Question

A polypeptide chain containing three lysine and four arginine residues is treated with sufficient concentration of trypsin in a reaction mixture with optimum condition for reaction of trypsin yield Consider no arginine and lysine residues are present in carboxy and amino terminal end of that polypeptide

• A. Four peptides
• B. Five peptides
• C. Seven peptides
• D. Eight peptides

Hint:

For enzymatic cleavage of linear proteins:
$\text{Number of fragments} = \text{Internal cleavage sites} + 1$.
If there were a Lys or Arg at the very C-terminal, that cleavage would not generate an additional fragment, but since the prompt excludes terminal residues, the formula holds true.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Trypsin is a highly specific serine protease that catalyzes the cleavage of peptide bonds on the carboxyl-terminal side of basic amino acid residues, specifically Lysine (Lys, K) and Arginine (Arg, R).
Detailed Explanation:
- The given polypeptide chain contains:
- $3$ Lysine residues
- $4$ Arginine residues
- Since trypsin cleaves at the C-terminal of both Lysine and Arginine, the total number of cleavage sites within this polypeptide is:
\[ \text{Total cleavage sites} = 3\text{ (Lysine)} + 4\text{ (Arginine)} = 7\text{ sites} \] - The problem specifies that no Lysine or Arginine residues are located at the extreme amino ($\text{N}$-) or carboxyl ($\text{C}$-) terminals of the polypeptide.
- This ensures that all $7$ cleavage sites are located internally within the chain.
- For any linear polypeptide, cleaving it at $N$ internal sites will divide the chain into exactly $N + 1$ fragments.
- Therefore, the number of resulting peptide fragments is:
\[ \text{Number of peptides} = 7\text{ (cleavage sites)} + 1 = 8\text{ peptides} \]

Step 2: Final Answer:
The treatment of this polypeptide with trypsin will yield eight peptides, corresponding to option (D).