Question

Ala-Val-Gly-Val-Met-Try-Lys-Ala-Glu-Ala-Arg-Ala-Val-Try-Gly-His-Gly is the amino acid sequence of a peptide. If this peptide is treated with enzyme trypsin in a proper reaction condition then the maximum number of the smaller peptides that could be produced is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Trypsin: cleaves after Lys (K) and Arg (R).
Number of peptides = Number of cleavage sites + 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Trypsin is a protease that cleaves peptides at specific sites.
It cleaves at the C-terminal side of lysine (K) and arginine (R).

Step 2: Key Formula or Approach:
No specific formula.
Need to identify the trypsin cleavage sites.

Step 3: Detailed Explanation:
The peptide sequence is:
Ala-Val-Gly-Val-Met-Try-Lys-Ala-Glu-Ala-Arg-Ala-Val-Try-Gly-His-Gly.
Trypsin cleaves after Lys (K) and Arg (R).
Cleavage sites:
After Lys at position 7: Ala-Val-Gly-Val-Met-Try-Lys | Ala-Glu-Ala-Arg-Ala-Val-Try-Gly-His-Gly.
After Arg at position 11: Ala-Glu-Ala-Arg | Ala-Val-Try-Gly-His-Gly.
Thus, the peptide is cleaved into 3 smaller peptides.

Step 4: Final Answer:
The maximum number of smaller peptides produced is 3.
Hence, the correct option is (B).