Question 1

Table sugar is classified as a

Accepted Answer

disaccharide made up of D-glucose and D-fructose. Table sugar is sucrose, a disaccharide formed from one molecule of D-glucose and one of D-fructose joined by a glycosidic bond.

Question 2

Which pair of compounds will both produce a positive result with Tollens’ reagent?

Accepted Answer

Glucose and fructose. Both glucose and fructose give a positive Tollens test. Glucose has a free aldehyde; fructose (a ketose) tautomerises to an aldose under the basic conditions and is also oxidised. Sucrose is non-reducing (no free anomeric OH).

Question 3

The process of breaking down sucrose by hydrolysis is known as

Accepted Answer

inversion. Hydrolysis of sucrose gives an equimolar mixture of glucose and fructose; the optical rotation changes sign from dextro to laevo, so the process is called inversion (and the product is invert sugar).

Question 4

Identify the statement that is incorrect.

Accepted Answer

The human body is able to synthesize every protein it requires.. The body cannot synthesise all the amino acids it needs; the essential amino acids must come from the diet, so the body cannot make every protein it requires. The other statements are correct.

Question 5

How many amino acids present in proteins can be produced by the human body itself?

Accepted Answer

10. Of the 20 standard amino acids, 10 are essential (must come from diet) and the other 10 (non-essential) can be synthesized by the human body.

Question 6

Amylose is chemically a ________ made up of about 200–1000 α-D-(+)-glucose units joined through ______ glycosidic bonds.

Accepted Answer

long unbranched chain, C1–C4. Amylose is a type of starch that consists of a long unbranched chain of glucose units, specifically linked by C1–C4 glycosidic bonds, which allows for a linear structure essential for its function in energy storage.

Question 7

Identify the incorrectly paired item.

Accepted Answer

Insulin – steroid hormone. Insulin is a polypeptide (protein) hormone, not a steroid hormone, so 'Insulin - steroid hormone' is the incorrectly paired item. The other pairings (estrone, oxytocin, potassium metabisulphite) are correct.

Question 8

If a DNA strand reads ATGCTTGA, what sequence would appear on the complementary strand?

Accepted Answer

TACGAACT. Reading ATGCTTGA and pairing A-T, T-A, G-C, C-G gives the complementary strand TACGAACT.

Question 9

In nucleic acids, the repeating order of components is:

Accepted Answer

sugar - base - phosphate. The correct order of components in nucleic acids is sugar, base, and phosphate, as this sequence reflects the structure of nucleotides, which are the building blocks of nucleic acids like DNA and RNA.

Question 10

Which of the following statements about the peptide bond is not correct?

Accepted Answer

In proteins, the C–N bond is longer than a normal C–N single bond. Because of resonance/partial double-bond character, the peptide C-N bond is shorter than a normal C-N single bond and the -C(=O)-NH- linkage is planar. Therefore the statement that the C-N bond is longer than a normal C-N single bond is incorrect.

Question 11

An amino acid has a carboxyl group with pKₐ1 = 2.34 and an ammonium group with pKₐ2 = 9.60. At what pH will this amino acid have its isoelectric point?

Accepted Answer

5.97. For a neutral amino acid the isoelectric point is the mean of the two relevant pKa values: pI = (2.34 + 9.60)/2 = 5.97.

Question 12

In osazone formation, which structural unit must be present for the reaction to occur effectively?

Accepted Answer

–CHO. Osazone formation with phenylhydrazine requires a carbonyl adjacent to a CHOH, i.e. an alpha-hydroxy aldehyde/ketone. Of the choices the -CHO group is the structural unit that allows osazone formation.

Question 13

Which reagent is used to identify the N-terminal amino acid residue of a peptide?

Accepted Answer

2,4-Dinitrofluorobenzene. The N-terminal residue is identified with Sanger's reagent, 2,4-dinitrofluorobenzene (DNFB), which reacts with the free alpha-amino group to give a yellow DNP-amino acid that survives hydrolysis and can be identified.

Question 14

Which pair of functional groups is commonly found in a typical carbohydrate molecule?

Accepted Answer

>C=O and –OH. The correct option includes a carbonyl group (>C=O) and a hydroxyl group (–OH), which are essential components of carbohydrates, specifically in the structure of sugars where these functional groups contribute to their reactivity and properties.

Question 15

Enzymes act as biocatalysts and are known to work in which of the following conditions?

Accepted Answer

An aqueous environment at about 30–35°C and pH 7. Enzymes are biocatalysts that function optimally in an aqueous medium at moderate temperature (about 30-35 C) and near-neutral pH (~7). Extreme acidity/alkalinity or organic solvents denature them, so the correct condition is the aqueous, ~30-35 C, pH 7 environment.

Question 16

Deficiency of essential amino acids in the diet results in which disease?

Accepted Answer

Kwashiorkor. Deficiency of essential amino acids (dietary protein) causes Kwashiorkor. Night blindness is vitamin A, pernicious anaemia is vitamin B12, and sickle cell anaemia is a genetic haemoglobin disorder.

Question 17

Which part of DNA serves as the functional instruction set for making a protein?

Accepted Answer

gene. A gene is a specific sequence of DNA that contains the instructions for synthesizing proteins, making it the fundamental unit of heredity and the functional blueprint for protein production.

Question 18

Which of the following statements about enzymes is correct?

Accepted Answer

Enzymes are specific biological catalysts that have distinct, well-defined active sites.. Enzymes function as biological catalysts by binding to specific substrates at their active sites, which are uniquely shaped to facilitate chemical reactions, ensuring high specificity in their action.

Question 19

In nucleic acids, the nitrogenous base and the phosphate group are attached to which positions of the sugar in DNA and RNA, respectively?

Accepted Answer

At the 1′ carbon and the 5′ carbon of the sugar, respectively. In a nucleotide the nitrogenous base is attached by an N-glycosidic bond to the 1' carbon of the (deoxy)ribose sugar, while the phosphate group esterifies the 5' carbon. Hence base -> 1', phosphate -> 5'.

Question 20

How many molecules of phenylhydrazine are required for one molecule of glucose to form osazone?

Accepted Answer

three. Glucose reacts with 3 molecules of phenylhydrazine to form the osazone: one forms the hydrazone at C1, a second oxidizes C2 (acting as oxidant, reduced to aniline + NH3), and a third condenses with the resulting C2 keto group. Net: 3 phenylhydrazine per glucose.

Question 21

Out of the following, which type of interaction is responsible for the stabilisation of α-helix structure of proteins?

Accepted Answer

Hydrogen bonding. Hydrogen bonding is crucial for stabilizing the α-helix structure in proteins, as these bonds form between the carbonyl oxygen of one amino acid and the amide hydrogen of another, creating a helical shape that is essential for protein structure.

Question 22

What is the term used for the change observed in the optical rotation of a freshly prepared glucose solution?

Accepted Answer

Mutarotation. A freshly prepared glucose solution slowly changes its optical rotation until equilibrium between the alpha and beta anomers is reached. This phenomenon is called mutarotation.

Question 23

RNA is different from DNA because RNA contains

Accepted Answer

ribose sugar and uracil. RNA is distinct from DNA primarily because it contains ribose sugar instead of deoxyribose and uses uracil in place of thymine, which is found in DNA.

Question 24

Complete hydrolysis of cellulose gives

Accepted Answer

D-glucose. Cellulose is a polysaccharide made up of long chains of D-glucose units linked by β-1,4-glycosidic bonds. When cellulose undergoes complete hydrolysis, these bonds are broken, resulting in the release of individual D-glucose molecules.

Question 25

The reason for double helical structure of DNA is operation of

Accepted Answer

hydrogen bonding. The double helical structure of DNA is primarily stabilized by hydrogen bonding between complementary nitrogenous bases, which allows the two strands to coil around each other while maintaining the necessary distance and orientation.

Question 26

Which nitrogenous base is found in RNA but is absent from DNA?

Accepted Answer

Uracil. Uracil is a nitrogenous base that replaces thymine in RNA, allowing RNA to have a distinct structure and function compared to DNA.

Question 27

Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories?

Accepted Answer

A hormone. Insulin is classified as a hormone because it is produced by the pancreas and plays a crucial role in regulating blood sugar levels in the body.

Question 28

In both DNA and RNA, heterocyclic base and phosphate ester linkages are at –

Accepted Answer

C1' and C5' respectively of the sugar molecule. The correct option is right because in both DNA and RNA, the heterocyclic base is attached to the C1' carbon of the sugar, while the phosphate group is linked to the C5' carbon, establishing the backbone structure of nucleic acids.

Question 29

The term anomers of glucose refers to

Accepted Answer

isomers of glucose that differ in configuration at carbon one (C-1). Anomers are a specific type of stereoisomer that differ in configuration at the anomeric carbon, which is carbon one (C-1) in glucose. This distinction is crucial for understanding the different forms of glucose, such as alpha and beta anomers.

Question 30

Which of the following compounds can be detected by Molisch's test ?

Accepted Answer

Sugars. Molisch's test is specifically designed to detect the presence of carbohydrates, which include sugars. The test relies on the reaction of sugars with alpha-naphthol in the presence of sulfuric acid, resulting in a purple ring, indicating a positive result.

Question 31

Which one of the following statements is correct?

Accepted Answer

All amino acids except glycine are optically active. All standard amino acids have a chiral alpha-carbon except glycine (its alpha-carbon bears two H atoms), so all amino acids except glycine are optically active.

Question 32

Which one of the following bases is not present in DNA?

Accepted Answer

Quinoline. DNA contains adenine, guanine, cytosine and thymine. Quinoline is not a nucleobase at all, so it is the base not present in DNA.

Question 33

Which of the vitamins given below is water soluble ?

Accepted Answer

Vitamin C. Vitamin C (ascorbic acid) is water soluble, whereas vitamins D, E and K are fat soluble. Hence the water-soluble vitamin here is Vitamin C.

Question 34

Thiol group is present in:

Accepted Answer

Cysteine. The thiol (-SH) group is found in cysteine. Methionine contains a thioether (-S-CH3), and cystine has a disulfide (-S-S-) linkage. Answer: option 0.

Question 35

Glucose on prolonged heating with HI gives:

Accepted Answer

n-Hexane. On prolonged heating with HI, all the -OH and the carbonyl groups of glucose are reduced, and the straight C6 chain survives, giving n-hexane. This established the unbranched six-carbon skeleton of glucose.

Question 36

The increasing order of pKa of the following amino acids in aqueous solution is:
Gly Asp Lys Arg

Accepted Answer

Asp < Gly < Lys < Arg. Ordering by characteristic pKa: Asp side chain ~3.7 < Gly (alpha-NH3+ ~9.6) < Lys epsilon-NH3+ ~10.5 < Arg guanidinium ~12.5. Since the guanidinium group of arginine has the highest pKa, the increasing order is Asp < Gly < Lys < Arg.

Question 37

Match the following:
(i) Riboflavin
(ii) Thiamine
(iii) Pyridoxine
(iv) Ascorbic acid
(a) Beriberi
(b) Scurvy
(c) Cheilosis
(d) Convulsions

Accepted Answer

(i) – (c), (ii) – (a), (iii) – (d), (iv) – (b). Deficiency diseases: riboflavin (B2)->cheilosis, thiamine (B1)->beriberi, pyridoxine (B6)->convulsions, ascorbic acid (C)->scurvy. This gives (i)-c, (ii)-a, (iii)-d, (iv)-b.

Question 38

Which one of the following bases is not present in DNA ?

Accepted Answer

Quinoline. Quinoline is not a nucleotide base found in DNA; instead, DNA contains adenine, cytosine, and thymine as its primary nitrogenous bases.

Question 39

Among the following the incorrect statement is -

Accepted Answer

Sucrose and amylase have 1,2-glycosidic linkage. The statement is incorrect because sucrose is a disaccharide formed from glucose and fructose linked by a 1,2-glycosidic bond, while amylase is an enzyme that breaks down starch and does not contain any glycosidic linkages.

Question 40

Glucose on prolonged heating with HI gives:
(1) n-Hexane
(2) 1-Hexene
(3) Hexanoic acid
(4) 6-iodohexanal

Accepted Answer

n-Hexane. Prolonged heating of glucose with hydrogen iodide (HI) leads to the reduction of the sugar, resulting in the formation of n-hexane through the complete deoxygenation and saturation of the carbon chain.

Question 41

Which of the following test cannot be used for identifying amino acids ?

Accepted Answer

Barfoed Test. Ninhydrin, xanthoproteic and biuret tests are all used for amino acids/proteins. The Barfoed test distinguishes monosaccharides from reducing disaccharides (a carbohydrate test) and cannot be used to identify amino acids.

Question 42

The increasing order of pKa of the following amino acids in aqueous solution is: Gly Asp Lys Arg

Accepted Answer

Asp < Gly < Lys < Arg. The pKa values of amino acids reflect their side chain acidity; as the side chain becomes more acidic, the pKa decreases. Aspartic acid has the lowest pKa due to its carboxyl side chain, followed by glycine, lysine, and arginine, which have progressively higher pKa values due to their basic side chains.

Question 43

Fructose and glucose can be distinguished by -

Accepted Answer

(2) Seliwanoff's test. Seliwanoff's test gives a fast cherry-red color with ketoses like fructose but is slow/negative with aldoses like glucose. Fehling's, Benedict's, and Barfoed's react with both, so they cannot distinguish them.

Question 44

The peptide that gives positive ceric ammonium nitrate and carbylamine tests is:

Accepted Answer

Ser - Lys. Ceric ammonium nitrate test is positive for -OH (serine side chain); carbylamine test is positive for a primary amine (lysine side chain). The dipeptide Ser-Lys gives both positive tests.

Question 45

Among the following compounds most basic amino acid is
(1) Lysine
(2) Asparagine
(3) Histidine
(4) Serine

Accepted Answer

(1) Lysine. Lysine carries an additional amino (-NH2) group on its side chain, giving it the highest basicity among the listed amino acids. Hence lysine is the most basic.

Question 46

The correct statement(s) among I to III with respect to potassium ions that are abundant within the cell fluid is/are
I. They activate many enzymes
II. They participate in the oxidation of glucose to produce ATP
III. Along with sodium ions they are responsible for the transmission of nerve

Accepted Answer

I and III only. Potassium ions play a crucial role in activating various enzymes, which is essential for numerous cellular processes, and they are also involved in the transmission of nerve signals alongside sodium ions. However, while potassium is important for cellular functions, it does not directly participate in the oxidation of glucose to produce ATP, which primarily involves other molecules.

Question 47

Consider the following reactions:
(i) Glucose + ROH dry HCl Acetal x eq. of (CH3CO)2O acetyl derivative
(ii) Glucose Ni/H2 A y eq. of (CH3CO)2O acetyl derivative
(iii) Glucose z eq. of (CH3CO)2O acetyl derivative
'x', 'y' and 'z' in these reactions are respectively.

Accepted Answer

4, 6 & 5. (i) Forming the methyl acetal (glycoside) consumes the anomeric OH, leaving 4 OH to acetylate, so x=4. (ii) Ni/H2 reduces glucose to sorbitol, an open-chain hexitol with 6 OH, so y=6. (iii) Glucose itself has 5 OH, so z=5. Thus x,y,z = 4, 6 & 5.

Question 48

The correct observation in the following reactions is:
Sucrose —(Glycosidic bond cleavage / Hydrolysis)→ A + B —(Seliwanoff's reagent)→ ?

Accepted Answer

Formation of red colour. The correct option is right because when sucrose is hydrolyzed, it breaks down into glucose and fructose. Fructose reacts with Seliwanoff's reagent to produce a red color, indicating the presence of a ketose sugar.

Question 49

Which of the following is not an essential amino acid:

Accepted Answer

Tyrosine. Tyrosine is classified as a non-essential amino acid because the body can synthesize it from phenylalanine, whereas valine, leucine, and lysine are essential amino acids that must be obtained through diet.

Question 50

Which one of the following statements not true ?

Accepted Answer

Lactose contains α-glycosidic linkage between C1 of galactose and C4 of glucose.. In lactose the linkage between C1 of galactose and C4 of glucose is a beta-1,4-glycosidic bond, not alpha. The other statements (8 hydroxyl groups, hydrolysis to glucose + galactose, reducing sugar giving Fehling's test) are all true. So the untrue statement is the one claiming an alpha-linkage.

JEE Main Chemistry: Biomolecules questions with solutions

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