Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
You are characterizing a new enzyme isolated and purified in the laboratory. If the
maximum v elocity of the enzyme is 18 00 µmoles L -1 min-1 and the total
concentration of the enzyme in the reaction mixture is 1.5 µM, then the turnover
number of the enzyme is _______ s-1. (answer in integer)
Official Answer Key:
20 to 20
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2026 Q.32)**
**1. Governing Concept & Definition**:
- **Topic**: Enzyme Kinetics & Turnover Number ($k_{\text{cat}}$)
- The catalytic turnover number ($k_{\text{cat}}$) is the maximum number of substrate molecules converted to product per active site per unit time:
$k_{\text{cat}} = \frac{V_{\max}}{[E]_t}$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- $V_{\max} = 1800\text{ }\mu\text{moles L}^{-1}\text{ min}^{-1} = \frac{1800}{60}\text{ }\mu\text{mol L}^{-1}\text{ s}^{-1} = 30\text{ }\mu\text{M s}^{-1}$
- Total enzyme concentration $[E]_t = 1.5\text{ }\mu\text{M}$
- **Calculation of $k_{\text{cat}}$**:
$k_{\text{cat}} = \frac{30\text{ }\mu\text{M s}^{-1}}{1.5\text{ }\mu\text{M}} = 20\text{ s}^{-1}$
**3. Final Answer**:
- **Calculated Value**: **20 s⁻¹**
- **Official GATE Accepted Range**: **20 to 20**
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