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GATE 2025 BT Q.31: solution and answer

GATE 2025 · BTQ.31 · 1 mark · NAT

The contour length of a B-DNA molecule that encodes a bacterial protein of 33 kDa is ______ nm.

Consider the average molecular weight of an amino acid as 110 Da and helix rise per base pair for B-DNA as 0.34 nm.

(Round off to the nearest integer)

See the question in the paper
GATE 2025 BT question 31Tap to open full size

Answer and solution

ANSWER

The contour length of the B-DNA is 306 nm.

Official answer: 300 to 310Source: Official answer key, GATE 2025 BT (IIT Roorkee)

  1. Number of amino acid residues in the protein = protein mass ÷ average residue mass = 33{,}000\text{ Da} \div 110\text{ Da} = 300 residues.

  2. The genetic code is triplet: each amino acid residue is specified by one codon of 3 nucleotides, i.e. 3 base pairs of the coding B-DNA. Number of coding base pairs = 300 \times 3 = 900\text{ bp}.

  3. Contour length of B-DNA = (number of base pairs) \times (rise per base pair) = 900 \times 0.34\text{ nm} = 306\text{ nm}.

  4. 306 is already an integer, so the rounded answer is 306 nm.

KEY IDEA

Convert protein mass to number of amino acid residues (mass ÷ average residue mass), use the triplet genetic code to get the number of coding base pairs (residues × 3), then multiply by the B-DNA rise per base pair (0.34 nm) to get contour length.

WATCH OUT FOR

Forgetting that each amino acid is encoded by a codon of 3 nucleotides (3 base pairs), and instead multiplying the number of amino acids directly by the rise per base pair (300 × 0.34 = 102 nm), which omits the triplet code entirely and gives a value far outside the key range.

From the GATE 2025 BT paper. Every question of this paper · Download the full paper · All 9 Mutation & repair questions, with notes

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