Vidaara.orgClass 12 · Chemistry
CodeVID-C12-14-CH-01
Chapter Assignment — Biomolecules
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
Which is a non-reducing disaccharide?
- A.maltose
- B.lactose
- C.sucrose
- D.cellobiose
2.
The anomeric carbon of glucose is:
- A.C1
- B.C2
- C.C4
- D.C6
3.
An amino acid exists mainly as a:
- A.free radical
- B.zwitterion
- C.carbocation
- D.diradical
4.
Denaturation does not break:
- A.hydrogen bonds
- B.disulphide bridges
- C.peptide bonds
- D.ionic bonds
5.
Scurvy is caused by deficiency of vitamin:
- A.A
- B.C
- C.D
- D.K
6.
In DNA, adenine pairs with:
- A.guanine
- B.cytosine
- C.thymine
- D.uracil
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
Why is sucrose called invert sugar after hydrolysis?
8.
Define the isoelectric point of an amino acid.
9.
Distinguish a nucleoside from a nucleotide.
10.
Name two fat-soluble vitamins and a deficiency disease of each.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Explain why glucose forms a pyranose ring and define its anomers.
12.
Compare the α-helix and β-pleated sheet secondary structures of proteins.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Describe the four levels of organisation of protein structure and explain denaturation.
14.
Describe the double-helix structure of DNA and explain how complementary base pairing ensures accurate replication.
Answer Key
Section A — Multiple Choice Questions
- (C) sucrose
- (A) C1
- (B) zwitterion
- (C) peptide bonds
- (B) C
- (C) thymine
Section B — Short Answer (2 marks)
- Its rotation changes sign from +66.5° to about −20° because the strongly laevorotatory fructose outweighs the dextrorotatory glucose in the hydrolysis mixture.
- The pH at which the amino acid is present almost entirely as the zwitterion with zero net charge and does not migrate in an electric field.
- Nucleoside = nitrogenous base + pentose sugar; nucleotide = nucleoside + a phosphate group (the building block of nucleic acids).
- Vitamin A — night blindness; vitamin D — rickets (others: E, K).
Section C — Short Answer (3 marks)
- The C5–OH adds across the C1 aldehyde, forming a six-membered (pyranose) ring and making C1 a new chiral centre. The two products, α-D-glucose (C1–OH down) and β-D-glucose (C1–OH up), are anomers that interconvert by mutarotation.
- Both use backbone H-bonds. The α-helix is a single chain coiled right-handed with intramolecular H-bonds (keratin); the β-sheet has extended chains side by side held by intermolecular H-bonds, forming a pleated sheet (silk fibroin).
Section D — Long Answer (5 marks)
- Primary: amino-acid sequence linked by peptide bonds. Secondary: local folding by H-bonds into α-helix or β-sheet. Tertiary: full 3-D folding (fibrous/globular) held by H-bonds, –S–S– bridges, ionic and van der Waals forces. Quaternary: assembly of two or more sub-units (e.g. haemoglobin). Denaturation by heat/acid/heavy metals breaks the weak forces, unfolding the 2° and 3° structures and destroying activity, while the primary peptide bonds remain intact (e.g. boiled egg white).
- DNA is a right-handed double helix of two antiparallel polynucleotide strands with a sugar–phosphate backbone outside and bases inside. The strands are held by complementary H-bonds: A=T (two bonds) and G≡C (three bonds), so A equals T and G equals C (Chargaff's rule). During replication the helix unwinds and each strand templates a new complementary strand (A with T, G with C), giving two identical daughter helices each retaining one parent strand — semi-conservative copying that transmits genetic information faithfully.
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