19

CHAPTER

Biomolecules

or ketonic group and reduce Fehling’s solution and Tollen’s

INTRODUCTION reagent are referred to as reducing sugars. All monosaccharides,
whether containing aldehydic group or ketonic group behave
The molecules involved directly with the life processes are called
as reducing sugars. The disaccharides may be reducing if the
bio-molecules. This chapter aims at discussing important bio-
aldehyde or ketone group of the monosaccharides are free or non-
molecules including the roles they play in living organisms.
reducing if the reducing groups of monosaccharides are bonded.
For example, maltose and lactose are reducing sugars but sucrose
CARBOHYDRATES is a non-reducing sugar.

‘Chemically, the carbohydrates may be defined as optically active

polyhydroxy aldehydes or ketones or “the compounds which Key Note
produce such units on hydrolysis.” Carbohydrates are also called
saccharides (Greek: sakcharon means sugar). ™ Fehling’s solution is alkaline solution of Cu2+ ion
complex with tartrate ion. Upon reaction with compound
Some common examples are cane sugar, glucose, starch, etc. Most
containing free aldehydic or ketonic group, a red
of them have a general formula, Cn(H2O)n, and were considered as
precipitate is produced.
hydrates of carbon from where the name carbohydrate was derived.
For example, the molecular formula of glucose (C6H12O6) fits into ™ Tollen’s reagent is ammonical AgNO3 which is reduced
this general formula, C6(H2O)6. to silver by free aldehydic and ketonic groups.

Classification of Carbohydrates
The carbohydrates can be divided into three major classes,
depending on their constitution:
MONOSACCHARIDES
(i) Monosaccharides: These cannot be hydrolysed to simpler Monosaccharides are further classified on the basis of number
carbohydrates. They are crystalline solids, soluble in water of carbon atoms and the functional group present in them. If a
and sweet in taste. About 20 monosaccharides are known monosaccharide contains an aldehydic group, it is known as an
to occur in nature. Some common examples are glucose,
aldose and if it contains a keto group, it is known as a ketose.
fructose, ribose, etc.
Number of carbon atoms containing the monosaccharide is also
(ii) Oligosaccharides: The oligosaccharides (oligo means few)
introduced in the name as is evident from the examples given in
are carbohydrates which yield a few but definite number
following table.
(2–10 units) of monosaccharide molecules on hydrolysis.
These are also crystalline solid, soluble in water and sweet Different Types of Monosaccharides
in taste. The most common oligosaccharide is disaccharide.
Carbon atoms General term Aldehyde Ketone
(iii) Polysaccharides: These are high molecular mass
carbohydrates which yield many molecules of 3 Triose Aldotriose Ketotriose
monosaccharides on hydrolysis. They are amorphous,
insoluble in water and tasteless. For example, starch and 4 Tetrose Aldotetrose Ketotetrose
cellulose both have a general formula (C6H10O5)n and yield
several monosaccharide molecules. 5 Pentose Aldopentose Ketopentose
Classification of Carbohydrates Based on Nature
6 Hexose Aldohexose Ketohexose
(Oxidizing or Reducing) of Sugars
The carbohydrates may also be classified as either reducing or 7 Heptose Aldoheptose Ketoheptose
non-reducing sugars. Carbohydrates which contain free aldehydic
 +
H
GLUCOSE AND FRUCTOSE C12 H 22 O11 + H 2 O  → C6 H12 O6 + C6 H12 O6

Sucrose Glucose Fructose
Glucose and fructose are the specific examples of an aldohexose
(an aldehyde containing 6 carbons) and of a ketohexose (a ketone 2. From Starch: Commercially, glucose is obtained by
containing 6 carbons) respectively. hydrolysis of starch by boiling with dilute H2SO4 at 393 K
Glucose under pressure.
Glucose occurs freely in nature as well as in the combined form.
( C6 H10 O5 )n + nH 2 O 
+
H
It is present in sweet fruits and honey. Ripe grapes also contain 393K;2 − 3atm
→ nC6 H12 O6
glucose in large amounts. It is prepared as follows: Starch or cellulose Glucose

Preparation of Glucose
1. From Sucrose (Cane sugar): When alcoholic solution of Structure of Glucose
sucrose is boiled in dil. HCl or dil. H2SO4, glucose and It was assigned the structure on the basis of the following
fructose are obtained. evidences:
Chemical Reactions of Glucose
COOH

— —
(Tollen’s reagent)
COOH [Ag(NH3)2]OH Ag + (CHOH)4
— —

(Silver mirror)
HNO3
(Glucaric acid) (CHOH)4 CH2OH
(C6H10O6) or I
(Saccharic acid) COOH
COOH

— —
COOH (Fehling's solution) or
— —

Br2 + H2O (Benedict's solution)

Cu2O(Cuprous oxide) + (CHOH)4
(CHOH)4 Gluconic acid (C6H12O7) (Red ppt)
II CH2OH
CH2OH CHO
— — — — —

CH2OH
CHOH
— —

H2/Ni
(C6H14O8) Glucitol (sorbitol) (CHOH)4 5HIO4
III CHOH 5HIO3 + 5HCOOH + HCHO
CH2OH
CHOH
CH=N—OH H2NOH CHOH
— —

(–H2O)
(CHOH)4 (Glucose oxime)
IV CH2OH
CH2OH (+) – Glucose
CH=NNHC6H5
CN
— — —

3C6H5NH—NH2
— — —

C=NNHC6H5 (Glucose
*CH—OH (Glucose cyanohydrin) HCN
–H2O,–NH3 Osazone)
V
–PhNH2 (CHOH)3
(CHOH)4
CH2OH CH2OH

CHO O
— —

—
—

(CH–O–C–CH3)4
O (CH CO) O
3 2
—
—

CH2–O–C–CH3 Glucose penta-acetate VI NaHSO3 or DNP or

No reaction
Schiff's reagent
Red P + HI
n-Hexane
VII
Key Note

Reaction VII proves presence of six-carbon straight chain. On the ™ The exact spatial arrangement of different — OH groups
basis of above sequence of reactions, we can observe that –CHO was given by Fischer after studying other properties. The
(by step-II) at one end and CH2OH at the other end (step-I and structure of D-(+)-glucose is:
step-II) may be present. Reaction IV and V prove that definitely CHO

–CH=O is present along with 5 –OH groups (step-VI). Therefore, H OH

a likely structure may be written as CHO(CHOH)4CH2OH. HO H

H OH
H OH
CH2OH

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D and L Notation Epimers Epimers
‘D’ before the name of glucose represents the relative configuration CHO
1
CHO CHO

— — — — —

— — — — —
— — — — —
whereas ‘(+)’ represents dextrorotatory nature of the molecule. 2
H—C—OH H—C—OH HO—C—H
‘D’ and ‘L’ have no relation with (d, l) nomenclature or the optical 3
activity of the compound. The meaning of D and L–notations is HO—C—H HO—C—H HO—C—H
given as follows.
4
HO—C—H H—C—OH H—C—OH
™ The letters ‘D’ or ‘L’ before the name of any compound indicate H—C—OH H—C—OH
5
H—C—OH
the relative configuration of a particular stereoisomer. This refers 6
CH2OH CH2OH CH2OH
to their relation with a particular isomer of glyceraldehyde.
D (+) galactose D (+) glucose D (+) mannose
Glyceraldehyde contains one asymmetric carbon atom and C4 is epimeric carbon C2 is epimeric carbon
exists in two enantiomeric forms as shown below.
A Note on Cyclic Structure of Glucose
CHO CHO
The following findings could not be explained on the basis of open
chain structure.
H OH HO H
1. Despite having an aldehydic group, glucose does not
give 2, 4-DNP test, Schiff’s test and it does not react with
CH2OH CH2OH sodium bisulphite and ammonia.
(+)-Glyceraldehyde (–)-Glyceraldehyde 2. The pentaacetate of glucose does not react with
hydroxylamine indicating absence of free –CHO group.
™ All those compounds which can be chemically 3. Glucose is found to exist in two different crystalline forms
correlated to (+) isomer of glyceraldehyde are said to have which are named as α and β. When glucose was crystallized
D-configuration whereas those which can be correlated to (–) from a concentrated solution at 30°C, it gave a form of
isomer of glyceraldehyde are said to have L-configuration. glucose called the α-form having m.p. 146°C. The other
CHO form called the β–form having m.p 150° C is obtained on
crystallization of glucose from a hot saturated aqueous
solution at a temperature above 98°C.
H OH
If was found that glucose forms a six–membered ring in which —
OH at C–5 is involved in ring formation. This explains the absence
CH2OH
of —CHO group and also existence of glucose in two cyclic forms
D–(+)–Glyceraldehyde as shown below. These two cyclic forms exist in equilibrium with
open chain structure.
™ For assigning the configuration of monosaccharides, it is the
lowest asymmetric carbon atom (as shown below) which is
compared. (For this comparison, the structure is written in a
way that most oxidised carbon e.g., –CHO is towards the top
of fischer projection.)
CHO
H CHO
OH CHO

HO H H OH HO H
H HO OH H H OH
H OH HO H
H OH
H OH HO H Key Note
CH2OH
CH OH CH2OH
2
™ The two cyclic hemiacetal forms of glucose differ only
D–(+)–Glucose
D-(+)Glucose L-(–)Glucose
in the configuration of the hydroxyl group at C1, called
anomeric carbon (the aldehyde carbon before cyclisation).
™ Such isomers, i.e., α-form and β-form, are called anomers.
EPIMERS ™ The six membered cyclic structure of glucose is called pyranose
A pair of diastereomers that differ only in the configuration structure (α– or β–), in analogy with pyran. Pyran is a cyclic
about a single chiral carbon atom are said to be epimers. D(+)- organic compound with one oxygen atom and five carbon
atoms in the ring. The cyclic structure of glucose is more
glucose is epimeric with D(+)-mannose and D(+)-galactose as
correctly represented by Haworth structure as given below.
shown below.

Biomolecules 149
 O
6
CH2OH CH2OH
H 5 O H HO Anomeric
O
carbon
4 H 1
HO OH OH
OH H
HO 3 2 OH
Pyran –D (+) glucose

H OH ™ In α-D-Glucose, –OH group at anomeric cabon is in axial position.

–D–(+)–Glucopyranose
Mutarotation
6 When pure α-D-glucose is dissolved in water, its specific rotation
CH2OH is found to be +112°. With time, however the specific roation of the
O solution decreases ultimately and reaches equilibrium value of +52.7°.
H 5 OH When β-D-glucose is dissolved in water, it has a specific rotation of
4 H 1 19°. The specific rotation of this solution increases with time to +52.7°.
OH H This change of optical rotation with time is called mutarotation. It
HO 3 2 H
is caused by the conversion of α and β glucopyranose anomers into
an equilibrium mixture of both. Mutarotation is catalyzed by both
H OH acid and base, but also occurs even in pure water. Mutarotation is
–D–(+)–Glucopyranose characteristic of the cyclic hemiacetal form of glucose.
All reducing sugars will mutaroate. Mutaroation occurs first by
Chair Forms (Conformation) of α- and β-D-Glucose: opening of the pyranose ring to the free aldehyde form.
O
FRUCTOSE
CH2OH Fructose is an important ketohexose.
OH
HO Anomeric Preparation
carbon 1. By acid hydrolysis of cane sugar.
OH H O/H +
HO C12 H 22 O11 
2
→ C6 H12 O6 + C6 H12 O6
Sucrose α− glucose β− fructose
–D (+) glucose 2. By enzymatic action of sucrose.
™ In β-D-Glucose (most stable form of glucose) all bulky- invertase
C12 H 22 O11  → Glucose + Fructose
groups are in equatorial position. Sucrose

CHEMICAL REACTIONS OF FRUCTOSE

COOH COOH COOH
— —

— —

—
HNO3
(CHOH)3 + (CHOH)2 + CH2OH
COOH (Glycolic acid)
COOH COOH
— —

Fehling's solution or Trihydroxy glutaric acid (Tartaric acid)

Cu2O(Cuprous oxide) + (CHOH)4
Benedict’s solution
(Red precipitate) HCN
CH2OH Fructose
cyanohydrin
CH2—OH H2NOH
5HIO4 Fructose oxime
—

2HCH=O + 3HCOOH + CO2 + H2O –5HIO3 C=O H2NNHPh (3 mole)

CH2OH CH2OH HO H Osazone
H OH NaHSO3 or DNP
H OH HO H No reaction
HO H HO H H OH or Schiff's regent
H2 + Ni or Na–Hg + H2O
H OH + H OH CH2OH or Br2 + H 2O
or NaBH4 or LiAlH4
D-(–)-fructose COOH
H OH H OH
— —

Tollen’s reagant
(CHOH)4 + Ag 
CH2OH CH2OH
Sorbitol Mannitol CH2OH
(C6H14O6)
Red P + HI
n-Hexane
D

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 7. Calcium Forms calcium Forms calcium
Key Note hydroxide glucosate, fructosate,
soluble in water insoluble in
™ Both glucose & fructose same osazone. water

8. Molisch’s Forms a violet Forms a violet

Cyclic Structure of Fructose reagent ring ring
It also exists in two cyclic forms which are obtained by the addition
of —OH at C-5 to the ( C = O) group. 9. Fehling’s Gives red Gives red
™ The ring, thus formed is a five membered ring and is named solution precipitate precipitate
as furanose with analogy to the compound furan. Furan is a
five membered cyclic compound with one oxygen and four 10. Tollen’s reagent Forms silver Forms silver
carbon atoms. mirror mirror
11. Phenyl Forms osazone Forms osazone
1 2 2 1
HOH2C—C—OH HO—C—CH2—OH hydrazine

O HO
3
H O HO
3
H O 12. Resorcinol Faint pink color Gives red or
4 + HCl (dil.) brown colour
4 H OH
H OH (Seliwanoff’s
H 5 test)
Furan H 5
13. Alcoholic No colouration A purple colour
6
6
CH2OH CH2OH a-naphthol + (violet) on
–D–(–)–Fructofuranose –D–(–)–Fructofuranose HCl (conc.) boiling
™ The cyclic structures of two anomers of fructose are (Rapid
represented by Haworth structure as given: Furfural test)

6 1 6
HOH2C O CH2OH HOH2C O OH Kiliani-Fischer Synthesis-Chain Elongation
5
H H
2 5 2 Kiliani-Fischer synthesis results in formation of two new sugars with
OH OH H H
4 3 4
OH
3
CH2OH one more carbon atom (epimeric at C–2) introduced in the ring.
1
OH H OH H
–D–(–)–Fructofuranose –D–(–)–Fructofuranose
– BaSO4

Comparison between Glucose and Fructose 3

+
D-Glucose
S.No. Property Glucose Fructose D-Arabinose

1. Molecular C6H12O6 C6H12O6

formula
2. Nature Polyhydroxy Polyhydroxy
aldehyde ketone
D-Mannose
3. Melting Point 146°C/150°C 102°C/ 105°C

4. Optical activity Dextrorotatory Laevorotatory

of natural form
5. Oxidation Ruff Degradation: Chain Shortening
(a) With bromine Gluconic acid No reaction Ruff degradation leads to the sugar with one carbon less than
water starting sugar.
(b) With nitric Saccharic acid Mixture of
acid (Glucaric acid) glycolic acid, CHO
tartaric acid H OH CHO
and trihydroxy H OH
1. Br2, H2O H OH
glutaric acid 2. CaCO3
H OH
H OH
6. Reduction Sorbitol Mixture of 2+
3. Fe /H2O2
sorbitol and CH2OH CH2OH
mannitol D-Ribose D-Erythrose

Biomolecules 151
 free aldehyde group can be produced at C-l of second glucose in
DISACCHARIDES solution and it shows reducing properties, so it is a reducing sugar.
6 6
Disaccharides on hydrolysis with dilute acids or enzymes yield two CH2OH CH2OH
molecules of either the same or different monosaccharides. The two H 5
O
H H 5
O
H
monosaccharides are joined together by an oxide linkage formed by the 4
H
1 4
H
1
loss of a water molecule. Such a linkage between two monosaccharide HO
OH H O OH H
OH
3 2 3 2
units through oxygen atom is called glycosidic linkage.
H OH H OH
Sucrose (Cane Sugar) (I) Maltose (II)
Sucrose on hydrolysis gives equimolar mixture of D–(+)–glucose –D–Glucose –D–Glucose
and D–(–) fructose. Lactose (Milk Sugar)
It is known as milk sugar since it occurs in milk. It is composed
C12 H 22 O11 + H 2 O 
→ C6 H12 O6 + C6 H12 O6 of β-D-galactose and β-D-glucose. The linkage is between C-1 of
Sucrose D − ( + ) − Glucose D − ( − ) − Fructose
galactose and C-4 of glucose. Free aldehyde group can be produced
These two monosaccharides are held together by a glycosidic at C-l of glucose unit, hence it is also a reducing sugar.
6 6
linkage between C-1 of α-glucose and C-2 of β-fructose. Since the CH2OH CH2OH
reducing groups of glucose and fructose are involved in glycosidic 5
O 5
O
bond formation, sucrose is a non-reducing sugar. HO H
H
H OH
4 1 O 4 1
Sucrose H OH H OH H
3 2 H 3 2 H

H OH H OH
(I) Lactose (II)
–D–Galactose –D–Glucose

POLYSACCHARIDES
Inversion of Sucrose These are carbohydrates in which hundreds or even thousands of
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory monosaccharide units are joined by glycosidic linkages. They mainly
glucose and laevorotatory fructose. Since the laevorotation of act as the food storage or structural materials. Starch, cellulose,
fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), glycogen, dextrins, etc. are some examples of polysaccharides.
the product mixture is overall laevorotatory. Thus, hydrolysis of Starch (C6H10O5)n
sucrose brings about a change in the sign of rotation, from dextro ™ Starch is the main storage polysaccharide of plants.
(+) to laevo (–) and the product is named as invert sugar. ™ It is the most important dietary source for human beings.
™ High content of starch is found in cereals, roots, tubers and
Maltose (Malt Sugar) some vegetables.
Maltose is composed of two α-D-glucose units in which C-l of ™ It is a polymer of α-glucose and consists of two components—
one glucose (I) is linked to C-4 of another glucose unit (II). The Amylose and Amylopectin.

Amylose
Amylose is water soluble component which constitutes about 15-20% of starch. Chemically, amylose is a long unbranched chain with 200
–1000 α–D–(+)–glucose units held by C1– C4 glycosidic linkage.

6 6 6
CH2OH CH2OH CH2OH
O H H 5 O H H 5 O H
H 5
4 H H H
O OH H 1 O 4 OH H 1 O 4 OH H 1O
3 2 3 2 3 2
H OH –Link H OH –Link H OH
Amylose
Amylopectin
Amylopectin is insoluble in water and constitutes about 80-85% of starch. It is a branched chain polymer of α-D-glucose units in which
chain is formed by C1-C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage.

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 6 6
CH2OH CH2OH
O H H 5 O H
H 5
4 H 1 H
4 1
O OH H O OH H
3 2 3 2
OH H OH –Link
H
O

—
6 6 Branch at C6 6
CH2OH CH2 CH2OH
O H H 5 O H H O H
H 5 5
4 H 4
H H
1 4
O OH H 1 O OH H O OH H 1
O
3 2 3 2 3 2
H OH –Link H OH –Link H OH
Amylopectin

Cellulose (C6H10O5)n
™ Cellulose occurs exclusively in plants and it is the most abundant
organic substance in plant kingdom. It is a predominant constituent
of cell wall of plant cells. Example 1: Which of the following indicates the presence
™ Cellulose is a straight chain polysaccharide composed only of β-D- of 5–OH groups in glucose?
glucose units which are joined by glycosidic linkage between C-1
(a) Formation of penta-acetyl derivative of glucose.
of one glucose unit and C-4 of the next glucose unit.
(b) Cyanohydrin formation of glucose.
CH2OH CH2OH CH2OH
(c) Reaction with Fehling’s solution.
H H O O O O
H H H H (d) Reaction with Tollens reagent.
OH H O OH H O OH H (CH3CO)2 O
O H H H Sol. Glucose  → glucose penta acetate.
H OH H OH H OH Example 2: Which is correct structure of β-D-glucopyranose?
-links Cellulose -links
Glycogen (a) HO
OH
O
The carbohydrates are stored in animal body as glycogen. It is also H OH
known as animal starch because its structure is similar to amylopectin H
and is rather more highly branched. It is present in liver, muscles and H
OH
brain. When the body needs glucose, enzymes break the glycogen HO H
down to glucose. Glycogen is also found in yeast and fungi. (b) OH
H O
Importance of Carbohydrates H OH
Carbohydrates are essential for life in both plants and animals. HO H
™ They form a major portion of our food. OH
HO H
™ Honey has been used for a long time as an instant source of energy
by ‘Vaids’ in ayurvedic system of medicine. Carbohydrates are used OH
H O
as storage molecules as starch in plants and glycogen in animals. (c)
H OH
™ Cell wall of bacteria and plants is made up of cellulose.
HO H
™ We build furniture, etc. from cellulose in the form of wood OH
OH
and clothe ourselves with cellulose in the form of cotton
fibre. OH
™ They provide raw materials for many important industries (d) H
OH
O
like textiles, paper, lacquers and breweries.
OH OH
™ Two aldopentoses viz. D–ribose and 2–deoxy–D–ribose are
present in nucleic acids. HO H
HO H
™ Carbohydrates are found in biosystem in combination with H
many proteins and lipids.

Biomolecules 153
 Sol. In chair form of β-D-Glucopyranose, all –OH groups
are equatorial.
Example 3: Ketones do not reduce Tollen’s reagent, but
fructose with a keto group reduces it. Which one of the 1. CH2OH

— — —
following is a correct explanation? C=O
(a) Enolisation of keto group of fructose and NaBH4
A+B
transformation into aldehyde group in presence (CHOH)3
of OH – (present in Tollen’s reagent).
CH2OH
(b) CHOH group is also oxidised to keto group.
—

Fructose
—

(c) Both statements are correct. The product A and B in the above reaction are:
(d) None of the statement is correct. (a) Diastereomers
Sol. Enolisation of keto group of fructose and transformation (b) Enantiomers
into aldehyde group in presence of OH – is known as (c) Anomers
Lobry de Bruyn-van Ekenstein rearrangement. (d) Optically active hexahydroxy compounds
Example 4: The reaction of D-glucose with ammoniacal
2. 3 molecules of phenylhydrazine are used in osazone
AgNO3 produces:
formation. The correct statement about the use of
(a) CHO (b) CO2H phenylhydrazine is:
H OH H OH (a) All the three molecules react in similar manner.
HO H HO H (b) Two molecules react in similar manner whereas
H OH H OH the third reacts in different way.
H OH H OH (c) All the three molecules react in different way.
CHO CO2H (d) Only two react in same manner but the third
molecule remains unreacted.
(c) CO2H (d) CO2H
3. Glucose, fructose and mannose form the same osazone
H OH H OH because they have the same configuration at all carbon
HO H HO H atoms except:
H OH H OH (a) C1 and C2 (b) C2 and C3
H OH H OH (c) C3 and C4 (d) C4 and C5
CH2OH CHO 4. In amylose and maltose, the glucose units are linked
Sol. between:
(a) C1 and C1 (b) C1 and C2
H OH
(c) C1 and C4 (d) C4 and C5
—

C=O C=O 5. The wrong statement(s) among the following is/are:

H OH H OH (a) Glucose does not react with Grignard reagent.
HO H Ammonical HO H (b) Hydrolysis of maltose can be called inversion of
AgNO3
H OH H OH sugar.
H OH H OH (c) Amylopectin gives blue colour with iodine.
CH2OH CH2OH (d) Ribose is an aldopentose.
D-Glucose
Tollen’s reagent oxidise aldehyde group —C—H into
PROTEINS
carboxylic acid —C—OH . O
The word protein is derived from Greek word, “proteios” which
O
means primary or of prime importance. Proteins are high molecular
Example 5: Which of the following carbohydrates undergo
mass complex biopolymers of amino acids present in all living
mutarotation?
cells. Proteins perform a variety of functions in the body, namely:
(a) Maltose (b) Sucrose
(i) They catalyze biochemical reactions as enzymes.
(c) Cellulose (d) Fructose
Sol. All reducing sugars will mutarotate. (ii) They regulate metabolic processes as hormones.
(iii) They protect the body against toxic substances as
antibodies.

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a-Amino Acids Classification of a-Amino Acids
™ Amino acids characteristically contain amino (–NH2) and α-amino acids can be classified on the basis of the relative number
carboxyl (–COOH) functional groups apart from other of amino groups and carboxylic groups such as:
functional groups.
Neutral - Containing equal number of amino and carboxyl groups.
™ Only α-amino acids (amino group on a α carbon from the
Example: glycine, tyrosine.
carboxylic group) are obtained on hydrolysis of proteins.
Basic - Containing greater number of amino groups than carboxyl
™ Proteins yield a-amino acids which have a general formula as given:
groups. Example: lysine, arginine, histidine.
O
Acidic - Containing greater number of carboxyl groups as compared
R—CH—C—O—H
— to amino groups. Example: Glutamic acid, Aspartic acid.

NH2 Essential and Non-essential Amino Acids

™ α-amino acids are conveniently known by their trivial names Non-essential amino acids- The amino acids which can be
(e.g. glycine etc.) although they can be named as 1– amino synthesized in the body, are known as non-­essential amino acids
alkanoic acid as per IUPAC nomenclature. (e.g. glycine, alanine, glutamic acid etc).
COOH
Essential amino acids- The amino acids which cannot be
H2 N H
synthesized in the body but must be obtained through diet, are
H known as essential amino acids (e.g. valine, leucine, isoleucine
Glycine
(1-Aminoethanoic acid)
etc).

Name of the amino Characteristic feature of side Three letter symbol One letter code
acids chain (–R)

1. Glycine –H Gly G

2. Alanine –CH3 Ala A

3. Valine* (H3C)2CH– Val V

4. Leucine* (H3C)2CH–CH2– Leu L

5. Isoleucine* CH3–CH2–CH– Ile I

—

CH3

6. Arginine HN=C—NH—(CH2)3— Arg R

—

NH2

7. Lysine* H2N–(CH2)4– Lys K

8. Glutamic acid HOOC–CH2–CH2– Glu E

9. Aspartic acid HOOC–CH2– Asp D

10. Glutamine O Gln Q

H2N—C—CH2—(CH2)—

11. Asparagine O Asn N

H2N—C—CH2—

12. Threonine* H3C—CH— Thr T

—

OH

Biomolecules 155
 13. Serine HO–CH2– Ser S

14. Cysteine HS–CH2– Cys C

15. Methionine* H3C–S–CH2–CH2– Met M

16. Phenylalanine* C6H5–CH2– Phe F

17. Tyrosine (p)HO–C6H4–CH2– Tyr Y

18. Tryptophan* —CH2 Trp W

N
H

19. Histidine* —H2C His H

NH

20. Proline COOHa Pro P

NH H

CH2

* Essential amino acid, a = entire structure of amino acid

Preparation of α-Amino Acids (3) From α-ketoacid:

(1) Amination of α-haloacid:
O
NH 3 H+
R—CH—COOH excess R—CH—COOH R—C—COOH
NH3
R—C—COOH
Na–C2H5OH
–H3O
—

Cl NH2 NH NH2
–amino acid —
(2) Gabriel phthalimide synthesis: R—CH—COOH
The product of reaction of α-Halogenated acids or esters with
(4) The strecker synthesis: Treating an aldehyde with ammonia
potassium phthalimide on hydrolysis gives α-amino acid. .
and HCN yields α-amino nitrile, which on hydrolysis gives
α-amino acid.

NH3 H 3O +
R—CHO HCN
R—CH—CN R—CH—COOH
—

NH2 NH2

(5) From Natural protein:

H O/H +
Natural protein 
2
→ A mixture of α-amino acid

+
Physical Properties of α-Amino Acids
™ They are colourless, crystalline water soluble solids with high
melting points.

156 P JEE Dropper Module-4 CHEMISTRY

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 ™ In solutions, they behave more like salts rather than as amines For example:
or carboxylic acids. This behaviour is due to the presence of
(i) For neutral amino acid: Isoelectric point (pI) varies
both acidic(carboxylic group) and basic amino group in the
between 5.1 to 6.5. e.g. Glycine has pI value of 6.0.
same molecule.
™ Except glycine, all other naturally occurring α–amino acids (ii) For acidic amino acid: Isoelectric (pI) is around 3. e.g.
are optically active, since the α-carbon atom is asymmetric. Aspartic acid has pI value of 2.77.
These exist both in ‘D’ and ‘L’ forms. Most naturally (iii) For basic amino acid: Isoelectric point (pI) varies
occurring amino acids have L-configuration. L-Aminoacids
between 7.6 to 10.8. e.g. Lysine has pI value of 9.8.
are represented by writing the –NH2 group on left hand side
in fischer representation.
™ In aqueous solution the carboxyl group can lose a proton and
Key Note
amino group can accept a proton, giving rise to a dipolar ion
known as zwitter ion. This is neutral but it contains both
™ At isoelectric point, the amino acids are least soluble in
positive and negative charges. In zwitter ionic form, amino
water. This property is utilized in separation of various
acids show amphoteric behaviour as they react with both
amino acids.
acids and bases.

O O
Calculation of pI (Isoelectric Point)
R—CH—C—O—H R—CH—C—O ™ The pI of an amino acid that does not have an ionisable side
—

 chain such as alanine is average of pKa value of the carboxyl

NH2 NH3 group and protonated amino group.
(Zwitter ion)
O
O O CH3CHCOH pKa = 2.34

—
OH– H+
R—CH—C—O– R—CH—C—O– NH3 pKa = 9.69
—

 O
NH2 NH3 2.34 + 9.69
=pI = 6.02
R—CH—C—OH 2
—

 ™ If an amino acid has an ionisable side chain, its pI is

NH3 the average of the pKa values of the similarly ionising
Amphoteric nature of amino acids
groups.

Isoelectric Point of a-Amino Acids (pI) In lysine:

Because of amphoteric nature, in acidic solution, it exist as the +ve O
ion. Hence, it migrates towards cathode while in basic solution, it 
exist as –ve ion and migrates towards anode. pKa = 10.79 H3NCH2CH2CH2CH2CHCOH pKa = 2.18
—

At some intermediate pH, amino acids exist as a neutral dipolar  NH pKa = 8.95
ion i.e., the concentration of the cation and anions are equal and it 3

does not show migration towards any electrode. This pH is called 8.95 + 10.79
isoelectric point of amino acid which is different for different =pI = 9.87
2
amino acids.
In glutamic acid:
  pH<pI 
H3N—CH—COO + H(aq.) H3N—CH—COOH O O
—

R R pKa = 4.25HOCCH2CH2CHCOHpKa = 2.19

—


NH3 pKa = 9.67
 
H3N—CH—COO + H(aq.) pH>pI H2N—CH—COO =pI
2.19 + 4.25
= 3.22
—

2
R R
Anode

Biomolecules 157
General Reactions of Amino Acids
(1) Reactions due to –NH2 group (2) Reactions due to –COOH group
H H
HBr ROH
R COOH 
R COOR
H
  NH
NH3Br 3

H
MeCOCl NaOH
R COOH R—CH2—NH2
CaO/
H
NHCOMe
R COOH H
H
NaNO2/HCl NH2 LAH
R COOH R CH2OH
–N2
OH NH2 O
O
H H
NOCl NH3/ OH Heat N —H
R COOH NHR –H2O
2
CONH2
Cl NH2 –lactam

(3) Heating Effect O

™ Heating of α-amino acids leads to intermolecular dehydration O
to form cyclic diamides. C—OH Heat
O N—H –H2O N—H

—
CH2—C—OH O H –lactam
— H
NH
Structure of Proteins
— —
—

NH 
H –2H2O
HN NH
CH2 Peptides
HO—C
O Proteins are the polymers of α-amino acids and they are connected
O to each other by peptide bond or peptide linkage.
Key Note Chemically, peptide linkage is an amide linkage formed between
–COOH group and –NH2 group.
™ When alanine is heated, then two diastereomers are ™ Dipeptide-The reaction between two molecules of similar or
obtained. One of them (trans) is non-resolvable. different amino acids, proceeds through the combination of the
amino group of one molecule with the carboxyl group of the
other. This results in the elimination of a water molecule and
formation of a peptide bond –CO–NH–. The product of the
reaction is called a dipeptide because it is made up of two amino
acids. For example, when carboxyl group of glycine combines
with the amino group of alanine, we get a dipeptide named
glycylalanine.
™ When β-amino acids are heated, α, β-unsaturated acid is
formed.
 RCH=CH—COOH
RCHCH2COOH –NH
—

NH2
™ When γ or δ-amino acid are heated, it gives γ or δ-lactam.
O
O

OH Heat N —H
NH2 –H2O ™ Tripeptide: If a third amino acid combines to a dipeptide, the
product is a tripeptide and contains three amino acids linked
–lactam
by two peptide linkages.
O
158 O P JEE Dropper Module-4 CHEMISTRY
C—OH Heat W
N—H –H2O N—H
—

H –lactam
 ™ Similarly when four, five or six amino acids are linked, 1. Primary Structure of Protein: Each polypeptide chain in
the respective products are tetrapeptide, pentapeptide or a protein has amino acids linked with each other in a specific
hexapeptide, respectively. sequence and it is this sequence of amino acids that is said to be
™ Polypeptide: When the number of such amino acids is more the primary structure of that protein. Any change in this primary
than ten, then the products are called polypeptides. structure i.e., the sequence of amino acids creates a different
protein.
2. Secondary Structure: The secondary structure of a protein
Key Note refers to the shape in which a long polypeptide chain can exist.
There are two different conformations of the peptide linkage
™ The distinction between a polypeptide and a protein is not present in proteins viz. α-helix and β-pleated sheet structure.
very sharp. Polypeptides with fewer amino acids are likely These structures arise due to the regular folding of the backbone
to be called proteins if they ordinarily have a well defined of the polypeptide chain due to hydrogen bonding between
conformation of a protein such as insulin which contains 51
C=O and –NH– groups of the peptide bond. α-Helix is
amino acids.
one of the most common ways in which a polypeptide chain
™ A polypeptide with more than hundred amino acid
forms all possible hydrogen bonds by twisting into a right
residues, having molecular mass higher than 10000 u is
handed screw (helix) with the –NH group of each amino acid
generally called a protein.
residue hydrogen bonded to the C=O of an adjacent turn of
the helix.
Writing and Naming of Polypeptides
The structure of polypeptides are written in such a way that
amino acid with free –NH2 group is written on the left hand side
(N-terminal) of the polypeptide chain, while the amino acid with the
free carboxyl group is written on the right hand side (C-terminal)
of the chain. Thus, a tripeptide alanylglycylphenylalanine (Ala-
Gly-Phe) is represented as follows:
O O
H2N—CH—C—NH—CH2—C—NH—CH—COOH
—

Glycine
CH3 CH2—C6H5
Alanine Phenylalanine
Ala–Gly–Phe or A–G–F
The name of any polypeptide is written starting from the
N-terminal residue. While writing the name, the suffix ‘ine’ in
the name of the amino acid is replaced by ‘yl’ (eg. glycyl for
glycine alanyl for alanine etc) for all the constituent amino acids
except the C- terminal residue. This nomenclature is usually not
used these days, instead the three letter abbreviation or one letter In β - pleated sheet structure all peptide chains are stretched
codes for various α-amino acids present in the chain is used. For out to nearly maximum extension and laid side by side, which
example, the above tripeptide is named as Ala-Gly-Phe or A-G-F. are held together by inter molecular hydrogen bonds. These
Classification of Proteins based on their Molecular structures resembles the pleated folds of drapery.
Shape 3. Tertiary Structure of Proteins: The tertiary structure of
Proteins can be classified into two types on the basis of their proteins represents overall folding of the polypeptide chains
molecular shape. i.e., further folding of the secondary structure. It gives rise
1. Fibrous proteins: When the polypeptide chains run parallel and to two major molecular shapes viz. fibrous and globular.
are held together by hydrogen and disulphide bonds, then fibre- The main forces which stabilise the 2° and 3° structures of
like structure is formed. Such proteins are generally insoluble proteins are hydrogen bonds, disulphide linkages, vander
in water. Some common examples are keratin (present in hair, Waals and electrostatic forces of attraction.
wool, silk) and myosin (present in muscles), etc. 4. Quaternary structure of proteins: Some of the proteins
2. Globular proteins: This structure results when the chains are composed of two or more polypeptide chains referred to
of polypeptides coil around to give a spherical shape. These as sub-units. The spatial arrangement of these subunits with
are usually soluble in water. Insulin and albumins are the
respect to each other is known as quaternary structure.
common examples of globular proteins.

Structure of Proteins Denaturation of Proteins

Depending on the complexity, the structure of proteins may be ™ Native Protein- Protein found in a biological system with a
distinctly studied as primary structure, secondary structure and unique three-dimensional structure and biological activity is
tertiary structures. called a native protein.

Biomolecules 159
 ™ When a protein in its native form, is subjected to physical Classification of Vitamins
change like change in temperature or chemical change like Vitamins are generally classified into two broad types based on
change in pH, the hydrogen bonds are disturbed. Due to their solubility, i.e., fat soluble and water-soluble. These two
this, globules unfold, helix get uncoiled and protein loses its groups discharge different functions.
biological activity. This is called denaturation of protein.
1. Fat soluble vitamins: These are oily substances not readily
™ During denaturation, 2° and 3° structures are destroyed soluble in water. The group includes vitamins A, D, E and K.
but 1° structure remains intact. The coagulation of egg Liver cells are rich in fat soluble vitamins e.g. Vitamin A and
white on boiling is a common example of denaturation. Vitamin D. This group of hydrophobic, lipid soluble vitamins
Another example is curdling of milk which is caused due to as a class are not absorbed in the body unless fat digestion
the formation of lactic acid by the bacteria present in milk. and absorption proceed normally. Their deficiency can cause
malabsorptive diseases. Excess intake of these vitamins may
VITAMINS cause hypervitaminosis.
2. Water soluble vitamins: This group includes the remaining
Vitamins can be defined as essential dietary factors required by an
vitamins e.g., vitamins of B group (B-Complex), vitamin
organism in minute quantities and their absence causes specific
C, etc. The water soluble vitamins are stored in much lesser
deficiency diseases. Vitamins are designated by alphabets A, B, C,
amounts in the cells. Water soluble vitamins must be supplied
D, etc. Some of them are further named as sub-groups e.g. B1, B2,
regularly in diet because they are readily excreted in urine
B6, B12, etc. Excess of vitamins is also harmful and vitamin pills
and cannot be stored (except vitamin B12 in our body).
should not be taken without the advice of a doctor.
The term “Vitamine” was coined from the word vital + amine since Key Note
the earlier identified compounds had amino groups. Later work
showed that most of them did not contain amino groups, so the letter ™ Vitamin H (Biotin) is an exception, since it is neither
soluble in water nor in fat.
‘e’ was dropped and the term “vitamin” is used these days.

Some important vitamins, their sources and the deficiency diseases related to them may be summed up as under:
S. Name of vitamin Sources Deficiency diseases
No.
1. Vitamin A Fish liver oil, carrots, butter and milk Xerophthalmia i.e., hardening of cornea of eye,
Night blindness
2. Vitamin B1 (Thiamine) Yeast, milk, green vegetables and cereals Beri-Beri (loss of appetite, retarded growth)
3. Vitamin B2 (Riboflavin) Milk, egg white, liver and kidney Cheilosis (fissuring at corners of mouth and lips),
digestive disorders and burning sensation of the skin.
4. Vitamin B6 (Pyridoxine) Yeast, milk, egg, yolk, cereals and grams Convulsions
5. Vitamin B12 Meat, fish, egg and curd Pernicious anaemia (RBC deficient in hemoglobin)
6. Vitamin C (Ascorbic acid) Citrus fruits, amla and green leafy vegetables Scurvy (bleeding gums)
7. Vitamin D Exposure to sunlight, fish and egg yolk Rickets (bone deformities in children) and
osteomalacia (soft bones and joint pain in adults)
8. Vitamin E Vegetable oils like wheat germ oil, sunflower Increased fragility of RBCs and muscular
oil, etc. weakness.
9. Vitamin K Green leafy vegetables Increased blood clotting time

DNA molecules, the sugar moiety is β-D-2-deoxyribose whereas

NUCLEIC ACIDS in RNA molecule, it is β-D-ribose.
5 5
The particles in nucleus of the cell, responsible for heredity, are HOH2C O OH HOH2C O OH
called chromosomes, which are made up of proteins and another 4 1 4 1
type of biomolecules called nucleic acids. These are mainly of H H H H
H H H H
two types, the deoxyribonucleic acid (DNA) and ribonucleic acid 3 2 3 2
(RNA). Nucleic acids are long chain polymer nucleotides, so they OH OH OH H
are also called polynucleotides. –D–ribose –D–2–deoxyribose
Chemical Composition of Nucleic Acid: Complete hydrolysis DNA contains four bases viz. adenine (A), guanine (G), cytosine
of DNA (or RNA) yields a pentose sugar, phosphoric acid and (C) and thymine (T). RNA also contains four bases, the first three
nitrogen containing heterocyclic compounds (called bases). In bases are same as in DNA but the fourth one is uracil (U).

160 P JEE Dropper Module-4 CHEMISTRY

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 O
NH Base
NH
N NH2 N Hence nucleic acid chain can be written as: Sugar–Phosphate n
H2 N N
N N NH
Primary structure of Nucleic Acid
Adenine (A) Guanine (G) Information regarding the sequence of nucleotides in the chain of
NH2 a nucleic acid is called its primary structure.
—

CH3 Nucleic acids have a secondary structure also. DNA has double
N NH NH strand helix structure. Two nucleic acid chains are held together
O O O O O
by hydrogen bonds between pairs of bases. Adenine forms two
N N N
H H H hydrogen bonds with thymine whereas cytosine forms three
Cytosine (C) Thymine (T) Uracil (U) hydrogen bonds with guanine.
In secondary structure of RNA, these are only single stranded and
Structure of Nucleic Acid sometimes they fold back on themselves to form double helix
Nucleosides: A unit formed by the attachment of a base to 1′ structure.
position of sugar is known as nucleoside. In nucleosides, the sugar 5'
3'
carbons are numbered as 1′, 2′, 3′ etc. in order to distinguish these
from the bases.
–GC–
General formula of nucleosides is Nitrogenous Base + Sugar =
Nucleoside
–––A = T––––
––T = A––––
––A=T––
Base Abbreviation Nucleoside
Adenine A Adenosine ––C
G––
Guanine G Guanosine ––G
––T= C–––
A–––
Cytosine C Cytidine ––CG
–
Thymine T Thymidine
Uracil U Uridine –
––T =A–
– – G C–––
––
O ––A=T–
5' 5' –T=A––
HO—H2C O Base O—P—O—H2C O Base
4' 1' 4' 1' ––A
O =T
H H H H H H H H ––T= ––
3' 2' 3' 2' A––
––CG –
–––
OH OH OH OH –GC––
Nucleoside Nucleotide

Nucleotides: Nucleotides are joined together by phosphodiester 5'

3'
linkage between 5′ and 3′ carbon atoms of the pentose sugar. The Double strand helix structure for DNA
formation of a typical dinucleotide is shown below:
–
O
– Tests of Protein
O
Base – 5' Base (a) Biuret test
–
O—P—O—CH2
O—P—O—CH2 O O (i) On adding a dillute solution of copper sulphate to
O Sugar O Sugar an alkaline solution of protein, a bluish violet colour is
3'
developed.
+ O
OH OH O OH
(ii) This test is due to the presence of peptide (—C—NH—)
–
O Phosphodiester
–
– Base linkage O—P=O linkage.
O—P—O—CH2 O
O (b) Millon’s test
O Sugar
Base (i) Millon’s reagent consist of mercury dissolved in nitric acid
5' CH
2
O (forming a mixture of mercuric and mercurous nitrates).
OH OH Sugar (ii) When millon’s reagent is added to a protein, a white ppt
forms which turn brick red on heating.
3'
(iii) This test is given by protein which yield tyrosine on
OH OH
Formation of a dinucleotide hydrolysis (due to the presence of phenolic group).

Biomolecules 161
(c) Ninhydrin test
(i) This test is given by all proteins. Example 8: Amino acids are classified as acidic, basic or
(ii) When protein is boiled with a dilute solution of ninhydrin, neutral depending upon the relative number of amino and
a deep blue or violet colour is produced. carboxyl groups in their molecule. Which of the following
O
H O are acidic amino acids?
(a) (CH3)2CH—CH—COOH

— —
C OH

—
C +R—C—C—OH
OH NH2
C NH2
-amino acid
(b) HOOC–CH2–CH(NH2)COOH
O
Ninhydrin (c) H2N–CH2–CH2–COOH
O O–
(d) HOOC—CH2—CH2—CH—COOH

—
C C
=N—C NH2
C C Sol. Acidic amino acids contains more number of –COOH
O O group than –NH2 group.
Violet complex Example 9: Which of the following B group vitamins can
be stored in our body?
Test of Carbohydrates
(a) Vitamin B1 (b) Vitamin B2
Molisch test (c) Vitamin B6 (d) Vitamin B12
On adding a few drops of alcoholic solution of α-naphthol and
concentrated sulphuric acid to the given solution, a violet ring is Sol. Vitamin B12 can be stored in our body because it is
formed. water insoluble.
Uses of Proteins Example 10: Dinucleotide is obtained by joining two
nucleotides together by phosphodiester linkage. Between
(i) Protein consitute an essential part of our food. Meat,
eggs, fish, cheese provide protein to human beings. which carbon atoms of pentose sugars of nucleotides are
(ii) Casein (a milk protein) is used in the manufacture of these linkages present?
artificial wool and silk. (a) 5′ and 3′ (b) 1′ and 5′
(iii) Amino acid needed for medicinal use and feeding (c) 5 and 5′ (d) 3′ and 3′
experiment, are prepared by hydrolysis of proteins.
Sol. Phosphodiester links the two nucleotides together to
(iv) Gelatin is used in deserts, salads, candies, bakery goods, etc.
form a dinucleotide. Between the pentose sugars of
(v) Leather is obtained by tanning the protein of animal hides.
nucleotides, there are 5′ and 3′ connections.
(vi) Haemoglobin present in blood is responsible for carrying
oxygen and CO2.
(vii) Harmones control various body processes.
(viii) Enzymes are the proteins produced by living system and
catalyse specific biological reactions.

COOH
6. HS
Example 6: Which of the following terms are correct about NH2
⊕
enzyme? The pKa values of –COOH, –SH and – NH3 groups
(a) Proteins (b) Dinucleotides present in the amino acid, cysteine are 1.8, 8.2 and
10.8, respectively. The structure of cysteine at a pH =
(c) Nucleic acids (d) Biocatalysts
5 will be:
Sol. Enzymes are protein molecules that operate as O O
biocatalysts in the body, facilitating chemical processes. C—OH C—O
Example 7: Which of the following acids is a vitamin? (a) HS (b) HS
 
NH3 NH3
(a) Aspartic acid (b) Ascorbic acid
(c) Adipic acid (d) Saccharic acid O O
Sol. 
Ascorbic acid is the chemical name for vitamin C—O C—O
C. Aspartic acid is an amino acid, adipic acid is a (c) S (d) S
dicarboxylic acid with an eight-carbon chain, and

NH3 NH2
saccharic acid is a dicarboxylic acid.

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 7. Proteins can be classified into two types on the basis 9. Amino acid contains an acidic group –COOH as well
of their molecular shape i.e., fibrous proteins and as a basic group –NH2, the structure of alanine is
globular proteins. Examples of globular proteins are: 
NH3—CH—COO
(a) Insulin

—
(b) Keratin CH3
Which of the following statements is/are incorrect
(c) Albumin about it?
(d) Myosin (a) It is an acidic amino acid.
8. The name of the tripeptide (b) It is an essential amino acid.
(c) The isoelectric point (pI) of alanine is 6.02.

H2NCH2CONHCHCONHCHCOOH is — (d) It is an optically active amino acid.
—

10. Correct statement about peptide linkage in a protein

CH2C6H5 CH3
molecule is/are–
(a) Glycylphenylalanylalanine (a) It is an amide linkage.
(c) Alanylglycylphenylalanine (b) It has partial double bond character.
(b) Alanylphenylalanylglycine (c) It is hydrophilic in nature.
(d) Phenylalanylglycylalanine (d) It connects protein molecules through H-bonds.

AARAMBH (SOLVED EXAMPLES)

1. How will you separate a ternary mixture of arginine, alanine (b) CH2COO
and aspartic acid?
—

Sol. A few drops of a solution of an amino acid mixture are H—C—NH3

—

applied to the middle of a piece of filter paper. When the SH

paper is placed in a buffer solution (pH=5) between the two (c) CH2—CH—COO
electrodes and an electric field is applied, then arginine and
alanine with pI > pH moves towards the cathode and aspartic +
NH3
acid with pI < pH moves towards the anode. Out of arginine HN
and alanine, alanine will move slowly towards the cathode
N
due to lesser positive charge. + –
Before electrophoresis After electrophoresis (d) NH3(CH2)4CH(NH2)COO
Sol. H ydrolysis of protein (natural molecules), yields
α-aminoacids, but the option (b) is β-aminoacid.

Buffer A B C Buffer
solution solution
(pH = 5) (pH = 5)

Mixture of arginine
+ alanine + aspartic acid

(a) Arginine (pI = 10.76) Therefore, option (b) is the correct answer.
(b) Alanine (pI = 6.02) 3. Tyrosine is an α-amino carboxylic acid shown below:
(c) Aspartic acid (pI = 2.98)
2. The amino acid that cannot be obtained by hydrolysis of HO CH2CHCOOH
—

protein is: NH2

(a) Write the most stable structural formula:
(a) In it’s cationic form. (b) In it’s anionic form.
(c) In it’s dianionic form. (d) In it’s Zwitter ionic form.

Biomolecules 163
 Sol. NH2
Sol. (a) O
NH–C COOH

(b) HO CH2—CH—COO– H2N COOH

—
NH2
NH2
O
(c) –O CH2—CH—COO–
NH–C

—
COOH
NH2

H2N COOH
(d)
NH2 NH2
O

4. Aspartame, an artificial sweetener is a peptide and has the HOOC NH–C COOH
following structure:
NH2 NH2
CH2—C6H5 O
—

H2N—CH—CONH—CH—COOCH3 NH–C

HOOC COOH
—

CH2—COOH 6. Consider the following compounds and indicate whether

(a) Identify the four functional groups. each is a reducing sugar or a non-reducing sugar.
(b) Write the zwitter ionic structure. HOH2C
(c) Write the structures of the amino acids obtained from O
(a) HO
the hydrolysis of aspartame.
OCH2CH2CH3
(d) Which of the two amino acids is more hydrophobic?
Sol. (a) Amine, carboxylic acid, Amide, Ester. OH
OH
(b) Zwitterionic structure
HOH2C
CH2—C6H5 O
(b) HO
—


H3N—CH—CONH—CH—COOCH3
—

HO
CH2COO OH
Me
O O
OH
Hydrolysis
(c) H2N—CH—C—NH—CH—C—OCH3 HOH2C
O
—
—

HO H HO H
(c) HO
CH2COOH CH2C6H5
HO
O
H2N—CH—COH + CH2C6H5 + CH3OH OH
—
—

(d) HOH2C O OCH2CH3

CH2COOH NH2—CH—COOH
Aspartic acid Phenyl alanine Methanol
H H
(d) Phenyl alanine is more hydrophobic because its side H CH2OH
chain is highly non polar with respect to others.
5. Following two amino acids lysine and glutamic acid form OH OH
dipeptide linkage. What are the four possible dipeptides? Sol. (a) It is an acetal, hence a non-reducing sugar.
NH2 NH2 (b) It is an ether, hence a non-reducing sugar.
(c) It is a hemiacetal, hence a reducing sugar.
H2N COOH + HOOC COOH (d) It is an acetal, hence a non-reducing sugar.

164 P JEE Dropper Module-4 CHEMISTRY

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 7. All sugars with the hemiacetal structure exhibit mutarotation.
For example, α-D-galactopyranose has an [α]D of +150.7°
and β-D-galactopyranose has an [α] D of +52.8°. If
either anomer is dissolved in water and allowed to reach
equilibrium, the [α]D of the resulting solution is +80.2°.
Calculate the percentages of each in solution.
Sol. Let x be the fraction of α-D-galactopyranose in the solution

⇒150.7x + 52.8(1 – x) = 80.2

⇒97.9x = 27.4
27.4

⇒
= x = 0.28 (b) 10 moles of HIO4 are needed for the complete cleavage
97.9 of compound (A).
Hence, α-D-galactopyranose = 28% and β-D-galactopyranose (c) 10 moles of formic acid and 2 moles of a formaldehyde
= 72% are formed in the reaction of compound (A) with
8. Draw fischer projections of L-Glucose and L-fructose. periodic acid.
10. In the following periodate oxidation reaction, identify the
Sol. L-glucose and L-fructose are enantiomers of their
products formed and their molar equivalent ratios.
corresponding D-sugar.
CH3
Mirror
H3C OH
CHO CHO H OH HIO4
H OH HO H H OH
HO H H OH H OH
H OH HO H
H OH HO H CH3
CH3
CH2OH CH2OH
Sol.
H3C OH
D(+)-glucose L(–)-glucose
H OH 3HIO4
Mirror CH3COCH3 + 2HCOOH
H OH
CH2OH CH2OH + CH3CHO
H OH
O O
CH3
HO H H OH
H OH HO H 11. Predict products you expect to obtain by the reaction of
H OH HO H β-D-fructose with
(a) CH3COCl (b) CH3OH/H+
CH2OH CH2OH
L(+)-fructose Sol.
D(–)-fructose

9. A disaccharide (A), C 12 H 22 O 11 gives a negative test

with Benedict’s solution and does not mutarotate.
Disaccharide (A) is hydrolyzed by α-glucosidases but
not by β-glucosidases. Methylation of (A) followed
by hydrolysis gives two molar equivalents of 2,3,4,6
-tetramethyl-D-glucose.
(a) Identify the structure of disaccharide (A).
(b) How many moles of periodic acid will react with (A)? 12. Count the total number of hydrogen bonds possible for
(c) How many moles each of formaldehyde and formic acid the following DNA sequence(of a stretch of double helical
are formed in the reaction of (A) with periodic acid? structure).

Sol. (a) The given disaccharide (A) is a non-reducing sugar with

an α-linkage at the anomeric carbon. Since, compound G C T A T T
(A) gives only 2,3,4,6-tetramethylated-D-glucose, Sol. A–T pair forms two H-bonds while G–C pair forms three
possible structure of disaccharide (A) is: H-Bonds.

Biomolecules 165
 G C T A T T
H-Bonds
C G A T A A
No. of H-Bonds = 14
Therefore, [14] is the correct answer.
13. Predict products you expect to obtain by the reaction of β-D-glucose with
(a) CH3COCl + Pyridine (b) (i) NaH (ii) CH3Br(excess, SN2) (c) CH3OH/H+
Sol.
CH2OAc

H O OAc
CH3COCl H
OAc H
Py
AcO H
CH2OH [Acylation]
H OAc
H O OH Glucose pentaacetate
H
OH H CH2OCH3
HO H
H O OCH
(i) NaH 3
H OH H
(ii) CH3Br(SN2) OCH3 H
-D-glucose
H3CO H

H OCH3
CH2OH CH2OH

H O H H O OMe
CH3OH/H+ H H
OH + [Formation of acetal]
H OH H
HO OMe HO H

H OH H OH
-and -methylglucoside

14. Which of the following pairs is/are correctly matched? (b) Among the a-amino acids that constitute proteins, glycine
(a) α-D (+) glucose and β-D(+) glucose → C-2 epimers is the only one that does not possess chiral center.
(b) Glucose and fructose → C-3 epimers (c) An important and sensitive test for the detection of
(c) Glucose → Mutarotation L-amino acid is the ninhydrin colour test.
(d) Sucrose → Glucose + fructose (d) HNO2 liberates nitrous oxide from amino acid.
Sol. �   
Glycine is H2N—CH2—COOH. So, it doesn’t have any
Sol. Glucose shows mutarotation, sucrose gives glucose and
Chiral center.
fructose on hydrolysis.
 L-amino acids constitute protein.
Therefore, option (c,d) are the correct answers.  Ninhydrin colour test is used to detect the L-amino acid.
15. Consider the following statements about amino acids:  HNO2 liberates N2 gas from amino acid rather than
(a) The amino acids that constitute proteins are all L-amino nitrous oxide.
acids. Therefore, option (a,b,c) are the correct answers.

166 P JEE Dropper Module-4 CHEMISTRY

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 PRARAMBH (TOPICWISE)

CARBOHYDRATE 12. Five membered ring structure of fructose is known as

(a) Pyranose (b) Furanose
1. Amylopectin is
(c) Both (a) and (b) (d) None of these
(a) Water soluble
13. Sucrose in presence of invertase on hydrolysis gives
(b) Water insoluble
(a) Glucose (b) Fructose
(c) Forms colloidal solution with water (c) Ethyl alcohol (d) Both (a) and (b)
(d) Both (b) and (c) 14. Which of the following are polysaccharides?
2. Which of the following is invert sugar? (A) Starch (B) Cellulose
(a) Mixture of glucose and galactose in equimolar ratio. (C) Dextrins (D) Glycogen
(b) Mixture of glucose and fructose in equimolar ratio. (a) A, B and C (b) A, B, D
(c) A type of cane sugar. (c) A and C (d) A, B, C, D
(d) Optically inactive form of sugar.
15. Number of chiral C-atoms in glucose and fructose are
3. The number of ring atoms in the cyclic structure of
(a) 4 in each
D-fructose is
(b) 3 in each
(a) 5 (b) 6 (c) 4 (d) 7
(c) 4 in glucose and 3 in fructose
4. Reducing sugars are the one which
(d) 3 in glucose and 4 in fructose
(a) Reduce Fehling’s solution.
(b) Reduce Tollen’s reagent. 16. Which of the following monosaccharides is a pentose?
(c) Have free aldehydic or ketonic groups. (a) Glucose (b) Fructose
(d) All of these (c) Ribose (d) Galactose
5. Which is correct amongst the following statements? 17. Ring structure of glucose is due to formation of hemiacetal
(a) Monosaccharides are also known as sugars. and ring formation between
(b) Polysaccharides are non-sugars. (a) C1 and C5 (b) C1 and C4
(c) Maltose and lactose are reducing sugars. (c) C1 and C3 (d) C2 and C4
(d) All of these 18. Which of the following is not a reducing sugar?
6. General formula for carbohydrates is (a) Sucrose (b) Galactose
(a) CnH2nO2n+2 (b) Cn(H2O)2n (c) Glucose (d) Lactose
(c) Cn(H2O)n (d) None of these
7. Hydrolysis of sucrose brings about a change in sign of PROTEIN AND AMINO ACID
rotation from dextro(+) to laevo(–) and such a sign change 19. Insulin is
is known as (a) An amino acid (b) A protein
(a) Racemization (b) Inversion (c) A carbohydrate (d) A lipid
(c) Mutarotation (d) None of these
20. How many peptide bonds will be present in a tripeptide?
8. In fructose, the number of possible optical isomers are (a) 3 (b) 2
(a) 12 (b) 8 (c) 16 (d) 4
(c) 1 (d) 4
9. In glucose
21. In amino acids, more number of amino groups than carboxyl
(a) Five –OH groups are present. groups make it
(b) Four secondary and one primay alcoholic groups are present. (a) Acidic (b) Basic
(c) One –CHO group is present. (c) Neutral (d) None of these
(d) All the above statements are correct.
22. Which amino acids are called non-essential?
10. Which one of the following is a disaccharide? (a) Those which can be synthesized in the body.
(a) Glucose (b) Fructose (b) Those which have more amino groups as compared to
(c) Xylose (d) Sucrose carboxyl groups.
11. Six membered cyclic structure of glucose is known as (c) Those which have equal number of amino acid and
(a) Pyranose (b) Furanose carboxyl groups.
(c) Both of these (d) None of these (d) None of these

Biomolecules 167
 23. Which of the following is not an essential amino acid? VITAMIN AND NUCLEIC ACID
(a) Serine (b) Lysine 30. A nucleotide consists of
(c) Threonine (d) Tryptophan (a) Base and sugar (b) Base and phosphate
(c) Sugar and phosphate (d) Base, sugar and phosphate
24. Which of following are essential amino acids?
31. Which of the following is responsible for heredity character?
(A) Aspartic acid (B) Leucine (a) DNA (b) RNA
(C) Valine (D) Glycine (c) Proteins (d) Hormones
(E) Alanine 32. Complete hydrolysis of DNA or RNA yields following:
(a) Ribose in RNA and deoxyribose in DNA.
(a) A, B and C (b) A, B, D and E
(b) Heterocyclic nitrogenous purine bases.
(c) B and C (d) A, C, D and E (c) Heterocyclic nitrogenous pyrimidines.
25. Biuret test is used for the detection of (d) All of these
(a) Saturated oils (b) Sugars 33. DNA contains following purine bases
(c) Proteins (d) Fats (A) Adenine (B) Guanine
(C) Thymine (D) Cytosine
26. The linkage present in proteins and peptides is (a) A, C (b) A, B
O O O (c) A, C, D (d) A, B, C, D
(a) —C—O—C— (b) —C—O— 34. RNA contains following pyrimidine bases
(A) Thymine (B) Uracil
O (C) Cytosine (D) Adenine
(c) –NH– (d) —C—NH— (a) B, C and D (b) B & C
(c) A, B, D (d) All of these
27. If a native protein is subjected to physical or chemical 35. Nucleosides are
treatment which may disrupt its conformers without affecting (a) Base + sugar = Nucleoside
its primary structure, are called (b) N–glycosides of purine or pyrimidine bases with
(a) Inactive protein (b) Denatured protein pentose sugar
(c) Both (a) and (b) (d) None of these (c) Both of these
(d) None of these
28. A pigment protein in animals is
(a) Chlorophyll (b) Insulin 36. Ascorbic acid is a
(a) Vitamin (b) Enzyme
(c) Keratin (d) Haemoglobin
(c) Protein (d) Carbohydrate
29. a-helical structure refers to the 37. Vitamin B takes part in
(a) Primary structure of protein. (a) Increasing in blood pressure.
(b) Secondary structure of protein. (b) Decreasing in blood pressure.
(c) Tertiary structure of protein. (c) Activating bone marrow.
(d) Quaternary structure of protein. (d) Maturation of RBC’s.

PRABAL (JEE MAIN LEVEL)

1. The name of the following dipeptide is: NH2
—

H 2 NCHCONHCH 2 COOH (c) CH2—CH2—CH—COOH

|
—

CH3
COOH
(a) Glycylglycine (b) Glycylalanine NH2
—

(c) Glycine alanine (d) Alanylglycine (d) HOOC—CH2—CH—COOH

2. Which of the following is a basic amino acid?
3. Which of the following is an α-amino acid?
NH2
—

(a) H2N—C—NH(CH2)3—CH—COOH (a) —NH—CH2—CH2—COOH

NH NH2
(b) HOH2C—CH—NH2
—

(b)
—CH—COOH
—

COOH

168 P JEE Dropper Module-4 CHEMISTRY

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 NH2 15. In DNA, guanine pairs with

—
(a) Cytosine (b) Thymine
(c) —CH—CH2—COOH
(c) Adenine (d) Uracil
(d) CH3—CH2—CH—CH2—NH2 16. Peptide bond is a key feature in

—
(a) Polysaccharide (b) Proteins
COOH
(c) Nucleotide (d) Vitamins
4. ‘Kwashirokor’ is a disease caused by the deficiency of­
(a) Vitamins (b) Hormones 17. A hexapeptide with the composition Arg, Gly, Leu, Pro has
(c) Blood (d) Essential amino acids proline at both C-terminal and N-terminal position. The
5. The end product of protein digestion is partial hydrolysis of the hexapeptide gives
(a) Peptides (b) Peptones Gly-Pro-Arg, Arg-Pro and Pro-Leu-Gly. The hexapeptide is
(c) Protones (d) α-Amino acids (a) Pro-Gly-Leu-Pro-Arg-Pro
6. The nucleotides of one polynucleotide chain are joined (b) Pro-Leu-Gly-Pro-Arg-Pro
together by (c) Pro-Leu-Gly-Arg-Pro-Pro
(a) Weak hydrogen bonds (d) Pro-Arg-Pro-Leu-Gly-Pro
(b) Disulphide bonds 18. Synthesis of amino acids that will be affected due to
(c) Phospho-diester bonds prolonged uptake of food with no sulfur content is
(d) Glycosidic bonds (a) Lysine (b) Tyrosine
7. Hair, finger, nails, hoofs etc. are all made of (c) Aspartic acid (d) Cysteine
(a) Fat (b) Vitamins 19. Titration of the amino acid lysine has three pKa values viz., pKa
1
(c) Proteins (d) Iron (2.18), pKa (8.95) and pKa (10.53). The pH at which this amino
2 3
8. Mark the globular protein in the following acid will show no net migration in an electric field is
(a) Collagen (a) 5.57 (b) 9.74
(b) Myoglobin or Haemoglobin (c) 6.35 (d) 7.22
(c) Myosin 20. Observe the pKa values (P1–P3) of the given amino acid
(d) Fibroin CH2—CH—COOH
9. Continuous bleeding from an injured part of body is due to

—
 P1=1.82
deficiency of  NH3
HN NH
(a) Vitamin A (b) Vitamin E P3=9.14

P 2 =6.04
(c) Vitamin B (d) Vitamin K
Which form of this amino acid will exist in aqueous solution
10. Scurvy is a disease caused by
at pH = 8?
(a) A virus
(a) Dication (b) Monocation
(b) Deficiency of vitamin E
(c) Zwitter ion (d) Monoanion
(c) Deficiency of ascorbic acid
(d) Deficiency of vitamin D 21. Find true and false from the following statements regarding
carbohydrates.
11. The base present in RNA but not found in DNA, is
S1: A ll monosaccharides whether aldoses or ketoses
(a) Thymine (b) Uracil are reducing sugars.
(c) Adenine (d) Guanine
S2: Bromine water can be used to differentiate between
12. Which pairing is found in DNA? aldoses and ketoses.
(a) Adenine with thymine (b) Thymine with guanine S3: A pair of diastereomeric aldoses which differ only in
(c) Guanine with adenine (d) Uracil with adenine configuration at C-2 are anomers.
13. If a portion of DNA code is CAT, the anticodon for this code S4: O sazone formation destroys the configuration at
on tRNA will be C-2 of an aldose, but does not affect the configuration
(a) GUA (b) GTA of the rest of the molecule.
(c) ATG (d) AUG (a) TTTT (b) TFTF
14. The similarity between DNA and RNA is that both (c) TTFT (d) FTTT
(a) Are polymers of nucleotides. 22. Ultra violet light absorption occurring in protein is due to
(b) Are always double stranded. the presence of
(c) Have similar kind of sugar. (a) Alanine (b) Cysteine
(d) Have similar type of pyrimidine bases. (c) Glutamic acid (d) Tryptophan

Biomolecules 169
 23. Match the Column-I with Column-II. (a) CH2OH (b) CHO

—
Column-I Column-II (CHOH)3 (CH—OH)3

—
A. RCHCO2– p. Acidic amino acid
CH2OH CH2—OH
+
NH3
(c) COOH (d) COOH
B. Arginine q. Neutral amino acid

—
—
(CH—OH)3 (CHOH)3
C. Valine r. Zwitter ion

—
—
D. Aspartic acid s. Basic amino acid CH2—OH COOH

(a) A → r; B → s; C → q; D → p 29.
(b) A → p; B → q; C → s; D → r
(c) A → q; B → s; C → r; D → p
(d) A → s; B → p; C → q; D → r
24. Periodic acid splits glucose and fructose into formic acid The final product of the reaction, is:
and formaldehyde. Ratio of formic acid and formaldehyde
(a) COOH (b) CHO
from glucose and fructose is

—
—
(a) 5/1 and 4/2 (b) 5/1 and 3/2 (CHOH)3 (CHOH)4

—
(c) 4/2 and 4/2 (d) 3/2 and 4/2

—
CH2—OH CH2—OH
25. An aldose is converted into its next higher homologue by
(a) Ruff’s method (b) Amadori rearrangement (c) CHO (d) COOH
(c) Kiliani synthesis (d) Wohl’s method

—
26. Which of the following gives an optically inactive (CHOH)2 (CHOH)3

—
dicarboxylic acid on oxidation with dilute HNO3 acid? CH2OH COOH
(a) CHO (b) CHO
H OH H OH 30. Which of the following statement is correct?
H OH HO H
(a) The Ruff procedure lengthens an aldose chain and gives
HO H H OH
a single product.
CH2OH CH2OH
(b) The Ruff procedure shortens an aldose chain and gives
(c) (d) CHO two epimers.
CHO
H OH HO H (c) The Kiliani-Fischer procedure shortens an aldose chain
HO H H OH and gives a single product.
HO H H OH
CH2OH (d) The Kiliani-Fischer procedure lengthens an aldose chain
CH2OH
and gives two epimers.
27. Which of the following statements most correctly defines
the isoelectric point? 31. Which of the following compounds will not show mutarotation?
(a) The pH at which all molecular species are ionised and (a) Methyl-α-D-glucopyranoside
carry the same charge. (b) α-D-(+)-glucopyranose
(b) The pH at which all molecular species are neutral and
(c) β-D-(+)-glucopyranose
are uncharged.
(c) The pH at which half of the molecular species are (d) β-D-(+)-galactopyranose
ionised and the other half unionised. 32. Identify structure of Thymine deoxyribo-nucleotide.
(d) The pH at which all molecular species carry charge but
net charge on any molecule is zero. (a) O

28. CHO N NH
—

O
CH—OH N
HO—P—O O N NH2
—

(1) HCN
CH—OH Product
—

(2) H2/Pd–BaSO4
—

+ OH
CH2—OH (3) H3O
—

The final product of the reaction, is: OH

170 P JEE Dropper Module-4 CHEMISTRY

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 (b) O 37. Which of the following statement(s) is/are true?
(i) All amino acids contain one chiral center.
NH
(ii) Some amino acids contain one, while some contain more
O
chiral centers or even “NO” chiral center.
N O
HO—P—O O (iii) All amino acids found in proteins have L configuration.

—
—
OH (iv) All amino acids found in proteins have 1° amino group.
(a) (ii), (iii) and (iv) (b) (ii) and (iii)
—
OH (c) (i), (iii) and (iv) (d) (i) and (iv)
(c) NH2
38. In DNA, the complimentary bases are:
(a) Adenine and thymine; guanine and cytosine
NH
(b) Adenine and thymine; guanine and uracil
N O (c) Adenine and guanine; thymine and cytosine
O
HO (d) Uracil and adenine; cytosine and guanine
—

39. Which of the following is correct about H-bonding in

nucleotide?
—

OH
(a) A==A and T==T (b) G==T and A C
(d) NH2
(c) A G and T C (d) A==T and G C
N 40. Assertion(A): Vitamin D cannot be stored in our body.
N
Reason(R): Vitamin D is a fat soluble vitamin.
O N N
HO (a) Assertion is correct, reason is correct; reason is a correct
—

explanation for assertion.

(b) Assertion is correct, reason is correct; reason is not a
—

OH correct explanation for assertion.

33. The specific rotation of a freshly prepared solution of (c) Assertion is correct, reason is incorrect.
α-D-glucose changes from a value of x° to a constant value (d) Assertion is incorrect, reason is correct.
of y°. The value of x and y are respectively
41. Match the Column-I with Column-II.
(a) 112° and 52.5° (b) 19° and 52.5°
(c) 52.5° and 19° (d) 52.5° and 112° Column-I Column-II (Deficiency
34. The symbols D and L represent diseases)
(a) The optical activity of compounds. A. Vitamin A p. Scurvy
(b) The relative configuration of a particular stereoisomer. B. Vitamin B12 q. Hemorrhagic condition
(c) The dextrorotatory nature of molecule. C. Vitamin C r. Sterility
(d) The laevorotatory nature of molecule.
D. Vitamin E s. Xeropthalmia
35. Which of the following properties of glucose cannot be
explained by its open chain structure? E. Vitamin K t. Pernicious anaemia
(i) Glucose does not form hydrogen sulphite with NaHSO3.
(a) A → t; B → s; C → p; D → r; E → r
(ii) On oxidation with HNO3, glucose gives saccharic acid.
(iii) Glucose is found to exist in two different crystalline (b) A → s; B → t; C → p; D → q; E → r
forms which are named as α and β. (c) A → s; B → t; C → p; D → r; E → q
(a) (ii) only (b) (i) and (iii)
(d) A → t; B → s; C → p; D → r; E → q
(c) (ii) and (iii) (d) (i) and (ii)
42. Which of the following statement(s) is/are correct?
36. An electric current is passed through an aqueous solution
(buffered at pH=6.0) of alanine (pI=6.0) and arginine (i) Vitamins A, D, E and K are insoluble in water.
(pI=10.2). The two amino acids can be separated because (ii) Vitamins A, D, E and K are stored in liver and adipose
(a) Alanine migrates to anode and arginine to cathode. tissues.
(b) Alanine migrates to cathode and arginine to anode. (iii) Vitamin B and vitamin C are water soluble.
(c) Alanine does not migrate, while arginine migrates to (iv) Water soluble vitamins should not be supplied regularly
cathode. in diet.
(d) Alanine does not migrate, while arginine migrates to (a) (i), (ii) and (iv) (b) (i), (ii) and (iii)
anode. (c) (i) and (iv) (d) (ii) and (iv)

Biomolecules 171
 43. Match the Column-I with Column-II. CH3O H
Column-I (Enzymes) Column-II (Reactions) H OH
         and HO H O
Decomposition of urea into H OH
A. Invertase p.
NH3 and CO2. H

Conversion of glucose into CH2OH

B. Maltase q.
ethyl alcohol.         (III) Methyl-β-D-glucoside

Hydrolysis of maltose into S1: The glucosides do not reduce Fehling’s solution.
C. Pepsin r.
glucose. S2: The glucosides do not react with hydrogen cyanide or
hydroxylamine.
D. Urease s. Hydrolysis of cane sugar S3: Behaviour of glucosides as stated in S1 and S2 indicates
the absence of free –CHO group.
Hydrolysis of proteins into
E. Zymase t. S4: The two forms of glucosides are enantiomers.
peptides.
47. How many of these amino acids are negatively charged at
(a) A → s; B → r; C → t; D → p; E → q pH = 7.0?
(b) A → r; B → q; C → s; D → p; E → t Alanine, Lysine,   Cysteine, Glutamic acid
(c) A → q; B → p; C → r; D → s; E → t Glycine, Leucine,   Aspartic acid, Arginine
(d) A → s; B → p; C → t; D → q; E → r
48. Number of correct statements among the following are:
INTEGER TYPE QUESTIONS (i) Glucose is the only aldose that shows mutarotation.
44. Number of chiral centres present in glucopyranose and (ii) The smallest aldose to form a hemiacetal is a tetrose.
fructofuranose are p and q respectively. Here (p + q) = ____. (iii) α-D-glucose  aldehyde form  β-D-glucose. This
conversion is known as mutarotation.
45. Consider an amylose chain of 4000 glucose units. How many
(iv) The anomers of D-glucose have specific rotations of
cleavage are required to lower the average length to 400 units?
same magnitude but opposite sign.
46. How many of the following statements are correct?
(v) Sucrose contains 1,4-glycosidic linkage.
D-glucose, on treating with methanol in presence of dry HCl
(vi) In an alkaline medium, fructose is a reducing sugar.
gives methyl glucosides according to the following reaction
(vii) Cellobiose is a polysaccharide.
H OCH3
CHO (viii) Most natural sugars have D-configuration.
H OH H OH CH2OH
HO H CH3OH HO H O
H OH
H OH dry HCl H OH Br2
H H 49. H Compound (A)
OH OH H H 2O
HO OH
CH2OH CH2OH
H OH
(I) D-glucose (II) Methyl-α-D-glucoside Find number of chiral centre in compound A is_____.

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 PARIKSHIT (JEE ADVANCED LEVEL)

SINGLE CORRECT TYPE QUESTIONS 4. Two aldopentoses X and Y give the same osazone
derivative. X is oxidised to an optically active aldaric
1. The average energy of each hydrogen bond in A–T pair is x acid by dilute nitric acid. Ruff degradation of Y gave a
kcal mol–1 and that in G–C pair is y kcal mol–1. Assuming tetrose which was similarly oxidised to an optically active
that no other interaction exists between the nucleotides, aldaric acid. Assign the structures of X and Y from the
the approximate energy required in kcal mol–1 to split the following list.
following double stranded DNA into two single strands is
A—T—A—T—G—C—A—G (I) CHO (II) CHO
H OH HO H
H OH HO H
T—A—T—A—C—G—T—C OH H OH
H
[Each dashed line may represent more than one hydrogen CH2OH CH2OH
bond between the base pairs]
(a) 10x + 9y (b) 5x + 3y (III) CHO (IV) CHO
(c) 15x + 6y (d) 5x + 4.5y HO H H OH
H OH HO H
2. A nonapeptide in rat on hydrolysis gave the following H OH H OH
identifiable tripeptides: CH2OH CH2OH
Gly-Ala-Phe, Ala-Leu-Val, Gly-Ala-Leu, Phe-Glu-His and
His-Gly-Ala. The sequence in the nonapetide is: (a) X = (I) and Y = (III)
(a) Gly-Ala-Leu-Val-Phe-Glu-His-His-Gly (b) X = (IV) and Y = (II)
(b) Ala-Phe-Leu-Val-Gly-Leu-Phe-Glu-His (c) X = (II) and Y = (IV)
(c) Gly-Ala-Phe-Glu-His-Gly-Ala-Leu-Val
(d) X = (III) and Y = (I)
(d) Phe-Ala-Leu-Val-Gly-Glu-His-Gly-Ala
3. Nitrous acid (HNO2) converts amino acids into hydroxy acids 5. Which of following compound give negative Tollen’s test?
with retention of configuration. Estimation of dinitrogen gas (a) OH
evolved in the reaction is the basis of Van Slyke estimation of OH
amino acids: O
NH2 OH
OH
—
—

HNO2
R—CH—COOH R—CH—COOH+N22 OH
Which of the following amino acids cannot be analysed by OH
Van Slyke method? (b) OH
NH2 OH
—

O
(I) HS—CH2—CH—COOH (Cysteine)
OCH3

(II) (Proline) OH
COOH OH
N
—

H NH2 (c)
—

—CH2—CH—COOH

(III) H—N N (Histidine)

NH2
—

(d)
(IV) CH3—CH—CH—COOH (Valine)
—

CH3
(a) Only (I) (b) Only (II)
(c) (I) and (III) (d) (I), (III), (IV)

Biomolecules 173
MULTIPLE CORRECT TYPE QUESTIONS 11. The correct statement(s) about starch is/are:
(a) It is a pure single compound.
6. (I) CHO (II) CHO
(b) It is mixture of two polysaccharides of glucose.
HO H H OH
HO H (c) It involves the (C1–C4) α-glycosidic linkage between
HO H H OH two α-D-glucose units.
CH2OH CH2OH (d) It involves branching by (C1–C6) glycosidic linkage.
(III) CH2OH 12. The correct structures of glycine at given pH are:
—


C=O (a) H3NCH2—C—OH at pH = 2.0
HO H O
H OH 
(b) H3NCH2—C—O at pH = 6.0
CH2OH
O
The correct statement about the sugars given above are:
(c) H2NCH2—C—O at pH = 9
(a) I and II are L-Sugars (b) II and III are D-Sugars
(c) I and III are D-sugars (d) I is L-sugar O
7. CH2OH (d) at pH = 12
H O OH
H
OH H Glucose 13. The final product of which of the following reactions
HO H furnishes evidence that glucose has unbranched carbon
H OH chain?
The correct statement(s) about above structure of glucose (a) Glucose 
1.Br ,H O
2 2 →
2.Red P + HI
is/are:
(a) It is a pyranose form. (b) It is a furanose form. (b) Glucose 
1. NaBH
4 →
2.Red P + HI
(c) It is a β-anomer. (d) It is a D-sugar.
1.HCN
8. D-Mannose differs from D-glucose in its stereochemistry at (c) Glucose  ⊕
→
2.H3O
C-2. The pyranose form of D-Mannose is
3. Red P + HI
(a) CH2OH (b) CH2OH
3 CH OH,H⊕
H OH H O OH (d) Glucose →
H H
OH HO OH H 14. Which of the following statements are correct for glucose?
HO OH HO H
(a) It gives positive test with Schiff’s reagent.
H H H HO (b) It reacts with NaHSO3 and NH3.
(c) CH2OH (d) CH2OH (c) Pentaacetate derivative of glucose does not react with
H O OH H OH H2N–OH.
H H
OH HO OH H (d) It gives positive test with Fehling solution.
HO H HO OH 15. The phenomenon of mutarotation is shown by:
H H H HO (a) Glucose (b) Fructose
(c) Cellulose (d) Starch
9. The correct statements about anomers is/are
(a) Anomers have different stereochemistry at C-1 16. Identify the compound that gives same osazone:
(anomeric carbon). (a) CHO (b) CHO
(b) α-D-glucopyranose and β-D-glucopyranose are anomers. H OH HO H
HO H HO H
(c) Both anomers of D-glucopyranose can be crystallised H OH
HO H
and purified. H H OH
OH
(d) When pure α-D-glucopyranose is dissolved in water,
its optical rotation slowly changes. CH2OH CH2OH
10. Which of the following pair form the same osazone with (c) CH2OH (d) CHO
phenylhydrazine? O HO H
(a) D-Glucose and D-Fructose HO H HO H
HO H HO H
(b) D-Fructose and D-Mannose H H
OH OH
(c) D-Glucose and D-Mannose
(d) D-Glucose and D-Galactose CH2OH CH2OH

174 P JEE Dropper Module-4 CHEMISTRY

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 17. Identify the compounds which are inter-convertible in basic 19. What is the structure of L-Glucose?
medium?
(a) CHO (b) CHO
(a) CHO (b) CHO HO H H OH
H OH HO H H OH HO H
H OH H OH HO H H OH
H OH H OH HO H H OH
H OH H OH CH2OH CH2—OH
CH2OH CH2OH CHO
(c) (d) None of these
H OH
(c) CH2OH (d) HO H
O H OH
H OH HO H
H OH
CH2—OH
H OH
20. HC—OH
CH2OH
C—OH
HO H
COMPREHENSION BASED QUESTIONS H OH
Comprehension (Q. 18 to 20): It is convenient at times to H OH
represent the cyclic structure of a monosaccharide without CH2OH
specifying whether the configuration of the anomeric carbon atom
The given structure is enol form of
is α or β. When we do this, we shall use formulas such as the
following: (a) D-glucose (b) D-mannose
(c) D-fructose (d) All of these

Comprehension (Q. 21 to 23): Answer the following questions

by appropriately matching the information given in the three
columns of the following table.
Column-I Column-II Column-III
Indicates a or β (three-dimensional view not specified) Reactant in question Group present in No. of stereoisomers
gives positive test reactant in question of reactant in
18. Which of the following represents the anomer of the (Lab Test) question (including
compound shown? given structure)
(I) Tollen’s Reagent (i) Aldehyde (p) 8
HOCH2 O H
(II) 2, 4-DNP Test (ii) a-hydroxy ketone (q) 16
H H
H OH (III) Schiff Reagent (iii) Acetal linkage (r) 4
(IV) NaHSO3 Test (iv) Glycosidic linkage (s) 2
HO OH

(a) HOCH2 O 21. Which of the following will be true for D-Glucose?
H
(a) (I) (i) (q)
OH H
H OH (b) (II) (i) (q)
H OH (c) (I) (ii) (q)
(d) (I) (iv) (p)
(b) HOCH2 O OH
22. A monosaccharide reducing sugar (X) having molecular
H H formula C6H12O6, does not give Br2 + H2O test. Which will
H H
be correct for X?
HO OH
(a) (I) (i) (q) (b) (I) (ii) (p)
(c) (c) (I) (i) (s) (d) (I) (iii) (r)
23. A compound (Z), is epimer of D-glucose at C2. Which of
the following will be correct for Z?
(a) (I) (iv) (q) (b) (I) (ii) (p)
(d) None of these (c) (I) (i) (q) (d) (III) (ii) (s)

Biomolecules 175
Comprehension (Q. 24 to 26): Proteins are biomolecules 25. In the given trimer, if R1 = H; R2 = CH3 and R3 = Ph, then
composed of α-amino acids. An α-amino acid has a general formula the amino acids present in the trimer are:
R–CH–COOH. The amino acids polymerise and form an amide (a) Glycine, Alanine and Phenyl Alanine
(b) Glycine, Leucine and Phenyl Alanine
NH2
linkage (peptide linkage) between two monomeric amino acid units. (c) Alanine, Valine and Phenyl Alanine
The polymerisation takes place as follows: (d) Alanine, Leucine and Lysine
26. Which statement is incorrect about the given trimer?
R1 R2 R3
(a) It will liberate CO2 with NaHCO3.
—

—
H2N—CH—C—OH+H—NH—CH—C—OH+H—NH—CH—C—OH (b) It will liberate N2 with NaNO2/HCl.
O O O (c) It will give yellow precipitate with 2, 4-Dinitrophenyl-
hydrazine.
R1 R2 R3 (d) It will rotate plane polarized light.
—

—
Polymerisation
H2N—CH—C—NH—CH—C—NH—CH—C—OH
(–2H2O)
MATCH THE COLUMN TYPE QUESTIONS
O O O 27. The disaccharides are composed of two molecules of
[A Tripeptide]
Peptide linkage monosaccharides. On hydrolysis with dilute acids or enzymes,
Two or more similar amino acids can also polymerise, for example they yield two molecules of either the same or different
a dimer will be like monosaccharides. Find the correct match from the following:
R R Column-I Column-II
A. Sucrose p. Glucose + Glucose
—

H2N—CH—C—NH—CH—C—OH B. Lactose q. Glucose + Fructose

C. Maltose r. Glucose + Galactose
O O
D. Cellobiose s. Fructose + Galactose
24. In the above trimer, if R1 = H; R2 = CH3 and R3 = Ph, then (a) A→p; B → r; C → s; D → p
total number of optically active stereoisomers will be: (b) A→p; B → q; C → s; D → r
(a) 8 (b) 6 (c) A→q; B → r; C → p; D → p
(c) 4 (d) 2 (d) A→q; B → p; C → r; D → s

28.
Column-I Column-II
CH2OH
H OH
A. H
OH H p. It will undergo osazone formation.
HO OH
H OH
CH2—OH
H O OH
When undergo acetylation reaction with acetic anhydride
B. H
q.
OH H
H molecular weight increases by 210.
HO
H OH
CH2—OH
H O OCH
3
C. H
OH H
r. It is a reducing sugar.
HO H
H OH
CH2—OH CH2—OH
H O O OH
D. H
OH H
H
OH H
s. It is known as a-D-Glucopyranose.
HO O H
H OH H OH
t. It is not a reducing sugar and does not show mutarotation

176 P JEE Dropper Module-4 CHEMISTRY

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 (a) A → p, q, r ; B → p, r ; C → t; D → p, r Find the sum of total number of moles of PhNHNH2 used
(b) A → p, q, r, s; B → p, q, r; C → t; D → p, r with A and B.
(c) A → p, q, r, s; B → p, q, r; C → q; D → p, r
(d) A → p, q; B → p, q, r; C → t; D → p
29. Match the items of Column-I to items of Column-II. There 32. For aspartic acid , the pKa1,
can be single or multiple matches.
Column-I Column-II
A. D(+) glucose p. Lobry de Bruyn-van Ekenstein pKa and pKa are 2, 4 and 10 respectively and for arginine
2 3
rearrangement
B. D(–) fructose q. Red precipitate with Fehling’s
solution , pK a , pK a
1 2
C. (+) maltose r. Exhibits mutarotation
D. (+) sucrose s. Oxidized by Br2/H2O
t. C1–C2 glycosidic linkage and pKa are 2, 9 and 13 respectively. Calculate sum of pI
3
(a) A → p, q, r, s; B → p, q, r; C → q, r, s; D → t of aspartic acid and arginine.
(b) A → p, q, r; B → p, q, s; C → q, r, t; D → t 33. Identify the total number of carbohydrates among the
(c) A → p, r, s; B → p, q, r; C → q, r, s; D → r following which give mutarotation.
(d) A → p, q, r, s; B → p, q, r; C → r; D → t OH OH
OH OH

—
INTEGER TYPE QUESTIONS — O H — O H H O H
H H OH H H HO
30.  

—
H OMe H OH OH
HO

—
Mixture Separation
of amino of amino OH OH H OH HOOH H
acids acids (P) (Q) (R)
HO— O OH H O OH
— —
Buffer Buffer OH H H HO
solution solution H OH H OMe
(pH = 5.0) (pH = 5.0)
H OH OH H
(S) (T)
Filter paper
H OMe HOH2C OMe
Amino acids pKa Values O H OH H OH O
S. H OH H OH
⊕ Side
No. H H
α–COOH α–NH3 chain H OH
H
1. Arginine 2.17 9.04 12.48 CH2OH
2. Alanine 2.34 9.69 –––– (U) (V)
3. Aspartic acid 2.09 9.82 3.86 HOH2C OH HOH2C OMe
HO H O
4. Lysine 2.18 8.95 10.79 H OH
HO
H
H O
OH
5. Glutamic acid 2.19 9.67 4.25 H H
6. Valine 2.32 9.62 –––– H CH2OH
Calculate (W) (X)
Number of non-essential amino acids moving towards cathode = x 34. Identify total number of carbohydrates among the following
Number of essential amino acids moving towards anode = y which give positive Tollen’s test.
Number of amino acids moving towards cathode = z (a) CH2OH (b) CH2OH
Find the value of (x + y + z). OH O OCH
H H
31. 1
CH=O 1 H
3
CH=O H
OH H
OH H
H 2 H H 2
OH HO OH HO H
HO 3 H PhNHNH2 H 3 H PhNHNH2
4 H OH H OH
H OH H 4 OH
H 5
6
OH H 5 OH (c) HOH2C O CH2OH (d) HOH2C O CH2OH
6
CH2OH CH2OH H HO H HO
H OH H OCH3
(A) (B)
(A)
D-2-Deoxy glucose D-3-Deoxy glucose OH H OH H

Biomolecules 177
 OH O O
O
(e) OMe (h) HOCH2—C—CH2OH
O
HO HO OH
HO OH (i) HOH2C O CH2OH

OMe O
O OCOCH3

—
O OH
(f ) HO HO OH OH
HO OH 35. What is the approximate number of glucose unit in the sample
—CH2OH of starch if its aqueous solution of 10 g/L has an osmotic
—

O O pressure of 5 × 10– 3 atm at 25°C?

OH HO
—

—
(g) HO O
—

CH2OH
—

OH OH

PYQ's (PAST YEAR QUESTIONS)

CARBOHYDRATES Choose the most appropriate answer from the options given
below: [27 July, 2022 (Shift 2)]
1. L-isomer of tetrose X (C4H8O4) gives positive Schiff’s (a) A - (IV); B - (I); C - (II); D - (III)
test and has two chiral carbons. On acetylation. ‘X’ yields
(b) A - (IV); B - (III); C - (II); D - (I)
triacetate. ‘X’ also undergoes following reactions
(c) A - (III); B - (I); C - (IV); D - (II)
HNO NaBH

' A ' ←
3
 ' X ' 
4
→ ' B' (d) A - (I); B - (III); C - (IV); D - (II)
Chiral compount
4. Which one among the following chemical tests is used to
‘X’ is [11 April, 2023 (Shift-I)]
distinguish monosaccharide from disaccharide?
CHO  [27 July, 2021 (Shift-I)]
H OH
H OH (a) Seliwanoff’s test (b) Iodine test
(a) H OH (b) (c) Barfoed test (d) Tollen’s test
O HO H
H 5. Which one of the following statements is not true?
CH2OH CH2OH  [6 Sept, 2020 (Shift-II)]
CHO (a) Lactose contains α-glycosidic linkage between C1 of
H OH galactose and C4 of glucose.
HO H (b) Lactose is a reducing sugar and it gives Fehling’s test.
(c) HO H (d)
O HO H (c) On acid hydrolysis, lactose gives one molecule of D(+)-
H glucose and one molecule of D(+)-galactose.
CH2OH (d) Lactose (C11H22O11) is a disaccharide and it contains 8
CH2OH
hydroxyl groups.
2. A sugar ‘X’ dehydrates very slowly under acidic condition 6. Amylopectin is composed of: [10 April, 2019 (Shift-I)]
to give furfural which on further reaction with resorcinol (a) α-D-glucose, C1–C4 and C1–C6 linkages
gives the coloured product after sometime. Sugar ‘X’ is (b) α-D-glucose, C1–C4 and C2–C6 linkages
 [27 July, 2022 (Shift 1)] (c) β-D-glucose, C1–C4 and C2–C6 linkages
(a) Aldopentose (b) Aldotetrose (d) β-D-Glucose, C1–C4 and C1–C6 linkages
(c) Oxalic acid (d) Ketotetrose 7. A disaccharide X cannot be oxidised by bromine water. The
acid hydrolysis of X leads to a laevorotatory solution. The
3. Match Column-I with Column-II.
disaccharide X is [JEE Adv. 2023]
Column-I Column-II (a)
(A) Glucose + HI (I) Gluconic acid
(B) Glucose + Br2 /water (II) Glucose pentacetate
Glucose + acetic Saccharic acid
(C) (III)
anhydride
(D) Glucose + HNO3 (IV) Hexane

178 P JEE Dropper Module-4 CHEMISTRY

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 (b) 9. Given: [JEE Adv. 2021]
CHO
H OH
HO H HNO3
P
H OH
H []D = +52.7º
OH
CH2OH
D-(+)-glucose
(c)
The compound(s), which on reaction with HNO3 will give
the product having degree of rotation, [α]D = –52.7° is/are:
(a) CHO (b)
HO H
HO H
H OH
H OH
CH2OH
(d)
(c) CHO (d) CHO
HO H H OH
H OH H OH
HO H HO H
HO H H OH
CH2OH CH2OH
10. Which of the following statements(s) is/are true?
8. Treatment of D-glucose with aqueous NaOH results in a  [JEE Adv. 2019]
mixture of monosaccharides, which are
(a) The two six-membered cyclic hemiacetal forms of
 [JEE Adv. 2022] D-(+)-glucose are called anomers.
CHO CHO
CHO (b) Hydrolysis of sucrose gives dextrorotatory glucose and
HO H H OH HO H laevorotatory fructose.
HO H HO H HO H
(a) H OH , HO H and HO H (c) Monosaccharides cannot be hydrolysed to give
H OH H OH H OH polyhydroxy aldehydes and ketones.
CH2OH CH2OH CH2OH (d) Oxidation of glucose with bromine water gives glutamic
acid.
CH2OH CHO CHO
O HO H HO H PROTEINS
HO H H OH HO H
(b) H OH , H OH and HO H 11. The naturally occurring amino acid that contains only one
H OH H OH H OH basic functional group in its chemical structure is
 [13 April, 2023 (Shift-II)]
CH2OH CH2OH CH2OH (a) arginine (b) lysine
(c) asparagine (d) histidine
12. The structure of protein that is unaffected by heating is:
 [29 June, 2022 (Shift-II)]
(c) (a) Secondary structure (b) Tertiary structure
(c) Primary structure (d) Quaternary structure
13. The secondary structure of protein is stabilised by:
[16 March, 2021 (Shift-II)]
(d) CHO CHO CHO (a) Peptide bond (b) Glycosidic bond
H HO H
OH H OH
HO H (c) Hydrogen bonding (d) Van der Waals forces
HO H HO H
HO H , H OH and H OH 14. Which of the following is not an essential amino acid?
H OH H OH H OH  [5 Sept, 2020 (Shift-I)]
CH2OH CH2OH CH2OH (a) Tyrosine (b) Valine
(c) Lysine (d) Leucine

Biomolecules 179
 15. The correct match between Item-I and Item-II is: 19. Which one of the following statements is not true about
 [11 Jan, 2019 (Shift-II)] enzymes? [20 July, 2021 (Shift-II)]
Item-I Item-II (a) The action of enzymes is temperature and pH specific.
(A) Ester test (I) Tyr (b) Enzymes are non-specific for a reaction and substrate.
(B) Carbylamine test (II) Asp (c) Enzymes work as catalysts by lowering the activation
(C) Phthalein dye test (III) Ser energy of a biochemical reaction.
(IV) Lys (d) Almost all enzymes are proteins.
(a) (A)-(II); (B)-(IV); (C)-(I) 20. Which of the given statements is INCORRECT about
(b) (A)-(III); (B)-(II); (C)-(I) glycogen? [12 April, 2019 (Shift-II)]
(c) (A)-(III); (B)-(IV); (C)-(II)
(a) It is present in some yeast and fungi
(d) (A)-(II); (B)-(IV); (C)-(III)
(b) It is present in animal cells
16. The structure of a peptide is given below:
(c) Only a-linkages are present in the molecule
NH2
(d) It is a straight chain polymer similar to amylase
HO
VITAMINS
H
O 21. All structures given below are of vitamin C. Most stable of
N OH them is: [1 Feb, 2023 (Shift-II)]
H2N N
O
H
O (a) (b)
CO2H

If the absolute values of the net charge of the peptide at pH
= 2, pH = 6, and pH = 11 are |z1|, |z2|, and |z3|, respectively,
then what is |z1| + |z2| + |z3|? [JEE Adv. 2020]
(c) (d)
ENZYMES AND HORMONES
17. Testosterone, which is a steroidal hormone, has the following
structure.

22. Which of the following is a water soluble vitamin, that is

not excreted easily? [26 June, 2022 (Shift-II)]
(a) Vitamin B2 (b) Vitamin B1
(c) Vitamin B6 (d) Vitamin B12
23. Which of the following vitamin is helpful in delaying the
blood clotting? [26 April, 2021 (Shift-I)]
The total number of asymmetric carbon atom/s in testosterone (a) Vitamin B (b) Vitamin K
is __________. [1 Feb, 2023 (Shift-II)]
(c) Vitamin E (d) Vitamin C
18. Match Column-I with Column-II.
24. Match the following: [7 Jan, 2020 (Shift-I)]
Column-I (Enzyme) Column-II (Conversion)
Column-I Column-II
(A) Invertase I. Starch into maltose
(A) Riboflamin I. Beriberi
(B) Zymase II. Maltose into glucose
(B) Thiamine II. Scurvy
(C) Diastase III. Glucose into ethanol
(D) Maltase IV. Cane sugar into glucose (C) Pyridoxine III. Cheilosis
Choose the most appropriate answer from the options given (D) Ascorbic acid IV. Convulsions
below: [26 June, 2022 (Shift-II)] (a) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(a) (A) - (III), (B) - (IV), (C) - (II), (D) -(I) (b) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(b) (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
(c) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(c) (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(d) (A) - (IV), B - (II), C - (III); D - (I) (d) (A)-(d), (B)-(II), (C)-(I), (D)-(III)

180 P JEE Dropper Module-4 CHEMISTRY

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NUCLEIC ACIDS 28. Which of the following will react with CHCl3 + alc. KOH?
 [4 Sept, 2020 (Shift-I)]
25. Uracil is base present in RNA with the following structure.
(a) Adenine and thymine (b) Thymine and proline
% of N in uracil is ________. [24 Jan, 2023 (Shift-I)]
(c) Adenine and lysine (d) Adenine and proline
O
29. Among the following compounds, which one is found in RNA?
NH  [11 Jan, 2019 (Shift-I)]
NH2
N O
H N
(a)
Given: N O
Molar mass N = 14 g mol–1; O = 16 g mol–1; C = 12 g
H
mol–1; H = 1 g mol–1;
O
26. Sugar moiety in DNA and RNA molecules respectively are
 [29 June, 2022 (Shift-I)] NH
(b)
(a) β-D-2-deoxyribose, β-D-deoxyribose N O
(b) β-D-2-deoxyribose, β-D-ribose
H
(c) β-D-ribose, β-D-2-deoxyribose
O
(d) β-D-deoxyribose, β-D-2-deoxyribose CH3
NH
27. Out of following isomeric forms of uracil, which one is (c)
present in RNA? [27 Aug, 2021 (Shift-I)] N O
OH H
O
O
N
(a) (b) HN N Me
(d)
O N N O
H HO N
Me
O 30. Which of the following statements is not true about RNA?
OH  [12 April, 2019 (Shift-I)]
HN (a) It has always double stranded α-helix structure.
(c) (d) N (b) It usually does not replicate.
O N (c) It is present in the nucleus of the cell.
H HO N (d) It controls the synthesis of protein.

PW CHALLENGERS

PASSAGE I DNA code for proteins. (Each amino acid in a protein is

encoded by a group of three nucleotides).
Deoxyribonucleic acid (DNA) represents the genetic program of
all living beings. The human genetic program is subdivided into 3. The DNA of the bacteriophage M13 shows the following
23 chromosomes. base composition (in mole %):
1. Calculate the mass of a DNA thread in milligrams, which A: 23 %, T: 36 %, G: 21 %, C: 20 %
reaches from earth to the moon (340,000 km). A mass of 1 g What does the base composition tell about the structure of
represents 1,000 nucleotide pairs. One nucleotide pair (base the DNA?
pair) has a length of 0.34 nm.
2. Give estimation (in terms of 109) on how many nucleotide PASSAGE II
pairs are stored in the chromosome set of a human being. The sequence of the amino acids in a peptide can be determined by
Human cells can synthesize 50,000 different proteins, which a combination of chemical and enzymatic methods. The peptide in
are on the average 300 amino acids long. Only 2 % of the question functions in the human body as a pain reliever.

Biomolecules 181
 (a) Hydrolysis of the peptide in 6 M HCl at 110°C followed by
NH2
an analysis of the liberated amino acids, resulted in a molar Na+ –O P O

ratio of Gly, Leu, and aromatic amino acids Phe, Tyr = 2 : O

N
Cytosine
1 : 1 : 1. O
CH2 N
O
(b) Reacting the peptide with 2,4-dinitrofluorobenzene
H H
(DNFB), followed by hydrolysis and chromatographic
H H
analysis, yielded the tyrosine derivative. O H

(c) Partial hydrolysis with chymotrypsin yielded Leu, Tyr Na+ –O P O

O
and a smaller peptide. After hydrolysis of this peptide HN
O Thymine
Gly and Phe were identified in a ratio 2 : 1. Chymotrypsin
is a protease which cleaves a peptide bond following an CH2
O
O N
aromatic amino acid. H H

4. Determine the amino acid sequence from the given H H

O H
information.
5. In a similar peptide which shows the same biological activity,
leucine is replaced by methionine. Explain from the chemical 7. Assuming that equimolar amounts of the four bases are
structure of both amino acids why the replacement is possible present in DNA, write the number of H atoms per P atom.
without loss of biological activity.
8. Calculate (to 3 significant figures) the theoretical weight
PASSAGE III percentage of H expected upon elemental analysis of DNA.
The following reaction scheme represents part of anaerobic 9. Chargaff extracted the separated bases and determined their
degradation of saccharides, i.e. the glycolysis, involving concentrations by measuring UV absorbance. The Beer-
equilibrium constants K1 and K2: Lambert law was used to obtain the molar concentration.
glucose-1-phosphate  glucose-6-phosphate, K1 = 19 Chargaff discovered the following molar ratio for bases in
glucose-6-phosphate  fructose-6-phosphate, K2 = 0.50 DNA:
6. In the beginning of the reaction the reaction mixture contained adenine to guanine = 1.43, thymine to cytosine = 1.43
1 mmol of glucose-6-phosphate. Calculate the moles of
adenine to thymine = 1.02, guanine to cytosine = 1.02
glucose-6-phosphate, glucose-1-phosphate and fructose-6-
phosphate respectively in the mixture at equilibrium. (As Chargaff’s discovery suggested that the bases might exist
the reaction take place in a constant volume, the ratio of the as pairs in DNA. Watson and Crick mentioned in their
moles of substances equals that of their concentrations.) celebrated 1953 paper in Nature: “It has not escaped
our notice that the specific pairing we have postulated
PASSAGE IV
immediately suggests a possible copying mechanism for the
The Chemistry of DNA
genetic material.” Draw structures of the specific pairing
In 1944 Oswald Avery isolated a genetic material and showed by
found in DNA. Indicate hydrogen bonds. Omit the sugar-
elemental analysis that it was a sodium salt of deoxyribonucleic
acid. A segment of DNA with formula mass of 1323.72 is shown. phosphate backbone.

O 10. The plausibility of the formation of purine and pyrimidine

bases in the prebiotic atmosphere of the Earth from HCN,
Na+ –O P O NH2 NH3, and H2O has been demonstrated in the laboratory.
N
O N
Adenine Write the minimum number of HCN and H2O molecules
CH2 O N N
respectively required for formation of the following
H H compounds.
H H
O H

Na+ –O P O O
N
O NH
Guanine
CH2 O N N NH2

H H
H H
O H I II III IV

182 P JEE Dropper Module-4 CHEMISTRY

W
 Answer Key

CONCEPT APPLICATION
1. (a) 2. (b) 3. (a) 4. (c) 5. (b,c) 6. (b) 7. (a,c) 8. (a) 9. (a,b) 10. (a,b,c,d)

PRARAMBH (TOPICWISE)
1. (b) 2. (b) 3. (a) 4. (d) 5. (d) 6. (c) 7. (b) 8. (b) 9. (d) 10. (d)
11. (a) 12. (b) 13. (d) 14. (d) 15. (c) 16. (c) 17. (a) 18. (a) 19. (b) 20. (b)
21. (b) 22. (a) 23. (a) 24. (c) 25. (c) 26. (d) 27. (b) 28. (d) 29. (b) 30. (d)
31. (a) 32. (d) 33. (b) 34. (b) 35. (c) 36. (a) 37. (d)

PRABAL (JEE MAIN LEVEL)

1. (d) 2. (a) 3. (b) 4. (d) 5. (d) 6. (c) 7. (c) 8. (b) 9. (d) 10. (c)
11. (b) 12. (a) 13. (a) 14. (a) 15. (a) 16. (b) 17. (b) 18. (d) 19. (b) 20. (c)
21. (c) 22. (d) 23. (a) 24. (b) 25. (c) 26. (b) 27. (d) 28. (b) 29. (c) 30. (d)
31. (a) 32. (b) 33. (a) 34. (b) 35. (b) 36. (c) 37. (b) 38. (a) 39. (d) 40. (d)
41. (c) 42. (b) 43. (a) 44. [9] 45. [9] 46. [3] 47. [6] 48. [4] 49. [4]

PARIKSHIT (JEE ADVANCED LEVEL)

1. (a) 2. (c) 3. (b) 4. (c) 5. (c) 6. (b, d) 7. (a,c,d) 8. (a, c) 9. (a,b,c,d) 10. (a,b,c)
11. (b,c,d) 12. (a,b,c) 13. (a,b,c) 14. (c,d) 15. (a,b) 16. (a,c,d) 17. (a,b,c) 18. (b) 19. (a) 20. (d)
21. (a) 22. (b) 23. (c) 24. (c) 25. (a) 26. (c) 27. (c) 28. (b) 29. (a) 30. [5]
31. [4] 32. [14] 33. [4] 34. [5] 35. [302]

PYQ’ (PAST YEAR QUESTIONS)

1. (b) 2. (a) 3. (a) 4. (c) 5. (a) 6. (a) 7. (a) 8. (c) 9. (c,d)
10. (a,b,c) 11. (c) 12. (c) 13. (c) 14. (a) 15. (a) 16. [5] 17. [6] 18. (c) 19. (b)
20. (d) 21. (a) 22. (d) 23. (b) 24. (a) 25. [25] 26. (b) 27. (c) 28. (c) 29. (b)
30. (a)

PW CHALLENGERS
1. [1] 2. [2.25] 4. [Tyr-Gly-Gly, Phe-Lev] 6. [0.644, 0.034, 0.322] 7. [11.3] 8. [3.43]
10. [I-5,0; II-5,1; III-4,2; IV-4,1]

Biomolecules 183