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CHEMISTRY BIOMOLECULES

BIOMOLECULES

Introduction:

In this Unit, Structures and functions of some of biomolecules will be discuss. The structure and
functions of biomolecules inside the living being is studied in biochemistry. Living systems are
made up of various complex biomolecules such as carbohydrates, proteins, enzymes, lipids,
vitamins, hormones, nucleic acids and compounds for storage and exchange of energy such as
ATP, etc.

Carbohydrates:

Classification of Carbohydrates
On the basis of their behaviour upon hydrolysis, carbohydrates can be divided into three main
groups :
i. Monosaccharides : A carbohydrate which cannot be hydrolyzed into simpler unit of
polyhydroxy aldehyde or ketone is called monosaccharide. About 20 monosaccharides are
known to occur in nature. e.g., glucose, fructose, ribose etc.
ii. Oligosaccharides : A carbohydrate which upon hydrolysis yields 2–10 unit of monosaccharide
is called oligosaccharide. They are further classified as disaccharides, trisaccharides, etc.,
depending upon the number of monosaccharides, they provide on hydrolysis. For example,
sucrose is a disaccharide which on hydrolysis yields two unit of monosaccharides i.e., glucose
and fructose whereas raffinose is a trisaccharide which on hydrolysis yields three unit of
monosaccharides i.e., glucose, fructose and galactose.
iii. Polysaccharides: A high molecular mass carbohydrate which upon hydrolysis yields a large
number of monosaccharide units is called polysaccharide e.g., starch, cellulose, glycogen,
gums, etc.
(C6H10O5)n + nH2O ⟶ nC6H12O6

Sugar and non-sugars : In general monosaccharides and oligosaccharides, are crystalline solids,
soluble in water and sweet to taste, are collectively known as sugars. The polysaccharides, on
the other hand, are amorphous insoluble in water and tasteless, are known as non-sugars.
Reducing and non-reducing carbohydrates : The carbohydrates containing free aldehydic or
ketonic group can reduce Fehling’s solution and Tollen’s reagent are known as reducing
carbohydrates. All monosaccharides whether aldose or ketose are reducing in nature. The
carbohydrates in which the reducing parts are not free cannot reduce Fehling’s solution and

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Tollen’s reagent are known as non-reducing carbohydrates. All polysaccharides like starch,
cellulose, glycogen etc. are non-reducing carbohydrates.

i. Monosaccharides
If a monosaccharide contains an aldehyde group, it is known as an aldose and if it contains a
keto group, it is known as a ketose.

Glucose
Glucose occurs in nature in free as well as in the combined forms. It is present in sweet fruits and
honey. Ripe grapes contain ~20% of glucose.

Preparation of Glucose
1. From Sucrose (Cane Sugar): When sucrose is boiled with dilute HCl or H2SO4 in alcoholic
solution, glucose and fructose are obtained in equimolar proportion.
C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6
2. From Starch : When starch is boiled with dilute H2SO4 at 393 K under pressure, glucose is
obtained. This is the commercial method for the preparation of glucose.
(C6H10O5)n + nH2O ⟶ nC6H12O6

Structure of Glucose: Glucose is an aldohexose and is the monomer of many larger

carbohydrates like starch, cellulose etc. It is the most abundant organic compound on the Earth.

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Cyclic Structure of Glucose: It was proposed that glucose can form a six-membered ring in which
–OH at C-5 can add to the –CHO group and can form a cyclic hemiacetal structure. This explains
the absence of –CHO group and also the existence of glucose in α and β-anomeric forms as

The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group
at C-1, called anomeric carbon and the corresponding α and β-forms are called anomers. It
should be noted that α and β-forms of glucose are not mirror images of each other, hence are
not enantiomers.

Fructose

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Fructose is an important ketohexose. It is obtained by the hydrolysis of sucrose. On the basis of

molecular weight determination, elemental analysis and various reaction its molecular formula is
found to be C6H12O6 and open chain structure of it can be written as

Fructose also exists in two cyclic forms like glucose i.e., α-D-(–) - fructose and β-D- (–) - fructose.
The five membered cyclic structure of fructose is formed by the involvement of –OH at C-5 and
carbonyl group. The five-membered ring of fructose is named as furanose with analogy to the
compound furan.

ii. Disaccharides

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The disaccharides are composed of two units of monosaccharides. On hydrolysis with dilute
acids or specific enzymes they give the corresponding monomers.
C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6
In disaccharides the two monosaccharides units are joined together by an oxide linkage formed
by the loss of a water molecule and the linkage is known as glycosidic linkage.
a) Sucrose
Sucrose is formed by the glycosidic linkage between C-1 of α-D-(+)-glucose and C-2 of β-D-(–)
fructose:

b) Maltose
Maltose is formed by the glycosidic linkage between C-1 of one glucose unit to the C-4 of
another glucose unit.

c) Lactose
Lactose is found in milk so it is also known as milk sugar. It is formed by the glycosidic linkage
between C-1 of α-D-galactose unit and C-4 of β-D-glucose unit. Lactose is a reducing sugar.

iii. Polysaccharides

Polysaccharides are long chain polymer of monosaccharides joined together by glycosidic

linkages. For example, starch, cellulose, glycogen etc. They mainly act as the food storage or

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structural materials.
Starch (C6H10O5)n
Starch is the main storage polysaccharide of plants. High content of starch is found in cereals,
roots, tubers and some vegetables. Starch is a polymer of α-D-(+) Glucose coming of two
components namely Amylose and Amylopectin.
Amylose is water soluble component, which constitutes about 15 - 20% of starch. It is a straight
chain polysaccharide containing α-D-(+)-glucose units joined together by β-glycosidic linkage
involving C-1 of one glucose unit and C-4 of the next.

Amylopectin is a branched chain polysaccharide insoluble in water. It constitutes about 80 – 85%

starch. It is a branched chain polymer of α-D-glucose units in which chain is formed by C-1 - C-4
glycosidic linkage whereas branching occurs by C-1 – C-6 glycosidic linkage.

Cellulose
Cellulose is a straight chain polysaccharide composed of only β-D-glucose units. In cellulose there
is β-glycosidic linkages between C-1 of one glucose unit and C-4 of the next glucose unit.
Cellulose occurs mainly in plants and it is the most abundant organic substance in plant kingdom.

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Glycogen
Its structure is similar to amylopectin with more branching than in amylopectin. It is also known
as animal starch. In body, carbohydrates are stored as glycogen and when the body needs
glucose, enzymes break the glycogen down to glucose. Glycogen is present in liver, muscle and
brain.
Note : Carbohydrates are essential for life in both plants and animals. Carbohydrates are stored
in plant as starch and in animals as glycogen.

Proteins:

Proteins are high molecular mass complex biopolymer of α-amino acids present in all living cells.
They occur in every part of the body and form the fundamental basis of structure and functions
of life. The term protein is derived from the Greek word “proteios” which means of prime
importance. Proteins are the most abundant biomolecules of the living system. Chief sources of
proteins are milk, cheese, pulses, peanuts, fish etc.
Amino Acids : The compound containing –NH2 and –COOH functional groups are known as
amino acid, depending upon the relative position of –NH 2 group with respect to –COOH group,
amino acids are classified into α, β, γ, δ and so on amino acid. Hydrolysis of proteins gives only α-
amino acids represented as

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Essential and non-essential amino acids : The amino acids which cannot be synthesized in the
body are known as essential amino acids which must be taken through diet. The amino acids,
which can be synthesized in the body are known as non-essential amino acids.
Peptides : When amino acids are joined together by amide bonds, they form larger molecules
called peptides and proteins.

Polypeptide : A dipeptide contains two amino acids linked by one peptide linkage, a tripeptide
contains three amino acids linked by two peptide linkages and so on. When number of such
amino acids is more than ten, then the products are called polypeptides.

Classification of Protein

On the basis of molecular shape, proteins are classified into two types :
1) Fibrous Proteins : When the polypeptide chains run parallel and are held together by
hydrogen and disulphide bonds, then fibre-like structure is formed, known as fibrous
proteins. Such proteins are insoluble in water. For example: Keratin, Myosin etc.
2) Globular Proteins : When the polypeptide chains coil around to give a spherical shape, the
formation of globular protein takes place. Such proteins are usually soluble in water. For
example : Insulin, Albumins etc.
Primary, Secondary, Tertiary & Quaternary Structures of Proteins :
1) Primary Structure : Proteins may have one or more polypeptide chains. Each polypeptide
in a protein has amino acids linked with each in a specific sequence and it is this sequence
of amino acids that is said to be the primary structure of that protein.
2) Secondary Structure : The secondary structure of protein refers to the shape in which a
long polypeptide chain can exist. They are found to exist in two different types of structure
namely α-helix and β-pleated sheet structure.

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3) Tertiary Structure : The tertiary structure of proteins represents overall folding of the
polypeptide chains i.e., further folding of secondary structure. It gives rise to two major
molecular shapes namely fibrous and globular.
4) Quaternary Structure : Some of the proteins are composed of two or more polypeptide
chains referred to as sub units. The spatial arrangement of these subunits with respect to
each other is known as quaternary structure.
Denaturation of Proteins : The loss in biological activity of a protein due to unfolding of globules
and uncoiling of helix is called denaturation of protein. During denaturation secondary and
tertiary structures are destroyed but primary structure remains intact. The coagulation of egg
white on boiling is a common example of denaturation.

Enzymes:

Colloidal solution of protein which works as biological catalyst is known as enzyme. All enzymes
are globular proteins. Zymase, Invertase, Maltase, Lactase, Emulsin, Urease, Pepsin, Trypsin, α-
Amylase etc are the example of enzyme.
Note : The enzymes work best at an optimum temperature range of 298 K to 313 K. Their activity
decreases with decrease or increase in temperature and stops at 273 K.

Vitamins:

Vitamins are organic compounds which are essential for normal growth of life for animals, some
bacteria and micro organism. Vitamins are not synthesized by animals (except vitamin D).
Vitamins are supplied to the organism through food. They are essential dietary factor.

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Classification
Vitamins are classified in two categories :
1) Water Soluble Vitamins : Vitamin-B-complex and vitamin-C are water soluble.
2) Fat Soluble Vitamins : Vitamin-A (Retinol), Vitamin-D (Calciferol), Vitamin-E (Tocopherol),
Vitamin-K (Phylloquinone).

Nucleic Acid:

The particles in nucleus of the cell, responsible for heredity are called chromosomes which are
made up of proteins and another type of biomolecules called nucleic acid. These are natural
biopolymers made of nucleotide units i.e., polynucleotides. Nucleic acid contain the elements
carbon, oxygen, nitrogen and phosphorous.

Hormones:

Hormones are molecules that act as intercellular messengers. These are produced by endocrine
glands in the body and are poured directly in the blood stream which transports them to the site
of action. Hormones have several functions in the body. They help to maintain the balance of
biological activities in the body. Testosterone is the major sex hormone produced in males.

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 Mind map : learning made simple Chapter - 14
Optically active polyhydroxy aldehydes or ketones or compounds which produce such
(Polymers of –amino acids)
units on hydrolysis.
Amino acids contain –NH2 and –COOH group.
Classification:
Classification:
(I) Monosaccharides : (Aldehyde group – aldose, keto group –ketose)
On the basis of relative number of –NH2 and –COOH group
Glucose : Preparation :
(I) Neutral–equal number of –NH2 and –COOH group. +
(a) From sucrose : C12H22O11 + H2O H C6H12O6 + C6H12O6
(ii) Basic – more number of –NH2 than –COOH group. Sucrose Glucose Fructose
+
(iii) Acidic – more number of –COOH than –NH2 group. (b) From starch : (C6H10O5)n + nH2O H nC6H12O6
On the basis of place of synthesis 393K; 2-3 atm HI, CH3CH2CH2CH2CH3
(i) Essential – cannot be synthesized in the body. Structure:

es
CHO
(ii) Non-essential – synthesized in the body. CH=N–OH

rat
(CHOH)4

yd
On the basis of shape
NH 2
OH (CHOH)4
(I) Fibrous – fibre –linke structure CH2OH

boh
(ii) Globular – spherical Peptide linkage CHO CH2OH

Car
Structure : H2N – CH2 – CO–NH – CH –COOH Cylcic Structure CN
CH

–
(CHOH)4
OH
CH3 CH2OH CH2OH
Denaturation of proteins : O CH2OH (CHOH)4
H H H O OH HCN
When a protein in its native form is subjected to physical change, globules OH OH H
H CH2OH
unfold, helix get uncoiled and protein loses its biological activity. OH OH OH H
Pyran H OH H OH COOH
–D–(–) –D–(+)– Br2Water (CHOH)4
Glucopyranose Glucopyranose
(CH2OH)

Structure of Fructose CHO O

Biomolecules
=

Acetic
O HOH2C O CH2OH HOH2C O OH
(CH–O–C–CH3)4
Anhydride O
=

H H OH OH H H OH CH2OH CH2–O–C–CH3
Organic compounds required in diet in small amounts to perform specific biological Furan OH H OH H COOH
functions for maintenance and growth. –D–(–)– –D–(–)– Oxidation
Classification: fructofuranose fructofuranose (CHOH)4
(i) Fat soluble : Soluble in fats and oils but insoluble in water. (vitamins A,D,E and K) COOH

En
(ii) Water soluble : B group and vitamin C are soluble in water. (ii) Disaccharides : Linkage between 2 monosaccharides– Glycosidic linkage (Sucrose, maltose)

zym
(iii) Polysaccharides : Large number of monosaccharides units joined by glycosidic linkages.

es
(a) Starch : Polymer of –glucose with two components amylase and amylopectin
(b) Cellulose
(c) Glycogen
Importance:
Chromosomes : Particles in nucleus responsible for heredity. Chromosomes are Form a major portion of food.
made up of proteins and nucleic acid. As storage molecules.
Two types : Deoxyribonucleic acid (DNA), ribonucleic acid (RNA) Cellulose forms cell wall of bacteria and plants.
Composition : In DNA, sugar is –D–2–deoxyribose whereas in RNA is –D–ribose. Raw materials for industries like textiles, paper, lacquers and breweries.
DNA contains A,G,C,T whereas RNA has A,G,C,U.
Structure : –
Nucleoside : Formed by attachment of a base to 1’ of sugar’ Globular proteins specific for particular
Nucleotide : Formed by link to phosphoric acid at 5’ of sugar. reaction and for particular substrate.
Base Base Base Mechanism : Reduces the magnitude

–
–
–
–Sugar–Phosphate– Sugar–Phosphate –Sugar–
  of activation energy

Types of RNA : m–RNA, r–RNA, t–RNA

Biological Functions :
 Chemical basis of heredity.
 Responsible for identity of different species of organisms.
 Nucleic acids are responsible for protein synthesis in cell.

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Important Questions
Multiple Choice questions-
1.The linkage which holds various amino acid units in primary structure of proteins is
(a) glycosidic linkage
(b) hydrogen bond
(c) peptide linkage
(d) ionic bond
2. Vitamin A is called
(a) Ascorbic acid
(b) Retinol
(c) Calciferol
(d) Tocoferol
3.The deficiency of vitamin B1 causes which disease?
(a) Beriberi
(b) Rickets
(c) Anaemia
(d) Xerosis
4.Deficiency of vitamin C causes
(a) Scurvy
(b) Rickets
(c) Anaemia
(d) None of these
5.An example of non-reducing sugar is
(a) Sucrose
(b) Lactose
(c) Maltose
(d) None
6.Which of the following is not an essential amino acid?
(a) Glycine
(b) Lysine
(c) Phenyl alanine
(d) Valine
7.Which of the following is a water-soluble vitamin?
(a) Vitamin E

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(b) Vitamin K
(c) Vitamin B
(d) Vitamin A
8.Vitamin B1 is
(a) Riboflavin
(b) Cobalamin
(c) Thiamine
(d) Pyridoxine
9.Which is sweetest of the following:
(a) Sucrose
(b) Glucose
(c) Fructose
(d) Maltose
10.Rickets may be caused by the deficiency of which vitamin?
(a) Vitamin D
(b) Vitamin C
(c) Vitamin A
(d) Vitamin B

Very Short Question:

1. Give some examples of bimolecules
2. What are carbohydrates?
3. Give one example of each- Monosaccharide, disaccharide and polysaccharide
4. Which disaccharides are non – reducing sugars?
5. Classify the following as monosaccharides disaccharides and polysaccharides Glucose,
Sucrose, maltose, ribose, glycogen, lactose, fructose.
6. What is the meaning of statement- Glucose is an aldohexose.
7. Why are polysaccharides considered non- sugars?
8. Give two examples of reducing sugars.
9. Which sugar is present in milk?
10.Name the reagents used to check the reducing nature of carbohydrates.
Short Questions:
1. What are the expected products of hydrolysis of lactose?
2. The melting points and solubility in water of amino acids are generally higher than
that of the corresponding halo acids. Explain.

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3. When RNA is hydrolysed, there is no relationship among the quantities of different
bases obtained. What does this fact suggest about the structure of RNA?
4. What are monosaccharides?
5. What do you understand by the term glycosidic linkage?
6. What are the hydrolysis products of (i)sucrose and (ii)lactose?
7. What happens when D-glucose is treated with the following reagents?
8. Enumerate the reactions of D-glucose which cannot be explained by its open chain
structure.
9. Differentiate between globular and fibrous proteins.
10.How do you explain the amphoteric behavior of amino acids?
Long Questions:
1. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
2. What is the basic structural difference between starch and cellulose?
3. Define the following as related to proteins
4. What are the common types of secondary structure of proteins?
5. Write the important structural and functional differences between DNA and RNA.
Assertion and Reason Questions:
1. In these questions, a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices.
a) Assertion and reason both are correct statements and reason is correct explanation for
assertion.
b) Assertion and reason both are correct statements but reason is not correct explanation
for assertion.
c) Assertion is correct statement but reason is wrong statement.
d) Assertion is wrong statement but reason is correct statement.
Assertion: Uracil occurs in DNA.
Reason: DNA undergoes replication.
2. In these questions, a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices.
a) Assertion and reason both are correct statements and reason is correct explanation for
assertion.
b) Assertion and reason both are correct statements but reason is not correct explanation
for assertion.
c) Assertion is correct statement but reason is wrong statement.
d) Assertion is wrong statement but reason is correct statement.

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Assertion: Cysteine can cross link peptide chains.
Reason: Amino acids are classified as essential and non-essential amino acids.
Case Study Questions:
1. Read the passage given below and answer the following questions:
When a protein in its native form, is subjected to physical changes like change in
temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due
to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is
called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures
remains intact. Examples of denaturation of protein are coagulation of egg white on boiling,
curdling of milk, formation of cheese when an acid is added to milk.
The following questions are multiple choice questions. Choose the most appropriate
answer:
(i) Mark the wrong statement about denaturation of proteins.
a) The primary structure of the protein does not change.
b) Globular proteins are converted into fibrous proteins.
c) Fibrous proteins are converted into globular proteins.
d) The biological activity of the protein is destroyed.
(ii) Which structure(s) of proteins remains(s) intact during denaturation process?
a) Both secondary and tertiary structures.
b) Primary structure only.
c) Secondary structure only.
d) Tertiary structure only.
(iii) α-helix and β-pleated structures of proteins are classified as:
a) Primary structure.
b) Secondary structure.
c) Tertiary structure.
d) Quaternary structure.
(iv) Cheese is a:
a) Globular protein.
b) Conjugated protein.
c) Denatured protein.
d) Derived protein.
(v) Secondary structure of protein refers to:
a) Mainly denatured proteins and structure of prosthetic groups.
b) Three-dimensional structure, especially the bond between amino acid residues that
are distant from each other in the polypeptide chain.
c) Linear sequence of amino acid residues in the polypeptide chain.

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d) Regular folding patterns of continuous portions of the polypeptide chain.
2. Read the passage given below and answer the following questions:
Carbohydrates are polyhydroxy aldehydes and ketones and those compounds which on
hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not
hydrolysed are called monosaccharides. Monosaccharides with aldehydic group are called
aldose and those which free ketonic groups are called ketose. Carbohydrates are optically
active. Number of optical isomers = 2n
Where n = number of asymmetric carbons. Carbohydrates are mainly synthesised by plants
during photosynthesis. The monosaccharides give the characteristic reactions of alcohols
and carbonyl group (aldehydes and ketones). It has been found that these monosaccharides
exist in the form of cyclic structures. In cyctization, the -OH groups (generally C5 or C4 in
aldohexoses and C5 or C6 in ketohexoses) combine with the aldehyde or keto group. As a
result, cyclic structures of five or six membered rings containing one oxygen atom are
formed, e.g., glucose forms a ring structure. Glucose contains one aldehyde group, one IO
alcoholic group and four 2° alcoholic groups in its open chain structure.
The following questions are multiple choice questions. Choose the most appropriate
answer:
(i) First member of ketos sugar is:
a) Ketotriose.
b) Ketotetrose.
c) Ketopentose.
d) Ketohexose.
(ii) In CH2OHCHOHCHOHCHOHCHOHCHO, the number of optical isomers will be:
a) 16
b) 8
c) 32
d) 4
(iii) Some statements are given below:
1. Glucose is aldohexose.
2. Naturally occurring glucose is dextrorotatory.
3. Glucose contains three chiral centres.
4. Glucose contains one 1° alcoholic group and four 2° alcoholic groups.
Among the above, correct statements are:
a) 1 and 2 only
b) 3 and 4 only
c) 1, 2 and 4 only
d) 1, 2, 3 and 4
(iv) Two hexoses fonn the same osazone, find the correct statement about these hexoses.
a) Both of them must be aldoses.
b) They are epimers at C-3.

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c) The carbon atoms I and 2 in both have the same configuration.
d) The carbon atoms 3, 4 and 5 in both have the same configuration.
(v) Which of the following reactions of glucose can be explained only by its cyclic structure?
a) Glucose forms cyanohydrin with HCN.
b) Glucose reacts with hydroxylamine to form an oxime.
c) Pentaacetate of glucose does not react with hydroxylamine.
d) Glucose is oxidised by nitric acid to gluconic acid.

Answers key
MCQ answers:
1. Answer: (c) peptide linkage
2. Answer: (b) Retinol
3. Answer: (a) Beriberi
4. Answer: (a) Scurvy
5. Answer: (a) Sucrose
6. Answer: (a) Glycine
7. Answer: (c) Vitamin B
8. Answer: (c) Thiamine
9. Answer: (c) Fructose
10. Answer: (a) Vitamin D

Very Short Answers:

1. Examples of biomolecules –
carbohydrates, proteins, Nucleic acids, Lipids, enzymes etc.
2. Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds
which produce such units on hydrolysis.
3. Answer:
 Monosaccharide – Glucose, Fructose etc.
 Disaccharide – Sucrose, maltose etc.
 Polysaccharide – Cellulose, starch etc.
4. Answer: In disaccharides, if the reducing groups of monosaccharides, i.e. aldehydic or
ketonic groups are bonded eg. In sucrose, these are non- reducing.
5. Answer:
Monosaccharide Disaccharides Polysaccharides

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Glucose Sucrose Glycogen

Fructose Maltose

Ribose Lactose
6. Glucose is an aldohexose means that it contains six carbon atoms and aldehyde group.
7. Answer: Polysaccharides are not sweet in taste & hence are called non – sugars.
8. Answer: Examples of reducing sugars: Maltose and Lactose.
9. In milk, lactose is present.
10. Tollen’s reagent and Fehlings solution can be used to check reducing nature of sugars.
Short Answers:
1. Answer
Lactose is composed of -D-galactose and -D-glucose. Thus, on hydrolysis, it gives -D-
galactose
and -D-glucose.

2. Answer:
Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of
amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino
group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion.

Due to this dipolar behaviour, they have strong electrostatic interactions within them and
with water. But halo-acids do not exhibit such dipolar behaviour.
For this reason, the melting points and the solubility of amino acids in water is higher
than those of the corresponding halo-acids.
3. Answer:
A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always
pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of

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DNA, the quantity of adenine produced is equal to that of thymine and similarly, the
quantity of cytosine is equal to that of guanine.
But when RNA is hydrolyzed, there is no relationship among the quantities of the
different bases obtained. Hence, RNA is single-stranded.
4. Answer :

Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler

units of polyhydroxy aldehyde or ketone.
Monosaccharides are classified on the bases of number of carbon atoms and the
functional group present in them. Monosaccharides containing an aldehyde group are
known as aldoses and those containing a keto group are known as ketoses.
Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and
heptoses according to the number of carbon atomsthey contain. For example, a ketose
containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is
called aldotriose.
5. Answer
Glycosidic linkage refers to the linkage formed between two monosaccharide units
through an oxygen atom by the loss of a water molecule.
For example, in a sucrose molecule, two monosaccharide units, -glucose and -
fructose, are joined together by a glycosidic linkage.

6. Answer:
(i) On hydrolysis, sucrose gives one molecule of -D glucose and one molecule of -
fructose.

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(ii) The hydrolysis of lactose gives -galactose and -glucose.

7. Answer:
(i) When D-glucose is heated with HI for a long time, n-hexane is formed.

(ii) When D-glucose is treated with water, D- gluconic acid is produced.

(iii) On being treated with , D-glucose get oxidised to give saccharic acid.

8. Answer :
(1) Aldehydes give 2, 4-DNP test, Schiff’s test, and react with to form the
hydrogen sulphite addition product. However, glucose does not undergo these reactions.
(2) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a
free -CHO group is absent from glucose.
(3) Glucose exists in two crystalline forms – and . The form (m.p. = 419 K)
crystallises from a concentrated solution of glucose at 303 K and the form (m.p = 423 K)
crystallises from a hot and saturated aqueous solution at 371 K. This behavior cannot be
explained by the open chain structure of glucose.

9. Answer :

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Fibrous protein Globular protein
It is a fibre-like structure formed
The polypeptide chain in this
by the polypeptide chain. These
protein is folded around
1. proteins are held together by 1.
itself, giving rise to a
strong hydrogen and disulphide
spherical structure.
bonds.
2. It is usually insoluble in water. 2. It is usually soluble in water.
Fibrous proteins are usually used
All enzymes are globular
for structural purposes. For
proteins. Some hormones
3. example, keratin is present in nails 3.
such as insulin are also
and hair; collagen in tendons; and
globular proteins.
myosin in muscles.
10.Answer:
I n aqueous solution, the carboxyl group of an amino acid can lose a proton and the
amino group can accept a proton to give a dipolar ion known as zwitter ion.

Therefore, in zwitter ionic form, the amino acid can act both as an acid and as a base.

Thus, amino acids show amphoteric behaviour.

Long Answers:
1. Answer:
D-glucose reacts with hydroxylamine(NH2OH) to form an oxime because of the presence of
aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose
forms an open chain structure in an aqueous medium, which then reacts with NH2OH to
give an oxime.

But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate
does not form an open chain structure.

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2. Answer:
Starch consists of two components – amylose and amylopectin. Amylose is a long linear
chain of∝-D-(+)-glucose units joined by C1-C4 glycosidic linkage (∝--link).

Amylopectin is a branched-chain polymer of ∝-D-glucose units, in which the chain is formed

by C1-C4 glycosidic linkage and the branching occurs by C1-C6 glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of 𝛽 -D-glucose units joined

by C1-C4 glycosidic linkage (𝛽 -link).

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3. Answer:
(i) Peptide linkage:
The amide formed between -COOH group of one molecule of an amino acid and -NH2 group
of another molecule of the amino acid by the elimination of a water molecule is called a
peptide linkage.

(ii) Primary structure:

The primary structure of protein refers to the specific sequence in which various amino
acids are present in it, i.e., the sequence of linkages between amino acids in a polypeptide
chain. The sequence in which amino acids are arranged is different in each protein. A
change in the sequence creates a different protein.
(iii) Denaturation:
In a biological system, a protein is found to have a unique 3-dimensional structure and a
unique biological activity. In such a situation, the protein is called native protein. However,
when the native protein is subjected to physical changes such as change in temperature or
chemical changes such as change in pH, its H-bonds are disturbed. This disturbance unfolds
the globules and uncoils the helix. As a result, the protein loses its biological activity. This
loss of biological activity by the protein is called denaturation. During denaturation, the
secondary and the tertiary structures of the protein get destroyed, but the primary
structure remains unaltered.
One of the examples of denaturation of proteins is the coagulation of egg white when an
egg is boiled.

4. Answer:
There are two common types of secondary structure of proteins:
(i) -helix structure
(ii) pleated sheet structure

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– Helix structure:
In this structure, the -NH group of an amino acid residue forms H-bond with the
group of the adjacent turn of the right-handed screw ( -helix).

pleated sheet structure:

This structure is called so because it looks like the pleated folds of drapery. In this structure,
all the peptide chains are stretched out to nearly the maximum extension and then laid side
by side. These peptide chains are held together by intermolecular hydrogen bonds.

5.Answer:
The structural differences between DNA and RNA are as follows:
DNA RNA
The sugar moiety in DNA molecules is 𝛽 -D-2 The sugar moiety in RNA molecules is 𝛽 -D-
1. 1.
deoxyribose. ribose.
DNA contains thymine (T). It does not contain RNA contains uracil (U). It does not contain
2. 2.
uracil (U). thymine (T).
3.The helical structure of DNA is double – 3.The helical structure of RNA is single-

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stranded. stranded.
The functional differences between DNA and RNA are as follows:
DNA RNA
1 DNA is the chemical basis of heredity. 1 RNA is not responsible for heredity.
2 DNA molecules do not synthesise proteins, 2 Proteins are synthesised by RNA
but transfer coded message for the synthesis molecules in the cells.
of proteins in the cells.

Assertion and Reason Answers:

1. (d) Assertion is wrong statement but reason is correct statement.
Explanation:
Uracil occurs in RNA
2. (b) Assertion and reason both are correct statements but reason is not correct
explanation for assertion.
Explanation:
Cysteine can cross link peptide chains through disulphide bridge. Cross linking by disulphide
bridge can occur either between the distant, properly oriented parts of the same
polypeptide chain (as in oxytocin or vasopressin) or between different polypeptide chains.
Case Study Answers:
1. Answer :
(i) (c) Fibrous proteins are converted into globular proteins.
(ii) (b) Primary structure only.
(iii) (b) Secondary structure.
(iv) (c) Denatured protein.
Explanation:
Cheese is a denatured protein.
(v) (d) Regular folding patterns of continuous portions of the polypeptide chain.
2. Answer :
(i) (a) Ketotriose.
(ii) (a) 16
(iii) (c) 1, 2 and 4 only
Explanation:
Glucose contains four chiral centres.
(iv) (d) The carbon atoms 3, 4 and 5 in both have the same configuration.

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Explanation:
In the formation of osazone, C-1 and C-2 react with phenylhydrazine to form
phenylhydrazone. If C-3, C-4, C-5 have same configuration they will form same osazone
even if they differ in configuration at C-1 or C-2.
(v) (c) Pentaacetate of glucose does not react with hydroxylamine.
Explanation:
Pentacetate of glucose does not react with hydroxylamine showing absence of free -
CHO group. This cannot be explained by open structure of glucose.

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