9.biomolecules 140mcq Aakash
9.biomolecules 140mcq Aakash
9.biomolecules 140mcq Aakash
Biomolecules
Solutions
SECTION - A
Objective Type Questions
1. Which of the following is not strictly a biomacromolecule?
(1) Proteins (2) Lipids (3) Polysaccharides (4) Nucleic acid
Sol. Answer (2)
Because the molecular weight of lipids does not exceed 800 Da but they come under the acid insoluble fraction.
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2 Biomolecules Solution of Assignment
9. Proteins which catalyse biochemical reactions in the living world are known as
(1) Enzymes (2) Hormones (3) Antibodies (4) Receptor
Sol. Answer (1)
Because almost all enzymes are protein which catalyse the biochemical reaction.
Hormones can be steroid and amino-acid derivative also.
Antibodies are glycoprotein.
Receptors are associated with sensory reception.
13. Lecithin is a
(1) Type of wax (2) Phospholipid (3) Oil (4) Simple fatty acid
Sol. Answer (2)
Because lecithin made up of a molecule of glycerol, a phosphate group, 2 fatty acid molecules, choline
(N-containing alcohol molecule)
O
O CH2 O C R1
R2 C O CH O CH3
CH2 O P O CH2 CH2 N CH3
OH CH3
Lecithin
14. Lipids that insulate the nerve fibre are
(1) Lecithin (2) Cholesterol (3) Suberin (4) Glycolipids
Sol. Answer (4)
The glycolipids are present in myelin sheath of nerve fibres.
16. Nucleoside is
(1) Sugar + Nitrogenous base (2) Sugar + Phosphate
(3) Nitrogenous base + Phosphate (4) Purine + Pyrimidine
Sol. Answer (1)
Nucleoside = Sugar + Nitrogenous base
Nucleotide = Sugar + Nitrogenous base + Phosphate
18. In a DNA molecule adenine of one strand base pair with ____ on the other strand
(1) Guanine (2) Thymine (3) Cytosine (4) Both (1) & (3)
Sol. Answer (2)
In DNA, A T, C G
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4 Biomolecules Solution of Assignment
20. The nitrogenous bases of the two strands of DNA are joined by
(1) Phosphodiester bond (2) Hydrogen bond (3) Glycosidic bond (4) Peptide bond
Sol. Answer (2)
Hydrogen bond : The bond formed between two polynucleotide strands of DNA.
Glycosidic bond : Formed between the two monosaccharides (sugar).
Peptide bond : Formed between 2 amino acids.
Phosphodiester bond : Bond formed in nucleic acid i.e. DNA or DNA, between the phosphate and hydroxyl
group of sugar.
25. The primary precursor for the production of cholesterol in our body is
(1) Acetic acid (2) Citric acid (3) Ethyl alcohol (4) Methanol
Sol. Answer (1)
In biosynthetic pathway or anabolic pathway, the acetic acid is converted into cholesterol in liver.
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Solution of Assignment Biomolecules 5
Sol. Answer (4)
Because ATP is made up of adenine, ribose sugar and three phosphate.
31. What is the fate of pyruvic acid under anaerobic conditions in our body?
(1) It gets converted into methyl alcohol
(2) It gets converted into acetyl CoA
(3) It gets converted into lactic acid
(4) It gets converted into glycogen
Sol. Answer (3)
Under anaerobic condition, the glucose is converted into lactic acid in muscle.
Enzyme (Holoenzyme)
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6 Biomolecules Solution of Assignment
SECTION - B
Objective Type Questions
CH2OH CH2OH
O H H O H
H
H H
1 4
OH H OH H
OH O
OH
H OH H OH
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Solution of Assignment Biomolecules 7
2. The given structure represents a monosaccharide known as
O
HOCH2
OH OH
OH OH
(1) Ribose (2) Glucose (3) Fructose (4) Raffinose
Sol. Answer (1)
Ribose is a pentose sugar.
Because chitin is a polymer of nitrogen containing glucose derivative known as N-acetyl glucosamine.
4. The given amino acid is _______ in nature.
COOH
H2N–C–H
CH2.COOH
CH2OH HOOC–R1
CHOH + HOOC–R2
CH2OH HOOC–R3
Glycerol Fatty acid
(1) Monoglyceride (2) Diglyceride (3) Triglyceride (4) Both (1) & (3)
Sol. Answer (3)
Because given structure has three fatty acids.
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8 Biomolecules Solution of Assignment
P = Phosphate
S A T S S = Sugar
A = Adenine
P P T = Thymine
G = Guanine
S G C S C = Cytosine
P P
S T A S
5' P 3'
10. If the sequence of bases in one of the DNA strand is A G G A G A A, then the sequence of bases in the other
complementary strand of DNA would be
(1) C C T T C T T (2) T C T C T C C (3) T C C T C T T (4) C C T C T C T
Sol. Answer (3)
Because A = T, G C
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Solution of Assignment Biomolecules 9
12. In which of the following energy is released?
(1) Conversion of glucose into pyruvate (2) Formation of proteins from amino acids
(3) Conversion of glucose into lactic acid (4) Both (1) & (3)
Sol. Answer (4)
Anaerobic combustion of glucose releases energy in the form of ATP.
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10 Biomolecules Solution of Assignment
17. Enzymes catalysing the breakdown of larger molecules into smaller molecules are
(1) Hydrolases (2) Isomerases (3) Ligase (4) Both (1) & (3)
Because, hydrolases are the enzyme which catalyse the breakdown of larger molecules into smaller molecules
with the addition of water.
Isomerases : They are the enzymes which catalyse the rearrangement of molecular structure to form isomers.
18. Michaelis constant (Km) value of enzyme is substrate concentration at which velocity of reaction is
(1) Vmax (2) One third Vmax (3) Half Vmax (4) One fifth Vmax
Vmax
Velocity Vmax
of reaction 2
Km [s]
20. The enzyme that catalyses the conversion of glucose-6-phosphate into fructose-6-phosphate is
isomerase
Glucose-6-phosphate
fructose-6-phosphate
(c) The active site of enzyme breaks the chemical bonds of the product
(1) (a) & (c) (2) (c) & (d) (3) (b) & (c) (4) (a) & (d)
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Solution of Assignment Biomolecules 11
Sol. Answer (4)
(b) Statement is wrong, because thermophillic enzymes don't get denatured at high temperature, they work
efficiently at high temperature (750ºC).
(c) Statement is wrong, because enzyme don't break chemical bonds of product but of substrate.
22. One of the following is correct sequence of carbohydrates in the order of increasing complexity of chemical
structure
(1) Sucrose, starch, oligosaccharide, maltose, triose (2) Triose, maltose, sucrose, oligosaccharide, starch
(3) Triose, glucose, maltose, oligosaccharide, starch (4) Oligosaccharide, triose, starch, sucrose, maltose
Glucose – Monosaccharide (6 C)
Starch – Polysaccharides
24. A cellulose molecule is formed by the polymerisation of glucose. The number of glucose molecules present
in a cellulose is
(1) 600 (2) 6000 (3) 60,000 (4) 60
Sol. Answer (2)
A cellulose molecule is formed by polymerisation of 6000 molecules of glucose.
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12 Biomolecules Solution of Assignment
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Solution of Assignment Biomolecules 13
Sol. Answer (1)
A = T (2 hydrogen bonds between adenine and thymine)
G C (3 hydrogen bonds between cytosine and guanine)
SECTION - C
Previous Years Questions
1. In sea urchin DNA, which is double stranded 17% of the bases were shown to be cytosine. The percentages
of the other three bases expected to be present in this DNA are [AIPMT-2015]
(1) G 8.5%, A 50%, T 24.5% (2) G 34%, A 24.5%, T 24.5%
(3) G 17%, A 16.5%, T 32.5% (4) G 17%, A 33%, T 33%
Sol. Answer (4)
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14 Biomolecules Solution of Assignment
3. Select the option which is not correct with respect to enzyme action: [AIPMT-2014]
(2) Addition of lot of succinate does not reverse the inhibition of succinic dehydrogenase by malonate
(3) A non - competitive inhibitor binds the enzyme at a site distinct from that which binds the substrate
(1) Only an unsaturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached
(2) A saturated or unsaturated fatty acid esterified to a glycerol molecule to which a phosphate group is also
attached
(3) A saturated or unsaturated fatty acid esterified to a phosphate group which is also attached to a glycerol
molecule
(4) Only a saturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached
7. Transition state structure of the substrate formed during an enzymatic reaction is : [NEET-2013]
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Solution of Assignment Biomolecules 15
9. Which one out of A – D given below correctly represents the structural formula of the basic amino acid?
A B C D
NH2 NH2 CH2OH
| | NH2
|
H C COOH H C COOH CH2 H C COOH
| | |
CH2 CH2
CH2 CH2 |
| | | CH2
CH2 OH NH2 |
| CH2
C |
CH2
OH |
O
NH2
[AIPMT (Prelims)-2012]
(1) A (2) B (3) C (4) D
Sol. Answer (4)
Because in 'D', it has an extra amino group, because of which it carry +ve charge which comes under basic
amino acid.
10. Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds
in the living tissues. Identify the category shown and the one blank component “X” in it
HOCH2 O "X"
OH OH
[AIPMT (Prelims)-2012]
Category Component
(1) Nucleotide Adenine
(2) Nucleoside Uracil
(3) Cholesterol Guanin
(4) Amino acid NH2
Sol. Answer (2)
Because the given structure doesn't have phosphate group so it is nucleoside and "X" is uracil because it
consists ribose sugar.
12. Which one of the following biomolecules is correctly characterised? [AIPMT (Mains)-2012]
(1) Alanine amino acid – Contains an amino group and an acidic group anywhere in the molecule
(2) Lecithin – a phosphorylated glyceride found in cell membrane
(3) Palmitic acid – an unsaturated fatty acid with 18 carbon atoms
(4) Adenylic acid – adenosine with a glucose phosphate molecule
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16 Biomolecules Solution of Assignment
13. The curve given below show enzymatic activity with relation to three conditions (pH, temperature and substrate
concentration)
y-axis
x-axis
What do the two axes (x and y) represent? [AIPMT (Prelims)-2011]
x-axis y-axis
(1) Enzymatic activity Temperature
(2) Enzymatic activity pH
(3) Temperature Enzyme activity
(4) Substrate concentration Enzymatic activity
Sol. Answer (3)
y
Enzymatic
activity
x
pH or Temperature
14. Which one of the following structural formulae of two organic compound is correctly identified along with its
related function
O
O CH2—O—C—R
A R2—C—O—CH O
CH2—O—P—O—CH2—CH2
OH N+
CH3 CH3
NH2 CH3
N
B N
[AIPMT (Prelims)-2011]
N NH
(1) A : Lecithin – a component of cell membrane
(2) B : Adenine – a nucleotide that makes up nucleic acids
(3) A : Triglyceride – major source of energy
(4) B : Uracil – a component of DNA
Sol. Answer (1)
Given structure 'A' is Leathin and 'B' is Adenine.
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Solution of Assignment Biomolecules 17
15. The figure given below shows the conversion of a substrate into product by an enzyme. In which one of the
four options (1–4) the components of reaction labelled as A, B, C and D are identified correctly?
A
C
D
B Substrate
Product
Progress of Reaction [AIPMT (Mains)-2010]
A B C D
(1) Potential energy Transition state Activation energy with Activation energy
enzyme without enzyme
(2) Transition state Potential energy Activation energy Activation energy with
without enzyme enzyme
(3) Activation energy Transition state Activation energy with Potential energy
without enzyme enzyme
(4) Activation energy with Transition state Activation energy Potential energy
enzyme without enzyme
16. Three of the following statements about enzymes are correct and one is wrong. Which one is wrong?
[AIPMT (Mains)-2010]
(1) Enzymes require optimum pH for maximal activity
(2) Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even
at temperatures 80°-90°C
(3) Enzymes are highly specific
(4) Most enzymes are proteins but some are lipids
Sol. Answer (4)
Because most enzymes are protein but some are RNA i.e. Ribonuclease-P and ribozyme.
17. Which one of the following pairs is wrongly matched? [AIPMT (Prelims)-2009]
(1) Alcohol – Nitrogenase
(2) Fruit juice – Pectinase
(3) Textile – Amylase
(4) Detergents – Lipase
Sol. Answer (1)
18. Carbohydrates are commonly found as starch in plant storage organs. Which of the following five properties
of starch (a - e) make it useful as a storage material? [AIPMT (Prelims)-2008]
a. Easily translocated b. Chemical non-reactive
c. Easily digested by animals d. Osmotically inactive
e. Synthesized during photosynthesis
The useful properties are
(1) Both a & e (2) Both b & c (3) Both b & d (4) a, c & e
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18 Biomolecules Solution of Assignment
19. An organic substance bound to an enzyme and essential for its activity is called [AIPMT (Prelims)-2006]
(1) Coenzyme (2) Holoenzyme (3) Apoenzyme (4) Isoenzyme
Sol. Answer (1)
Organic compound which bound to an enzyme is either coenzyme or prosthetic group.
21. Which of the following is the simplest amino acid? [AIPMT (Prelims)-2005]
(1) Tyrosine (2) Asparagine (3) Glycine (4) Alanine
Sol. Answer (3)
Because in glycine, the R-group is replaced by hydrogen.
H
HOOC C NH2
H = R group
22. Enzymes, vitamins and hormones can be classified into a single category of biological chemicals, because all of
these [AIPMT (Prelims)-2005]
(1) Enhance oxidative metabolism
(2) Are conjugated proteins
(3) Are exclusively synthesized in the body of a living organism as at present
(4) Help in regulating metabolism
Sol. Answer (4)
(1) is incorrect, because enzymes can both enhance and inhibit the oxidative metabolism.
(3) is incorrect, different enzymes, vitamins and hormones are synthesized in the body at different situations.
(2) is incorrect, because all hormones and enzymes are not conjugated protein.
23. Carbohydrates, the most abundant biomolecules on earth, are produced by [AIPMT (Prelims)-2005]
(1) All bacteria, fungi and algae (2) Fungi, algae and green plant cells
(3) Some bacteria, algae and green plant cells (4) Viruses, fungi and bacteria
Sol. Answer (3)
Autotrophic organism produce glucose by photosynthesis which is a carbohydrate. Heterotrophic are dependent
on autotrophes and don't produce carbohydrate and fungi are saprobic organisms.
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Solution of Assignment Biomolecules 19
24. Which of the following statements regarding enzyme inhibition is correct ? [AIPMT (Prelims)-2005]
(1) Non-competitive inhibition of an enzyme can be overcome by adding large amount of substrate
(2) Competitive inhibition is seen when a substrate competes with an enzyme for binding to an inhibitor protein
(3) Competitive inhibition is seen when the substrate and the inhibitor compete
(4) Non-competitive inhibitors often bind to the enzyme irreversibly
Sol. Answer (3)
(1) is wrong, because competitive inhibition of an enzyme can be overcome by adding large amount of
substrate.
(2) is wrong, because competitive inhibtion is seen when a substrate competes with an inhibitor for binding
to the active site of enzyme.
(4) is wrong, because non-competitive inhibitor often bind to enzyme irreversibly.
25. The catalytic efficiency of two different enzymes can be compared by the [AIPMT (Prelims)-2005]
(1) The Km value (2) The pH optimum value
(3) Formation of the product (4) Molecular size of the enzyme
Sol. Answer (1)
26. The four elements that make up 96% of all the elements found in a living system are
(1) C, H, O and P (2) C, N, O and P (3) H, O, C and N (4) C, H, O and S
Sol. Answer (3)
% weight of human body : C = 18.5; O = 65.0; N = 3.3; H = 0.5
Essential amino acids are leucine, isoleucine, valine, tryptophane, phenylalanine, lysine and methionine.
Lipids are hydrophobic that's why they are not soluble in water.
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20 Biomolecules Solution of Assignment
30. The major role of minor elements inside living organisms is to act as
Minor elements basically includes Zn, Mg, K, Ni, Co, NAD+, NADP+, they all come under the category of
cofactor.
31. Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite molecule formed by
32. About 98 percent of the mass of every living organism is composed of just six elements including carbon,
hydrogen, nitrogen, oxygen and
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Solution of Assignment Biomolecules 21
36. Which of the following is incorrect regarding the amino acids and their functions?
(1) Tyrosine : Converted into epinephrine hormone and used in the synthesis of melanin pigment
(2) Glycine : Involved in the formation of heme
(3) Tryptophan : Helps in the synthesis of auxin hormone
(4) Histidine : Can be converted into histamine by the removal of amino group
Sol. Answer (4)
Histamine is derived from amino acid histidine through decarboxylation.
O O
CH2 – O – C – R1 CH2 – O – C – R1
O O
(1) CH – O – C – R2 (2) CH – O – C – R2
O O
CH2 – O – P – O – CH2 – CH2 – NH3
+
CH2 – O – C – R3
OH
O O
CH2 – O – C – R1 CH2 – O – C – R1
O O
(3) CH – O – C – R2 (4) R2 – C – O – CH
O CH3 O
OH CH3 OH
Sol. Answer (3)
Lecithin is a phospholipid present in cell membrane
Carbohydrate = 3%
Lipid = 2%
Protein = 10–15%
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22 Biomolecules Solution of Assignment
(1) Carbohydrate (2) Respiratory pigment (3) Vitamin (4) Skin pigment
Because haemoglobin get binds to oxygen and help in the transportation of O2.
43. Collagen is
(1) Fibrous protein (2) Globular protein (3) Lipid (4) Carbohydrate
Because collagen are thread like proteins, which are insoluble in water.
Fructose = Monosaccharides
(1) Red algae (2) Blue-green algae (3) Brown algae (4) Green algae
Agar is a type of mucopolysaccharide and is obtained from red algae. It is used as culture medium in
laboratory.
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Solution of Assignment Biomolecules 23
47. Which of the following groups consists of polysaccharides only?
(1) Sucrose, glucose and fructose (2) Maltose, lactose and fructose
(3) Glycogen, sucrose and maltose (4) Glycogen, cellulose and starch
Sol. Answer (4)
Sucrose Disaccharide; Fructose Monosaccharide
Glucose Monosaccharide; Maltose Disaccharide
Lactose Disaccharide; Glycogen Polysaccharide
Cellulose Polysaccharide; Starch Polysaccharide
49. Cellulose, the most important constituent of plant cell wall is made up of
(1) Branched chain of glucose molecules linked by 1, 4 glycosidic bond in straight chain and
1, 6 glycosidic bond at the site of branching
(2) Unbranched chain of glucose molecules linked by 1, 4 glycosidic bond
(3) Branched chain of glucose molecules linked by 1, 6 glycosidic bond at the site of branching
(4) Unbranched chain of glucose molecules linked by 1, 4 glycosidic bond
Sol. Answer (2)
Glycogen and starch are branched polymer of glucose while cellulose is unbranched polymer of glucose.
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24 Biomolecules Solution of Assignment
52. An important step in the manufacture of pulp for paper industry is the
(1) Preparation of pure cellulose
(2) Treatment of wood with chemicals that break down cellulose
(3) Removal of oils present in the wood by treatment with suitable chemicals
(4) Removal of water from the wood by prolonged heating at approximately 50°C
Sol. Answer (1)
Because the raw material for paper is cellulose.
55. If the total amount of adenine and thymine in a double-stranded DNA is 60%, the amount of guanine in this
DNA will be
(1) 15% (2) 20% (3) 30% (4) 40%
Sol. Answer (2)
A + T = 60%
Then, C + G = 40%, when 20% C and 20% = G
56. DNA has equal number of adenine and thymine residues (A = T) and equal number of guanine and cytosine
(G = C). These relationships are known as
(1) Chargaff's rule (2) Coulomb's law
(3) Le Chatelier's principle (4) Van't Hoff plot
Sol. Answer (1)
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Solution of Assignment Biomolecules 25
57. Which of the following pairs of nitrogenous bases of nucleic acids is mismatched with the category mentioned
against it?
(1) Adenine, Thymine – Purines (2) Thymine, Uracil – Pyrimidines
(3) Uracil, Cytosine – Pyrimidines (4) Guanine, Adenine – Purines
Sol. Answer (1)
Adenine and Guanine Purines
Thymine, uracil, cytosine Pyrimidines
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26 Biomolecules Solution of Assignment
63. The 3-5 phosphodiester linkages inside a polynucleotide chain serve to join
(1) One DNA strand with the other DNA strand (2) One nucleoside with another nucleoside
(3) One nucleotide with another nucleotide (4) One nitrogenous base with pentose sugar
Because nucleotide is a monomeric unit of nucleic acid which are joined together by 3' – 5' phosphodiester bond.
64. ATP is
Because ATP (Adenosine triphosphate) is made up of adenine, ribose sugar and 3 phosphate groups.
Product
Progress of reaction
66. Which of the following factor(s) do(es) not affect enzyme activity?
A. Temperature B. pH
C. Enzyme concentration D. Product concentration
E. Substrate concentration F. Activation energy
(1) C only (2) C & D (3) D only (4) F only
Sol. Answer (4)
Because enzyme activity is affected by temperature, pH, enzyme concentration, product concentration and
substrate concentration.
Enzyme are not affected by activation energy but it lowers down the activation energy.
(1) Malate (2) Malonate (3) Oxaloacetate (4) Both (2) & (3)
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Solution of Assignment Biomolecules 27
68. Which of the following is a typical example of ‘feedback inhibition’?
(1) Cyanide and cytochrome reaction
(2) Sulpha drugs and folic acid synthesizer bacteria
(3) Allosteric inhibition of hexokinase by glucose 6-phosphate
(4) Reaction between succinic dehydrogenase and succinic acid
Sol. Answer (3)
(1) Cyanide and cytochrome reaction Example of non-competitive inhibition.
(2) Sulpha drugs and folic acid synthesizer bacteria are example of non-competitive inhibition.
(4) Reaction between succinic dehydrogenase and succinic acid are example of competitive inhibition.
69. Which factor is responsible for inhibition of enzymatic process during feedback?
(1) Substrate (2) Enzyme (3) End product (4) Temperature
Sol. Answer (3)
Feedback inhibition is also konwn as End product inhibition or allosteric modulation.
Vmax
No
Inhibitor
With inhibitor
Reaction
velocity Vmax
2
Km K'm [S]
vmax
Velocity vmax
of reaction 2
Km [s]
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28 Biomolecules Solution of Assignment
72. If an enzyme has been given the EC code 5.2.1.7, it is likely to be involved in
(1) Digestion (2) Redox reaction (3) Isomerization (4) Molecular breakdown
Sol. Answer (3)
EC code 5.2.1.7, in this the first digit represents class of enzyme.
And the class V of enzyme is isomerases, which catalyse the isomerisation reaction.
Protein Non-protein
(Apoenzyme) (Cofactor)
76. Apoenzyme is
(1) Always a protein (2) Often a metal
(3) Always an inorganic compound (4) Often a vitamin
Sol. Answer (1)
Protein part of enzyme is known as apoenzyme and non-protein part of enzyme is known as cofactor.
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Solution of Assignment Biomolecules 29
SECTION - D
Assertion-Reason Type Questions
1. A : Cofactor of an enzyme may be a prosthetic group.
R : NAD derived from niacin is a co-enzyme.
Sol. Answer (2)
Organic compound firmly attach to protein part of enzyme is prosthetic group.
5. A : Dextrins are intermediate polysaccharides formed during hydrolysis of starch into sugar.
R : Ascorbic acid is a sugar derivative.
Sol. Answer (2)
Ascorbic acid is sugar acid.
7. A : The polypeptide coil of collagen helix is strengthened by the estabilishment of hydrogen bond between >
NH - group of glycine residue of each strand with –CO group of other two strand.
R : In collagen helix locking effect also occurs with the help of proline and hydroxyproline amino acid.
Sol. Answer (2)
In collagen, there are generally three polypeptides coil around one another.
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30 Biomolecules Solution of Assignment
10. A : Tertiary structure of protein is absolutely necessary for many biological activities of proteins.
R : In protein, only right handed helices are observed.
Sol. Answer (2)
Enzymes have tertiary structure of protein.
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