Q.43 The total number of fragments generated by the complete and sequential cleavage of the
polypeptide given below by Trypsin followed by CNBr is _________
Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln

Polypeptide Cleavage: Trypsin and CNBr Generate 5 Fragments

Sequential cleavage first by trypsin then CNBr on the polypeptide Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln produces 5 fragments total. This biochemical technique maps protein sequences by exploiting enzyme and chemical specificities.​

Trypsin Cleavage Specifics

Trypsin hydrolyzes peptide bonds at the C-terminal side of lysine (Lys, K) and arginine (Arg, R), unless followed by proline. In the sequence, Lys appears after Ala (position 6), and Arg after Gly (position 12). Cleavages occur after Lys6 and Arg12, yielding three fragments: Phe-Trp-Met-Gly-Ala-Lys, Leu-Pro-Met-Asp-Gly-Arg, and Cys-Ala-Gln.​

CNBr Cleavage Specifics

Cyanogen bromide (CNBr) cleaves specifically at the C-terminal side of methionine (Met, M) residues. The trypsin fragments contain Mets: the first has one Met (position 3), and the second has one Met (position 9 in original, after Leu-Pro). No Met in the third fragment. Thus, CNBr further splits the first into Phe-Trp-Met and Gly-Ala-Lys, and the second into Leu-Pro-Met and Asp-Gly-Arg.​

Final Fragment Count

The third trypsin fragment (Cys-Ala-Gln) remains intact. Combined post-CNBr: Phe-Trp-Met, Gly-Ala-Lys, Leu-Pro-Met, Asp-Gly-Arg, Cys-Ala-Gln—exactly 5 distinct fragments. The correct answer is 5.​