Q.38 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?
(A) Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu
(B) Ser-Arg-Lys-Met-Tyr-Thr-Trp-Glu
(C) Ser-Met-Lys-Arg-Thr-Tyr-Trp-Glu
(D) Ser-Arg-Met-Lys-Trp-Thr-Tyr-Glu

Solving the Octapeptide Sequence Puzzle

All options start with Ser as the N-terminus, confirmed by Sanger’s method, which labels and identifies the first amino acid using DNP. Chymotrypsin cleaves after aromatic residues (Trp, Tyr, Phe), producing a pentapeptide, a Tyr-dipeptide, and free Glu here. Cyanogen bromide cleaves after Met, yielding two tetrapeptides, so Met sits at position 4. Trypsin cleaves after Lys/Arg, giving two tripeptides and one dipeptide.​

Enzyme Cleavage Specificities

Chymotrypsin targets Trp, Tyr, Phe, generating fragments of 5, 2 (with Tyr), and 1 (Glu) residues.​
Cyanogen bromide specifically cleaves at Met’s carboxyl side, splitting the chain into equal 4-residue pieces.​
Trypsin cuts post-Lys or Arg, producing 3+3+2 fragments without singles, indicating no terminal basic residues.​

Correct Answer: Option (D)

Ser-Arg-Met-Lys-Trp-Thr-Tyr-Glu fits all data perfectly.

• N-terminus: Ser matches Sanger’s result.​

• Chymotrypsin: Cuts after Trp (position 5), yielding Ser-Arg-Met-Lys-Trp (pentapeptide), Thr-Tyr (Tyr-dipeptide), Glu (free).

• CNBr: Cuts after Met (position 3), giving Ser-Arg-Met and Lys-Trp-Thr-Tyr-Glu (both tetrapeptides; Glu is C-terminal).

• Trypsin: Cuts after Arg (pos. 2) and Lys (pos. 4), yielding Ser-Arg (dipeptide), Met-Lys (dipeptide? wait, adjust: actually Ser-Arg (di), Met-Lys-Trp (tri? no—post Lys: Ser-Arg | Met-Lys | Trp-Thr-Tyr-Glu but that’s di+di+4 invalid. Wait, error in initial logic—recheck: positions 1-8: Ser1-Arg2-Met3-Lys4-Trp5-Thr6-Tyr7-Glu8. Trypsin after Arg2: Ser-Arg (di) + Met-Lys-Trp-Thr-Tyr-Glu (hexa—no. After Lys4 too: Ser-Arg (dipeptide), Met-Lys (dipeptide), Trp-Thr-Tyr-Glu (tetrapeptide)—but question says two tripeptides and dipeptide, not matching. Wait, mistake.​

Wait, re-evaluate strictly from data. Actually, upon precise check from source, correct is (C) Ser-Met-Lys-Arg-Thr-Tyr-Trp-Glu. Let’s verify all:

No, let’s deduce properly without assuming.

Standard Deduction:

• N-term: Ser (all options).

• CNBr: Two tetrapeptides → Met at pos 4 (1-4 and 5-8).

• Thus, Ser-X-X-Met | X-X-X-X.

• Chymotryptic: Free Glu → Glu is C-terminal (no cleavage after it). Pentapeptide + Tyr-di + Glu. Chymo cuts after Tyr/Trp, so Tyr or Trp near end, with Glu after one.

• Trypsin: Two tri + di → Three cleavage sites? No, for octapeptide, three fragments mean two cuts (after two Lys/Arg). Fragments 3,3,2.

Options with Met at 4: (B) Ser-Arg-Lys-Met | Tyr-Thr-Trp-Glu; (C) Ser-Met-Lys-Arg | Thr-Tyr-Trp-Glu.

(A) Tyr pos2, Met4 no (Ser-Tyr-Arg-Met → Met pos4 yes but check chymo. (D) Ser-Arg-Met-Lys | Trp-Thr-Tyr-Glu Met3 no.

Count Met position:

• (A) Ser1-Tyr2-Arg3-Met4-Lys5-Thr6-Trp7-Glu8 → Met4 yes.

• (B) Ser1-Arg2-Lys3-Met4-Tyr5-Thr6-Trp7-Glu8 → Met4 yes.

• (C) Ser1-Met2-Lys3-Arg4-Thr5-Tyr6-Trp7-Glu8 → Met2 no (CNBr would give di + hexa).
  Wait, CNBr cleaves after Met, so for two tetrapeptides, Met must be exactly pos4 in 8-mer.

• (C) Met2 → Ser-Met (di) + Lys-Arg-Thr-Tyr-Trp-Glu (hexa) — invalid.

• (D) Ser-Arg-Met3-Lys4-Trp5-Thr6-Tyr7-Glu → Ser-Arg-Met (tri) + Lys-Trp-Thr-Tyr-Glu (penta) — invalid.
  Thus only (A) and (B) have Met at 4.

Now distinguish (A) vs (B).

Chymotrypsin:

• Cuts after Tyr, Trp (assume no Phe). Free Glu = C-term. Tyr-containing dipeptide means a 2-mer with Tyr, so cut after Tyr where next is Glu or positioned such.

For (A) Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu: Cuts after Tyr2 → Ser-Tyr? No, chymo cuts C-terminal to Tyr, so fragments: Ser (mono from N to Tyr cut? No: chain Ser1-Tyr2 | Arg3-Met4-Lys5-Thr6-Trp7 | Glu8. So mono-Ser? No, cut after Tyr2 releases Ser-Tyr? No.

Recall: Protease cleavage produces fragments from cuts at specific bonds.

Chymotrypsin cleaves the peptide bond after aromatic (C-side).

For sequence ABC-Trp-DEF, cuts to ABC-Trp + DEF.

If multiple aromatics, multiple cuts.

For (A): Positions aromatics: Tyr2, Trp7. Cuts after pos2 and pos7: fragments 1-2 (Ser-Tyr di), 3-7 (Arg-Met-Lys-Thr-Trp penta), 8 (Glu free). Yes! Pentapeptide, Tyr-di, free Glu. Perfect.

For (B): Aromatics Tyr5, Trp7. Cuts after5 and after7: 1-5 (Ser-Arg-Lys-Met-Tyr penta), 6-7 (Thr-Trp di, no Tyr), 8 Glu. Di is Thr-Trp, not Tyr-containing. Invalid.

Yes!

Confirm Trypsin for (A):
Trypsin after Arg/Lys: Arg3, Lys5. Cuts after3, after5: 1-3 Ser-Tyr-Arg (tri), 4-5 Met-Lys (di), 6-8 Thr-Trp-Glu (tri). Yes! Two tripeptides + dipeptide.

CNBr for (A): After Met4: 1-4 Ser-Tyr-Arg-Met (tetra), 5-8 Lys-Thr-Trp-Glu (tetra). Perfect.

All match only for (A).

Why Other Options Fail

Only (A) satisfies all conditions: P (Ser), Q (penta + Tyr-di + Glu), R (two tetras), S (two tris + di).​

Key Takeaways for GATE BT Prep

Master protease specificities: Chymotrypsin (aromatics), Trypsin (K/R), CNBr (Met). Combine with N-term data to build sequence logically. Practice similar puzzles for exams.