CBSE Class 12 Chemistry Biomolecules NEET JEE Questions Set 03

Single Correct Answer

Question 1. The hormone which controls the processes of burning of fats, proteins and carbohydrates and liberates energy in the body is (AIIMS 2009)
(a) thyroxine
(b) adrenaline
(c) insulin
(d) cortisone.
Answer:
Ans. (c)
Since insulin maintains the level of glucose in blood, therefore, it controls the processes of burning of fats, proteins and carbohydrates.

Teacher's Note:
1. Insulin lets cells take up and use glucose; without it the body burns fat and protein for energy instead.

Question 2. The hormone that helps in the conversion of glucose into glycogen is
(a) cortisone
(b) bile acids
(c) adrenaline
(d) insulin
Answer:
Ans. (d)
Insulin lowers blood glucose level by helping the conversion of glucose into glycogen (glycogenesis) in the liver and muscles.

Teacher's Note:
1. Insulin stores glucose as glycogen; glucagon and adrenaline do the reverse.

Question 3. Which of the following hormones is produced under the conditions of stress which stimulate glycogenolysis in the liver of human beings? (AIPMT 2014)
(a) Thyroxin
(b) Insulin
(c) Adrenaline
(d) Estradiol
Answer:
Ans. (c)
Adrenaline (the ‘fight or flight’ hormone) is released under stress and stimulates the breakdown of liver glycogen into glucose.

Teacher's Note:
1. Glycogenolysis = glycogen → glucose; adrenaline triggers it in emergencies.

Question 4. Hormones are secreted by ductless glands of human body. Iodine containing hormone is (Karnataka CET 2017)
(a) Insulin
(b) adrenaline
(c) testosterone
(d) thyroxine.
Answer:
Ans. (d)
Thyroxine, secreted by the thyroid gland, is an iodinated derivative of the amino acid tyrosine.

Teacher's Note:
1. Lack of iodine in diet reduces thyroxine and causes goitre.

Question 5. The deficiency of vitamin C causes
(a) scurvy
(b) rickets
(c) pyorrhea
(d) pernicious anaemia.
Answer:
Ans. (a)
Deficiency of vitamin C (ascorbic acid) causes scurvy (bleeding gums).

Teacher's Note:
1. Rickets: vitamin D; pernicious anaemia: vitamin B₁₂; scurvy: vitamin C.

Question 6. Deficiency of vitamin B₁ causes the disease (AIPMT Prelim 2012)
(a) convulsions
(b) beri-beri
(c) cheilosis
(d) sterility
Answer:
Ans. (b)
Deficiency of vitamin B₁ (thiamine) causes beri-beri.

Teacher's Note:
1. B₁ → beri-beri, B₂ → cheilosis, B₆ → convulsions, E → sterility.

Question 7. Cheilosis and digestive disorders are due to the deficiency of (Karnataka CET 2015)
(a) ascorbic acid
(b) pyridoxine
(c) thiamine
(d) riboflavin
Answer:
Ans. (d)
Deficiency of riboflavin (vitamin B₂) causes cheilosis and digestive disorders.

Teacher's Note:
1. Cheilosis (cracks at the corners of the lips) is the classic sign of riboflavin deficiency.

Question 8. Biotin is an organic compound present in yeast. Its deficiency in diet causes dermatitis and paralysis. It is also known as
(a) vitamin H
(b) vitamin B₁
(c) vitamin B₁₂
(d) vitamin D
Answer:
Ans. (a)
Biotin is also known as vitamin H; its deficiency causes dermatitis and paralysis.

Teacher's Note:
1. Biotin = vitamin H (also called B₇).

Question 9. Which of vitamins given below is water soluble? (JEE Main 2015)
(a) Vitamin E
(b) Vitamin K
(c) Vitamin C
(d) Vitamin D
Answer:
Ans. (c)
Vitamin C is water soluble.

Teacher's Note:
1. Vitamins A, D, E and K are fat soluble; B-complex and C are water soluble.

Question 10. Which one of the following is a water soluble vitamin that is not excreted easily? (JEE Main 2022)
(a) Vitamin B₂
(b) Vitamin B₁
(c) Vitamin B₆
(d) Vitamin B₁₂
Answer:
Ans. (d)
Vitamin B₁₂ contains cobalt. Unlike other water soluble vitamins, it is stored in the liver and is not excreted easily in urine.

Teacher's Note:
1. Water soluble vitamins must be taken daily because they are excreted; vitamin B₁₂ is the exception.

Question 11. Deficiency of which vitamin causes osteomalacia? (NEET 2020)
(a) Vitamin D
(b) Vitamin K
(c) Vitamin E
(d) Vitamin B
Answer:
Ans. (a)
Deficiency of vitamin D causes osteomalacia (soft bones and joint pain).

Teacher's Note:
1. Vitamin D deficiency gives rickets in children and osteomalacia in adults.

Question 12. The RBC deficiency is deficiency of (NEET 2021)
(a) vitamin B₁₂
(b) vitamin B₆
(c) vitamin B₁
(d) vitamin B₂
Answer:
Ans. (a)
Deficiency of B₁₂ causes pernicious anaemia which is due to deficiency of red blood cells (RBC) in haemoglobin.

Teacher's Note:
1. Pernicious anaemia (low RBC) is linked to vitamin B₁₂.

Question 13. Which among the following pairs of vitamins is stored in our body relatively for longer duration? (JEE Main 2021)
(a) Vitamin A and vitamin D
(b) Thiamine and Ascorbic acid
(c) Thiamine and vitamin A
(d) Ascorbic acid and vitamin D
Answer:
Ans. (a)
Vitamin A and vitamin D being insoluble in H₂O are stored in our body relatively for longer durations.

Teacher's Note:
1. Fat-soluble vitamins are stored in the liver and adipose tissue; water-soluble ones are excreted.

Question 14. Which of the following will react with CHCl₃ + alc. KOH? (JEE Main 2020)
(a) Adenine and proline
(b) Thymine and proline
(c) Adenine and lysine
(d) Adenine and thymine
Answer:
Ans. (c)
Adenine and Lysine contain free amino groups (–NH₂) and hence react with CHCl₃ + alc KOH to give carbylamine test.
Proline is a secondary amine (ring NH) and thymine has no free –NH₂ group, so they do not give this test.

Teacher's Note:
1. The carbylamine (isocyanide) test is given only by primary amino groups.

Question 15. The nitrogenous base present only in RNA is (Kerala PET 2018)
(a) guanine
(b) adenine
(c) cytosine
(d) uracil
(e) thymine
Answer:
Ans. (d)
Uracil is present only in RNA.

Teacher's Note:
1. Thymine is found only in DNA and uracil only in RNA; A, G and C are common to both.

Question 16. ADP and ATP differ in the number of (West Bengal JEE 2017)
(a) phosphate units
(b) ribose units
(c) adenine base
(d) nitrogen atom.
Answer:
Ans. (a)
ADP (Adenosine Diphosphate) contains two while ATP (Adenosine Triphosphate) contains three phosphate units.

Teacher's Note:
1. Energy is released when ATP loses its terminal phosphate to form ADP.

Question 17. Adenosine is an example of a (Karnataka CET 2015)
(a) purine base
(b) nucleoside
(c) nucleotide
(d) pyridoxine base
Answer:
Ans. (b)
Adenosine is a nucleoside since it contains the base adenine and the sugar ribose.

Teacher's Note:
1. Nucleoside = base + sugar; nucleotide = base + sugar + phosphate.

Question 18. Which of the following is the correct structure of Adenosine? (JEE Main (online) 2018)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (a)
Adenine is 6-aminopurine, and in nucleosides, 9-position of the purine is linked to C₁ of the sugar by β-linkage. Hence adenosine has the NH₂ group at C-6 and ribose at N-9, as in structure (a).

Teacher's Note:
1. In purine nucleosides the sugar is on N-9; in pyrimidine nucleosides it is on N-1.

Question 19. RNA and DNA are chiral molecules, their chirality is due to (AIPMT 2007)
(a) D-sugar component
(b) L-sugar component
(c) chiral bases
(d) chiral phosphate ester unit
Answer:
Ans. (a)
D-sugar component is optically active due to the presence of chiral carbons.

Teacher's Note:
1. The bases and phosphate are achiral; the chirality of nucleic acids comes from D-ribose or D-2-deoxyribose.

Question 20. The correct statement regarding RNA and DNA, respectively is (NEET Phase-I 2016)
(a) the sugar component in RNA is a arabinose and the sugar component in DNA is ribose
(b) the sugar component in RNA is 2′-deoxyribose and the sugar component in DNA is arabinose
(c) the sugar component in RNA is arabinose and the sugar component in DNA is 2′-deoxyribose
(d) the sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose.
Answer:
Ans. (d)
RNA contains β-D-ribose while DNA contains β-D-2-deoxyribose as the sugar component.

Teacher's Note:
1. Arabinose does not occur in either nucleic acid; it is a common distractor.

Question 21. In both DNA and RNA, heterocyclic base and phosphate ester linkages are at
(a) C₅′ and C₂′ respectively of the sugar molecule
(b) C₂′ and C₅′ respectively of the sugar molecule
(c) C₁′ and C₅′ respectively of the sugar molecule
(d) C₅′ and C₁′ respectively of the sugar molecule
Answer:
Ans. (c)
The base is attached to C₁′ of the sugar and the phosphate is esterified at C₅′ of the sugar in a nucleotide.

Teacher's Note:
1. Base at 1′, phosphate at 5′; in the chain, phosphodiester links run from 3′ of one sugar to 5′ of the next.

Question 22. Chargaff’s rule states that in an organism.
(a) Amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine (C)
(b) Amount of adenine (A) is equal to that of guanine (G) and the amount of thymine (T) is equal to that of cytosine (C)
(c) Amount of adenine (A) is equal to that of cytosine (C) and the amount of thymine (T) is equal to that of guanine (G)
(d) Amounts of all bases are equal.
Answer:
Ans. (a)
According to Chargaff’s rule, in DNA the amount of A = T and G = C because A pairs with T and G pairs with C.

Teacher's Note:
1. Chargaff’s rule follows from base pairing: a purine always pairs with a specific pyrimidine.

Question 23. In DNA, the complimentary bases are (AIPMT 2008)
(a) adenine and guanine; thymine and cytosine
(b) uracil and adenine; cytosine and guanine
(c) adenine and thymine; guanine and cytosine
(d) adenine and thymine; guanine and uracil
Answer:
Ans. (c)
In DNA, adenine pairs with thymine (two H-bonds) and guanine pairs with cytosine (three H-bonds).

Teacher's Note:
1. A=T has two H-bonds and G≡C has three; uracil appears only in RNA.

Question 24. The central dogma of molecular genetics states that the genetic information flows from (NEET Phase-II, 2016)
(a) Amino acids → Proteins → DNA
(b) DNA → Carbohydrates → Proteins
(c) DNA → RNA → Proteins
(d) DNA → RNA → Carbohydrates
Answer:
Ans. (c)
DNA first codes three types of RNAs which, in turn, synthesize proteins: DNA → RNA → Proteins.

Teacher's Note:
1. Transcription (DNA → RNA) is followed by translation (RNA → protein).

Question 25. Sugar moiety in DNA and RNA molecules, respectively, are (JEE Main 2022)
(a) β-D-2-deoxyribose, β-D-deoxyribose
(b) β-D-2-deoxyribose, β-D-ribose
(c) β-D-ribose, β-D-2-deoxyribose
(d) β-D-deoxyribose, β-D-2-deoxyribose
Answer:
Ans. (b)
DNA contains β-D-2-deoxyribose and RNA contains β-D-ribose.

Teacher's Note:
1. Mind the order asked (“DNA and RNA, respectively”) before picking the option.

Question 26. The Fischer presentation of D-glucose is given below.The correct structure(s) of β-L-glucopyranose is (are) (JEE Advanced 2018)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (d)
L-Glucose is the mirror image of D-glucose: its –OH groups at C-2, C-3, C-4 and C-5 are on the left, right, left and left in the Fischer projection, so in the Haworth formula the –OH groups at C-2 and C-4 are above and that at C-3 is below the plane.
Please note that in Haworth projection formula of L-glucose, the C₆–CH₂OH is below the plane. Similarly, in β-anomer, OH group is below the plane.
Thus, option (d) is correct.

Teacher's Note:
1. For an L-sugar the CH₂OH is down; β means the anomeric OH is on the same side as CH₂OH, so here it is also down.

One or More Correct Answers

Question 27. Globular proteins are present in
(a) blood
(b) keratin
(c) eggs
(d) muscles
Answer:
Ans. (a, c)
Albumins of blood (and haemoglobin) and egg albumin are globular proteins. Keratin (hair, nails) and myosin of muscles are fibrous proteins.

Teacher's Note:
1. Globular proteins are water soluble (albumin, insulin, haemoglobin); fibrous ones are insoluble (keratin, myosin, collagen).

Question 28. Purine and pyrimidine bases present in both DNA and RNA are
(a) uracil
(b) thymine
(c) cytosine
(d) adenine
Answer:
Ans. (c, d)
Adenine (purine) and cytosine (pyrimidine) are present in both DNA and RNA; thymine occurs only in DNA and uracil only in RNA.

Teacher's Note:
1. A, G and C are common to both nucleic acids; T and U are the only bases that differ.

Passage-Based Questions

Question 29. Comprehension 2: α-Amino acids are the building blocks of proteins. About 20 α-amino acids have been isolated by the hydrolysis of proteins. All these amino acids except glycine are chiral and have L-configuration. Ten amino acids (valine, leucine, isoleucine, phenylalanine, methionine, tryptophan, threonine, lysine, arginine and histidine) which the body cannot synthesize are called essential amino acids. The remaining ten are called non-essential amino acids. All α-amino acids exist as zwitterions each of which has a specific isoelectric point. Above isoelectric point, a α-amino acid exists as an anion. Two, three or many α-amino acids join together to form di-, tri- or polypeptides or proteins respectively. Each polypeptide or protein has a free amino group at one end called the N-terminal end and a free carboxyl group at the other end called the C-terminal end. Sanger’s reagent (i.e., 2,4-dinitrofluorobenzene) is used to determine the N-terminal end while C-terminal end is determined by hydrazinolysis.
The acid showing salt-like character in aqueous solution is
(a) acetic acid
(b) benzoic acid
(c) formic acid
(d) α-aminoacetic acid.
Answer:
Ans. (d)
α-Aminoacetic acid (glycine) has both a basic –NH₂ and an acidic –COOH group and exists in solution as the dipolar ion (zwitterion) H₃N⁺–CH₂–COO⁻, which is an internal salt.

Teacher's Note:
1. The zwitterion explains the high melting points and water solubility of amino acids.

Matrix Match Questions

Teacher's Note:
1. Only sugars with a free anomeric carbon (reducing sugars) form osazones; sucrose and starch do not.

Teacher's Note:
1. Purines (A, G) have two fused rings; pyrimidines (C, T, U) have one ring.

Integer and Numerical Answer Questions

Question 32. The total number of basic groups in the following form of lysine is
H₃N⁺–CH₂–CH₂–CH₂–CH₂–CH(NH₂)–COO⁻ (IIT Paper 2010)
Answer:
Ans. 2
The number of basic groups in lysine are COO⁻ and NH₂, i.e., 2.
H₃N⁺CH₂CH₂CH₂CH₂–CH(NH₂)–COO⁻ \( \xrightarrow{2\mathrm{H^+}} \) H₃N⁺CH₂CH₂CH₂CH₂–CH(N⁺H₃)–COOH

Teacher's Note:
1. A basic group is one that can accept a proton: –NH₂ and –COO⁻; the –N⁺H₃ group is already protonated.

Question 33. The number of oxygen atoms present in a nucleotide formed from a base that is present only in RNA is
Answer:
Ans. 9
The base present only in RNA is uracil; its nucleotide is uridine monophosphate (ribose + uracil + phosphate).
4 (Phosphate) + 3 (Ribose) + 2 (Uracil) = 9

Teacher's Note:
1. Count oxygens part by part: phosphate group 4, ribose part 3, uracil 2 (two C=O groups).

Question 34. When a solution of 0.094 g of an unknown amino acid was treated with HNO₂, there was obtained 10.8 cm³ of nitrogen gas at NTP. What is the minimum molecular mass of the acid?
Answer:
Ans. 390
H₂N–CHR–COO⁻ (1 mole) \( \xrightarrow{\mathrm{HNO_2}} \) HO–CHR–COOH + H₂O + N₂ (1 mole, 22400 cm³ at NTP)
Since 1/2 of N₂ evolved comes from amino acid and the other half comes from HNO₂, therefore, actual volume of N₂ coming from amino acid = 10.8/2 = 5.4 cm³
Now, 5.4 cm³ of N₂ at NTP are produced from amino acid = 0.094 g
∴ 22400 cm³ of N₂ at NTP will be produced from acid \( = \frac{0.094}{5.4}\times 22400 \) = 390
Thus, the molecular mass of the acid = 390 g mol⁻¹

Teacher's Note:
1. In the reaction with HNO₂, one N atom of N₂ comes from the amino group and one from HNO₂; “minimum” means assume one NH₂ group per molecule.

Assertion–Reason Questions

Question 35. Assertion: A solution of sucrose in water is dextrorotatory but on hydrolysis in presence of little hydrochloric acid, it becomes laevorotatory.
Reason: Sucrose on hydrolysis gives unequal amounts of glucose and fructose as a result of which change in sign of rotation is observed. (AIIMS 2015)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Sucrose on hydrolysis gives equal amounts of glucose and fructose. Since fructose is more laevorotatory than glucose is dextrorotatory, therefore, the resulting mixture becomes laevorotatory.

Teacher's Note:
1. Invert sugar: equal moles, but fructose (−92°) outweighs glucose (+52°), giving −20°.

Question 36. Assertion: Sucrose is a non-reducing sugar.
Reason: It has glycosidic linkage. (AIIMS 2004)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. Glucose and fructose in sucrose are connected through their reducing centres.

Teacher's Note:
1. Maltose and lactose also have glycosidic links yet are reducing; what matters is which carbons are linked.

Question 37. Assertion: Cellulose is not digested by human beings.
Reason: Cellulose is a polymer of β-D-glucose.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. Enzymes which hydrolyse β-glycosidic linkages are not present in human digestive system.

Teacher's Note:
1. Both statements are true, but the reason for indigestibility is the missing enzyme (cellulase).

Question 38. Assertion: Alpha (α) amino acids exist as internal salt in solution as they have amino and carboxylic acid groups in near vicinity.
Reason: H⁺ ion given by carboxyl group (–COOH) is captured by amino group (–NH₂) having lone pair of electrons. (AIIMS 2007)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
The proton lost by –COOH is taken up by the lone pair of the –NH₂ group, giving the dipolar ion H₃N⁺–CHR–COO⁻ (internal salt or zwitterion). Thus the reason explains the assertion.

Teacher's Note:
1. The zwitterion is an intramolecular acid–base reaction.

Question 39. Assertion: Carboxypeptidase is an exopeptidase.
Reason: It cleaves N-terminal bond. (AIIMS 2004)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Carboxypeptidase is an exopeptidase since it cleaves the peptide chain at the C-terminal end.

Teacher's Note:
1. The name tells the end: carboxy-peptidase works at the C-terminal (carboxyl) end; aminopeptidase at the N-terminal end.