GATE BT 2023 • Q52
NAT • 2 Marks
Section 2: General Biotechnology
Biomolecules, enzyme kinetics
Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
The number of different possible ways of forming five intramolecular disulfide
bonds with ten cysteine residues of a protein is ___________.
Official Answer Key:
945 to 945
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.52)**
**1. Governing Concept & Definition**:
- **Topic**: Protein Structure & Combinatorics of Disulfide Bond Formation
- The number of distinct ways to partition $2n$ distinct cysteine residues into $n$ pairs (disulfide bonds) is given by the double factorial of $(2n - 1)$:
$\text{Ways} = \frac{(2n)!}{2^n \cdot n!} = (2n - 1)!! = (2n - 1) \times (2n - 3) \times \dots \times 3 \times 1$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Total cysteine residues $= 10$ ($2n = 10 \implies n = 5$ disulfide bonds)
- **Calculation**:
$\text{Ways} = (2 \times 5 - 1)!! = 9!! = 9 \times 7 \times 5 \times 3 \times 1$
- $9 \times 7 = 63$
- $63 \times 5 = 315$
- $315 \times 3 = 945$
- $945 \times 1 = 945$
**3. Final Answer**:
- **Calculated Value**: **945**
- **Official GATE Accepted Range**: **945 to 945**
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