Suppose that the peptide Ala-Met-Lys-Ser is reacted with phenylisothiocyanate at pH 8.0,followed by mild acidification,which is the first cycle of Edman method? Which of the following would be released?
A)the labelled peptide Ala-Met-Lys-Ser-PTH
B)PTH-Ala,PTH-Ser,PTH-Lys,and PTH-Met
C)PTH-Ser and the peptide Ala-Met-Lys
D)PTH-Ala and the peptide Met-Lys-Ser

Question

Quizplus · Accepted Answer

In the first cycle of the Edman method, only one amino acid is released, and it is the N-terminal amino acid, which reacts with phenylisothiocyanate to form a phenylthiocarbamoyl derivative. In this case, the N-terminal amino acid is Ala, so it will react with phenylisothiocyanate to form PTH-Ala, which will be released upon mild acidification. The remaining peptide is then subjected to a second cycle of Edman degradation to release the next N-terminal amino acid. Therefore, only PTH-Ala and the peptide Met-Lys-Ser will be released in the first cycle.

Suppose That the Peptide Ala-Met-Lys-Ser Is Reacted with Phenylisothiocyanate at PH