Class 12 Chemistry Chapter 10 Biomolecules – Extra Questions with Answers

These extra practice questions for Class 12 Chemistry Chapter 10 – Biomolecules go beyond the NCERT textbook exercises to reinforce carbohydrate chemistry, protein structure, vitamin deficiency diseases, and nucleic acid base-pairing. Useful for board exam revision and quick concept checks.

Last Updated: September 23, 2026

Very Short Answer Type Questions (1 Mark)

Q1. Name the deficiency disease caused by a lack of vitamin B₁ and vitamin D.
Ans: Vitamin B₁ (thiamine) deficiency causes beriberi; vitamin D (calciferol) deficiency causes rickets in children (and osteomalacia in adults).

Q2. What is the anomeric carbon in glucose?
Ans: The anomeric carbon is C1 of glucose — the carbon that becomes a new stereocentre when the open-chain structure cyclises into the pyranose (hemiacetal) ring, giving rise to the α and β anomers.

Q3. Name the two components of starch.
Ans: Amylose (unbranched, water-soluble, forms a helix) and amylopectin (branched, water-insoluble).

Q4. Which base is present in RNA but not in DNA, and which is present in DNA but not in RNA?
Ans: Uracil is present in RNA but not DNA; thymine is present in DNA but not RNA.

Short Answer Type Questions (2–3 Marks)

Q5. What is meant by mutarotation? Illustrate with reference to glucose.
Ans: Mutarotation is the spontaneous change in optical rotation of a freshly-prepared solution of a sugar’s pure anomeric form towards a fixed equilibrium value, as the two anomers interconvert through the open-chain form. Freshly dissolved α-D-glucose (specific rotation +111°) and freshly dissolved β-D-glucose (specific rotation +19.2°) each gradually change, on standing, to the same equilibrium value of +52.5° — corresponding to an equilibrium mixture of both anomeric forms via the open-chain aldehyde.

Q6. Explain why fibrous proteins are generally insoluble in water while globular proteins are usually soluble.
Ans: In fibrous proteins, long polypeptide chains lie parallel to one another and are cross-linked by hydrogen bonds and disulphide bonds into extended, rope-like or sheet-like structures with relatively little hydrophilic surface exposed to water, making them water-insoluble (e.g. keratin, collagen, myosin). In globular proteins, the chain folds up so that hydrophobic side chains are buried in the interior and polar/hydrophilic side chains face outward on the compact spherical surface, allowing extensive interaction with surrounding water molecules and making them water-soluble (e.g. insulin, haemoglobin, albumin).

Q7. Distinguish between a nucleoside and a nucleotide, and name the sugar and one purine base found in DNA.
Ans: A nucleoside is a base+sugar unit (no phosphate); a nucleotide is a base+sugar+phosphate unit, i.e. a phosphorylated nucleoside. DNA’s sugar is β-D-2-deoxyribose, and one of its purine bases is adenine (the other purine base is guanine; DNA’s pyrimidines are cytosine and thymine).

Higher Order Thinking Skills (HOTS)

Q8. A student argues: “Since DNA and RNA are both nucleic acids built from very similar nucleotide units, they should be equally stable and equally suited to long-term storage of genetic information.” Evaluate this claim with reference to the structural differences between DNA and RNA.
Ans: The claim is not correct. Although DNA and RNA share the same basic nucleotide architecture, two structural differences make DNA far better suited to long-term genetic storage. First, DNA’s sugar is 2-deoxyribose (lacking the 2′-OH group that ribose has); RNA’s extra 2′-OH makes its backbone chemically more reactive and more prone to hydrolytic cleavage, so RNA degrades faster than DNA. Second, DNA is normally double-stranded, with each strand’s base sequence protected and “backed up” by its complementary partner strand (A–T, G–C base pairing) — damage to one strand can be repaired using the other as a template. RNA is usually single-stranded, with no such built-in redundancy. Together, these differences explain why DNA, not RNA, serves as the durable, error-correctable long-term store of hereditary information in most organisms, while RNA is generally used for shorter-lived, working copies of genetic instructions.

Q9. Two carbohydrate samples, both of molecular formula C₁₂H₂₂O₁₁, are tested with Fehling’s solution. Sample X gives a positive test (red precipitate); Sample Y gives no reaction. Both hydrolyse under acid conditions to give monosaccharides. Identify a plausible identity for X and Y, and explain the difference in their Fehling’s test behaviour in terms of their glycosidic linkages.
Ans: Sample X (Fehling’s-positive, a reducing disaccharide) is plausibly maltose (glucose+glucose via an α-1,4 glycosidic link) or lactose (glucose+galactose via a β-1,4 link) — both leave one anomeric carbon free to open into a reactive aldehyde form. Sample Y (Fehling’s-negative, a non-reducing disaccharide) is plausibly sucrose (glucose+fructose), whose glycosidic bond forms between the anomeric carbons of both monosaccharide units (C1 of glucose and C2 of fructose), leaving no free anomeric –OH on either unit. Since Fehling’s test requires a free, openable aldehyde (or an α-hydroxy ketone that can isomerise to one), only X, with its unused anomeric centre, can reduce Cu²⁺ to the red Cu₂O precipitate; Y cannot.

Written by Satish

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