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HORIZON Publication
Std. XII Sci. Success Chemistry - II
14



BIOMOLECULES
Contents
14.0 Prominent Scientists
14.1 Introduction
14.2 Carbohydrates
14.3 Proteins
14.4 Enzymes
14.5 Lipids
14.6 Hormones
14.7 Vitamins
14.8 Nucleic acids
Short Test
Multiple Choice Questions
14.0 Prominent Scientist
Name
1. Andreas Marggraf
2. Hermann Fischer
3. Har Gobind Khorana
4. Francis Crick
5. Frederick Sanger
Contributions
i.
Pioneer in analytical chemistry.
ii. Discovered formic acid, phosphoric acid, isolated zinc by heating
calamine.
iii. Discovered sugar in beetroots.
i.
Discovered hydrazine base, phenyl hydrazine.
ii. Established relation between glucose, fructose and mannose.
iii. Synthesized glucose, fructose and mannose from glycerol.
iv. Prepared polypeptide containing 18 amino acids.
v. Synthesized dipeptides, tripeptides and polypeptides.
v. Awarded Nobel prize.
i.
Took interest in both proteins and nucleic acid at Cambridge in UK
ii. Shared Nobel prize for medicine and physiology for cracking the
genetic code.
i.
Most noted for one of two co-discoveries of the structure of DNA
molecule.
ii. Awarded Nobel prize
i.
Two times awarded Nobel prize.
ii. First on structure of protein.
iii. Second for determination of base sequence in nucleic acids.
14.1 Introduction
*1.
What are biomolecules ? Give examples. OR
Define biomolecule and give two examples.
Ans. Biomolecules are the lifeless, molecules which combine in a specific manner to produce life or control
biological reactions.
Eg. : Carbohydrates, proteins, lipids (fats & oils), nucleic acids, enzymes, vitamins, mineral salts, etc.
Biomolecules
14. 1
Std. XII Sci. Success Chemistry - II
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2.
What are the different types of biomolecules ? Give two examples of each type.
Ans. Two types of biomolecules are i. Simple biomolecules and ii. Macromolecules.
i.
Simple biomolecules : Water, mineral salts, vitamins, etc.
ii. Macromolecules : Carbohydrates, proteins, lipids (fats and oils), etc.
3.
Write a note on importance of biomolecules ?
Ans. Biomolecules are simple molecules or macromolecules which are derived from reactions of simple
molecules.
i.
ii.
Biomolecules
Carbohydrates
Proteins
iii.
iv.
Lipids
Nucleic acids
Functions
Major constituents of food and source of energy.
i.
Help in proper functioning of living beings.
ii. Important constituents of skin, hair, muscles.
iii. Enzymes that catalyse chemical reactions taking place in cells
are proteins.
Are the storehouses of energy.
RNA (Ribonucleic acid) and DNA (Deoxyribonucleic acid) are
responsible for genetic characteristics and synthesis of proteins.
14.2 Carbohydrates
*4. What are carbohydrates ? Give their general formula and one example.
Ans. Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds
that can be hydrolysed to polyhydroxy aldehydes or polyhydroxy ketones,.
Carbohydrates are also called saccharides.
Eg. : Glucose C6H12O6 , Maltose C12H22O11
5.
How is the name carbohydrates derived ?
Ans. i.
According to earlier chemists the carbohydrates had molecular formula as Cx(H2O)y.
ii. They considered carbohydrates as the compounds in which carbon atom is surrounded by water
molecules i.e. hydrates of carbon, hence the name was given as carbohydrates.
iii. Eg. : Compounds like glucose (C6H12O6) was considered as C6(H2O)6 and sucrose C12H22O11
as C12(H2O)11.
6.
Give examples of the compounds which do not justify the earlier formula of carbohydrates.
Ans. Earlier formula suggested for carbohydrates was Cx(H2O)y. However the compound like acetic acid,
CH3COOH can be represented as C2(H2O)2 but cannot be classified as carbohydrate, but it is an acid.
The compound rhamnose, CH3(CHOH)4CHO cannot be represented as Cx(H2O)y but it is a
carbohydrate.
Hence, the carbohydrates now have a different meaning or definition.
14.2.1 Classification of Carbohydrates
7.
Give classification of carbohydrates. OR
How are carbohydrates classified?
Ans. On the basis of their behaviour on hydrolysis, carbohydrates are mainly classified into two groups as
i.
Simple carbohydrates and
ii. Complex carbohydrates
Biomolecules
14. 2
HORIZON Publication
Std. XII Sci. Success Chemistry - II
i.
Simple carbohydrates :
a.
Are known as monoscharides.
b. A carbohydrates that cannot be further hydrolyzed to simple sugar is called
monosaccharide.
c.
Monosaccharides are the basic units of all carbohydrates.
d. Depending upon the type of carbonyl group present are further classified as
i. Alodses (containing an aldehyde group) and
ii. Ketoses (containing keto group)
e.
Depending upon the number of carbon atoms and the functional group present are
further classified as
i. Triose
ii. Tetrose
iii. Pentose
iv. Hexose etc.
Sr.
No.
Type of
Monosaccharide
1.
2.
3.
4.
Triose
Tetrose
Pentose
Hexose
Class
Triose
(C3H6O3)
CHO
Aldoses
H
OH
CH2OH
D  (+)  Glyceraldehyde
(Aldotriose)
Ketoses
CH2OH
|
C=O
|
CH2OH
Dihydroxyacetone
(Ketotriose)
ii.
Number of
‘C’ atoms
present
3
4
5
6
Tetrose
(C4H8O4)
CHO
OH
H
H
OH
CH2OH
D  () Erythrose
(Aldotetrose)
CH2OH
|
C=O
H
OH
CH2OH
D  Erythrulose
(Ketotetrose)
Pentose
(C5H10O5)
Hexose
(C6H12O6)
CHO
OH
CHO
OH
H
H
H
OH
HO
H
H
OH
H
OH
H
OH
CH2OH
D  ()  Ribose
(Aldopentose)
CH2OH
D  (+) Glucose
(Aldohexose)
CH2OH
|
C=O
OH
H
HO
CH2OH
|
C=O
H
OH
H
OH
H
OH
H
CH2OH
D  Ribulose
(Ketopentose)
CH2OH
D ()Fructose
(Ketohexose)
Complex carbohydrates. :
a.
Two categories of complex carbohydrates are
i. Oligosaccharides
ii. Polysaccharides.
b. Oligosaccharides : A carbohydrate that yield two to ten monosaccharide units is called
Oligosaccharide.
c.
Depending on the number of monosaccharides derived, oligosaccharides are further
classified as :
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14. 3
Std. XII Sci. Success Chemistry - II
d.
e.
8.
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Sucrose 
 Glucose + Fructose
Hydrolysis
Maltose 
 Glucose + Glucose
Hydrolysis
Lactose 
 Glucose + Galactose
Hydrolysis
Raffinose 
 Glucose + Fructose + Galactose
Hydrolysis
i.
Disaccharides (C12H22O11) :
ii.
iii.
Trisaccharides (C18H32O16) :
Tetrasaccharides (C24H42O21) :
Hydrolysis
 Glucose + Fructos + Galactose + Galactose
Stachyose 
Polysaccharides : A carbohydrate which on hydrolysis yields a large number of
monosaccharide units is called Polysaccharides.
Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally
occurring polymers of carbohydrates.
Eg. Cellulose, starch, glycogen.
On the basis of solubility carbohydrates are classified as :
i.
Sugars : Crystalline, sweet to taste and soluble in water. Monosaccharides and
Oligosaccharides are sugars.
ii. Non-sugars : Amorphous, tasteless and insoluble in water. Polysaccharides are nonsugars and further classified as :
a.
Reducing Sugars : They reduce Fehling’s solution and Tollen’s reagent.
Eg. : Maltose
b. Non-reducing Sugars : They do not reduce Fehling’s solution and Tollen’s
reagent. Eg. : Sucrose
Referring to the table give below answer the following questions.
[Intext Question text book page no. 513]
Ans.
Class
Triose
(C3H6O3)
CHO
Aldoses
H
OH
CH2OH
D  (+)  Glyceraldehyde
(Aldotriose)
Ketoses
CH2OH
|
C=O
|
CH2OH
Dihydroxyacetone
(Ketotriose)
Biomolecules
Tetrose
(C4H8O4)
Pentose
(C5H10O5)
Hexose
(C6H12O6)
CHO
OH
H
CHO
OH
H
CHO
OH
H
H
OH
H
OH
HO
H
H
OH
H
OH
H
OH
CH2OH
D  ()  Erythrose
(Aldotetrose)
CH2OH
|
C=O
H
OH
CH2OH
D  Erythrulose
(Ketotetrose)
CH2OH
D  ()  Ribose
(Aldopentose)
CH2OH
D  (+) Glucose
(Aldohexose)
CH2OH
|
C=O
OH
H
HO
CH2OH
|
C=O
H
OH
H
OH
H
OH
H
CH2OH
D  Ribulose
(Ketopentose)
CH2OH
D ()Fructose
(Ketohexose)
14. 4
HORIZON Publication
Std. XII Sci. Success Chemistry - II
i.
Identify the monosaccharides that contain only two chiral carbon atoms.
Ans.
CH2OH
CHO
H
OH
H
OH
|
C=O
OH
H
CH2OH
D  ()  Erythrose
OH
H
CH2OH
D  Ribulose
ii. Write the Fischer projection formulae of (a) L – (+) – erythrose (b) L – (+) – ribulose.
Ans. a.
b.
CHO
CH OH
2
H
HO
OH
|
C=O
OH
H
H
CH2OH
L  (+)  Erythrose
HO
H
CH2OH
L  (+)  Ribulose
iii.
Is the followings sugar, D - sugar or L – sugar ?
HO
CHO
H
HO
H
HO
H
CH2OH
Ans. The compound is L – sugar.
9.
Explain D and L configuration in sugars. OR Explain Fischer projection formula.
Ans. i.
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L
configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and
exist in two enantiomeric forms as,
CHO
H
CHO
OH
CH2OH
D-Glyceraldehyde
(+) Glyceraldehyde
ii.
iii.
iv.
v.
HO
H
CH2OH
(L) Glyceraldehyde
() Glyceraldehyde
The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration
i.e., D-glyceraldehyde.
The laevo enantiomer of glyceraldehyde is represented as () glyceraldehyde and is corelated as
L-configuration i.e., L-glyceraldehyde.
In Fischer projection formula, a monosaccharide is assigned D-configuration if the ( OH)
hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is
assigned L-configuration if the  OH group on the last chiral carbon atom and lies on the left
hand side.
In monosaccharides, the most oxidised carbon (i.e.,  CHO) is at the top.
Biomolecules
14. 5
Std. XII Sci. Success Chemistry - II
Eg. :
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OH
H
CH2OH
D(+)Glyceraldehyde
CHO
CHO
CHO
CHO
H
OH
H
OH
H
OH
CH2OH
D  (+) Ribose
H
OH
HO
H
HO
H
H
OH
H
OH
HO
H
H
OH
CH2OH
HO
H
D  (+) Glucose
CH2OH
L  () Glucose
10. Identify D & L configurations of following monosaccharides.
Ans.
Monosaccharides
Configuration
CHO
i.
H
HO
L
CH2OH
ii.
HO
CHO
H
HO
H
HO
H
L
CH2OH
CH3
iii.
OH
H
D
COOH
COOH
HO
H
H
OH
iv.
D
COOH
CH3
v.
HO
H
L
CH2OH
Biomolecules
14. 6
HORIZON Publication
Std. XII Sci. Success Chemistry - II
11. What is monosaccharide ? Explain how are monosaccharides classified ?
Ans. The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further
is known as monosaccharide. The monosaccharide is crystalline and soluble in water.
Eg. Glucose, fructose, ribose.
Monosaccharides are classified as :
 H

 group in their structure. Eg. : Glucose.
i.
Aldoses : Aldoses contain aldehydic  |
 C  O 


Aldoses are further classified depending upon the number of carbon atoms present in the
monosaccharide. The number of carbon atoms present in the molecule is indicated by the prefix
tri for 3-carbon, tetra for 4-carbon, etc. The aldose accordingly is called Aldotriose, Aldotetrose,
Aldopentose, Aldohexose, etc.
Eg. :
Glyceraldehyde
C3H6O3
Aldotriose
Erythose
C4H8O4
Aldotetrose
Ribose, Arabinose, Xylose C5H10O5
Aldopentose
Glucose, Mannose
C6H12O6
Aldohexose
ii.
Ketoses : Ketoses contain ketonic, C = O group in their structures.
Eg. : Fructose.
Ketoses are further classified depending upon the number of carbon atoms present in the
monosaccharide.
The number of carbon atoms present in the molecule is indicated by the prefix, tri for
3-carbons, tetra for 4-carbons, etc. The Ketoses accordingly are called Ketotriose, Ketotetrose,
Ketopentose, Ketohexose, etc.
Eg. :
Dihydroxy acetone
C3H6O6
Ketotriose
Erythrulose
C4H8O4
Ketotetrose
Ribulose
C 5H10O5 Ketopentose
Fructose
C6H12O6
Ketohexose
*12. Classify the following into monosaccharides, oligosaccharides and polysaccharides.
i.
Starch
ii. Glucose
iii. Stachyose
iv. Maltose
v.
Raffinose
vi. Cellulose
vii. Sucrose
viii. Lactose.
Ans.
Glucose
Monosaccharides
Stachyose, maltose, raffinose, sucrose,
Oligosaccharides
lactose
Starch, cellulose
Polysaccharides
14.2.2 Preparation of Glucose
*13. How is glucose prepared in the laboratory ?
OR
Give laboratory preparation of glucose.
Ans. Preparation of glucose from sucrose (cane sugar) : Laboratory method
i.
When cane sugar or sucrose is boiled with dilute hydrochloric acid or sulphuric acid it undergoes
hydrolysis to give glucose and fructose in equal amounts.
ii. While cooling, alcohol is added as glucose is insoluble in alcohol it crystallized out first.
Fructose being soluble in alcohol it remains in the solution.
iii. Crystals of glucose are separated by filtration.
dil HCl or H 2SO 4
C12 H 22O11  H 2O 
 C6 H12O 6  C6 H12O 6

Sucrose
Glu cos e
Fructose
Biomolecules
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Std. XII Sci. Success Chemistry - II
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14. Give reason: At the time of cooling in laboratory method used for glucose preparation, alcohol is
added.
Ans. i.
Hydrolysis of cane sugar gives a mixture of glucose and fructose.
ii. Glucose is insoluble in alcohol and crystallized out first while Fructose being soluble in alcohol
it remains in the solution.
iii. Hence, in order to separate glucose from fructose at the time of cooling alcohol is added in
laboratory method.
*15. How is glucose prepared on commercial scale ?
Ans. Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid.
Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under pressure.
Starch is hydrolysed to give glucose.
dil H 2SO 4
(C 6 H10 O 5 ) n  nH 2 O 
 nC 6 H12 O 6
393K , 2  3 atm
Starch glucose
glucose
When hydrolysis is complete, calcium carbonate (chalk powder) is added. Chalk powder neutralizes
excess of acid.
To the solution, activated charcoal is added which removes coloured impurities by adsorption. The
solution is then filtered to remove insoluble impurities, The solution is then concentrated, when
glucose is obtained in the crystalline form. Pure crystals of glucose are obtained by recrystallization.
16.
Give reason :
i.
Chalk powder is added when the hydrolysis of starch, in the manufacture of glucose, is
complete.
ii. Activated charcoal is added at the end of the hydrolysis in the manufacture of glucose in
commercial method.
Ans. i.
a.
Hydrolysis of starch with dil. H2SO4 at 393 K under pressure gives glucose.
b. Chalk powder is calcium carbonate (CaCO3). Chalk powder reacts with excess of sulphuric
acid forming precipitate of calcium sulphate which is removed by filtration.
Neutralization
H2SO4  CaCO3 
 CaSO4   H2O  CO2 
Thus to remove excess of H2SO4 chalk powder is added.
ii. a.
The coloured impurities are removed by adding activated charcoal.
b. Coloured impurities are adsorbed on the charcoal.
c.
Thus activated charcoal is added at the end of the hydrolysis in the manufacture of glucose
in commercial method.
14.2.3 Open Chain Structure of Glucose
*17. What is the action of following reagents on glucose ?
i.
HI
ii. Hydroxyl amine (NH2OH)
iii. Hydrogen cyanide
iv. Bromine water
v.
Nitric acid
vi. Acetic anhydride.
Ans. i.
Action of HI :
Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are
linked in straight chain.
CHO
|
hot HI
(CHOH) 4 
 CH 3  CH 2  CH 2  CH 2  CH 2  CH 3
|
CH 2 OH
n- Hexane
Glu cos e
Biomolecules
14. 8
HORIZON Publication
Std. XII Sci. Success Chemistry - II
ii. Action of hydroxyl amine :
Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime.
CHO
|
(CHOH) 4
|
CH 2 OH
Glucose
NH 2  OH


hydroxyl amine
CH  NOH
|
(CHOH) 4
+ H2O
|
CH 2 OH
Glucoxime
iii.
Action of hydrogen cyanide :
Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.
CHO
CN
|
|
HCN
(CHOH) 4 
 CHOH
|
|
CH 2OH
(CHOH) 4
|
CH 2OH
Glucose
Glucose cyanohydrin
iv.
Action of bromine water :
Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which
shows that the carbonyl group in glucose is aldehyde group.
COOH
CHO
|
|
Bromine water
(CHOH)
(CHOH) 4  (O) 
4
|
|
CH 2 OH
CH 2 OH
Glu cos e
Gluconic acid
v.
Action of dil. nitric acid :
Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This
indicates the presence of a primary alcoholic group ( CH2OH) in glucose.
CHO
COOH
COOH
COOH
|
|
|
|
dil HNO 3
dil HNO 3
(CHOH)
(CHOH)
(CHOH) 4 

(CHOH) 4 

4
4
;
|
|
|
|
COOH
COOH
CH 2 OH
CH 2 OH
Saccharic acid
Saccharic acid
Glu cos e
Gluconic acid
vi.
Action of acetic anhydride :
When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta
acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl
groups.
CHO
O
CHO
O
O
|
||
|
||
||
Pyridine
(CHOH) 4  5(CH 3  C) 2 O 
(CH  O  C  CH 3 ) 4  5 CH 3  C  OH
|
|
CH 2 OH
CH 2  O  C  CH 3
Glucose
||
O
Glucose pentaacetate
Biomolecules
14. 9
Std. XII Sci. Success Chemistry - II
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18. A monosaccharide on acetylation gives triacetate. It is not oxidized by bromine water but on
oxidation by dil. HNO3 gives dicarboxylic acid. Write the possible structure of the
monosaccharide.
[Intext Question text book page no. 515]
Ans. The compound C4H8O4 i.e. D – Erythrulose on acylation gives triacetate and on treatment with dil.
HNO3 gives dicarboxylic acid. Hence, the structure of monosaccharide is
CH2OH
C=O
CHOH
CH2OH
D – Erythrulose
19. What are the expected products of hydrolysis of lactose?
Ans. Lactose on hydrolysis forms an equimolar mixture of D-glucose and D-galactose.
20. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose ?
Ans. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free –CHO.
14.2.4 Cyclic Structure of Glucose
21. List down the facts and reactions that are not explained by open chain structure of glucose.
Ans. Open chain structure of glucose does not explain the following reactions and facts:
i.
Glucose in spite of having aldehyde group, does not give condensation reaction with 2,4 dinitro
phenyl hydrazine, addition reaction with sodium bisulphite and gives negative test with Schiff base.
ii. Glucose penta – acetate does not condense with hydroxylamine, indicating the absence of free
aldehyde group.
iii. Glucose is found to exist in two different crystalline forms α and β, called anomers. On
crystallization from hot and saturated aqueous solution, α-glucose (m.p. 423 K) is obtained at
303 K while -glucose (m.p. 423 K) is obtained at 371 K.
CH2OH
CH2OH
O
H
H
OH
H
OH
H
HO
H
OH
OH
H
HO
OH
H
O
H
H
H
OH
( –)
( –)
Write the two cyclic structures of -D-( + )-Glucose pyranose and -D-( + )-Glucose pyranose
exist in equilibrium with open chain structures.
Ans.
O
1
1
1 ||
H  C  OH
HO  C  H
HC
2
2
H  C  OH
H  C  OH
O
H

C  OH
O
3
3
3
HO  C  H
HO  C  H
HO  C  H
4
4
4
H  C  OH
H

C  OH
H  C  OH
22.
5
HC
6
CH2OH
D(+)Glucose pyranose
Biomolecules
5
5
H  C  OH
6
CH2OH
HC
6
CH2OH
D(+)Glucose pyranose
14. 10
HORIZON Publication
Std. XII Sci. Success Chemistry - II
*23. Write the structures of -D-(+)-glucopyranose and -D-(+)-glucopyranose.
Ans.
1C OH
2 OH
H
H
HO
H
C H
2 OH
3 H
4
OH
5
6
CH2OH
H
HO
H
H
O
3 H
4 OH
5
6
CH2OH
H
1
HO
  D  (+)  Glucopyranose
O
  D  (+)  Glucopyranose
24. Show how is Fischer projection formula of glucose converted into Haworth projection formula.
Ans. The cyclic structures of glucose is converted into Haworth projection formula as follows :
1 CHO
2
H
H
4
4
CH2OH
H
H
H
C
H1 1
O
2
3
OH
H
6
OH
OH
HO
5
6
5
H
≡
OH
H
CH2OH
OH
3
HO
6
H
H
5
H
OH
4
HO
3
H
OH
O
H
C1
1
H
OH
2
+
OH
CH2OH
H+
D(+)Glucose
6
CH2OH
H
4
CH2OH
5
H
O H
H
OH
HO
H
O
H
OH
1
OH
H
H
HO
2 OH
3
OH
H
 D  (+)  Glucopyranose (cis)
OH
H
 D  (+)  Glucopyranose (trans)
25.
H
Following are the structures of four stereoisomeric erythrofuranoses. Name each isomer using
proper D, L and ,  notations.
[Intext Question text book page no. 517]
i.
H
O
H
H
ii.
H
OH
OH
Biomolecules
OH
OH
O
H
iii.
OH
H
OH
OH
H
O
H
iv.
OH
OH
H
OH
OH
O
H
H
H
OH
14. 11
Std. XII Sci. Success Chemistry - II
Ans. i.
D – erythrofuranose
ii. L – erythrofuranose
iii. –D– (–) erythrofuranose
iv. –D– (–) erythrofuranose
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14.2.5 Structure of Fructose
26. Explain the structure of fructose.
Ans. Fructose has molecular formula C6H12O6. It contains ketonic functional group at carbon number 2 and
six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is
written as D-(  )-fructose.
CH2OH
|
C=O
H
HO
H
OH
H
OH
CH2OH
D ()  Fructose
27.
Write the Haworth projection formulae for -D-( )-Fructofuranose (trans) and -D-() fructofuranose (cis).
Ans.
1
HOH2C
C H 2 OH
O
5
HO
H
OH
OH H
D() Fructofuranose
(trans)
CHO
HO
OH
ii.
H
HO
H
HO
CH2OH
Fructose
CHO
OH
H
OH H
D() Fructofuranose
(cis)
*28. Draw mirror images of glucose and fructose.
Ans. i.
Glucose
H
O
2
H
H
HOH2C
H
OH
HO
CH2OH
CH2OH
C=O
C=O
H
H
OH
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
CH2OH
D(+)Glucose
Biomolecules
CH2OH
L()Glucose
CH2OH
D(+)Fructose
CH2OH
L(–)Fructose
14. 12
HORIZON Publication
Std. XII Sci. Success Chemistry - II
29. Write the conversion of Fischer projection formula into Haworth projection formula for
fructose.
Ans.
CH2OH
H
C=O
6

OH
O:
HO H 2C
HO
H
HOH 2 C
:OH
O
2
6
H
5 H
HO C
H
HO C
5

H
OH
2
4
3
H
H
4
3
C H 2 OH
CH 2OH
H
OH
1
1
OH
H
OH
H
CH2OH
H+
1
6
HOH 2 C
O
5
HOH 2 C
HO
4
3
OH
H
OH H
D() Fructofuranose
(trans)
30.
OH
O
5
2
H
H
6
C H 2 OH
2
H
HO
4
3
C H 2 OH
1
OH H
D() Fructofuranose
(cis)
Refer to the Fischer projection formula of ribose. Furanose form of ribose is produced by
addition of – OH group at C-4 to aldehyde group. Draw Haworth projection formulae of four
stereoisomeric ribofuranoses using D, L and ,  notations.
[Intext Question text book page no. 518]
Ans.
CH2OH
O
H
H
H
OH
OH
OH
O
H
OH
-D-(-)-Ribofuranoses
OH
-D-(-)-Ribofuranoses
OH
O
H
CH2OH
OH
HOH2C
H
OH
OH
OH
-L-(-)-Ribofuranoses
OH
H
OH
H
OH
OH
-L-(-)-Ribofuranoses
Adjacent is the Haworth projection formula of -D-ribulofuranose.
Write its Fischer projection formula.
[Intext Question text book page no. 518]
5
Ans. Fischer projection formula is as follows :
H
H
H
HOH2C
31.
CH2OH
|
C=O
O
O
1 CH
2OH
2
H
3 OH
4
OH
OH
CH2OH
Biomolecules
14. 13
Std. XII Sci. Success Chemistry - II
14.2.6 Disaccharides
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*32. Write the hydrolysis products of (i). lactose and (ii). sucrose.
Ans. i.
Lactose on hydrolysis in presence of an acid or enzyme lactose gives one molecule each of
glucose and galactose
C12 H 22 O11

H 3O
 H 2 O 

or lactose
Lactose
ii.
C 6 H12 O 6

C 6 H12O 6
D  (  ) Glucose
D  (  )  Galactose
Sucrose on hydrolysis in the presence of dil. acid or the enzyme invertase gives one molecule
each of glucose and fructose .
C12 H 22 O11

H 3O
 H 2 O 

or invertase
Sucrose
C 6 H12 O 6

C 6 H12O 6
D  (  ) Glucose
D  (  )  Fructose
33.
Explain the structures of following molecules :
i.
Sucrose
ii. Maltose
iii. Cellobiose
iv. Lactose
Ans. i.
Sucrose :
a.
It is known as cane sugar or common table sugar.
b. In sucrose C – 1 of   D – Glucopyranose is linked to C – 2 of  – D – Frouctofuranose
by glycosidic linkage.
c.
The reducing groups of glucose and fructose are involved in glycosidic bond formation and
hence sucrose is non reducing sugar.
6
C H 2 OH
5
H
O
H
OH
4
H
1
H
Glycosidic linkage
OH
2
3
H
OH
6
O
HOH2C
5
H
2
H
4
OH
ii.
O
HO
3
CH2OH
1
H
Maltose :
a.
In Maltose C – 1 of one   D – Glucopyranose is linked to C – 4 of another   D –
Glucopyranose molecule by glycosidic linkage.
b. Thus in maltose 1  4  glycoside bond is present.
c.
It is reducing sugar as it gives – CHO group at C – 1 in second glucose molecule.
Biomolecules
14. 14
HORIZON Publication
Std. XII Sci. Success Chemistry - II
6
6
CH2OH
CH2OH
H
5
O
H
OH
4
HO
O
2
3
H
OH
 - D - Glucopyranose
O
H
OH
4
1
H
5
H
H
H
1
H
OH
2
3
6H
OH
 - D - Glucopyranose
Maltose ( – anomer), 4 – O – ( – D – Glucopyranosyl) – D – glucopyranose
iii.
Cellobiose :
a.
In this C – 1 of the  – D Glucopyranose is linked to C – 4 of another  – D Glucopyranose
molecule by glycosidic linkage.
b. It contains 1  4  glycoside bond.
c.
It is a reducing sugar as – CHO group is produced at C – 1 in second glucose molecule.
6
H
CH2OH
H
5
O
H
OH
4
4
H
2
3
OH
H
1
HO
H
H
OH
 - D - Glucopyranose
2
3
O
H
OH
H
H
1
OH
O
5
6 CH2OH
 - D - Glucopyranose
Cellobiose ( – anomer) 4 – O – ( – D – Glucopyranosyl) – D – glucopyranose
iv.
Lactose :
a.
In lactose C – 1 of  – D galactopyranose is linked to C – 4 of  – D – glucopyranose by
glycosidic linkage.
b. Thus it contains 1  4  glycoside bond.
c.
It is a reducing sugar as free – CHO group is produced at C – 1 of  – D – glucopyranose
molecule.
6
C H 2 OH
HO
5
O
H
OH
4
H
O
H
OH
H
1
H
2
H
OH
 - D - Galactopyranose
OH
2
H
4
H
3
3
H
5
H
O
1
OH
CH2OH
 - D - Glucopyranose
6
lactose ( anomer) 4 – O – ( – D – Galactopyranosyl) – D – glucopyranose
Biomolecules
14. 15
Std. XII Sci. Success Chemistry - II
14.2.7 Polysaccharides
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34. Define polysaccharides and give general formula and example.
Ans. The molecules in which a large number of same or different monosaccharides are linked together by
glycosidic linkages is called polysaccharide.
General formula (C6H10O5)n
Eg. : Starch, Cellulose
35. Explain the structure of cellulose.
Ans. Cellulose mainly occurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain
polymer. In cellulose,   D  glucopyranose molecules are linked by glycosidic linkage between
C1of one unit of glucose and C4 of another glucose unit. Thus cellulose contains 1  4 glycosidic
linkages like those in cellobiose.
6
CH2OH
6
H
5
H
CH2OH
4
5
O
4 H
O
1
H
OH
O
3
2
H
OH
O
H
OH
H
3
2
H
OH
O
1
H
 - link
H
Cellulose (–1,4’ polymer of D–glucose)
36. Explain the structure of starch.
Ans. Starch is found in cereal grains, roots, tubers, potatoes etc. It is a polymer of   D  glucopyranose
and consists of two components, amylose and amylopectin.
Amylose is water soluble component and constitutes about 20% of starch. Amylose contains 200 to
1000   D glucopyranose molecules linked together by glycosidic linkage between C1 of one unit
and C4 of another unit.
6
H
5
4
H
OH
O
3
CH2OH
CH2OH
H
H 5
H
O
4
1
H
OH
O
H
6
6
CH2OH
2
OH
3
O
H
2
H 5
H
4
1
O
H
OH
O
3
H
H
1
H
2
O
OH
H
OH
 links
Amylose (D  Glucopyranose)
Amylopectin is insoluble in water and constitutes about 80% starch. It is a branched chain polymer.
In amylopectin,   D  glucopyranose molecules are linked together by glycosidic linkage between
C1 of one unit and C4 of another unit to form long chain and branching occurs by glycosidic linkage
between C1 and C6 glycosidic linkage.
Biomolecules
14. 16
HORIZON Publication
Std. XII Sci. Success Chemistry - II
6
CH2OH
H
4
O
6
5
O
H
OH
H
H
H
CH2OH
5
O
H
OH
H
4
1
O
2
3
H
H
O
6
CH2
6
CH2OH
4
O
Branch at C-6(-1,6 glycosidic linkage)
OH
 link
H
1
2
3
OH
H
5
O
H
OH
H
3
H
H
H
4
2
OH
1
6
CH2OH
5
O
H
H
5
O
H
OH
H
1
4
H
OH
H
O
3
H
2
OH
O
3
H
2
H
1
O
OH
Amylopectin
37. What is the basic structural difference between starch and cellulose ?
Ans. Starch is a polymer of -glucose and consists of two components-amylose and amylopectin. In
amylose   D  (+)  glucose units held by C1  C4 glycosidic linkage and in amylopectin,   D 
glucose units held by C1  C4 glycosidic linkage whereas branching occurs by C1  C6 glycosidic
linkage.
Cellulose is a straight chain polysaccharide composed only of   D  glucose units held by C1C4
glycosidic linkage.
38. Write a note on structure of glycogen.
Ans. i.
Glucose is stored in the form of glycogen in animal body.
ii. Glycogen is known as animal starch, as its structure is similar to amylopectin.
iii. The branching in glycogen is more extensive than in amylopectin.
iv. Many end groups present in highly branched structure of glycogen, are available for quick
hydrolysis to provide the glucose needed for metabolism.
v.
So it is a source of instant energy.
*39. State the importance of carbohydrates.
OR
Why are carbohydrates important in diet ?
Ans. i.
Carbohydrates are essential for life in both plants and animals.
ii. They are the important constituent of supporting tissues in plants e.g. cellulose found in wood,
cotton, etc.
iii. They form a major part of our food and store chemical energy in plants.
iv. They act as the major source of energy for animals and human beings.
v. Animals store glucose in the form of glycogen which is the main source of energy.
vi. Carbohydrates satisfy the three basic needs i.e. food, clothes and shelter.
Biomolecules
14. 17
Std. XII Sci. Success Chemistry - II
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40. How is glycogen different from starch ?
Ans. Starch is the main storage molecule of plants whereas glycogen is the main storage molecule of
animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.
*41. Write the structures of cellobiose, lactose, cellulose and starch.
Ans.
Compound
Structure
i.
Cellobiose
6
C H 2 OH
H
5
H
O
O
OH
H
1
H
H
OH
 - D - Glucopyranose
1
OH
O
5
2
3
H
H
H
HO
2
H
4
H
OH
4
OH
3
CH2OH
 - D - Glucopyranose
6
Cellobiose (-anomer) 4  O  (  D  Glucopyranosyl)  D 
glucopyranose
ii.
Lactose
6
C H 2 OH
5
HO
O
H
O
OH
H
H
H
H
OH
 - D - Galactopyranose
1
OH
O
5
2
3
2
H
1
H
H
OH
3
4
H
OH
4
H
CH2OH
 - D - Galactopyranose
6
Lactose (-anomer) 4  O  (  D  galactopyranosyl  D 
glucopyranose
iii.
Cellulose
6
C H 2 OH
H
6
C H 2 OH
H
4
5
4
O
H
OH
O
3
H
H
OH
H 1
O
O
1
H
H
3
H
H
O
5
2
OH
-link
2
OH
Cellulose (  1, 4’ polymer of D-glucose)
Biomolecules
14. 18
HORIZON Publication
Std. XII Sci. Success Chemistry - II
iv. Starch : It is a mixture of two compounds (i) a water soluble : amylase (15  20%) &
(ii) water insoluble : amylopectin (80  85%)
6
6
CH2OH
H 5
H
4
OH
O
3
H
O
H
6
H
CH2OH
H 5
O
1
4
H
OH
H
O
2
OH
H
H
CH2OH
H 5
O
H
1
4
H
OH
H
1
3
H
2
OH
O
2
OH
3
O
 - links
Amylose (D – Glucopyranose)
CH2OH
CH2OH
O
H
H
OH
H
H
H
O
H
O
H
OH
H
H
OH
O
H
OH
Branch at C6 (-1,6 glucosidic linkage)
 - links
O
6
CH2OH
5
H
H
4
OH
O
3
H
6
O
H
CH2OH
H 5
H
1
4
6
CH2
O H
H
2
OH
H 5
H
OH
4
1
O
3
H
2
OH
O
O H
H
OH
H
3
H
2
OH
1
O
Amylopectin
14.3 Proteins
42. Give reason ‘Proteins are ranked first of all the organic compounds’.
Ans. i.
Proteins are the most abundant biomolecules of the living system.
ii. Proteins are the substances of life.
iii. They are found in living cell and also are present in skin, hair, muscle, nerves, enzymes,
antibodies and hormones.
iv. Hormone that are proteins, regulate metabolic processess.
v. Without proteins life would not be possible.
vi. Thus proteins are ranked first of all the organic compounds.
How is the word ‘Protein’ derived ? and mention the basic unit or hydrolysis product of
proteins.
Ans. i.
The name protein is derived from the Greek word ‘proteios’ meaning first.
ii. Hydrolysis product of proteins is  amino acid.
R
|
H 2O
Proteins 
 H 2 N  CH  COOH
Mixture of  - amino acids
43.
Biomolecules
14. 19
Std. XII Sci. Success Chemistry - II
www.horizonpublication.com
*44. Define protein.
Ans. Proteins are the biopolymers of a large number of -amino acids. They are naturally occurring
polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages
( CO  NH ).
45. What are the sources of proteins.
Ans. The sources of proteins are milk, egg, fish, pulses, cereals, peanuts, cheese etc.
*46. What are -amino acids ?
Ans. i.
-amino acids are the derivatives of carboxylic acids obtained by replacing -H by amino
group.
ii. They are bifunctional as they have 2 functional groups i.e. – NH2, – COOH.
iii. Proteins on hydrolysis give -amino acids.
Note :
Name of amino acid
2.
3.
4.
5.
Neutral amino-acids
Glycine
Alanine
Valine
Leucine
Isoleucine
6.
7.
8.
Phenylalanine
Tyrosine
Tryptophan
1.
3 Letter symbol
(abbreviation)
Side chain R of amino acids
H–
CH3 –
(CH3)2CH –
(CH3)2–CH – CH2 –
CH 3  CH 2  CH 
|
CH 3
C6H5 – CH2 –
–HO–C6H4–CH2–
Gly
Ala
Val
Leu
Ile
Phe
Tyr
Try
CH2 –
N
H
9. Serine
10. Proline
11. Glutamine
HO–CH2–
–CH2–CH2–CH2–
O
||
H 2 N  C  CH 2  CH 2 
12. Cysteine
13. Cystine
SH – CH2 –
14. Methionine
CH3–S–CH2–CH2–
Met
15. Threonine
CH 3 – CH –
|
OH
Thr
16. Asparagine
H2N–CO–CH2–
Asn
Biomolecules

Ser
Pro
Gln

O OC  CH  CH 2S  CH 2S  CH  CO O
|
|
 NH 3
 NH 3
Cys
Cys.cys
14. 20
HORIZON Publication
Std. XII Sci. Success Chemistry - II
Acidic amino acids
17. Glutamic acid
HOOC–CH2–CH2–
18. Aspartic acid
HOOC–CH2–

Basic amino acids
H
N
 C  NH  (CH 2 )3 
2
19. Arginine
|
NH 2
20. Lysine
H2N–(CH2)4 –
21. Histidine
CH = C – CH2 –
|
|
N
NH
Glu
Asp
Arg
Lys
His
C
|
H
*47. What is zwitter ion.
Ans. i.
Amino acids due to presence of both acidic and basic groups in the same molecule behave like
salts.

ii.
H 2 N  CH  COOH
H 3 N  CH  COO  zwitter ion (dipolar ion)
|
|
R
R
Such doubly charged ion is known as zwitter ion.
*48. Write the classification of amino acids, giving examples.
Ans. Amino acids are of three types :
i.
Acidic amino acids : If carboxyl groups are more in number than amino groups then amino
acids are acidic in nature.
COOH
Eg. : Glutamic acid : HOOC  CH2  CH2 – CH
NH2
Aspartic acid :
HOOC  CH2 CH
COOH
NH2
ii.
Basic amino acids : If amino groups are more in number than carboxyl groups then amino acids
are basic in nature.
COOH
+
Eg. : Arginine :
H 2 N  C  NH  (CH 2 ) 3  CH
|
NH 2
NH2
COOH
Lysine :
H2N – (CH2)4 – CH
NH2
iii.
Neutral amino acids : The amino acids having equal number of amino and carboxyl groups are
called neutral amino acids.
COOH
Eg. : Alanine :
CH3 – CH
NH2
Biomolecules
COOH
Valine :
(CH3)2 – CH – CH
NH2
14. 21
Std. XII Sci. Success Chemistry - II
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*49. What are essential and non-essential amino adds ? Give two examples of each.
Ans. i.
The amino acids, which cannot be synthesised in the body and are supplied through diet are
called essential amino acids.
Eg. : Lysine, Valine
ii. The amino acids which are synthesized in the body are called non-essential amino acids.
Eg. : Glutamic acid, Serine
*50. Explain the amphoteric behaviour of amino acids.
Ans. i.
Amino acids are bifunctional compounds containing acidic carboxyl group (COOH) and basic
amino group ( NH2) within the same molecule. In aqueous solution, the carboxyl group loses a
proton while the amino group accepts it. as a result, a dipolar or zwitter ion is formed.
H 2 N  CH  C
|
R
ii.
O


+
H 3 N  CH  C
|
R
Zwitter ion (dipolar ion)
OH
O
O–
-amino acids show amphoteric behaviour as they react with both acids and bases.
In the acidic medium, COO ion of the zwitter ion accepts a proton to form the cation (I) while in

the basic medium, N H 3 ion loses a proton to form the anion (II).
+
+
H
H 
H 3 N  CH  COOH 
 H 3 N  CH  COO   H 2 N  CH  COO 
|
|
|
R
R
R
(I)
Zwitter ion
(II)

Thus, N H 3 group acts as the acid while COO groups acts as the base.
51. Define peptide linkage.
Ans. Peptide linkage is an amide formed between – COOH and – NH2 group by elimination of water
O
molecule. ||
is peptide linkage.
C  NH 
*52. Write a note on peptide linkage.
OR
How is peptide linkage formed ?
Ans. i.
Peptide linkage is an amide formed between – COOH and – NH2 group by elimination of water
O
molecule. ||
is peptide linkage.
 C  NH 
ii. Same or different amino acid molecules loose water molecule and get linked through peptide
linkage.
O
H
O
O H
O
||
|
||
|| |
||

H 2 N  CH  C  OH  H  N  CH  C  OH 
 H 2 N  CH  C  N  CH  C  OH  H 2O
|
|
|
|
R
R
R
R
Peptide bond
iii.
Many amino acid molecules combine together through peptide linkages and results in dipeptide,
tripeptide, tetrapeptide and finally a protein molecule.
Eg. Insulin contains 51 amino acids.
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14. 22
HORIZON Publication
Std. XII Sci. Success Chemistry - II
*53. Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Ans. i.
Dipeptide from glycine :
Carboxylic group of glycine reacts with amino group of another molecule of glycine to form
dipeptide.
O
O
||
||
H 2O
H 2 N  CH 2  C  OH  H  N  CH 2  COOH 
 H 2 N  CH 2  C  NH  CH 2  COOH
|
Glycine
Glycine
Dipeptide
H
ii. Dipeptide from alanine
Carboxylic group of alanine reacts with amino group of another molecule of alamine to form
dipeptide
CH 3 O
CH 3
CH 3 O
CH 3
|
||
|
|
||
|
H 2O
H 2 N  CH  C  OH  H  N  CH  COOH 
 H 2 N  CH  C  NH  CH  COOH
|
H Alanine
Alanine
Dipeptide
iii.
Dipeptide from glycine and alanine.
Carboxylic group of glycine reacts with amino group of another molecule of alanine to form
dipeptide
CH 3
CH 3
O
O
||
|
||
|
 H 2O
H 2 N  CH 2  C  OH  H  N  CH  COOH 
 H 2 N  CH 2  C  NH  CH  COOH
|
H Alanine
Glycine
Dipeptide
*54. How are proteins classified ?
Ans. On the basis of their molecular shape proteins are classified into two types as follows :
i.
Fibrous proteins
ii. Globular proteins
i.
Fibrous proteins :
a.
Long thread like and lie side by side to form fibres.
b. Insoluble in water.
c.
Polypeptide chains are held together by hydrogen bonds.
Eg. Collagen in tendons, keratin in hair, skin, nail etc.
ii. Globular proteins :
a.
They are folded to form spherical shape and have intramolecular hydrogen bonding.
b. Soluble in water and aqueous solutions of bases, acids and salts.
c.
Hydrogen bonding of these proteins in weak as compared to fibrous proteins.
Eg. Haemoglobin (in blood), albumin (in egg)
*55. Distinguish between globular and fibrous proteins.
Ans.
Globular proteins
i.
The proteins are folded to form i.
spherical structure, are globular
proteins.
ii. Globular proteins are soluble in water.
ii.
iii. They are sensitive to small changes of iii.
temperature and pH.
iv. They possess biological activity.
iv.
Biomolecules
Fibrous proteins
The proteins in which the polypeptide
chains lie parallel to form fibre like
structure, are fibrous proteins.
Fibrous proteins are insoluble in water.
They are stable to moderate changes of
temperature and pH.
They do not possess biological activity.
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*56. Explain the structure of proteins.
Ans. The structure and shape of proteins can be studied at four different levels called primary, secondary,
tertiary and quaternary. Each level is more complex than the previous one.
i.
Primary structure :
In primary structure, proteins contain polypeptide chain. Each protein has amino acids linked
with each other in a specific sequence and this sequence of amino acids is termed as primary
structure.
O
O
||
||
 H  CH  C  NH  CH  C  NH  CH 
|
|
|
R1
R2
R3
Any change in the primary structures creates different protein molecule.
ii.
Secondary structure :
When long amino acid chain is coiled, looped or folded, it gives particular shape to a protein
molecule, it is referred as secondary structure. There are two different types of structures -helix
and -pleated sheets.
a.
 -Helix structure :
In -Helix structure, a poly peptide chain gets coiled by twisting into right handed spiral
known as -helix.
|
The hydrogen bonding between  C  O and  NH groups occurs in different parts of the
same chain resulting in folding of polypeptide chain.
Right Handed -helix
-helix structure
-helix
b.
-pleated structure :
The polypeptide chain lie side by side and are held together by intermolecular hydrogen
bonding. The poly peptide chains are stretched out resulting in a flat sheet. The contraction
results in a pleated sheets called -pleated sheet.
HRO
H
H
R
O
N
C
C
N
H
H R
C
O
H
C
N
O
C
C
H R
Less than 7.2 A
Contracted peptide chain
Biomolecules
-pleated structure
Pleated sheet structure
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iii. Tertiary structure :
The secondary structure on folding gives rise to molecular shapes i.e., fibrous and globular. The
polypeptide in tertiary structure held by disulphide or hydrogen bonds or Van Der Waals forces
or electrostatic forces of attraction.
iv. Quaternary structure :
Two or more amino acid chains or polypeptide chains forms complex protein. The spatial
arrangement of these polypeptide chains with respect to each other is known as quaternary
structure.
57. Write a note on (i) -helix (ii) -pleated.
Ans. Ref. Q. 56 (ii) (a) and (b).
*58. What is denaturation of proteins ? How is it brought about ?
Ans. i.
The irreversible process of precipitation of proteins which takes place very easily is known as
denaturation of proteins.
ii. This process is brought about by heating the protein with alcohol, concentrated inorganic acids
or by salts of heavy metals.
iii. Denaturation uncoils the protein and destroys the shape and thus the characteristic biological
activity.
iv. Eg. a. Conversion of milk to curd
b. Boiling of egg coagulates egg white
14.4 Enzymes
*59. Define enzymes.
Ans. Enzymes are the biological catalysts for various biochemical reactions in living organisms.
60. Write a note on enzymes and give example.
Ans. i.
All biological or bio-catalysts which catalyse the reactions in living organisms are called
enzymes.
ii. Chemically all enzymes are proteins.
iii. They are required in very small quantities as they are catalyst also they reduce the activation
energy for a particular reaction.
iv. Eg. : Enzyme maltase converts maltose to glucose.
maltase
C12 H 22 O11 
Maltose
C6 H12 O 6
Glucose
*61. Betaine C5H11O2N, occurs in beet sugar molasses. It is water soluble solid that melts with
decomposition at 300 °C. It is unaffected by a base but reacts with hydrochloric acid to form a
crystalline product C5H12O2NCl. It can be made from glycine with methyl iodide or treatment of
chloroacetic acid with trimethyl amine. Draw structure for betaine which will account for all
properties given above.
HCl
Ans. i.
C5H11O 2 N

C5H12O 2 NCl
Betaine present in beetroot
Crystalline product
ii. Preparation of Betaine :
CH 3
CH 3
|
|

H 2 N  CH 2  COOH  3CH3I 
 H 3C  N  CH 2  COOH 
H 3C  N   CH 2  COO 
 HI
2HI
|
|
CH 3
CH 3I 
Glycine
Methyl iodide
Betaine
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Std. XII Sci. Success Chemistry - II
CH 3
|
H 3C  N  CH 2  COOH 
 H 3C 
|
|
CH 3 Cl
Trimethyl
amine
iii.
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CH 3
|
 H 3C 
N   CH 2  COOH 
 HI
|
CH 3 Cl 
Chloroaceticacid
CH 3
|
N   CH 2  COO 
|
CH 3
Betaine
CH 3
|
The structure of Betaine is H 3C  N   CH 2  COO 
|
CH 3
14.5 Lipids
62. What are lipids?
Ans. Lipids are naturally occurring biomolecules having limited solubility in water and isolated from
organisms by extraction with non-polar solvents.
*63. How are lipids classified ?
Ans. Lipids are classified into two categories :
i.
Complex lipids
ii. Simple lipids
i.
Complex lipids :
They are the esters of long chain fatty acids. It can be hydrolysed. Complex lipids includes
triglycerides, glycolipids, phospholipids and waxes.
a.
Triglycerides are triesters of glycerol with higher fatty acids.
O
CH 2  OH
||
|
H2C – O – C – R1
CH  OH
O
|
||
CH 2  OH
HC – O – C – R2
Glycerol
O
||
H2C – O – C – R3
Triacyl glycerol
b.
c.
d.
ii.
Fats and oils are mixtures of triacyl glycerols. R1, R2 and R3 may be same or different and
may be saturated or unsaturated.
Glycolipids : When lipids are associated with sugars, it forms glycolipids. The simplest
animal glycolipids are cerebrosides. In plant glycolipids, the sugar group is commonly
galactose.
In phospholipids, two of the hydroxyl groups in glycerol are esterified by fatty acids and
one by phosphate group.
Waxes are mainly esters of long chain carboxylic acids with long chain alcohols, usually
they are secreted by plants and animals.
Simple lipids :
These lipids have ester linkages and cannot be hydrolysed. These include steroids, terpenes and
prostaglandins.
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a.
Steroids are derived from cyclopentanoperhydrophenanthrene, which has nucleus of four
rings.
OH
HO
Steroid nucleus
b.
c.
d.
e.
f.
O
Cholesterol (sterol)
Testosterone (androgen)
Sterols exist as free sterols or esters of fatty acids. Animal sterols include cholesterol and
lanosterol.
Terpenes are unsaturated hydrocarbons. Terpenoids, which are derivatives of terpenes
include geraniol, menthol and vitamin A.
Prostaglandins are group of C20 lipids. It contains five members ring with two long side
chains. It is detected in many body tissues.
Testosterone and androsterone is a male sex hormone.
Esrtone and estradiol are female sex hormones.
Note :
Number of
Carbon atom
Class
Example
10
Monoterpenes
-Phellandrene
15
20
Sesquiterpenes Diterpenes
Abscisic acid
Cembrene
30
Triterpenes
Squalene
40
Tetraterpenes
-Carotene
64. State the functions of lipids in living organism.
Ans. i.
Food energy is stored as fats and oils.
ii. Components of cell membrane are glycolipids. In plants, glycolipids are the main lipid
constituents of chloroplasts.
iii. Phospholipids and sterols like cholesterol are major components of cell membrane. Lipo proteins
are found in cell membrane.
iv. Waxes are also lipids and form protective covering on leaves, fruits, feathers of birds.
v. Steroids contain adrenal hormones, sex hormones and bile acids. Bile acid helps digestion of fat
in intestine.
vi. Terpenes include vitamin E and K and phytol. These are biological active compounds useful for
the growth and maintenance of the structure in human beings.
vii. Prostaglandin can lower blood pressure, gastric secretion and stimulate uterine contractions
during child birth.
*65. What are sterols and give their classification.
Ans. i.
Sterols are free esters of fatty acids.
ii. Sterols are classified into animal sterols and plant sterols.
iii. Animal sterols include cholesterol and lanosterol.
iv. Plant sterol is sitosterol. Fungal sterols (mycosterols) include ergosterol.
*66. What are triacyl glycerols ?
Ans. Triacyl glycerols are the triesters of glycerol with higher fatty acids. Fats and oils are mixtures of
triacyl glycerols. R1, R2 and R3 may be same or different and may be saturated or unsaturated. The
fatty acids in triacyl glycerols contains an even number of carbon atoms and an unbranched carbon
chain.
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*67. Mention the examples of terpenes.
Ans. Terpenes are unsaturated hydrocarbons and consist of isoprene units. i.e. H2C = C(CH3)CH = CH2
Eg. : -phellandrene, abscisic acid, cembrene, squalene, -carotene.
14.6 Hormones
*68. What are hormones and how is the name derived ?
OR
Write a note on hormones.
Ans. i.
Hormones are the secretions of endocrine glands.
ii. Hormone word is derived from Greek word ‘Hormaein’.
iii. The part where hormones secreted in the body are called effectors and the cells where they act on
are called targets.
iv. Hormones are easily diffusible, have low molecular weight of affect biological processes.
v.
Hormones are derived from amino acid derivatives and proteins or steroids.
Eg. : Thyroid gland secrete Thyroxine.
69.
State the various functions of hormones.
OR
Give different examples, body parts where the hormones are secreted in human body and their
functions.
Ans. Functions of hormones :
1.
2.
3.
4.
Hormone
Thyroxine
Body part
Function
Thyroid glands in the i. To increase the rate of energy exchange.
neck
ii. To increase consumption of O2.
Insulin
Pancreas
i. To control carbohydrate metabolism by
(Peptide hormone)
increasing glycogen in muscles.
ii. Oxidation of glucose in tissue.
iii. To lower the blood sugar.
Sex hormones
Ovaries (females)
i. To help in reproduction development.
estrogen, progesterone
ii. To
develop
secondary
sexual
(Steroids)
characteristics.
iii. To differentiate males from females
Androgen
Testes (Males)
i. To help in reproduction, development.
(Steroids)
ii. To differentiate males from females.
14.7 Vitamins
70. What are vitamins ?
Ans. i.
Vitamins are the organic substances that must be supplied to permit proportionate growth in
living beings (humans) or for the maintenance of the structure.
ii. Plants synthesize nearly all the vitamins but most of them cannot be synthesized by our body.
iii. Our daily diet like milk, vegetables, fruits etc, contain all necessary vitamins. However excess or
lack of vitamins result in certain diseases. These vitamins perform specific biological functions.
iv. Vitamins are designated by alphabets A, B, C, D, H etc. Some of them have subgroups, they are
designated as B1, B2, B3, B12, etc.
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*71. Give classification of vitamins with examples.
Ans. Vitamins are classified into two groups depending upon their solubility in water or fat and their
chemical structure.
i.. Depending upon their solubility vitamins are classified into two types :
a.
Water soluble vitamins : Vitamins B and C are soluble in water. They are readily excreted
in urine, have low toxicity and cannot be stored in body.
b. Fat soluble vitamins : Vitamin, A, D, E, K and P are soluble in oils and fats are stored in
the body in liver and in tissues.
ii.
Vitamins are also classified according to their chemical structures as follows :
a.
Aliphatic series : Vitamins of this series contain a long chain of aliphatic compounds
e.g. Vitamin C
b. Aromatic series : Vitamins of this series contain long chains of aromatic compounds
e.g. Vitamin K
c.
Alicyclic series : Vitamins of this series contain alicyclic rings in their structure.
e.g. Vitamin A
d. Heterocyclic series : Vitamins of this series contain rings containing hetero atoms,
e.g. Vitamin B complex. It is mixture of vitamins B1 + B2 + B3 + B5 + B6 + B12 +
Mesoinsositol + Folic acid + -Biotins.
72. Mention different types of vitamins, their sources and diseases due to deficiencies.
Ans.
Vitamins
Sources
Diseases due to deficiencies
i.
Milk, fish liver oil, tomatoes,
Night blindness, retardation of
A
carrots, sweet potatoes
growth, dryness of skin and hair.
ii.
B1
Rice,
wheat,
meat,
green Causes the disease called Beriberi.
(Thiamine)
vegetables.
iii.
Egg yolk, fist, yeast, liver
Inflammation of tongue, drying of
B2
lips and at corners of mouth, cheilosis
(Riboflavin)
(retarding the growth and digestion)
iv.
Barley, liver, maize, wheat, rice
Pellagra-pigmentation of the skin,
B5
degeneration of spinal cord, mental
(Nicotinamide)
confusion.
v.
B6
Wheat, fish, maize, liver, milk, Convulsions, failure to gain weight,
(A mixture of cereals, yeast
mental changes, derangement of
Pyriodoxine,
enzymes (which control carbohydrate
Pyridoxal and
metablosim)
pyridoxamine)
vi.
B12
Egg, curd, fish, liver of pig, sheep Degradation of spinal cord, anaemia
vii.
Oranges, grapes, lemon, green Scurvy (bleeding, spongy, swollen
C
vegetables
like
cabbage, gums.)
tomatoes, onion, all citrus fruits.
viii.
Rice, liver of cattle, seed oils, Weakness of muscles, abnormal
E
soya bean oil, palm oil, wheat growth and deposition of tissue
germ oil, cotton seed oil.
ix.
All green leafy vegetables like Increases blood clotting time
K
spinach, cauliflower, fish, meat
etc.
x.
Orange, grapes
Haemorrhage, decrease in capillary
P
resistance.
Biomolecules
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73. Write the names of all the sources of Vitamin A and C. [Intext Question text book page no. 529]
Ans. Vitamins
Sources
A
Milk, fish liver oil, tomatoes, carrots, sweet potatoes
C
Oranges, grapes, lemon, green vegetables like cabbage, totatoes,
all citrus fruits.
74. Write the names of vitamins present in fish and eggs.
Ans. Source
Vitamins
Fish
A, B6, B12, K
Eggs
B2, B12
[Intext Question text book page no. 529]
14.8 Nucleic acids
75. What are nucleic acids ?
Ans. Nucleic acids are bimolecules which are found in the nuclei of all living cells in the form of nucleo
proteins or chromosomes.
(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)
Nucleic acids are esters of phosphoric acid with sugar.
Base
|
O
|
Base
|
O
|
Base
|
O
|
Sugar  O  P  O  Sugar  O  P  O  Sugar  O  P  O
|
O–
|
O–
|
O–
Polynucleotide chain
DNA is Deoxyribose nucleic acid.
RNA is Ribonucleic acid.
76. Explain the chemical composition of nucleic acids.
Ans. The complete hydrolysis of DNA or RNA gives mixture of three different components :
i.
Pentose sugar
ii. Phosphoric acid
iii. Nitrogen containing heterocyclic compounds (nitrogeneous bases)
iv. Nucleoside and nucleotide
i.
Pentose Sugar :
In RNA the sugar is D-ribose and in DNA D-2-deoxyribose, the sugars are found in furanose
form.
2-Deoxy means no-OH group at C2 position.
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ii. Phosphoric acid :
The sugar units are joined to phosphate through C3 and C5 hydroxyl groups.
iii.
Nitrogen containing heterocyclic compounds :
The heterocyclic bases in DNA are adenine and guanine containing purine ring, cystosine and
thymine which contains pyrimidine ring. The heterocyclic bases in RNA are adenine, gaunine,
cystosine and uracil.
Purines :
Pyrimidines :
iv.
Nucleoside and Nucleotide :
A base-sugar unit is called nucleoside and base-sugar phosphoric acid unit is called nucleotide.
*77. A nucleotide from DNA containing thymine is hydrolysed. What are the products formed ?
Ans. When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and
phosphoric acid is obtained.
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Std. XII Sci. Success Chemistry - II
78. Draw the structures of DNA and RNA.
Ans. Structure of DNA :
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Structure of RNA :
–
*79. Write the structures of (i) nucleoside and (ii) nucleotide ?
Ans. i.
Structure of nucleoside :
A nucleoside contains two basic components of nucleic acids i.e. a
pentose sugar and a nitrogenous base. It is represented a sugar-base.
A nucleoside is formed when 1-position of a pyrimidine (cytosine,
thymine or uracil) or 9-position of guanine or adenine base is
attached to C  1 of sugar by -linkage.
ii.
Structure of nucleotide :
A nucleotide contains all three basic components of nucleic
acids i.e. a pentose sugar, a phosphoric acid and a
nitrogenous base. These are obtained by esterification of
C5  OH group of the pentose sugar by phosphoric acid.
Nucleotides are joined together through phosphate ester
linkage.
O
*80. What are the types of RNA ? How do they function ?
Ans. i.
There are three different types of RNA found in the cell.
a.
The messenger RNA which carries the message to the ribosome
b. Ribosomal RNA where synthesis of protein takes place
c.
The transport RNA.
ii. The messenger RNA calls up the series of transport RNA molecules which contain particular
amino acid.
iii. The order in which the transport RNA are called up by messenger RNA depends upon the
sequence of bases of messenger RNA chains.
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HORIZON Publication
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81. Give differentiating points between DNA and RNA
Ans.
DNA
RNA
i.
Sugar present in -D-2-deoxyribose.
Sugar present in -D-Ribose.
ii.
Pyrimidine bases present in DNA are Pyrimidine bases present are cystosine
cystosine, thyamine.
and uracil.
iii.
In secondary structure helixes are In secondary structure helixes are
double stranded.
single stranded.
iv.
DNA is the repository of the heredity RNA usually do not replicate and play
information and uses it by replication important role in protein preparation.
and duplicate the identical DNA.
v.
There are no types of DNA present.
RNA has three types :
i.
Messenger RNA
ii. Ribosomal RNA
iii. Transport RNA
82.
RNA on hydrolysis give four bases but there is no relationship between their quantities, similar
to that observed for the bases obtained from DNA. From this what can be suggested about the
strucutre of RNA ?
[Intext Question from Textbook Page 531]
Ans. i.
This suggests that there are different types of RNA molecules which contain different quantities
of bases.
ii. The secondary structure of RNA is single stranded.
iii. All the three types of RNA function on co-ordination basis with each other.
Higher Order
83.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring
compounds) are insoluble in water. Explain.
Ans. Glucose and sucrose have –OH groups while cyclohexane and benzene do not. Thus glucose and
sucrose can form hydrogen bonds with water and are soluble in water. While cyclohexane and benzene
do not form hydrogen bonds with water and thus insoluble in water.
The  and -glucose have different specific rotations. When either is dissolved in water their
rotation changes until the same fixed value results.
i.
What is this phenomenon called ?
ii. Explain the chemistry behind it.
Ans. i.
Mutarotation.
ii. The change in specific rotation of an optically active compound in solution with time, to an
equilibrium value, is called mutarotation.
When -D-glucose solution has a specific rotation of +111o is allowed to stand, the specific
rotation gradually falls to +52.7o and remains constant at this value. Similarly when -D glucose
solution has a specific rotation of +19o is allowed to stand, its specific rotation gradually
increases and remains constant at +52.7o.
In mutarotation, when either forms of glucose ( or ) is dissolved in water, it is converted into
other anomer and an equilibrium mixture is formed together with small amount of open chain
form.
84.
  D glucose
Open chain form
  D glucose
36%
0.02%
63%
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85. List the reactions of glucose which cannot be explained by its open chain structure.
Ans. Limitation of the open chain structure : Although the open chain structure of D – (+) – glucose
explains most of its reactions, yet it fails to explain the following facts.
i.
D (+) – glucose does not undergo certain reactions of aldehydes. For example, glucose does not
form NaHSO3 addition product, aldehyde – ammonia 2, 4 – DNP derivative and does not
respond to Schiff’s reagent test.
ii. Glucose reacts with NH2OH to form an oxime but glucose penta acetate does not.
86.
The melting points and solubility in water of amino acids are generally higher than that of the
corresponding halo acids. Explain.
Ans. – NH2 group present in amino acids has the ability to form intermolecular H-bonds. Thus melting
points and solubility in water of amino acids are generally higher than that of the corresponding halo
acids.
87. Where does the water present in the egg go after boiling the egg ?
Ans. About three-fourth of an egg is plain water in which the albumin protein and fats are suspended. At
room temperature, the protein strands are tightly folded in a complex three dimensional shape.
Because the individual proteins are able to move freely, both the egg white and yolk remain liquid. But
when egg is boiled, the protein strands straighten out. The protein ends, normally protected within the
folds, become exposed and the open ends join together in bridge like bonds forming a giant structure.
These new bonds prevent the molecules to move thereby making the egg material solid and the water
seems to be disappeared.
88. Why cannot vitamin C be stored in our body ?
Ans. Vitamin C is water soluble and is thus readily excreted in urine and cannot be stored in our body. It is
due to this reason that it must be supplied regularly in diet.
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HORIZON Publication
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Quick Review
Carbohydates
O
||
Optically active, Polyhydroxy and – CHO or – C –
On the basis of hydrolysis
1.
Simple carbohydrates known
as Monosaccharides
Complex carbohydrates
Aldoses
– CHO group
Triose
3 ‘C’
Tetrose
4 ‘C’
Ketoses
O
||
 C  group
Pentose
5 ‘C’
Hexose
6 ‘C’
In all functional group is CHO
Triose
3 ‘C’
Tetrose
4 ‘C’
Pentose
5 ‘C’
Hexose
6 ‘C’
In all functional group is Keto
Oligosaccharides on hydrolysis give two
to ten monosaccharides
Disaccharide give 2
monosaccharide
Trisaccharide give 3
monosaccharides
Polysaccharides give large no. of
monocasccharides on hydrolysis
Tetrasaccharide give 4
monosaccharides
On the basis of solubility
Sugars
Soluble in H2O, monosaccharides
oligosaccharides
Non-sugars
Insoluble in H2O, Polysaccharides
On the basis of reducing property
Reducing sugars reduce Fehling’s
solution, Tollen’s reagent
Biomolecules
Non-reducing sugars do not reduce
Fehling solution and Tollen’s reagent
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Std. XII Sci. Success Chemistry - II
2.
Glucose known as dextrose
Preparation
Laboratory method from cane sugar
alcohol is added to remove fructose
3.
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Commercial method from starch, chalk powder
is added to remove excess H2SO4, charcoal
powder to remove coloured impurity
n-Hexane
Glucose + CH3COOH
pentaacetate
 HI
Gluconic acid
Oxidation
(CH3CO)2O
NH2OH
Glucoxime
Reactions
of glucose
HCN
Oxidation
dil. HNO3
Bromine water
Oxidation
dil. HNO3
Saccharic acid
Glucose cyanohydrin
Haworth projection formulae
structures of
4.
Monosaccharides
i. Glucose
ii. Fructose
5.
Disaccharides
i. Sucrose
ii. Maltose
iii. Cellobiose
iv. Lactose
Polysaccharides
i. Starch
ii. Cellulose
Proteins
Basic units -amino acids
Fibrous proteins
Eg. Keratin
Globular proteins
Eg. Haemoglobin
6.
Structures of proteins
Primary
Biomolecules
Secondary
Tertiary
Quaternary
14. 36
Std. XII Sci. Success Chemistry - II
7.
HORIZON Publication
Classification of amino acids
Acidic
more – COOH groups
8.
Basic
more – NH2 groups
Neutral
equal – COOH and – NH2
groups
Enzymes : eg. Maltase Behaves mainly as catalysts in different reactions.
Lipids
9.
Complex
Eg. Triacyl glycerol
Simple
Eg. Cholesterol
10.
Hormones : Secretions of endocrine glands regulate vital body functions.
Eg. : Thyroxin, Insuline
11.
Vitamines :
Based on solubility
Water soluble
vit. B, C
Fat soluble
vit. A, D
Based on chemical structures
Vitamins of aliphatic series
vit. C
Aromatic series
vit. K
12.
Heterocyclic series
vit. B-complex
Nucleic acids
DNA
Double helix
Replicate, duplicate
identical DNA
Biomolecules
Alicyclic series
vit. A
RNA
Single strand
No replication and duplication
Perform functions through 3 types
of RNA
14. 37
Std. XII Sci. Success Chemistry - II
www.horizonpublication.com
SHORT TEST
Time : 1 Hour
Max. Marks : 20
1.
Answer the following questions.
i.
What happens when glucose is treated with
[3]
a.
Bromine water
b. Dilute nitric acid
c.
Hydrogen cyanide
ii. Define complex lipids and give two functions of lipids.
[2]
iii. What are hormones and write the structure of simple triglycerides.
[2]
iv. What is meant by peptide linkage and biocatalysts ?
[2]
v.
What is glycogen ? How is it different from starch ?
[2]
vi. What is essentially the difference between -glucose and -glucose ? what is meant by pyranose
structure of glucose and give structures.
[3]
vii. Write the structures of nucleotide and nucleoside.
[2]
viii. Write the full forms of DNA and RNA.
[2]
2.
Choose the correct option :
[2]
i.
Inflammation of tongue is due to the deficiency of
(a) vit. B1
(b) vit. B2
(c) vit. B5
(d) vit. B6
ii. How many moles of acetic anhydride will be required to form glucose penta-acetate from 2M of
glucose ?
(a) 2
(b) 5
(c) 10
(d) 2.5
ANSWER
1.
i.
a.
Refer to Q.17 (iv)
b.
Refer to Q.17 (v)
c.
Refer to Q.17 (iii)
ii.
Refer to Q.63 and Q.64.
iii.
Refer to Q.63 and Q.68.
iv.
Refer to Q.51 and Q.59.
v.
Refer to Q.40.
vi.
Refer to Q.21.
vii. Refer to fig. of Q.76 (iv)
viii. Refer to Q.75.
2.
i.
(b) Refer to Q.72 (iii)
ii.
(c)
Biomolecules
1 Glucose + 5 acetic anhydride 
 1 Glucose Pentacetate
14. 38