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GATE 2025 BT Q.38: solution and answer

GATE 2025 · BTQ.38 · 2 marks · MCQ

An octapeptide composed of these L-amino acids – Lys, Thr, Ser, Met, Arg, Trp, Tyr, Glu – was subjected to analyses with the following outcomes:
P. The N-terminal sequencing analysis by Sanger's method yielded 'Ser' at the N-terminus
Q. Chymotrypsin treatment gave a pentapeptide, a 'Tyr' containing dipeptide and a free 'Glu'
R. Cyanogen bromide treatment gave two tetrapeptides
S. Trypsin treatment gave two tripeptides and a dipeptide
Which one of the following is the correct octapeptide sequence?

  1. (A)

    Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu

  2. (B)

    Ser-Arg-Lys-Met-Tyr-Thr-Trp-Glu

  3. (C)

    Ser-Met-Lys-Arg-Thr-Tyr-Trp-Glu

  4. (D)

    Ser-Arg-Met-Lys-Trp-Thr-Tyr-Glu

See the question in the paper
GATE 2025 BT question 38Tap to open full size

Answer and solution

ANSWER

The octapeptide is Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu, which is option (A).

Official answer: ASource: Official answer key, GATE 2025 BT (IIT Roorkee)

  1. Sanger's reagent (FDNB) labels the N-terminal residue, so Ser is residue 1. All four options begin with Ser, so clue P does not separate them.

  2. Cyanogen bromide cleaves on the C-terminal side of Met. Two tetrapeptides means a single cut after residue 4, so Met must be residue 4. Option D has Met at position 3, which gives a tripeptide and a pentapeptide, so D is out. Option C has Met at position 2, which gives a dipeptide and a hexapeptide, so C is out.

  3. Option B has Met at position 4 and passes the CNBr test. Chymotrypsin cuts after Tyr and Trp. In B, Tyr is at position 5 and Trp at position 7, giving Ser-Arg-Lys-Met-Tyr (pentapeptide), Thr-Trp (dipeptide) and free Glu. The dipeptide contains Trp, not Tyr, so B contradicts clue Q. Trypsin, which cuts after Arg and Lys, would also give Ser-Arg, Lys and Met-Tyr-Thr-Trp-Glu on B, which contradicts clue S. B is out.

  4. Check A, Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu. Chymotrypsin cuts after Tyr(2) and Trp(7), giving Ser-Tyr (a Tyr-containing dipeptide), Arg-Met-Lys-Thr-Trp (a pentapeptide) and free Glu. This matches Q.

  5. In A, CNBr cuts after Met(4), giving Ser-Tyr-Arg-Met and Lys-Thr-Trp-Glu, which are two tetrapeptides. This matches R.

  6. In A, trypsin cuts after Arg(3) and Lys(5), giving Ser-Tyr-Arg (tripeptide), Met-Lys (dipeptide) and Thr-Trp-Glu (tripeptide). This matches S. Option A satisfies all four clues.

KEY IDEA

Use the specific cleavage sites. CNBr cuts after Met, trypsin after Lys and Arg, and chymotrypsin after Tyr, Trp and Phe. Start with the clue that fixes one position, here Met at residue 4, to eliminate options, then check the remaining clues on the survivor.

WATCH OUT FOR

Cutting on the wrong side of the residue, for example before Met instead of after it. Another mistake is forgetting that Trp is also cleaved by chymotrypsin, which makes B look acceptable.

From the GATE 2025 BT paper. Every question of this paper · Download the full paper · All 12 Biomolecules questions, with notes

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