Experinent |

Measurement of the Trypsin Activity by Folin

CivcalteauReagent
Principle:
Typsin (EC 3.4.21.4) is aserinc proteusc, found in the digcstive system of many
vertebrates, where it hydrolyses proteins. Trypsin is formed in the sInall intestine when its
prvcnzyne ton, the trypsinogen produccd by tlhc pancrcas, is uclivaled. Trypsin cleaves
pepide chains mainly at the carboxyl side of thc amino acids Iysine or arginine, cxcept when
cither is tollowed by prolinc. It is uscd for numerous
biotechnological processes. The process
is commonly referred to as trypsin protcolysis or trypsinisation, and proteins that have been
digested/reated with trypsin arc said to have been trypsinized. Proteolytic activity of Trypsin
is frequently measured with casein as substrate. The
hydrolysis
results in the formation of TCA soluble peptides. These Peptides are of peptide bonds of casein
measured
reagent. The blue color generated is read at 578 nm within 10 minutes of phenol by phenol
addition. reagent
The phenolic group of tyrosine and trytophan residues (amino acid) in a
produce a blue purple color complex, with maximum absorption in the protein will
region
wavelength, with Folin- Ciocalteau reagent which consists of sodium tungstate molybdateof 660 nm
phosphate. Thus the intensityof color depends on the amount of these aromatic amino acids and
present and will hus vary lor different proteins. Most proteins estimation techniqucs usc
Senum AJbumin (BSA) universally as a standard protein, because Bovin
of its low cost, high purity
and ready availability. The method is sensitive down to about 10 ug/ml and is
probably the
most widely used protein assay despitc its being only a relative method, subject to interference
from Tris buffer, EDTA, nonionic and cationic detergents, carbohydrate, lipids and some salts.
The incubation time is very critical fora reproducible assay. Thc reaction is also dependent on
pH and a working range of pH9 to 10.5 is essential.

Reagents:
(1) 0.IM phosphate buffer, pH -7.6 (0.2 M Sodium dibydrogen orthophosphate, 0.2M
Disodium hydrogen phosphate ixed in equalvolume to get 0.! M)
(2) Casein - lg/100 ml of Phosphate buffer
(3) Trypsin -2 mg /ml in 0.001 N HCI

4) Trichloroacetic acid (TCA) - S%

(5)0.5 MNaOH

(6) Folin- Ciocaltcau reagent diluted threc times with water

(7) Bovine Serum Albumine (BSA) Solution- 1.0 mg/ml

6
 Procedure:
I. Take 0.2,0.4, 0.6, 0.8 and 1.0ml of trypsin solution and make up the final
volume to I
ml with phosphate buffer.
2. Addlml casein solution, mix it and incubate for 30
min.
3. Stop the reaction by adding 3 ml of
TCA.
4. Again leave it for 30 min., and
then filter it.
J. From the filtrate take 1.5 ml of solution and
add 3 ml of NaOH solution, followed by
addition of 0.9 ml of Folin- Ciocalteau rcagent.
6. Mix and read at 578 nm
within a period of 10 minutes.
I. Tne control is prepared by substituting the enzyme with
phosphate buffer solution.
Draw a plot of absorbance vs. enzyme
8. Use BSA for making a standard curve,
concentration.
based on which products generated from trypsin
hydrolysis were estimated.
Trypsin Water (ml) Casein (ml) TCA (ml) Total vol.
(ml)
0.0 I.0 3 5 ml
0.2 0.8
Incubation3 Leave atS ml
0.4 0.6 1 at 37°C 3 room S ml
0.6 0.4 for 30 min 3
temperature 5 ml
0.8 0.2 1 for 30 min S ml
1.0 3
S ml

Take 1.5 ml from each experimental tube into a fresh tube and add following addition

Exp. sample vol. (ml) NaOH (ml) Folin- Ciocalteau reagent (ml)
1.5 3 0.9
1.5 3 0.9 Mix well and

1.5 3 0.9
take absorbance
at 578 nm within
1.5 3 0.9
10 min
1.5 3 0.9
1.5 3 0.9

7
 Standard curve lor BSA: Stundard BSA stock 2 myml

BSA (nl) Water (ml) NaOH (ml) Folin- Ciocalteau rcagant (ml)
2.0 3 0.9
0.4 1.6 3 0.9 Mix well and
0.8 .2 3 0.9 take
1.2 0.8 3 0.9 absorbance at
1.6 0.4 3 0.9 578 nm within
2.0
3 0.9 10min

Exercises:
I. What are the reacting
2. What willbe the effectcompounds present in Folin- Ciocalteau reagent?
on casein hydrolysis by Trypsin
in its catalytic site? with a phosphorylated serine
3. Which among the following
peptides will respond to trypsin digestion test?
(a) Ala-Gly-Ser-Ser-His-Arg-Leu-Ile-Cys-Ser-His-Asp-Glu-Gln-Ser-Ser-Ala-Thr
His-Asp
(b) Ala-G!y-Tyr-Tyr-His-Tyr-Trp-lle-Cys-Trp-His-Asp-Glu-Gln-Tyr-Tyr-Ala-Thr
His-Asp
(c) Ala-Gly-Trp-Trp-His-Trp-Tyr-lle-Cys-Tyr-His-Asp-Glu-GIn-Trp-Trp-Ala-Thr
His-Asp
References
1. Lowry, O.H., Rosebrough, N.J., FarT, A.L., and Randall, R.J.
(The original method). (1951) J.Biol.Chem 193: 265
2. Hartree E.E. (1972). Anal.
Biochem. 48:422 (This modification makes the assay linear
larger range than the original assay). over a

8
 Experimnent 2
Determination of Km and Vmax Of a-amylase
Principle:
Enzymes arc protcin molecules that act as biological catalysts by increusing the rate of
reactions without changing the overall process. They are long chain amino acids bound together
by peptide bonds. Each enzyme is quite specific in character, acting on a
to produce a particular products. The central approach particular substrates
for studying the mechanism of an
enzyme-catalysed reaction is to determine the rate of the reaction and its changes in response
wilh the changes in paramctcrs such as substrate
temperaturc ctc. concentration, enzyme concentration, pHl,
Onc of the important
Is the substrate conceutration,parameters afTccting the rale of a reaction
(S). During enzyme substrate reaction,catalysed by an enzyme
gradually increases with increasing concentration of the the initial velocity Vo
beyond which the increase in Vowill not depend on the (S].substrate. Finally a point is reached,
Concentration on the X axis and corresponding velocity on When we plot a graph with substrate
Y axis. It can be observed from the
graph that as the
the Vo. However concentration
of the substrate increases, there is a
beyond a particular substrate concentration, the corresponding increase in
velocity
velocity of an enzyme catalysedremains
without any further increase. This maximum constant
substrate saturation is called the Vmuk, Maximum velocity. reaction under
(uM/min) Vax

Vo
veodty,

1/2 Vmx
Iaítlal

Km
Substrate concentration, [S] (mM)
Michaelis-Menten Equation
Leonor Michaelis and Maud Menten postulated that the enzyme frst combines
reversibly with its substrate to form an enzyme-substrate complex in a relatively fast
reversible step:

E-S ES

In the next step, this ES complex is

product breaks down in to the free cnzyme and the
reaction

ES ’ E -P

9
 Sincc the second step is the rate limiting step, the rate of overall reaction must bc proportional
to the concentration of'the ES that reucts in the second stcp. The relationship between substrate
concentration, [S] and Initial veloeity of enzyme, vo (Fig 1) has the same general shape for
most enzyncs (it approaches a rectongular hypei bulu). This can be expressed algebraically by
the Michaelis-Menten equation. Bascd on their basic hypothesis that the rate limiting step in
cnzymatic reactions is the brcakdownof the EScomplex to frce cnzymc and product, Michaclis
and Menten derived an cquation which is;
Vinax (S]
Km |S

Lineweaver-Burke Plot
In 1934, Lineweaver and Burke madea
by plotting a duuble inverse of simple mathematical alteration in the process
substrate concentration and reaction rate.

1 Km 1
+

Vmax [S] Vmax

Estimation of Gucose
This method tests for the
presence of free carbonyl group (C-0), the
sugars. This involves the oxidation of the aldehyde so-called reducing
glucose and the ketone functional group in functional group present in, for example,
(DNS) is reduced to fructose. Simultaneously,
3-amino,5-nitrosalicylic
oxidation
3,5-dinitrosalicylic acid
acid under alkaline conditions:
aldehyde group -> carboxyl group

reduction
3,5-di nitrosalicylic acid ->
Because dissolved 3-amino, 5-nitrosalicylic acid
not necessary for the colouroxygen can interfere with glucose oxidation, sulfite, which itself is
The above reaction scheme reaction, is added in the
shows that one mole of reagent to absorb the dissolved oxygen.
dinitrosalicylic acid. sugar will rcact with one mole of3.5
reaction stoichiometryHowever,
it is suspecied that there are
is more many side reactions, and the actual
complicated than that previously described. The type of side
reaction depends on the exact nature of the reducing sugars.
vield different colour intensities; it is necessary to Differcnt reducing
the oxidation of thc thus,
carbonyl groups in the sugar, calibrate for cach sugar.sugars gcnerally
In addition to
other side rcactions such as the decomposition
10
 of sugar also compctes for the availability of
carboxymethyl ccllulose 3,5-dinitrosnlicylic ucid. As a conscquence,
can affcct the calibration curvç by cnhancing the intensity of the
developed colour.
Although this is a convenicnt and relatively
low speciticity, one must run blanks diligcntly if theincxpensive method, due to the relatively
colorimetric results are to be interpreted
corectly and accuratcly, One can determine the backyround
cellulosc substrate solution by adding absorption on the original
measuring the ahsorbance, i.e. fullowing cellulasc,
exactly the
immediately stopping the reaction, and
same procedures for the actual samples.
When the effccts of extrancous
compounds arc not known, one can effectively include
called internal standard by first fully developing the colour for a so
known amount ofsugar is added to this samplc. The increase in the unknown sample; then, a
color development is equivalent to the the absorbance upon the second
incremental amount of sugar added.
Reagents:
1. Starch solution-2%
2. a-amylase
3. Dinitrosalicylic Acid Reagent Solution, 1%
a.
b. Dinitrosalicylic acid: 10 g
Phenol: 2 g (optional, sce Note 1)
C. Sodium sulfite: 0.5 g
d. Sodium hydroxide: 10 g
e. Add water to: 1lit
4. Potassium sodium tartrate solution, 40%
Procedure:
1. Make different concentrations of soluble starch (0.2%, 0.4%,
0.6%, 0.8%, 1%, 1.2%,
1.4%, 1.6%, 1.8%, 2%).
2. Add Sml starch solutions of all
concentralions in respcctive test hubes and incubate vith
0.5 ml a-amylase enzyme for 10 miautes at
37°C.
3. After 10 min incubation, ] ml of aliquot was
taken out and added to the test tubes
containing DNS solution.
4. Keep in boiling water þath for 5 minutes at 100°C and
cool it.
5. Dilute all the tubes by adding 8 ml of distilled water.
6. Read the absorbance of test solutions against the
control at 540 nm
7. A standard curve will be
preparcd using glucose, bascd on which the
anylase hydrolysed reducing sugar was estimated. concentration of

11
 Volumcs arc in ml
|Starch Water Amylasc DNSA
S.0 0.5
0,5 4.5 0.5 Incubate Cool it

1.0 10 min at down to

4.0 0.5 1

1.5 37°C Kcep in room

Read
3.5 0.5
2.0 and take boiling temperaure
Absorbance
3.0 0.5
ml
water at 540 nm
2.5 2.5 0.5 and add 10
bath for
3.0 2.0 0.5
aliquot ml of water

3.5 from to each test

1.5 0.5
each minutes
4.0 1.0 0.5 tubes
tube to
4.5 0.5 0.5
fresh
S.0 0.5
tubes

Notes

Phenol, up to 2g/1, intensifies the colour density. It

of absorbance versus glucose concentration but doeschanges the slope of the calibration curve
not
procedure yields an absorbance of 1 for lg/l of glucose inaffect the linearity. The above
the original sample in the abscnce
of phenol in the reagent, as opposed to an absorbance of 2.5 for l
g/l of glucose in 2 g/l of
phenol. This property can be exploited to achieve the maximum sensitivity for dilute samples.

Questions
1. How much time was needed for the complete color
development? Justify your
answer with a plot of color intensity as a function of time.
2. Obtain an absorption spectrum over wavelengths in the visible range (i.c. 400
700 nm). Justify the use of 540 nm chosen in the Procedure.
3. Find the proccdurcsfor at least two other methods
commonly employed to
measure sugar concentrations. List the advantages and disadvantages of these
methods.

Reference:

Miller, G.L, Use of dinitrosalicylic acid reagent for detcrmination of reducing sugar, Anal.
Chem., 31, 426, 1959.

12
 Experimnent 3
Determination of acid value and saponification
value of fat
Acid Value:
Principle:
Duing storage, fats may become rancid as a result of
Donds by atmospheic oxygen and peroxide formation at the double
acid. The amount of free acid presenthydrolysis by microorganisms with the liberation of free
Ine acid value (AV) is a common therefore gives an indication of age and quality of the fat.
as the wcight of KOH in mg neededparameter in the specification of fats and oils. It
is
a measure of the free fatty to neutralize the organic acids present in lg of fat defined
acids (FFA) present in the fat or oil. An and it is
FFA in a sample of oilor fat indicates increment in the amount of
action of lipase enzyme and it is an hydrolysis of triglycerides. Such reaction occurs by
the
(.e., high temperature and relative indicator of inadequate processing and storage conditions
the tissue from which the oil or fathumidity, tissue damage). The source of the enzyme can be
was
including microorganisrns. Besides FFA,extracted or it can be a
hydrolysis contaminant from other
of triglycerides produces glycerol. cells
table below shows the acid value of The
some common oils and bee's wax.
FFA are a source of
tend to be water soluble and flavors and aromas. On one side, we have short
chain FFA
volatile
chain saturated and unsaturated fatty with characteristíc smell. On the other side, we havewhich
long
form and their breakdown products acids. The later are more prone to oxidation in their free
characteristic flavors and aromas. In(aldehydes, ketones, alcohols, and organic acids) provide
most cases these flavors and arormas
defect oils, fats, and foods that contain them.
in are considered a
of triglycerides and oxidation of FFA are key in However, therc are instances wbere hydrolysis
the development of
cheeses and some processed meats.desirable flavor and aroma
in foods. This is the case of aged

Materials:
1. Buter (use a fresh sample and one that has been stored at several monts at room
temperature)
2. Fat solvent (equal volumes of 95% v/v alcohol and ether
neutralized to
phenolphthalein)
3. Phenolphthalcin
4. KOH (0.I mol/lit)
5. Burettes
Procedure:
1. Accurately weigh out 10g of test compound and suspend the melted fat in about 50
ml of fat solvent.
2. Add few drops of phenolphthalcin solution and mix thoroughly.
3. Titrate with 0. l mo/lit KOH until the faint pink colour persists.

13
 10
(8 g
(6

4. The number of ml of standard KOH required and calculate the acid value of the fat.
(Nole- 0.I moVlit KOH contains 5.6 g/lit)

Saponification Value:
Principle:
Fats and oils are the principle stored forms of energy in many organisms. They are
highly reduced compounds and are derivatives of faty acids. Fatty acids are carboxylic acids
with hydrocarbon chains of4 to 36 carbons, they can be saturated or unsaturated. The simplest
lipids constructed from fatty acids are triacylglycerols or triglycerides. Triacylglycerols are
composed of three fatty acids each in ester linkage with a single glycerol. Since the polar
hydroxyls of glycerol and the polar carboxylates of the fatty acids are bound in cster linkages,
triacyl glycerols are nonpolar, hydrophobic molecules, which are insoluble in water.
Saponification
is the hydrolysis
of fats or oils H,C-0C-R
under basic H,C
conditions to
HC-OH + HC-0-C-R
afford glycerol 3RC-OH
and the salt of
the correspond H,C-OH H,C-0-C-R
ing fatty Glycerol Fats Triacylgycerol
acid. Saponification literally means "soap making". It is important to the
know the amount of free fatty acid present, since this determines in large industrial user to
loss. The amount of free fatty acid is estimatcd by determining the mcasure the retining
be added to the fat to render it neutral. This is donc by warming a quantity of alkali that must
known amount of the fat with
strong aqueous caustic soda sohution, which converts the free fatty acid into
then removed and the amount of fat remaining is then determined. The loss soap. This soap is
is estimated by
subtracting this amount from the amount of fat originally taken for the test.
The saponification numbcr is the number of milligrams of
bydroxide required to neutralize the fatty acids resulting from the complete hydrolysispotassium
of lg of
fat. It gives information concerning the character of the fatty acids of the fat- the longer the
carbon chain, the less acid is liberated per gram of fat hydrolysed. It is also considered as
a measure of the average molecular weight (or chain length) of all the fatty acids present. The
long chain fatty acids found in fats have low saponification value because they have a relatively
fewer number of carboxylic functional groups per unit mass of the fat and therefore high
molecular weight. Fats (triglycerides) upon alkalinc hydrolysis (cither with KOH or NaOFH)
yield glycerol and potassium or sodium salts of fatty acids (soap).
Materials Required:

1) Fats and Oils (coconut oil, sunflower oil]

2) Conical Flask
3) 100ml beaker
4) Weigh Balance
5) Dropper
6) Reflux condenser
7) Boiling Waier bath
14
 8) Glass pipette (25nl)
9) Burctte

Reagents Requircd:
1) Ethanolic KOH (95%
2) Potassium hydroxide ethanol, v/v)
[0.5N
3) Fat solvent (cqual vol
4) of95% ethanol and ether mixturel]
Hydrochloric acid [0.5N)
5) Phenolphthalein indicator

Procedure:
1. Weigh lg of fat in a tared
beaker and dissolve in about 3ml of the fat
lether mixture]. solvent ethanol
2. Quantitatively trangfer the contents of the beaker three times with a
solvent. further 7ml of the
3. Add 25ml of 0.5N
alcoholic KOH and mix well, attach this to a refhux
4. Set up another reflux condenser.
condenscr as the blank with all other reagents present
5. Place both the
flasks in a boiling water bath for 30 except the fat.
6. Cool the flasks to minutes.
room temperature.
7. Now add
phenolphthalein indicator to both the flasks and titrate with 0.5N HCI.
8. Note down the
endpoint of blank and test.
9. The differencc between the
blank and test reading gives the number of
KOH required to saponify lg of fat. millilitres of 0.5N
10. Calculate the
saponification value using the formula:
Saponification value or number of fat = mg of KOH consumed by lg of fat.
Weight of KOH = Normality of KOH *
Bquivalent weight* volume of KOH in litres
Volume of KOH consumcd by lg fat = (Blank - test] ml
Questions:
1. How many molecules of fatty acids
2. What will be the are releascd from a triglyceride?
mg as 400?
saponification value of an oil, given the weight of KOH consumed in

Reference:
Plummer, D. T. An Introduction to Practical
Third edition pp 195-197. Biochemistry. Tala MacGraw-Hill Edition.

15
 Experiment 4
Identification of sugars in fruit juices using thin
layer chromatography. Microbiological media
preparation.
Thin Layer Chromatography
Principle:
Separation of compounds on a thin layer is similar to many ways to
chromatography, but has the added advantage thata variety of supporting media can be paperused
so that scparation can be by adsorption, ion cxchangc, parlition
fillration depending on the nature of the medium cmployed. The methudchromatograpgy, or gel
is very rapid and many
Separation can be completed in an hour. Compounds can be detected at lower concentration
than on paper as the spots are very compact. Furthermore, seperated
detected by corrosive sprays and clevated temperatures with some compounds can be
of course is not possible with thin layer materials, which
paper.
How does thin layer
The stationary phasechromatography
- silica gel
work?
Silica gel is a form of silicon dioxide (silica), The silicon atoms are
atoms in a gianl covalcnt structure. However. at the surface ot the silica gel,joined via oxygen
the silicon atoms
are attached to -OH groups. So, at the surface of the silica gel you have
of Si-0-Si bonds. The diagram shows a small Si-0-H bonds instead
part of the silica surface
OH OH OH

O-S 0- Si- 0 - 9-O

main body of silica structure

The surface of the silica gel is very polar and, because of the -OH groups, can forn hydrogen
bonds with suitable compounds around it as well as van der Waals dispersion forces and dipole
dipole attractions. The other commonly used stationary phase is alumina -aluminiun oxide.
The aluminium atoms on the surface of this also have -OH groups attached. Anything we say
about silica gel thereforc applies equally to alumina.
Materials:
1. Thin layer plates
2. Separation chamber
3. Solvent (ethyl acetate: isopropanol: water: pyridine, 26: 14:7:2)
4. Fruit juiccs from Iresh fruit
5. Absolute ethanol

16
 6. Oven at 105°C.
7. Standard sugar solutions (glucose, fructose, lactose, xylose,
ribose,
8.
Anilinc-diphenylaminc
aniline, and S vol of10 g/llocation rcagent. (prepare tresh by mixing 5galactose)
acid). This is toxic and shoulddiphenylamine in acelune with l vol of 85%volumes of 10g/
phosphoric
be handled very carefully.
Procedure:
1. Add 3 mlof ehanol to l ml
of fruit juice and centriuge at
denatured proteins. 10000 rom to remove the
L. Corefully spot
the
sugar solutions. supernatent to a thin layer plate together with some standard
on
3. Place the plate in a
chainber
untilthe solvent front is closesaturated
to the
with the solvent and develop the
lup chromatogram
4. DraW a line accress the of tho plate.
plate at this point and remove
reaches the mark. Dry the plate in a stream of cold air the chromatogramn when solvent
spraying the plates with and locate the sugar by
S. Heat briefly at
6. Note the 100°C. aniline-diphenylamine in fume chamber.
colour of each
7. Calculate Rf value and sugar.
identify the sugars present in the fruit juice.

Microbiological
Prtinciple: media preparation
The survival and growth of
environment. Culture media microorganisms depend on
are nutrient solutions available
used in
and a favourable growth
microorganisms. For the successful cultivation of a
given
laboratories to grow
understand nutritional requirements and then supply themicroorganism, it is necessary to
its
form and proportion in a culture
medium. The essential autrients in the proper
H-donors and acceptors (approximatelygeneral composition of a mediun is as follows:
C-source (approximately 1-20 g/L) 1-15 g/L)
N-source (approximately 0.2-2 g/L)
Other inorganic nutrients e.g S, P (50 mg/L)
Trace elements (0.1-1 ug/L)
Growth factors (amino acids, purines, pyrimidines,
occasionally 0.1-1 mg/L) occasionally S0 mg/L, vitamins
Solidifying agent (e.g. agar 10-20 g/L)
Solvent (usually distilled water)
Buffer chemicals

Material:
1. Glucose
2. Yeast extract (the common name for various
by extracting the cellcontents (removing the cellforms of processed yeast products made
walls); they are used as food additives
or flavorings, or as nutrients for bacterialculture medial
3. Tryptone [Tryptone is the assort1nent of peptides
the protcase trypsin. Tryptone is commonly used informed the digestion
by of casein by
microbiology
broth for the growth of E. coli and other microorganisms. It to produce lysogeny
amino acids for the growing bactcria] provides source of
a
4. NaCl

17
 S. Agar [Agar is derived from the polysaccharide agarosc, which forms the
supporting
stnicturc in the ccll walls of certain species of algae, and which is released on boiling.
Thesc algac arc
phyum,l²0) Agarknuwn us uyurophytes and belong to the Rhodophyta (red algae)
is actually the resulting mixture of two componcnts: the lincar
polysaccharide agarose, and a hcterugeneous mixturc of simaller molecules called
agaropectin)
Procedure:
In the laboratory, gram-negative
liquid midium, named Luria-Bertani Escherichia coli cells will be grown in enriched
broth with following composítion
Tryptone 1.0 g
Yeast extract- 0.5 g
NaCI- 1.0 g
Ghucose- 0.5 g
H0- 100 ml
Liquid culture (LB broth) can be taken either in test
ubes or in conical flask and has to be
sterilized in an autoclave [Sterilization autoclaves are widely
autoclaves are used to sterilize equipment and used in microbiology, medicin.
lb) saturated steam at 121 °C (249 F) for ~
supplies by subjecting them to high-pressure (15
After sterilization, cool it down to room 30 minutes)
for bacterial inoculation. tempcrature and at that condition media can be used
For the growth of bacteria on solid
plate (each petri medium, add 2% agar before sterilization and pour on petri
plate contain nearly 20 ml of medium.
Prepare LB medium and Agar petriplates for
future experiment.

Reference:
I.Plummer, D. T. An Introduction to Practical Biochemistry. Tata
Edition. Third edition pp 195-197. MacGraw-Hill
2. Pelczar, M. J. Microbiology 5h edition.
Tata MacGraw-Hill Edition.

18
 Experiment 5
Determination of Myoglobin Stability by UV
Visible Spectroscopy
inciple: Myoglobin (M, 16,700) is a
muscle cclls. Myoglobin (Mb) containssinple monomeric oxygen binding
a single polypeptide chain of 153 protein found within
and a heme prosthetic group. The heme group is amino acid residues
protein"s interior. The interaction with the heme withburricd in a hydrophobic pocket within the
Sorct band (or B-band), a strong absorbance peak in the protein environment results in the
the visible spectrum at 409 nm
In addition. two low-lying peaks at
~506 nm and ~637 nm are also observed for Mb.for Mb.
denaturation of myoglobin, a decrease in the absorbance at 409 nm Upon
Cxposure of the heme to the polar aqueous occurs owing to the
shows an absorption maximum at 409 nm solvent. For native myoglobin, the intact heme
(As09) while the broad absorption band with a
maxiraum at ~360 nm (Ao) has often been attributed
loosely attached (non-covalently and/or out of the herne to that of dissociated heme remaining
in pH leads to significant pocket) to the protein matrix. Decrease
changes in the Soret band. Under conditions wherein heme
dissociation is minimal, the Soret band shifls to the blue with
concomitant
signifying extensive water coordination as the heme pocket opens up. A broadening,
infornative means of monitoring the effect convenient and
the A4osA360 ratio as a function of the pH. of pH on heme stability/loss, is therefore, to plot
For example, for native myoglobin, at pH =7,
ratio of bound to free heme (Aa09/A360) is the
~4.85.
Materials:
1. Myoglobin
2. 50 mM Acetate buffer (pH 4, 5)
3. 50 mM Phosphate Buffer (pH6, 7, 8)
4. 50 mM Tris-HCl buffer (pH 9, 10)
Metbods:
Make l mM stock solution of myoglobin in 50 mM phosphate
buffer pH 7.0.
" Dilute the myoglobin stock solution with buffers of
different pHs (buffer systems
used: 50 mM acetate buffer pH 4 and 5; 50 mM Phospahte
MM Tris-HCI buffer pH 9 and 10) to make a final buffer ph 6, 7 and 8; 50
volume of 3 ml. concentration of 10 uM in Mb in a

Incubate at room temperature for 30 min.

Collect the absorption spectrum of' each diluted sample in their

with pH varying from 4 to J0. The blank for cach sample is to berespective buffers
carried
respectivepH buffer only. (Plcase make sure to take the data files and plotoutthcwith the
spectra
yourself using a graphing software)
" Note downthe changes in absorbances at 409 nm and 360 nm. Plot A4/A360 Versus
pH.

19
 Comment on the changes of AowA with pH interns of myoglobin stability.

Reference:

Kuntu, J., Kar, U., Gautam, S.,Kar1nakar, S., Chowdhury P.K. (2015). Unusual effects of
crowders on heme retention in myoglobin. FEBS Lett. 589, 3807-38 15.

20
 Experiment 6
Isolation of DNAfrom Escherichia coli
Principle:
DNA, or deoxynbonucleic acid, is the
CVery ccll in a person's body has the same DNA, hereditary material in alldomains of life. Nearly
(where it is called nuclear Most
DNA), but a small amount ofDNA is located in the cell
nucleus
mitochondria (where it is callcd
stored as a code made up of four mitochondrial DNA mtDNA), The
or
DNA can also be found in the
thymine (T). Human DNA consists chemical bases: adenine (A), guanine information
(G),
in DNA is
of
bases are th.e same in all people. about billion bases, and more than 99
3 cytosine (C), and
information The order, or sequence. of percent of
these bases determinesthose
available for building and
letters of the alphabet appear in a maintaining an organiso, similar to tbe the
up with each other, A with T certain order to form words and sentences. way in which
and C
to a sugar molecule and a with G, to form units called base pairs. Each base
DNA bases pair
attached phosphate is also
are called a nucleotide.
Nucleotides are molecule. Together, a base, sugar, and
double helix. The of the doublearrangcd
in two long
helix is somewhat strands that form aspiralphosphate
forming the ladder'structure
s rungs and the sugar and
like a called a
ladder, with thc base pairs
sidepieces of the ladder. An important property of phosphate molecules forming the vertical
of itself. Each strand of DNA in the DNA is that it can replicate, or make
sequence bases. This is critical when cells
of double helix can serve as a copies
exact copy of the DNA present divide because each pattern
new
for duplicating the
cell needs to have an
in
Phenol:chloroform extraction method.the old cell. Genomic DNAfrom bacteria is isolated by
Intoductlon
A
phenol-chloroform extractlon is a
that seporates
mixtures ol liquid-llquid extaction. A
two differentmolecules based on the
lecues in liquid-liquid extroction is a
widely used to isolale RNA. DNA. Immiscible iquids (28). differential solubililies of Ihe
or proteins. Liqui-liquid exlroctions are
Brief History
VoIkin &
Corter reported the first Use of
(30). In 1953,
GrOssmann & guanidinium chloride in
Ihe efficOcy of the isolation of RNA in 1951
from aqueoUS soltion (16). Defner described
Utilizíng this find. phenol at extracting proteins
seoorote nUcleic acids from proteins in 1956 (18). Kirby demonstroted the
of guanidiniym Cox ond others renewed use of phenol to
0L)2. 13). From chloride in the isolation of
1hen on, guanidiniun RNA from interest in the use
Durificolion. extraclions were ribonUcleoproteins in
the method of choicethefor 196Os
QUanidiniumn replacing phenol
chloido was firsl exlraction.
brlefly
The Use of guaniciniumn
Ihiocyanale
RNA
Instead of
SUCcessfully employed by Chirgwin et mentioned by Ullrich et al. in 1977 (29). ond ater
thiocvangte to isolatc yndegruded RNAol. frorm in 1979 (8).
Chirgwin et al. UScd guünidlnium
Combinotion of guanidinium
Feramisco et at. in 1981 (14).thiocyonate rlbonuclease-rich
and hot phenol for RNA tissues like poncregs. A
& Sacchi isolation was reported by
In
thiocyangte1he with phenol-chloroform1987. Chomczynski combined
oxtraction under ocidi Conditions (9).gUanidinium
inceolion, Chomczynski & Sacchi methnod has been the mnehod of Since its
from Cultured cells and most animol tissues choice to isolote RNA
(|0).
Extractlon of Nuclelc Aclds
The extraction of nUcleic acids involves adding an equal
aqueOUs sotution of lysed cells or volune of phenol-chlorolorm to on
the phases to seporole by homOgenized issue, nixing Ihe wo phoses. and ollowiog
centrlfugotion
phos6s: the lower orgonic plhase (Figure 1). Centrifugoion of Ihe
und the uppr dqueOUs mnixture yields two
phase.
Chloroform nnlxed with phenol is moreelflclen! denoturing proteins than elther
of
glone. The
phenol-chloroform combingtion reagent is
reduces the par titioning of poly(A)+ mRNA Into
21
 The organic phase ond reducos the formgtion of insoluble RNA
Cells
protein complexes at lhe interphase (24). Moreover. phenol relains
about 10-15% of the aquoous phose. which results in u similar loss of
RNA: chlorofomm prevernts this relention of waler and thus
RNA yield !22). Iypical mixlures of phenol to chloroform oreimproves
I:l und
5:1 (v/v). Al acidic pH, o 5:l ratiorasulls In the absence of DNA from Lyse and
the uppor oqueoUS phase: whereas a l:1 ratio. wile homogenlze
provldlng
maximal recovery of al RNAS, will maintain somne DNA present in
upper aqueous phase (17). Isoomyl alcohol is somelimes addedthe to
prevent foaming (typicaly in a rallo of 24 parts chlorolorm to lpart
isoomyl alcohol). Guanidinium solts are uSed to reduce the effect of
nucleses.
Add chloroform
and shake
Puified phenol has a density of L.07 g/cm? and
lower phase when mixed with water (1.00 g/cm)therefore (21).
forms the
ensures phase separation of the wo liguids because Chloroform
miscible with phenol and it has ahigher density (1.47 chloroform is Separate phases
phenol (21): it forces o sharper separation of the g/cm?) thon
OqueOUS phases thereby organic ond
phase with minimolcross assisling in the removal of the aqueous
contamination from the organic phase.
In general, a solute
dissolves best in o solvent that is most similar in
chemical structure to itself. The overall solvotion capacity of o Transter
solvent depends primarily on its
polar Solute sUch as ureg is very polarity (20). For example. a very aqueous phase
soluble in faity polar methonol, andsoluble in highly polar water, less
solvents such as chloroform ond ether almost
(21).
insoluble in non-polar
Nucleic
because of their negatively charged phosphate acids ore polar
therefore nucleic gcids are soluble backbone, and
in the upper aqueoUS phase
instead of the lower organic phase (water is more polar than phenol)
(20). Conversely, proteins contain
and unchorged domains. producingvarying proportions of charged FIGURE 1. The procedure
hydrophobic
regions (3). In the presence of phenol. the hydrophobicand hydrophilic of pheno-CHCb
interoct with phenol, causing precipitation of proteins cores extraction. Afterward,
and polymers the product is cleaned
(including carbohydrates) to collect at the
wo phases (often as gwhite flocculent) or forinterface between the
lipids to dissolve in the
by alcohol precipitation
or column puificotion.
lower oganic phose (3).
The pH of phenol determines the partitioning of DNA and RNA
beween the organic phose
and the aqueoUs phase (6.23). Af neutral or slightly alkoline pH (pH 7-8). the phosphote
diesters in nucleic ocids are negotively charged, ond thus DNA and RNA both partition into
the gqueous phase. DNA is removed from the dqueoUs layer with
the pH is lowered with g maximum efficiency at pH 4.8. Af this acidicincreasing efficiency as
pH, most proteins gnd
small DNA fragments (<I0 kb) fractionate into the organic phase and lorge DNA
and some proteins remain ot the interphase beween the organic and fragments
aqueoUs phases

(6.9.25). Acidic phenol retains RNA in the aqueOUS phdse, but moves DNA into the phenol
phase, because the phosphate groups on fhe DNA are more easily neuralized thon those in
RNA (i.e.. DNA is less acidic/has a greater pko than RNA) (Figure 2] (5.26). An ocid pH also
mininizes RNase activity (7).

22
 (A)

pH 7.0

DNA
Protein RNA

(B)

pH 4.8

Negative charges
on carboryic acid
side-chains romaln
Proteins precipitate

Negt charge on
PO,is noutralzed

DNA becomes uncharged

4 +

ANAemains waler soluble

HM H H H

FIGURE 2, Acid phenol specificolly legves RNA in the

the aqueous phase. A: the pH decreases.
concentration of protons increases. DNA Comes a negative
phosphote groups in its sugar-phosphate backbone. which are charge becOUse
neutralized of the
in acid by
protonalion. in this case. DNA dissolves in the organic phase (ike issolves like). RNA, on
other hgnd, is not neutralized in acid becOUse, even the
has exposed nilrogenou boses (it is single-stranded).though il also has a negotive charge. it
which con form hydrogen bonds with
water, keeping it in the aqueous phase. (3)

23
Materials and Reagents
1. Tris base
2. Proteinase K
3, Phenolchlorofom (1: 1)
4. 200 proof cthanol
5. RNAasc
6. Ethanol
7. SDS
8, EDTA
9. TE butfer (sce Rccipes)
10. Lysis buffer (see Recipes)

Equipment
I. Tabletop centrifuge (Eppendor)
2. 1.5 ml Eppendorf tube
3. Incubator
4, Gloves

Procedure:
S. Trans fer 1.5 ml of the overnight E. coli culure (grown in LB medium) to a 1.5
ml Eppendorf tube and centrifuge at max specd for Imin to pellet the cells.
Discard the supermatant. Note: Remove as much of the supernatant as you can
without disturbing the cell pellet.
6. Resuspend the cellpellet in 600 l lysis bufer and vortex to completely
resuspend cell pellet.
7. Incubate 1 h at 37 °C.
8. Add an equal volume of pheno/chloroform and mix well by inverting the tube
untilthe phases are complctely mixed.
Note: Do not vertex the tube-it can shear the DNA. CAUTION: Phenol is a
very strong acid that causes severe burns. Chloroform is a carcinogen. Wear
gloves, goggles and lab coat, and keep tubes capped tightly. To be safe, work
in the hood if possible.
9. Spin at max speed for 5 min at RT (all spins are performed at RT, unless
indicated otherwise). There is awhite layer (protein layer) in the aqueous:
phenol/chloroform interface.
10. Carefully transfer the upper aqueous phase to a new tube by using l ml
pipetman (to avoid sucking the interface, use 1l ml tip with wider mouth-cut I
ml tip-mouth about ~2 mm shorter).
11. Steps 3-6 can be repeated until the white protein layer disappears.
12. To remove phenol, add an equal volume of chloroform to the aqueous layer.
Again, mix wellby inverting the tube.
13. Spin at max speed for 5 min.
14. Remove aqueous layer tonew tube. proof ethanol (store
15. To precipitate the DNA, add 2.5 or3 volume of cold 200 visible).
ethanol at -20 °C freezer) and mix gently (DNA precipitation can be the tube at
Note: DNA precipitation may simply diffiuse, which is normal. Keep
and then spin it down
-20 degree for at least 30 min (the longer the betler)
(see Steps I5-16). You should see DNApellet. It looks transparency when it is
becomes dry.
wet and turns tO while when it
more.
16. Incubate the rube at -20 °C for30 min or
24
 I7. Spin nt max spccd tor 15 min ut 4
18. Diseard the supematant und °C.
rinsc thc DNA pcllct withlml 70%
(stored at RT). cthanol
19, Spin at mux spced for 2 min.
DNA pellet (tilt the tube a littleCaretfully discard the
bit on paper towel),supernatant and air-dry the
at 37 °C
incubator. To be faster, dry the tube
20. Resuspend DNA in TE
butfer.
Note: Lurge amounts of RNA will be
subsequent present in the DNA sample. So, Jor
m') RNAasereuctions,
for example, to digest plusmid DNA, udd I-5
to the digestion ul
to completely remove RNA. Or, (I mg
RNAase directly to lysis buffersolution
with a final concentration of l mg m'. add
21. Check isolated
Note: we expectGemonic
to
DNA on an agarose gel.
see bands with smear patterns from
although most of DNA fragments are high to low MW range,
ývou see most ofDNA fragments are accumulated at high MW on the gel. So,
degraded. small, very likely your DNA got
Recipes
1. TE buffer
10 mM Tris-CI (pH
l mM EDTA(pH 8.0)
2. Lysis buffer (10 ml)8.0)
9.34 ml TE buffer
600 ul of 10% SDS
60 ul of proteinase K (20 mg ml:)
Questions:
1. In an attempt to isolate total RNA from
the cell, a student in
mistakenly added KOH solution instead ofK-acetatc. Biochemistry Laboratory
of addition of KOH on RNA? Describe what would be the effect
2. Why addition of chloroform to phenol
bacterial cell? improves the extraction of nucleic acids
from
3. Why acidic phenol is used to
purify total
HCI(pH 8.0) saturated phenol is used forRNA from cell? What will happen when Tris
RNA isolation?
Reference:
J. Maniatis T., E.F. Fritsch, and J. Sambrook (1982).
Manual, Cold Spring Harbor Laboratory, Molecular Cloning A Laboratory
Cold Springs Harbor, NY.

B. Viable count of bacteria

Take a LBagar platc and place it cither on
bacterial viable count in air. Open the lid for working table or by the window to get a
10 min and incubate (facc down) at 37°C
overnight.
Nextday, count the bacterial colony on the plate. Note it down.

25