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JEE • CHEMISTRY • BIOMOLECULES

Biomolecules practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 159 questions

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121 Medium
121 of 159
Remembering +4 | -1
is/are a nonreducing sugar.
View answer and explanation
Correct option: B
Sucrose
Option B (Sucrose) is correct because sucrose does not have a free aldehyde or ketone group to act as a reducing agent.
122 Medium
122 of 159
Remembering +4 | -1
Cane sugar is made of:
View answer and explanation
Correct option: C
6 membered glucose ring and 5 membered fructose ring
Option C (6 membered glucose ring and 5 membered fructose ring) is correct because cane sugar (sucrose) consists of a 6-membered glucose ring and a 5-membered fructose ring.
123 Medium
123 of 159
Remembering +4 | -1
The purine base present in RNA is
View answer and explanation
Correct option: A
Guanine
Option A (Guanine) is correct because Guanine is a purine base found in both DNA and RNA.
124 Medium
124 of 159
Understanding +4 | -1
Vitamin C must be supplied regularly in diet because
View answer and explanation
Correct option: A
it is water soluble hence excreted in urine and can’t be stored in the body
Option A (it is water soluble hence excreted in urine and can’t be stored in the body) is correct because Vitamin C is water-soluble and cannot be stored in the body, requiring regular intake.
125 Medium
125 of 159
Remembering +4 | -1
On what basis are the manosaccharides classified?
View answer and explanation
Correct option: D
Number of C\mathrm{C}-atoms and the functional group present in them
Option D (Number of C\mathrm{C}-atoms and the functional group present in them) is correct because monosaccharides are classified based on the number of carbon atoms they contain and the type of carbonyl group (aldehyde or ketone) present.
126 Medium
126 of 159
Remembering +4 | -1
The base adenine occurs in which among the following.
View answer and explanation
Correct option: C
DNA and RNA both
Option C (DNA and RNA both) is correct because adenine is one of the five main nucleobases used in forming nucleic acids, and it is found in both DNA and RNA.
127 Medium
127 of 159
Remembering +4 | -1
Non-essential amino acid among the following is
View answer and explanation
Correct option: B
Glutamine
Option B (Glutamine) is correct because glutamine is a non-essential amino acid, meaning the human body can synthesize it.
128 Medium
128 of 159
Remembering +4 | -1
Essential oils are:
View answer and explanation
Correct option: B
Pleasant smelling liquids occurring in plants
Option B (Pleasant smelling liquids occurring in plants) is correct because essential oils are volatile, aromatic liquids extracted from plants.
129 Medium
129 of 159
Understanding +4 | -1
Which of the following statements about enzymes are true?

1. Enzymes lack in nucleophilic groups.

II. Pepsin is proteolytic enzyme.

III. Enzymes catalyse chemical reactions enhances the rate of reaction by lowering the activation energy.

IV. Enzymes are highly specific both in binding chiral substrates and in catalysing their reactions.
View answer and explanation
Correct option: D
II, III and IV
Option D is correct because enzymes are indeed specific, catalyze reactions by lowering activation energy, and pepsin is a proteolytic enzyme.
130 Medium
130 of 159
Understanding +4 | -1
Assertion (A): α\alpha-helix structure of proteins is in the shape of right handed screw.

Reason (R): There is hydrogen bond between the >NH> \mathrm{NH} group of each amino acid to the >=O> =\mathrm{O} of an adjacent turn of the helix.
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is the correct explanation of A\mathrm{A}.
Option A (Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is the correct explanation of A\mathrm{A}.) is correct because the alpha-helix structure is a right-handed screw stabilized by hydrogen bonds between the NH group of each amino acid and the CO group of an adjacent turn.
131 Medium
131 of 159
Application +4 | -1
The maximum concentration of zwitter ion of given amino acid will be at
NH3(pKa=9.8)⊕−CH2−CO2H(pK a1=2.3)\mathop {N{H_3}}\limits_{\left( {{\text{pKa}} = 9.8} \right)}^ \oplus - C{H_2} - \mathop {C{O_2}H}\limits_{\left( {{\text{pK }}{{\text{a}}_{\text{1}}} = 2.3} \right)}
View answer and explanation
Correct option: C
6 (app.)
The correct answer is C (Increases by factor of four) because the rate is proportional to the square of the carbon monoxide concentration.
132 Medium
132 of 159
Remembering +4 | -1
The hormone produced by pancreas
View answer and explanation
Correct option: B
Glucogen
Option B (Glucogen) is correct because the pancreas produces glucagon, a hormone that increases blood glucose levels.
133 Medium
133 of 159
Remembering +4 | -1
The RBC deficiency is deficiency disease of:
View answer and explanation
Correct option: C
VitaminB12\mathrm{Vitamin }{B_{12}}
Option C (VitaminB12\mathrm{Vitamin }{B_{12}}) is correct because a deficiency of Vitamin B12 causes pernicious anemia, which is characterized by a deficiency in red blood cells (RBC).
134 Medium
134 of 159
Remembering +4 | -1
One mole of sucrose on hydrolysis gives
View answer and explanation
Correct option: B
One mole of glucose and 1 mole of fructose
Option B (One mole of glucose and 1 mole of fructose) is correct because sucrose, upon hydrolysis, yields equimolar amounts of glucose and fructose.
135 Medium
135 of 159
Remembering +4 | -1
Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives
View answer and explanation
Correct option: C
1 molecule of glucose +1 molecule of fructose
Option C (1 molecule of glucose +1 molecule of fructose) is correct because sucrose, upon hydrolysis, yields one molecule of glucose and one molecule of fructose.
136 Medium
136 of 159
Understanding +4 | -1
Assertion(A): Sucrose is a reducing sugar.

Reason(R): On hydrolysis sucrose bring the change in sign of rotation from dextro(+) to laevo(-)
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: C
A is not correct but R is correct
Option C (A is not correct but R is correct) is the correct answer because sucrose is a non-reducing sugar, but the reason regarding the change in sign of rotation upon hydrolysis is correct.
137 Medium
137 of 159
Remembering +4 | -1
Which functional group participates in disulphide bond formation in proteins?
View answer and explanation
Correct option: B
Thiol
Option B (Thiol) is correct because thiol groups (-SH) are involved in the formation of disulfide bonds in proteins.
138 Medium
138 of 159
Remembering +4 | -1
The disaccharide held together by a glycosidic linkage between C1 of α−D\alpha \mathrm{ - D} -glucose and C2 of β−D\beta \mathrm{ - D} -fructose is
View answer and explanation
Correct option: A
Sucrose
Option A (Sucrose) is correct because sucrose is formed by a glycosidic linkage between C1 of α−D\alpha\mathrm{ - D} -glucose and C2 of β−D\beta\mathrm{ - D} -fructose.
139 Medium
139 of 159
Understanding +4 | -1
Which one group of biological molecule are micro and macromolecule respectively?
View answer and explanation
Correct option: B
Lipids, RNA and proteins
Option C (Lipids and polysaccharides) is correct because lipids are generally considered micromolecules, while polysaccharides are macromolecules.
140 Medium
140 of 159
Understanding +4 | -1
Which of the statements about "Denaturation" given below are correct ?

(A) Denaturation of proteins causes loss of secondary and tertiary structures of the protein.

(B) Denturation leads to the conversion of double strand of DNA into single strand.

(C) Denaturation affects primary structure which gets distorted.

Options :
View answer and explanation
Correct option: C
(A) and (B)
Option C ((A) and (B)) is correct because denaturation disrupts secondary and tertiary structures and can convert double-stranded DNA to single-stranded DNA.
141 Medium
141 of 159
Remembering +4 | -1
Chemically amylose is a with 200−1000200-1000 α\alpha-D-(+)-glucose units held by glycosidic linkage
View answer and explanation
Correct option: C
long unbranched chain, Cl−C4\mathrm{Cl}-\mathrm{C} 4.
Option C (long unbranched chain, Cl−C4\mathrm{Cl}-\mathrm{C} 4) is correct because amylose is a linear polymer of glucose units linked by α\alpha-1,4-glycosidic bonds.
142 Medium
142 of 159
Understanding +4 | -1
Glucose reacts with acetyl chloride to form penta acetyl glucose, it indicates presence of:
View answer and explanation
Correct option: D
Five ─OH groups
Option D is correct because the formation of penta-acetyl glucose indicates the presence of five hydroxyl (-OH) groups in the glucose molecule, each reacting with acetyl chloride.
143 Medium
143 of 159
Remembering +4 | -1
Which of the following bases is not present in DNA?
View answer and explanation
Correct option: B
Uracil
Option B (Uracil) is correct because Uracil is a nitrogenous base found in RNA, not DNA. DNA contains Thymine instead of Uracil.
144 Medium
144 of 159
Analyzing +4 | -1
Assertion (A): Glucose does not form the hydrogen bisulphite addition product.

Reason (R): Glucose is not so reactive to form the product with NaHSO3\mathrm{NaHSO}_3.
View answer and explanation
Correct option: C
A\mathrm{A} is correct but R\mathrm{R} is incorrect.
Option C is correct because glucose exists predominantly in a cyclic form, making the aldehyde group less available for reaction with NaHSO3.
145 Medium
145 of 159
Remembering +4 | -1
The change in optical rotation of a freshly prepared solution of glucose with time is called ______.
View answer and explanation
Correct option: C
Mutarotation
Option C (Mutarotation) is correct because mutarotation describes the change in optical rotation of a freshly prepared glucose solution over time as it equilibrates between its alpha and beta anomers.
146 Medium
146 of 159
Remembering +4 | -1
Starch can be used as an indicator for the detection of the traces of:
View answer and explanation
Correct option: C
Iodine in aqueous solution
The correct answer is C (Iodine in aqueous solution) because starch forms a deep blue complex with iodine, which is used as an indicator for the presence of iodine.
147 Medium
147 of 159
Analyzing +4 | -1
Assertion (A): A solution of sucrose on hydrolysis becomes laevorotatory from dextrorotatory.

Reason (R): Sucrose on hydrolysis gives more number of glucose units due to which sign of rotation is changed.
View answer and explanation
Correct option: C
Assertion (A) is true, but Reason (R) is false
Option C (Assertion (A) is true, but Reason (R) is false) is correct because sucrose does become levorotatory upon hydrolysis, but this is due to fructose's higher negative rotation, not the number of glucose units.
148 Medium
148 of 159
Remembering +4 | -1
Which of the following sets of bases is present both in DNA and in RNA?
View answer and explanation
Correct option: B
Adenine, guanine, cytosine
Option B (Adenine, guanine, cytosine) is correct because these three bases are present in both DNA and RNA.
149 Medium
149 of 159
Understanding +4 | -1
Assertion:- Sucrose is dextrorotatory but its aqueous solution is laevorotatory.

Reason:- Dextro rotation of glucose is less than laevorotation of fructose.
View answer and explanation
Correct option: A
If both assertion & reason true & the reason is a correct explanation of the assertion
Option A (If both assertion & reason true & the reason is a correct explanation of the assertion) is correct because sucrose is initially dextrorotatory, but after hydrolysis, the resulting solution is laevorotatory due to the greater laevorotation of fructose compared to the dextrorotation of glucose.
150 Medium
150 of 159
Remembering +4 | -1
Which of the following is not an example of fibrous protein?
View answer and explanation
Correct option: B
Albumin
Option B (Albumin) is correct because Albumin is a globular protein, while Keratin, Collagen, and Myosin are fibrous proteins.
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