Biomolecules 307

14. Biomolecules

Section–I : Carbohydrates Definition : Carbohydrates are optically

active polyhydroxy aldehydes or polyhydroxy
14.0 GENERAL INTRODUCTION
ketones or other substances which produces
The complex organic molecules which form these on hydrolysis are known as carbohydrates.
the basis of life i.e. which build up living organisms They contain hydroxy group, ketone group,
and also required for their growth and aldehyde group, hence these are polyfunctional
maintenance are known as biomolecules. Some compounds.
common examples are carbohydrates, proteins,
fats, enzymes, amino acids, nucleic acids, lipids, 14.1.2 Classification of carbohydrates
steroids, hormones, vitamins etc. I] Classification on the basis of smaller unit
Some of them biomolecules are obtained on hydrolysis :
micromolecules and polymeric or non polymeric 1. Monosaccharides or simple carbohydrates:
macromolecules. These are simplest carbohydrates. They does not
Micromolecules : They includes small undergo further hydrolysis or does not divided into
molecules containing up to 30 carbon atoms and smaller unit. These are optically active (except
having molecular mass 100 to 1000. These are dihydroxy acetone). These are basic unit of
present in free state in cell. carbohydrates having general formula (CH2O)n.
Macromolecules : These are large where n = 3 to 7 or 10. These are further classified
molecules having very high molecular mass. They on the basis of number of carbon atoms and
may be polymeric or nonpolymeric. functional group present in compounds.
e.g. Proteins (polymeric), Chlorophyll (non i) Aldoses – These contains aldehyde group
polymeric), Haemoglobin (non polymeric), Starch a) Aldotriose – C3H6O3
(polymeric) e.g. Glyceraldehyde (glycerose)
14.1 SECTION–I CARBOHYDRATES b) Aldotetrose – C4HgO4
e.g. Erythrose
14.1.1 Introduction
c) Aldopentose C5H10O5
Carbohydrates are considered as hydrates
of carbon. They have general formula Cn(H2O)n. e.g. Ribose
But however this formula does not hold good for d) Aldohexose – C6H12O6
deoxyribose C 5 H 10 O 4 , rhamnose C 6 H 12 O 5 , e.g. Glucose
rhamno heptose C7H14O6, mannitol C6H14O6. ii) Ketoses – These contains ketone group
Some compounds which fit in this formula but a) Ketotriose – C3H6O3
not carbohydrates. e.g. Dihydroxy acetone
e.g. H–CHO, CH3–COOH, H–COOCH3. b) Ketotetrose – C4HgO4
These are also termed as saccharides (in e.g. Erythrulose
Lattin Saccharum–sugar). They contains in the c) Ketopentose C5H10O5
ratio 1 : 2 : 1. Carbohydrates are widely distributed
e.g. Ribulose
in plants and animals. In the from of cellulose
d) Ketohexose – C6H12O6
carbohydrates form the wood structures and
fibres of plants. In the from of starch e.g. Fructose (laevulose)
carbohydrates serves as the reserve food material 2. Complex carbohydrates:
for animals and humans. Carbohydrates are a) Oligosaccharides : These on hydrolysis gives
abundantly in rice, maize, potato, tuber, etc. two to ten monosaccharide units. e.g. C12H22O11
Sucrose, Maltose, Lactose (Milk sugar),

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 Biomolecules 308
Cellobiose. 14.1.3 Preparation of glucose:
These are divided by, It is prepared by following methods,
i) Disaccharides: These on hydrolysis gives two 1. In laboratory from cane sugar (sucrose)
monosaccharide units. e.g. C12H22O11 sucrose, 2. On large scale from starch (commercial method)
maltose, lactose, cellobiose 1. From sucrose (beet sugar or cane sugar) :
Sucrose 
hydrolysis
  –D–(+)– Glucose + These on hydrolysis gives equimolar mixture of
 –D– (+) – glucose and  –D–(–) fructose
 –D–(–) – Fructose (Hetero disaccharide)
(Invert sugar). Sucrose is dextro rotatory but
Maltose 
hydrolysis
  –D– (+)–Glucose + resulting mixture is laevorotatory. This is becaues
laevo rotation of fructose(– 92.4°) is more than
 –D–(+) – Glucose (Homo disaccharide)
that of dextro rotation of glucose (+53°). The
Lactose 
hydrolysis
  –D–(+)– Glucose + product is called invert sugar because on
hydrolysis of sucrose, the sign of rotation is
 –D–(+) – Galactose (Hetero disaccharide)
changed from (+)to (–).
Cellobiose 
hydrolysis
  –D–(+)– Glucose +
C12H22O11 + H2O 
90% C2 H 5 OH + conc.HCl
323K
 –D–(+) –Glucose (Homo disaccharide) C6H12O6 + C6H12O6
ii) Trisaccharides: These on hydrolysis gives three 2. From starch (potato or barley starch or
monosaccharide units. dextrin) :
e.g. C18H32O16 raffinose. When starch is hydrolysed by dil. H2SO4 under
Raffinose 
hydrolysis
 glucose + fructose + 5 atm. pressure at 413K. gives glucose. It is a
commercial method to prepare glucose.
galactose
[C6H10O5] + nH2O   nC6H12O6
dil.H 2SO 4 413K
iii) Tetrasaccharides: These on hydrolysis gives 5atm. pressure

four monosaccharide units e.g. C 24 H 42 O 21 14.1. 4 Physical properties of glucose:

stachyose. 1. It is white crystalline solid and sweet in taste.
Stachyose 
hydrolysis
 glucose + fructose + 2. It has M.P. 419K.
galactose + galactose. 3. It is soluble in water.
b) Polysaccharides: These are neutral polymeric 4. It is optically active and dextrorotatory. Hence
compounds, on hydrolysis gives large number of named as dextrose. It has specific rotation + 53°.
monosaccharide units. These are the polymer of 5. All reducing sugar shows mutarotation. It is a
monosaccharides. The common polysaccharides phenomenon in which change in angle of rotation
have general formula (C6H10O5)n, where n = 100 in neutral medium. The  –d glucose (+113°) and
to 3000. e.g. [C 6H 10 O 5] n Starch, cellulose,  –d glucose (+19°) to a constant final value of
glycogen, gums. d–glucose is + 53°. Only reducing sugar show
Starch, glycogen 
hydrolysis
 n  –O–(+)– mutarotation.
Glucose 14.1.5 Open chain structure of glucose:
Cellulose 
hydrolysis
 n  –O– (+)–Glucose From elemental analysis and molecular
weight detremination experiments the molecular
II] Classification on the basis of reducing
formula of glucose is C6H12O6. Following reaction
property:
suggest the open chain structure of glucose.
i) Reducing sugar : They reduce Tollen's reagent,
1. Reduction: Glucose is reduced by HI gives n–
Fehling's solution. e.g. Maltose, lactose, celloboise,
hexane.
glucose, fructose.
ii) Non reducing sugar: They do not reduce CHO
|
Tollen's reagent, Fehling's solution and Schiff's (CHOH) 4  14(H) 
HI
 CH3(CH2)4CH3 + 6H2O
reagent. e.g. Sucrose. |
CH 2 OH

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 Biomolecules 309
glucose n–hexane present in glucose.
This reaction shows that all six carbon atoms are 5. Acetylation (reaction due to OH group) :
linked in straight chain. Acetylation of glucose is carried out by using well
2. Oxidation: known acetylating agents i.e. acetyl chloride and
i) By using mild oxidising agent i.e. bromine acetic anhydride. When glucose is reacted with
water or Ag2O or Cu++ or NaOBr or NaOI. acetyl chloride or acetic anhydride in the presence
Glucose is oxidised by bromine water (mild of anhydrous ZnCl2 gives penta acetyl derivative
oxidising agent) gives gluconic acid. i.e. glucose penta acetate or penta–o–acetyl
glucose.
CHO COOH
| | CHO
(CHOH) 4  (O) 
Br2  H 2 O
 (CHOH) 4 |
(CHOH) 4  5CH 3COCl  
Unhydrous
| | ZnCl2
CH 2 OH CH 2 OH |
CH 2 OH
glucose gluconic acid
This reaction shows that carbonyl group present CHO
in glucose is aldehyde group. |
ii) Byusing strong oxidising agent: Glucose is (CHOOC  CH 3 ) 4  5HCl
oxidised by dil. or conc. nitric acid (strong |
CH 2 OOC  CH 3
oxidising agent) gives saccharic acid or glucaric
acid. This reaction confirm the presence of five –OH
groups present on different carbon atoms in
CHO COOH
| | glucose and also confirm open chain structure of
(CHOH) 4  3(O) 
HNO3
 (CHOH) 4  H 2 O glucose.
| | 14.1.6 D and L configuration of monosaccharides:
CH 2 OH COOH
The sugars are divided in to two types i.e. D–
This reaction shows that primary alcoholic group family and L–family.
present in glucose. In 1906 Rosanoff classify the two families
3. Reaction with hydroxyl amine: Glucose is on the basis of configuration by taking example
reacted with hydroxyl amine gives glucose oxime. of gluceraldehyde. He observed that
glyceraldehyde can exist in two enatiomeric
CHO CH  NOH forms.
| | 1. D – configuration : When –OH group is
(CHOH) 4  H
3(O)N – OH 
 (CHOH) 4  H 2O
|
2
| attached to right side adjacent to –CH2–OH group
CH 2 OH CH 2 OH or last asymmetric carbon atom.
2. L–configuration: When –OH group is attached
glucose glucose oxime to left side adjacent to –CH2–OH group or last
4. Reaction with hydrogen cyanide : Glucose is asymmetric carbon atom.
reacted with HCN gives glucose cyanohydrine.
CHO CHO
CN | |
| H  C  OH HO  C  H
CHO CH  OH | |
| | CH 2  OH CH 2  OH
(CHOH) 4 + HCN 
 (CHOH) 4
| | D– glyceraldehyde L– glyceraldehyde
CH 2 OH CH 2  OH It has been found that all naturally occurring
sugars i.e. glucose and fructose belong to D–
glucose glucose cyanohydrine series.
Reaction 3 and 4 confirm that aldehyde group

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 Biomolecules 310

CHO CH 2 OH OR
| | |
H  C  OH CO H  C  OR ' H 2 O
| | |
HO  C  H HO  C  H acetal
| |
CHO H  C  OH H  C  OH
| Monosaccharides contain a number of –OH
| |
H  C  OH H  C  OH H  C  OH groups aldehyde or ketone group. Therefore they
| | | can undergoes intramolecular reactions to form
CH 2  OH CH 2  OH CH 2  OH cyclic structures. For example, glucose form a
D–(+)–Glycerladehyde D–(+)–Glucose D–(–) Fructose six membered ring of five carbon atoms and one
oxygen atom like pyran and fructose form five
CHO CHO membered ring with four carbon atoms and one
| |
H  C  OH H  C  OH oxygen atom like furan.
| |
HO  C  H HO  C  H
| |
H  C  OH H  C  OH
| |
H  C  OH H  C  OH Anomers : These are cyclic structure of
| | monosaccharides differ in configuration of –OH
CH 2  OH CH 2  OH
group around anomeric carbon atom.
D–(+)–Glucose D–(+)–mannose If –OH group is at right side on anomeric
Note : It may be noted that D and L do not carbon atom called as –  , while –OH group is
represent dextro and laevorotatory. The optical at left side on anomeric carbon atom called
activity of molecule is represented by (+) and as –  .
(–) sign.
i) Anomers of glucose : Glucose is found to exist
14.1.7 Cyclic structure of monosaccharides: in two crystalline form.
Open chain structure of monosaccharides do not
i.e.  –D– glucose and  –D–glucose called
give following reactions.
anomers.
1. They does not gives condensation reaction with
2, 4 –ONP though the presence of aldehyde a) Aqueous solution of glucose on crystalline at
group. 303 K produces  – glucose
2. They does not give addition reaction with b) Aqueous solution of glucose on crystalline at
NaHSO3 though the presence of aldehyde group. 371 K produces  –glucose
3. Glucose penta acetate do not react with NH2OH The pair of optical isomers which differs in the
though the presence of aldehyde group. configuration around Cl carbon atoms. Glucose
4. Glucose does not reduce Schiff's reagent though form hemiacetal between. –CHO group and
the presence of aldehyde group. –OH group on C5 atom. As a result Cl become
All above reaction indicates the absence of asymmetric and form two isomers called
>C = O group in monosaccharides.  –D–glucose and  –D–glucose. These two
Cyclic structure can be studied when isomers differs in the orientation of Hand OH
monosaccharides reacts with alcohols to form around Cl atom called anomers.When –OH group
hemiacetals and acetals. present on anomeric carbon atom at right side
called  , while –OH group present on anomeric
OH
| | carbon atom at left side called  .
 H  C  OR ' 
H  C  O  R ' OH 
R  OH

|
hemiacetal
aldehyde

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 Biomolecules 311

Note : Anomers are five membered four carbon

atom, one oxygen atom cyclic structure, in which
–OH of C5 involved in ring formation.
Conversion of Fischer projection formula of
sugers into Haworth projection formula:
Rules for Haworth projection formula :
i) Draw the hexagonal or pentagonal ring with its
oxygen atom at top.
ii) The terminal –CH2–OH group is always placed
above the plane of hexagonal or pentagonal ring
in D–series and below the hexagonal or
Note: pentagonal ring in L–series
1. Anomers are six membered five carbon atom iii) Place all the groups (on C2, C3 and C4) which
with one oxygen atom cyclic structures of glucose, are present left hand side in Fischer projection
in which –OH of C5 is involved in ring formation. formula above the plane of the hexagonal or
2. This ring structure explain the absence of –CHO pentagonal ring in both the series sugar.
group in glucose. iv) Place all the group (on C2, C3 and C4) which are
3. Cl carbon in aldehyde carbon before cyclilisation present right hand side in Fischer projection
is called anomeric carbon. formula, below the plane of the hexagonal or
ii) Anomers of fructose: pentagonal ring in both the series sugar.
Fructose has molecular formula C6H12O6 and v) For D–series carbohydrates: (a) If –OH
ketonic group at a carbon 2. It is belongs to group is up the configuration of anomeric carbon
D–series and laevorotatory compound, hence it is  and (b).
is written as D– (–) – fructose. Fructose form If OH group is down the configuration of
hemiketal between > C = O and –OH group of anomeric carbon is  .
C5 atom. As a result C2 atom become asymmetric vi) For L–series carbohydrates: (  ) If –OH
and form two isomers called  –D–fructose and
group is up the configuration of anomeric carbon
 –D–fructose. These two isomers differs in the is a and (b) If –OH group is down the
configuration of H and –OH around C2 atom configuration of anomeric carbon is  .
called anomers.
e.g. 1. Hawoth Structure of glucose:

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 Biomolecules 312
Sucrose  –D–(+)  –D–(–)
– glucose fructose
In cyclic structure, two monosaccharides are held
together by a glycosidic linkage (acetal bond)
between C1 of  –glucopyranose (  –glucose)
and C2 of  –fructofuranose (  –fructose). Since
the reducing groups of glucose and fructose are
involved in glycosidic bond formation, Hence
2. Hawoth Structure of Fructose : Fructose sucrose is non reducing sugar. (+) sucrose is
is made by isomerisation of glucose. named equally well as either.
 –D–glucopyranosyl  –D–fructofuranoside or
 –D–fructofuranosyl  –D–glucopyranoside.

Thus sucrose contain 1  2  –  glycosidic

14.1.8 Structures of disaccharides : linkage.
All disaccharides have molecular formula 2. Structure of maltose (4–O–  –D–
C 12 H 22 O 11 (Sucrose, maltose, lactose and Glucopyranosyl D–glucopyranose) : It is
celloboise). These on hydrolysis gives two obtained by partial hydrolysis of starch. These
monosaccharide units. Sugar whose name ends on hydrolysis gives two molecules of  –D–
with suffix 'oside' is non reducing sugar while glucose.
ends with 'ose' is reducing sugar.
C12H22O11 + H2O 
 2C6H12O6
1. Structure of sucrose: These on hydrolysis gives
equimolar mixture of  – D– (+) – glucose and  –D–glucose
 – D–(–) fructose. Sucrose is dextro rotatory In cyclic structure, two  –D–glucopyranose
molecules are held together by glycosidic linkage
but resulting mixture is laevorotatory.
between C1 of one glucopyranose molecule and
C12H22O11 + H2O 
 C6H12O6 + C6H12O6 C4 of another glucopyranose molecule. The free
aldehyde group can be produced at C1 of second

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 Biomolecules 313
glucopyranose molecule. Hence maltose is C 1 of  –D–galactopyranose and C 4 of  –
reducing sugar. Dglucopyranose (C4–C1)
Thus lactose contain 1  4  –  glycosidic
linkage.
OR
According to Board Book

 –D–glucopyranose  –D–glucopyranose
Structure of maltose
C 1 of (I) glucopyranose and C 4 of (II)
glucopyranose (C1–C4)
Thus maltose contain 1  4  –  glycosidic 4. Structure of celloboise (4–O–  –D–
linkage. Glucopyranosyl D–glucopyranose) :
3. Structure of lactose (4–O–  –D– Celloboise is obtained by partial hydrolysis of
Galactopyranosyl D–glucopyranose) : These cellulose.
on hydrolysis gives  –D–glucose and  –D– C12H22O11 + H2O 
 C6H12O6 + C6H12O6
galactose. It is known as milk sugar.
celloboise  –D–glucose  –D–glucose
C12H22O11 + H2O 
 C6H12O6 + C6H12O6
In cyclic structure of celloboise C1 of one  –D–
 –D  –D glucopyranoseis linked to C4 of another  –D–
–glucose –galactose glucopyranose by glycosidic linkage. Thus
In cyclic structure, two monosaccharide units are celloboise contain 1  4  –  glycosidic linkage.
held together by glycosidic linkage between C4 It is a reducing sugar because –CHO group at
of  –glucopyranoseand C 1 of  –D– C1 in second glucose molecule.
galactopyranose. The free aldehyde group
produce at  –D–glucopyranose molecule.
Hence lactose is reducing sugar.

Structure of celloboise
C 1 of  –D–glucopyranose and C 4 of  –
Dglucopyranose i.e C1–C4 bond
 –D–galactose  –D–glucose
Structure of lactose

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 Biomolecules 314
OR
According to Board Book

14.1.9 Structure of polysaccharides

All polysaccharides have molecular formula (C 6H10O5) n. These on hydrolysis gives number of
monosaccharide units. In polysaccharides large number of monosaccharides units are joined together by
glycosidic linkage.
1. Structure of starch: It is main storage polysaccharide of plants. High content of starch is found in
cereals, roots, tubers and in some vegetables. Which on hydrolysis gives number of  –D–glucose.
(C6H10O5)n + nH2O 
 n C6H12O6
 –D–glucose
Starch is polymer of  –D–glucose and consist of two components amylose and amylopectin. Amylose :
It is water soluble, which consist about 15 to 20% starch. Chemically amylose is long unbranched chain
polymer with 200–1000  –D–glucopyranose units held by C1–C4 glycosidic linkage.

Amylose
(C1–C4 glycosidic linkage)
Thus amylose contain 1  4  –  glycosidic linkage like maltose.
Amylopectin: It is water insoluble, which consist about 80 to 85% starch. It is a branched chain polymer
of  –D–glucopyranose units, held together by C1–C4 glycosidic linkage, whereas branching occurs at
C1–C6 glycosidic linkage.

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 Biomolecules 315
Thus amylopectin contain 1  4  –  glycosidic linkage in long chain and 1  6  –  glycoside in
branched
2. Structure of cellulose: It is most abundant substance in plant. It is present in cell wall of plant cell. It
isstraight chain polysaccharide, which on hydrolysis gives number of  –D–glucose. Hence it is a polymer
of  –D–glucose.

(C6H10O5)n + n H2O 
 n C6H12O6
Cellulose  –D–glucose
In cyclic structure of cellulose ,number of  –D–glucopyranose units are joined by glycosidic linkage between
C1 of one  –D–glucopyranose and C4 of another  –D–glucopyranose molecule.

Cellulose
(C1 of (I)  –D–glucopyranose and C4 of another  –D–glucopyranose)
Thus cellulose contain 1  4  –  glycosidic linkage.
3. Glycogen: It is a animal polysaccharide found in brain muscle and liver. It serves as reserve carbohydrates
for animal and hence known as animal starch. It's structure is similar to amylopectin and highly branched.
It is also found in yeast and fungi.
14.1.10 Importance of carbohydrates:
1. It act as main source of energy
2. It act as storage of energy for the functional of living organism.
3. They form structural material for cell walls.
4. Cellulose in the form of cotton, used in textiles, papers.
5. Two aldopentose i.e. D–ribose and 2–deoxy D–ribose are present in nucleic acid, which involves biosystem
in combination with proteins and lipids.

Section– II : Proteins Proteins are condensation biopolymer of  –

amino acids and have molecular weight above
14.2.1 Definition:
10,000.
Proteins are naturally occurring nitrogenous

polypeptide compounds, which on hydrolysis gives –[–HN–CHR–CO–]n + nH2O 
H

various number of L–  –amino acids.
nH 2N–CHR–COOH
14.2.2 Acid hydrolysis of proteins:
14.2.3 Elementary idea of amino acids:
In 1900 German chemist Emil Fisher studied the
1. Amino acids : These are derivatives of
structure of protein. He observed that, proteins
carboxylic acids obtained by replacing H atom
are completely hydrolysed by acids (25 % HCl
by –NH2 group.
or 35% H2SO4) or alkalies gives mixture of
These are organic compound containing both
various number of L–  –amino acids.
–COOH and –NH2 functional group are known
It is clear that  –amino acids are fundamental
as amino acids.
unit of proteins.

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 Biomolecules 316
H v) In zwitter ionic form,  –amino acid shows
| amphoteric behavior as they react with both acids
R  C  COOH and bases.
|
NH 2 a) When acid is added to  –amino acid –COO–
accept proton and therefore, the basic nature is
2. Classification: Depending upon position of amino
due to – COO– group. In acidic medium a–amino
group. These are classified as  ,  ,  ,  and acid act as cation. When electric field applied,
so on amino acids. Only  –amino acids obtained they migrate towards cathode.
by hydrolysis of proteins. Generally all naturally
 
occurring amino acids are a–amino acids. (NH2 H 3 N  CH  COO  
H 
 H 3 N  CH  COOH
group present on  –carbon atom). | |
3. Classification of  –amino acids : These are R R
classified as b) When alkali is added to  –amino acid, –NH3+
i) Neutral  –amino acids : They contains equal group release proton and therefore acidic
number of –COOH and –NH2 group. character is due to –NH3+ group. In basic medium
ii) Acidic  –amino acids : They contains more  –amino acid act as anion. When electric field
number of carboxyl group than amino group. applied, they migrate towards anode.
iii) Basic  –amino acids: They contains more 
number of amino group than carboxyl groups. H 3 N  CH  COO  
H

iv) Essential  –amino acids: These are not |
R
synthesized in the body but obtained through diet.
e.g. Valine, lysine etc.
H 2 N  CH  COO   H 2 O
v) Non–essential  –amino acids: These are |
synthesized in the body. R
e.g. Glycine, serine, proline
Note:
4. L–family  –amino acids : All  –amino acids
1) Thus in aqueous solution the basic character of
are optically active except glycine. They exist both
 –amino acids is due to –COO– group and
D and L–forms.The 'D' refer to the isomer with
acidic character is due to –NH3+ group.
–NH2 group at right side and 'L' refer to the
isomer with –NH2 group at left side. All naturally 2) Aqueous solution of neutral  –amino acid is
occurring  –amino acids are belong to L–series. slightly acidic because acidic character of
–NH3+ group is more than that of basic character
COOH COOH –COO– group.
| |
H  C  NH 2 H2N  C  H 6. Isoelectric point of  – amino acids : In
| | certain H+ ion concentration (pH), the dipolar ion
CH 3 CH3 exist as a neutral ion and does not migrate towards
D–alanine L–alanine anode and cathode is known as isoelectric point.
5. Physical properties of  –amino acids: At isoelectric point, amino acids are less soluble
i) These are colour less crystalline solid. in water, and this property is used for separation
ii) All are optically active except glycine. of different amino acids obtained from hydrolysis
iii) These are soluble in water. of proteins.
iv) In aqueous solution, the –COOH group can lose 14.2.4 Classification on the basis of structure and
a proton and –NH2 group can accept proton solubility:
giving rise to dipolar ion (zwitter ion). a) Fibrous proteins (structural proteins or
 sclero proteins) : Long helical structure (  –
H 3 N  CH  COO  form) insoluble in water, acids and alkalies. They
|
R serve as, chief structural material for tissue. e.g.
Keratin in hair, nail and wool, myosin in muscle,
(zwitter ion)

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 Biomolecules 317
collagen in tendons and bones, silk, feathers, horns, 14.2.6 Structure of proteins:
hooves, skin, cartilage, elastin in ligaments etc. Structure and shape of proteins can be
b) Globular proteins: These are spherical, elliptical studied at four different level. i. e. primary,
or oval shape and soluble in water, base, acid and secondary, tertiary and quaternary. Each level
salt.They are involved in the maintenance and being more complex than previous one.
regulation of life process. They are folded to form 1. Primary structure of proteins :
spherical shape and have weak intramolecular
In primary structures of proteins number of
hydrogen bonding and weak intermolecular forces
 –amino acid are linked together in linear
than fibrous proteins e.g. Albumin, globulin,
sequence by peptide bond.
protamine, prolamine, histone, glutelin, thyroglobin,
Each molecule of  –amino acid of a given
insulin, heamoglobin, casein of milk, all enzymes,
protein has the same sequence along with
venoms of snakes, scorpion, bees etc.
polypeptide chain.
14.2.5 Peptide linkage (peptide bond) : Any change in this primary structure i.e. the
Definition: sequence of  –amino acids creates a different
–(CO–NH)– linkage in protein is known as protein.
peptide linkage. It is planar in which oxygen and The primary structure of protein is determined
hydrogen are at trans position. by its successive hydrolysis with enzymes or acids
Formation of peptide linkage: or alkalies.
Peptide linkage in protein is formed by free Proteins  1° – proteoses  2°–proteoses 
–NH2 group of one  –amino acid is joined with pep tones  polypeptides  simple peptide 
free –COOH group of another  –amino acid  –amino acids.
by elimination of water molecule.
R R
R R | |

| | H 2 N  C  COOH + HHN  C  COOH  

H 2 N  C  COOH + HHN  C  COOH   | |
| | H H
H H   a min o acid
  a min o acid
R R
R R | |
| | H 2 N  C  CONH  C  COOH + H O
2
H 2 N  C  CONH  C  COOH + H O | |
| | 2 H H
H H
dipeptid
dipeptid 2. Secondary structure of proteins
Depending upon the number of amino acids Secondary structure is arise due to folding or
residues per molecule, the peptides are called as arrangement of polypeptide chain.
dipeptide (one peptide linkage), tripeptide (two Secondary structure of protein is divided in to two
peptide linkage), tetrapeptide (three peptide types depending upon size of R–group.
linkage), etc. The formation of peptide bonds can a)  –helix structure:
continue until a molecule containing thousands of When intramolecular hydrogen bonding takes
amino acids. Relatively shorter pep tides are place between –NH group of one unit and
called, oligopeptides. While, larger polymer are carbonyl oxygen atom of fourth units then the
called, polypeptides or proteins. By convention a chain of  –amino acids coils at right hand side
peptide having molecular mass up to 10,000 is called  –helix.
called polypeptide. While, a peptide having
Intramolecular hydrogen bonding between same
molecular mass more than 10,000 is called
unit is responsible for holding helix in a position.
proteins.

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 Biomolecules 318
b)  –pleated structure: Section–III: Lipids
This structure is formed when polypeptide chains 14.3 : LIPIDS
are arranged in a zig–zag manner with alternate 1. Definition:
bulky R–group on same side. The chains are held
These are oily ,fatty ,waxy substancse of plants,
together by a very large number of intermolecular
animals and tissues which are insoluble in water
as well intramolecular hydrogen bonds between
and soluble in organic solvent like CHCl3, CCl4,
> C = O and –NH of different chains. This results
ethers, benzene etc are called lipids.
the formation of flat sheet.
e.g. Oils, fats, phospholipids, steriods, glycolipids,
These sheets can slide over each other to form
lecithin, cephalin, cholestrol, lanosterol etc.Lipids
three dimensional structure called as  –pleated are mainly made from carbon, hydrogen and
structure e.g. Fibroin of silk. oxygen. The number of oxygen molecules
3. Tertiary structure of proteins: molecules are always small as compare to carbon.
The tertiary structure is formed due to folding, They also contain P, N, S.
refolding, coiling and bonding of polypeptide chain Lipids serve as energy reserve for use in
producing three dimensional structure. This metabolism and as a major structural material in
structure gives the overall shape of proteins e.g. cell membranes for regulating the activities of cell
Fibrous and globular proteins. and tissues.
The main forces which stabilize the secondary 2. Classification of lipids:
and tertiary structure of proteins are hydrogen These are classified on the basis of product
bonds, disulphide linkage, Vander Waals force and obtained on hydrolysis,
electrostatic force of attraction (ionic bond). I) Complex or compound lipids,
4. Quaternary structure of proteins: II) Simple lipids
Proteins that have more than one peptide chain I) Complex or compound lipids:
are known as oligomers. The individual chains They have ester linkage and on hydrolysis gives
are called subunits. fatty acids, alcohols, phosphoric acid and nitrogen
The subunits are held together by hydrogen containing base e.g. Oils, fats, phospholipids,
bonding, electrostatic attractions, hydrophobic glycolipids, waxes.
interactions etc. Quarternary structure explains (a) Oils and Fats:
the way the sub units are arranged in space. Introduction:
14.2.7 Denaturation of proteins: These are naturally occurring tasteless, odourless,
The structure of natural proteins is responsible colourless, nonvolatile liquids or solids, lighter than
for their biological activity. These structures are water, lesspolar compounds. These are insoluble
maintained by various attractive forces i.e. in water and soluble in nonpolar solvent like CCl4,
hydrogen bonding, Vander Waals force, sulphide ethers. These are termed as natural lipids.
bond, ionic bond etc. Fats are stored energy source and act as heat
The breaking of these attractive forces by insulator for the loss of heat from body.
physical or chemical method, proteins lose its Preparation of triglycerides (glyceride or
biological activity called as denaturation of triacyl glycerol (TAG) or oils or fats) :
proteins. When glycerol (glycerine) is heated with fatty
During denaturation secondary and tertiary acid gives triglycerides. Triglyceride has wide
structures are destroyed but primary structure application in preparation of soap, paints,
remain intact.The peptide bond does not break. varnishes, ink, ointments and cream.
e.g. i) Coagulation of egg white by heat.
ii) Curdling of milk which is caused due CH 2  OH HOOCR CH 2  OOCR
| |
to formation of lactic acid by the CH  OH  HOOCR 
 CH  OOCR  3H 2 O
bacteria present in milk. | |
iii) Formation of chease from milk. CH 2  OH HOOCR CH 2  OOCR

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 Biomolecules 319
Important definitions:
CH 2  OOC  C17 H 35
1) Glycerol: It is trihydroxy alcohol, obtained by |
replacing three hydrogen atoms from propane CH  OOC  C17 H 35
by three – OH groups. It is water soluble, |
nontoxic, viscous, hygroscopic, high boiling point CH 2  OOC  C17 H 35
liquid.
Glycerol trioleate or triolein
2) Fatty acids : Long chain higher monocarboxylic
3. Classification of triglycerides (oils or fats) :
acids are known as fatty acids.
i) Simple triglycerides : When three –OH groups
3) Saturated fatty acids (CnH2nO2) : Long chain
of glycerol are replaced by three same fatty acids
higher monocarboxylic acids containing
then these are known as simple triglycerides.
carboncarbon single bond.
e.g. Stearic acid C17H35COOH CH 2  OOC  C17 H 23 CH 2  OOC  C17 H 31
4) Unsaturated fatty acids : Long normal chain | |
higher monocarboxylic acids containing CH  OOC  C17 H 23 CH  OOC  C17 H 31
| |
carboncarbon multiple bond. CH 2  OOC  C17 H 22 CH 2  OOC  C17 H 31
e.g. Oleic acid C17H33COOH (1 C = C bond)
5) Triglycerides : These are triesters of glycerol Glycerol trilaurate Glycerol tripalmitate
with fatty acids. The fatty acid in triglyceride
CH 2  OOC  C17 H 35
contain an even number of carbon atoms and an |
unbranched carbon chain. CH  OOC  C17 H 35
6) Oils: These are triglycerides of unsaturated fatty |
acids. CH 2  OOC  C17 H 35
e.g. ii) Mixed triglycerides : When three –OH groups
of glycerol are replaced by three different fatty
CH 2  OOC  C17 H 33 CH 2  OOC  C17 H 31
| | acids then these are known as mixed triglycerides.
CH  OOC  C17 H 33 CH  OOC  C17 H 31 e.g.
| |
CH 2  OOC  C17 H 33 CH 2  OOC  C17 H 31 CH 2  OOC  C15 H 23
|
Glycerol trioleate Glycerol trilinoleate CH  OOC  C17 H 23
or triolein |
CH 2  OOC  C17 H 22
CH 2  OOC  C17 H 29
| Glycerol palmito stero oleate.
CH  OOC  C17 H 29 (b) Phospholipids: It is a mixed glyceride of higher
| fatty acid and phosphoric acid in which two
CH 2  OOC  C17 H 29
–OH group of glycerol are esterified by fatty acids
Glycerol trilinoleate while third –OH group of glycerol is esterified by
7) Fats: These are triglycerides of saturated fatty phosphoric acid. e.g. Lecithins, cephalins etc.
acids. (c) Glycolipids: Lipids may be associated with sugar
to form glycolipids. The sugar is typically glucose
CH 2  OOC  C17 H 23 CH 2  OOC  C17 H 31 or galactose.The animal glycolipids is
| | glycocerebroside. The plant glycolipid is
CH  OOC  C17 H 23 CH  OOC  C17 H 31 galactocerebroside.
| |
CH 2  OOC  C17 H 22 CH 2  OOC  C17 H 31 (d) Waxes: They provide waterproofing on leaves,
fruits, berries, animal fur and feather of birds.
Glycerol trilaurate Glycerol tripalmitate These are mainly esters of long chain carboxylic
acid with long chain monohydric alcohols.
II) Simple lipids:

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 Biomolecules 320
They donot have ester linkage and can not be hydrolysed.
e.g. Steroids like cholesterol, terpens, prostaglandins, fats soluble vitamins like
A, D, E, K.
These are divided in to three types.
1. Steroids: These are derived from cydopentaperhydrophenanthrene, which has nucleus of four rings.

steroid nucleus cholesterol (a sterol) testosterone (an androgen)

e.g. Animal sterols (Zoosterols) : It includes male sex hormone (cholestrol, lansosterol, testosteron,
androsteron), Female sex hormons ( estrogen, estron, estradiol)
Plant sterols (Phytosterols) : It includes  – Sisosterol.
Fungal sterols (Mycosterols) : It includes ergosterol
2. Terpens: These are unsaturated hydrocarbon which consist of number of isoprene units.
e.g. Geraniol, menthol, phytol, vitamin A, D, E, K.
CH 3
|
CH 2  C  CH  CH 2
These are classified on the basis of number of isoprene units (Each isoprene units contain 5 carbon atoms)

No. of C - atoms 10 15 20 30 40
Class Monoterpene Sesquiterpene Diterpene Triterpene Tetraterpene
Examples -phellandrene Abscisic acid Cembrene Squalene -Carotene
3. Prostaglandins: These are group of C20 lipids that contain a five membered ring with two long side chain.
They may be detected in many body tissues.
Uses of lipids:
1) Fats and oils : Have a convenient and concentrated storing food energy in plants and animals. Although
carbohydrates also serve as source of readily available energy, equal mass of fat produce over twice the
amount of energy than carbohydrates.
2) Glycolipids: These are components of cell membrane. Glycolipids occur in bacterial cell wall. In plants,
glycolipids are principal lipid constituents of chloroplasts. Cerebrosides are animal glycolipids that are found
in plasma membranes of neural tissues and are abundant in myelin sheath of neurons.
3) Phospholipids: They form membrane like structure in water. Phospholipids and sterols like cholesterol
are major components of cell membranes.
4) Waxes: They provide vital waterproofing for body surfaces. Waxes are water repelling solids that are
protective coatings on leaves, fruits, berries, animal fur and feather of birds.
5) Steroids: It include adrenal hormones, sex hormones and bile acids. Lipids can combine with proteins to
form lipoproteins, found in cell membranes. Bile acids are steroids which related to digestion of fat in
intestine. Cholic acid an example of bile acids. Prostaglandins have a wide range of biological effects.
6) Terpenes: It include vitamin A, E and K and phytol. Terpenes occur in essential oils such as menthol and
camphor. Terpenes are the main constituents of the essential oils secreted by the glands of certain aromatic
plants.

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 Biomolecules 321
e.g. Myrcene (oil of bayberry), Limonene (oil Chemical composition of nucleic acids
of lemon),  –pinene (oil of turpentine), geraniol Nucleic acid (DNA or RNA) on complete
(oil of roses), menthol (peppermint), zingiberene hydrolysis gives pentose sugar, phosphoric acid,
(oil of ginger), caryophyllene (oil of cloves) and nitrogen containing heterocyclic compound (called
squalene (shark liver oil). base).
7) Prostaglandins can lower blood pressure, affect In nucleic acid the sequence is Base–
blood platelet aggregation during clotting, lower Sugar– Phosphate.This sequence is known as
gastric secretion and stimulate uterine nucleotide.
contractions during child birth.
Section–IV Enzymes

14.4 ENZYMES
Definition: These are proteins, produced by living
system and catalyse specific biological reactions
1. Deoxyribonucleic acid (DNA)
are called enzymes.The enzymes differs from
other catalyst in being highly selective and It on hydrolysis gives  –D–2–deoxyribose
specific. (pentose sugar), four bases i.e. adenine
e.g. Maltose is hydrolysed in to glucose by (A),cytosine (C), guanine (G) and thymine (T)
maltase enzyme, i.e (ACGT) and phosphoric acid.The  –D–2–
C12H12O11 
maltase
 2 C6H12O6 deoxyribose means no –OH group at C2 position.
maltose glucose
Mechanism of enzyme action:
Enzymes are needed in small quantity for the
progress of reaction. Similar to the action of
chemical catalyst, enzyme reduce the magnitude
of activation energy and increase the rate of
reaction mechanism.
Step–i: Binding of enzyme to substrate to form
activated complex
 ES
E + S 
Step–ii: Formation of product in the activated
complex
ES 
 EP
Step–iii: Decomposition of EP in to enzyme and
product
EP 
 E+P

Section–V: Nucleic Acids

14.5 NUCLEIC ACIDS

Nucleus of the living cell is responsible for
a) Primary structure of DNA:
the transmission of inherent character known as
heredity. The nucleus of a cells are made from When base is attached to position 1 of the pentose
nucleoprotein which contain two types of nucleic sugar is known as Nucleosides. When nucleosides
acids. link to phosphoric acid at position 5 of a pentose
sugar is known as Nucleotides.
1) Deoxyribonucleic acid (DNA)
2) Ribonucleic acid (RNA)

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The two nucleotides are joined together by phospho diester linkage between C–3 of one nucleotide and C–
5 of another nucleotide to form dinucleotides.

When number of nucleotides are condense together to form very long chain of polynucleotide or DNA
b) Secondary structure of DNA:
When two nucleic acid chains are wound about each other and held together by hydrogen bond between
pairs of base. Adenine (A) form hydrogen bond with thymine (T) i.e. A– T, and cytosine (C) form hydrogen
bond with guanine (G). i.e.C–G. This double strand helix structure is known as DNA.
The simplified version of nucleic acid chain is shown as,

2. Ribonucleic acid (RNA) :

It on hydrolysis gives  –D–ribose (pentose sugar), four bases i.e. adenine (A), cyctosine (C), guanine (G)
and uracil (U) i.e. (ACGU) and phosphoric acid

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 Biomolecules 323

 –D–ribose Uracil (U) Pyrimidine base

When base is attached to position 1of the pentose sugar is known as Nucleosides.
When nucleosides link to phosphoric acid at position 5 of a pentose sugar is known as Nucleotides.

Nucleoside Nucleotide
a) Primary structure of RNA : The two nucleotides are joined together by phosphodiester linkage between
C–3 of one nucleotide and C–5 of another nucleotide to form dinucleotides.

Nucleotide Dinucleotides
When number of nucleotides are condense together to form very long chain of polynucleotide or RNA.
b) Secondary structure of RNA: The secondary structure of RNA are also present which are only single
strand. Sometime they fold back themselves to form a double helix structure of RNA. RNA molecules are
three types and they perform different functions these are named as
i) Messenger RNA (m–RNA)
ii) Ribosomal RNA (r–RNA)
iii) Transfer RNA (t–RNA)

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Section–VI: Vitamins

Definition:
Organic substances required for regulating some body process are known as vitamins.
These compound can not be produced by organism. Vitamins are required in small amounts and
deficiency of anyone vitamins causes diseases. Vitamins are denoted by alphabets
A, B, C, D, E, K, H, P some of these are further named as B1, B2, B3, B5, B6, B12, etc.
Classification:
Depending upon solubility in water or oils,fats these are classified in three types,
1) Water soluble: B group and C group vitamins are water soluble. Water soluble vitamins supplied regularly
in diet because they are readily exerted in urine and can not be stored in our body except vitamin B 12
2) Fats soluble: A, D, E, and K are insoluble in water and soluble in fats and oils. These are stored in liver
and adipose tissue (fat storing tissues).
3) Vitamin H (Biotin ): It is insoluble in water, fats and oils.
Classification:
Depending upon their chemical structure. These are classified as,
1) Vitamins of aliphatic series: They contain long chain of aliphatic compounds e.g. Vitamin–C
2) Vitamins of aromatic series: They contain long chain of aromatic compounds e.g. Vitamin–K
3) Vitamins of alicyclic series: They contain alicyclic ring e.g. Vitamin–A
4) Vitamins of heterocyclic series : They contain heterocyclic ring e.g. Vitamin–B (B1, B2, B3, B5, B6,
B12), mesoinsositol, folic acid, vitamin –H.
Some vitamins, their sources and diseases due to deficiencies
No Vitamins Sources Diseases due to deficiencies
1. Vit–A Milk, fish liver,tomatoes, Night blindness, retardation of growth, dryness
Retinol/Axerophthol carrot, sweet potatoes of skin and hair
2. Vit–B1 Rice, wheat, meat, Beriberi
Thiamine green vegetable
3. Vit–B2 or G Egg yolk, fishes, yeast, Inflammation of tongue, dryness of lips and mouth,
Riboflavin liver. cheilosis( retarding the growth and digestion
4. Vit–B3 Yeast, liver, tomatoes, Dermatitis, graying of hair,
Panthothermic acid egg, meat, retard body and mental growth, reproductive
deability
5. Vit–B5 Barely, liver, maize, Pigmentation of skin (pellagra),
Nicotinamide wheat, rice degeneration of spinal cord, mental confusion.
6. Vit–B6 Milk, liver,maize, Convulsion, loss of weight, mental change,
Pyriodoxine or wheat, fish, yeast derangement of enzymes ( which control
pyridoxamine carbohydrates metabolism)
7. Vit–B12 Egg, liver of pig, sheep Degradation of spinal cord, anaemia
Cyanocobalamin
8. Vit–C Orange, grapes, lemon, Scurvy( bleeding, spongy, swollen) gums
Ascorbic acid tomatoes, onion, cabbage
9. Vit–D Butter, liver, egg, fish oil, milk, Rickets, osteomalacia
Ergocalciferol meat, in skin cell in sun
10. Vit–E light Rice, liver of cattle, Weakness of muscles, abnormal growth and
Tocopherol seed oils, wheat deposition of tissue, decrease reproductive power
11. Vit–H Biotin Yeast, eggs, fruits, wheat Skin lesions, loss of apatite, hair fall, paralysis
12. Vit–K Green leafs of spinach, fish, Increase blood clotting time (hemorrhage),
Phylloquinine meat, cauliflower poor coagulation of blood
13. Vit–P Orange, grapes Haemorrhagia, decrease in capillary resistance

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Section–VII: Hormones

14.7 Hormones:
These are chemical substances which are secreted by ductless gland and control different physiological
function–of the body.
Hormones control growth of tissues, heart beat, blood pressure, secretion of digestive enzymes, kidney
function, the reproductive system and lactation etc.
In mammals, the secretion of hormones is controlled by interior lobe of pituitary gland located at the
base of brain. These hormones are then carried to other gland such as adernal cortex and sex gland to
stimulate the production of other hormones.
Classification of hormones:
These are classified on the basis of structure and composition.
The are mainly classified in to three types:
1) Steroid hormones
2) Polypeptide hormones
3) Amino acid hormones

Sr. Name of hormones Sources/origin Function

1) Steroid hormones
Sex hormones
a) Androgens (Male sex hormon)
e.g. Testosterone Testis Regulate and stimulate male sex organs
b) Female sex hormon
e.g. (i) estrogens Overy It maintain the function of
female sex organs
(ii) progesterone Overy It control the development
and maintenance of pregnancy
2) Polypeptide hormones :
i) Insulin Pancreas Maintain glucose level in blood,
control carbohydrate metabolism by
increasing glycogen in muscles and
oxidation of glucose in tissue.
3) Amino acid hormones
i) Thyroxine Thyroid gland Increase the rate of energy exchange
and consumption of oxygen and also
regulate metabolism of lipids,
carbohydrates and proteins.
ii) Adrenaline and Adernal medulla They reinforce the function of
non–adrenaline symphatic nervous system like
increase glucose level in blood
and lactic acid in muscles.



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MULTIPLE CHOICE QUESTIONS

SECTION – I : CARBOHYDRATES d) maltose and lactose

12. Which of the following carbohydrate is used in
1. Choose the correct relationship for glucose and
silvering of mirrors?
fructose
a) Glucose b) Sucrose
a) these are functional isomers
c) Cellulose d) Starch
b) these are chain isomers
13. The function of glucose is to
c) these are position isomers
a) provides energy b) promote growth
d) all of these
c) prevent diseases d) perform all above
2. Which one of the following compounds is different
from the rest? 14. Which is the disaccharide present in the milk?
a) Sucrose b) Maltose a) Sucrose b) Maltose
c) Lactose d) Glucose c) Galactose d) Lactose
3. Sugar which will not reduce Fehling's solution is 15. Carbohydrates are stored in the body as
a) maltose b) lactose a) sugars b) starch
c) sucrose d) glucose c) glucose d) glycogen
4. An invert sugar is 16. Sucrose hydrolyses readily in acids to give
a) isorotatory b) dextrorotatory a) two molecules of glucose
c) laevorotatory d) optically inactive b) two molecules of fructose
5. The open–chain glucose, (an aldohexose) and c) one molecules of glucose and fructose
fructose (an 2–oxohexose) have .... and .... chiral d) one molecules of glucose and galactose
carbons respectively 17. Glucose is also known as
a) 4, 4 b) 4, 3 a) grape sugar b) blood sugar
c) 3, 3 d) 3, 4 c) dextrose d) all of these
6. The total number of optical isomers in open–chain 18. Which of the following is polysaccharide?
aldohexose (such as glucose) is a) Glucose b) Ribose
a) 8 b) 8 c) Sucrose d) Starch
c) 16 d) 2 19. All carbohydrates contain
7. Ribose is an example of an a) –CHO group b) >C = O group
a) aldopentose b) keto hexose c) –COO group
–
d) –CONH– group
c) aldohexose d) disaccharide 20. Which of the following statements concerning
8. Digestible carbohydrate, which is also a glucose is incorrect?
constituent of our diet, is a) It has 4 asymmetric C – atoms
a) cellullose b) galactose b) It is an aldehyde
c) maltose d) starch c) It is optically active
9. Which of the following is a aldohexose? d) It is a disaccharide
a) Fructose b) Sucrose 21. Starch is
c) Glucose d) Raffinose a) C 12H22O11 b) C6H10O5
10. Why? Chalk powder is added after complete c) (C6H10O5) n d) (C6H12O6)n
hydrolysis of starch 22. On hydrolysis of starch by dilute acids we get
a) to solidify glucose b) to remove CaSO4 finally
c) to neutralise H2SO4 d) to crystalise starch a) glucose and fructose
11. Common table sugar is a disaccharide of b) glucose
a) glucose and fructose c) fructose
b) glucose and galactose d) sucrose
c) fructose and galactose 23. A carbohydrates insoluble in water is

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 Biomolecules 327
a) glucose b) fructose 36. Which of the following is a animal
c) cellulose d) sucrose polysaccharide?
24. Which of the following monosaccharide is a) Amylose b) Cellulose
pentose? c) Glycogen d) Pectin
a) Ribose b) Fructose 37. Common sugar is
c) glycogene d) Glucose a) glucose b) fructose
25. Carbohydrates may be regarded as c) sucrose d) both 'a' and 'b'
a) aromatic compounds 38. The intermediate compound in the conversion of
b) alicyclic compounds starch to glucose is
c) polyfunctional compounds a) maltose b) lactose
d) all of these c) sucrose d) fructose
26. Glucose cannot be classified as 39. The carbohydrates which reduce Tollen's reagent
a) carbohydrate b) hexose and Fehling's solution are termed as
c) aldose d) olieosaccharide a) non–reducing sugars
27. The colour of precipitate formed when a reducing b) reducing sugars
sugar is heated with Fehling's solution? c) oxidised sugars
a) Yellow b) Red d) both 'b' and 'c'
c) Blue d) Green 40. Glucose and fructose are
28. Glucose is an example of a) optical isomers b) tautomers
a) aldohexose b) ketohexose c) metamers d) functional isomers
c) disaccharide d) non–reducing sugar 41. To become a carbohydrate, a compound must
29. Which of the following is not an essential contain atleast
constituent of carbohydrate? a) 3 carbons b) 6 carbons
a) N b) O c) 4 carbons d) 2 carbons
c) C d) H 42. Which of the following is laevo rotatory?
30. Gum is a) Fructose b) Glucose
a) disaccharide b) monosaccharide c) Sucrose d) All of these
c) polysaccharide d) trisaccharide 43. A solution of d–glucose in water rotates the plane
31. Cane sugar is converted into a mixture of glucose polarised light towards
and fructose by a) right b) left
a) aq. KOH b) H3PO4 c) either side d) none of these
c) ale. NaOH d) dil. HCl 44. Aldotetroses consist of two different chiral
32. A carbohydrate that cannot be hydrolysed into carbon atoms and they exist in
simpler units is called a) 2 optically active forms
a) polysaccharides b) trisaccharides b) 4 optically active forms
c) disaccharides d) monosaccharides c) 6 optically active forms
33. Hydrolysis of sucrose is called d) 8 optically active forms
a) saponification b) inversion 45. Some statement are made below
c) esterification d) hydration 1) glucose is aldohexose
34. Which of the following carbohydrates is a 2) naturally occurring glucose is dextro rotatory
disaccharide? 3) glucose contain three chiral centre
a) Raffinose b) Glucose 4) glucose contain one 10 alcoholic group and
c) Maltose d) Fructose four 20 alcoholic groups
35. Raffinose is an example of Among the above correct statement(s) is/are
a) trisaccharide b) disaccharide a) 1 and 2 b) 3 and 4
c) monosaccharide d) polysaccharide c) 1, 2 and 4 d) all are correct

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46. The term invert sugar refers to an equimolar 3. polymer of glucose is starch
mixture of 4. fatty acids are aliphatic saturated higher
a) glucose and galactose monocarboxylic acids.
b) glucose and fructose Among the above, correct statement(s) is / are
c) glucose and mannose a) only 1 and 3 b) only 4
d) glucose and ribose c) only 1, 3 and 4 d) only 1 and 4
47. Sugar present in fruits is 57. Monosaccharides usually contains carbon atoms
a) glucose b) galactose a) C3 to C10 b) C1 to C6
c) fructose d) sucrose c) C4 to C10 d) C5 to C8
48. Fructose contains 58. All of the statements concering monoraccharides
a) one ketonic group are correct except
b) two primary and three secondary alcoholic a) the number of optical isomers is 2n where 'n' is
groups the number of asymmetric carbon atoms
c) five hydroxy groups b) monoraccharides with 5 to 6 carbons are
d) all of these carbohydrates
49. Glucose gives silver mirror test with Tollen's c) the monoraccharides with C3 to C10 carbons
reagent. It shows the presence of are carbohydrates
a) acidic group b) alcoholic group d) sorbitol is not carbohydrates
c) aldehydic group d) ketonic group 59. Plants produces glucose by the process of
50. Cellulose is a polymer of a) respiration b) autolysis
a) galactose b)  –glucose c) photosynthesis d) dialysis
c) fructose d)  –glucose 60. Glucose is said to have CHO group. Which of
the following reaction is not expected with glucose
51. Starch is a polymer of
a) it form oxime
a)  –glucose b)  –glucose b) it react with NaHSO3
c) fructose d) mannose c) it reduce Tollen's reagent
52. The common source of carbohydrates, fats and d) form n–hexane with HI
proteins is 61. Blood sugar and grape sugar are repectively
a) rice b) milk a) glucose, fructose b) fructose, glucose
c) egg d) ghee c) glucose (of both) d) fructose (of both)
53. Stachyose has formula 62. Consider the following statements about mono
a) C 12H22O11 b) C24H42O21 saccharides
c) C 18H32O16 d) C24H42O24 1) they are optical active compound except
54. In the preparation of glucose from cane sugar, dihydroxy acetone
alcoholic medium is necessary to 2) fructose is ketose sugar but it is reducing sugar
a) get more yield of glucose 3) glucose and fructose are functional isomers
b) effect of separation of product 4) fructose and glucose have same molecular
c) act as catalyst formuls
d) to make reaction faster Among the above correct statements is/are
55. Which one of the following is isomeric with a) 1 and 2 b) 2 and 3
sucrose? c) 3 and 4 d) 1, 2, 3 and 4
a) Lactose b) Ribulose 63. Which is correct statement?
c) Glucose d) Fructose a) Starch is polymer of  –glucose
56. Some statements are given below b) Amylose is component of cellulose
1. glucose is penta hydroxy aldehyde c) Protein are composed of only one type of  –
2. fructose is ketohexose contain four chiral amino acid
center d) celloboise is polysaccharides

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64. Sucrose on treatment with cone. HCl produces c) sucrose d) starch
a) glucose b) fructose 76. Chalk powder is added to hydrolysed solution of
c) invert sugar d) gluconic acid starch during the manufacture of glucose
65. All monosaccharides are defined as a) for hydrolysis it is necessary
a) non reducing sugars b) for cooling of sulphuric acid
b) reducing sugars c) for alkylation of sulphuric acid
c) hydrolysing sugars d) because the hydrolysed solution of starch
d) non–hydrolysing sugars contains excess of sulphuric acid which as
66. When glucose react with bromine water the main neutralised by chalk powder
product is 77. Ethanolic hydrochloric acid is added in the
a) acetic acid b) saccharic acid preparation of glucose from sucrose because
c) gluconic acid d) n–hexane a) hydrochloric acid provides acidic medium
67. To detect reducing and non reducing sugar b) glucose is insoluble in ethanol
following test is used c) fructose is soluble in ethanol
a) Millon test b) Biuret test d) all of these
c) Tollen's test d) Xanthoproteic test 78. Gluconic acid is prepared by
68. The sugar that is disaccharide among the a) oxidation of sucrose with bromine water
following is b) reduction of sucrose with sodium
a) glucose b) maltose c) reduction of glucose with sodium amalgam and
c) xlose d) stachyose water
69. Glucose is reacted with HI gives d) oxidation of glucose by cone. HNO3
a) sorbitol b) n–hexane 79. Oxidation products of glucose are
c) saccharic acid d) gluconic acid a) sucrose b) glucaric acid
70. In the acetylation of glucose, which group is c) gluconic acid d) 'b' and 'c' both
involved in the reaction 80. Which one of the following is the reagent used to
a) CHO group b) >C = O group identify glucose?
c) alcoholic OH group d) all of these a) Neutral ferric chloride
71. Glucose is oxidised by strong oxidising agent b) Chloroform and alcoholic KOH
gives c) Ammoniacal silver nitrate
a) saccharic acid b) gluconic acid d) Sodium ethoxide
c) n–hexane d) sorbitol 81. On complete hydrolysis of starch, we finally get
72. Rhamnose has formula a) Glucose
a) C6H12O5 b) C5H10O4 b) Fructose
c) C5H12O5 d) C5H9O5 c) Glucose and fructose
73. Biomolecules are d) Sucrose
a) aldehydes and ketones 82. Sucrose on hydrolysis gives
b) acids and esters a) Two molecules of glucose
c) carbohydrates, proteins and fats b) Two molecules of fructose
d) alcohols and phenols c) One molecules each of glucose and fructose
74. A glycogen is d) One molecule each of glucose and mannose
a) a polysaccharide found in animals 83. Which one of the following compounds is found
b) a polysaccharide found in plants abundantly in nature?
c) a polysaccharide found in fruits a) Fructose b) Starch
d) an enzyme c) Glucose d) Cellulose
75. The carbohydrate which is found in cotton is 84. Glucose is a
a) glycogen b) cellulose a) Monosaccharide b) Disaccharide

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c) Trisaccharide d) Polysaccharide c) pentahydroxy ketone
85. The most common disaccharide has the molecular d) an hexahydric aldehyde
formula 96. Maltose and glucose are
a) C 10H18O9 b) C10H20O10 a) oxidising sugar
c) C 18H32O11 d) C12H22O11 b) reducing sugar
86. Which one of the following give positive Fehling c) first is oxidising and second is reducing sugar
solution test? d) both are non–reducing sugar
a) Sucrose b) Glucose 97. On heating glucose and Fehling's solution, use get
c) Fats d) Protein precipitate whose colour is
87. Sugars have the suffix a) yellow b) white
a) o1 b) ose c) red d) pink
c) oside d) one 98. The disaccharides that gives only glucose on
88. Carbohydrates have hydrolysis is
a) bitter taste a) Lactose b) maltose
b) sour test c) sucrose d) xylose
c) sweet test 99. The letter D and L in carbohydrates represent
d) some have sweet test and some are tasteless a) it's optical rotation b) its mutarotation
89. The carbohydrates which cannot be hydrolysed c) its direct synthesis d) its configuration
by human digestive system. 100. Sucrose is made up of
a) starch b) cellulose a) D–glucose + L–fructose
c) glycogen d) glucose b) D–glucose + D–fructose
90. Which carbohydrates is an essential constituent c) L–glucose + L–fructose
of plant cells d) L–glucose + D–fructose
a) starch b) cellulose 101.  –D– glucose and  –D– glucose differ from
c) sucrose d) glycogen
each other due to the difference in one of the
91.  – D(+) glucose and  – D(+) glucose are carbon atom with respect to
a) an timers b) epimers a) configuration
c) anomers d) tautomers b) number of OH–groups
92. Glucose is heated with CH3–OH in presence of c) conformation
dry HCl gas  –and  –methyl glucosides are d) size of hemiacetal ring
formed. 102. Which of the following is leavorotatory?
This is because it contain a) glucose b) sucrose
a) aldehyde group b) ketone group c) fructose d) lactose
c) CH2–OH group d) a cyclic structure 103. An example of non–reducing sugar is
93. The number of chiral carbon atoms present in a) maltose b) lactose
cyclic structure  –D(+)) glucose c) cellobiose d) cane sugar
a) 3 b) 4 104. Cellulose is linear polymer of
c) 5 d) 6 a)  – (D) glucose b)  – (D) glucose
94. The reagent can be used to distinguish between
c)  – (D) fructose d) amylose
cane sugar and lactose is
a) Bayer's reagent b) Iodine solution 105. Which is correct statement
c) Millon's reagent d) Tollen's reagent a) starch is polymer of  –glucose
95. Reaction of glucose with (CH3CO)2O suggest b) Amylose is component of cellulose
that c) In cyclic structure of fructose there are five
a) pentahydroxy aldehyde carbons and one oxygen atom
b) hydrate of carbon d) glucose and galactose are anomers

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106. Complete hydrolysis of cellulose gives a)  –D–glucopyranose
a) D–fructose b) D–glucose b)  –D–glucofuranose
c) D–ribose d) D–glucose c)  –D–glucopyranose
107. The two form of glucopyranose obtained from
d)  –D–glucofuranose
D–glucose are known as
a) epimers 114. The monomer unit of starch are
b) anomers a)  –glucose b)  –glucose
c) enantiomers c) pyranose d) galactose
d) geometrical isomers 115. Maltose is made up of
108. Glucose has different from fructose is that a)  – D – glucose
a) does not undergoes hydrolysis b) D – fructose
b) is a monosaccharides c)  – D– glucose and  – D – glucose
c) gives silver mirror test with Tollen's reagent d) glucose and fructose
d) none of the above 116. Reduction of glucose by HI suggest that
109. The term anomers of glucose refer to a) presence of OH groups
a) isomer of glucose that differs in configuration b) presence of –CHO group
at carbon one and four (C – 1 and C– 4) c) cyclic structure of glucose
b) a mixture of D–glucose and L–glucose d) six carbon atoms are arranged in straight chain
c) enantiomers of glucose 117. Glucose is reduced by HI gives
d) isomers of glucose that differ in configuration a) sorbitol
at carbon one (C–1)
b) glucitol
110. Which of the following indicate open chain
c) n–hexane
structure of glucose
d) gluconic acid
a) penta–acetyl derivative of glucose
118. Reaction of bromine water with glucose suggest
b) cyanohydrin formation with HCN
that
c) reaction with hydroxyl amine
a) 1° alcoholic group present in glucose
d) reaction with Br2 water
b) 2° alcoholic group present in glucose
111. Which of the following does not form oxime?
c) aldehyde group present in glucose
a) glucose penta–acetate
d) cyclic structure of glucose
b) glucose
119. Oxidation of glucose by dil.HNO3 gives saccharic
c) xylose acid. This reaction suggest that the presence of
d) galactose a) aldehyde group
112. Which of the terms correctly unidentified the b) 1° – alcoholic group
carbohydrates shown.
c) 2° – alcoholic group
d) ketone group
120. Structure of D–fructose is

CH 2 OH
| CH 2 OH
CO |
| CO
OCH |
1) Pentose 2) Hexose | HO  C  H
3) Aldose 4) Ketose OH  C  OH |
a) b) H  C  OH
5) Pyranose 6) Furanose |
H  C  OH |
a) 2, 3, 4 b) 1, 3, 6 | H  C  OH
CH 2  OH |
c) 1, 3, 5 d) 2, 4, 6 CH 2  OH
113. Cellulose is made up of

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126. Which sets of terms correctly identifies the
CH2 OH CH 2 OH carbohydrates shown
| |
CO CO
| |
OCH HCH
| |
c) OH  C  OH d) H  C  OH
| |
OH  C  OH OH  C  OH
| |
CH2  OH CH 2  OH

121. Which of the following is L–fructose?

1) Pentose 2) Hexose
3) Aldose 4) Ketose
5) Pyranose 6) Furanose
a) 1, 3, 5 b) 2, 4, 5
c) 1, 3, 5 d) 2, 3, 5
127. The following cyclic structure represent

a)  – (D)–glucofuranose

122. Glucose form hemiacetal between CHO group b)  – (D)–glucopyranose

and –OH group on c)  – (D)–glucofuranose
a) C–2 b) C–3 d)  – (D)–glucopyranose
c) C–4 d) C–5 128. Isomerization of glucose produces
123. Anomer of glucose is a) galactose b) fructose
a) six membered five carbon atoms and one c) mannose d) Allose
oxygen atom cyclic structure 129. Fructose form hemiketal between >C = O group
b) five membered five carbon atoms and one and –OH group of
oxygen atom cyclic structure a) C–3 b) C–4
c) six membered six carbon atoms and one c) C–5 d) C–6
oxygen atom cyclic structure 130. Formation of hemiketal in fructose between
d) five membered four carbon atoms and one > C = O group and OH group of C–5 atom, which
oxygen atom cyclic structure carbon atom become chiral
124. Cyclic structure of D–glucose resembles with a) C–1 b) C–2
a) furan b) pyran c) C–3 d) C–4
c) THF d) oxiran 131.  –(D)–(–)fructose and  –(D)–(–)–fructose
125. Aqueous solution of glucose on crystalline at 303k are
produces a) anomers b) epimers
a) anomers b) epimers c) diastereoisomers d) tau tomers
c) enantiomers d) polymers 132. In anomeric forms of fructose which carbon atom

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 Biomolecules 333
involved in ring formation c) 1–2  –  – acetal bond
a) C–2 and C–5 b) C–3 and C–5
d) 1–2  –  – acetal bond
c) C–2 and C–4 d) C–1 and C–5
141. In cyclic structure of cane sugar glycosidic bond
133.  –(D)–(–) fructose and  –(D)–(–)–fructose is formed in between
differs in orientation at a) C–1 of  – D–glucopyranose and
a) C–1 b) C–2
C–5 of  –D–fructofuranose
c) C–3 d) C–4
b) C–5 of  – D–glucopyranose and
134. Cyclic structure of fructose resembles with
a) pyran b) furan C–1 of  – D–glucofuranose
c) pyridine d) oxiran c) C–1 of  – (D)–glucopyranose and
135. Which set of terms correctly identifies the C–2 of  –(D)–fructofuranose
carbohydrate shown. d) C–2 of  – (D)–glucopyranose and
C–1 of  –(D)–fructofuranose
142. Dextro rotatory sucrose is named equal as either
a)  – D–glucopyranosyl  –D–fructofuranoside
b)  – D–glucopyranoside  –D–fructofuranosyl
c)  – D–fructopyronoside
1) Pentose 2) Hexose d) Both a and b
3) Aldoses 4) Ketoses 143. Maltose an hydrolysis produces
5) Pyranose 6) Furanose
a)  –D–glucose b)  –D glucose
a) 2, 3, 5 b) 1, 3, 5
c) 2, 4, 6 d) 1, 4, 6 c)  –D–fructose d)  –D–fructose
136. Non reducing sugar end with suffix 144. In cyclic structure of maltose, acetal bond is
a) oside b) ose formed between
c) one d) al a) C–2 of one glucopyranose and C–2 of another
137. Sucrose on hydrolysis produces equimolar mixture glucopyranose
of b) C–1 of one glucopyranose and C–2 of another
a) D(+)–glucose and D(+)– fructose glucopyranose
b) D(+)–glucose and D(–)– fructose c) C–1 of one glucopyranose and C–4 of another
glucopyranose
c) D(–)–glucose and L(+)– fructose
d) C–1 of one glucopyranose and C–4 of
d) D(–)–glucose and L(–)– fructose
fructofuranose
138. In disaccharide and poly saccharides two or more
145. Maltose contain
monosaccharides units are held together by
a) 2–4–  –acetal bond
a) acetal bond b) glycosidic linkage
b) 1–2–  –acetal bond
c) ether linkage d) all of these
139. Sucrose molecule is formed by monosaccharide c) 1–4–  acetal bond
of d) 1–4–  acetal bond
a)  –D–glucofuranose and  –D–fructopyranose 146. Lactose on hydrolysis produces
b)  –D–glucopyranose and  –D–fructofuranose a)  –D–glucose and  –D–galactose
c)  –D–glucopyranose and  –D–fructofuranose b)  –D–glucose and  –D–galactose
d)  –D–glucopyranose and  –D–fructofuranose c)  –D–glucose and  –D–galactose
140. Sucrose contain d)  –D–glucose and  –D–galactose
a) 1–2  –B – acetal bond 147. In cyclic structure of maltose glycosidic linkage
b) 1–2  –  – acetal bond present between

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a) C–1 of  –D–glucopyranose and consist of
a) amylose and agar
C–2 of  –D–galactopyranose
b) amylopectin and agar
b) C–4 of  –D–glucopyranose and c) amylose and amylopectin
C–1 of  –D–galactopyranose d) amylose and cellobiose
c) C–1 of  –D–glucopyranose and 155. Amylopectin is
a) liner polymer of  –D–glucopyranose
C–4 of  –D–galactopyranose
b) branched polymer of  –D–glucopyranose
d) C–4 of  –D–glucopyranose and c) Linear polymer of  –D–glucopyranose
C–1 of  –D–galactopyranose d) branched polymer of  –D–glucopyranose
148. Maltose contain 156. In amylopectin glycosidic long chain and
a) 1–4–  –glucoside bond branching occurs in between
b) 2–4–  –glucoside bond a) C–1 of one  –D–glucopyranose –C–4 of
c) 1–3–  –glucoside bond another  –D–glucopyranose and branching at
C–1 of one glucopyranose C–6 of another
d) 2–3–  –glucoside bond glucopyranose
149. Cellobiose is obtained by b) C–1 of one  –D–glucopyanose –C–3 of
a) complete hydrolysis of cellulose another  –D–glucopyranose and branching at
b) partial hydrolysis of cellulose C–1 of one glucopyranose and c–s of another
c) complete hydrolysis of glycogen glucopyranose
d) partial hydrolysis of raffinose c) C–1 of one b–D–glucopyranose –C–4 of
150. Cellobiose on hydrolysis produces another b–D–glucopyranose and branching at
a)  –D–glucose b)  –D–fructose C–1 of one P–D– glucopyranose and C–S of
another  –D glucopyranose
c)  –D–glucose d)  –D–fructose
d) C–2 of one  –D–glucopyranose –C–4 of
151. In cyclic structure of cellobiose acetal bond is
another  –D–glucopyranose and branching at
formed between
C–1 of  –D–glucopyranose and C–6 of another
a) C–1 of one  –D–glucopyranose and  –D–glucopyranose
C–2 of another  –D–glucopyranose 157. In amylopectin glycosidic branching present in
between
b) C–1 of one  –D–glucopyranose and
a) 1–4  –D–glucopyranose
C–4 of another  –D– glucopyranose
b) 1–4–  –D–glucopyranose
c) C–1 of one  –D–glucopyranose and c) 1–6  –D–glucopyranose
C–4 of another  –D fructofuranose d) 1–6  –D– glucopyranose
d) C–1 of one  –d–glucopyranose and 158. Amylose contain
C–4 of another  –D–glucopyranose a) C–1–C–4  –D–glycosidic bond
152. Cellobiose contain b) C–1–C–4  –D–glycosidic bond
a) C–1–C–4 glycosidic bond c) C–1–C–6  –D–glycosidic bond
b) C–1–C–3 glycosidic bond
d) C–1–C–4  –D–glycosidic bond
c) C–2–C–4 glycosidic bond
d) C–3–C–4 glycosidic bond 159. Which of following has similar glycosidic bond
153. Starch on hydrolysis produces a) maltose and lactose
b) maltose and cellobiose
a)  –D–glucose b)  –D–glucose
c) Amylose and amylopectin
c)  –D–fructose d)  –D–fructose d) maltose and amylose
154. In starch molecule  –D–glucose molecule 160. Amylose and amylopectin are constituent of

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 Biomolecules 335
a)  –D– fructose
b)  –D – glucose
c)  –D– fructose
d)  –D– fructose
161. Starch composed of
a) amylose and amylopectin
b) sucrose and maltose
c) maltose and lactose
d) amylose and celloboise
162. In cyclic structure of cellulose glycosidic bond
present in between
a) C–1 of one  –D–glucopyranose and
C–4 of another  –D–glucopyranose
b) C–1 of one  –D–glucopyranose and
C–4 of another  –L–glucopyranose
c) C–1 of  –D–glucopyranose and
C–4 of  –D–glucopyranose
d) C–1 of  –D–glucopyranose and
C–4 of  –D–fructofuranose
163. Cellulose contain
a) C–1  C–4  –D glucopyranose glycosidic
bond
b) C–1  C–S  –D glucopyranose glycosidic
bond
c) C–1  C–4  –D glucopyranose glycosidic
bond
d) C–1  C–4  –D fructofuranose glycosidic
bond
164. Glycogen is also known named as 168. How many chiral carbon atoms are present in
a) Plant starch ribulose?
b) animal starch a) 2 b) 3
c) celloboise c) 4 d) 5
d) Dextrin
169. A 
H O
 glucose + fructose
3


165. Which of the following has highly branched of

structure like amylopectin? B 
H O
3
 glucose + galactose


a) cellulose C   glucose + glucose


H O
3

b) maltose
The A, B, C are respectively
c) fructose
a) sucrose, lactose, maltose
d) glycogen
b) maltose, lactose, sucrose
166.  –D–galactose and  –D– galactose are c) sucrose, maltose, lactose
a) epimers b) metamers d) maltose, sucrose, lactose
c) anomers d) tautomers 170. Lactose can be names as
167. Which of following is maltose. a)  –D–glucospyranosyl  –D–galactospyranose

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 Biomolecules 336
b)  –D–glucospyranosyl  –D–galactospyranose a) 3 b) 4
c)  –D–galactopyranosyl  –D–glucopyranose c) 5 d) 6
d)  –D–galactopyranosyl  –D–glucopyranose 173. Which of the following statement is / are correct?
171. Hydrolysis of sucrose with dilute aqueous I. Glucose is reducing sugar
sulphuric acid yields II. Sucrose is reducing sugar
a) 1:10 (+)–Glucose; 0–(–)– fructose III. Maltose is non reducing sugar
b) 1:20 –(+)–Glucose; 0–(–)– fructose IV.Lactose is reducing sugar
c) 1:10 –(–)–Glucose; 0–(+)– fructose a) I and II only
d) 1:20 –(–) Glucose; 0–(+)– fructose b) I and III only
172. The number of chiral centres in the pyranose form c) I and IV only
of glucose is d) All


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MULTIPLE CHOICE QUESTIONS

SECTION – II : PROTEINS a) Wool b) Hair

c) Cellulose d) Nail
1. Amino acids are produced by the hydrolysis of
11. A peptide bond joins two amino acids together.
a) fats b) proteins
What atoms are linked by this bond in chain?
c) nucleic acids d) carbohydrates
a) C–O b) C–H
2. Which among the following statements are true
c) C–N d) N–S
for glycine?
12. Which one of the following elements is not found
1) It exists in crystalline form
in proteins?
2) It is optically active
a) N b) F
3) It is soluble in water
c) C d) O
4) It can form Zwitter ions
13. Which one of the following is the general
a) 1, 2, and 3 b) 1, 2, and 4
structural formula of an amino acid?
c) 1, 3 and 4 d) 2, 3 and 4
a) RCH2CONH2 b) RCH(NH2)OH
3. Which among the following peptide linkage?
c) RCH2NH2 d) RCH(COOH)NH2
a) C  NH  b) C  O  N  14. The functional group CONH found in protein is
|| | | called as
O R H
a) amide group b) carboxylic acid group
c) C  N d)  N  C  O c) peptide d) both 'a' and 'c'
| | 15. Which of the following is a fibrous protein?
R R a) Haemoglobin b) Keratin
4. Amino acids are the building block of c) Albumin d) Enzymes
a) fats b) vitamins 16. Which of the following is globular protein?
c) proteins d) carbohydrates a) Collagen b) Heamoglobin
5. An essential amino acid is one that c) Myosin d) Fibroin
a) must be included in the diet 17. Magnesium is present in
b) occurs in all types of protein a) haemoglobin b) chlorophyll
c) contains no sulphur c) casein d) keratin
d) the body synthesis 18. Iron present in haemoglobin is an
6. The simplest amino acid is a) ferrous state
a) alanine b) valine b) ferric state
c) tyrosine d) glycine c) partly in ferrous and partly in ferric
7. An amino acid with a phenolic hydroxyl group is d) elemental state
a) alanine b) tyrosine 19. Polymers of more than 10000 amino acids are
c) valine d) phenyl glycine termed
8. Which of the following statements is not true? a) proteins b) tripeptide
a) Protein is polypeptide c) dipeptide d) oligopeptide
b) Two peptides can form two different amino 20. Proteins are used as
acids a) enzymes b) antivirus vaccines
c) Peptides are not a–amino acids c) food d) all of these
d) Peptides have amide linkage 21. Proteins are hydrolysed by enzymes into
9. Which one of the following is a protein? a) hydroxy acids b)  –amino acids
a) Rayon b) Nylon c) dicarboxylic acid d) none of these
c) Natural silk d) Dacron 22. Proteins contain
10. Which one of the following is not a protein? a) C, H, O b) only N

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 Biomolecules 338
c) C, H d) C, H, O and N 34. Fabroin is term related to
23. Who proved that in proteins the amino acids are a) hair b) milk
linked together by peptide linkage? c) horn d) silk
a) Emil Fisher b) Cannizzaro 35. The main structural feature of protein is
c) Kekul d) Hoffman a) peptide linkage b) ester linkage
24. Which of the following food stuffs contain c) ether linkage d) all of these
nitrogen? 36. Which of the following is structural protein?
a) Glucose b) Fats a) Albumin b) Insulin
c) Proteins d) None of these c) Thyroglobulin d) Albumin
25. Insulin is a 37. Protein on hydrolysis gives
a) hormone b) enzyme a)  –aminoacids b)  –aminoacids
c) carbohydrate d) fat
c)  –aminoacids d) o–aminoacids
26. Keratin present in hair is,
38. The peptide bond joining amino acid into proteins
a) Fibrous protein b) Globular protein
is a specific example of
c) Denaturad protein d) Lipo protein
a) ester b) carbonyl
27. Which of the following molecules is capable of
c) glycosidic d) amide
forming Zwitter ion?
39. Two functional group that are present in all amino
a) NH2CH2COOH b) CH3CH2NH2
acids are the
c) CH3CH2COOH d) All of these
a) hydroxy, amine b) hydroxy, amide
28. Which one of the following is an example of
c) carboxyl, amino d) carboxyl, amide
fibrous proteins?
40. Collagen is a example of
a) Collagen in bone b) Myosin in muscles
a) carbohydrates b) oils
c) Fibroin in silk d) All of these
c) fats d) proteins
29. Enzymes belong to
41. Consider the following statements about proteins
a) synthetic polymers b) polysaccharides
1) All natural amino acids which constituents of
c) polypeptides d) polyesters
proteins are L – amino acids
30. The protein that transports oxygen in the blood
2) glycine is optically active
stream is
3)  – amino acids are connected by ester
a) haemoglobin b) insulin
linkage
c) collagen d) albumin
4) Myosin is structural protein
31. Antibodies are
Among these statements
a) enzymes b) hormones
a) only 1 and 4 are correct
c) proteins d) amino acids
b) only 2 and 3 are correct
32. Polymer of  –amino acid is
c) only 3 and 4 are correct
a) acetamide b) ammonia
d) only 4 is correct
c) protein d) fatty acids
42. Simplest proteins has one peptide linkage. It is
33. Some statements are given below
a) tripeptide b) dipepetide
1. –CONH– linkage present in all proteins
c) tetrapeptide d) oligopeptide
2. proteins are addition polymer of  –amino
43. Consider the following compound
acids
1) tyrocine 2) terephthalic acid
3. proteins are condensation polymer of  –
amino acids 3) adipic acid 4) glucanic acid
4. all polyamide are called proteins which can form zwitter ion?
Among the above, correct statement(s) is / are a) only 2 b) 1, 2, 3
a) only 1 b) only 3 c) only 1 d) 1, 2, 3, 4
c) only 1 and 4 d) only 1 and 3 44. Following acid can't not from a–amino acid
a) succinic acid b) tryptophane

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 Biomolecules 339
c) phenyl alanine d) tyrosine a) Amino acid residues join together to make a
45. Peptides are amino acid polymer in which the protein molecule
individual amino acid units are called b) Proteins are polymers with formula
a) monomer b) residue (C 6 H 10O 5 ) n
c) epimer d) amide c) Eggs are rich in protein
46. A tripeptide has, how many peptide bond d) Pulses are good source of proteins
a) 1 b) 2 56. Which one of the following proteins transports
c) 3 d) 4 oxygen in the blood stream ?
47. Most of the amino acid have chiral centres but a) Myoglobin b) Insulin
not in c) Albugmin d) Haemoglobin
a) phenyl alanine b) tryptophane 57. Mg is an important component of which
c) tyrocine d) glycine biomolecule occurring extensively in living world?
48. Select correct statement a) Haemoglobin b) Chlorophyll
a) Valine is essential amino acid c) Vitamin d) ATP
b) in peptide linkage oxygen and hydrogen are at 58. Which is not true statement?
trans positions a) Protein is polymer of  –amino acids
c) molecular mass up to 10,000 are called b) All proteins are found in L–form
polypeptide c) Human body can synthesize all proteins they
d) all are correct need
49. All of the following are example of fibrous proteins d) At pH–7 both amino and carboxylic groups
except exist in ionised form
a) wool b) silk 59. The functional group, which is found in amino
c) horn d) insulin acid, is
50. The amino acids, which build up proteins, have a) –CH3 group b) –NH2 group
both the COOH and NH2 groups. These amino c) –COOH group d) both 'b' and 'c'
acids are 60. Amino acids usually exist in the form of zwitter
a)  –amino acids b)  –amino acids ions. This means that it consists of
a) no acidic or basic group
c)  –amino acids d) o–amino acids
b) basic group – NH3+ and acidic group – CO2
51. Tyrosin contains
c) basic group – CO2 and acidic group – NH3+
a) alcoholic OH group
d) basic group – COO– and acidic group – NH3+
b) phenolic OH group
61. Aqueous solution of  –amino acid is slightly
c) aldehyde group
acidic, which is due to
d) ketonic group
a) Acidic character of NH3+
52. Thyroglobin is an example of
b) basic character of COO–
a) scieroproteins b) structural proteins
c) Acidic character of COO–
c) fibrous proteins d) globular proteins
d) basic character of NH3+
53. Large molecules can be formed by the
62. Which of the following have coiled helical
combination of a number of smaller molecules.
structure
These smaller molecules are called
a) Lipids b) carbohydrates
a) isomers b) monomers
c) vitamins d) proteins
c) epimer d) polymers
63. Helical structure of protein is stabilized by
54. Polypeptides are the chains of
a) ionic bond
a) amino acids b) nitrogen atoms
b) covalent bond
c) hydrogen atoms d) oxygen atoms
c) Vander Waals forces
55. Which of the following statements about proteins
is not true? d) hydrogen bond
64. Coagulation of protein is known as

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 Biomolecules 340
a) dehydration b) decay prosthetic group
c) deamination d) denaturing d) linear sequence of amino acid in polypeptide
65. Point out wrong statement about protein chain
a) They are nitrogenous organic compound with 75. One of essential  –amino acid is
high molecular weight a) Lysin b) serine
b) On hydrolysis by enzymes give  –L–amino c) glycine d) proline
acids 76. Which of the following biomolecule contain
c) Many of them are enzymes nontransition metal ion?
d) They do not contain polypeptide chain a) insulin b) chlorophyll
66. The only  –amino acid which is achiral is c) haemoglobin d) vitamin B–12
a) Lysin b) glycine 77. Which of the of following shows aromatic
c) proline d) alanine properties
67. The number of amino acids which form proteins a) valin b) serine
in nature is c) leucine d) tyrosine
a) 10 b) 15 78. The helical structure of protein is stabilized by
c) 20 d) 30 a) dipeptide bond
68. Which one is not the essential constituent of b) glycosidic bond
balance diet c) intramolecular hydrogen bond between–NH
a) carbohydrates b) vitamins and carbonyl oxygen
c) fats d) hormones d) Intermolecular hydrogen bond between –NH
69. The PH value of solution in which the polar amino and carbonyl oxygen
acid does not migrate under the influence of 79. Which one of the bio–rnolecule is insoluble in
electric field is called water
a) neutralization point a) keratin b) haemoglobin
b) isoelectronic point c) insulin d) globulin
c) isoelectric point 80. Which of the following exist as Zwitter ion?
d) iso–merisation point a) salicylic acid b) sulphanilic acid
70. The human body does not produce c) ethanamine d) p–aminoacetophenone
a) vitamins b) proteins 81. Which of the following is not globular protein?
c) enzymes d) hormones a) keratin b) haemoglobin
71. Point out wrong statement about proteins c) insulin d) thyroglobulin
a) These are polymeric macromolecules 82. Which of the following statement about a–amino
b) They are present in food stuff acid is not true?
c) Many of them are hormones and enzymes a) They are present in all protein
d) They do not contain CONH group b) Most naturally occurring amino acid have D–
72. Fibrous protein are present in configuration
a) wool b) haemoglobin c) They are characterized by isoelectric point
c) albumin d) thyroglobulin d) Glycine is the only naturally occurring  –
73. Globular protein is present in amine acid which is achiral
a) silk b) horn 83. A tripeptide is written– glycine–alanine–glycine.
The correct structure of tripeptide is
c) keratin d) blood
74. Secondary structure of protein refer to
a) Three dimensional structure, specially the bond
between amino acid residue that are distant from
each other in the polypeptide chain
b) regular folding pattern of polypeptide chain
c) mainly denatured proteins and structure of

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 Biomolecules 341
2) Sulphur containing B) Histidine
amino acid
3) Basic amino acid C) Cystine
4) Optically inactive amino acid D) Tyrosine
a) 1 – A, 2 – B, 3 – C, 4 – D
b) 1 – D, 2 – B, 3 – A, 4 – C
c) 1 – B, 2 – D, 3 – A, 4 – C
d) 1 – D, 2 – C, 3 – B, 4 – A
90. The bond in protein structure, that are notbroken
in denaturation is
a) hydrogen bond b) ionic bond
c) peptide bond d) sulphide bond
91. The amino acid which has a nonpolar side chain
is
a) isoleusine b) aspartic acid
84. Denaturation of protein c) serine d) lysine
a) disrupts the 1° and 2° structure of proteins 92. Non essential  –amino acid is
b) disrupts the 2° and 3° structure of proteins a) Alanine b) Valine
c) disrupts 1°, 2°, 3° structure of proteins c) Leucine d) Lysine
d) is reversible process 93. Proteins are polypeptide of
85. Which of the following consist of only essential
a)  –amino acid b)  –hydroxy acid
 –amino acids
a) glycine, serine, proline c) D–  –amino acid d) L–  –amino acid
b) valine, glycine, lucine 94. Which of the following is a protein
c) serine, Tryptophan, proline a) serine b) glycogen
d) valine, lucine, tryptophan c) alanine d) keratin
86. A nanopeptide contain how many peptide bond 95. Enzymes belong to which class of compound
a) 7 b) 9 a) polysaccharide
c) 8 d) 10 b) polypeptide

c) hydrocarbon
87. An  –amino acid exist as H3 N – CH2 – COOH d) nitroheterocyclic compound
at PH(2) and its isoelectric point is 6. The amino 96. Which of the following structural unit found in
acid at PH 10.97 will exist as enzymes or hormones

a) H3 N –CH2 –COOH–
b) H2N –CH2 –COO–

c) H3 N –CH2 –COOH
d) H2N –CH2 –COOH
88. Basic  –amino acids are
a) aspartic acid and histidine 97. The acid showing salt like character in aqueous
b) arginine and histidine solution is
c) lysin and histidine a) Acetic acid b) citric acid
d) serine and histidine c) proline d) fumaric acid
89. Match the list, and select correct answer from 98. Which of following  –amino acid is achiral
given codes a) Alanine b) proline
List I List – II c) Valin d) glycine
1) Cheese like amino acid is A) Glycine 99. Formation of cheese from milk is not

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 Biomolecules 342
a) denaturation
b) breaking of hydrogen bond
c) breaking of ionic bond
d) breaking of peptide bond
100. Which of the following statement is true for
proteins?
a) They act as antibodies
b) They act as hormones
c) They catalyze the biochemical reaction
d) all of these
101. Which bond is not present in a–helix structure of 107. Which of the following provide chief structural
proteins material for tissues
a) intermolecular hydrogen bond a) myosin b) insulin
b) intramolecular hydrogen bond c) albumin d) pepsin
c) sulphide bond 108. During coagulation of egg which change occurs
d) Vander Waals forces 1) breaking of peptide bond
2) breaking of hydrogen bond
102. In  –pleated structure polypeptide chain is
3) breaking of ionic bond
a)  –helix b)  –helix 4) breaking of sulphide bond
c) zig–zag d) linear a) only 1 b) 1,2,4
103. Which of the following has  –pleated structure c) 2,3,4 d) 1,2,3,4
a) oxytocin b) mucin 109. Curdling of milk is
c) fibroin of silk d) insulin a) naturation of protein
104. Tertiary structure of protein is arises due to b) denaturation of protein
a) folding of polypeptide chain c) folding of polypeptide chain
b) folding, coiling and bonding of polypeptide chain d) coiling of polypeptide chain
c) linear sequence of amino acid in polypeptide 110. Tertiary structure of protein is stabilized by
chain 1) hydrogen bond
d) denatured proteins 2) ionic bond
105. Linear sequence of polypeptide bond refer in 3) sulphide bond
a) secondary structure 4) Vander Waals force
b) primary structure a) 2, 4 b) 1, 2
c) tertiary structure c) 1, 2, 4 d) 1, 2, 3, 4
d) quaternary structure 111. The PH – 5, glycine exist as
106. A tripeptide is written phenyl alanine and glycine. a) H3N+ – CH2 – COO–
The correct structure of tripeptide is b) H3N+ – CH2 – COOH
c) H2N – CH2 – COO–
d) H2N – CH2 – COOH
112. The PH –7 alanine exist as
 
a) H3 N  CH  COO
|
CH 3


b) H3 N  CH  COOH
|
CH3

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 Biomolecules 343

c) H 2 N  CH  COO  b) H3 N   CH  COOH
| |
CH 3 CH3

c) H 2 N  CH  COO 
d) H 2 N  CH  COOH |
| CH 3
CH 3
d) H 2 N  CH  COOH
113. At PH–9, alanine exist as |
CH 3
a) H3 N   CH  COO 
| 114. Which of the following is not protein?
CH 3 a) wool b) hair
c) nail d) starch


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MULTIPLE CHOICE QUESTIONS

SECTION – III : LIPIDS d) enzyme

12. Which of the following is an ester?
1. The group linkage present in fats is
a) Soap b) Seed oil
a) peptide linkage b) ester linkage
c) Glycerine d) Kerosene oil
c) glycosidic linkage d) none of these
13. Which of the following reaction takes place during
2. A distinctive and characteristic functional group
the preparation of triglyceride?
of fats is
a) If–atom from –OH group of glycerol is
a) an ester group b) a peptide group
replaced by acetyl group
c) a ketonic group d) an alcoholic group
b) H–atom from –OH group of glycerol is
3. The alcohol obtained by the hydrolysis of oils and
replaced by acyl group
fats is
c) –OH–group of glycerol and H–atom of from
a) glycol b) glycerol
carboxylic group of the acid are eliminated as
c) propanol d) pentanol H2O molecule.
4. Most concentrated source of energy in human d) H–atom from –OH group of glycerol is
body is replaced by alkyl group
a) nucleic acid b) sugars 14. Fats and oils are formed from respectively.
c) fats d) proteins a) glycerol and long chain unsaturated acids only
5. Lipids are b) glycerol and long chain saturated acids only
a) amino acids c) glycerol and long chain saturated acids and
b) carbohydrates unsaturated acids
c) enzymes d) ethylene glycol and long chain unsaturated and
d) esters of long chain fatty acids and alcohols saturated acids
6. Oils and fats are esters of higher fatty acids with 15. The molecular formula of saturated fatty acid is
a) glycerol b) glycol a) CnH2nO2 b) CnH2n–1O2
c) alcohol d) ethers c) C nH2n+2O2 d) CnH2n+1O2
7. Fats and oils are 16. Fats contain higher percentage of
a) aldehydes b) esters a) unsaturated fatty acids
c) acids d) alcohols b) saturated fatty acids
8. Glycerol tristearate (stearin) can not undergo, c) free fatty acids
which of the following reaction? d) glycerol
a) Saponification b) Acid hydrolysis 17. Which of the following are lipids?
c) Hydrogenation d) None of these a) Only oils b) Only fats
9. The most important reserves food of animals are c) Oils and fats d) Sugars
a) carbohydrates b) proteins 18. Oils are
c) vitamins d) fats a) triglycerides of saturated fatty acids
10. Vegetable oils are b) triglycerides of unsaturated fatty acids
a) glycerides of unsaturated fatty acids c) diglycerides of saturated fatty acids
b) glycerides of saturated fatty acids d) diglycerides of unsaturated fatty acids
c) sodium salts of higher fatty acids 19. Glycerides are
d) mixture of sodium and potassium salts of lower a) esters of fatty acids and glycol
acids b) esters of fatty acids and glycerol
11. The function of fat in the body is to act as c) esters of fatty acids and sorbitol
a) thermal insulator d) esters of fatty acids and glucose
b) an absorber of minerals 20. Which of the following compound does not
c) catalyst

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 Biomolecules 345
belongs to liquids? a) vitamin A b) prostaglandians
a) Fats b) Ethanol c) Lecithin d) vitamin O
c) Ethanoic acid d) Oils 32. In phospholipids
21. Some statements are given below about oils and a) two –OH group of glycerol are esterified
fats b) One OH group of glycerol is esterified
1. oils can be converted into fats and vice versa c) Three OH groups of glycerol are esterified
2. oils and fats are triesters d) No any OH group of glycerol is esterified
3. oils have high melting point than fats 33. In plant glycolipids sugar is
4. fats have strong Vander Waals force of a) glucose b) fructose
attraction than oils, c) galactose d) mannose
Among the above, correct statement(s) is / are 34. The typical animal glycolipids is
a) only 2 b) only 4 a) lecithin b) cephalin
c) only 2 and 4 d) only 1 c) prostaglandins d) cerebrosides
22. Monoterpene contain how many carbon atoms? 35. Waxes are
a) 10 b) 12 a) ester of long chain carboxylic acids and long
c) 14 d) 16 chain monohydric alcohols
23. The characteristic functional group of fats is b) polypeptides of long chain nitrogen base
a) an ester group b) ether c) long chain fatty acid
c) a peptide group d) an alcoholic group d) esters of long chain aldehydes and ketones
24. A glyceride is 36. Steroids are derived from
a) an ether formed by glycerol a) highly branched glycerides
b) an ester of glycerol with fatty acids b) long chain fatty acids
c) a molecular compound of glycerol with a metal c) cyclopenta perhydrophenanthrene
salt d) galacto cerebrosides
d) none of these 37. Which of the following is simple lipids?
25. Which alcohol reacts with fatty acids to form fats? a) Fat soluble vitamins
a) Ethanol b) Glycerol b) prostaglandins
c) Methanol d) Isopropanol c) anomer of D–glucose
26. Which is not essential constituent of diet? d) both a and b
a) Soap b) Glucose 38. Which of the following do not contain ester
c) Carbohydrate d) Protein linkage
27. Main elements present in lipids are a) lecithin b) oils
a) C b) H c) fats d) cholesterol
c) O d) C, H, O 39. Testosterone is
28. Which of the following is lipids a) animal steroid
a) fats b) glycogen b) plant steroid
c) blood d) pepsin c) ester of long chain fatty acid
29. Lipids serves d) triolein
a) biocatalyst b) transport oxygen 40. Terpenes are
c) provide energy d) provide immunity a) four ring cyclic structure
30. Complex lipids contains b) unsaturated hydrocarbon
a) phosphoric acid c) fatty acids
b) phosphorous acid d) containing heterocyclic ring
c) hyphophosphoric acid 41. Which of the following is plant steroid?
d) metaphosphoric acid a) estrogen b) testosterone
31. Which of the following is phospholipids? c) androsterone d) sitosterol

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42. Ergosterol is 48. Which of the following is steroid nucleus
a) animal sterol b) fungal sterol
c) terpenes d) simple lipids
43.  –carotin is
a) monoterpene b) sequiterpene
c) diterpene d) tetra terpene
44. Prostaglandins is
a) a group of C20 lipids
b) a group of C10 lipids
c) a group of C50 lipids 49. Isoprene unit present in
d) a group of C10D lipids a) terpenes b) waxes
45. Which of the following is detected in body tissues c) phospholipids d) glycolipids
a) testosterone b) estrogen 50. Phytol is
c) sitosterol d) prostaglandins a) oils b) fats
46. Abscisic acid is c) terpenes d) glycolipids
a) triglyceride of long chain alcohol 51. The glycolipids abundantly found in
b) sesquiterpenes a) oils
c) diterpenes b) fats
d) glycoli pids c) keratin in hair
47. Fat soluble vitamins are d) myelin sheath of neurons
a) Proteins b) Complex lipids

c) Simple lipids d) Carbohydrates

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 Biomolecules 347

SECTION – IV : ENZYMES b) nucleic acids

c) proteins and nucleic acids
1. Which catalyzed biological reaction. d) carbohydrates and nucleic acids
a) hormones b) enzymes 2. The relation between the nucleotide triplets and
c) glycogen d) fats the amino acid is called
2. Enzymes are a) enzymes b) replication
a) carbohydrates b) lipids c) genetic code d) mutation
c) fats d) polypeptides 3. Bases common to RNA and DNA are
3. The function of enzymes in living system is to a) adenine, guanine, cytosine
a) transport oxygen b) adenine, uracil, cytosine
b) provide immunity c) adenine, guanine, thymine
c) catalyze biochemical reaction d) guanine, uracil, thymine
d) provide energy 4. In nucleotide phosphate group is attached to
4. Which of the following statement about enzymes a) C – 1 b) C – 2
is / are true c) C – 4 d) C – 5
1) enzymes lacks in nucleophilic group 5. In nucleoside adenine is attached to
2) enzymes are highly specific and selective a) C – 2 b) C – 1
3) Enzyme catalyze the chemical reaction by c) C – 3 d) C – 4
lowering the activation energy
6. In nucleic acid the sequence is
4) pepsin is enzymes
a) base–phosphate–sugar
a) 2, 3, 4 b) 1, 4
b) base–sugar–phosphate
c) 2, 3 d) 1, 2, 3, 4
c) sugar–base–phosphate
5. Which of the following is not correct for enzymes
d) phosphate–base–sugar
a) It acts as biocatalyst
7. A base–sugar–phosphate unit in nucleic acid is
b) It can catalyze any chemical reaction called as
c) It increase rate reaction by lowering activation a) base phosphate b) nucleotide
energy
c) phosphotide d) nucleoside
d) Maltose convert glucose by using maltase
8. Nucleic acids are
enzyme
a) polymer of nucleoside
6. Enzymes are made up of
b) polymer of purine base
a) carbohydrates
c) polymer of nucleotides
b) nitrogen containing carbohydrates
d) polymer of pyrimidine base
c) edible proteins
9. The function of DNA is
d) protein with specific structure
a) to synthesis RNA
7. The effect of enzymes on a biological reaction is
that the b) to synthesis necessary proteins
a) rate of forward reaction is increased but the c) to carry the hereditary character
rate of backward reaction is not altered d) all are correct
b) rate of backward reaction is decreased but 10. RNA is
rate of forward reaction is not altered. a) single helix strand b) double helix strand
c) rate of forward reaction and backward c) triple helix strand d) all of these
reaction are altered by the same factor so that 11. Which of the following is responsible for the
d) neither rate of forward reaction nor that of heredity character of cell
backward reaction is altered a) RNA b) DNA
SECTION – V : NICLEIC ACID
c) proteins d) hormones
12. The reason for helical structure of DNA is
1. Chromosomes are made from operation of
a) proteins a) hydrogen bond

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 Biomolecules 348
b) electrostatic attraction 23. In DNA complimentary bases are
c) Vander Waals forces a) Adenine and guanine, thymine and cytosine
d) dipole–dipole attraction b) Adenine and thymine, cytosine and guanine
13. The purine base present in RNA is c) Adenine and cytosine, guanine and thymine
a) adenine b) cytosine d) Thymine and uracil, cytosine and guanine
c) uracil d) thymine 24. Which of the following is not present in
14. Nucleoside on hydrolysis gives nucleotide?
a) an aldopentose and heterocyclic base a) cytosine b) adenine
b) an aldopentose and orthophosphoric acid c) guanine d) tyrosine
c) an aldopentose, heterocyclic base and or tho 25. Which of the following is not present in
phosphoric acid nucleoside?
d) heterocyclic base and orthophosphoric acid a) phosphoric acid b) cytosine
15. Which of the following statement is true for c) uracil d) guanine
protein synthesis (translation) 26. Consider the double helix structure of DNA. The
a) Amino acids are directly recognize by m–RNA base pair are
b) The third base of codon is less specific a) part of the back bone structure
c) Only one codon codes for an amino acids b) in side the helix
d) every t–RNA molecule has more than one c) out side the helix
amino acid attachment d) all of these
16. DNA multiplication is called as 27. Mutation in DNA occurs due to change in the
a) translation b) transduction sequence of
c) transcription d) replication a) nitrogen base b) ribose unit
17. Pyrimidine base present in DNA are c) phosphate unit d) all of these
a) Adenine and cytosine 28. DNA consist of
b) Guanine and thymine a)  – D – ribose sugar
c) cytosin and thymine b)  –D deoxyribose sugar
d) Adenine and guanine
c)  –D ribose sugar
18. Thymine is
d)  –0 – deoxyribose sugar
a) 1 – methyl uracil b) 3 – methyl uracil
29. Polynucleotide chain is
c) 4 – methyl uracil d) 5 – methyl uracil
a) polyamide chain b) polyester chain
19. RNA differ from DNA in respect to base
c) polypeptide chain d) polyglycosidic chain.
a) Thymine b) Cytosine
30. Uracil pyrimidine base is present in
c) Adenine d) Guanine
a) DNA b) RNA
20. DNA differ from RNA in respect to base
c)  –D – ribose d)  –D – deoxyribose
a) uracil b) cytosine
c) adenine d) guanine 31. In nucleoside base unit is attached at
21. RNA and DNA are chiral molecule, their chirality a) position one of pentose sugar unit
is due to b) position two of pentose sugar unit
a) chiral phosphate ester linkage c) position three of pentose sugar unit
b) D – sugar component d) position of four of pentose sug~r unit
c) L – sugar component 32. Nucleoside consist of
d) chiral base a) sugar and H3PO4
22. A sequence of how many nucleotides in b) sugar and base
messenger RNA makes a codon for an amino c) H3PO4 and base
acid? d) only pentose sugar unit
a) 2 b) 3 33. In nucleotide phosphonic acid link at position.
c) 4 d) 5 a) one of pentose sugar

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 Biomolecules 349
b) one of base unit 6. Structural unit of vit–B is
c) five of pentose sugar
d) five of base unit
34. The linkage present in two nucleotide is
a) amide linkage
b) peptide linkage
c) phosphodiester linkage 7. Structural unit of vitamin–A is
d) glycosidic linkage
35. Phospho diester linkage present between
a) 1 and 2 carbon atoms of two pentose sugar
b) 3 and 5 carbon atoms of two pentose sugar
c) 2 and 3 carbon atoms of two pentose sugar
d) 1 and 3 carbon atoms of two pentose sugar 8. The vitamins stored in body is
36. Pentose sugar present in RNA is a) vit–C b) vit–B1
a)  – D – ribose c) vit–B1 d) vit–D
9. Vitamin – C is
b)  – D – ribose
a) citric acid b) ascorbic acid
c)  – D –2– deoxyribose
c) lactic acid d) tartaric acid
d)  – D – deoxyribose 10. Ascorbic acid is
37.  – D – 2 – deoxyribose means a) protein b) vitamin
a) no H – atom at C – 2 position c) enzyme d) oil
b) no OH – group at C – 2 position 11. Chemical name of vitamin–A is
c) no – H – atom at C – 3 position a) thiamine b) axerophthol (retinol)
d) no – OH – group at C–3 position c) thiamine d) nicotinamide
SECTION – VI : VITAMINS 12. Vitamin that is most readily produced in our
body is
1. Which biomolecule doesn't produce in human a) vit–C b) vit–B
a) protein b) glycogen c) vit–D d) vit–P
c) testosterone d) vitamins 13. Vitamin A deficiency leads to disease known as
2. Which of the following are water soluble vitamins a) beri–beri b) T.B.
1) vit–B, 2) vit–C c) Join pain d) night blindness
3) vit–E 4) vit–D 14. Which of the following is found in cod–liver oil?
a) 1, 3 b) 1, 2 a) vit–A b) vit–C
c) 3, 4 d) 1, 4 c) vit–E d) vit–B1
3. Which of the following are fats soluble 15. Deficiency of vit–E causes
vitamins? a) scurvy b) beri–beri
1) vit–A 2) vit–D c) antifertility d) TB
3) vit–H 4) vit–K 16. Vit–B2 is also known as
5) vit–C a) Tocopherols b) Retinol
a) 1, 2, 4 b) 2, 4, 5 c) riboflavin d) pyridoxine
c) 3, 4, 5 d) 1, 2, 3, 4 17. Vit–B1 is known as
4. The vitamin which contain aromatic ring is a) Retinol b) thiamine
a) vitarnin–K b) Vitamin–C c) riboflavin d) ascorbic acid
c) vitamin–A d) vitamins–B 18. The vitamin which is water soluble and antioxidant
5. Vitamin–C is a) vit–A b) vit–B
a) aliphatic vitamin b) aromatic vitamin c) vit–C d) vit–D
c) alicyclic vitamin d) heterocyclic vitamin

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19. Rickets is caused due to the deficiency of c) B6 d) B12
a) vit–A b) vit–B 33. Nicotinamide is named for
c) vit–C d) vit–D a) vit–B2 b) vit–B5
20. Vitarnin–D is also known as c) vit–B d) vit – B12
a) ascorbic acid b) reproductive vitamin 34. Cyanocobalamin is
c) growth vitamin d) sunshine vitamin a) vit–A b) vit–B
21. Identify the vitamin whose deficiency out food c) vit–B1 d) vit–B12
decrease reproductive power 35. Deficiency of poor coagulation of blood is due to
a) vit–A b) vit–D lack of
c) vit–E d) vit–P a) vit–A b) vit–C
22. Which of the following is provitamin –A c) vit–E d) vit–K
a) citric acid b) riboflavin SECTION – VII : HORMONES
c)  –carotene d) calciferol
1. The hormones which controls the presence of
23. Vitamin– D is burning of fats, proteins, and carbohydrate and
a) tocopherol b) ergosterol liberates energy in the body is
c) tocopherols d) calciferols a) thyroxine b) insulin
24. Beri–beri is caused due to c) adrenaline d) cortisone
a) vit–A b) vit–C 2. Which of the following is not sex hormones?
c) vit–B d) vit–D a) Testosterone b) Estrogen
25. Scurvy is caused due to c) Progesteron d) Thyroxin
a) vit–A b) vit–K 3. Insulin is secreted from
c) vit–E d) vit–C a) thyroid b) adrenal body
26. Vitamin which play role in coagulation property c) pancreas d) liver
of blood is 4. The hormone which transport glucose form blood
a) vit–A b) vit–K to tissue is
c) vit–E d) vit–D a) glycogen b) thyroxin
27. Two vitamins absorbed from intestine along with c) insulin d) heparin
fats are 5. Hormones which regulate metabolism of lipids,
a) A and D b) A and C carbohydrates and protein is
c) A and B d) D and C a) epinephrine b) thyroxin
28. Lack of vit–P causes c) oxytocin d) estrone
a) Beri–beri b) weakness of muscles 6. Insulin regulate the metabolism of
c) Hemorrhage d) Scurvy a) minerals b) amino acids
29. Biotin is an organic compound present in yeast. c) glucose d) vitamins
It's deficiency in diet causes paralysis. It is also 7. Hormones that help in the conversion of glucose
known as to glycogen is
a) vit–A b) vit–B3 a) cortisone b) adrenaline
c) vit–B12 d) vit–H c) bile acid d) insulin
30. The vitamin which is neither soluble in water nor 8. Which of the following is female sex hormones?
in fats is
a) Adrenaline b) Non–adrenaline
a) vit–A b) vit–H
c) Estrogen d) Testosterone
c) vit–P d) vit–D
9. Hormones are secreted from
31. Convulsions is caused due to deficiency of
a) plant cell wall b) nerve tissues
a) vit–B b) vit–P
c) duct less gland d) heart
c) vit–H d) vit–D
10. Which control the secretion of all hormones
32. The vitamin present in liver of pig in
a) kidney b) liver
a) B2 b) B3

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 Biomolecules 351
c) heart d) pituitary gland 15. Which of the following is polypeptide hormones?
11. Hormones are a) gestogens b) insulin
a) steroid b) peptide c) nor–adrenaline d) progestron
c) amino acid d) all of these 16. Which of following is/are steroid hormones?
12. Which amine hormone control function of a) testosterone b) progestogen
sympathetic nervous system? c) estrogene d) all of these
a) thyroxin b) progestron 17. Which of the following is protein hormones
c) Adrenaline d) insulin a) insulin b) testosteron
13. Hormones which control the development and c) thyroxin d) progestron
maintenance of pregnancy is 18. Which of the following are amino acid hormones
a) estrone b) cortisone 1) thyroxin 2) Adrenaline
c) progesterone d) vasopressin 3) insulin 4) estrogen
14. Which hormones increase lactic acid in muscles? a) 1, 3 b) 1, 2
a) progestron b) estrogene c) 2, 3 d) Only 1
c) nor–adrenaline d) androgen
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