An octapeptide was determined to have the following amino acid composition: Lys (2),Phe (2),Gly (1),His (1),Leu (1),Met (1).The native peptide was run through one cycle of the Edman degradation and the PTH-leucine derivative was identified by HPLC.When the native peptide was exposed to cyanogen bromide (BrCN),a heptapeptide and free glycine were recovered.Incubation of the native protein with trypsin gave a tetrapeptide,a tripeptide,and free lysine.The peptides were separated and each run through one cycle of the Edman degradation.The tetrapeptide yielded the PTH-leucine derivative,and the tripeptide yielded the PTH-phenylalanine derivative.Incubation of the native protein with pepsin produced a dipeptide and two tripeptides.The amino acid composition of the tripeptides (not the order)were determined to be (Phe,Gly,Met)and (Phe,Lys,Lys).What is the sequence of the octapeptide?
Specificities of Proteases
BrCN cuts at the C-terminal side of Met
Trypsin cuts at the C-terminal side of Lys or Arg
Pepsin cuts at the N-terminal side of Phe,Trp or Tyr

Question

Quizplus · Accepted Answer

The sequence of the octapeptide can be deduced by analyzing the results of the treatments with different enzymes and chemicals. The first clue comes from the Edman degradation, which identifies the N-terminal amino acid as leucine, eliminating options that do not start with Leu. The action of cyanogen bromide (BrCN) cleaves at the C-terminal side of Met, producing a heptapeptide and free glycine, indicating that Gly is the C-terminal residue. The results from trypsin digestion, which cleaves at the C-terminal side of Lys, combined with the Edman degradation of the resulting peptides, further refine the sequence. The action of pepsin, which cuts at the N-terminal side of Phe, and the composition of the resulting peptides, provide additional constraints. Considering all these data, the sequence that fits all the given experimental outcomes is Leu-His-Phe-Lys-Lys-Phe-Met-Gly.

An Octapeptide Was Determined to Have the Following Amino Acid