CBSE Class 12 Chemistry Biomolecules NEET JEE Questions Set 02

Single Correct Answer

Question 1. The correct observation in the following reaction is
Sucrose \( \xrightarrow[\text{cleavage (Hydrolysis)}]{\text{Glycosidic bond}} \) A + B \( \xrightarrow{\text{Seliwanoff reagent}} \) ? (JEE Main 2020)
(a) formation of a blue colour
(b) gives no colour
(c) formation of red colour
(d) formation of violet colour
Answer:
Ans. (c)
Sucrose \( \xrightarrow[\text{cleavage (Hydrolysis)}]{\text{Glycosidic bond}} \) Glucose (A) + Fructose (B) \( \xrightarrow{\text{Seliwanoff’s test}} \) Red colour
Sucrose on hydrolysis gives an equimolar mixture of glucose (A) and fructose (B). Seliwanoff’s test is used for fructose which gives red colour when a solution of resorcinol in dil. HCl (Seliwanoff’s reagent) is added and heated in a water bath.

Teacher's Note:
1. Seliwanoff’s test (resorcinol + HCl) gives a quick red colour with ketoses such as fructose.

Question 2. The glycosidic linkage present in sucrose is between (Karnataka CET 2017)
(a) C–1 of α-glucose and C–2 of β-fructose
(b) C–1 of β-galactose and C–4 of α-glucose
(c) C–1 of α-glucose and C–4 of α-glucose
(d) C–1 of α-glucose and C–4 of β-fructose.
Answer:
Ans. (a)
C–1 of α-glucose and C–2 of β-fructose are joined through an oxygen atom in sucrose.

Teacher's Note:
1. Remember the pairs: sucrose C1(α)–C2(β), maltose C1(α)–C4, lactose C1(β) of galactose–C4 of glucose.

Question 3. Which of the following sets of monosaccharides forms sucrose? (AIPMT Prelim 2012)
(a) α-D-galactopyranose and α-D-glucopyranose
(b) α-D-glucopyranose and β-D-fructofuranose
(c) β-D-glucopyranose and α-D-fructofuranose
(d) α-D-glucopyranose and β-D-fructopyranose
Answer:
Ans. (b)
In sucrose, C₁–α of D-glucopyranose is connected to C₂–β of fructofuranose.

Teacher's Note:
1. In sucrose, glucose is a six-membered pyranose but fructose is a five-membered furanose ring.

Question 4. Maltose on treatment with dilute HCl gives (JEE Main 2019)
(a) D-glucose and D-fructose
(b) D-fructose
(c) D-glucose
(d) D-galactose
Answer:
Ans. (c)
Maltose is made of two α-D-glucose units, so on hydrolysis it gives D-glucose (two molecules).

Teacher's Note:
1. Maltose = glucose + glucose; sucrose = glucose + fructose; lactose = glucose + galactose.

Question 5. Which of the following statements is not true? (JEE Main 2020)
(a) Lactose contains α-glycosidic linkage between C₁ of galactose and C₄ of glucose
(b) Lactose is a reducing sugar
(c) Lactose (C₁₂H₂₂O₁₁) is a disaccharide and contains 8 hydroxyl groups
(d) On acid hydrolysis, lactose gives one molecule of D(+)-glucose and one molecule of D(+)-galactose
Answer:
Ans. (a)
Lactose contains β-glycosidic linkage between C₁ of galactose and C₄ of glucose and not α-glycosidic linkage.

Teacher's Note:
1. Lactose has a β(1→4) link; its glucose unit keeps a free anomeric OH, so it is reducing.

Question 6. Identify the incorrect statement from the following: (JEE Main 2021)
(a) β-Glycosidic linkage makes cellulose polymer
(b) Glycogen is called an animal starch
(c) Starch is a polymer of α-D-glucose
(d) Amylose is a branched chain polymer of glucose
Answer:
Ans. (d)
Option (d) is wrong since amylose is a linear chain polymer of glucose.

Teacher's Note:
1. Amylose is unbranched (only C1–C4 links); amylopectin and glycogen are branched through C1–C6 links.

Question 7. Starch is a mixture of two components, a water soluble component amylose (15–20%) and a water insoluble component amylopectin (80–85%). The aqueous solution of amylose gives a blue colour with iodine solution due to the formation of (AIIMS 2013)
(a) amylose iodide
(b) amylose iodate
(c) inclusion complex
(d) amylose tetraiodide complex
Answer:
Ans. (c)
Iodine molecules fit inside the helical coils of amylose, forming an inclusion complex which is blue.

Teacher's Note:
1. The blue colour is physical trapping of I₂ in the amylose helix, not a covalent iodide.

Question 8. The difference between amylose and amylopectin is (NEET 2018)
(a) amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage
(b) amylose have 1 → 4 α-linkage and 1 → 6 β-linkage
(c) amylopectin have 1 → 4 α-linkage and 1 → 6 β-linkage
(d) amylose is made up of glucose and galactose
Answer:
Ans. (a)
Amylopectin have 1–4 α-linkage and 1 → 6 α-linkage, while amylose has only 1 → 4 α-linkages.

Teacher's Note:
1. All links in starch are α; the β-links belong to cellulose.

Question 9. Fructose and glucose can be distinguished by (JEE Main 2019)
(a) Seliwanoff’s test
(b) Benedict’s test
(c) Barfoed’s test
(d) Fehling’s test
Answer:
Ans. (a)
Fructose and glucose is distinguished by Seliwanoff’s test. Fructose (a ketose) gives a red colour quickly; glucose does not.

Teacher's Note:
1. Benedict’s, Fehling’s and Barfoed’s tests are given by both glucose and fructose, so they cannot tell them apart.

Question 10. Which of the following is a basic amino acid? (NEET 2020)
(a) Alanine
(b) Tyrosine
(c) Lysine
(d) Serine
Answer:
Ans. (c)
Since Lysine contains two NH₂ groups and one –COOH group, therefore, it is a basic amino acid.
H₃N⁺–CH(COO⁻)–CH₂(CH₂)₃NH₂ (Lysine)

Teacher's Note:
1. Basic amino acids (lysine, arginine, histidine) have an extra nitrogen base in the side chain.

Question 11. Which of the following is not an essential amino acid? (JEE Main 2020)
(a) Tyrosine
(b) Leucine
(c) Valine
(d) Lysine
Answer:
Ans. (a)
Tyrosine is not an essential amino acid while all others are essential amino acids.

Teacher's Note:
1. Tyrosine can be made in the body from phenylalanine, so it is non-essential.

Question 12. Thiol group is present in (JEE Main 2016)
(a) cytosine
(b) cystine
(c) cysteine
(d) methionine
Answer:
Ans. (c)
Cysteine HSCH₂CH(NH₂)COOH contains the thiol (SH) group.

Teacher's Note:
1. Cysteine has –SH; cystine is two cysteines joined by an –S–S– bridge; methionine has –S–CH₃.

Question 13. Lysine is least soluble in water in the pH range
(a) 3 to 4
(b) 5 to 6
(c) 6 to 7
(d) 8 to 9
Answer:
Ans. (d)
Lysine is a basic amino acid, therefore, its isoelectric point lies in the basic range 8-9. At the isoelectric point, an amino acid has the lowest solubility.

Teacher's Note:
1. At the isoelectric point the net charge is zero, so the molecules attract each other most and solubility is lowest.

Question 14. Within the list shown below, the correct pair of structures of alanine in pH ranges 2-4 and 9-11 is
I. H₃N⁺–CH(CH₃)CO₂H
II. H₂N–CH(CH₃)CO₂⁻
III. H₃N⁺–CH(CH₃)CO₂⁻
IV. H₂N–CH(CH₃)CO₂H (West Bengal JEE 2015)
(a) I, II
(b) I, III
(c) II, III
(d) III, IV
Answer:
Ans. (a)
In the acidic medium, alanine exists as a cation (I) and in the basic medium, it exists as anion (II).
H₃N⁺–CH(CH₃)CO₂⁻ \( \xrightarrow{\mathrm{H^+}} \) H₃N⁺–CH(CH₃)CO₂H (I)
H₃N⁺–CH(CH₃)CO₂⁻ \( \xrightarrow{\mathrm{OH^-}} \) H₂N–CH(CH₃)CO₂⁻ (II)
Thus, option (a) is correct.

Teacher's Note:
1. Acid protonates the COO⁻ (cation); base removes H⁺ from NH₃⁺ (anion). Structure IV never exists in water.

Question 15. Which of the following will not exist in zwitter-ionic form at pH = 7? (JEE Main (online) 2018)
(a) o-H₂N–C₆H₄–SO₃H (2-aminobenzenesulphonic acid)
(b) CH₃CO–NH–CH(CH₃)–CO₂H (N-acetylalanine)
(c) 2-Aminocyclohexane-1-sulphonic acid
(d) 3-Aminocyclohexane-1-carboxylic acid
Answer:
Ans. (b)
In structure of option (b), nitrogen atom is not basic because it is part of an amide. Thus, it cannot form a zwitterion.

Teacher's Note:
1. A zwitterion needs a basic N and an acidic group; an amide N has its lone pair delocalised on C=O and is not basic.

Question 16. Which of the following compounds can form a zwitterion? (NEET 2018)
(a) Aniline
(b) Acetanilide
(c) Benzoic acid
(d) Glycine
Answer:
Ans. (d)
Glycine has both a basic –NH₂ group and an acidic –COOH group, so it exists as the zwitterion H₃N⁺–CH₂–COO⁻.

Teacher's Note:
1. Only molecules with both an acidic and a basic group in the same molecule can form a zwitterion.

Question 17. In a protein molecule various amino acids are linked together by (NEET Phase-I 2016)
(a) peptide bond
(b) dative bond
(c) α-glycosidic bond
(d) β-glycosidic bond
Answer:
Ans. (a)
Amino acids are joined by peptide bonds (–CO–NH–) formed between the –COOH of one amino acid and the –NH₂ of the next.

Teacher's Note:
1. Peptide bonds join amino acids in proteins; glycosidic bonds join sugar units in carbohydrates.

Question 18. Which of the following structures represents the peptide chain?
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (c)
The peptide chain contains the sequence –NH–CH–CO–NH–CH–CO–, i.e., each α-carbon lies between an N–H and a C=O, as in structure (c). Thus, option (c) is correct.

Teacher's Note:
1. In a peptide (from α-amino acids) the repeating unit is always N–C_α–C(=O): exactly one carbon between N and C=O.

Question 19. The number of tripeptides formed by three different amino acids are
(a) Three
(b) Four
(c) Five
(d) Six.
Answer:
Ans. (d)
If A, B and C are three amino acids, the tripeptides formed are six. These are : A–B–C, A–C–B, B–A–C, B–C–A, C–A–B and C–B–A.

Teacher's Note:
1. Using each amino acid once, the number of sequences is 3! = 6; A–B–C and C–B–A are different because a peptide has an N-end and a C-end.

Question 20. Which of the following statements is not correct? (NEET 2017)
(a) Ovalbumin is a simple food reserve in egg-white.
(b) Blood proteins thrombin and fibrinogen are involved in blood clotting.
(c) Denaturation makes the proteins more active.
(d) Insulin maintains sugar level in the blood of a human body.
Answer:
Ans. (c)
Denaturation makes proteins less active.

Teacher's Note:
1. Denaturation destroys the 2° and 3° structure; the protein loses its biological activity.

Question 21. Subunits present in haemoglobin are:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
Ans. (c)
Haemoglobin has quaternary structure made of four polypeptide subunits (two α and two β chains).

Teacher's Note:
1. Haemoglobin is the standard example of quaternary structure: 2α + 2β = 4 subunits.

Question 22. During denaturation of proteins, which of these structures will remain intact?
or
Structure of protein that is unaffected by heating is (JEE Main 2022)
(a) Primary
(b) Secondary
(c) Tertiary
(d) Quaternary
Answer:
Ans. (a)
During denaturation, the hydrogen bonds and other weak forces are broken, but the peptide bonds are not hydrolysed. So the primary structure (amino acid sequence) remains intact.

Teacher's Note:
1. Denaturation breaks weak forces, not covalent peptide bonds; so only the sequence survives.

Teacher's Note:
1. The biuret test needs at least two peptide bonds, so a single amino acid such as alanine gives it negative.

Question 24. The correct structure of Rhumann’s purple, the compound formed in the reaction of ninhydrin with proteins is: (JEE Main 2021)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (d)
2 Ninhydrin + H₂N–CHR–COOH \( \xrightarrow{\mathrm{OH^-}} \) Rhumann’s purple (deep blue) + R–CHO + CO₂ + 4H₂O
In Rhumann’s purple, two indane-1,3-dione units are joined through a =N– bridge: one unit keeps both C=O groups and is joined by C=N, the other is an enolate (C–O⁻ with a ring C=C), as in structure (d).
Ninhydrin is useful for identification of proteins and α-amino acids. It reacts with α-amino acids to give deep blue colour.

Teacher's Note:
1. The N of Rhumann’s purple comes from the amino acid; the R group leaves as the aldehyde R–CHO.

Question 25. The incorrect statement regarding enzymes is: (NEET 2022)
(a) Like chemical catalysts enzymes reduce the activation energy of bio processes.
(b) Enzymes are polysaccharides.
(c) Enzymes are very specific for a particular reaction and substrate.
(d) Enzymes are biocatalysts.
Answer:
Ans. (b)
Enzymes are globular proteins but not polysaccharides.

Teacher's Note:
1. Almost all enzymes are globular proteins; their names usually end in -ase.

Question 26. How many amino acids are present in insulin? (Kerala PET 2013)
(a) 25
(b) 51
(c) 20
(d) 22
(e) 52
Answer:
Ans. (b)
Insulin contains 51 amino acids (21 in chain A and 30 in chain B).

Teacher's Note:
1. Insulin: two chains, A (21) + B (30) = 51 amino acids, held by disulphide links.

One or More Correct Answers

Question 27. For ‘invert sugar’, the correct statement(s) is(are). (Given: specific rotations of (+)-sucrose, (+)-maltose, L-(−)-glucose and L-(+)-fructose in aqueous solution are +66°, +140°, −52° and +92°, respectively) (JEE Advanced 2016)
(a) ‘invert sugar’ is prepared by acid-catalyzed hydrolysis of maltose
(b) ‘invert sugar’ is an equimolar mixture of D-(+)-glucose and D-(−)-fructose
(c) specific rotation of ‘invert sugar’ is −20°
(d) on reaction with Br₂ water, ‘invert sugar’ forms saccharic acid as one of the products
Answer:
Ans. (b, c)
Invert sugar is an equimolar mixture of D-(+)-glucose and D-(−)-fructose. Since specific rotations, i.e., [α] of L-(−)-glucose and L-(+)-fructose are −52° and +92° respectively, therefore, [α] of D-(+)-glucose and D-(−)-fructose are +52° and −92° respectively.
∴ Specific rotation of invert sugar = 0.5 × [α] of D-(+)-glucose + 0.5 × [α_D] of D-(−)-fructose = 0.5 (+52) + 0.5 (−92) = −20°.
With Br₂–H₂O, invert sugar forms gluconic acid as one of the products because Br₂–H₂O oxidises glucose to gluconic acid while fructose is not oxidised by it. Thus, options (b) and (c) are correct.

Teacher's Note:
1. Enantiomers have equal and opposite rotations: use this to convert the given L-values into D-values before averaging.

Question 28. Which of the following statements about α-amino acids are true?
(a) In α-amino acids, the acidic group is –N⁺H₃ while the basic group is –COO⁻
(b) All the α-amino acids which constitute proteins have D-configuration
(c) The isoelectric point of glycine is 6.1
(d) Valine is an essential amino acid
Answer:
Ans. (a, c, d)
(a) In the zwitterion, –N⁺H₃ donates a proton (acidic group) and –COO⁻ accepts a proton (basic group). (b) is wrong because natural α-amino acids of proteins have L-configuration. (c) The isoelectric point of glycine is about 6.1. (d) Valine cannot be synthesised in the body and is an essential amino acid.

Teacher's Note:
1. In the dipolar ion the roles are reversed from the names: NH₃⁺ is the acid and COO⁻ is the base.

Passage-Based Questions

Question 29. Comprehension 1: Monosaccharides containing an aldehyde group are called aldoses while those containing a keto group are called ketoses. The aldehyde group is always present at C₁ while the keto group is usually present at C₂. All monosaccharides containing five and six carbon atoms have cyclic structures, furanose (five membered) and pyranose (six-membered). During ring formation, C₁ in aldoses and C₂ in ketoses becomes chiral and hence all these monosaccharides exist in two stereoisomeric forms called the α-anomer and the β-anomer while C₁ and C₂ are called glycosidic or anomeric carbon atoms. In contrast, stereoisomers, which differ in configuration at any other chiral carbon other than the glycosidic carbon are called epimers. Two molecules of the same or different monosaccharides combine together through glycosidic linkage to form disaccharides. All monosaccharides (aldoses and ketoses) and disaccharides except sucrose reduce Tollens’ reagent and Fehling’s solution, undergo mutarotation and form osazones.
In disaccharides, the linkage connecting monosaccharide units is called
(a) glycoside linkage
(b) nucleoside linkage
(c) glycogen linkage
(d) peptide linkage.
Answer:
Ans. (a)
The two monosaccharide units in a disaccharide are joined through an oxygen atom by a glycosidic (glycoside) linkage.

Teacher's Note:
1. Glycosidic link: sugars; peptide link: amino acids; phosphodiester link: nucleotides.

Matrix Match Questions

Teacher's Note:
1. Br₂ water oxidises only CHO; HNO₃ also oxidises the terminal CH₂OH.

Integer and Numerical Answer Questions

Question 31. The number of chiral carbons in β-D(+)-glucose is.
Answer:
Ans. 5
In the pyranose ring, HOCH₂–C₅H–C₄HOH–C₃HOH–C₂HOH–C₁HOH, with ring O between C-1 and C-5: C-1, C-2, C-3, C-4 and C-5 are all chiral. Hence the number is 5.

Teacher's Note:
1. Open-chain glucose has 4 chiral carbons; the cyclic form has 5 because C-1 becomes chiral.

Question 32. The number of disulphide linkages present in insulin is.
Answer:
Ans. 3
Insulin has three disulphide linkages: two between the A and B chains and one within the A chain.

Teacher's Note:
1. Remember 51 amino acids, 2 chains and 3 disulphide bridges for insulin.

Assertion–Reason Questions

Question 33. Statement-1: Maltose has two α-D-glucose units linked at C₁ and C₄ and is a reducing sugar.
Statement-2: Maltose has two monosaccharides. α-D-glucose and β-D-glucose linked at C₁ and C₆ and it is a non-reducing sugar. (JEE Main 2020)
(a) Both statement-1 and statement-2 are correct.
(b) Both statement-1 and statement-2 are incorrect.
(c) Statement-1 is correct but statement-2 is incorrect.
(d) Statement-1 is incorrect but statement-2 is correct.
Answer:
Ans. (c)
Statement-1 is correct but Statement-2 is incorrect. Maltose has two α-D-glucose units joined by a C1–C4 α-glycosidic linkage, and the free anomeric carbon of the second unit makes it a reducing sugar.

Teacher's Note:
1. Maltose: α(1→4), reducing. A C1–C6 link is found at the branch points of amylopectin and glycogen.

Question 34. Assertion: Deoxyribose, C₅H₁₀O₄ is not a carbohydrate.
Reason: Carbohydrates are hydrates of carbon so compounds which follow C_x(H₂O)_y formula are carbohydrates. (AIIMS 2013)
(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. (d)
Correct assertion. Deoxyribose (C₅H₁₀O₄) is a carbohydrate.
Correct reason. Optically active polyhydroxyaldehydes or ketones or substances which give these on hydrolysis are called carbohydrates.

Teacher's Note:
1. The formula C_x(H₂O)_y is neither necessary (deoxyribose) nor sufficient (acetic acid, C₂(H₂O)₂) for a carbohydrate.

Question 35. Assertion: Fructose does not contain an aldehyde group but still reduces Tollens’ reagent.
Reason: In presence of a base, fructose undergoes a rearrangement to form glucose and mannose.
(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)
In the basic medium of Tollens’ reagent, fructose rearranges (through an enediol) into glucose and mannose, which contain –CHO and reduce the reagent. So the reason explains the assertion.

Teacher's Note:
1. This base-catalysed change is the Lobry de Bruyn–van Ekenstein rearrangement.

Question 36. Assertion: Glycosides are hydrolysed in acidic conditions.
Reason: Glycosides are acetals. (AIIMS 2003)
(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)
Glycosides are cyclic acetals. Acetals are stable in base but are readily hydrolysed by aqueous acid, so the reason explains the assertion.

Teacher's Note:
1. Acetals (and so glycosides and disaccharides) are cleaved by dilute acid, not by alkali.