Question

An octapeptide composed of these L-amino acids – Lys, Thr, Ser, Met, Arg, Trp, Tyr, Glu – was subjected to analyses with the following outcomes: 
P. The N-terminal sequencing analysis by Sanger’s method yielded ‘Ser’ at the N-terminus
Q. Chymotrypsin treatment gave a pentapeptide, a ‘Tyr’ containing dipeptide, and a free ‘Glu’
R. Cyanogen bromide treatment gave two tetrapeptides
S. Trypsin treatment gave two tripeptides and a dipeptide
Which one of the following is the correct octapeptide sequence?

• A. Ser-Tyr-Arg-Met-Lys-Thr-Trp-Glu
• B. Ser-Arg-Lys-Met-Tyr-Thr-Trp-Glu
• C. Ser-Met-Lys-Arg-Thr-Tyr-Trp-Glu
• D. Ser-Arg-Met-Lys-Trp-Thr-Tyr-Glu

Hint:

When analyzing peptide sequencing, pay attention to the results of cleavage by different enzymes like chymotrypsin, cyanogen bromide, and trypsin. These enzymes cleave at specific amino acid residues, which can help reconstruct the full sequence.

Answer 1

The Correct Option is A

We are given an octapeptide with the amino acid sequence of Lys, Thr, Ser, Met, Arg, Trp, Tyr, and Glu, and four types of analyses. Let’s go step-by-step to determine the correct sequence.

Step 1: N-terminal sequencing by Sanger’s method yielded ‘Ser’ at the N-terminus. This means Ser is the first amino acid in the peptide sequence.

Step 2: Chymotrypsin treatment gave a pentapeptide, a ‘Tyr’-containing dipeptide, and a free ‘Glu’. This implies that Tyr is located close to the C-terminus and Glu is at the C-terminus, since it is free after the chymotrypsin treatment.

Step 3: Cyanogen bromide treatment cleaves peptides at methionine residues. Since cyanogen bromide treatment produced two tetrapeptides, the sequence likely has two Met residues in it.

Step 4: Trypsin treatment gave two tripeptides and a dipeptide. Trypsin cleaves peptides at the C-terminal side of Lys and Arg. This implies that Lys and Arg are separated from other amino acids, and these positions help us deduce the correct sequence.

From the above analyses, we can conclude that the correct octapeptide sequence is:

\[ \text{Ser–Tyr–Arg–Met–Lys–Thr–Trp–Glu} \]

This sequence satisfies all the conditions provided in the question.