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UNDERSTANDING WHICH ARE

HIGHLY EXPECTED IN JEE MAINS
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EXAM
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 Chapter - 2
BIOMOLECULES, BIOLOGICAL PROCESSES &
CHEMISTRY IN ACTION
Introduction
Biochemistry may be defined as the study of the chemical composition and structure of living
matter and the chemical changes that take place in them during the life processes.

Complex organic molecules (carbohydrates, lipids, proteins and nucleic acids) which govern the
common activitites of a living organism are called as biomolecules.

2.1 Carbohydrates
Carbohydrates is defined as hydrates of carbon is which the ratio of hydrogen and oxygen in these
compounds is the same as present in water.
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Carbohydrates are also defined as optically active polyhydroxy aldehydes or polyhydroxy ketones
and other large molecules, like starch, glycogen or cellulose, which produce these compounds on
hydrolysis.
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Classification of carbohydrates
Carbohydrates are classified into two major groups on the basis of their physical properties.
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(a) Sugars (monosaccharides and Oligosaccharides)

(b) Non sugars or Polysaccharides

(a) Sugars: Sugars are crystalline substances, sweet in taste, and readily soluble in water. These
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have a fixed molecular weight and thus have a sharp melting point. Examples are glucose,
fructose, sucrose and lactose etc.
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Sugars are further classified as monosaccharides and oligosaccharides depending upon

their behaviour towards hydrolysis.

Monosaccharides: These are the simplest carbohydrates which do not undergo further
hydrolysis, i.e., glucose, fructose, arabinose, ribose, etc.

Classification of monosaccharides
Molecular Common Number of Nature of Classification
formula name carbon atom carbonyl group and nomenclature

C 3H 6O 3 Dihydroxy acetone 3 Ketonic group Ketotriose

C 4H 8O 4 Erythrose 4 Aldehydic group Aldotetrose
C5H10O5 Ribose 5 Aldehydic group Aldopentose
C5H10O5 Xylulose 5 Ketonic group Ketopentose
C6H12O6 Glucose 6 Aldehydic group Aldohexose
C6H12O6 Fructose 6 Ketonic group Ketohexose

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Biomolecules, Biological Process & Chemistry in Action

Oligosaccharides : Oligosaccharides are sugars which on hydrolysis give two or more

molecules of monosaccharides. These are further classified as di, tri tetra saccharides, etc.

Disaccharides : Disaccharides are sugars which on hydrolysis produce two molecules of the
same or different monosaccharides. These are sucrose, maltose and lactose. They have the
same chemical formula C12H22O11.

C12 H 22 O11  H 2 O Invertase

  C 6 H12 O 6  C 6 H12 O 6
Sucrose or H  Glucose Fructose
maltase
C12 H 22 O11  H 2O  2 C6 H12 O 6
Maltose or H  Glucose
Lactase
C12 H 22 O11 + H 2 O  C6 H11O 6 + C6 H12O 6
Lactose or H + Glucose Galactose

Trisaccharides : Sugars which yield three molecules of the same or different

monosaccharides on hydrolysis are called tri-sacchardies. e.g., raffinose C18H32O16

C18 H 32 O16  2H 2 O  C 6 H12 O 6  C 6 H12 O 6  C 6 H12 O 6

Raffinose Glucose Fructose Galactose
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(b) Nonsugar or polysaccharides: Polysaccharides are tasteless, amorphous solids. These are
either insoluble in water or form colloidal Solution (suspension). Polysaccharides or nonsugars
are naturally occuring polymeric compounds and produce a large number of monosaccharides
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on hydrolysis. For example, starch, cellulose, dextrin, glycogen, etc.


(C 6 H10 O 5 ) n  nH 2 O H n C 6 H12 O 6
Starch Glucose
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Difference between monosaccharides, disaccharides and polysaccharides.

Monosaccharides Disaccharides Polysaccharides

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1. These are sweet in taste These are generally These are tasteless
sweet in taste
2. Soluble in water Soluble in water Either less soluble insoluble in
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water
3. They have general These are formed from These are polymers of
moecular formula as two monosaccharides monosaccharides and formed
CnH2nOn where n=3 to 7 by the condensation of large
number of monosaccharides
4. They can not be further These are hydrolysed to These on hydrolysis produce 100
hydrolysed form a mixture of two to 3000 monosaccharides
similar or different molecules
monosaccharides
5. These have hemiacetal These have acetal linkage These do not have acetal or
or hemiketal linkage in their constitution hemiacetal structure because of
in their structure polymeric nature
6. No glycosidic linkage is They have on ( or ) These have large number of
glycosidic linkage per glycosidic linkage per molecule

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Biomolecules, Biological Process & Chemistry in Action

Some of the characteristic properties shown by monosaccharides are listed below.

Reducing nature: All monosaccharides are strong reducing sugars as evident from their reaction
with Tollen’s reagent or Fehling solution.

Cyclic Structures of Monosaccharides

Open chain strucutre of monosaccarides is only correct for those which have upto 3 or 4 carbon
atoms are present in the backbone. Monosaccharides having 5 or 6 carbon atoms in the main
chain exist in cycle structure in the aqueous solution. the cyclic structure is due to intramolecular
hemiacetal formation between aldo/keto group and OH of any one carbon. The ring formed are
generally six membered (pyranose) or five membered (furanose). Due to cyclization there in
creation of a new asymmetric centre in addition to existing ones. The isomers thus formed due
to cyclizations are called anomers. For example, when D-glucose (open structure) cyclise, it gives
-D-glucose and -D-glucose.

The anomers only in the orientation of the –OH group at C 1.

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CHO
H OH H OH HO H
H OH
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H OH HO H
HO H O H OH HO H O
H OH H OH H OH
H CH2OH H
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CH2OH CH2OH
-D-Glucose D-Glucose -D-Glucose
(pyranose structure) (Open structure) (pyranose structure)
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Ordinary -D-glucose has specific rotation of +111° in aqueous solution which gradually change
until it reaches a constant value of +52.7°. This phenomenon is known as mutarotation. Similar
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the case with -D-glucose. Mutarotation is observed because of equilibrium between open chain
structure and cyclic structures of monosaccharides in solutions. The specific rotation of -D-
glucose is +19° and equilibrium mixture contains 36.2% and 63.8% -D-glucose and -D-glucose
respectively.

These two anomers of glucose, i.e., -D glucose and -D glucose have different melting points
and exist in different crystalline forms. They also give different optical rotations. These anomers
in pyranose ring structure are best written by Haworth projection formula as follows.
6 6
CH2OH CH2OH
H 5 O H 5 O
H OH
H H
4 1 4 1
OH H OH H
HO OH HO H
3 2 3 2
H OH H OH

-D-Glucopyranose -D-Glucopyranose
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Biomolecules, Biological Process & Chemistry in Action

6 6
O 1 O
HOCH2 CH2OH HOCH2 OH
5 2 5 2
H HO H HO
H OH H CH2OH
4 3 4 3 1
OH H OH H

-D-Fructofuranose -D-Fructofuranose

Disaccharides
A glycosidic bond is developed when hydroxyl group of the hemiacetal carbon of one
monosaccharide condenses with a hydroxyl group of the second monosaccharide molecule just as
hemiacetal reacts with alcohol to give acetal and water.

H H
C  O  H + H O  R  C  OR + H2O
Alcohol
OR OR
Hemiacetal Acetal
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The three important naturally occuring disaccharides are Sucrose, Maltose, and Lactose.
(i) Sucrose : Sucrose is a disaccharide obtained from sugar-cane. Its molecule is composed of
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two monosaccharides namely -D (+) glucose and -D (–) fructose. During formation of
sucrose, –OH group at C1 at -D-glucose unit is involved in condensation with –OH group
at C2 of -D-fructose. Thus sucrose have 1, 2 linkage between two monosaccharides and
acquires the given structure.
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CH2OH
H O H
H O CH2OH
H
OH H O OH H
HO H
CH2
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H OH H OH
-D (+) Glucose -(–) Fructose.
(ii) Maltose : Maltose is malt sugar and one molecule of it is formed by condensation of two
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molecules of -D (+) glucose in which –OH group at C1 of one -D-glucose unit is
condensed with –OH group at C4 of second -D-glucose unit. So in maltose molecule, two
-D-glucose molecules are linked together by -C1 –C4 glucosidic bond as depicted below.
Since in maltose one of the -(D)+glucose units has a free hemi acetal group, it is a reducing
disaccharide.
-glycosidic linkage
6 6
CH2OH CH2OH CH2OH CH2OH
H 5 O H H 5 O H H O H H O H
H H –H2O H H
4 1 4 1
OH H OH H OH H OH H
HO OH HO OH HO O OH
3 2 3 2
H OH H OH H OH H OH

-D(+) Glucose -D(+) Glucose Maltose

(iii) Lactose : It consists of two monosaccharides [-D (+) glucose and -D(+) galactose]
linked together by C1 – C4 O-glycosidic linkage.
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Biomolecules, Biological Process & Chemistry in Action

Relative sweetness of sugar

Sugar Relative sweetness Sugar Relative sweetness
Sucrose 100 (standard) Glucose 74
Lactose 16 Invert sugar 130
Galactose 32 Fructose 173
Maltose 33

Polysaccharides
Polysaccharides are naturally occuring, colourless, tasteless, amorphous powders. These have the
general formula (C6H10O5)n where value of n varies between 12 to 200.
(i) Starch : Starch is the most abundant carbohydrate found in vegetable kingdom. On hydrolysis,
starch is initially converted to disaccharide (maltose) and subsequently to monosaccharide
(glucose). Therefore, starch is made up of -D-(+)glucose units joined through o-glycosidic
linkage. Starch consists of two derivatives: Amylose and Amylopectin.
(a) Structure of amylose : Amylose is a straight chain polysaccharide composed entirely of
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-D-glucopyranose units joined to each other by C1 –C4 O-glycosidic linkage involving

C1 of one glucose unit and C 4 the other, in amylose range from 60-300 and thus its
molecular mass has a range of 10,000 to 50,000 amu. It is water soluble component of
starch.
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CH2OH CH2OH CH2OH CH2OH

H O H O H H O H H O
H
H H H H
O OH H OH H O OH H O OH H O
O
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H OH H OH H OH H OH
Amylose
-glycosidic linkage

Structure of amylose
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(b) Structure of amylopectin : The water insoluble component of starch is known as

amylopectin and has a branched chain structure. It is a highly branched polymer containing
250–300 glucose units and has molecular mass in the range of 50,000 to 1,00,000 amu.
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First carbon atom, C1, of terminal glucose unit in each amylose chain is linked to the C4-of
the other glucose unit is linear or chain and branched in the next chain through C1– C6
linkage by O-glycosidic bond. The structure of amlylopectin is given below.
CH2OH CH2OH
H O O H
H H
H H -1, 6 glycosidic
OH H OH H linkage
O O

H OH H OH
n
CH2 CH2
O H H O H
-1, 4 glycosidic H
linkage H H
OH H O OH H
HO

H OH H OH n

Structure of amylopectin.
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Biomolecules, Biological Process & Chemistry in Action

Differences between amylose and amylopectin

S.No. Amylose Amylopectin

1. It is water soluble fraction of starch It is water insoluble fraction of starch.

2. It is 20% of the quantity of starch It is 80% of starch.
3. It is a straight chain polymer of It is branched chain polymer of
-D-glucose units. -D-glucose units.
4. In amylose, the -D(+)glucose units In amylopectin, the -D(+) glucose
are joined together by C1 – C4 units are joined by C1–C 4 O-glycosidic
O-glycosidic linkage bond in linear chain and by C 1 –C 6
o-glycosidic bond is branching.
5. Its molecular mass lies in the range Its molecular mass is in the range of
of 10,000 – 50,000 50,000 – 1,00,000.
(ii)\ Cellulose: Cellulose is the main structural material of cell walls of plants and vegetable
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tissues. Cellulose is mainly present in wood (75%), cotton (90 – 95%), jute, flex and hemp,
ete. In fact, 50% of all the living matter is composed of cellulose. The cellulose is a linear
polymer of nearly 2000 -glucopyranose rings and its molecular formula is (C6H10O5)n.
(glucose  glucose  glucose  glucose)n 
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Linear polymer C1 – C4 -linkage

The better representation of the structure of cellulose molecule is given below.

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–6
CH2OH CH2OH CH2OH
H O H 5 O H O
H 4
H H
O O 1 O O
OH H OH H OH H
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H H H
3 2
H OH H OH H OH
n–2
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-linkage

Cellulose is the most important natural polymer whose chemical treatment gives various useful
derivatives e.g.,
a) Rayon: Cellulose acetate and cellulose xanthate are used as a fibre.
b) Celluloid: Cellulose dinitrate also known as pyroxylin, mixed with plasticizer and alcohol,
is used for the manufacturing of photographic film, spectacle frames, piano keys, etc. It
is known as artificial ivory.
c) Explosive: Cellulose trinitrate is used extensively as a blasting and propellant explosive.
d) Lacquer: Collodion is used for manufacturing washable cellulose paints.
e ) Water proofing : Solution of cellulose acetate is used to provide antishrink property to
textile fabric.
f) Methyl cellulose is used in fabric sizing, paste and cosmetics. Ethyl cellulose is used for
manufacturing of rain coats and plastic films.
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Biomolecules, Biological Process & Chemistry in Action

(iii) Glycogens : It is known as animal starch (reserve polysaccharide) and has the same
molecular formula (C 6H10O5)n. This carbohydrate is stored in the cells of the liver and
muscles of the animal. Glycogen is a linear polymeric molecule in which the average chain
length consists of about 10 – 14 glucose units, contrary to vegetable starch in which
amylopectin is having 20 – 25 glucose units. This shows that glycogen is more branched
polysaccharide than starch. It is hydrolysed by -amylase to maltose.
(iv) Gums: Gums are complex polysaccharides excreted by trees or shrubs to seal wounds in
the bark. These on hydrolysis yield a number of monosaccharides such as arabinose,
rhamnose, galactose, etc. This differentiate gums from starch, cellulose and glycogen which
gives only D-glucose on hydrolysis. Gums swell in water to form gels and are employed as
thickening material for improving the texture in food industry. Gum arabic is completely soluble
in water and thus finds its extensive use in pharmacy and as adhesive.
(v) Pectins: Pectins are polysaccharides which are mainly present in the fruit skins. It is obtained
by boiling fruit skins or stems with aqueous alcohol. Pectins are methyl esters
of polygalacturonic acid and on hydrolysis gives pectic acid. Solutions of pectins in fruit juice
have the power of setting to a jelly and thus responsible for the setting of jams. Pectins
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find their applications in drug preparation where they are used as inert agents to
suspend an active drug.

Functions of carbohydrates
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(i) They act as biofuels and provide energy for the normal functioning of the living systems. In
this unit we have studied that polysaccharides (starch, glycogen, etc.) and disaccharides
(maltose, lactose and sucrose) are hydrolysed to glucose by enzymes responsible for digestive
systems. Glucose is then transported from cell to cell by the blood in human body and by cell
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sap in plants. Oxidation of glucose to carbon dioxide and water and is catalysed by enzymes
and provides necessary energy for the functioning of the living cells.

Enzyme
C 6 H12 O 6  
 6CO2 + 6H2O + 2868 kJ/mol
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Glucose

(ii) Carbohydrates are the structural material for the cell walls e.g., cellulose acts as the chief
constituent of the plant cell walls.
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(iii) Carbohydrate is also stored in the liver as glycogen known as animal starch. It serves as
instant food, quickely digested and readily converted into the required fuel for producing energy.

SAMPLE PROBLEMS 2.1 (MCQ)

Problem 1: Which of the following is a polysaccharide ?
(a) Glucose (b) Galactose
(c) Sucrose (d) Pectines

Solution: (d) Pectines is a polymer of monosaccharide and a derivative of cellulose.

Problem 2: Star ch can be used as an indicator for the detection of traces of

(a) Glucose in aqueous solution (b) Proteins in blood
(c) Iodine in aqueous solution (d) Urea in blood

Solution: (c) Iodine forms amylum starch molecule which is blue black in colour.

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Biomolecules, Biological Process & Chemistry in Action

Problem 3. Which of the following statements about ribose in incorrect ?

(a) It is a polyhydroxy compound (b) It is an aldehyde sugar
(c) It has six carbon atoms (d) It exhibits optical activity

Solution: (c) Ribose is a five carbon containing sugar.

Problem 4: -D-Glucose and -D Glucose differ from each other due to difference in one of carbon
with respect to its
(a) Size of hemiacetal ring (b) Number of—OH groups
(c) Configuration (d) Conformation

Solution: (c) In -D-Glucose –OH group is axial position and In b-D-Glucose –OH is present in
equitorial position

Problem 5: The letter ‘D’ in carbohydrates represents

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(a) Its direct synthesis (b) Its dextrorotation
(c) Its mutarotation (d) Its configuration

Solution: (d) It is a methord of nomenclature of monosaccharide.

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Problem 6: Which of the following is the structure of D-Xylulose?

CHO CHO
HO — — H HO — — H
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(a) H — — OH (b) HO — — H
H — — OH H — — OH
CH2OH CH2OH
CHO CHO
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H — — OH H — — OH
(c) HO — — H (d) H — — OH
H — — OH H — — OH
CH2OH CH2OH
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Solution: (c) A structural formula.

Problem 7: Glucose gives the silver mirror test with ammoniacal solution of silver nitrate because it
contains the group
(a) Aldehyde (b) Ester
(c) Ketone (d) Amide

Solution: (a) –CHO group gives silver mirror test.

Problem 8: It is best to carry out reactions with sugars in neutral or acidic medium not in alkaline
medium. This is because in alkaline medium sugar undergoes one of the following changes.
(a) Decomposition (b) Inversion
(c) Rearrangement (d) Racemization

Solution: (b) Conversion of d-form of sugar into l form on hydrolysis is called inversion.

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Problem 9: The calorific values of fats, carbohydrates and proteins vary in the order
(a) Fats > carbohydrates > proteins (b) Fats > proteins > carbohydrates
(c) Carbohydrates > proteins > fats (d) None of these

Solution: (b) Amount of energy liberated by the combursion of one mole of molecule.

Problem 10: Molich’s test is done for the detection of

(a) Alkyl halide (b) Carbohydrate
(c) Alkaloid (d) Fat

Solution: (b) When few drops of alcoholic solution of alpha nephthol is added to 2 ml of Glucose
solution and 1 ml of conc. H2SO4 a violet ring is formed is called Molich’s test.

2.2 Proteins
Proteins are complex, nitrogenous organic substance which occur in all animals and plants. They
are so named because proteins are the most vital chemical substances of primary importance
neccessary for the normal growth and maintenance of life. Protein serves following functions in
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our body.

a) To promote growth.
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b) To supply essential amino acids to blood.

c) To maintain body tissues.
d) To synthesize various enzymes.
e) To protect body from infection.
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Protein content of food stuffs

Food stuff % Protein Food stuff % Protein

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Milk 5 Meat 24 – 26
Wheat 14 Egg yolk 16
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Peas 21 Egg (white) 12.5

Maize 10 Cheese 33

Classification of Proteins
Proteins are classified by two different methods. According to first mode of classification proteins
are of two types depending upon their shape and functions.

(a) Fibrous proteins

(b) Globular proteins

(a) Fibrous proteins: These have thread like molecules which lie side by side to form fibres.
The various molecules are held together by hydrogen bonds. These are insoluble in water but
soluble in concentrated acids and alkalis e.g., hair, nails, wood, feathers and horn, etc, are made
up of keratin. Muscles have myosin. Silk is composed of fibroin. Bones and cartilage have
collagen.

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(b) Globular proteins: This type of protein has molecule folded into compact units which often
acquire spheroidal shape. Such proteins are soluble in water, dilute acids and alkalis e.g., insulin,
haemoglobin, albumin, etc.

-Amino acids
-Amino acids are substituted carboxylic acids in which one -hydrogen atom of alkyl group is
substituted by amino (–NH2) group. These may be represented by the general formula.

R  CH  COOH
where R = H or alkyl group
NH2

Structure of -amino acids

The amino acids containing one carboxylic group and one amino group behave like a neutral
molecule. This is due to the fact that in aqueous solution the acidic carboxylic group and basic-
amino group neutralise each other intramolecularly to form an internal salt structure, known as
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zwitter ion or dipolar ions.

+
NH2 NH3
R  C  COOH R  CH  COO–
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Zwitter ion

However, the neutral zwitter ion (dipolar ions) changes to cation in acidic solution and exist as
anion in alkaline medium. Thus amino acids exhibit amphoteric character.
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Alkali Acid
H2N  CH  COO– H 3N+  CH  COO– H 3N+  CH  COOH
Acid Alkali
R R R
Zwitter ion
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Anion Cation
I II III

Therefore, amino acid exist as zwitter ion when the solution is neutral or its pH ~ 7. The pH
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at which the structure of an amino acid has no net charge is called its isoelectric point. At this
pH, the concentration of cationic form (III) and anionic form (I) will become equal and hence
there is no net migration of the -amino acids under the influence of an electric field.

Classification of amino acids

 -Amino acids are broadly classified in three main groups based on the relative number of
–NH2 and –COOH group:

(i) Neutral amino acids : Amino acids containing one –NH2 group and one –COOH group
are called neutral amino acids, e.g., glycine, valine, alanine, etc.

(ii) Basic amino acids: Amino acids which contain one –COOH group and two –NH 2
groups are called basic amino acids, e.g., lysine and arginine.

(iii) Acidic amino acids : Amino acids containing two –COOH groups and one –NH2 group
are classified as acidic amino acids, e.g., aspartic acid and glutamic acid, etc.

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List of 20 naturally occuring -amino acids

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-Amino acids have also been classified as Essential and Non-Essential amino acids.

 Out of 20 -Amino acids required for protein synthesis, human body can synthesize only 10 out of
them (e.g., Gly, Ala, Ser, Cys, Gln, Tyr, Pro, Asp, Asn, Glu) and are called as non-essential amino
acids. The other 10 amino acids (e.g. Val, Leu, Ile, Phe, Met, Trp, Thr, Lys, Arg and His) which
cannot be synthesized by human body are called as essential amino acids.

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Structure of Proteins

The structure of proteins is quite complex. Study of its structure is carried out under the following
headings.

Primary structure of protein : The primary structure of protein refers to its covalent structure,
i.e., the sequence in which various -amino acids are arranged in protein or in the polypeptide
structure of protein.

The linkage –CO – NH – is known as peptide linkage.

H2N – CH – COOH + H2N – CH – COOH

R R
H2N – CH – CONH  CH – COOH + H2O
Peptide bond
R R

The dipeptide still has free amino and carboxyl groups through which it can react with other
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molecules of amino acid resulting in polypeptide formation

O H R O H R
C N CH C N CH
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CH C N CH C N
R O R R O H

C-terminal N-terminal
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Linear polypeptide chain

In polypeptide chain, the free amino end is termed as N-terminal and the free carboxyl end is
said to be C-terminal end.
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Secondary structure of proteins ..

.. H ..
Secondary structure of protein refers to the arrangement of polypeptide H N CH
C O.. N
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C N
chains into a definite three dimensional structure which protein assumes
O.
. O. H
as a result of hydrogen bonding. Depending upon the size of the R-group CH . N
H
C O.
. N C
of the amino acids in polypeptides, two different types of secondary N
O.
. H O.
structure are possible as discussed below H N C H
C N C O.. N
(a) -Helical structure (b) -Pleated sheets O. O. H
. .
H H N
-Helical structure : This type of secondary structure is acquired when C N
N
O.
the R groups in amino acids are large and involve in coiling of the .
polypeptide chains. The helical pattern in right handed coil and the shape
is stabilised by the intra molecular hydrogen bond between the >C = O group of one amino
acid and –NH group of the fourth amino acid.

-Pleated structure: This type of secondary structure is acquired when the R groups of amino
acids are small. In this structure the linear polypeptide chains are arranged side by side and held
together by intermolecular hydrogen bond between the C = O and –NH group.
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Biomolecules, Biological Process & Chemistry in Action

H R O H R
H
C C N C
C N C C N
O H H R O H
H O H R H O
N C C N C
C N C C
H R O H R
H
7.2Å

Tertiary Structure of Proteins

The tertiary structure of protein is the most stable shape that a protein assumes under the normal
conditions of temperature and pH. During acquiring of tertiary structure, various types of attractive
forces between the polypeptide chains are involved. These attractive forces, are hydrogen bond,
disulphide bonds, ionic, chemical and hydrophobic bonds, result in a complex and compact structure
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of protein. The two important tertiary structures of proteins are fibrous structures and globular
structure. Fribrous proteins have largely helical structure and are rigid molecules of rod like
shape. Globular proteins, on the other hand, have a polypeptide chain which consist partly of helical
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sections and partly -pleated structure and remaining in random coil form. These different segment
of secondary structure then fold up to give protein a spherical shape.

Quaternary structure of proteins

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The quaternary structure of protein is developed when the polypeptide chains, which may or may
not be identical are held together by hydrogen bonds. It results in the increase of molecular mass
of protein greater than 50,000 amu. For example, haemoglobin contains four submits, two identical
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-chains containing 141 amino acids each and the other two identical -chains containing 161
amino acids each.

Denaturation of proteins
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When the proteins are subjected to the action of heat, mineral acids or alkali, the water soluble
form of globular protein changes to water insoluble fibrous protein resulting in the
precipitation or coagulation of protein. This is called denaturation of proteins.

Biological Functions of Proteins

 As Structural materials: (e.g., Keratin in skin, hair, nails, wool, etc. Collagen in tendons. Myosin
in muscles and Fibroin in silk).

 As transport agents: (e.g., Haemoglobin in blood).

 As enzymes which catalyse all biochemical reactions.

 As metabolic regulators.(e.g., Insulin controls glucose level in blood).

Thyroglobulin helps to synthesize thyroxine horomone. Nucleoproteins carry genetic information

form parents to the offsprings.

[ 43 ]
Biomolecules, Biological Process & Chemistry in Action

Enzymes
Enzymes are biological catalysts and are chemically globular proteins.

Co-factors: Certains enzymes are associated with some non-protein components for its biological
activity and are called as co-factors. e.g., Inorganic ions like Na+, K+, Fe2+, Cu2+, Co2+, Mg2+, Zn2+,
Mn2+, etc.

Organic molecules: Some vitamins, coenzymes and prosthetic groups.

The protein co-factor complex is usually called as holoenzyme while the inactive proteins part left
after the removal of co-factor is called as apoenzyme.

Deficiency diseases:

 Phenylketone urea: by deficiency of enzyme Phenylalanine hydroxylase.

 Albinism: by deficiency of enzyme Tyrosinase.

JE

Hormones
Hormones are biomolecules produced in the endocrine glands and are circulated through the blood
ES

stream throughout the body.

They control various metabolic processes.

These are of 3 types mainly:

an

Steroid hormones

(a) Sex hormones:

ka

(i) Androgens: Androsterone and Testosterone are male sex hormones.

(ii) Estrogens: Estrone, Estradial and Estriol are female sex hormones.
lp

(iii) Gestogens: (Progesterone, Pregnandiol, control the development and maintenance of

pregnancy):

(b) Adrenal cortex hormones:

Cortisone, Aldosterone, etc. regulate the metabolism of fats, proteins and carbohydrates.

Peptide hormones: e.g., Oxytocin, Vasopressin, Angiostensin-II, Insulin, etc.

Amine hormones: e.g., Adrenalin or Epinephrine and Thyroxine.

Vitamins
Vitamins are the biomolecules which are required in small amounts for the normal metabolic processes
and for the life, growth and health of human beings. Vitamins neither supply energy nor help in
building tissue structures. They are just needed for keeping good health of human beings and animals.
Their deficiency causes many diseases.

[ 44 ]
Biomolecules, Biological Process & Chemistry in Action

Vitamins cannot be synthesized in the body and hence must be supplied in the food. Vitamin-D,
however may be either supplied in food or may be produced in the skin by the irradiation of steroids
(present in the oils and fats) with sunlight.

There are about 25 vitamins known. They are broadly divided into two categories:

(i) Fat soluble vitamins – Vitamin -A, D, E and K.

(ii) Water soluble vitamins – Vitamin-B1, B2, B6, B12 and C.

Vitamins, their sources and deficiency diseases

Vitamin Sources Deficiency diseases

1. Vitamin-A (Retinol) Milk, cod liver oil, butter, eggs, Night blindness, xerophthalmia
carrots, green leaves, tomatoes, etc. and xerosis.

2. Vitamin -B1 Pulses, nuts, green vegetables, Beriberi and loss of apetite
or Thiamine polished rice, whole cereals, yeast
JE

or Lactoflavin and egg yolk.

3. Vitamin B2 Milk, meat, green vagetables Inflammation of tongue and

or Riboflavin and yeast. cheilosis (cracking of the lips and
ES

or Adermine corners of the mouth)

4. Vitamin - B6 Rice, bran, yeast, meat, fish, egg, Specific dermatitis called
or Pyridoxine yolk, maize, spinach and lettuce acrodynia, pellagra (shrivelled
an

or Adermine skin) and anaemia.

5. Vitamin - B12 Milk, liver, kidney and eggs, Inflammation of tongue, mouth,
or Cyanocobalamine etc., and pernicious anaemia
ka

6. Vitamin - C Citrous fruits (oranges, lemons) Scurvy and brittleness of bones,

or L-Ascorbic acid amia, sprouted pulses, germinated swelling and bleeding of gums and
lp

grains and leafy vegetables loosening of teeth

7. Vitamin-D Fish liver oil, cod liver oil, milk Rickets (softening and bending of
Ergocalciferol and eggs bones) in children, controls Ca and
or antirachtic vitamin P metabolism
or sunshine vitamin

8. Vitamin - K or Cabage, alfalfa, spinach and Haemorrhage and lengthens time

Tocopherols (, , ) cotton seed oil, etc. of reproduction
or Antisterility Eggs, milk, fish, wheat germ oil
factor cotton seed oil etc.

9. Vitamin-K or Cabage, alfalfa, spinach and Haemorrhage and lenthens time

Phylloquinones or Carrot tops of blood clotting
Anthihaemorrhagic
vitamin

[ 45 ]
Biomolecules, Biological Process & Chemistry in Action

SAMPLE PROBLEMS 2.2 (MCQ)

Problem 1: The secondary structure of a protein refers to:
(a) fixed configuration of the polypeptide backbone
(b) -helical backbond
(c) hydrophobic interactions
(d) sequence of -mino acids

Solution: (a) A secondary structure protein is formed due to inter or intra molecular hydrogen bonding
and by the formation of sulphide bridge.

Problem 2: Denaturation of protein

(a) disrupts the primary or secodnary or tertiary structure of protein
(b) disrupts the secondary and tertiary structures only
(c) disrupts all the primary, secondary and tertiary
(d) will not affect the original biological activity
JE

Solution: (b) On heating hydrogen bonds and sulphide bridge clived in the protein molecule.

Problem 3: Insulin production and its action in human body are responsible for the level of diabetes.
ES

This compound belongs to which of the following categories.

(a) a coenzyme (b) a hormone
(c) an enzyme (d) an antibiotic
an

Solution: (b) Harmones are protienous in nature.

Problem 4: At the isoelectric point of an amino acid, the species present are
ka

(a) R–CH–COOH (b) R–CH–COOH

| |
NH2 +NH
3

(c) R–CH–COO– (d) R–CH–COO–

lp

| |
NH2 +NH
3
Solution: (d) The pH at which is no net migration of amino acid during electrolysis is called isoelectric
point.

Problem 5: The iron in haemoglobin is bounded by

(a) Hydrogen bonds (b) Chelation
(c) Ionic bonds (d) Covalent bonds

Solution: (b) In haemoglobin Fe chelation with prophrine ring chelation.

Problem 6: Coordination polymerization was developed by

(a) Zeigler and Natta (b) Linus Pauling
(c) Beckmann (d) None of these
Solution: (a) Factual

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Biomolecules, Biological Process & Chemistry in Action

Problem 7: The hormone, epinephrine, is secreted by the organ

(a) Testis (b) Ovary
(c) Adrenal Medulla (d) Pancreas
Solution: (c) A biological fact.

Problem 8: For Amino acid having the structure R—CH—CO2H

|
NH2

Which of the following statements are true?

(A) Water solubility is maximum at a pH when concentration of anions and cations are
equal.
(B) They give ninhydrin test
(C) On reacting with nitrous acid give off N2
(a) All (b) B and C
(c) A and B (d) None of these
Solution: (b) (2,2-Dihydroxyindane-1,3-dione) is used to detect ammonia 1°,2° amines and some amino
acids.
JE

OH
OH
ES

O
Problem 9: Which of the following statements is true of proteins?
(a) They catalyse the biochemical reactions (b) They act as antibodies
an

(c) They act as hormones (d) They perform all these functions
Solution: (d) Factual
Problem 10: An example of a water soluble vitamin is
ka

(a) Vitamin A (b) Vitamin C

(c) Vitamin D (d) Vitamin E

Solution: (b) Vitamin B and C are water soluble.

lp

2.3 Nucleic acids

Nucleic acids are vital biomolecules which are present in the nuclei of all living cells in the form
of nucleoproteins. These are long chain polymers with a high molecular mass. These are also
known as biopolymer having nucleotide as a repeating structural unit (monomer). These play an
essential role in transmission of the heredity characteristics from one generation to the next and
also in the biosynthesis of proteins. Therefore, the genetic information coded in nucleic acid
governs the structure of protein during its biosynthesis and hence controls the metabolism in the
living system.
The nucleic acids are of two types differing mainly in the nature of carbohydrate present in them.
(i) DNA (Deoxyribonucleic acid) (ii) RNA (Ribonucleic acid)
Structure of nucleic acids
The nucleic acids are the posthetic component of the nucleoproteins. Nucleic acid on stepwise
hydrolysis gives following products as shown in the chart

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Biomolecules, Biological Process & Chemistry in Action

Nucleic acid
Hydrolysis
Nucleotides
Hydrolysis

Nucleosides Phosphoric acid

Hydrolysis

Pentose sugar Purine bases Pyrimidine bases

Ribose or (Adenine and Guanine) (Thymine and Cytosine) from DNA
Deoxyribose (Uracil and Cytosine) from RNA

N-containing bases

Difference between DNA and RNA

JE

The main points of difference between the two types of nucleic acids are given in the table.

Sl. DNA RNA

No.
ES

1. The pentose sugar present in it is It has D (–) ribose sugar.

2-deoxy D(–) ribose.
2. It contains cytosine and thymine as It contains cytosine and uracil as
an

pyrimidine bases pyrimidine bases.

3. DNA is double stranded and pairing of bases It is a single stranded molecule looped
is present throughout the molecule. back on itself. The pairing of bases is
ka

present only in helical part.

4. It occurs in the nucleus of the cell. It mainly occurs in the cytoplasm of the
cell.
lp

5. It is a very large molecule and the molecular It is a much smaller molecule and its
weight vary from 6 million to 16 million amu. molecular weight ranges from 20
thousand to 40 thousand.
6. It has a characteristic property of replication It does not replicate.
7. DNA controls the heredity character. RNA only governs the biosynthesis of
proteins.

Ribonucleic acids

1. The pentose sugar in RNA is ribose.

2. Purine bases of RNA are represented by adenine and guanine, the pyrimidine bases are uracil
and cytosine.

3. The thymine in DNA is replaced by uracil in RNA.

[ 48 ]
Biomolecules, Biological Process & Chemistry in Action

4. RNA is single stranded, but double stranded RNA is present in Rheovirus and wound tumour
virus.

5. There are three major classes of RNA; each with specific functions in protein synthesis

mRNA

6. Messenger RNA is produced by DNA by the process called as transcription.

7. Messenger RNA encodes the amino acid sequence of a protein in their nucleotide base sequence.

8. A group of three of nitrogenous bases specifying an amino acid in mRNA is called as triplet
codon.

tRNA

9. tRNA is also known as soluble RNA (sRNA). It is soluble in 1 molar solution of sodium chloride.

10. tRNA identifies amino acids in the cytoplasm and transports them to the ribosome.
JE

11. Molecules of tRNA are single-stranded and relatively very small.

12. Anticodon is a three-base sequence in a tRNA molecule that forms complementary base pairs
with a codon of mRNA.
ES

13. All transfer RNA possess the sequence CCA at their 3 ends; the amino acid is attached to the
terminal residue.
an

rRNA

14. Ribosomal RNA is found in ribosomes of cells and is also called insoluble RNA.

15. The main function of rRNA is to attract and provide large surface for spreading m-RNA over
ka

ribosomes during translocation process of protein synthesis.

Type Sedimentation Mol. Wt. Number of % of total cell

lp

coefficient nucleotide residues RNA

mRNA 6 to 25S 25,000-1000,000 75-3000 2

tRNA 4S 23,000-30,000 75-90 16
rRNA 5S 35,000 100
16S 550,000 1500 82
23S 1100,000 3100

Genetic Code

16. The relationship between the sequence of amino acids in polypeptide with base sequence of
DNA or mRNA is genetic code.

17. Genetic code determines the sequence of amino acids in a protein.

18. A triplet would code for a given amino acid as long as three bases are present in a particular
sequence.
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Biomolecules, Biological Process & Chemistry in Action

19. Later in a cell-free system, Marshall Nirenberg and Philip Leder (1964) were able to show that
GUU codes for the amino acid Valine.

20. The spellings of further codons were discovered by R. Holley, H. Khorana and M. Nirenberg.
They have been awarded the Nobel Prize in 1968 for researches in genetic code.

Mechanism of protein synthesis

21. The protein synthesis takes place in following steps :

(i) Transcription, (ii) Activation of amino acids, (iii) Attachment of activated amino acids with tRNA
and (iv) Translation.

22. The mode by which DNA passes its genetic information to mRNA is known as transcription.

Transcription
23. Transcription is the process by which a RNA molecule is synthesised on a DNA strand.

24. The molecules tRNA and rRNA are also transcribed from DNA templates.
JE

25. Wherever A,. T, G or C is present in the DNA template. U, A, C or G is incorporated into the
mRNA molecule.

26. A segment of the DNA helix unwinds and unzips and complementary RNA nucoeotides pair
ES

with DNA nucleotides of the strand that is to be transcribed.

27. Transcription begins when RNA polymerase attaches to a region of DNA called a promoter.

28. A promotor defines the start of a gene, the direction of transcription and the strand to be copied.
an

29. Finally RNA polymerase comes to a terminator sequence at the end of the gene being transcribed.

30. The terminator causes RNA polymerase to stop transcription and to release to stop transcription
and to release the mRNA molecule, now called as RNA transcript.
ka

31. Transcription copy of mRNA from DNA–AGT CCT TGG AAT will be UCA GGA ACC UUA.

32. Base sequence of DNA strand is CAT TAG CAT CAT GAC. The base sequence on RNA
lp

strand will be GUA AUC GUA GUA CUG.

33. The process of inhibition of RNA synthesis on a DNA template is called repression.

34. RNA synthesis (transcription) is inhibited by Actinomycin D.

Lipids
The term lipids represent a group of biomolecules which are insoluble in water but
soluble in organic solvents of low polarity such as chloroform, toluene, ether, carbon
tetrachloride, etc. Lipids are the constituent of all cell membranes and thus regulate the activities
of cells and tissues. They also serve as the energy reserve for living cells.

Lipids are broadly classified in three groups:

(i) Triglyceride esters of higher fatty acids or oils and fats

(ii) Phospholipids
(iii) Waxes
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Biomolecules, Biological Process & Chemistry in Action

(i) Triglyceride esters of higher fatty acids or oils and fats: The lipids which are triesters
of glycerol with long chain monocarboxylic acid are called triglycerides. In nature, oils and
fats most abundantly present as triglycerides in plants and animals.

Oils are the triglyceride esters of glycerol with longchain unsaturated monocarboxylic acids.
They are liquid at ordinary temperature. The most widely observed unsaturated fatty acids
in oils are

Oleic acid CH3 – (CH2)7 – CH = CH – (CH2)7 – COOH

Linoleic acid CH3 – (CH2)4 – CH = CH – CH2 – CH = CH – (CH2)7 – COOH
Linolenic acid CH3 – CH2 – CH = CH – CH2 – CH = CH – CH2 – CH = CH – (CH2)7 – COOH

Contrary to oils, the fats are solid at room temperature and have long chain saturated fatty
acids, e.g.,

Myristic acid CH3 – (CH2)12 – COOH

JE

Palmitic acid CH3 – (CH2)14 – COOH

Stearic acid CH3 – (CH2)16 – COOH

Oils or fats on hydrolysis with NaOH produces glycerol and the sodium salt of long chain fatty
ES

acids (soap). This process is known as saponification. Soaps are generally manufactured by
heating oil or fat with NaOH.

CH2 – O CO – R CH2 – OH
an

CH2 – O CO – R + 3NaOH CH – OH + 3R – COONa

Sodium salt of
CH2 – O CO – R CH2 – OH
higher fatty acids
Oil or fat Glycerol (soap)
ka

(ii) Phospholipids : Phospholipids are the major constituents of cell membrane. These are mixed
glycerides of higher fatty acids and phosphoric acid. In this, two –OH groups of glycerol are
esterified with long chain fatty acids while the third –OH group is esterified with some
lp

derivatives of pshosphoric acid. The phosphoric acid derivative contains a nitrogen containing
alcohol. e.g., ethanol amine, choline, serine of inositol. The simplest example of phospholipid
is lecithin which is found in egg yolk, brain and nerve tissue

(iii) Waxes: Structurally waxes are the esters of long chain fatty acids with long chain monohydric
alcohols, e.g.,
O
Supermaceti (Cetyl palmitate) CH3 – (CH2)14 – C – O – (CH2)15CH3

O
Bee wax (Myricyl palmitate) CH3 – (CH2)14 – C – O – (CH2 )29 – CH3

O
Carnauba wax (Myricyl cerotate) CH3 – (CH2)24 – C – O – (CH2 )29 – CH3

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Biomolecules, Biological Process & Chemistry in Action

Metabolism
Metabolism is the combination of sum total of catabolism and Anabolism. One of these aspects
concerns the degradation of nutrients and various components of cells and tissues,
e.g., carbohydrates., proteins, lipids and nucleic acids.

The degradative processes are called as catabolism, and the synthetic as anabolism. A third
consideration, which involves both anabolism and catabolism,, is the derivation and utilisation of
energy.

Catabolism. The degradation of complex organic molecules (e.g. carbohydrates, lipids, proteins,
vitamins, steroids, steroids, etc.) into smaller ones (e.g. CO2, NH3, urea, lactic acid, etc.) by a
series of enzymatic reactions within body is known as catabolism.

Anabolism. The synthesis of complex organic molecules (like carbohydrates, proteins, lipids,
nucleic acid, etc.) from simpler molecules in living organisms by a series of enzymatic reactions
is called anabolism.
JE

Digestion
Digestion may be defined as the process which converts relatively large organic molecules into
smaller molecules capable of absorbed by the gastrointestinal tract (GIT).
ES

Role of enzymes in digestion of foods: The three important constituents of food, viz.
carbohydratres, fats (lipids) and proteins are subjected to the action of various enzymes to form
simple constituents capable of being absorbed by the small intestine (digestion).
an

(a) Polysaccha rides Ptyalin

  Disacchari des Maltase
  Monosaccha rides
(e.g. Starch) (Amylase) (Maltose) etc. (Glucose, etc.)

Lipases
(b) Fats bile
ka
 Emulsified fat   Fatty acids  Glycerol
salts

(c) Proteins Pepsin


Peptidases
 Proteoses and peptones     α  Amino acid
lp

SAMPLE PROBLEMS 2.3 (MCQ)

Problem 1: The presence of absence of hydroxyl group on which carbon atom of sugar differentiates
RNA and DNA
(a) 2nd (b) 3rd
(c) 4th (d) 1st

Solution: (a) In DNA, molecule, on ribose sugar’s second carbon a deficient of oxygen.

Problem 2: If one strand of DNA has the sequence ATGCTTGA, the sequence in the complementary
strand would be
(a) TAGGAACT (b) TCCGAACT
(c) TACGTACT (d) TACGTAGT

Solution: (a) Adenine pairs with tymine and cytosine with guanine.

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Biomolecules, Biological Process & Chemistry in Action

Problem 3: Which of the following statements about “Denaturation” given below are correct?
Statements:
(i) Denaturation of proteins causes loss of secondary and tertiary structures of the proteon.
(ii) Denaturation leads to the conversion of double strand of DNA into single strand/
(iii) Denaturation affects primary structure which gets distorted.
(a) (i) and (iii) (b) (ii) and (iii)
(c) (i) and (ii) (d) (i), (ii), (iii)

Solution: (d) Factual

Problem 4: Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage.

Between which carbon atoms of pentose sugars of nucleotides are these linkages present?
(a) 5 and 2 (b) 3 and 5
(c) 5 and 5 (d) 3 and 3

Solution: (b) Linkage always established either at poisition 3 or position 5 whichever –OH group is
free.

Problem 5: Chargaff’s rule states that in an organism

JE

(a) amount of adenine (A) is equal to that of thymine (T) and amount of guanine (G) is equal
to that of cytosine (C)
(b) amount of adenine (A) is equal to that of guanine (G) and the amount of thymine (T) is
ES

equal to that of cytosine (C)

(c) amount of adenie (A) is equal to that of cytosine (C) and the amount of thymine (T) is
equal to that of guanine (G)
(d) amount of all bases are equal
an

Solution: (a) Adinine pairs with thymine and cytosine with guanine.

Problem 6: Mark the incorrect statement about ATP

(a) It is a nucleotide
ka

(b) It contains the purine adenine

(c) The enzyme catalysed hydrolysis of ATP to ADP and AMP is accompanied by absorption
of energy
lp

(d) Energy is stored in the cell in form of ATP

Solution: (c) On the hydrolysis of ATP energy is released not observed.

Problem 7: Which of the following structures for a nucleotide is not correct?

(a) Cytosine-ribose-phosphate (b) Uracil-2-deoxyribose-phosphate
(c) Uracil-ribose-phosphate (d) Thymine-2-deoxyribose-phosphate

Solution: (b) Urazine and deoxyribose sugar at position no. 2 of ribose sugar no linkage is possible at
position 2has no –OH group.

Problem 8: Two samples of DNA, A and B have melting points 340 K and 350 K respectively. This is
because
(a) B has more GC content than A (b) A has more GC content than B
(c) B has more AT content than B (d) Both have same AT content

Solution: (a) Between guanine and cytosine three hydrogen bonds are present but between A and T
two hydrogeb bonds are present.

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Biomolecules, Biological Process & Chemistry in Action

Problem 9: Which of the following statements is correct?

(a) Some tranquillisers function by inhibiting the enzymes which catalyse the degradation of
noradrenaline
(b) Tranquillisers are narcotic drugs
(c) Tranquillisers are chemical compounds that do not affect the message transfer from
nerve to receptor
(d) Tranquillisers are chemical compounds that can relieve pain and fever.

Solution: (a) Medical fact

Problem 10: Which of the following sets of reactants is used for preparation of paracetamol from phenol?
(a) HNO3, H2/Pd, (CH3CO)2O (b) H2SO4, H2/Pd, (CH3CO)2O
(c) C6H5N2Cl, SnCl2/HCl, (CH3CO)2O (d) Br2/H2O, Zn/HCl, (CH3CO)2O

Solution: (a)
OH OH OH
NO2
JE

HNO3

 

phenol
NO2
ES

reduce para and ortho

nitrophenol
OH OH
an



ka

NHCOCH3 NH2
paracetamol p--aminophenol

2.4 Antiseptics
lp

Antiseptics are the chemicals which prevent or destroy the growth of the harmful micro-organisms
They do not harm the living tissues when applied to wounds, cuts, ulcers and diseased skin
surfaces. Some of the common antiseptics are savlon, cetavelon, acriflavin, gention, violet,
iodine, methylene blue, mercurochrome and potassium permaganate.

CH3
Cl

CH3 OH OH

Chloroxylenol -Terpineol

Disinfectants
The chemical compounds capable of completely destroying the microorganisms are termed as
disinfectants. These are toxic to living tissues and thus can’t be applied on the skin. Therefore,

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Biomolecules, Biological Process & Chemistry in Action

these are used for sterillization of inanimate objects like floor, toilets, instruments and clothes, etc.
Depending upon the concentration, the same substance can act as disinfectant as well as
antiseptic. For example, 1% solution of phenol is a disinfectant whereas 0.2% solution of phenol
is a antiseptic.

Analgesics
Analgesics are substances which are used to get relief from pain. These are of two types

(a) Narcotics or habit forming drugs

(b) Non-narcotics

(a) Narcotics: These are alkaloids and mostly opium products. These drugs produce sleep and
unconsciousness when taken in higher doses. Common examples of narcotics are morphine,
codeine, heroin (morphine diacetate), methadone and pethidine hydrochloride.

(b) Non-narcotics : Analgesics beloging to this category are effective antipyretics also. Examples
are aspirin and novalgin.
JE

CH3
–
CO CH – N – CH2 SO3Na+
COOH
O – C – CH3 C6H5 – N C – CH3
ES

N
O
CH3
Acetylsalicylic acid Novalgin
(Aspirin)
an

Antipyretics
The chemical compounds used to bring down the body temperature in high fevers are called
ka

antipyretics. These drugs cause excessive sweating and thus lower down the body temperature.
Some of the common antipyretics are:

(a) Aspirin (Acetyl salicylic acid) (b) Analgin (Novalgin)

lp

(c) Paracetamol (d) Phenacetin

OH
OCOCH3 OC2H5
COOH
NH – C – CH3
O NHCOCH3
Aspirin 4-Acetamidophenol Phenacetin
(Paracetamol)

Antimalarials
The chemical substances used to bring down the body temperature during malarial fever are called
antimalarials. Malarial fever is developed when the female mosquito Anopheles bites the man
and transfer the malarial parasite (plasmodium) in the body. The medicines used for its treatment
are quinine, chloroquine, paraquine and primaquine, etc.

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Biomolecules, Biological Process & Chemistry in Action

Tranquilizers
The chemical substances which act on the central nervous system and has a calming
effect to reduce anxiety are specified as tranquilizers. Since these are used for the treatment
of mental diseases, so these are also known as psychotherapeutic drugs. They are basically
of two types:

(a) Sedative or hypnotics

(b) Mood elevators or antidepressants

(a) Sedative and hypnotics : Sedatives are central nervous system depressants that reduce
nervous tensin and promote relaxation without inducing sleep whereas hypnotics are central
nervous system depressants that induce sleep. Synthetic compounds used as sedative and
hypnotics are barbituric acid and its derivatives such luminal and seconal.
CH3
O O O
C3H7 – CH
C2H5
NH NH CH2 = CHCH2
JE
C2H5
N O N O O N O
O O
H H H

Barbituric acid Luminal Seconal

ES

Calmpose (dizepam) and equanil (meprobamate) are also effective sedatives.

O CH3 O
H2N – C – O – CH2 – C – CH2 – O – C – NH2
an

CH3
Equanil
(b) Antidepressants : A drug used for the treatment of highly depressed patient, who has lost
ka

his confidence, to improve his efficiency is known as antidepressant or

mood elevator.
lp

CH3 CH3
–CH2 – CH – NH2 –CH2 – CH – NH – CH3
Benzedrine Methamphetamine
(Amphetamine)
Anaesthetics
These are chemical substances administered for producing general or local insensibility to pain and
other sensation. As evident from the definition, anaesthetics are of two types.

(a) General anaesthetics

(b) Local anaesthetics

(a) General anaesthetics : These produce unconsciousness and are given when major surgical
operations are to be performed. Some of the common general anaesthetics are listed below.

Gaseous form : Nitrous oxide, cyclopropane, ethylene, etc.

Liquid form : Chloroform, divinyl ether and sodium pentothal, etc.

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Biomolecules, Biological Process & Chemistry in Action

(b) Local anaesthetics : These produce loss of sensation on a small portion of the body where
the drug is applied. Local anaesthetics are used for minor operations.
Jelly form : Oxylocain
Spray form : Ethyl chloride
Injection form : Procain

Dyes
1. Dyes are coloured compounds used to impart colour to the textiles, silk, wool, foodstuffs, etc. If
a compound absorbs light in the visible region, its colour will be that of the reflected light, i.e.,
complementary to that absorbed. For example, if a dye absorbs blue colour, it will appear yellow
which is the complementary colour of blue.
2. Chromophores are groups which impart colour to a compound, i.e., NO2, NO, N=N, quinonoid
structures.
3. Auxochromes are groups which themselves do not absorb light (i.e., are not chromophores)
JE

but deepen the colour when introduced into the coloured compounds, i.e., OH, NH 2, Cl,
CO2H etc.
4. Classification based upon source:
ES

(i) Natural dyes are obtained from plants. For example, allzarin, indigo etc.
(ii) Synthetic dyes are prepared in the laboratory. For example, martius yellow, malachite
green, orange-I, orange-II, congo red, aniline yellow, etc.
an

5. Classification based upon their structure

(i) Nitro dyes–martius yellow, (ii) Azo dyes – aniline yellow, methyl orange, congo red etc.
(iii) Triphenylmethane dyes – malachite green, (iv) Indigoid dyes – Indigo, (v)
ka
Anthraquinone dyes–alizarin and (vi) Phthalein dyes – phenolphthalein.
6. Classification based upon their application
(i) Acid dyes are sodium salts of azo dyes containing sulphonic acid or carboxylic acid groups,
lp

e.g. orange–I, orange–II, and congo red.

(ii) Basic dyes are the salts of azo and triphenylmethane dyes containing amino groups as
auxochromes, e.g., aniline yellow, butter yellow and malachite green.
(iii) Direct dyes are the dyes which are directly applied to the fabric from an aqueous solution.
These dyes are the most useful for fabrics which can form H-bonds such as cotton, wool,
silk, rayon and nylon. Some examples of direct dyes are : congo red and martius yellow.
(iv) Disperse dyes are applied to the fabric in form of their dispersion in a soap solution in
presence of some stabilizing agent such as phenol, cresol or benzoic acid. e.g., Celliton fast
pink B and Celliton fast blue B.

(v) Fibre reactive dyes contain a reactive group which combines directly with the hydroxyl or
the amino group of the fibre. Cotton, wool and silk can be dyes with this type of dyes. Dyes
which are derivatives of 2, 4-Dichloro-1, 3, 5-triazine are important example of fibre
reactive dyes.

[ 57 ]
Biomolecules, Biological Process & Chemistry in Action

(vi) Ingrain dyes are water insoluble azo dyes which are produced in situ on the surface of the
fabric by means of coupling reactions, e.g., para red.

(vii)Vat dyes are insoluble dyes which are first reduced to a colourless soluble from (jeuco
compound) in large vats with a reducing agent such as alkaline sodium hydrosulphite and
applied to the fabric and then oxidised to the insoluble coloured form by exposure to air or
some oxidising agents such as chromic acid or perboric acid, e.g., indigo.

(viii) Mordant dyes are applied to the fabric after treating them with a metal ion (mordant)
which acts as a binding agent between the dyes and the fabric. Depending upon the metal
ion used, the same dye can a give different colours. Thus, alizarin gives a rose red (turkey
red) coloured with Al3+ ions and blue colour with Ba2+ ions.

Propellants

In general, a rocket propellant consists of a fuel and an oxidiser:

JE

1. Solid propellants consist of a solid fuel and solid oxidiser. These are of two types:

(i) Composite propellants use a polymeric binder such as polyurethane or polybutadiene as a

fuel and ammonium perchlorate as an oxidiser. The efficiency of these propellants can be
ES

further enhanced by using additives like finely divided Mg or Al along with the fuel.

(ii) Double base propellants consist of nitroglycerine and nitrocellulose.

The main disadvantage of solid propellants is that they continue burning at the same rate without
an

any stop and start capability.

2. Liquid propellants
ka

(i) Monoliquid propellants are a single chemical substance which acts both as a fuel as well
as an oxidiser. Examples are : Methyl nitrate, Nitromethane, Hydrogen peroxide etc.

(ii) Biliquid propellants consist of two liquids, one of which acts as a fuel while the other acts
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as the oxidiser. Most commonly used liquid fuels are :keroseneoil, alcohol, hydrazine,
monomethylhydrazine, unsymmetrical dimethylhydrazine or liquid H2O2 while the most
commonly used liquid oxidisers are liquid oxygen, liquid nitrogen tetraoxide or nitric acid.

3. Hybrid propellants consist of a solid fuel and a liquid oxidiser, e.g., acrylic rubber (solid fuel)
and liquid N2O4 (liquid oxidiser)

4. Energy of a propellant is measured in terms of its specific impulse (Is), which is related to the
flame temperature (Tc) inside the rocket motor and average molecular mass (M) of the product
gases coming out of the rocket nozzle, i.e.,

Ic
Is 
M

Evidently, higher the value of Ic and lower the molecular mass of the product gases, better is the
rocket propellant.

[ 58 ]
Biomolecules, Biological Process & Chemistry in Action

SAMPLE PROBLEMS 2.4 (MCQ)

Problem 1: The most useful classification of drugs for medicinal chemists is ____
(a) on the basis of chemical structure (b) on the basis of nature of animals
(c) on the basis of molecular targets (d) on the basis of pharmacological effect

Solution: (c) Medicinal fact.

Problem 2: Which of the following is employed as antihistamine?

(a) Omeprazole (b) Chlorampheicol
(c) Diphenylhydramine (d) Norethindrone

Solution: (c) Anti histamins are those medicines which affect the activty of histamine
(an amino acid).

Problem 3: Equanil is_____

(a) artificial sweetner (b) tranquillizer
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(c) antihistamine (d) antifertility drug

Solution: (b) Tranquillizers are sedative drugs.

Problem 4: A sulpha drug used for the treatment of pneumonia is

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(a) sulphadizine (b) sulphaguanidine

(c) sulphanilamide (d) sulphapyridine

Solution: (d) Sulphapyridine is used for treatment of pneumonia.

an

Problem 5: Which of the following is an example of non-biodegradable detergent?

(a) H3C (CH2 )11 SO3 Na

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H3C (CH2 )9 CH SO3 Na

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(b)
CH3

CH3 CH3
(c) H3C CH CH2 CH SO3 Na
3

(d) (CH3(CH2)10CH2OSO\3Na

Solution: (c) Greater the branching, greater is the non-biodegradability.

Problem 6: Which of the following dye has a nitro group ?

(a) Malachite (b) Indigo
(c) Aniline yellow (d) Martius yellow
Solution: (d) Factual

[ 59 ]
Biomolecules, Biological Process & Chemistry in Action

Problem 7: ‘Placebo’ is often given to patients. It is

(a) an antidepressant (b) a broad spectrum antibiotic
(c) a sugar pill (d) a tonic

Solution: (c) It is matter of treatement in which a pseudomedication is given to the patient.

Problem 8: Which of the following biomolecules will give blue colouration with Cu2+ and purple colur
with ninhydrin solution?
(a) Aldehyde (b) -Amino acid
(c) Carbohydrate (d) DNA

Solution: (b) Only amino acids and protein exhibit ninhydrin test.

Problem 9: Prosthetic group of glycoprotein is:

(a) fat (b) carbohydrates
(c) protein (d) nucleic acid
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Solution: (b) Non proteinous part of the protein is called prosthetic group.


HOH/ H
Problem 10: A   glucose + fructose
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
HOH/H
B   glucose + glucose

HOH/ H
C   glucose + galactose
the disacchardies A, B and C respectively are:
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(a) lacrtose, scurose, maltose (b) sucrose, maltose, lactose

(c) sucrose, lactose, maltose (d) maltose, sucrose, lactose

Solution: (b) Sucrose is a disaccharides of glucose and fructose, maltose is a diasaccharide of glu-
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cose and glucose and lactose disaccharide glucose and lactose.

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[ 60 ]
Biomolecules, Biological Process & Chemistry in Action

SOLVED EXAMPLES
Example 1: (a) What is the average molecular mass of starch?
(b) What is the approximate average numebr of glucose units in the sample of starch?
(Given: 20 gm L–1 of starch has an osmotic pressure = 10–2 atm at 25°C

Soltuon: (a)  = MRT (M = molarity of solution)

π 102 atm
M   4 104 M
RT (0.082 L atm mol1K 1 )(298K)

W2 1000
M
MW2  vol. of solution in ml
20  1000
MW2   5 104 gm mol1
4 104  1000
(b) Starch in a polymer of Glucose; each glucose unit (MW of glucose = 180 gm mol–1)
joined to another glucose unit with loss of one H2O unit (18 gm mol–1) giving a MW of
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180 – 18 = 162 gm mol–1 per unit.

 Number of glucose units
5  104 gm mol1
  309  310 glucose unit
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162 gm mol1 unit 1

Example 2: The specific rotation of -glucose is +112° and -glucose is +19° and the specific rotation
of the constant equilibrium mixture is +32.7°. Calculate the percentage compositon of anomers
( and ) in the equilibrium mixture.
an

Solution: Let  and -glucose have a and b%, respectively

Equilibrium specific rotation
ka

( a specific rotation of α) + (b  specific rotation of β) a  112  b  19

  52.7 
100 100
112a  (100  a) 19

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100
Solving for a and b: a = 36.2%, b = 63.8%

Example 3: One mole of a naturally occurring fat on hydrolysis with NaOH gave 1 mol of glycerol
together with sodium palmitate and sodium stearate in 1:2 molar ratio. The molecule of the
fat is symmetric. Write down the structure of the fat.

Solution: The 1:2 molar ratio of sodium palmitate and sodium stearate obtained on hydrolysis with
NaOH indicates that two stearic acid molecules and one palmitic acid are esterified with 1
mol of glycerol. Since the molecule of fat have symmetry, the two steric acid molecules
must be linked to two terminal primary alcoholic groups. Therefore, the structure of the fat
must be:
H2C O CO C17 H35
HC O CO C15 H31
H2C O CO C17 H35

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Biomolecules, Biological Process & Chemistry in Action

Example 4: Differentiate between o-erythrose and o-threose by (a) mild oxidation and (b) reduction

Solution: (a) 1COOH

1CHO

2 OH 2 OH
[O]

HNO3

3 OH 3 OH
4CH OH 4COOH
2
D-Erythrose Meso-tartaric acid
Plane of symmetry
H2 / Ni (Optically inactive)
[H] or
NaBH4

1COOH

2 OH Meso-butan-1,2,3,4-tetraol
(Optically inactive)
3 OH
4COOH
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(b) 1CHO 1COOH

2 OH 2 OH
[O]
 
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HNO3
3 OH 3 OH
4CH OH 4COOH
2
D-Tthreose D-(-) Tartaric acid
Optically active
H2 / Ni enantiomer
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[H] or
NaBH4

1CH2 OH
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HO 2 Meso-butan-1,2,3,4-tetraol
(Optically active
3 OH enantiomer)
4CH OH
2
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Example 5: Give the structure and IUAC name of disaccharide, which gives the reaction given below:

H 2O
Disaccharide   Glucose + Fructose
(A)
Hydrolyse both by
maltase or emulsin
Methylation and hydrolysis

2,3,4,6-Tetra-metyhyl-D-glucopyrase (B)

1,3,4,6-Tetra-methyl-d-fructofurnose (C)
Given, (A) does not reduce Fehling’s solution and does not mutarotate.

Solution: (i) It does not reduce Fheling’s solution and does not mutarotate, which suggests the
hemiacetic OH group at C-1 of glucose and hemiketalic OH group at C-2 of fructose
are not free and are involved in glycosidic linkage.
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Biomolecules, Biological Process & Chemistry in Action

(ii) Hydrolysis by maltase suggests -linkage and hydrolysis by emulsin suggests -linkage.
(iii) Products obtained after the methylation and hydrolysis of (A) suggest (C-1) (OH) ()
of glucose and (C-2) (OH) () of fructose are involved in glycosidic linkage.
(iv) Glucose is in pyranose form which suggests a six membered ring.
(v) Fructose is in furanose form and this suggests a five membered ring.
All the observations suggest that (A) is sucrose.

6 CH2OH

5 1
4 OH
 (α  D  Glucopyranose)
HO
3 2 -link
OH O
( i) M e 2 S O 4 /  O H ( ii) H 
HOH2C -link  (β  D  Fructofuranose)
O
HO
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CH2OMe
CH2OH
O
OH 5 1 OH
4 OMe
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IUPAC H
MeO 2
-D-Glucopyranosyl--D 3
-fructofuranoside OMe
(B)
(Equilibrium mixture of α  and β  anomers)
an


MeOH2C
O OH
MeO
CH2 OMe
ka

OMe
(C)
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Example 6: (a) Give a chemical test of system with I2.

(b) What is the change in colour, when the test is performed of high temperature?
(c) What structural changes occurs during the change in reaction conditions?
(d) Do amylose and amylopectin give the same colour?

Solution: (a) Starch with I2 gives a deep blue-black colour.

(b) On heating, the colour changs to reddish brown.
(c) Amylose part in starch traps I2 molecule in its helix and forms a charge transfer com-
plex which has blue-black colour.
At high temperature, the helix partially breaks and some I2 molecules escape and some
I2 molecules get trapped. On cooling, the helix reforms, trapping more I2 molecules and
the original colour returns.
(d) Amylopectin gives a less intense red-brown colour because the helilcal stucture is dis-
rupted by the branching of the chain.

[ 63 ]
Biomolecules, Biological Process & Chemistry in Action

Example 7: Complete the following reactions

Kilani HNO 3
D-Glucose   Pair   Meso-Heptaldaric acid
(i) synthesis
(A) (B) (C)

Kilani HNO3
D- Allose   Pair   Meso-Heptaldaric acid
(ii) (D) synthesis
(E) (F)

Explain whether the acids (C) and (F) are same or different. Which pair out of (B) and
(E) give meso-acids (C) and (F)?

Solution: (i) 1 CHO CHO CHO

2 OH 2 OH HO 2
Kilani
HO 3 
synthesis
 3 OH 3 OH

4 OH HO 4  HO 4
5 OH 5 OH 5 OH
6 CH2 OH 6 CH2 OH 6 CH2 OH
D-Glucose
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(A) (B) (C-2) Epimer of

(B)
HNO 3 HNO3
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COOH COOH
2 OH 2 OH
Plane of symmetry
3 CH3 3 CH3

OH 4 H HO 4
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5 OH 5 OH
6 CH3 6 CH3
7COOH 7COOH
Meso-acid (It is optically active)
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(C)
1 CHO CHO CHO
(ii)
2 OH 2 OH HO 2
Kilani
3 OH   3 OH 3 OH
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synthesis

4 OH 4 OH  4 OH

5 OH 5 OH 5 OH
6 CH2 OH 6 CH2OH 6 CH2OH
D-allose
(D) (E) (C-2) Epimer of
(E)
HNO3 HNO3

COOH COOH

2 OH HO 2
Plane of symmetry
3 CH3 3 CH3

H 4 OH 4 OH

5 OH 5 OH
6 CH3 6 CH3
7COOH 7COOH
Meso-acid (It is optically active)
(F)

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