GATE 2019 BT Q.11: solution and answer

Official answer: BSource: Official answer key, GATE 2019 BT (IIT Madras)

1. A mutation in a gene produces a variant polypeptide that is shorter by exactly the last three amino acids — nothing else about the protein (reading frame, downstream identity) is stated to differ, since translation simply stops earlier than normal.

2. This pattern — reading frame preserved, chain ending early at a fixed point — is the signature of a premature stop (termination) codon appearing in place of what was originally a sense codon three residues before the natural end.

3. Check each option against this fact: (A) A synonymous mutation changes a codon to a different codon for the SAME amino acid, so the protein sequence does not change at all; it cannot shorten the polypeptide. Eliminated.

4. (B) A nonsense mutation changes a sense codon into a stop codon, causing translation to terminate at that point and truncating the polypeptide downstream of the mutation — this exactly matches losing the last three amino acids. Fits.

5. (C) A missense mutation changes a codon so that it codes for a DIFFERENT amino acid, giving a full-length polypeptide with one substituted residue, not a shortened one. Eliminated.

6. (D) A silent mutation has no effect on the amino acid sequence (most commonly synonymous, sometimes in a non-coding/regulatory region) — again, no truncation is possible. Eliminated.

7. Only a nonsense mutation explains a variant polypeptide that terminates prematurely, precisely removing the terminal three residues.