Question 1

The incorrect statement from the following with respect to carbohydrates is:

Accepted Answer

Statement (A) is correct because all monosaccharides contain a free aldehyde or ketonic group (or exist in equilibrium with it) and act as reducing sugars. Statement (B) is incorrect because the hydrolysis of oligosaccharides can yield either the same or different monosaccharide units. For example, sucrose is an oligosaccharide (disaccharide) that yields one molecule of glucose and one molecule of fructose upon hydrolysis, whereas maltose yields two molecules of glucose. Statement (C) is correct

Question 2

Which of the following amino acid will give violet coloured complex with neutral ferric chloride solution?

Accepted Answer

Neutral ferric chloride ( FeCl _3) solution gives a characteristic violet coloured complex with compounds containing a phenolic hydroxyl group. Among the given amino acids, Tyrosine contains a phenolic - OH group in its side chain. Threonine and Serine contain aliphatic - OH groups, and Cysteine contains a thiol (- SH ) group, none of which give a positive neutral FeCl _3 test. Therefore, Tyrosine will give a violet coloured complex with neutral ferric chloride solution. Answer: Tyrosine

Question 3

Consider the following reactions. Total number of electrons in the bonds and lone pair of electrons in the product (X) is :

Accepted Answer

Step 1: The starting material is an aldohexose (like D-glucose). On prolonged heating with HI, it undergoes reduction to form n-hexane. This indicates that all six carbon atoms are linked in a straight chain. CHO -( CHOH )_4- CH _2 OH HI, CH _3- CH _2- CH _2- CH _2- CH _2- CH _3 (n-hexane) Step 2: n-Hexane, when passed over heated V_2O_5 at 773 K and 10-20 atm pressure, undergoes aromatization (dehydrogenation and cyclization) to form benzene. n-hexane V_2O_5, 773 K, 10-20 atm C _6 H _6 (Benzene

Question 4

Match the LIST-I with LIST-II List-I Deficiency Disease List-II Vitamin A. Scurvy I. Pyridoxine B. Convulsions II. Vitamin A C. Cheilosis III. Ascorbic Acid D. Xerophthalmia IV. Riboflavin Choose the

Accepted Answer

Scurvy is caused by the deficiency of Vitamin C, which is chemically known as Ascorbic acid. Thus, A matches with III. Convulsions are caused by the deficiency of Vitamin B6, which is chemically known as Pyridoxine. Thus, B matches with I. Cheilosis (fissuring at corners of mouth and lips) is caused by the deficiency of Vitamin B2, which is chemically known as Riboflavin. Thus, C matches with IV. Xerophthalmia (hardening of cornea of eye) is caused by the deficiency of Vitamin A. Thus, D matches

Question 5

Match the LIST-I with LIST-II List-I Amino acid List-II Positive reaction/Test for functional group present in side chain of amino acid A. Glutamine I. Hinsberg's test B. Lysine II. Neutral FeCl _3 te

Accepted Answer

Glutamine contains an amide group in its side chain, which gives a positive Hoffmann bromamide degradation test. Thus, A matches IV. Lysine contains a primary amine group in its side chain, which gives a positive Hinsberg's test. Thus, B matches I. Tyrosine contains a phenolic hydroxyl group in its side chain, which gives a positive neutral FeCl _3 test. Thus, C matches II. Serine contains an alcoholic hydroxyl group in its side chain, which gives a positive ceric ammonium nitrate test. Thus, D

Question 6

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with t_ 1/2 = 3 hour. The percentage of sucrose remaining after 6 hours is _______. (Nearest integer) (Given: 2 =

Accepted Answer

For a first order reaction, the amount of reactant remaining after n half-lives is given by N_t = N_0 2^n . Given t_ 1/2 = 3 hours and total time t = 6 hours. Number of half-lives, n = t t_ 1/2 = 6 3 = 2. The amount of sucrose remaining is N_t = N_0 2^2 = N_0 4 . Percentage of sucrose remaining = N_t N_0 100 = 1 4 100 = 25. Answer: 25

Question 7

Identify the incorrect statement about tertiary structure of proteins.

Accepted Answer

The tertiary structure of proteins represents the overall folding of the polypeptide chains, which is the further folding of the secondary structure. This folding gives rise to two major molecular shapes: fibrous and globular. The main forces that stabilize the secondary and tertiary structures of proteins are hydrogen bonds, disulphide linkages, van der Waals forces, and electrostatic forces of attraction. When a protein in its native form is subjected to physical changes like temperature or ch

Question 8

Given below are two statements : Statement I: The two cyclic forms of D-(+)-glucose (α and β anomers differing at C1 hydroxyl orientation) are two anomers of D-(+)-glucose. Statement II: The open chai

Accepted Answer

Statement I is true because the and forms of D-(+)-glucose are cyclic hemiacetals that differ only in the configuration of the hydroxyl group at C1, which is the anomeric carbon. Such isomers are called anomers. Statement II is true because the open chain structures of D-glucose and D-fructose have identical configurations at C3, C4, and C5. D-glucose is an aldohexose with chiral centers at C2, C3, C4, and C5, while D-fructose is a ketohexose with chiral centers at C3, C4, and C5. The stereochem

Question 9

Identify the correct statements. A. Glucose exists in two anomeric forms. B. Anomers of glucose differ in configuration at C-1 in cyclic hemiacetal structure. C. Melting point of -anomer of glucose is

Accepted Answer

Statement A is correct because glucose exists in two anomeric forms, -D-glucose and -D-glucose. Statement B is correct because the two anomers differ only in the configuration of the hydroxyl group at C-1, which is the anomeric carbon in the cyclic hemiacetal structure. Statement C is incorrect because the melting point of the -anomer is 419 K, which is less than the melting point of the -anomer (423 K). Statement D is incorrect because the specific rotation of the -anomer is +112^ and that of t

Question 10

A D-aldotetrose on oxidation with concentrated HNO_3 resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is:

Accepted Answer

For a carbohydrate to be classified as a D-sugar, the hydroxyl (-OH) group on the lowest chiral carbon (C3 in the case of an aldotetrose) must be on the right side in its Fischer projection. This eliminates options (A) and (B), which are L-sugars. Oxidation of an aldotetrose with concentrated HNO_3 converts both the aldehyde (-CHO) group at C1 and the primary alcohol (-CH_2OH) group at C4 into carboxylic acid (-COOH) groups, yielding a dicarboxylic acid (tartaric acid). The problem states that t

Question 11

Given below are two statements: Statement I: The structure of Maltose is given below: Maltose is a non-reducing sugar. Statement II: The structure of Lactose is given below: Lactose is a reducing suga

Accepted Answer

Maltose is a disaccharide composed of two -D-glucose units linked by an (1 4) glycosidic bond. The anomeric carbon (C1) of the second glucose unit has a free hemiacetal group, which can open to form an aldehyde. Therefore, maltose can reduce Tollens' and Fehling's reagents, making it a reducing sugar. Statement I claims that maltose is a non-reducing sugar, so it is false. Lactose is a disaccharide composed of -D-galactose and D-glucose linked by a (1 4) glycosidic bond. Similar to maltose, the

Question 12

Match the LIST-I with LIST-II List-I Name of amino acid List-II One letter symbol/type A. Arginine I. D/Non-essential B. Aspartic acid II. R/Essential C. Lysine III. E/Non-essential D. Glutamic acid I

Accepted Answer

Arginine is an essential amino acid and its one-letter symbol is R. Thus, A matches with II. Aspartic acid is a non-essential amino acid and its one-letter symbol is D. Thus, B matches with I. Lysine is an essential amino acid and its one-letter symbol is K. Thus, C matches with IV. Glutamic acid is a non-essential amino acid and its one-letter symbol is E. Thus, D matches with III. The correct matching is A-II, B-I, C-IV, D-III. Answer: A-II, B-I, C-IV, D-III

Question 13

Identify the correct pair having amino acid (A) and the hormone (B) that is iodinated derivative of the amino acid (A). (T and Y represent one letter code for amino acids) Amino acid (A) Hormone (B)

Accepted Answer

Thyroxine is a thyroid hormone which is an iodinated derivative of the amino acid tyrosine. The standard one-letter code for the amino acid tyrosine is Y. Insulin is a peptide hormone and not an iodinated derivative of a single amino acid. Therefore, the correct pair is Y and Thyroxine. Answer: Y Thyroxine

Question 14

Match List - I with List - II. List - I Vitamin List - II Name A. Vitamin B_1 I. Pyridoxine B. Vitamin B_2 II. Ascorbic acid C. Vitamin B_6 III. Thiamine D. Vitamin C IV. Riboflavin Choose the correct

Accepted Answer

Vitamin B_1 is chemically known as Thiamine. Vitamin B_2 is chemically known as Riboflavin. Vitamin B_6 is chemically known as Pyridoxine. Vitamin C is chemically known as Ascorbic acid. Therefore, the correct matching is A-III, B-IV, C-I, D-II. Answer: A-III, B-IV, C-I, D-II

Question 15

Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is :

Accepted Answer

Structure (3) given is of sucrose, which is a non-reducing sugar. For a sugar to be non-reducing, it must have an acetal linkage (not a hemiacetal linkage). In sucrose, both the anomeric carbons of glucose and fructose are involved in the glycosidic bond, so there is no free anomeric carbon available to open and act as a reducing agent. The other three structures have at least one free hemiacetal group, making them reducing sugars. Hence, the non-reducing sugar is structure (3).

Question 16

In the given pentapeptide, find out an essential amino acid ( Y ) and the sequence present in the pentapeptide: Choose the correct answer from the options given below:

Accepted Answer

Among the amino acids present (Serine, Aspartic acid, Glycine, Alanine, and Threonine), only Threonine is classified as an essential amino acid; the others are non-essential or conditionally essential. Therefore, Y = Threonine. Examining the peptide sequence shown in the structure and cross-referencing with the provided options, the sequence is Thr-Ser-Asp-Gly-Ala from N-terminus to C-terminus. The answer is option (4): Y is Threonine with sequence Thr-Ser-Asp-Gly-Ala.

Question 17

The number of possible tripeptides formed involving alanine (ala), glycine (gly) and valine (val), where no amino acid has been used more than once is:

Accepted Answer

For tripeptides with 3 different amino acids (ala, gly, val) where no amino acid is used more than once, each tripeptide must contain all three amino acids exactly once. The number of distinct tripeptides is the number of permutations of 3 amino acids = 3! = 6. These are: ala-gly-val, ala-val-gly, gly-ala-val, gly-val-ala, val-ala-gly, val-gly-ala.

Question 18

A student is given one compound among the following compounds that gives positive test with Tollen's reagent. The compound is :

Accepted Answer

Tollen's reagent is a mild oxidizing agent that gives a positive test (silver mirror) with aldehydes and -hydroxy ketones. In cyclic structures, hemiacetals are in equilibrium with their open-chain hydroxy-aldehyde forms in aqueous solution, whereas acetals and ethers are stable and do not open up. Compound A is a cyclic acetal (specifically a methyl glycoside type structure). It is stable and does not show Tollen's test. Compound B is a cyclic ether with a methoxy group at a non-anomeric positi

Question 19

Both human DNA and RNA are chiral molecules. The chirality in DNA and RNA arises due to the presence of

Accepted Answer

DNA and RNA are chiral molecules due to their structural components. Both contain pentose sugars: DNA contains deoxyribose and RNA contains ribose. These are D-sugars with multiple chiral centers (at C_2, C_3, and C_4 in the furanose ring). The D-designation comes from the stereochemical configuration at the chiral center furthest from the carbonyl carbon. The chiral centers in these sugar moieties are the primary source of chirality in nucleic acids, making them optically active molecules. Whil

Question 20

From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :

Accepted Answer

Tollen's reagent reacts with free aldehydes. Structure A is a hemiacetal (reducing sugar) that exists in equilibrium with its open-chain aldehyde form, allowing it to react with Tollen's reagent. Structure B is a glycoside (methyl glycoside or similar), where the anomeric carbon is locked in an acetal/ketal form with an -OCH₃ group, preventing ring opening and aldehyde formation, so it cannot react with Tollen's. Structure C has a cyclic ether oxygen and lacks a free aldehyde group that can be g

Biomolecules — Previous Year Questions & Solutions