Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
The total number of fragments generated by the complete and sequential cleavage of the
polypeptide given below by Trypsin followed by CNBr is _________
Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln
Official Answer Key:
5 to 5
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2013 Q.43)**
**1. Governing Concept & Definition**:
- **Topic**: Protein Biochemistry & Enzymatic/Chemical Cleavage
- Number of peptide fragments generated by endopeptidase/cyanogen bromide cleavage equals:
$\text{Number of Fragments} = \text{Number of Cleavage Sites} + 1$
(for a linear polypeptide chain).
**2. Detailed Step-by-Step Calculation**:
- Counting specific recognition residues (e.g. Trypsin cuts after Lys/Arg; CNBr cuts after Met):
- Total internal cleavage sites $= 4$
- Number of resulting fragments $= 4 + 1 = 5$
**3. Final Answer**:
- **Calculated Value**: **5**
- **Official GATE Accepted Range**: **5 to 5**
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