Question

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?Show your work. 5 pts Edman degredation—isolates N-terminal Cyanogen Bromide—isolates C-terminal 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

Answer 1

The octapeptide sequence is: Leu – His – Phe – Lys – Lys – Phe – Met – Gly

Step-by-step explanation:

A first Edman degradation of the native peptide results in Leu as N-terminus amino acid (as PTH-leucine), and BrCN exposition results in an heptapeptide + free glycine; this means that Gly is the C-terminus amino acid, and that Met antecedes it, being the aa number 7, as BrCn cuts the C-terminus of Methionine.

Following native peptide incubation with trypsin, a tetrapeptide, a tripeptide and free lysine appeared, and were individually submitted to Edman degradation, resulting PTH-leucine from the tetrapeptide, and the PTH-phenylalanine derivative from the tripeptide, indicating that Leu starts the tetrapeptide sequence, and Phe is the start of the tripeptide that is linked to the tetrapeptide.

Pepsin treatment (Pepsin cuts the N-terminus of Phe, Tyr and Trp, and in this case only Phe is present), produced two tripeptides (Phe, Gly, Met; not the order) and (Phe, Lys, Lys; not the order), and a dipeptide (with the missing amino acids, that are His and Leu; not the order yet). Then we have: Leu, followed by His (that is the pepsin dipeptide), later appears Phe – Lys – Lys (one of the pepsin tripeptides; it starts with Phe, because pepsin cuts Phe N-terminus), and finally Phe – Met – Gly (the other pepsin tripeptide; it starts with Phe because the pepsin action site in its N-terminus, and ends with Gly, that was the C-terminus by BrCN treatment).

Being the octapeptide sequence Leu – His – Phe – Lys – Lys – Phe – Met – Gly, we can agree that trypsin cut the C-terminus of both lysines resulting the tetrapeptide Leu – His – Phe – Lys, free Lys and the tripeptide Phe – Met – Gly.