2019 H2 Chem P1 Q28 — Nitrogen Compound Solution

The answer is C.

Chymotrypsin cuts the peptide bond on the carboxyl (-CO-) side of a residue with an aromatic ring: here phenylalanine (benzene ring) and tryptophan (indole ring). A cut needs another residue after the aromatic one; an aromatic residue at the C-terminal end has no peptide bond after it.
Reading each tetrapeptide from the \(\text{H}_2\text{N}\) end:
• A: Ala-Val-Trp-Ser. One cut, after Trp (residue 3): a tripeptide (Ala-Val-Trp) + one amino acid (Ser).
• B: Val-Trp-Phe-Ala. Cuts after Trp and after Phe: a dipeptide (Val-Trp) + two amino acids (Phe, Ala).
• C: Lys-Trp-Ala-Phe. One cut, after Trp (residue 2); Phe is the C-terminal residue, so there is no bond after it to cut. Products: two different dipeptides, Lys-Trp and Ala-Phe.
• D: Phe-Trp-Ala-Ser. Cuts after Phe and after Trp: two amino acids (Phe, Trp) + a dipeptide (Ala-Ser).
Only C forms two different dipeptides.