Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
Which of the following statements is/are correct about an uncompetitive inhibitor
of an enzyme?
Options:
(A) It binds to the substrate binding site of the enzyme only
(B) It binds to the enzyme-substrate complex only
(C) It reduces the Vmax of the enzyme
(D) It binds to both free enzyme and enzyme-substrate complex
Official Answer Key:
B;C
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2024 Q.44)**
**1. Governing Concept**:
- **Topic**: Enzyme Kinetics & Uncompetitive Inhibition Mechanism
- In **uncompetitive inhibition**:
- The inhibitor binds **exclusively to the enzyme-substrate ($ES$) complex**, not to the free enzyme ($E$).
- Binding occurs at an allosteric site created only after substrate induces a conformational change upon binding.
- Since $ESI$ dead-end complex cannot turn over, apparent $V_{\max}$ decreases ($V_{\max}' = V_{\max} / \alpha'$).
- Depletion of $ES$ shifts the binding equilibrium toward more $ES$ formation, which artificially increases substrate affinity, lowering apparent $K_m$ ($K_m' = K_m / \alpha'$).
- Lineweaver-Burk double reciprocal plots display **parallel lines** with identical slopes ($K_m/V_{\max}$).
**2. Option Analysis**:
- **Option (A) is INCORRECT**: Competitive inhibitors bind to the free enzyme substrate-binding site; uncompetitive inhibitors do not.
- **Option (B) is CORRECT**: Uncompetitive inhibitors bind strictly to the enzyme-substrate ($ES$) complex.
- **Option (C) is CORRECT**: Uncompetitive inhibition lowers both apparent $V_{\max}$ and apparent $K_m$ by the same factor $\alpha'$.
- **Option (D) is INCORRECT**: Binding to both free enzyme and $ES$ complex characterizes non-competitive / mixed inhibition.
**3. Final Answer**:
- **Correct Options**: **B, C**
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