Weight
Specific gravity

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

We are told that the specific gravity of H2SO4 is 1.16 so that 1 mL weighs 1.16 g.
The volume which weighs 102.08 g is calculated as follows:

Volume (mL) = 10.21 = 8.8 mL

1.16

7. The following solutions were mixed together:

50 mL potassium chloride (5.0 g/L)

100 mL sodium chloride (50 g/L)

Calculate the molar concentrations of potassium, sodium, and chloride ions.

First calculate the molar concentration of each individual solution:

Potassium chloride (KCl). Atomic wt K = 39, atomic wt Cl = 35.5

Molecular weight of KCl = 39 + 35.5 = 74.5

KCl (mol/L) = KCl (g/L) = 5.0 = 0.0671 mol/L (3 sig figs)

Molecular wt 74.5

Sodium chloride (NaCl). Atomic wt Na = 23, atomic wt Cl = 35.5.

Molecular weight NaCl = 23 + 35.5 = 58.5

NaCl (mol/L) = NaCl (g/L) = 50 = 0.855 (3 sig figs)

Molecular wt 58.5

For potassium:

50 mL 0.0671M KCl + 100 mL 0.855 M NaCL → 150 mL mixture

0.0671 M KCl contains 0.06371 M K+. 50 mL is diluted to 150 mL.

Final K+ (mol/L) x Final vol (mL) = Initial K+ (mol/L) x Initial vol (mL)

Final K+ (mol/L) x 150 = 0.0671 x 50

Final K+ (mol/L) = 0.0671 x 50 = 0.022 mol/L (2 sig figs)

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 WORKED ANSWERS TO FURTHER QUESTIONS

150

For sodium:

100 mL NaCl 0.855 mol/L + 50 mL KCl → 150 mL mixture

Final Na+ (mol/L) x Final vol (mL) = Initial Na+ (mol/L) x Initial vol (mL)

Final Na+ (mol/L) x 150 = 0.855 x 100

Final Na+ (mol/L) = 0.855 x 100 = 0.57 mol/L

150

For chloride:

100 mL NaCL 0.855 mol/L + 50 mL KCl 0.0671 → 150 mL mixture

Final Cl- (mol/L) x Final vol (mL) = [Initial Cl- from KCl (mol/L) x

Final Cl- (mol/L) x 150 = [ 0.0671 x

vol KCl (mL)] + [Initial Cl- from NaCl (mol/L) x vol NaCl (L)]

50 ] + [ 0.855 x 100 ]

Final Cl- (mol/L) = [(0.0671 x 50) + (0.855 x 100)]

150

= 3.355 + 85.5
150

= 88.855
150

= 0.59 mol/L (2 sig figs)

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

8. If you have available 650 mL of 9 % ethanol, what is the maximum volume of 65

% ethanol you could prepare?

Initial vol (mL) x Initial concn (%) = Final vol (mL) x Final concn (%)

650 x 95 = Final vol (mL) x 65

Final vol (mL) = 650 x 95 = 950 mL

65

N.B. The expected volume of water to be added to the 95% ethanol (950 – 650 =
300 mL) will be insufficient because mixing an alcohol with water always results
in some contraction of the total volume. Therefore further water should be added
until a final volume of 950 mL is reached.

9. In order to prepare 1 L of a stock standard solution containing 0.2 mol/L, the

appropriate amount of sodium dihydrogen orthophosphate dihydrate should be
weighed out. Due to an error, the same weight of anhydrous sodium dihydrogen
orthophosphate used. Working standard was prepared by taking 5 mL of this
stock standard and diluting it to 250 mL. What is the phosphate concentration
(in mmol/L) of the working standard?

First calculate the weight of sodium dihdrogen orthophosphate dihydrate

(NaH2PO4.2H2O) which should have been used. Adding individual atoms
together gives the empirical formula: NaH6PO6

Atomic wt sodium = 23, therefore Na = 1 x 23 = 23

Atomic wt hydrogen = 1, therefore 6H = 6x1 = 6
Atomic wt phosphorus = 31, therefore P = 1 x 31 = 31
Atomic wt oxygen = 16, therefore 6O = 6 x 16 = 96

Molecular weight NaH2PO4.2H2O = 156

Therefore, 1L 1.0 mol/L contains 156 g NaH2PO4.2H2O

so that, 1 L 0.2 mol/L contains 156 g = 31.2 g NaH2PO4.2H2O

5

356
 WORKED ANSWERS TO FURTHER QUESTIONS

Next calculate the molar concentration if this weight (31.2 g) of anhydrous

sodium dihydrogen orthophosphate (NaH2PO4) was dissolved in 1 L of water.

Atomic wt sodium = 23, therefore Na = 1 x 23 = 23

Atomic wt hydrogen = 1, therefore 2H = 2x1 = 2
Atomic wt phosphorus = 31, therefore P = 1 x 31 = 31
Atomic wt oxygen = 16, therefore 4O = 4 x 16 = 64

Molecular weight NaH2PO4 = 120

Concentration (mol/L) = Concentration (g/L)

Molecular weight

Concentration (mol/L) = 31.2 = 0.26 mol/L

120

As a short cut the target concentration (0.2 mol/L) could be simply multiplied by
the ratio of the molecular weight of NaH2PO4.2H2O to that of NaH2PO4:

Actual concentration (mol/L) =

Target concentration (0.2 mol?L) x MW NaH2PO4.2H2O

MW NaH2PO4

= 0.2 x 156 = 0.26

120

Working standard was prepared by diluting 5 mL of this stock standard to

250 mL.

Initial concn (mol/L) x Initial vol (mL) = Final concn (mol/L) x Final vol (mL)
0.26 x 5 = Final concn (mol/L) x 250

Final concn (mol/L) = 0.26 x 5 = 0.0052 mol/L

250

Multiplication by 1,000 (since there are 1,000 mmol in a mol) converts this
concentration to mmol/L:

Working phosphate standard concentration = 0.0052 x 1,000 = 5.2 mmol/L

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

10. Solution A contains 12.0 g of anhydrous sodium dihydrogen phosphate per litre.
What is the phosphate concentration expressed as mmol/L? What volume of
solution A needs to be diluted to 1 L to give a phosphate concentration of
4 mmol/L.

First calculate the molecular weight of anhydrous sodium dihydrogen phosphate

(NaH2PO4):

Atomic wt sodium = 23, therefore Na = 1 x 23 = 23

Atomic wt hydrogen = 1, therefore 2H = 2x1 = 2
Atomic wt phosphorus = 31, therefore P = 1 x 31 = 31
Atomic wt oxygen = 16, therefore 4O = 4 x 16 = 64

Molecular weight NaH2PO4 = 120

Next calculate the molar concentration of a solution containing 12 g/L:

Concentration (mol/L) = Concentration (g/L)

Molecular weight

Concentration (mol/L) = 12 = 0.1 mol/L

120

Multiplication by 1,000 gives the concentration in mmol/L (since 1 mol = 1,000

mmol):

Phosphate concentration = 0.1 x 1,000 = 100 mmol/L

To prepare 1 L of 4 mmol/L phosphate:

Initial vol (mL) x Initial conc (mmol/L) = Final vol (mL) x Final conc (mmol/L)

Initial vol (mL) x 100 = 1,000 x 4

Initial vol (mL) = 1,000 x 4 = 40 mL

100

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 3

1. What is the pH of 0.5 per cent (w/v) hydrochloric acid (assume complete
dissociation, atomic weight Cl = 35.5)?

First calculate molar concentration of 0.5% HCl:

0.5 %w/v = 0.5 g/100 mL = 0.5 x 10 = 5.0 g/L

MW HCl = 1 + 35.5 = 36.5

Molar conc = g/L = 5.0 = 0.137 mol/L (3 sig figs)

MW 36.5

Next calculate pH assuming complete dissociation of HCl:

pH = - log10 [H+]

Substitute [H+] = 0.137 mol/L

pH = - log10 0.137 = - (-0.86) = 0.86 (2 sig figs)

2. The reference range for blood pH is often quoted as 7.35 – 7.45. Express this
range in terms of nannomoles of hydrogen ion per litre.

pH = - log10 [H+]

Rearranging: log10 [H+] = - pH

Therefore [H+] = antilog10 (-pH)

Substitute pH = 7.35:

[H+] = antilog10 (-7.35) = 4.47 x 10-8 mol/L

= 44.7 x 10-9 mol/L = 45 nmol/L (2 sig figs)

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Substitute pH = 7.45

[H+] = antilog10 (-7.45) = 3.55 x 10-8 mol/L

= 35.5 x 10-9 mol/L = 36 nmol/L (2 sig figs)

3. If the pH of urine is 6.0 and of blood 7.40, what is the gradient of hydrogen ion
concentrations across the tubular cell walls?

pH = - log10 [H+] therefore [H+] = antilog10 (-pH)

For urine substitute pH = 6.0:

[H+] = antilog10 (- 6.0) = 1.0 x 10-6 mol/L

= 1,000 x 10-9 mol/L = 1000 nmol/L

For blood substitute pH = 7.40:

[H+] = antilog10 (-7.40) = 3.98 x 10-8

= 39.8 x 10-9 mol/L = 40 nmol/L

Gradient = [H+] in urine = 1000 = 25:1

[H+] in blood 40

Another way of approaching this problem is to use the fact that the ration of two
values is equal to the antilog of the difference between their logarithms.

In other words, substitute antolog10 (- pH) for [H+]:

Gradient = [H+]urine = antilog10 (- pHurine)

[H+]blood (- pHblood)

= antilog10 { - pHurine – (- pHblood)}

= antilog10 (pHblood – pHurine)

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 WORKED ANSWERS TO FURTHER QUESTIONS

= antilog10 (7.4 - 6.0)

= antiolg10 1.4

= 25 (2 sig figs)

4. Determine the secondary dissociation constant of phosphoric acid if blood of pH

7.00 contains 12.85 mg disodium hydrogen orthophosphate and 6.88 mg sodium
dihydrogen orthophosphate per 100 mL of plasma.

The dissociation to be considered is:

H2PO4- ↔ HPO42- + H+

pH = pKa + log10 [Na2HPO4]

[NaH2PO4]

Rearranging: pKa = pH - log10 [Na2HPO4]

[NaH2PO4]

Next calculate the molar concentration of each phosphate.

[Na2HPO4] = 12.85 x 10
1,000 x MW

MW Na2HPO4 = (2 x 23) + 1 + 31 + (4 x 16) = 142

[Na2HPO4] = 12.85 x 10 = 0.000905 mol/L

1000 x 142

[NaH2PO4] = 6.88 x 10
1,000 x MW

MW NaH2PO4 = 23 + (2 x 1) + 31 + (4 x 16) = 120

[NaH2PO4] = 6.88 x 10 = 0.000573 mol/L

1000 x 120

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Next substitute these molar concentrations into the rearranged Henderson-

Hasselbalch equation and solve for pKa:

pKa = 7.0 - log10 0.000905

0.000573

pKa = 7.0 - log10 1.58 = 7.0 - 0.20 = 6.80

5. What weight of anhydrous sodium carbonate and sodium bicarbonate would be

required to prepare 500 mL of 0.2 M buffer pH 10.7 (pKa HCO3- = 10.3)?

The relevant dissociation is:

HCO3- → H+ + CO32-

pH = pKa + log10 [CO32-]

[HCO3-]

Rearrange, substitute pH = 10.7 and pKa = 10.3, then calculate ratio:

log10 [CO32-] = pH - pKa = 10.7 - 10.3 = 0.4

[HCO3-]

[CO32-] = antilog 0.4 = 2.51 ……… (i)

[HCO3-]

Since the total concentration of both bicarbonate and carbonate in the buffer is 0.2
mol/L:

0.2 = [HCO3-] + [CO32-] ………… (ii)

[HCO3-] = 0.2 - [CO32-]

Substitute for [HCO3-] in (i) and solve for [CO32-]:

[CO32-] = 2.51
0.2 - [CO32-]

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 WORKED ANSWERS TO FURTHER QUESTIONS

[CO32-] = (2.51 x 0.2) - 2.51 [CO32-]

[CO32-] + 2.51 [CO32-] = 0.502

3.51 [CO32-] = 0.502

[CO32-] = 0.502 = 0.143 mol/L

3.51

Substitute [CO32-] = 0.143 into (ii) and solve for [HCO3-]:

0.2 = [HCO3-] + 0.143

[HCO3-] = 0.2 - 0.143 = 0.057 mol/L

Calculate weights of both sodium carbonate and bicarbonate needed to prepare

500 mL of buffer:

Weight required (g) = Molar concentration (mol/L) x MW

2

MW Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106

MW NaHCO3 = 23 + 1 + 12 + (3 x 16) = 84

Wt Na2CO3 = 0.143 x 106 = 7.58 g (3 sig figs)

2

WT NaHCO3 = 0.057 x 84 = 2.39 g (3 sig figs)

2

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

6. Isotonic sodium lactate, pH 7.4, is commonly administered intravenously to

combat metabolic acidosis. How many ml of concentrated lactic acid (85% w/w,
density 1.2) and how many grams of anhydrous sodium lactate would be used to
prepare 2.5L of this solution (pKa lactic acid = 3.86)?

The relevant dissociation is:

LactH → Lact- + H+

pH = pKa + Log10 [Lact-]

[LactH]

Substitute pH = 7.4 and pKa = 3.86 then calculate ratio:

7.4 = 3.86 + Log10 [Lact-]

[LactH]

Log10 [Lact-] = 7.4 - 3.86 = 3.54

[LactH]

[Lact-] = antilog10 3.54 = 3467 ……..(i)

[LactH]

The concentration is not given but we are told that the solution must be isotonic.
Assuming physiological osmolarity is 285 mmol/L:

285 = [LactH] + [Lact-] + [Na+]

Where the concentrations of LactH, Lact- and Na+ are mmol/L.

Since the concentrations of Lact- and Na+ are equal:

285 = [LactH] + 2 [Lact-] ………… (ii)

Rearranging:

[LactH] = 285 - 2 [Lact-]

Substitute for [LactH] in (i) and solve for [Lact-]:

[Lact-] = 3467
285 - 2 [Lact-]

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 WORKED ANSWERS TO FURTHER QUESTIONS

[Lact-] = 3467 (285 - 2 [Lact-])

[Lact-] = 988095 - 6934 [Lact-]

[Lact-] + 6935 [Lact-] = 988095

6935 [Lact-] = 988095

[Lact-] = 988095 = 142 mmol/L (3 sig figs)

6935

Substitute [Lact-] = 142 into (ii) and solve for [LactH]:

285 = [LactH] + (2 x 142)

[LactH] = 285 - 284 = 1 mol/L

Calculate weight of sodium lactate:

Sodium lactate (g/2.5L) = [Lact-] mmol/L x MW x 2.5

1000

MW CH3CH(OH)COONa = 23 + (3 x 12) + (5 x 1) + (3 x 16) = 112

Wt sodium lactate = 142 x 112 x 2.5 = 39.8 g

1000

Calculate weight of lactic acid:

Lactic acid (g/2.5L) = [LactH] mmol/L x MW x 2.5

1000

MW CH3CH(OH)COOH = (3 x 12) + (6 x 1) + (3 x 16) = 90

Wt lactic acid = 1 x 90 x 2.5 = 0.225 g

1000

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Convert to mL 85% lactic acid SG = 1.2:

Vol lactic acid(mL) = Wt (g) x 100

% purity x SG

= 0.225 x 100 = 0.22 mL (2 sig figs)

85 x 1.2

7. A 24h urine collection has a pH of 5.5 and total phosphate content of 65 mmol. If
the arterial pH is 7.40 and the pKa for phosphate is 6.80, how many millimoles of
hydrogen ion are excreted as titratable acidity using HPO42- as buffer?

The reaction occurring when secreted hydrogen ions are buffered by phosphate in
the glomerular filtrate is:

HPO42- + H+ → H2PO4-

And the corresponding Henderson-Hasselbalch equation is:

pH = pKa + log10 [HPO42-]

[H2PO4-]

Calculate the ratio of the two phosphate ions in fresh glomerular filtrate (i.e. pH =
7.4):

7.4 = 6.8 + log10 [HPO42-]

[H2PO4-]

log10 [HPO42-] = 7.4 - 6.8 = 0.6

[H2PO4-]

[HPO42-] = antilog10 0.6 = 3.98 ………..(i)

[H2PO4-]

For simplicity assume a urine volume of 1 L so that the total phosphate

concentration is 65 mmol/L. (We are told the amount not the concentration but
since we are dealing with ratios and volume could be used).

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 WORKED ANSWERS TO FURTHER QUESTIONS

65 = [HPO42-] + [H2PO4-]

[H2PO4-] = 65 - [HPO42-]

Substitute for [H2PO4-] in (i) then solve for [HPO42-]:

[HPO42-] = 3.98
65 - [HPO42-]

[HPO42-] = 3.98 (65 - [HPO42-])

[HPO42-] + 3.98 [HPO42-] = 258.7

4.98 [HPO42-] = 258.7

[HPO42-] = 258.7 = 51.9 mmol/L (3 sig figs)

4.98

Repeat this procedure for acidified glomerular filtrate i.e. urine pH = 5.5

5.5 = 6.8 + log10 [HPO42-]

[H2PO4-]

log10 [HPO42-] = 5.5 - 6.8 = -1.3

[H2PO4-]

[HPO42-] = antilog10 -1.3 = 0.050

[H2PO4-]

Substitute [H2PO4-] = 65 - [HPO42-] and solve for [HPO42-]:

[HPO42-] = 0.050
65 - [HPO42-]

[HPO42-] = 0.050 (65 - [HPO42-])

[HPO42-] + 0.050[HPO42-] = 0.050 x 65 = 3.25

1.05 [HPO42-] = 3.25

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

[HPO42-] = 3.25 = 3.10 mmol/L

1.05

The titratable acidity is the concentration (or rather amount) of HPO42-

consumed:

Titratable acidity = 51.9 - 3.1 = 49 mmol (2 sig figs)

8. A buffer solution (pH 4.74) contains acetic acid (0.1 mol/L) and sodium acetate
(0.1 mol/L) i.e. it is a 0.2M acetate buffer. Calculate the pH after addition of 4
mL of 0.025 M hydrochloric acid to 10 mL of the buffer.

The dissociation to be considered is:

HAc → H+ + Ac-

And the relevant form of the Henderson Hasselbalch equation is:

pH = pKa + log10 [Ac-]

[HAc]

Determine pKa by substituting the pH (4.74) and concentrations of Ac- (0.1

mol/L) and HAc (0.1 mol/L):

4.74 = pKa + log10 0.1

0.1

Since 0.1/0.1 = 1 and log10 1 is 0, then pKa = 4.74

Calculate the adjusted cocnetrations of Ac- and HAc, and substitute into the
Henderson-Haseelbalch equation (using pKa = 4.74) then solve for pH:

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 WORKED ANSWERS TO FURTHER QUESTIONS

Addition of HCl to this buffer converts some of the acetate ions to acetic acid:

Ac- + H+ → HAc

Final [Ac-] = Initial [Ac-] - Added [HCl]

Allowance must be made for the dilution resulting from mixing 10 mL buffer with
4 mL HCl (total volume = 14 mL):

Initial [Ac-] = 0.1 x 10 = 0.071 mol/L

14

Added [HCl] = Initial [HCl] x 4 = 0.025 x 4 = 0.0071 mol/L

14 14

Final [Ac-] = 0.071 - 0.0071 = 0.0639 mol/L

Similarly:

Final [HAc] = Initial [HAc] + Added [HCl]

Since Initial [HAc] = Initial [Ac-]

Final [HAc] = 0.071 + 0.0071 = 0.0781 mol/L

Therefore: pH = 4.74 + log10 0.0639

0.0781

= 4.74 + log10 0.818

= 4.74 + (-0.087) = 4.65

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Chapter 4

1. An aqueous solution in a 1 cm cell has an absorbance of 0.23 when read against

a water blank at 500 nm. Assuming Beer’s Law is obeyed, what volume of this
solution would need to be added to 100 mL of water to give a solution which
absorbs 30% of the light entering it under the same measurement conditions?

Only absorbance is proportional to concentration, so the first step is to calculate

the absorbance of the final solution which absorbs 30% of the light entering into
it.

Absorbance (A) = log10 I0

I

Io = intensity of incident light = 100%

I = intensity of transmitted light. Since 30% was absorbed, 100 – 30 = 70%

is transmitted. Therefore, I = 70%

Substitute these values to obtain A:

A = log10 100 = log10 1.429 = 0.155 (3 sig figs)

70

Let x = vol of solution to be added to 100 mL water = x mL

Final volume = (100 + x) mL

x (mL) x 0.23 = (100 + x) x 0.155

x (mL) = (100 + x) x 0.155

0.23

x = 15.5 + 0.155 x
0.23

0.23 x = 15.5 + 0.155 x

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 WORKED ANSWERS TO FURTHER QUESTIONS

0.23 x - 0.155 x = 15.5

0.075 x = 15.5

x = 15.5 = 207 mL (3 sig figs)

0.075

2. Calculate the absorbances corresponding to the following percentage

transmittance readings:

a) 95 b) 75 c) 50 d) 25 e) 10 f) 1

If Io is the intensity of incident light and I the intensity of transmitted light, then:

transmittance (%T) = I x 100 and absorbance (A) = log10 Io

Io I

The expression for %T can be arranged to: Io = 100

I %T

Which can be substituted into the expression for A to give:

A = log10 100 = log10 100 - log10 %T

%T

Substituting 2 for log10 100 gives the following useful expression:

A = 2 - log10 %T

All that is required is to substitute values for %T into this expression to obtain A:

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

a) %T = 95

A = 2 - log10 95 = 2 - 1.978 = 0.022 (3 sig figs)

b) %T = 75

A = 2 - log10 75 = 2 - 1.875 = 0.125 (3 sig figs)

c) %T = 50

A = 2 - log10 50 = 2 - 1.699 = 0.301 (3 sig figs)

d) %T = 25

A = 2 - log10 25 = 2 - 1.398 = 0.602 (3 sig figs)

e) %T = 10

A = 2 - log10 10 = 2 - 1.000 = 1

f) %T = 1

A = 2 - log10 1 = 2 - 0 = 2

3. Calculate the % of incident light transmitted by solutions with the following

absorbances:

a) 0.1 b) 0.25 c) 0.50 d) 0.75 e) 1.00 f) 2.00

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 WORKED ANSWERS TO FURTHER QUESTIONS

The expression used to calculate A from %T can be rearranged to enable direct

calculation of %T from A:

A = 2 - log10 %T

A + log10 %T = 2

log10 %T = 2 - A

%T = antilog10 (2 - A)

Therefore substitute values for A into this expression then evaluate %T:

a) A = 0.1

%T = antilog10 (2 - 0.1) = antilog10 1.9 = 79% (2 sig figs)

b) A = 0.25

%T = antilog10 (2 - 0.25) = antilog10 1.75 = 56% (2 sig figs)

c) A = 0.50

%T = antilog10 (2 - 0.50) = antilog10 1.50 = 32% (2 sig figs)

d) A = 0.75

%T = antilog10 (2 - 0.75) = antilog10 1.25 = 18% (2 sig figs)

e) A = 1.00

%T = antilog10 (2 - 1.00) = antilog10 1 = 10%

f) A = 2.00

%T = antilog10 (2 - 2) = antilog10 0 = 1%

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

4. A solution of a compound (concentration 100 mmol/L) was placed in a cuvette

with a 1 cm light path and the percentage of incident light transmitted was 18.4.
Calculate the molar absorptivity of the compound.

First convert percentage transmittance (%T) to absorbance (A):

A = 2 - log10 %T

= 2 - log10 18.4

= 2 - 1.2648

= 0.735 (3 sig figs)

Use this absorbance to calculate molar absorptivitiy using the Beer-Lambert Law:

A = abc

Where A = absorbance reading = 0.735

a = molar absorptivity = unknown
b = light path length = 1 cm
c = concentration = 100 mmol/L

Since the question asks for calculation of molar absorptivity, the concentration
must be divided by 1,000 to convert it to mol/L (1 mol = 1,000 mmol):

Concentration (mol/L) = 100 (mmol/L) = 0.1 mol/L

1,000

Substitute A, b and c into the Beer-Lambert equation and solve for a:

0.735 = a x 1 x 0.1

a = 0.735 = 7.35
1 x 0.1

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 WORKED ANSWERS TO FURTHER QUESTIONS

The units can be derived by entering the individual units into the same equation
(remembering that absorbance is the logarithm of a ratio so has no units):

a = - = L = L/cm/mol (or L.mol-1.cm-1)

cm x mol/L cm x mol

Therefore molar absorptivity = 7.35 L.mol-1.cm-1

5. The transmittance of a solution of NADH at 340 nm is 45%. What is the

absorbance at 340 nm of a 1 in 5 dilution of this solution?

Since absorbance, not transmittance, is linearly proportional to concentration, the

first step is to convert the transmittance (%T) to absorbance (A):

A = 2 - log10 %T

= 2 - log10 45

= 2 - 1.6532

= 0.3468

Assuming NADH obeys Beer’s Law, the absorbance of a 1 in 5 dilution of this

solution will be a fifth of this value:

Absorbance of 1 in 5 dilution = 0.3468 = 0.069 (2 sig figs)

5

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

6. 75 mg of faeces were homogenised in 1 mL of concentrated hydrochloric acid,

then 3 mL diethylether added, mixed, 3 mL of water added and mixed again.
After centrifugation the aqueous phase (volume 4.5 mL) was scanned in a
spectrophotometer using a cell with a 1 cm pathlength and the peak height at
405 nm due to porphyrin, after applying a background correction, was 0.35
absorbance units. A separate 0.250 g portion of faeces was dried in a 100oC oven
for 3 hours after which it’s weight was 0.125 g. Given that the molar absorption
coefficient of porphyrin is 2.75 x 105 L/mol/cm calculate the porphyrin
concentration in nmol/g dry weight of faeces.

First use the Beer-Lambert equation to calculate the porphyrin concentration in

the extract:

A = abc

Where A = absorbance = 0.35

a = molar absorptivity = 2.75 x 105 L/mol/cm
b = path length = 1 cm
c = concentration = mol/L

0.35 = 2.75 x 105 x 1 x c

c = 0.35 = 1.273 x 10-6 mol/L

2.75 x 105 x 1

The answer is required in nmol not mol so this value must be multiplied by 109
(since 1 mol = 109 nmol):

c (nmol/L) = 1.273 x 10-6 x 109 = 1.273 x 103 nmol/L

Since the faecal sample produced 4.5 mL of extract, the amount of porphyrin in
the sample is obtained by dividing by 1,000 (to convert from nmol/L to
nmol/mL), the multiplying by the volume of extract (4.5 mL):

Porphyrin in sample = 1.273 x 103 x 4.5 = 5.729 nmol

1,000 (= 103)

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The porphyrin content (expressed as nmol/g fresh weight of faeces) is obtained

by dividing the porphyrin extracted (5.729 nmol) by the weight of sample used to
prepare the extract (75 mg = 0.075 g):

Porphyrin (nmol/g fresh stool) = 5.729 = 76.4 nmol/g fresh wt (3 sig figs)
0.075

To express this result on a dry weight basis, multiply by the fresh weight of faeces
used for the dry weight determination then divide by its dry weight:

Porphyrin content = 76.4 x 0.250 = 153 nmol/g dry faeces (3 sig figs)
0.125

7. A solution containing a substance of molecular weight 400 at a concentration of

3 g/L transmitted 75% of incident light of a particular wavelength in a 1 cm
cuvette. Calculate the % of incident light of the same wavelength that would be
transmitted by a solution of the same substance at a concentration of 4 g/L and
calculate the molar absorption coefficient for that substance at this wavelength.

First convert the % transmittance (T) to absorbance (A):

A = 2 - log10 %T

A = 2 - log10 75

= 2 - 1.875

= 0.125

Provided the substance obeys Beer’s Law over the range of concentrations (i.e.
absorbance is directly proportional to concentration), then the absorbance of a
different concentration (4 g/L) can be calculated from the relationship:

Absorbance2 = Absorbance1
Concentration 2 concentration1

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Rearranging: Absorbance2 = Absorbance1 x Concentration2

Concentration1

Substitute: absorbance2 = unknown ; absorbance1 = 0.125;

concentration2 = 4 g/L; concentration1 = 3 g/L

Absorbance(4g/L) = 4 x 0.125 = 0.1667 (4 sig figs)

3

Convert this absorbance to % transmittance:

Log10%T = 2 - A = 2 - 0.1667 = 1.833

%T = antilog10 1.833 = 68% (2 sig figs)

To calculate molar absorption coefficient use either pairs of concentration and

absorption:

A = abc

Where A = absorbance = 0.167 (for a concentration of 4 g/L)

a = molar absorption coefficient = ?
b = cell path length = 1 cm
c = concentration = 4g/L.
Since MW = 400
molar concentration = 4/400 = 1/100 = 1.0 x 10-2 mol/L

0.167 = a x 1.0 x 1.0 x 10-2

a = 0.167 = 0.167 x 102 = 16.7 L.mol-1.cm-1

1.0 x 10-2

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8. The absorbances of a solution containing NAD and NADH in a 1cm light path
cuvette were 0.337 at 340 nm and 1.23 at 260 nm. The molar extinction
coefficients are:

NAD: 1.8 x 104 at 260 nm, 1.0 x 10-3 at 340 nm

NADH: 1.5 x 104 at 260 nm, 6.3 x 103 at 340 nm

Calculate the concentrations of NAD and NADH in the solution.

Both NAD and NADH absorb at the two wavelengths used (260 nm and 340 nm).
Absorbances are additive, therefore at either wavelength:

Total absorbance = Absorbance of NAD + Absorbance of NADH

At any wavelength the absorbance of NAD or NADH is given by:

Absorbance = Molar extinction coefficient x Molar concentration x Cell path

Therefore for each wavelength equations can be set up relating measured total
absorbance to the sums of the individual absorbances of NAD and NADH:

Measured absorbance = (NADConc x NADCoeff) + (NADHConc x NADHCoeff)

At 340 nm: 0.337 = 1.0 x 10-3 [NAD] + 6.3 x 103 [NADH] .................... (i)

At 260 nm: 1.23 = 1.8 x 104 [NAD] + 1.5 x 104 [NADH] ................... (ii)

(The cell path is 1 cm and can be ignored)

These form a pair of simultaneous equations which can be solved for [NAD] and
[NADH] in the usual manner. However, solving a set of simultaneous equations
can be a lengthy process. Therefore we should look for approximations and short
cuts. In this particular example it is possible to considerably simplify the
calculation. The molar extinction coefficient of NAD at 340 nm is much lower
than that of NADH (by a factor of approx. 10-6) so that the contribution of NAD
to absorbance at this wavelength can be ignored. Equation (i) can then be
simplified to:

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

0.337 = 6.3 x 103 [NADH]

[NADH] = 0.337 = 5.35 x 10-5 M = 53.5 µmol/L

6.3 x 103

[NAD] can be calculated by substituting [NADH] = 5.35 x 10-5 into equation (ii):

1.23 = 1.8 x 104 [NAD] + (1.5 x 104 x 5.35 x 10-5 )

1.23 = 1.8 x 104 [NAD] + (8.03 x 10-1)

1.8 x 104 [NAD] = 1.23 - (8.03 x 10-1) = 0.427

[NAD] = 0.427 = 2.37 x 10-5 M = 23.7 µmol/L

1.8 x 104

9. 25 mg of bilirubin (C33H36O6N4) were dissolved in 4 mL of dimethyl sulphoxide;

200 µL of this solution was diluted to 250 mL with chloroform. This solution gave
an absorbance of 0.502 when measured in a 1 cm cell against a chloroform blank.

Given that the molar absorptivity of bilirubin under these conditions is 6.07 x 104,
calculate the percentage purity of the bilirubin.

First calculate the concentration of bilirubin in the final solution:

A = a x b x c

Where A = absorbance = 0.502

a = molar absorbivity = 6.07 x 104 L.mol-1.cm-1
b = path length = 1 cm
c = concentration in mol/L = ?
l = path length = 1 cm

0.502 = 6.07 x 104 x 1 x c

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 WORKED ANSWERS TO FURTHER QUESTIONS

Rearranging and solving for c:

c = 0.502 = 8.27 x 10-6 mol/L = 8.27 x 10-3 mmol/L

6.07 x 104

Use this concentration of the final solution to calculate the bilirubin content of the
weighed bilirubin:

The final solution was prepared by diluting 200 µL (i.e. 0.2 mL) of stock to
250 mL

Therefore actual concentration of stock

= 8.27 x 10-3 x 250 = 10.34 mmol/L

0.2

4 mL (the volume of DMSO the bilirubin was dissolved in) contains:

10.34 x 4 = 0.04136 mmol bilirubin

1000

Convert to wt of bilirubin:

Wt bilirubin (mg) = mmol bilirubin x MW

MW bilirubin = (33 x 12) + (36 x 1) + (6 x 16) + (4 x 14) = 584

Therefore, wt bilirubin = 0.0414 x 584 = 24.15 mg

% purity = Amount of bilirubin by assay x 100

Amount of bilirubin weighed

= 24.15 x 100 = 97% (2 sig figs)

25

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10. A method for creatinine determination based on the Jaffe reaction involved
mixing 0.1 mL of sample with 2.5 mL alkaline picrate reagent, incubating for 10
min at room temperature, then measuring the absorbance at 530 nm in a 1-cm
cuvette in a spectrophotometer set to read zero using a cuvette containing
distilled water. The following readings were obtained:

Blank (water as ample) 0.050

Creatinine standard (200 μmol/L) 0.250
Serum sample 0.125
Urine sample (prediluted 1 in 50 with water) 0.200

Calculate the creatinine concentration in the serum (in μmol/L) and urine (in
mmol/L).

First subtract the reagent blank (i.e. the reading obtained when using water as
sample) from each absorbance reading:

Absorbance Corrected
Absorbance

Blank (water as ample) 0.050 0.000

Creatinine standard (200 μmol/L) 0.250 0.200
Serum sample 0.125 0.075
Urine sample (prediluted 1 in 50 with water) 0.200 0.150

Corrected absorbance of unknown = Corrected absorbance of standard

Concentration of unknown Concentration of standard

Concentration of unknown =

Corrected absorbance of unknown x Concentration of standard

Corrected absorbance of standard

For serum:

Serum creatinine (μmol/L) = 0.075 x 200 = 75 μmol/L

0.200

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 WORKED ANSWERS TO FURTHER QUESTIONS

For urine the calculation is the same except that the result must be multiplied by
50 to allow for the predilution of the sample prior to assay, then divided by 1,000
to convert from μmol/L to mmol/L (1 mmol = 1,000 μmol):

Urine creatinine (mmol/L) = 0.150 x 50 x 200 = 7.5 mmol/L

0.200 x 1,000

11. A standard curve for a plasma glucose method was set up by preparing a series of
dilutions of a stock glucose standard (containing 50 mmol glucose/L) and
measuring the absorbance at 500 nm in a 1 cm cuvette using a blank with zero
glucose concentration to zero the instrument. The following readings were
obtained:

Glucose (mmol.L): 5 10 15 20 25 30
Absorbance: 0.102 0.203 0.305 0.375 0.410 0.432

Does the method obey Beer’s Law? What glucose concentration corresponds to
an absorbance reading of 0.250?

Plot the absorbance (vertical scale) against the standard concentration (horizontal
scale) including the zero as a point (since the blank was used to zero the
instrument):

0.5

0.45
Glucose standard curve
0.4

0.35
Absorbance

0.3

0.25

0.2

0.15

0.1

0.05

0
0 5 10 15 20 25 30 35

Glucose (mmol/L)

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Inspection of the curve shows that the method only obeys Beer’s Law up to a
concentration of 15 mmol/L (when absorbance = 0305).

The slope of the curve up to this point = 0.305 = 0.020 A/mmol

15

Therefore 1 A = 1 so that 0.250 A = 0.250 = 12.5 mmol/L

0.02 0.02

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Chapter 5

1. An aliquot of a 24 h urine (volume 1850 mL) has a creatinine concentration of

8500 μmol/L. Calculate the 24 h urinary creatinine excretion expressing the result
as mmol/24 h.

Creatinine excretion (mmol/24h) =

creatinine concentration (mmol/L) x 24 h urine volume (L)

Divide the creatinine concentration by 1,000 to covert from μmol/L to mmol/L

(1,000 μmol = 1 mmol).

Divide the urine volume by 1,000 to convert fro mL to l L (1,000 mL = 1 L).

Creatinine excretion (mmol/24 h) = 8,500 x 1,850 = 15.7 mmol/24 h

1,000 x 1,000

2. A patient has a GFR of 110 mL/min. If the plasma creatinine concentration is

180 μmol/L how many mmol of creatinine are filtered in 12 h?

First convert the filtration rate from mL/min to L/12 h. Multiply by 60 (to convert
from min to h), then by 12 (to convert from h to 12 h) and finally divide by 1,000
(to convert from mL to L):

Filtration rate (L/12 h) = 110 x 60 x 12 = 79.2 L/12h

1,000

The answer is required in mmol so divide the plasma concrentration by 1,000 to

convert from μmol/L to mmol/L (1,000 μmol = 1 mmol):

Plasma creatinine (mmol/L) = 180 = 0.18 mmol/L

1,000

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Creatinine filtered (mmol/12 h) =

Plasma creatinine (mmol/L) x Filtration rate (L/12 h)

Creatinine filtered (mmol/12 h) = 0.18 x 79.2 = 14.3 mmol/12 h

3. A urine collection (volume 3.2 L) was handed in by a patient which he said he had
collected over the previous day. Calculate the creatinine clearance given that the
urine was found to have a creatinine concentration of 7.2 mmol/L. The plasma
creatinine concentration taken during the collection was 94 µmol/L. Give the
most likely cause for this result.

Creatinine clearance(mL/min) =

Urine creatinine (mmol/L) x Urine flow rate (mL/min)

Plasma creatinine (mmol/L)

Urine creatinine concentration = 7.2 mmol/L

Plasma creatinine = 94 μmol/L = 94 mmol/L

1,000

Urine flow rate = 3.2 L/24 h = 3.2 L/h = 3.2 L/min = 3.2 x 1,000 mL/min
24 24 x 60 24 x 60

Creatinine clearance = 7.2 x 3.2 x 1,000 x 1,000 = 170 mL/min

24 x 60 x 94

This creatinine clearance is a little high. The most likely cause is that the 24
collection was made over a longer period than 24 h – perhaps the bladder was not
emptied at the start of the collection period (or if emptied it was added to the
collection instead of being discarded).

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 WORKED ANSWERS TO FURTHER QUESTIONS

4. The concentration of a compound in the plasma of a normal adult is 10 mg/L.

The GFR is 110 mL/min and 316.8 mg of the compound are excreted over 24 h in
a urine volume of 1584 mL. Comment on these findings.

First calculate the total amount of the compound filtered over a 24 h period (based
on the assumption that the compound is freely filtered at the glomerulus):

Amount filtered (mg) = GFR (L/24h) x Plasma concentration (mg/L)

GFR = 110 mL/min = 110 L/min = 110 x 60 L/h = 110 x 60 x 24 L/24 h

1,000 1,000 1,000

Amount filtered (mg/24 h) = 110 x 60 x 24 x 10 = 1584 mg/24 h

1,000

The rate of excretion (316.8 mg/24 h) is much less than this suggesting either that
the compound is either not freely filtered at the glomerulus or considerable
amounts are reabsorbed from the filtrate.

N.B. The urine volume was not used in this calculation. Another approach (which
would utilize urine volume) would be to calculate the clearance of the compound
(which comes out at 22 mL/min) then compare it with the GFR.

5. A subject with a GFR of 100 mL/min was infused with a 'drug' X at a rate of
100 µmol/min and the plasma concentration reached a steady state value of
200 µmol/L. It is known that this drug is not metabolized or excreted by organs
other than the kidney. What is the clearance of this drug? Comment on the
result.

When a steady state is reached the rate of excretion is equal to the rate of infusion
and the plasma concentration reaches a constant value.

Clearance (mL/min) = Excretion rate (μmol/min)

Plasma concentration (μmol/mL)

Excretion rate = infusion rate = 100 μmol/min

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Plasma concentration = 200 μmol/L = 200 μmol/mL

1,000

Clearance (mL/min) = 100 x 1000 = 500 mL/min

200

The clearance of the drug far exceeds the GFR suggesting that the mode of
excretion is predominantly tubular secretion.

6. A patient who is severely water depleted and excreted only 100 mL of urine in the
last 6 hours was a short time before, found to have a creatinine clearance of 100
mL/min with a plasma creatinine concentration of 100 µmol/L. If renal function
has remained unchanged what concentration of creatinine would you expect to
find in the latest 100 mL (6 h collection) specimen of urine?

This question involves calculating the urinary excretion when the plasma
concentration and clearance is known. The expression for clearance is:

Clearance (mL/min) =

Urine creatinine (mmol/L) x Urine flow rate (mL/min)

Plasma creatinine (mmol/L)

Rearranging this expression:

Urine creatinine (mmol/L) =

Clearance (mL/min) x Plasma creatinine (mmol/L)

Urine flow rate (mL/min)

Creatinine clearance = 100 mL/min

Plasma creatinine = 100 μmol/L = 100 mmol/L

1,000

Urine flow rate = 100 mL/6 h = 100 mL/h = 100 mL/min

6 6 x 60

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 WORKED ANSWERS TO FURTHER QUESTIONS

Urine creatinine (mmol/L) = 100 x 100 x 6 x 60 = 36 mmol/L

1,000 x 100

7. Estimate the effect on urinary sodium excretion in a person with a GFR of

95 mL/min and plasma sodium of 140 mmol/L, of a 1% decrease in the overall
reabsorption of sodium.

First calculate the amount of sodium filtered at the glomerulus in mmol/min:

Na filtered (mmol/min) = GFR (mL/min) x Plasma Na (mmol/mL)

Plasma Na = 140 mmol/L = 140 mmol/mL

1,000

Na filtered (mmol/min) = 95 x 140 = 13.3 mmol/min

1,000

If the amount reabsorbed decreases by 1% then the amount excreted in the urine
will increase by 1% of that filtered:

Increase in Na excretion = 13.3 x 1 = 0.133 mmol/min

100

Therefore the increase in urine Na over a 24 h period =

0.133 x 60 x 24 = 192 mmol/24 h

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8. The following data were obtained for a hypertensive patient on a low sodium diet:

Plasma: creatinine = 200 μmol/L sodium = 155 mmol/L

24 h Urine: creatinine = 12.5 mmol/L volume = 1250 mL

If the renal tubules reabsorb 90% of filtered sodium, how many grams of
sodium are excreted in the same 24 h period?

Fractional excretionNa = Na excreted in urine

Na filtered

If 90% of filtered Na is reabsorbed then 100 – 90 = 10% must excreted i.e.

fractional excretion (FENa) = 10%

FENa is calculated from the expression:

FENa (%) = (UrineNa x PlasmaCreatinine) x 100

UrineCreatinine x PlasmaNa

Which can be re-arranged to give an expression for urine sodium:

UrineNa = FENa x UrineCreatinine x PlasmaNa mmol/L

PlasmaCreatinine x 100

Substitute these values to obtain the urine sodium concentration N.B units must be
the same so convert plasma creatinine (μmol/L) to mmol/L by dividing it by
1,000.

UrineNa = 10 x 12.5 x 155 x 1,000 = 969 mmol/L

200 x 100

Since the 24 h urine volume is 1250 mL (= 1.25 L) the amount excreted in 24 h

is:
969 x 1.25 = 1211 mmol/24 h

Convert to g/24 h:

Na (g/24 h) = Na (mol/24 h) x MW

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 WORKED ANSWERS TO FURTHER QUESTIONS

Divide the sodium output by 1,000 to convert from mmol/24 to mol/24 h. MW

Na = 23.

Na (g/24 h) = 1211 x 23 = 28 g (2 sig figs)

1,000

Or expressed as NaCl (MW = 58.5):

NaCl (g/24 h) = 1211 x 58.5 = 71 g (2 sig figs)

1,000

Another approach to this problem would be to first calculate the GFR from the
creatinine results, then use this to calculate the Na filtered etc.

9. The following results were obtained in a 20 year old male admitted after a car
crash and found to be oliguric:

Plasma Na 135 mmol/L

K 5.0 mmol/L
Urea 25.0 mmol/L
Creatinine 250 µmol/l

Urine Na 90 mmol/L
Creatinine 2.4 mmol/L
Osmolality 200 mOsm/kg

Calculate the fractional excretion of sodium.

FENa = UrineNa x PlasmaCreatinine

UCreatinine x PlasmaNa

All units need to be the same, if mmol/L used then:

Plasma creatinine = 250 μmol/L = 250 mmol/L

1,000

FENa = 90 x 250 = 0.069 (or 6.9%)

1,000 x 2.4 x 135

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10. A 45 year old lady has a body weight of 56 kg and a height of 155 cm. If her
plasma creatinine is 150 μmol/L estimate her GFR expressing the result as
mL/min/1.73 m2.

Since body weight is given, the Cockcroft-Gault formula for females can be used:

GFR (mL/min) = (140 – age in yrs) x Body wt (kg) x 1.2 x 0.85

Plasma creatinine (μmol/L)

GFR (mL/min) = (140 – 45) x 56 x 1.2 x 0.85

150

= 36 mL/min

Next calculate the patient’s body surface area (A) using the body weight in kg (W)
and height in cm (H):

A = antilog10 [(0.425 x log10 W) + (0.725 x log10 H) – 2.144] m2

A = antilog10 [(0.425 x log10 56) + (0.725 x log10 155) - 2.144]

= antilog10 [(0.425 x 1.75) + (0.725 x 2.19) - 2.144]

= antilog10 [0.744 + 1.588 - 2.144]

= antilog10 0.188

= 1.54 m2

Corrected GFR (mL/min/1.73m2) = Measured GFR (mL/min) x 1.73

A (m2)

= 36 x 1.73
1.54

= 40 mL mL/min/1.73 m2

Alternatively the abbreviated MDRD formula can be used (height not required).

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 WORKED ANSWERS TO FURTHER QUESTIONS

11. Calculate the tubular maximum reabsorptive capacity (Tm/GFR) for glucose from
the following data:

Plasma glucose 10 mmol/L Plasma creatinine 120 μmol/L

Urine glucose 50 mmol/L Urine creatinine 6.0 mmol/L

The urine (volume 30 mL) was collected over a 15 minute period.

First calculate the fractional excretion of glucose (FEGlucose):

FEGlucose = UrineGlucose x PlasmaCreatinine

UrineCreatinine x PlasmaGlucose

All units must be the same so first correct plasma creatinine to mmol/L:

Plasma creatinine = 120 μmol/L = 120 mmol/L

1,000

FEGlucose = 50 x 120 = 0.1

1,000 x 6.0 x 10

This is the fraction of filtered glucose which is NOT reabsorbed by the tubules.
The fraction reabsorbed (TR) is next calculated:

TR = 1 - FE = 1 - 0.1 = 0.9

To convert this reabsorption fraction to the absolute amount reabsorbed (i.e. The
Tm/GFR), multiply by the plasma concentration:

Tm/GFR = TR x Plasma concentration

= 0.9 x 10

= 9 mmol/L glomerular filtrate

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12 A 6 h urine collection (volume 800 mL) has an osmolality of 200 mOsm/kg. If the
plasma osmolality is 260 mOsm/kg calculate the free water clearance in mL/min.

First calculate the osmolar clearance (Cosm):

Cosm = Uosm x V
Posm

Posm = 260 mOsm/kg

Uosm = 200 mOsm/kg
V = urine flow rate = 800 mL/6mh

= 800 mL/h = 800 mL/min = 2.22 mL/min

6 6 x 60

Cosm = 2 00 x 2.22 = 1.71 mL/min

260

The free water clearance (Cwater) is the difference between the urine flow rate and
the osmolar clearance:

Cwater = V - Cosm

Cwater = 2.22 - 1.71 = 0.51 mL/min

13. An estimation of glomerular filtration rate can be calculated using the

abbreviated MDRD (Modified Diet in Renal Disease) formula:

GFR (mL/min/1.73m2) = 186 x [serum creatinine x 0.011312]-1.154

x [age in years]-0.203 x 0.742 if female and/or x 1.21 if Afro American origin

(where serum creatinine is in μmol/L)

Calculate the GFR for a 57 year old Caucasian women whose serum creatinine is
130 μmol/L, and her creatinine clearance, given that a 24 h urine collection with
a volume of 1.1 L had a creatinine concentration of 4.7 mmol/L.

Comment critically on the two values.

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 WORKED ANSWERS TO FURTHER QUESTIONS

First calculate the GFR using the abbreviated MDRD formula by substituting
values for serum creatinine (130 μmol/L) and age (57 y) – remembering to
multiply by 0.742 since the patient is female:

GFR (mL/min/1.73m2) = 186 x [130 x 0.011312]-1.154 x [57]-0.203 x 0.742

= 186 x 1.471 –1.154 x 57 –0.203 x 0.74

= 186 x antilog10 [-1.154 x log10 1.471] x antilog10 [-0.203 x log10 57] x 0.742

= 186 x antilog10 [-1.154 x 0.1676] x antilog10 [-0.203 x 1.7559] x 0.742

= 186 x antilog10 (-0.1934) x antilog10 (-0.3565) x 0.742

= 186 x 0.6406 x 0.4400 x 0.742 = 39 mL/min/1.73 m2 (2 sig figs)

Next calculate the creatinine clearance:

Creatinine clearance (mL/min) =

Urine creatinine (mmol/L) x Urine flow rate (mL/min)

Serum creatinine (mmol/L)

Urine creatinine = 4.7 mmol/L

Serum creatinine = 125 μmol/L = 130 mmol/L

1,000

Urine flow rate = 1.1 L/24 h = 1.1 x 1,000 mL/24 h

= 1.1 x 1,000 mL/h = 1.1 x 1,000 mL/min

24 24 x 60

Creatinine clearance (mL/min) = 4.7 x 1.1 x 1,000 x 1,000

24 x 60 x 130

= 28 mL/min

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

There are several possible reasons for the discrepancy between the derived GFR and
the calculated clearance:

• Inaccuracy in the timed urine collection. This is potentially the greatest source
of error. Although the 24 h volume of 1.1 L seems reasonable the calculated
creatinine excretion seems low (1.1 x 4.7 = 5.2 mmol/24 h) - unless the lady has
a very low muscle mass – suggesting that the collection is incomplete.

• Failure to correct the creatinine clearance for body surface area (this would
require knowledge of weight and height). However, the MDRD formula does not
take into account individual variation in body surface area either, but just assumes
an average value based on the patient’s age and sex.

• Creatinine is secreted by tubules into the urine so that creatinine clearance

measurements always overestimates GFR.

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 6

1. Calculate the approximate osmolality of a glucose/saline infusion containing

equal proportions of 5% glucose and 0.9% sodium chloride.

First calculate the osmolalities due to glucose and sodium chloride individually.

Formula for glucose = C6H12O6

AW C = 12, therefore C6 = 6 x 12 = 72
AW H = 1, therefore H12 = 12 x 1 = 12
AW O = 16, therefore O6 = 6 x 16 = 96

MW = 180

OsmolalityGlucose = Glucose concentration (g/L)

MW

Glucose concentration = initially 5% = finally 2.5% (since mixed with an

equal volume of saline).

2.5% glucose = 2.5 g/100 mL = 25 g/L

OsmolalityGlucose = 25 = 0.139 Osm/kg = 139 mOsm/kg

180

First calculate mmolar sodium chloride concentration:

NaCl = 0.9% = 0.9 g/100 mL = 9 g/L

Final concentration (after mixing with an equal volume of 5% glucose) is one half
of this i.e. 4.5 g/L. MW of NaCl = 23 + 35.5 = 58.5.

NaCl (mol/L) = NaCL (g/L) = 4.5 = 0.077 mol/L = 77 mmol/L

MW 58.5

Sodium chloride dissociates to give two osmotically active species – Na+ and Cl-

Therefore, OsmolalityNaCl = 2 x 77 = 154 mOsm/kg

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OsmolalityTotal = OsmolalityGlucose + OsmolalityNaCl

= 139 + 154

= 293 mOsm/kg (i.e. essentially isosmolar)

2. Calculate the approximate osmolality of a solution containing 10% mannitol and

0.9% saline (MW mannitol = 182).

Calculate individual osmolalities separately.

Mannitol is undissociated so that OsmolalityMannitol = Molar concentration

Concentration of mannitol = 10% = 10 g/100 mL = 100 g/L

OsmolalityMannitol = Mannitol (g/L) = 100 Osm/kg

MW 182

= 100 x 1,000 mOsm/kg = 549 mOsm/kg

182

Concentration of NaCl = 0.9% = 0.9 g/100 mL = 9 g/L

MW NaCl = 23 + 35.5 = 58.5

NaCl (mmol/L) = NaCl (g/L) x 1,000 = 9 x 1,000 = 154 mmol/L

MW 58.5

Each molecule of NaCl dissociates into 2 ions (Na+ and Cl-).

OsmolalityNaCl = 2 x NaCl (mmol/L) = 2 x 154 = 308 mOsm/Kg

Adding these together gives the total osmolality:

OsmolalityTotal = OsmolalityMannitol + OsmolalityNaCl

= 549 + 308

= 857 mOsm/kg

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 WORKED ANSWERS TO FURTHER QUESTIONS

3. A patient was mistakenly given 500 mL 20% mannitol (C6H14O6) intended for the
patient in the next bed instead of the same volume of normal (0.9%) saline.
Calculate the extra osmolal load given over that which would have resulted from
isotonic saline.

First calculate the osmotic load of 500 mL of each solution.

For 20% mannitol:

Concentration = 20% = 20 g/ 100 mL = 200 g/L

AW C = 12, therefore C6 = 6 x 12 = 72
AW H = 1, therefore H14 = 14 x 1 = 14
AW O = 16, therefore O6 = 6 x 16 = 96
MW = 182

OsmolalityMannitol = Mannitol (g/L) = 200 Osm/kg

MW 182

= 200 x 1,000 mOsm/kg = 1099 mOsm/kg

182

Therefore osmotic load of 500 mL = 1099 = 550 mOsm

2

For 0.9% saline:

Concentration = 0.9% = 0.9 g/100 mL = 9 g/L

MW NaCl = 23 + 35.5 = 58.5

OsmolalityNaCl = NaCl (g/L) x 2 Osm/kg = 9 x 2 Osm/kg

MW 58.5

= 9 x 2 x 1,000 mOsm/kg = 308 mOsm/kg

58.5

(factor of 2 introduced since NaCl dissociates into 2 ions (Na+ and Cl-).

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Osmolar load due to 500 mL NaCl = 308 = 154 mOsm

2

Extra osmolal load = Osmolal loadMannitol - Osmol loadNaCl

= 550 - 154

= 396 mOsm

4. What increase in plasma osmoality would result from a plasma ethanol

concentration of 92 mg/dL?

First convert ethanol concentration to mmol/L:

Ethanol concentration = 92 mg/dL = 920 mg/L

Formula for ethanol = C2H5OH

AW C = 12, therefore C2 = 2 x 12 = 24
AW H = 1, therefore H6 = 6 x 1 = 6
AW O = 16, therefore O = 1 x 16 = 16
MW = 46

OsmolalityEthanol = Ethanol (mg/L) = 920 = 20 mOsm/kg

MW 46

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 WORKED ANSWERS TO FURTHER QUESTIONS

5. A 45-year old man is brought to casualty following a fit. He had been working
alone late in a garage, when he was found by the security guard who called an
ambulance. On admission, he has a large bruise on the left temple and is semi-
comatose, he smells of alcohol. The admitting team request urea and electrolytes,
glucose and an alcohol and blood gas estimation and arrange an urgent CT scan.
The results are as follows:

Sodium 141 mmol/L Urea 3.5 mmol/l

Ethanol 270 mg/dL Glucose 3.2 mmol/L

The CT scan does not show any bony injury or evidence of intracranial bleed.
The neurological registrar is called and asks for an osmolal gap to help provide a
quick estimation of whether there is a possibility that other toxic substances
present in the garage, such as antifreeze, have been taken in any quantity.

The measured osmolality is 330 mOsm/kg

Calculate the osmolal gap

Show whether the alcohol concentration explains the observed osmolal gap,
explaining any assumptions you make in the process.

a) First calculate osmolality due to Na+, glucose and urea:

Osmolality = 1.86 [Na+] + [glucose] + [urea] + 9

mOsm/kg mmol/L mmol/L mmol/L

Osmolality = (1.86 x 141) + 3.2 + 3.5 + 9 = 278 mOsm/kg

Osmolal gap = OsmolalityMeasured - OsmolalityCalculated

= 330 - 278

= 52 mOsm/kg

b) Calculate the expected contribution from ethanol:

Ethanol concentration = 270 mg/dL = 2,700 mg/L

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Formula of ethanol = C2H5OH

AW C = 12, therefore C2 = 2 x 12 = 24
AW H = 1, therefore H6 = 6 x 1 = 6
AW O = 16, therefore O = 1 x 16 = 16
MW = 46

OsmolalityEthanol = Ethanol (mg/L) = 2,700 = 59 mOsm/kg

MW 46

The osmolal gap is in reasonable agreement with the expected osmolal

contribution from ethanol. Therefore the ethanol concentration explains
the observed osmolal gap.

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 7

1. An antidepressant drug has a biological half-life of 30 hours. How long will it

take a plasma concentration of 50 mg/L to fall to 20 mg/L?

The first order elimination rate equation is:

ln Cpt = ln Cp0 - kd.t

Where Cpt = drug concentration at time t = 20 mg/L

Cp0 = initial drug concentration = 50 mg/L
kd = elimination rate constant

kd can be calculated from the half-life (t½ = 30 h):

kd = 0.693 = 0.693 = 0.023 h-1

t½ 30

Substitute these values into the rate equation and solve for t:

ln 20 = ln 50 - 0.023.t

3.00 = 3.91 - 0.023.t

0.023.t = 3.91 - 3.00 = 0.91

t = 0.91 = 40 h (2 sig figs)

0.023

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

2. A 15 year old boy presents to casualty following a convulsion. It turns out that he
had swallowed 30 of his mother’s lithium tablets about 10 hours previously.
On admission his lithium concentration is 4.1 mmol/L. A decision needs to be
made whether to haemodialyse him to reduce the lithium concentration. As this is
not going to be available quickly, the physicians want to know how long he will
have toxic levels just with endogenous clearance. Estimate the following,
indicating clearly any assumptions you have made:

a) The likely volume of distribution of the lithium at this stage in the situation,
given a body weight of 65 kg.

b) How long it will be before his lithium concentration drops to the relatively
safe level of 1.5 mmol/L below which toxicity is unlikely, given a clearance of
0.03 L/h/kg.

a) The volume of distribution of a drug is usually calculated by dividing the

total dose administered by the plasma concentration. In this question we
do not have a reliable estimate of the amount ingested. Since lithium is
readily water soluble its volume of distribution approximates to total body
water volume.

Total body water (L) = Body wt (kg) x % Body water

100

Assuming an average body water content of 60%:

Volume of distribution (Vd) = 65 x 60 = 39 L

100

c) Lithium is excreted from the body by glomerular filtration (with some

reabsorption by the proximal tubule which we shall ignore) and so its
elimination follows first order kinetics:

ln Cpt = ln Cp0 - kd.t

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 WORKED ANSWERS TO FURTHER QUESTIONS

Cp0 = initial concentration (before dialysis) = 4.1 mmol/L

Cpt = concentration at time t = 1.5 mmol/L

t = time taken (in hours) to reach the “safe” level of 1.5 mmol/L

kd = elimination rate constant

The clearance of the drug is given as 0.03 L/h/kg. Multiply by the patient’s
weight to give the total clearance:

Clearance = 0.03 (L/h/kg) x 65 (kg) = 1.95 L/h

The elimination rate constant (kd) can be calculated from the clearance (Cl)
and the volume of distribution (Vd):

kd = Cl = 1.95 = 0.050 h-1

Vd 39

Substitute for Cpt, Cp0 and kd then solve for t:

ln 1.5 = ln 4.1 - 0.050.t

0.405 = 1.411 - 0.050.t

0.050.t = 1.411 - 0.405 = 1.006

t = 1.006 = 20 h (2 sig figs)

0.050

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

3. A 60 mg dose of a drug is given to a male experimental subject who weighs 80 kg.

Assuming that the drug is completely absorbed and distributed evenly throughout
the total body water, estimate the potential peak plasma level. If the drug were
distributed only within the extracellular compartment, what would the plasma
level be?

Assuming distribution throughout total body water, the Vd = total body water vol:

Assume body water is 60% of body weight.

Total body water (L) = Body Wt (Kg) x 60%

= 80 x 60 = 48 L
100

Vd = Amount of drug in body (dose)

Plasma drug concentration

Plasma drug level (mg/L) = Dose (mg) = 60 = 1.25 mg/L

Vd (L) 48

If drug is only distributed throughout the ECF, the Vd must be adjusted. ECF is
normally 20% of body wt.

Vd (L) = Body wt (kg) x 20%

= 80 x 20 = 16 L
100

Plasma drug level (mg/L) = Dose (mg) = 60 = 3.75 mg/L

Vd (L) 16

Alternatively, since a third of body water is in the ECF, the drug level will be
3 times higher.

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 WORKED ANSWERS TO FURTHER QUESTIONS

4. A bolus of 6 g drug is given IV and 3 blood samples collected at intervals.

Time mg/L
2.5h 32
5h 10
7.5h 3

a) What is the half-life of the drug?

b) What is the volume of its distribution?

a) Assuming the clearance of the drug follows first order elimination kinetics then
the data should be described by the expression:

ln Cpt = ln Cp0 - kd.t

Therefore a plot of ln C versus t should be linear with an intercept on the ln C axis

of Cp0 and slope -kd:

Time (h) Conc (mg/L) lnC

2.5 32 3.47
5 10 2.30
7 .5 3 1.10

5
ln C0
4.5
Elimination curve
4

3.5

3
ln C

2.5

2
slope = - kd
1.5

0.5

0
0 1 2 3 4 5 6 7 8

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

This plot clearly demonstrates that elimination of the drug follows first order
kinetics so that Cp0 and kd could be determined directly from the graph.
Alternatively any 2 values can be substituted into the rate equation and solved
for kd:

Let level at time 2.5 h be Cp0 = 32 mg/L

Let level at time 5 h be Cpt = 10 mg/L
t = 2.5 h (the time difference between Cp0 and Cpt).

Therefore: ln 10 = ln 32 - kd.2.5

2.303 = 3.466 - 2.5 kd

2.5 kd = 3.466 - 2.303 = 1.163

kd = 1.163 = 0.465 h-1

2.5

a) Half-life (t½) can be calculated from kd:

t½ = 0.693 = 0.693 = 1.5 h (2 sig figs)

kd 0.465

b) First calculate the initial concentration (Cp0) using one other value (e.g.
2.5 h = 32 mg/L as Cpt and t = 2.5 h) and the value for kd:

ln 32 = ln Cp0 - (0.465 x 2.5)

3.466 = ln Cp0 - 1.163

ln Cp0 = 3.466 + 1.163 = 4.629

Cp0 = antiloge 4.629 = 102 mg/L

The Vd is then calculated from dose and Cp0:

Vd (L) = Dose (mg) = 6,000 = 59 L (2 sig figs)

Cp0 (mg/L) 102

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 WORKED ANSWERS TO FURTHER QUESTIONS

5. The plasma concentration of a drug is found to be 200 nmol/L at 9.00 am. It’s
elimination follows first order kinetics with a rate constant is 0.34/h. Calculate
the times at which the plasma concentrations will reach 100 nmol/L and
75 nmol/L.

The first order rate equation is:

ln Cpt = ln Cp0 - kd.t

where Cp0 = 200 nmol/L; kd = 0.34 h-1

Calculation of t when Cpt = 100 nmol/L:

ln 100 = ln 200 - 0.34.t

4.605 = 5.298 - 0.34.t

0.34.t = 5.298 - 4.605 = 0.693

t = 0.693 = 2.0 h (2 sig figs)

0.34

Calculation of t when Cpt = 75 nmol/L

ln 75 = ln 200 - 0.34.t

4.317 = 5.298 - 0.34.t

0.34.t = 5.298 - 4.317 = 0.981

t = 0.981 = 2.9 h (2 sig figs)

0.34

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

6. A patient in casualty with a suspected adrenal crisis is given an iv dose of

hydrocortisone at 18.00. The medical team on take wish to carry out a short
synacthen test to confirm the diagnosis but there will be a significant contribution
form the administered drug until its concentration has fallen to 10% of the peak
value. If the half-life of hydrocortisone is 2 h, what is the earliest time at which
the test can be carried out?

Assuming elimination follows first order kinetics:

ln Cpt = Cp0 - kd.t

Where Cpt = concentration at time t = 10%

Cp0 = initial concentration = 100%
t = time when Cpt reaches 10%

Calculate kd from t½:

kd = 0.693 = 0.693 = 0.347 h-1

t½ 2

Substitute these values into the rate equation and solve for t:

ln 10 = ln 100 - 0.347.t

2.303 = 4.605 - 0.347.t

0.347.t = 4.605 - 2.303 = 2.302

t = 2.302 = 6.6 h (2 sig figs)

0.347

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 WORKED ANSWERS TO FURTHER QUESTIONS

7. The SHO decides to treat a patient (weight 80 kg) with intravenous theophylline
(salt factor = 0.8). What loading dose would you recommend in order to achieve
a theophylline level of 12 mg/L given a volume of distribution of 0.5 L/kg and an
initial plasma theophylline level of 4 mg/L?

Loading dose (LD) = Vd x (Cptarget - Cpinitial)

S x F

Where Vd = volume of distribution = 0.5 L/kg

Cptarget = desired drug level = 12 mg/L
Cpinitial = starting drug level = 4 mg/L
S = salt factor = 0.8
F = bioavailability (not given so assume a value of 1)

Patients Vd = Body weight (kg) x 0.5

= 80 x 0.5 = 40 L

Substitute these values in order to obtain LD:

LD = 40 x (12 - 4)
0.8

= 40 x 8
0.8

= 400 mg

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

8. A patient, (body weight 55 kg) unable to take oral medication, had been receiving
intravenous valproate for a number of days and achieved an average steady state
level of 75 mg/L. After regaining consciousness the doctors wished to change to
an oral twice daily regimen. In order to maintain the same average steady state
concentration what dose would you recommend. Assume a clearance of
10 mL/h/kg, a bioavailability of 0.7 and a salt factor of 0.85.

The following expression allows calculation of the maintenance dose:

Maintenance dose = Cpss x Cl x τ

S x F

Where:

Cpss = steady state plasma concentration = 75 mg/L

Cl = clearance = 10 mL/h/kg
τ = dosing interval = 12 h (i.e. twice daily)
S = salt factor = 0.85
F = bioavailability = 0.7

First correct the clearance for the body weight and express it in litres (to be
compatible with the drug concentration which is given in mg/L):

Cl (L/h) = Cl (mL/h/kg) x Body wt (kg)

1,000

= 10 x 55 = 0.55 L/h
1,000

Substitute these values into the expression for maintenance dose:

Maintenance dose (mg) = 75 x 0.55 x 12

0.85 x 0.7

= 832 mg

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 8

1. Over a 24 h period a patient recovering from intestinal resection receives 2 L of

fluids intravenously and 750 mL orally but does not eat any solids over this
period. The urine output over the same period is 1.25 L and 600 mL of fluid is
lost via a fistula. Is the patient in positive or negative fluid balance and by how
much?

Draw up a table of fluid gains and losses then calculate the total of each. Assume
a value of 400 mL per day for net insensible losses.

Fluid gains Fluid losses

Oral 750 mL Urine output 1,250 mL

IV 2,000 mL Loss via fistula 600 mL
Net insensible loss 400 mL

Total 2,750 mL 2,250 mL

Fluid balance (mL) = Net fluid intake (mL) - Net fluid loss (mL)

= 2,750 - 2,250

= 500 mL

i.e. there is a net fluid gain of 500 mL.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

2. A patient known to have diabetes insipidus is admitted in coma. His plasma

osmolality is 324 mosm/kg. If his weight is 85 kg, estimate his body water deficit.

The average adult male has a body water content of approximately 60%. If the
body water deficit is x L, then the initial body water content can be calculated:

Initial body water (L) = (85 + x) x 60

100

= 5100 + 60x
100

= 51 + 0.6x

Assuming a normal initial osmolality (say 285 mOsm/kg) the total amount (in
mOsm) of osmotically active species present in the body can be calculated:

Osmolality (mOsm/kg) = Total solutes (mOsm)

Initial body water (kg)

285 = Total solutes (mOsm)

51 + 0.6x

Total solutes (mOsm) = 285 (51 + 0.6x)

= 14,535 + 171x

On presentation his body weight is 85 kg. Assuming the total amount of solutes
in the body is unchanged, then the body water volume can be calculated from the
current osmolality:

Final osmolality (mOsm/kg) = Total solutes (mOsm)

Final body water (kg)

324 = 14,535 + 171x

(51 + 0.6x) - x

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 WORKED ANSWERS TO FURTHER QUESTIONS

324 = 14,535 + 171x

51 - 0.4x

324(51 – 0.4x) = 14,535 + 171x

16,524 – 130x = 14,535 + 171x

171x + 130x = 16,524 - 14,535

301x = 1989

x = 1989
301

= 6.6 L

If it is assumed that the change in body wt is neglible (or that the initial body
water was the same as for an average 70 kg adult) then a simpler calculation
(using Eq. 8.3) can be used and gives a slightly different result which may be
adequate as a rough guide in clinical practice:

Fluid loss (L) = 42 - [ 12000 ]

Osmolality (mOsm/kg)

= 42 - [12000]
324

= 42 - 37

= 5L

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

3 A male adult insulin dependent diabetic forgot to take his insulin. His blood
glucose concentration, which was 5 mmol/L, rose to 15 mmol/|L in two hours.
Estimate the effect on his plasma sodium concentration, assuming that no other
water intake nor loss of water from the body takes place during this time, indicating
what assumptions you make.

Making a number of assumptions:

• That it is plasma glucose which is measured rather than whole blood glucose.

• That as a result of insulin deficiency there is no increase in glucose concentration

in the intracellular fluid (ICF).

• That the plasma glucose has equilibrated with interstitial fluid so that it’s
concentration in the extracellular fluid (ECF) is the same as in plasma.

• That there is negligible change in the concentrations of solutes other than glucose,
sodium and chloride.

• That the ratio of ICF:ECF volumes is 2 (i.e. ECF = 14L, ICF = 28L for average
adult male) and that the total body water is that of an average male i.e. 42 L

The effect of an increase in plasma (and hence ECF) glucose is to raise plasma (and
ECF) osmolarity. The body will retain water (stimulation of thirst increases intake
and stimulation of ADH reduces renal loss) until osmotic equilibrium is restored. If
there is a plentiful supply of water then the plasma osmolarity is returned to normal
and since the plasma glucose has risen by 10 mmol/L the plasma sodium must have
fallen by 10/2 = 5 mmol/L. However, this question states that there is no net loss or
gain of body water. Therefore, water will move, by osmosis, from the ICF
compartment (iso-osmolar) to the ECF (now hyper-osmolar) until osmotic
equilibrium is established. Since movement of water from the ICF leads to an
increase in ICF osmolarity, the movement of water is restricted and at equilibrium the
ECF will reach a value somewhere in-between normality and the original value i.e.
the osmotic load is shared between the ECF and ICF compartments, both of which
become hyperosmolar.

416
 WORKED ANSWERS TO FURTHER QUESTIONS

The plasma glucose has risen by 15 - 5 = 10 mmol/L

Rise in amount of glucose in ECF =

Rise in plasma glucose concentration (mmol/L) x ECF vol(L)

= 10 x 14 = 140 mmol

(a slight underestimate since there has been a small expansion in ECF vol)

At equilibrium, the rise in osmolarity (which is the same in the ECF and ICF) is
given by:

Increase in amount of glucose in body (mmol)

Total body fluid (ECF + ICF) volume (L)

= 140 = 3.33 mmol/L

42

Since the plasma osmolarity has risen by 3.33 mOsmol/L and the plasma glucose
by 10 mmol/L then the concentration of NaCl which has been displaced by
glucose is

10 - 3.33 = 6.67 mmol/L

and so the sodium has fallen by 6.67 = 3.34 mmol/L

2

i.e. the plasma sodium concentration has decreased by approximately 3 mmol/L.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

4. A plasma sample with a total protein content of 70 g/L gave identical sodium
results of 140 mmol/L when measured using either a direct-reading ion-selective
electrode or a flame photometer. What plasma sodium result would you expect
the ion-selective electrode to give with the same plasma sample if its total protein
concentration had been 90 g/L?

Flame photometry measures sodium as concentration in plasma i.e. 140 mmol/L

of plasma.

A direct-reading ion-selective electrode measures sodium as activity i.e.

140 mmol/L of plasma water. Large molecules such as proteins occupy
significant space in solution i.e. displace plasma water. If plasma contains 70 g/L
protein then this is equivalent to 0.070 kg/L. Assuming that 1 kg of protein
occupies a volume of 1 L then the volume of plasma water in which the sodium is
dissolved is (1.0 - 0.07) = 0.93 L. Assuming that the activity is the same as
concentration for sodium in plasma water (i.e. the activity coefficient is one), for a
plasma sodium of 140 mmol/L of plasma, the true concentration of sodium in
plasma water is:

Plasma sodium = 140 = 150.5 mmol/L water

0.93

There are two ways in which the ISE reading can be converted to the same as that
obtained by flame photometry (140 mmol/L):

• Subtraction of 10.5 mmol/L from the result

• Multiplication of the result by the factor 140/150.5 i.e. 0.930

At a protein concentration of 90 g/L (occupying 0.090 L plasma), the

concentration of sodium in plasma water will be:

140 = 140 = 153.8 mmol/L plasma water

(1.00 - 0.09) 0.91

Carrying out the two adjustments by the instrument:

• Subtraction of 10.5 gives 153.8 - 10.5 = 143.3 mmol/L.

• Multiplication by 0.930 gives 153.8 x 0.930 = 143.0 mmol/L

Therefore expected ISE reading = 143 mmol/L

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 9

1. An assay mixture for the measurement of lactate dehydrogenase constituted 2.7

mL of buffered NADH and 100 µL of serum. The reaction was started by adding
100 µL of sodium pyruvate. The absorbance change over 5 minutes was 0.150
when measured in a 0.5 cm light path at 340 nm. Assuming the molar
absorbtivity of NADH at 340 nm is 6.30 x 103 L.mol-1cm-1, calculate the enzyme
activity in international units per litre of serum.

One international unit of activity is the amount of enzyme present in 1 L of serum

which catalyses the conversion of 1 μmol substrate per min under the conditions
of the assay.

First calculate the absorbance change per min:

ΔA/min = ΔA/5min = 0.150 ΔA/min

5 5

Convert the absorbance change to concentration change per min:

ΔA = a.b.Δc

Where ΔA/min = 0.150

5

a = molar absorptivity = 6.30 x 103 L.mol-1.cm-1

b = light path length = 0.5 cm

Δc = change in concentration (mol/min)

Substitute these values then re-arrange to give an expression for Δc/min:

0.150 = 6.30 x 103 x 0.5 x Δc/min

5

Δc/min = 0.150 mol/min/L reaction mixture

5 x 6.30 x 103 x 0.5

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Multiply by 1,000,000 to convert the concentration units from mol to μmol (1 mol
= 1,000,000 μmol):

Δc/min = 0.150 x 1,000,000 μmol/min/L reaction mixture

3
5 x 6.30 x 10 x 0.5

The final step is to convert the activity to μmol/min/L serum. In the assay 100 μL
of serum was mixed with 2.7 mL buffer and 100 μl of substrate.

Total assay volume = 2.7 + 0.1 + 0.1 = 2.9 mL

Δc/min/L serum = Δc/min/L assay mixture x Total assay vol (mL)

Serum vol (mL)

= 0.150 x 1,000,000 x 2.9

5 x 6.30 x 103 x 0.5 x 0.1

= 276 IU/L serum

2. An assay for alkaline phosphatase activity involved mixing 0.05 mL of serum with
2.7 mL buffer, allowing temperature to reach equilibrium then starting the
reaction by adding 0.2 mL of substrate (4-nitrophenyl phosphate). The increase
in absorbance in a 1cm cuvette due to the liberation of product (4-nitrophenol)
was 0.180 over a 5-minute period. Calculate the alkaline phosphtase activity
expressing the result as a) international units per litre of serum, and b) katals per
litre of serum. Assume that the molar absorptivity of 4-nitrophenol is 1.88 x 104
L/mol/cm.

a) One international, unit is the amount of enzyme which liberates one μmol
of product per minute. Therefore to calculate the alk phos activity in IU/L
serum the following steps are involved:

Determine the rate of absorbance change in ΔA/min.

ΔA/min = ΔA/5min = 0.180

5 5

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 WORKED ANSWERS TO FURTHER QUESTIONS

Convert to the rate of change in concentration using the molar absorptivity

and pathlength:

ΔA = a.b.Δc

ΔA = rate of absorbance change = 0.180 A/min

5

a = molar absorptivity of 4-nitrophenol = 1.88 x 104 L/mol/cm

b = pathlength of cuvette = 1 cm

Δc = rate of change of concentration in mol/L/min = ?

Substitute these values and re-arrange to give an expression for Δc/min:

0.180 = 1.88 x 104 x 1 x Δc/min

5

Δc/min = 0.180 mol/L/min

5 x 1.88 x 104 x 1

Multiply by 1,000,000 to convert units from mol/L/min to μmol/L/min (1

mol = 1,000,000 μmol):

Δc/min = 0.180 x 1,000,000 μmol/min/L reaction mixture

5 x 1.88 x 104 x 1

To convert to activity per L serum multiply by the total volume of reaction

mixture and divide by the sample volume – using the same units:

Serum = 0.05 mL
Buffer = 2.70 mL
Substrate = 0.20 mL
Total = 2.95 mL

Alk phos activity = 0.180 x 1,000,000 x 2.95 μmol/min/L serum

5 x 1.88 x 104 x 1 x 0.05

= 113 IU/L

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

b) One Katal is the amount of enzyme that catalyses the reaction of

1 mol substrate per second

If alk phos activity = 113 IU/L (μmol/min/L)

Then divide by 1,000,000 to convert from μmol to mol, then by 60

to convert from min to seconds.

113 IU/L = 113 Katals/L

1,000,000 x 60

= 1.88 x 10-6 Kat/L

3. The Somogyi saccharogenic method for the assay of amylase involves measuring
the rate of release of glucose from substrate. One Somogyi unit is the amount of
enzyme catalysing the release of 1 mg of glucose in 30 min per 100 mL serum.
Derive a factor to convert Somogyi units to international units per litre of serum.

Somogyi units = mg glucose/30 min/100 mL serum

International units = μmol/min/L serum

Consider a sample with activity of x Somogyi units

First convert from mg glucose to μmol glucose:

Glucose formula = C6H12O6

AW C = 12, therefore C6 = 6 x 12 = 72
AW H = 1, therefore H12 = 12 x 1 = 12
AW O = 16, therefore O6 = 6 x 16 = 96
MW = 180

Activity (Somogyi units) = x mg/30 min/100 mL

Activity (mmol/30/min/100 mL) = x

180

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 WORKED ANSWERS TO FURTHER QUESTIONS

Multiply by 1,000 to convert from mmol to μmol:

Activity (μmol/30 min/100 mL) = x x 1,000

180

Divide by 30 to obtain the rate per minute:

Activity (μmol/min/100 mL) = x x 1,000

180 x 30

Multiply by 10 to obtain the activity per litre:

Activity (μmol/min/L) = x x 1,000 x 10

180 x 30

= x x 1.85

IU/L = 1.85 x Somogyi Units

4 One Wroblewski-laDue unit is the amount of lactate dehydrogenase which results in

an absorbance change (due to NADH) at 340 nm of 0.001 per minute per mL serum
in a reaction mixture with a total volume of 3 mL. Derive a factor to convert
Wroblewski-LaDue units to International units per litre of serum. Assume the molar
absorptivity of NADH is 6.3 x 103 L/mol/cm.

International units = μmol/min/L serum

Wroblewski-laDue (W-l-D units) = 0.001 ΔA/min/mL serum (total volume 3 mL)

Multiply by 3 to obtain absorbance change obtained with 1 mL of undiluted serum,

then by 1,000 to obtain the absorbance change due to 1 L serum:

1 W-l-D unit = 0.001 x 3 x 1,000 ΔA/min/L serum

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Next convert ΔA/min to Δc (i.e. mol/L/min):

ΔA = a.b.Δc

ΔA = absorbance change = 0.001 x 3 x 1,000 A/min

a = molar absorptivity of NADH = 6.3 x 103 L.mol-1.cm-1

b = cuvette pathlength = not given so assume 1 cm

Δc = rate of change of concentration in mol/L/min

Substitute these values and re-arrange to give an expression for Δc:

0.001 x 3 x 1,000 A/min = 6.3 x 103 x 1 x Δc

Δc = ΔA x 3 x 1,000 mol/min/L serum

6.3 x 103 x 1

Multiply by 1,000,000 to convert from mol to μmol:

1 W-l-D unit = 0.001 x 1,000 x 3 x 1,000,000 μmol/min/L reaction mixture

6.3 x 103 x 1

1 W-l-D unit = 476 μmol/min/L serum

i.e. 1 W-l-D unit = 476 IU/L

Therefore activity (IU/L) = 476 x Wroblewski-laDue Units

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 WORKED ANSWERS TO FURTHER QUESTIONS

5. If the Km of an enzyme which obeys simple Michaelis-Menten kinetics is 2.5

mmol/L, what velocity (expressed as a multiple of Vmax) would be obtained at a
substrate concentration of 10 mmol/L?

The Michaelis –Menten equation is:

v = Vmax [S]
Km + [S]

Where v = initial velocity

Vmax = maximal velocity
[S] = substrate concentration = 10 mmol/L
Km = Michaelis constant = 2.5 mmol/L

Substitute and solve for v:

v = 10 Vmax = 10 Vmax = 0.8 Vmax

2.5 + 10 12.5

6. What information can be obtained from the double-reciprocal plot for an enzyme
under the following conditions: a) 1/v = 0 when 1/[S] = -12.5 x 106 L/mol, b)
1/[S] = 0 when 1/v = 5.2 x 106 min/mol, c) 1/[S] = 0 when 1/v = 6.5 x 106
min/mol and the slope of the line is 100 min/L?

a) When 1/v = 0, the value for 1/[S] is –1/Km

Therefore -1 = -12.5 x 106 L/mol

Km

Which can be re-arranged to give the value of Km:

Km = -1 = 0.08 x 101-6 = 8.0 x 10-8 mol/L

- 12.5 x 106

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

b) When 1/[S] = 0, 1/v = 1/Vmax

Therefore 1 = 5.2 x 106 min/mol

Vmax

Rearrange and solve for Vmax:

Vmax = 1 = 0.19 x 10-6 = 1.9 x 10-7 mol/min

5.2 x 106

c) When 1/[S] = 0, 1/v = 1/Vmax

Therefore 1 = 6.5 x 106 min/mol

Vmax

Rearrange and solve for Vmax:

Vmax = 1 = 0.15 x 10-6 = 1.5 x 10-7 mol/min

6.5 x 106

The slope of the line gives Km/Vmax

Therefore Km = 100 min/L

Vmax

Substitute Vmax = 1.5 x 10-7 mol/min and solve for Km:

Km = 100
1.5 x 10-7

Km = 100 x 1.5 x 10-7 = 1.5 x 10-5 mol/L

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 WORKED ANSWERS TO FURTHER QUESTIONS

7. You carry out an enzyme experiment in which the substrate concentration is

expressed as mmol/L and the reaction velocity in μmol/L/min. What would be the
units for the axes of the three following plots: a) 1/[S] versus 1/v, b) [S]/v versus
[S], c) v versus v/[S]?

a) If [S] = mmol/L = 10-3 mol/L

1 = 1 = 103 L/mol
[S] 10-3 mol/L

If v = μmol/min = 10-6 mol/min

1 = 1 = 106 min/mol
v 10-6 mol/min

b) [S] = mmol/L = 10-3 mol/L = 103 min/L

v μmol/min 10-6mol/min

[S] = mmol/L = 10-3 mol/L

c) v = μmol/min = 10-6 mol/min

v = μmol/min = 10-6 mol/min = 10-3 L/min

[S] mmol/L 10-3 mol/L

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

8. Mucic acid is an inhibitor of β-glucuronidase. The following data were obtained

using phenolphthalein glucuronide as substrate, in the presence and absence of
mucic acid (concentration in the assay = 1.0 x 10-4 mol/L).

Substrate Reaction velocity

Concentration
(mmol/L) No inhibitor Mucic acid

0.5 33 9
1.0 50 17
2.0 67 29
4.0 80 44
10 91 67

Determine the type of inhibition and the enzyme-inhibitor dissociation constant.

The first step is to plot the data. Any linear transformation of the Michaelis-
Meneten equation can be used but the double-reciprocal plot is probably the
simplest. Calculated reciprocals are:

1/[S] 1/v
L/mmol No inhibitor Mucic acid

2.0 0.030 0.111

1.0 0.020 0.059
0.5 0.015 0.034
0.25 0.0125 0.023
0.10 0.011 0.015

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 WORKED ANSWERS TO FURTHER QUESTIONS

0.12

0.1
1/v + 1.0 x 10-4 mol/L mucic acid
0.08

0.06

0.04

0.02 No inhibitor

0
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
1/[S] L/mmol

Without inhibitor, when 1/v = 0, 1/[S] = -1/Km

Intercept on 1/[S] without inhibitor = - 0.947 L/mmol

Therefore Km = -1 = 1.06 mmol/L = 1.06 x 10-3 mol/L

- 0.947

Since the lines cross on the 1/v axis the type of inhibition is competitive.

With inhibitor, when 1/v = 0, 1/[S] = - 1/Kmapp

Intercept on 1/[S] with inhibitor = - 0.193 L/mmol

Therefore Kmapp = - 1 = 5.18 mmol/L = 5.18 x 10-3 mol/L

- 0.193

For competitive inhibition: Kmapp = Km (1 + [I]/Ki)

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Substitute Kmapp = 5.18 x 10-3 mol/L, Km = 1.06 x 10-3 mol/L,

[I] = 1.0 x 10-4 mol/L then solve for Ki:

5.18 x 10-3 = 1.06 x 10-3{1 + (1.0 x 10-4/Ki)}

5.18 x 10-3 = 1 + 1.0 x 10-4

1.06 x 10-3 Ki

4.89 - 1 = 1.0 x 10-4

Ki

Ki = 1.0 x 10-4 = 2.6 x 10-5 mol/L

3.89

9. An experiment was conducted to study the effect of pH on the activity of lactate

dehydrogenase. Using a histidine buffer at pH 5.5 and 7.4 the reaction was
monitored by following the increase in absorbance at 340 nm due to the reduction
of NAD. The following data were obtained:

Lactate Reaction velocity

concentration
mmol/L pH 7.4 pH 5.5

1 12 33
2 21 50
4 35 67
10 57 83
20 73 91

Stating any assumptions that you make determine the pH at which the enzyme has
greatest affinity for the substrate.

Calculate reciprocals then plot 1/v versus 1/[S] at each pH:

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 WORKED ANSWERS TO FURTHER QUESTIONS

1/[S] 1/v
L/mmol pH 7.4 pH 5.5

1 0.083 0.030
0.5 0.048 0.020
0.25 0.029 0.015
0.1 0.018 0.012
0.05 0.014 0.011

0.09

0.08

1/ v 0.07

0.06
pH 7.4
0.05

0.04

0.03

0.02
pH 5.5
0.01

0
-1 -0.5 0 0.5 1
1/[S] L/mmol

When 1/v = 0, 1/[S] = -1/Km

At pH 7.4, when 1/v = 0, 1/[S] = - 0.162 L/mmol

Therefore Km = - 1 = 6.2 mmol/L = 6.2 x 10-3 mol/L

-0.162

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

At pH 5.5, when 1/v = 0, 1/[S] = - 0.545 L/mmol

Therefore Km = - 1 = 1.84 mmol/L = 1.84 x 10-3 mol/L

- 0.545

Assuming that equilibrium conditions apply (i.e. that k+1>>K+2) then the Km is the
dissociation constant of the enzyme-substrate complex and is inversely
proportional to the affinity of the enzyme for the substrate. The Km is lower at
pH 5.5 than at pH 7.4. Therefore the enzyme has greatest affinity for its
substrate at pH 5.5.

10. The apparent Km and Vmax of an enzyme were measured over a range of inhibitor
concentrations and the following data obtained:

Inhibitor Apparent value

concentration Km Vmax
(mmol/L) (mmol/L) (μmol/min)

5 10 7.5
10 7 5
15 5 4
20 4 3

Determine the mode of inhibition and the inhibitor constant (Ki).

A competitive inhibitor causes an increase in the apparent Km. As the Km is

actually decreasing as inhibitor concentration increases this mode of inhibition
can be ruled out. The apparent Vmax is decreasing as inhibitor concentration is
increased; this behaviour is seen both with non-competitive and uncompetitive
inhibition. However, in non-competitive inhibition the Km is unaffected by the
inhibitor whereas in uncompetitive inhibition the apparent Km decreases with
increasing inhibitor concentration. Therefore these data are consistent with
uncompetitive inhibition.

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 WORKED ANSWERS TO FURTHER QUESTIONS

The value for Ki can be obtained from secondary plots of either 1/Km or
1/Vmax versus [I].

The relationship between Kmapp and [I] for an uncompetitive inhibitor is:

Kmapp = Km
(1 + [I]/Ki)

Inversion of this expression gives:

1 = (1 + [I]/Ki)
Km app
Km

Which can also be written:

1 = 1 x [I] + 1
Km app
KiKm Km

Therefore a plot of 1/Kmapp versus [I] is linear.

When 1/Kmapp = 0:

0 = 1 x [I] + 1
KiKm Km

Which can be rearranged to give:

- 1 = [I]
Km KiKm

Multiplying both sides by KmKi and changing the signs gives:

Km Ki = - [I]
Km
Cancelling Km:

Ki = - [I]

Therefore the intercept on the [I] axis is - Ki

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

[I] (mmol/L) 5 10 15 20
1/Kmapp (L/mmol) 0.1 0.14 0.2 0.25

0.3

0.25
app
1/K m L/mmol
0.2

0.15

0.1

0.05

0
-10 -5 0 5 10 15 20 25
[ I ] mmol/L

When 1/Kmapp = 0, [I] = - Ki

From graph, when Kmapp = 0, [I] = - 3.67 mmol/L

Therefore Ki = - (- 3.67) = 3.67 mmol/L = 3.7 x 10-3 mol/L (2 sig figs)

The relationship between Vmaxapp and Ki for an uncompetitive inhibitor is:

Vmaxapp = Vmax
(1 + [I]/Ki)

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 WORKED ANSWERS TO FURTHER QUESTIONS

Inversion gives:

1 = (1 + [I]/Ki)
Vmaxapp Vmax

Which can also be written:

1 = 1 x [I] + 1
Vmaxapp
KiVmax Vmax

When 1/Vmaxapp = 0:

0 = 1 x [I] + 1
KiVmax Vmax

Which can be rearranged to:

- 1 = [I]
Vmax KiVmax

- KiVmax = [I]
Vmax

Cancelling Vmax and changing the sign on both sides gives:

Ki = - [I]

Therefore the intercept on the [I] axis is – Ki.

Calculating 1/Vmax:

[I] mmol/L: 5 10 15 20
1/Vmaxapp 0.13 0.20 0.25 0.33

Then plotting 1/Vmax versus [I]:

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

0.35

0.3

0.25
app
1/Vmax
0.2

0.15

0.1

0.05

0
-5 0 5 10 15 20 25
[ I ] mmol/L

When 1/Vmaxapp = 0, [I] = - Ki.

From graph, when 1/Vmaxapp = 0, [I] = - 4.59 mmol/L

Therefore Ki = 4.59 mmol/L = 4.6 x 10-3 mol/L (2 sig figs)

The Ki s from the two plots do not agree exactly but the value is approximately
4 x 10-3 mol/L. This is due to errors inherent in manually constructing the plots
and reading off the values of the intercepts.

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 10

1. The following results were obtained for a QC sample:

Total protein (g/L): 70, 68, 71, 65, 68, 70, 73, 69, 75, 74, 69, 71

Calculate the mean, variance, standard deviation, coefficient of variation and 95

per cent confidence limits.

Construct a table with columns for protein result (x) and x2, then obtain the sum of
the results in each column:

Result (x) x2

70 4900
68 4624
71 5041
65 4225
68 4624
70 4900
73 5329
69 4761
75 5625
74 5476
69 4761
71 5041

Total: Σx = 843 Σx2 = 59307

Number of values of x (n) = 12

Mean (m) = Σx = 843 = 70.25 g/L

n 12

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Variance (s2) = Σ (x – m)2

n-1

As a short cut use the identity:

Σ(x – m)2 = Σx2 - (Σx)2

n

= 59,307 - 8432
12

= 59,307 - 59,221

= 86 g/L

Therefore, variance (s2) = 86 = 86 = 7.82 g/L

(12 - 1) 11

Standard deviation (s) = √ s2 = √ 7.82 = 2.80 g/L

Coefficient of variation (CV) = s x 100 = 2.80 x 100 = 4.0%

m 70.25

The confidence limits of the mean are:

Mean - (z x s) to mean + (z x s)

For 95% confidence limits z = 1.96 so that this expression becomes:

Mean - (1.96 x s) to mean + (1.96 x s)

Substituting mean = 70.25 g/L and s = 2.80 g/L gives the 95% confidence limits:

70.25 - (1.96 x 2.80) to 70.25 + (1.96 x 2.80)

= 70.25 - 5.49 to 70.25 + 5.49

= 64.8 to 75.7 g/L (3 sig figs)

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 WORKED ANSWERS TO FURTHER QUESTIONS

2. Serum thyroxine was measured in 10,000 healthy male adults. Assuming a

Gaussian distribution the normal range was calculated to be 50-150 nmol/L. How
many results are expected to be above 165 nmol/L?

Assume that the normal range is the mean ± 2 standard deviations.

The mean will be the mean of the upper and lower limits:

m = (50 + 150) = 200 = 100 nmol/L

2 2

The upper and lower reference limits will span 4 standard deviations:

s = (150 - 50) = 100 = 25 nmol/L

4 4

Next calculate the z value for a result of 165 nmol/L:

z = x - m = (165 - 100) = 65 = 2.6

s 25 25

From tables of z, the value of P when z is equal to 2.6 is 0.002. Therefore, 0.002
of results fall outside the range: mean ± 65 nmol/L and a half of these (0.001)
will be greater than 165 nmol/L.

Number of results >165 nmol/L = 0.001 x 10,000

= 10 results

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

3. Calculate the least significant difference for a change in cholesterol if the intra-
individual coefficient of variation for cholesterol is 4.7% and the analytical
coefficient of variation, 2.4%. A patient was changed from Atorvastatin 80 mg to
Rosuvastatin 40 mg and the total cholesterol fell from 6.9 to 5.9 mmol/L.
Calculate the percentage change in cholesterol and state whether this is
significant.

First calculate the total variation in terms of CV%:

CVTotal = √ (CVAnalytical2 + CVIntra-individual2)

= √ (2.42 + 4.72)

= √ (5.76 + 22.09)

= √ 27.85

= 5.28 %

Calculate the SD for the initial cholesterol value (6.9 mmol/L):

CV (%) = s x 100
m

s = CV (%) x m = 5.28 x 6.9 = 0.365 mmol/L

100 100

For a change to be significant the overall difference must be at least 2.8s:

Least significant change = 2.8 x 0.365 = 1.022 mmol/L

Convert this to a percentage change from the initial value:

% significant change = 1.022 x 100 = 14.8%

6.9

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 WORKED ANSWERS TO FURTHER QUESTIONS

Calculate the actual percentage change in the patient’s result:

Actual % change = (6.9 - 5.9) x 100

6.9

= 1.0 x 100
6.9

= 14.5%

Since this percentage change is not greater than 14.8%, the change is not quite
statistically significant at the 5% level of probability.

4. Your on-call laboratory service uses 30 different methods, each of which has a
1% probability of failing QC criteria during the course of a night. Assuming that
QC of any method is independent of that of the other methods, what is the
probability that on any one night all methods will pass the QC criteria?

The probability of a channel failing QC is 1% = 0.01

There are only two possible outcomes - pass or fail.

Therefore the probability of a channel passing QC is 1 - 0.01 = 0.99

This problem is analogous to flipping a coin. The joint probability of two

independent events is the product of their individual probabilities.

Thus if a coin is tossed once the probability of ‘heads’ is 0.5. If the coin is tossed
again then the probability of it landing ‘heads’ on both occasions is 0.5 x 0.5 =
0.25. Similarly if the probability of one channel passing QC is 0.99, then the
probability of two channels passing is 0.99 x 0.99 = 0.98. The chance of three
different channels passing is given by 0.99 x 0.99 x 0.99 = 0.97 i.e. (0.99)3.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

The general rule is:

Probability of event occurring on n occasions =

(probability of event occurring on a single occasion)n

Therefore the probability of 30 channels passing QC = (0.99)30 = 0.74

If your calculator does not have the facility to calculate x y then the result can be
easily calculated using logs:

Log10 (probability of 30 channels passing) = 30 x Log10 0.99

= 30 x 0.00436

= - 0.131

Probability of 30 channels passing = antilog (-0.131) = 0.74

5. You attempt to derive a reference range for TSH for an ethnic minority
population. The first 10 samples give the following results:

Result n
Between 0.5 and 1.49 5
Between 1.5 and 2.49 3
Between 2.5 and 3.49 0
Between 3.5 and 4.49 1
Between 4.5 and 5.49 1

On the basis of these results, what range of TSH values would encompass 95% of
the ethnic minority population?

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 WORKED ANSWERS TO FURTHER QUESTIONS

There are two problems with this set of data:

1. The individual results are not given, only the number of results falling into each
class interval. The easiest way to deal with this is to assume that the results fall in
the middle of the range i.e. there are 5 results within the range 0.5 to 1.49 so
assume there are 5 results of the mid-point value (1.0 mU/L), similarly there are 3
samples with a value of 2 mU/L. Using this approach 10 individual results are
produced which can be processed in the usual way.

2. The data are obviously skewed and do not form a Gaussian distribution. This can
be overcome to some extent by taking logarithms (to the base 10) of the results
then calculating the mean, SD and 95% confidence limits in the usual way.
Taking antilogarithms of the confidence limits then gives the reference range.

A table can be completed in the following way:

TSH result x = log10 TSH result x2

1.0 0 0
1.0 0 0
1.0 0 0
1.0 0 0
1.0 0 0
2.0 0.301 0.0906
2.0 0.301 0.0906
2.0 0.301 0.0906
4.0 0.602 0.3624
5.0 0.699 0.4886

n = 10 ∑x = 2.204 ∑ x2 = 1.123

Mean = ∑x = 2.204 = 0.220

n 10

s2 = ∑ x2 - (∑ x) 2 / n = 1.123 - 2.204 2 / 10 = 0.0708

n-1 10 - 1

s = √ 0.0708 = 0.266

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Alternatively the mean and s can be calculated directly on most modern pocket
calculators. The 95% confidence are given by mean – 1.96 s to mean + 1.96 s

= 0.220 – (1.96 x 0.266) to 0.220 + (1.96 x 0.266)

= -0.301 to 0.741 (these values are logs and so do NOT have units)

Taking antilogs (to the base 10) gives the 95% confidence limits in mU TSH/L:

0.50 to 5.51 mU/L

Although the original data may have been expressed to one or two decimal places,
this information has been lost by grouping the data into class intervals. Therefore
it would be more correct to quote a reference range of less then 6 mU/L.

6. You are required to pipette a 9ml volume and have available a 10 ml graduated
pipette which has a 2%CV associated with it’s delivery volume and 5 and 2 ml
volumetric pipettes each of which has a 1% CV associated with their delivery
volumes. What is the error of pipetting a 9 mL volume, expressed as plus/minus
mL volume?

a) using the graduated pipette

b) using the volumetric pipettes

Assume that the error is required as 95% confidence limits i.e. ± 2 s.

a) Using the graduated pipette:

Calculate s when mean = 9 mL and CV = 2%:

CV (%) = s x 100 =
m

Therefore: s = CV (%) x m
100

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 WORKED ANSWERS TO FURTHER QUESTIONS

s = 2 x 9 = 18 = 0.18 mL
100 100

Therefore 95% limits = ± 2s = ± 2 x 0.18 = ± 0.36 mL

Error = plus/minus 0.36 mL

b) Similarly calculate the error for each of the bulb pipettes:

For 5 mL bulb with CV = 1%

s = 1 x 5 = 5 = 0.05 mL
100 100

For 2 mL bulb with CV = 1 %:

s = 1 x 2 = 2 = 0.02 mL
100 100

To pipette 9 mL the 5 mL bulb is used once and the 2 mL bulb

twice. Calculate the overall s:

sTotal = √ (s5mL2 + s2mL2 + s2mL2)

= √ (0.052 + 0.022 + 0.022)

= √ (0.0025 + 0.0004 + 0.0004)

= √ 0.0033

= 0.0574 mL

Therefore total error (2s) = 2 x 0.0574 = 0.11 mL (2 sig figs)

Error = plus/minus 0.11 mL

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7. It has been suggested that a proposed analytical goal for an analyte is that the
between batch analytical coefficient of variation should not exceed one half of the
“true biological” inter-individual coefficient of variation. Calculate the
percentage “expansion” of the measured reference range over the true biological
reference range when this analytical goal is exactly met.

The relationship between the overall variation, analytical variation and biological
variation is:

CVTotal2 = CVAnalytical2 + CVBiological2

Both the analytical and biological CV’s share the same mean.

CVAnalytical = 0.5 CVBiological

Substitute this value for the analytical CV so as to obtain the total CV expressed in
terms of the biological CV:

CVTotal = √ [(0.5 CVBiological)2 + CVBiological2]

= √ [(0.25 x CVBiological2) + CVBiological2]

= √ (1.25 x CVBiological2)

= 1.118 CVBiological

The reference range encompasses a span of 4 CVs

Therefore biological reference range spans 4 CVs and the total reference range
spans 4 x 1.118 CVBiological = 4.47 CVBiological

Therefore the percentage expansion is:

(4.47 CVBiological - 4CVBiological) x 100

4CVBiological

CVBiological (4.47 - 4) x 100

4CVBiological

(4.47 - 4) x 100 = 0.47 x 100 = 11.8% (3 sig figs)

4 4

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 11

1. The following analytical results were obtained on the same QC sample: 109, 91,
105, 112, 90, 115, 89, 113, 93, 94. Calculate the mean, standard deviation and
standard error of the mean.

Construct a table with columns for result (x) and x2, then obtain the sum of the
results in each column:

x x2

109 11,881
91 8,281
105 11,025
112 12,544
90 8,100
115 13,225
89 7,921
113 12,769
93 8,649
94 8,836

Total: ∑x =1011 ∑x2 = 103,231

n = 10

Mean (m) = ∑x = 1011 = 101.1

n 10

Variance (s2) = ∑(x – m)2

n-1

∑(x – m)2 = ∑x2 - (∑x)2

n

= 103,231 - 10112
10

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

= 103,231 - 102,212

= 1019

s2 = ∑(x – m)2 = 1019 = 1019 = 113.2

n–1 (10 – 1) 9

Standard deviation (s) = √ s2 = √ 113.2 = 10.64

Standard error of the mean (SEm) = s = 10.64 = 10.64 = 3.37

√n √10 3.16

2. Two laboratories measured sodium in the same plasma sample ten times. One
laboratory obtained a mean of 145 mmol/L with an SD of 3 mmol/L; the other
obtained a mean of 147 mmol/L with an SD of 2 mmol/L. Do the laboratories
differ in their bias or imprecision?

First lab: m1 = 145 mmol/L; s1 = 3 mmol/L

2nd lab: m2 = 147 mmol/L; s2 = 2 mmol/L

n = 10 for each lab

To check for bias carry out a t-test:

t = m1 - m2
√ (s12/n + s22/n)

= 145 - 147
√ (32/10 + 22/10)

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 WORKED ANSWERS TO FURTHER QUESTIONS

= -2
√ (0.9 + 0.4)

= -2 = -2 = - 1.75
√ 1.3 1.14

Next calculate degrees of freedom (DF):

DF = (s12/n1 + s22/n2)2
[(s1 /n1) /(n1 – 1)] + [(s22/n2)2/(n2 – 1)]
2 2

= (0.9 + 0.4)2
0.92/9 + 0.42/9

= 1.32
0.09 + 0.018

= 1.69
0.108

= 15.6

From tables when t = 1.75 with 16 degrees of freedom, P = 0.10. Therefore there
is no significant difference between the means of the two set of results i.e.
no evidence of bias.

To compare imprecision perform an F ratio test:

F = s1 2 = 32 = 9 = 2.25
s2 2 22 4

From tables when F = 3.18 (with 9 degrees of freedom for both variances),
P = 0.05. Therefore there is no significant difference between the two variances.
i.e. no evidence of difference in imprecision.

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3. Serum thyroxine was measured in 500 healthy adults. Assuming a Gaussian

distribution, the normal range was calculated to be 50-150 nmol/L. What is the
probability that the mean of a set of 9 results taken at random from this
population is greater than 125 nmol/L?

Assume that the normal range is the mean ± 2s.

The mean is the average of the upper and lower reference limit:

Mean (m) = (50 + 150) = 200 = 100 nmol/L

2 2

The reference limits span 4s units so that that s is a quarter of the range:

Standard deviation (s) = (150 - 50) = 100 = 25 nmol/L

4 4

Standard error of the mean (SEm) for 9 results

= s = 25 = 25 = 8.33 nmol/L
√n √9 3

Calculate t for 9 results with m = 125 nmol/L, population mean (μ) = 100 nmol/L
and SEm = 8.33 nmol/L:

t = m - μ = (125 – 100) = 25 = 3.00 (DF = n - 1 = 8)

SEm 8.33 8.33

From tables, for t = 3.00 with 8 degrees of freedom P = approx. 0.02. Therefore
0.02 of results fall outside of the range mean ± 25 nmol/L and a half of these
results, 0.01, will be greater than 125 nmol/L.

Probability of mean of 9 results being greater than 125 nmol/L = 0.01.

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 WORKED ANSWERS TO FURTHER QUESTIONS

4. It is suspected that an instrument used for near patient measurement of

cholesterol is showing positive bias. The following data are the results of paired
analyses of samples from ten patients measured on the standard laboratory
analyser (A) and the instrument under investigation (B). Assuming that the results
from the main analyser are correct, is there any evidence of bias?

A B

6.8 7.2
4.2 4.5
5.0 4.8
5.6 5.9
8.5 8.7
2.9 2.8
4.8 4.9
7.6 8.1
6.5 6.4
5.0 5.2

Since these are paired samples the results should be compared using the paired
t-test. Construct a table with the individual differences between each pair of
results (d = A – B), d2, the difference between each d and the overall mean (md)
for all the values of d (i.e. d – md) and their squares i.e. (d - md)2.

A B d d2 d - md (d - md)2

6.8 7.2 -0.4 0.16 -0.24 0.058

4.2 4.5 -0.3 0.09 –0.14 0.020
5.0 4.8 0.2 0.04 0.36 0.130
5.6 5.9 -0.3 0.09 -0.14 0.120
8.5 8.7 -0.2 0.04 -0.04 0.002
2.9 2.8 0.1 0.01 0.26 0.070
4.8 4.9 -0.1 0.01 0.06 0.004
7.6 8.1 -0.5 0.25 -0.34 0.116
6.5 6.4 0.1 0.01 0.26 0.070
5.0 5.2 -0.2 0.04 -0.04 0.002

n = 10 ∑d = -1.6 ∑d 2 = 0.74 ∑(d - md)2 = 0.592

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Mean difference (md) = ∑d = -1.6 = - 0.16

n 10

Paired t = md
sd/√n

sd = √ [∑(d – md)2/(n – 1)]

= √ [0.592/9]

= √ 0.0658

= 0.256

Use this sd to calculate the paired t:

Paired t = md = md x √n
sd /√n sd

= -0.16 x √10 = - 0.16 x 3.16 = - 1.98

0.256 0.256

From tables, for t = 1.98 (degree of freedom = n - 1 = 9) P is greater than 0.05.

Therefore there is no significant bias between the two methods.

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 WORKED ANSWERS TO FURTHER QUESTIONS

5. Four laboratories in a managed network compared the performance of their

serum cholesterol assays by measuring the same sample 10 times with the
following results:

Lab
A B C D

7.6 7.5 7.0 7.7

7.3 7.6 7.4 7.8
7.5 7.2 7.7 7.4
7.7 7.5 7.5 7.5
7.5 7.7 7.4 7.6
7.6 7.4 7.2 7.5
7.4 7.8 7.5 7.3
7.8 7.5 7.2 7.8
7.2 7.3 7.5 7.6
7.5 7.4 7.3 7.6

Is there any significant difference in bias for serum cholesterol at this

concentration between the four laboratories?

Calculate n, ∑x, m, ∑x2, (∑x)2/n and ∑x2 - (∑x)2/n for each lab:

Lab
A B C D

7.6 7.5 7.0 7.7

7.3 7.6 7.4 7.8
7.5 7.2 7.7 7.4
7.7 7.5 7.5 7.5
7.5 7.7 7.4 7.6
7.6 7.4 7.2 7.5
7.4 7.8 7.5 7.3
7.8 7.5 7.2 7.8
7.2 7.3 7.5 7.6
7.5 7.4 7.3 7.6
Totals
∑x 75.1 74.9 73.7 75.8 299.5
n 10 10 10 10 40
m 7.51 7.49 7.37 7.58
∑x2 564.29 561.29 543.53 574.8 2243.91
(∑x)2/n 564.001 561.001 543.69 574.564 2242.735
∑x2 - (∑x)2/n 0.289 0.289 -0.16 0.236

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Number of groups (u) = 4, number in each group (v) = 10, uv = 40

Between groups sum of squares = ∑.(∑x)2/n - (∑.∑x)2/uv

= 2242.735 - 299.52/40

= 2242.735 - 2242.5063

= 0.2287

Within groups sum of squares = ∑.∑x2 - ∑.(∑x)2/n

= 2243.91 - 2242.735

= 1.175

Total sum of squares = ∑.∑x2 - (∑.∑x)2/uv

= 2243.91 - 299.52/40

= 2243.91 - 2242.5063

= 1.4037

Source Sum of squares DF s2 F

Between groups 0.2287 3 0.0762 2.34

Within groups 1.175 36 0.0326
Total 1.4037 39 0.0360

From tables the probability of obtaining an F value greater than 2.84 (for 3 and 40
degrees of freedom) is 0.05. Therefore the data is homogeneous and there is no
evidence for bias between the four laboratories.

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 12

1. Regression analysis of results using new standards (y) against old standards (x)
showed a linear relationship. The regression coefficient (slope) was 1.10 and the
intercept on the y axis 1.0 mmol/L. Calculate the results which would be expected
using new standards for the analysis of old standards containing (a) 15 mmol/L
and (b) 150 mmol/L.

a) Regression equation for new standards (y) upon old standards (x):

y = 1.10 x + 1.0

Substitute old standard containing 15 mmol/L for x then solve for y:

y = 1.10 x 15 + 1.0

= 16.5 + 1.0

= 17.5 mmol/L

b) Substitute old standard containing 150 mmol/L for x then solve for y:

y = 1.10 x 150 + 1.0

= 165 + 1.0

= 166 mmol/L

2. A laboratory changed its method for the assay of serum alkaline phosphatase
activity. Assay of a selection of patient’s samples by both methods yielded the
following data:

ALP (Old method), IU/L: 50 350 700 100 1500 2000 420 1200
ALP (New method), IU/L: 40 190 350 90 750 1500 280 600

A gastroenterologist has been using ALP to monitor patients on treatment. Use

these data to derive an expression to convert the new ALP results to the results
expected by the old method.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

The first step is to check that there is a linear relationship between the two
methods. This is best done by plotting the results using the new method (y-axis)
against those obtained using the old method (x-axis):

1600 Results for new versus old alk phos method

1400

1200
New ALP (IU/L)

1000
Line of best fit
800 drawn by eye

600

400

200

0
0 500 1000 1500 2000 2500
Old ALP (IU/L)

The data appear to fit a straight line so linear regression analysis is appropriate.

x y x2 y2 xy

50 40 2500 1600 2000

350 190 122500 36100 66500
700 350 490000 122500 245000
100 90 10000 8100 9000
1500 750 2250000 562500 1125000
2000 1500 4000000 2250000 3000000
420 280 176400 78400 117600
1200 600 1440000 360000 720000

∑x = 6,320 ∑y = 3,800 ∑x2 = 8,491,400 ∑y2 = 3,419,200 ∑xy = 5,285,100

Slope of regression line (b) = ∑(x – mx)(y – my)

∑(x – mx)2

= ∑xy - (∑x∑y/n)
∑x2 - (∑x)2/n

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 WORKED ANSWERS TO FURTHER QUESTIONS

b = 5,285,100 - (6,320 x 3,800/8)

8,491,400 - (6,3202/8)

= 5,285,100 - 3,002,000
8,491,400 - 4,992,800

= 2,283,100
3,498,600

= 0.653 (3 sig figs)

The value for the intercept (a) can be obtained by substituting the slope (b), the
mean of x for x and the mean of y for y into the linear expression y = bx + a,
then solving for a:

mx = ∑x = 6320 = 790 IU/L

n 8

my = ∑y = 3800 = 475 IU/L

n 8

475 = (0.653 x 790) + a

a = 475 - (0.653 x 790)

= 475 - 516

= - 41

Therefore regression equation of y (new results) upon x (old results):

New method = (Old method x 0.65) - 41

Rearranging to enable easy conversion of new to old results:

Old method x 0.65 = New method + 41

Old method = New method + 41

0.65

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

3. An endocrinologist has been using serum prolactin measurements to assess the

response of patients with prolactinoma to treatment with a new drug. The
following data were obtained for a series of patients:

Drug dosage (mg/kg body wt): 50 100 150 200 250 300 350 400
Prolactin (IU/L) 750 1500 350 400 2000 1250 500 1800

Do these data show a linear relationship between drug dosage and serum
prolactin concentration?

The first step is to plot the data with prolactin as the y-axis and drug dosage as the
x-axis:

2500 Plot of prolactin concnetration versus drug dosage

2000
Prolactin (IU/L)

1500

1000

500

0
0 50 100 150 200 250 300 350 400 450
Drug dose (mg/Kg body wt)

Visual inspection suggests that there is no significant relationship between serum

prolactin concentration and drug dosage. Further evidence could be obtained by
calculating the correlation coefficient:

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 WORKED ANSWERS TO FURTHER QUESTIONS

x x2 y y2 xy

50 2500 750 562500 37500

100 10000 1500 2250000 150000
150 22500 350 122500 52500
200 40000 400 160000 80000
250 62500 2000 4000000 500000
300 90000 1250 1562500 375000
350 122500 500 250000 175000
400 160000 1800 3240000 720000

∑x = 1,800 ∑x2 = 510,000 ∑y = 8,550 ∑y2 = 12,147,500 ∑xy = 2,090,000

n = 8

r = ∑xy - (∑x∑y/n)
√ {[∑x - (∑x)2/n] [∑y2 - (∑y)2/n]}
2

= 2,090,000 - (1,800 x 8,550/8)

√ {[510,000 - 1,8002/8] [12,147,500 - 8,5502/8]}

= 2,090,000 - 1,923,750
√ { [510,000 - 405,000] [12,147,500 - 9,137,813]}

= 166,250
√ {105,000 x 3,009,687}

= 166,250
562,154

= 0.30 (2 sig figs)

From tables, for r = 0.30 with 7 degrees of freedom, P >0.1. Therefore there is no
significant correlation between drug dosage and serum prolactin.

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

4. A research paper contains the following statement:

“A good correlation was obtained when 45 patient samples were analysed by

methods A and B (r = 0.90, B = 1.05A – 10)….” Comment on this statement.

No evidence is presented that the relationship between the two variables is linear.

Correlation analysis is not the best approach to comparing two analytical methods
– as they both measure the same analyte it would be surprising if there were no
correlation. Analysis of difference plots would be more appropriate.

The standard error of the slope (1.05) is not given.

The standard deviation of the residual (sres or syx) is not given – this is the best
indicator of the goodness of fit of the data to the regression line.

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 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 13

1. A test for a particular disease has a sensitivity of 95% and a specificity of 95%.
Calculate the predictive value of both a positive and a negative test result in a
population in which the prevalence of the disease is:

a) 1 in 2
b) 1 in 5000

a) It is easiest to work with proportions rather than percentages or absolute

numbers of results. The contingency table to use is:

Positive result Negative result Total

Patients with disease TP FN Prevalence

Patients without disease FP TN 1 – prevalence

If the prevalence of disease is 1 in 2 i.e. 0.5, then 1 – prevalence is also 0.5 so

this table becomes:

Positive result Negative result Total

Patients with disease TP FN 0.5

Patients without disease FP TN 0.5

The next task is determine values for TP, FN, FP and TN using the stated
sensitivity and specificity:

Sensitivity = TP = 0.95
TP + FN

Substitute (TP + FN) = 0.5, then solve for TP:

TP = 0.95 so that TP = 0.5 x 0.95 = 0.475

0.5

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

and FN = 0.5 - TP = 0.5 - 0.475 = 0.025

Similarly using specificity:

Specificity = TN = 0.95
TN + FP

TN = 0.95 so that TN = 0.5 x 0.95 = 0.475

0.5

and FP = 0.5 - TN = 0.5 - 0.475 = 0.025

Inserting these values into the contingency table gives:

Positive result Negative result Total

Patients with disease 0.475 0.025 0.5

Patients without disease 0.025 0.475 0.5

These values are then used to calculate positive and negative predictive values:

PV(+) = TP = 0.475 = 0.95 (95%)

TP + FP 0.475 + 0.025

PV(-) = TN = 0.475 = 0.95 (95%)

TN + FN 0.475 + 0.025

b) With a prevalence of 1 in 5,000 (=0.0002) the contingency table becomes:

Positive result Negative result Total

Patients with disease TP FN 0.0002

Patients without disease FP TN 0.9998

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 WORKED ANSWERS TO FURTHER QUESTIONS

TP = 0.0002 x 0.95 = 0.00019

FN = 0.0002 - 0.00019 = 0.00001

TN = 0.9998 x 0.95 = 0.94981

FP = 0.9998 - 0.94981 = 0.04999

So the contingency table becomes:

Positive result Negative result Total

Patients with disease 0.00019 0.00001 0.0002

Patients without disease 0.04999 0.94999 0.9998

Use these values to calculate positive and negative predictive values:

PV(+) = TP = 0.00019 = 0.004 (0.4%)

TP + FP 0.00019 + 0.04999

PV(-) = TN = 0.94999 = 1.00 (100%)

TN + FN 0.94999 + 0.00001

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2. The table shows data from two urinary screening tests for the detection of
phaeochromocytoma.

Test Sensitivity Specificity

VMA 96.7% 99.1%

Total metanephrines 100% 98%

Both tests were used to screen a population of 100,000 hypertensive patients in

which the incidence of phaeochromocytoma is known to be 0.5%.

a) How many patients with phaeochromocytoma were missed by the VMA

test?

b) How many patients were incorrectly diagnosed as having

phaeochromocytoma using the metanephrine test?

c) Which test would you use to screen a hypertensive population for

phaeochromocytoma? Give reasons for your choice.

a) The number of patients with phaeochromocytoma missed by the VMA test

is the number of false negatives using this test.

First calculate the proportion of false negatives i.e. use the sensitivity
expressed as a proportion (0.967) rather than percentage (96.7%) and the
prevalence calculated as follows:

Prevalence = 0.5 = 0.005

100

Sensitivity = TP
TP + FN

Substitute (TP + FN) = prevalence = 0.005, and sensitivity = 0.967

and solve for TP:

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 WORKED ANSWERS TO FURTHER QUESTIONS

Sensitivity = TP = 0.967
0.005

TP = 0.967 x 0.005 = 0.004835

Since TP + FN = 0.005

FN = 0.005 - 0.004835 = 0.000165

Multiply this proportion by the total number screened to obtain the

number of false negatives (i.e. cases of phaeochromocytoma missed):

Patients missed = 0.000165 x 100,000

= 16.5 (2 sig figs)

b) The proportion of patients incorrectly diagnosed with phaeochromocytoma

using the metanephrine test is the proportion of false positives which can
be calculated from the specificity and prevalence:

Specificity = TN = 0.98
TN + FP

TN + FP = 1 - prevalence = 1 - 0.005 = 0.995

Specificity = TN = 0.98
0.995

TN = 0.98 x 0.995 = 0.9751

FP = (1 – prevalence) - TN

FP = 0.995 - 0.9751 = 0.0199

Multiply the proportion of false positives by the total number tested to

give the absolute number of false positives i.e. the number of patients
incorrectly diagnosed with phaeochromocytoma by the metanephrine test:

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Number incorrectly diagnosed = 0.0199 x 100,000

= 1990

c) Probably the best way to decide which is the best test is to calculate the
positive and negative predictive values for each test:

For VMA:

Positive result Negative result Total

Patients with disease 0.004835 0.00165 0.005

Patients without disease 0.008955 0.986045 0.995

PV(+) = TP = 0.004835 = 0.004835 = 0.35

TP + FP 0.004835 + 0.008955 0.01379

PV(-) = TN = 0.986045 = 0.986045 = 0.998

TN + FN 0.986045 + 0.00165 0.987695

For metanephrines:

Positive result Negative result Total

Patients with disease 0.005 0.000 0.005

Patients without disease 0.0199 0.9751 0.995

PV(+) = TP = 0.005 = 0.005 = 0.20

TP + FP 0.005 + 0.0199 0.0249

PV(-) = TN = 0.9751 = 0.9751 = 1.00

TN + FN 0.9751 + 0.000 0.9751

466
 WORKED ANSWERS TO FURTHER QUESTIONS

To summarize:

Test PV(+) PV(-)

VMA 0.35 0.998

Metanephrines 0.20 1.00

Although the VMA test produces less false positives (i.e. higher PV+) this is
achieved at the expense of missing approximately 1 in 3 (FN/prevalence =
0.33) patients with phaeochromocytoma. Although the phaeochromocytoma
produces more false positives (i.e. lower PV+) this is achieved without
missing any cases of phaeochromocytoma (i.e. no false negatives). On
balance total metanephrines is the better test.

3. A new laboratory test has a sensitivity of 85% and a specificity of 90%. The
incidence of disease in a population considered at risk is 0.10. What is the
predictive value of

a) a positive result?
b) a negative result?

Start by drawing up a contingency table:

Positive result Negative result Total

Patients with disease TP FN Prev

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease FP TN 1 – prev

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total TP + FP TN + FN 1

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Using sensitivity and specificity expressed as proportions instead of percentages

i.e. sensitivity = 0.85 and specificity = 0.90 fill in the above table:

Positive result Negative result Total

Patients with disease 0.085 0.015 0.10

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease 0.09 0.81 0.90

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total 0.175 0.825

TP + FP TN + FN 1

a) Predictive value of a positive result:

PV(+) = TP = 0.085 = 0.49 (2 sig figs)

TP + FP 0.175

b) Predictive value of a negative result:

PV(-) = TN = 0.81 = 0.98 (2 sig figs)

TN + FN 0.825

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 WORKED ANSWERS TO FURTHER QUESTIONS

4. A proposed diagnostic serological test for coeliac disease was evaluated in 200
consecutive patients referred to a paediatric gastroenterology service in whom
the condition was suspected clinically. The test result was compared with the
diagnosis as established by biopsy, withdrawal of gluten and response to
re-challenge. On this basis 76 children had the condition of whom only 64 gave a
positive test result: 10 positive test results occurred in children who were shown
not to have coeliac disease. Calculate the sensitivity and specificity of the test
and the predictive value of a positive result.

The prevalence of disease amongst the population of 200 is 76, so that

(1 – prevalence) = 200 – 76 = 124

64 of these 76 gave a positive test result = true positives

10 were positive amongst those without celiac disease = false positives

Set up a 2 x 2 contingency table for these results:

Positive result Negative result Total

Patients with disease TP FN Prev

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease FP TN 1 – prev

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total TP + FP TN + FN 1

Fill in this table working with the data given in the question:

Positive result Negative result Total

Patients with disease 64 12 76

Patients without disease 10 114 124

Total 74 126 200

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CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

N.B. In a question like this when the sensitivity and specificity is not given and
actual numbers of results are supplied it is probably easiest to work with absolute
numbers rather than proportions.

Sensitivity = TP = 64 = 0.84 (or 84%) 2 sig figs

TP + FN 76

Specificity = TN = 114 = 0.92 (or 92%) 2 sig fgs

TN + FP 124

PV(+) = TP = 64 = 0.86 (or 86%) 2 sig figs

TP + FP 74

5. In a cancer clinic where the prevalence of ovarian malignancy is 40%, a tumour

marker has a specificity of 88% and a sensitivity of 92%. Calculate the predictive
value of a positive test result. If this test was used as a screening tool in all
patients attending a general gynaecological clinic with a cancer prevalence of
0.4%, what would be the predictive value of a positive test in this population?

Set up a 2 x 2 contingency table then fill in the gaps using a prevalence of 0.4,
sensitivity of 0.92 and specificity of 0.88:

Positive result Negative result Total

Patients with disease TP FN Prev

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease FP TN 1 – prev

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total TP + FP TN + FN 1

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 WORKED ANSWERS TO FURTHER QUESTIONS

Positive result Negative result Total

Patients with disease 0.368 0.032 0.4

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease 0.072 0.528 0.6

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total 0.44 0.56 1

PV(+) = TP = 0.368 = 0.84 or 84% (2 sig figs)

TP + FP 0.44

Recalculate the above table using a prevalence of 0.4 % (i.e. 0.004):

Positive result Negative result Total

Patients with disease 0.00368 0.00032 0.004

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease 0.12 0.876 0.996

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total 0.12368 0.87632 1

PV(+) = TP = 0.00368 = 0.030 or 3.0 % (2 sig figs)

TP + FP 0.12368

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6. A certain disease has a prevalence of 5 percent. A diagnostic test was applied to

a random sample of 400 individuals from this population and yielded 15 true
positive and 30 false positive results. Calculate: a) the positive predictive value
of the test applied to this population, b) the pre-test odds of disease, c) the
likelihood ratio positive; d) the post test odds of disease for a positive result, and
e) the post-test probability of disease for a positive result.

The 2 x 2 contingency table can be set up as follows:

Positive result Negative result Total

Patients with disease TP FN Prev

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease FP TN 1 – prev

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total TP + FP TN + FN 1

Complete the table using a prevalence of 5% = 0.05 which with a total of 400
individuals gives a prevalence in absolute numbers of 0.05 x 400 = 20.

Positive result Negative result Total

Patients with disease 15 5 20

(Sens x prev) (Prev – TP) (TP + FN)

Patients without disease 30 350 380

[(1 – prev) – TN] {Spec x (1 – prev)] (FP + TN)

Total 45 355 400

a) PV(+) = TP = 15 = 0.33 or 33% (2 sig figs)

TP + FP 45

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 WORKED ANSWERS TO FURTHER QUESTIONS

b) Pre-test odds = Prevalence

1 - prevalence

= 20
380

= 0.053 (2 sig figs)

c) LR+ = probability of +ve test with disease = sensitivity

probability of a +ve test without disease (1 – specificity)

Sensitivity = TP = 15 = 0.75
TP + FN 20

Specificity = TN = 350 = 0.92 (2 sig figs)

TN + FP 380

LR+ = 0.75 = 0.75 = 9.4 (2 sig figs)

(1 – 0.92) 0.08

d) Post-test odds = Pre-test odds x LR+

= 0.053 x 9.4

= 0.50 (2 sig figs)

e) Post-test probability = Post-test odds

(1 + post-test odds)

= 0.50
(1 + 0.50)

= 0.50
1.50

= 0.33

473
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

7. A two-stage sequential test strategy is used to screen for a rare inherited disease.
The prevalence of the disease is 0.0005. The initial test has a sensitivity of 98%
and specificity of 95%, the follow-up test a sensitivity of 95% and specificity of
99%. What is the probability of a patient with a positive result for the follow-up
test having the disease?

If the prevalence of disease is 0.0005 then the pre-test odds can be calculated:

Pre-test odds = Prevalence = 0.0005 = 0.0005 = 0.000500

(1 – prevalence) (1 – 0.0005) 0.9995

LR+ = probability of +ve test with disease = sensitivity

probability of a +ve test without disease (1 – specificity)

For the 1st test: LR+ = 0.98 = 0.98 = 19.6

(1 – 0.95) 0.05

For the 2nd test: LR+ = 0.95 = 0.95 = 95

(1 – 0.99) 0.01

Post test odds =

Pre-test odds x likelihood ratio (1st test) x likelihood ratio (2nd test)

Post test odds = 0.000500 x 19.6 x 95 = 0.931

Post-test probability = Post-test odds

(1 + post test odds)

= 0.931
1.931

= 0.48 or 48% (2 sig figs)

474
 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 14

1. A study into the effect of nutritional supplements on patients with Crohn’s disease
involved measuring serum albumin both before and after supplementation for a
four week period. During this period the mean serum albumin level increased
from 25 g/L to 30 g/L. The study involved 40 patients with a standard deviation
for albumin concentration of 10 g/L. What is the power of this study to detect a
5 g/L change in serum albumin at the 5% level of probability?

The power can be calculated from the following expression:

zα + zβ = ∆√n
s

∆ = difference between the means of the two groups = 30-25 = 5 g/L

n = number of subjects in the study = 40

s = standard deviation = 10 g/L

Substitute these values to obtain zα + zβ

zα + zβ = 5 √ 40 = 5 x 6.32 = 3.16
10 10

Since the probability (P) used as a decision level is 0.05, the corresponding
z value (obtainable from tables) is 1.96 (the question only requires detection of a
change – which could be either positive or negative – so both sides of the
distribution are being used). Therefore, α = 0.05 and zα is 1.96. Substitute this
value for zα and solve for zβ:

zβ = 3.16 - zα = 3.16 - 1.96 = 1.20

From tables of z, the value for β (i.e. proportion of area under the curve)
corresponding a to zβ of 1.20 is 0.1151 (single sided probability).

Therefore power = (1 - β) = (1 - 0.1151) = 0.88 or 88% (2 sig figs)

475
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

2. It is proposed to set up a study to determine the effect of dietary modification on

serum cholesterol. The population to be studied has a mean serum cholesterol of
7.5 mmol/L with standard deviation of 2.5 mmol/L. What number of participants
need to be recruited in order to demonstrate a lowering of serum cholesterol by
10% (using alpha = 0.05 as a critical value) with a power of 90%?

The expression for calculating sample size is:

n = [s (zα + zβ) / ∆]2

s = standard deviation = 2.5 mmol/L

∆ = difference between the means

= Final cholesterol - Initial cholesterol

= (90% x 7.5) - 7.5 (since cholesterol is required to be

lowered by 10%)

= 6.75 - 7.5

= - 0.75 mmol/L

The required power is 90%

Therefore (1 - β) = 0.9 and β = 1 - 0.9 = 0.1

From tables the corresponding z value (i.e. zβ) is 1.28 (one sided value).

The decision level used is a probability of 0.05 (α) with a corresponding z value
for one side of the distribution (since we are required to detect a decrease in
cholesterol) (zα) of 1.64.

Substitute these values and solve for n:

n = [2.5 (1.64 + 1.28) / -0.75]2

= [2.5 x 2.92 / -0.75]2

= 9.732

= 95 (2 sig figs)

476
 WORKED ANSWERS TO FURTHER QUESTIONS

Chapter 15

1. A 0.5 mL sample of urine is extracted into dichloromethane. An aliquot of the

extract is analysed by HPLC and found to give an apparent original
concentration of 320 nmol/L of analyte Y. 100 µL of Y standard with a
concentration of 880 nmol/L is added to a further 0.5 mL sample of the same
urine and the sample mixed. 0.5 mL of the mixed sample is then processed as
before, giving a measured concentration of 405 nmol/L. Calculate the recovery
of analyte Y.

Recovery % = Increase in concentration upon adding standard x 100

Concentration of standard added

Allowance must be made for dilution of both the sample and standard when they
are mixed – since only 0.5 mL of the mixture is used for the assay.

Concentration of Y from urine in the mixture

= Initial concentration x Volume of urine (mL)

Volume of mixture (mL)

Since initial concentration = 320 nmol/L

Mixture = 0.5 mL urine + 0.1 mL standard = 0.6 mL

Concentration of Y from urine = 320 x 0.5 = 266.7 nmol/L

0.6

Similarly concentration of standard in mixture = 880 x 0.1 = 146.7 nmol/L

0.6

Recovery (%) = (Measured concn – concn from urine) x 100

Standard added

= (405 - 266.7) x 100

146.7

= 138.3 x 100
146.7

= 94% (2 sig figs)

477
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

2. A new method for HCG in urine is being evaluated. The concentration in a

sample from a pregnant woman is measured at 8240 IU/L. A 50 µL aliquot of an
international standard containing 50,000 IU/L is added to 450 µL of the same
urine sample and the sample mixed. On measuring the mixed sample, the new
concentration is found to be 12100 IU/L. What is the recovery of HCG by this
method?

Calculate the expected concentrations in the mixture from the urine and the
standard separately:

Urine HCG in mixture = 8240 x 450 = 7,416 IU/L

500

Standard HCG in mixture = 50,000 x 50 = 5,000 IU/L

500

% recovery = HCG recovered x 100

HCG added

= (Measured HCG in mixture - Expected HCG from urine) x 100

HCG added

= (12,100 - 7,416) x 100

5,000

= 4,684 x 100
5,000

= 94% (2 sig figs)

478
 WORKED ANSWERS TO FURTHER QUESTIONS

3. Measurement of plasma AFP is used to monitor a patient with a teratoma. If the

initial concentration was 10,200 U/L what plasma level would you expect to find
21 days after successful surgery? Assume the half-life of AFP is 5.5 days.

Assuming the clearance of AFP follows first-order kinetics the rate equation is:

ln Cpt = ln Cp0 - kd.t

Cpt = concentration of AFP after 21 days = Cp21

Cp0 = initial concentration of AFP = 10,200 U/L
t = time period = 21 days

kd = elimination rate constant which can be calculated from the half-life (t½):

kd = 0.693 = 0.693 = 0.126 days-1

t½ 5.5

Substitute these values and solve for Cp21:

ln Cp21 = ln 10,200 - 0.126 x 21

ln Cp21 = 9.230 - 2.646 = 6.584

Cp21 = antiloge 6.584 = 723 U/L

479
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

4. A radioisotope has a half-life of 21 days. How long will it take for the activity to
fall to 10% of the initial value?

The decay of a radioisotope follows first-order kinetics:

ln At = ln A0 - kd.t

At = activity at time t = 0.1

A0 = initial activity = 1
t = time for activity to fall to 10% of initial value = ?

kd = decay constant which can be calculated from the given half-life (t½):

kd = 0.693 = 0.693 = 0.033 days-1

t½ 21

Substitute these values and solve for t:

ln 0.1 = ln 1 - 0.033.t

-2.303 = 0 - 0.033.t

0.033.t = 2.303

t = 2.303
0.033

= 70 days (2 sig figs)

An alternative approach is to use the expression:

log10 AR = - 0.30.N

AR = ratio of final to initial activity = 0.1

N = number of half-lives for this change to occur

480
 WORKED ANSWERS TO FURTHER QUESTIONS

Therefore log10 0.1 = - 0.30.N

N = 1 = 3.333 half-lives
0.30

As t½ = 21 days, t = 3.333 x 21 = 70 days

5. In normal pregnancy serum beta hCG has a doubling time of approximately

2 days. How long will it take for the serum level to increase ten-fold?

Exponential growth obeys the first-order rate equation:

ln Cpt = ln Cp0 + kd.t

If we take 1 as the initial concentration then a 10-fold increase will result in a

concentration of 10.

Cpt = concentration after time t = 10

Cp0 = initial concentration = 1
kd = specific growth rate, which can be calculated from the doubling time (td):

kd = 0.693 = 0.693 = 0.3465 day-1

td 2

t = time taken for concentration to increase 10-fold = ?

Substitute these values and solve for t:

ln 10 = ln 1 + 0.3465.t

2.303 = 0 + 0.3465.t

0.3465.t = 2.303

t = 2.303 = 6.6 days (2 sig figs)

0.3465

481
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Alternatively the following expression can be used:

log10 CR = 0.30 N

where CR is the concentration ratio = 10:1

N = number of doubling times required to achieve this ratio

Substitute CR = 10 and solve for N:

log10 10 = 0.30 N

1 = 0.30 N

N = 1 = 3.333
0.30

Therefore time taken = N x td

= 3.333 x 2

= 6.7 days

6. A patient receiving parenteral nutrition is receiving 11.8 g nitrogen/24 h as

amino acids. Urinary urea excretion is 580 mmol/24 h. Indicating what
assumptions you make, calculate whether she is in positive or negative nitrogen
balance.

Nitrogen excretion(g/24h) = Urea excretion (mmol/24 h) x 28

1,000

= 580 x 28
1,000

= 16.24 g/24 g

Nitrogen balance (g/24 h) = Nitrogen intake (g/24 h) - Nitrogen excretion (g/24 h)

= 11.8 - 16.24

= - 4.44 g/24h

482
 WORKED ANSWERS TO FURTHER QUESTIONS

If 20% is added to the urinary excretion to allow for other urinary losses and a
further 2 g/day added to allow for losses by other routes then the nitrogen
excretion becomes:

Corrected nitrogen excretion = [Urea nitrogen excretion (g/24 h) x 1.2] + 2.0

= (16.24 x 1.2) + 2.0

= 19.49 + 2.0

= 21.49 g/24 h

and the corrected nitrogen balance becomes:

Corrected nitrogen balance (g/24 h) = 11.8 - 21.49

= - 9.7 g (Negative balance)

7. A 30 min basal gastric secretion sample (total volume 27 mL) required 2.5 mL of
0.1 M NaOH to titrate 5 mL of the material to pH 7.4. Calculate the basal acid
secretion rate in mmol/h.

M1 x V1 = M2 x V2

M1 = molar concentration of HCl in gastric fluid = ?

V1 = volume of gastric fluid used in titration = 5 mL
M2 = molar concentration of NaOH = 0.1 M
V2 = titre of NaOH = 2.5 mL

M1 x 5 = 0.1 x 2.5

M1 = 0.1 x 2.5
5

= 0.05 M

483
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Since the answer is required in mmol multiply by 1,000:

HCl concentration = 0.05 x 1,000 = 50 mmol HCl/L gastric fluid

Divide by 1,000 to give the acid output per mL of gastric fluid then mutiply by the
total volume of gastric fluid collected (27 mL) to obtain the total output of acid:

Total HCl output = 50 x 27 = 1.35 mol HCl /27 mL gastric fluid

1,000

Since the gastric fluid was collected over 30 min, multiply this result by 2 to
obtain the amount of HCl secreted in 1 h:

Rate of HCl excretion = 1.35 x 2

= 2.7 mmol/h

8. A five-day faecal fat collection was homogenised and diluted to 1500 mL. A 10
mL aliquot of the homogenate was subjected to hydrolysis and the fatty acids were
extracted. The volume of 0.05 M sodium hydroxide required to effect
neutralisation was 48 mL. Calculate the fat excretion in mmol/24 h.

First calculate the fatty acid concentration in the homogenate:

M1 x V1 = M2 x V2

M1 = molar concentration of fatty acids in homogenate = ?

V1 = volume of homogenate titrated = 10 mL
M2 = molar concentration of NaOH used in titration = 0.05 M
V2 = titre of 0.05 M NaOH = 48 mL

M1 x 10 = 0.05 x 48

M1 = 0.05 x 48 = 0.24 mol/L

10

484
 WORKED ANSWERS TO FURTHER QUESTIONS

Multiply by 1,000 to convert this concentration to mmol/L:

Fatty acid concentration = 0.24 x 1,000

= 240 mmol fatty acid/L homogenate

Multiply by the total volume (in litres) of the homogenate to obtain the total fatty
acid output over the 5-day collection period:

Fatty acid output = 240 x 1.50 = 360 mmol fatty acid/5 days

Division by 5 gives the daily fatty acid output:

Daily fatty acid output = 360 = 72 mmol fatty acid/24 h

5

Assuming that all the fatty acids were liberated from triglyceride then division by
3 gives the total fat output:

Fat output = 72 = 24 mmol fat/24 h (as triglyceride)

3

9. Gas chromatography for a drug involves adding equal amounts of internal

standard to standard or sample prior to analysis. The following peak areas were
obtained:

Sample Peak area

Internal standard Drug

Standard (200 nmol/L) 50,000 200,000

Patient 40,000 150,000

Calculate the drug concentration in the sample.

485
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

Divide the drug peak area by the internal standard peak area to give the peak
height ratio (PHR) for both standard and patient:

Sample Peak area

Internal standard Drug PHR

Standard (200 nmol/L) 50,000 200,000 4.00

Patient 40,000 150,000 3.75

Assuming that the PHR is directly proportional to concentration then

concentration of the drug in the patient sample can be calculated from the
relationship:

PHRPatient = PHRStandard
ConcentrationPatient ConcentrationStandard

Substitute the PHR values and standard concentration to obtain the drug
concentration in the patient sample:

3.75 = 4.00
ConcentrationPatient 200

ConcentrationPatient = 3.75 x 200

4.00

= 188 nmol/L (3 sig figs)

486
 WORKED ANSWERS TO FURTHER QUESTIONS

10. Genotyping of a group of 100 unrelated individuals for a two-allele

polymorphism showed that the allele frequencies were:

A 0.65
B 0.35

Calculate the expected percentages of heterozygotes (AB) and homozygotes

(AA and BB) in the group.

Frequency of allele A = p = 0.65

Frequency of allele B = q = 0.35

The possible combinations are AA, AB and BB.

If the conditions for the Hardy-Weinberg equilibrium are met then the frequencies
of the three genotypes are:

AA = p2 = 0.652 = 0.4225

AB = 2pq = 2 x 0.65 x 0.35 = 0.455

BB = q2 = 0.352 = 0.1225

Therefore % heterozygotes (AB) = 0.455 x 100 = 45.5%

and % homozygotes (AA and BB) = (0.4225 + 0.1225) x 100 = 54.5%

487
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

11. The prevalence of an inherited metabolic disease (inherited in an autosomal

recessive manner due to a single allele) is 1 in 2,500. A survey identified 1 in 50
of the population as asymptomatic carriers. Is this finding consistent with a
population in a Hardy-Weinberg equilibrium?

Let the dominant gene be A and the recessive gene a. As the inheritance of the
disease is autosomal recessive only the homozygous recessive genotype (aa)
expresses the disease.

The incidence of the recessive disorder (aa) = 1 in 2,500 = 1 = 0.00040

2,500

Incidence of carriers (Aa) = 1 in 50 = 1 = 0.020

50

Since the total of all frequencies must equal 1, the frequency of the remaining
homozygous dominant genotype, AA (which does not express disease nor have
carrier status) can be calculated by difference:

Incidence of AA = 1 - (0.00040 + 0.020)

= 1 - 0.0204

= 0.9796

To summarize the observed frequencies of the three genotypes are:

Genotype AA Aa aa
Observed frequency 0.9796 0.020 0.00040

Next calculate the expected frequencies if the Hardy-Weinberg equilibrium is

operating starting with the frequency of aa which is the frequency of the disorder
i.e. 1 in 2,500.

Frequency of affected individuals (aa) = 0.00040 = q2

Therefore q = √ q2 = √ 0.00040 = 0.020

Since p + q = 1

p = 1 - q = 1 - 0.020 = 0.98

488
 WORKED ANSWERS TO FURTHER QUESTIONS

Using these values for p and q the frequencies of the other two genotypes can be
calculated:

Frequency of Aa = 2pq = 2 x 0.98 x 0.020 = 0.0392

Frequency of AA = p2 = 0.982 = 0.9604

Tabulate this data then calculate X2:

X2 = ∑ (O - E)2/E

Genotype Frequency (O - E) (O - E)2 (O - E)2/E

Observed Expected

AA 0.9796 0.9604 0.0192 0.00036864 0.000383840

Aa 0.02 0.0392 -0.0192 0.00036864 0.009404082

aa 0.0004 0.0004 0 0 0

Total: 1.0000 1.0000 0 0.0007372 0.0097878

X2 is the sum of all the values in the final column = 0.010 (2 sig figs)

Normally the degrees of freedom would be 3 - 1 = 2. However, since one of

the observations (frequency of disease) was used to estimate the expected values,
a further degree of freedom is lost leaving only one.

From tables, the value for P when X2 = 0.010 is somewhere between 0.95 and
0.99. Therefore there is no significant difference between the observed and
expected frequencies so that the data fit the Hardy-Weinberg equilibrium.

489
CALCULATIONS IN LABORATORY MEDICINE – A. DEACON

12. The following data were obtained for a digoxin radioimmunoassay employing
PEG precipitation of the primary antibody. The assay was performed in duplicate.
Calculate the digoxin concentration in the serum sample.

Sample Duplicate cpm

1 2
TC 15,100 15,900
NSB 320 380
TB 11,350 11,650
0.2 nmol/L standard 10,320 10,980
0.4 “ “ 9,250 8,340
0.8 “ “ 6,782 6,630
1.2 “ “ 5,104 5,890
2.4 “ “ 3,700 3,430
4.8 “ “ 1,350 1,650
Patient serum 4,350 5,000

Mean NSB = (320 + 380)/2 = 350 cpm

Mean TB = (11,350 + 11,650)/2 = 11,500 cpm

B0 = Mean TB - Mean NSB = 11,500 - 350 = 11,150 cpm

Calculate the mean for each pair of duplicates then B/B0 (%) using the formula:

B/B0 (%) = (Mean cpm standard/sample - Mean NSB) x 100

B0

These calculations are performed in the following table:

490
 WORKED ANSWERS TO FURTHER QUESTIONS

Sample Conc log conc Duplicate cpm Mean cpm Mean – NSB B/B0 (%)

NSB - - 320 380 350

TB 0 - 11,350 11,650 11,500 1,150 100
Standard 0.2 -0.70 10,320 10,980 10,650 10,300 92.3
“ 0.4 - 0.40 9,250 8,340 8,795 8,445 75.7
“ 0.8 -0.10 6,782 6,630 6,706 6,356 57.0
“ 1.2 0.08 5,104 5,890 5,497 5,147 46.2
“ 2.4 0.38 3,700 3,430 3,565 3,215 28.8
“ 4.8 0.68 1,350 1,650 1,500 1,150 10.3
Serum ? ? 4,350 5,000 4,675 4,325 38.8

100

90
Digoxin calibration curve
80

70

60

50

40
B /B 0 (%)

30

20

10

0
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8
log10 conc

From calibration curve log10 conc when serum B/B0 (%) = 0.21

Serum digoxin (nmol/L) = antilog10 0.21 = 1.6 nmol/L

491
INDEX

A
Absorbance measurements and analyte Correction of GFR for body surface area,
concentrations, 59 87-88
Absorption laws in spectrometry, 52-53 Correlation and regression, 239
Absorptivity, 60 Correlation coefficients, 241-244
Acid base, pH and buffers, 27-49 Creatinine clearance, 82-83
Analysis of variance (ANOVA), 229-233 calculation from plasma creatinine, 89-
Analytical imprecision, 193 Cumulative drug concentrations, 130
Antilogarithms, 29-30
Altman-Bland plots, 260-262 D
Dealing with mixtures in spectrometry, 69-71
B Deming regression, 258-259
Basis of statistics, 193-215 Dilutions
Beer’s law, 52 dealing with dilutions generally, 18-21
Beer-Lambert equation, 57 doubling dilutions, 23-24
Bioavailability, 113 linear dilutions, 22-23
Biological variability, 210-211 preparing a series of, 22-24
Blood urea nitrogen (BUN), 10 Dissociation constant of water, 28
Body surface area, 87 Distribution of body water, 139-140
Body fluids and electrolytes, 139-155
calculation of fluid balance, 141-142 E
composition of body fluids, 139-140 Eadie- Hofstee plot, 175-176
distribution of body water, 139-140 Effect of hyperglycaemia on plasma sodium,
effect of hyperglycaemia on plasma 147-148
sodium, 147-148 Eisenthal and Cornish-Bowden plot, 175-177
estimation of fluid losses, 141-146 Elimination rate constant, 118
extracellular fluid, 139-140 Enzymology, 157-189
intracellular fluid, 139-140 calculating enzyme activity, 161-162
Bouger’s law, 52 catalytic activity, 157-158
Buffers enzyme inhibition, 178-182
bicarbonate buffer system, 41 graphical solutions for Michaelis-Menton
calculations, 35 equation, 173-178
definitions, 34 interconversion of enzyme units, 164-165
quantitative properties, 38 measurement of activity, 157-160
urinary, 46 Michaelis-Menton equation, 167-172
Bunsen solubility coefficient, 42 Equivalent weights,
Estimation of fluid losses, 141-146
C Exponential growth, 306-307
Calculation of fluid balance, 141-142 Expression of degree of acidity/alkalinity,
Calibration curves in spectrometry, 62-68 28-29
Catalytic activity, 157-158 Extracellular fluid, 139-140
Clearance and GFR, 80-81
Clearance of a drug, 116-118 F
Clinical utility of laboratory tests, 265-283 Factors affecting enzyme actiity, 158-159
Cockcroft-Gault formula, 89-91 Fractional excretion, 94-98
Coefficient of variation, 197-198 Free water clearance, 99-100
Competitive binding immunoassays, 323-328
Composition of body fluids, 139-140
Correcting for purity, 16

492
 INDEX

G N
Gaussian distribution, 194-195 Naperian logarithms, 29
Glomerular filtration rate (GFR), 77-93 Nitrogen balance, 308-309
correcting for body surface area, 87-88 Numbers and logarithms, 30
Graphical solutions for Michaelis-Menton
equation, 173-178 O
Odds and likelihood ratios, 279-280
H Optical density, 53
Hanes plots, 173-174 Osmolal gap, 108-111
Hardy-Weinberg equilibrium, 315-322 Osmolality, 105-111
Henderson-Hasselbalch equation, 35-38 Osmolar clearance, 99-100
Henry’s law, 42 Osmotic pressure, 105
Hydrogen ion gradient in the kidney, 46
P
I Partial pressure of gaseous phase, 42
Internal standardisation in chromatography, Pascals, 42
312-314 Pearson correlation coefficient, 241-244
Inter- and intraindividual variation, 193-195 pH, definition and use, 31-33
Interconversion of enzyme units, 164-165 Pharmacokinetics, 113-135
Interconversion of mass and SI units, bioavailability, 113
7-8, 14 clearance of a drug, 116
Ionic product of water, 28 elimination rate constant, 118
Ion-specific electrodes, 152-154 practical applications, 124-135
salt conversion factor, 113
K volume of distribution, 113
Karmen units, 165-166 Population genetics, 314
Katals, 166 Predictive
King-Armstrong units, 164-165 values, 268-272
Kurtosis, 198-199 Prefixes for powers of 10, 3
Preparing a series of dilutions, 22-24
Preparing solutions from liquids, 17
L Preparing solutions from solids, 13
Laboratory manipulations, 13-24
Progress curves of an enzyme catalysed
Lambert’s law, 52
reaction, 159
Levy-Jennings QC charts, 207-208
Punnett’s square, 315
Linear dilutions, 22-23
Linear regression, 250-257
Lineweaver-Burk double reciprocal plot, R
173-174 Radioactive decay, 304-305
Loading dose of drugs, 124, 128 Rate of filtration, 79-81
Logarithms, 28-31 Rate of urinary excretion, 79-81
Receivor operator characteristic (ROC) curves,
276-278
M Recovery experiments, 297-299
Maintenance dose of drugs, 126
Relationship between clearance,
Measures of tubular function, 94-95
plasmaconcentration and urinary excretion,
Method comparison, 260
84-85
Michaelis-Menton equation, 135, 167-172
Renal function, 77-103
graphical solutions for, 173-178
Renal threshold, 96-98
Modification of Diet in Renal Disease
(MDRD) equations, 92-93

493
INDEX

S
Salt conversion factor, 113 probability points of a normal distribution,
Sensitivity and specificity, 266-268 204
SI units, 1 ROC curves, 276-278
Significant figures, 4-5 sensitivity and specificity, 266-267
Solubility coefficient, 43 skewness of data, 198-199
Specific gravity, 17 standard error, 219-220
Spectrophotometry, 51-73 statistical power, 287
absorbance, 53 tests for populations, 201
absortion laws, 52 t-tests, 219-227
basic principles, 51-52 variance ratio (F-test), 227-228
calibration curves, 62-68 variances, 210-211
transmittance of light, 51-55 Westgard rules, 207-208
Standard deviation, 197-198 Steady state of a drug, 129
Standard error of the mean (SEM), 219-221
Statistics T
Altman-Bland plots, 260-262 Titratable acidity, 46, 309-311
analysis of means and variance, 219-236 Total acid excretion, 46
analysis of variance (ANOVA), 229-233 t-tests, 219-227
basis of, 193-215 Tubular maximum, 94-99
coefficient of variation (CV), 196-198 Tubular reabsorption, 95
comparison of means, 219-236 Tumour marker kinetics, 301-303
correlation coefficients, 241-244
correlation and regression, 239 U
Gaussian or normal distribution, 194-195 Units and their manipulation, 1-12
in QC, 206-208 Units for expressing enzyme activity, 160
kurtosis, 198-199 Urinary buffers, 46
Levy-Jennings QC charts, 207-208
linear regression, 250-257
measures of central location, 196
V
Variance ratio (F-test), 227-228
measures of dispersion, 196-197
Volume of distribution,
median and interquartile range,
method comparison, 260
null W
hypothesis, 289-292 Westgard rules, 207-208
odds and likelihood ratios, 279-280
other modes of regression, 258-259 Z
predictive values, 268-272 Zollinger-Ellison syndrome, 309
presentation and descrition of laboratory
data, 194-195

494
Allan Deacon BSc PhD FRCPath was, until
his retirement in 2006, Consultant Clinical
Scientist in the Clinical Biochemistry
Department at Bedford Hospital.

Prior to that he held the position of

Consultant Clinical Scientist at Kings
College Hospital (London) and previously
trained in Clinical Biochemistry at
Northwick Park Hospital.

His main interest for many years was

porphyria and he provided a
Supraregional Analytical and Advisory
Service for porphyrins.

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