Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
An enzymatic reaction is described by the following rate expression: v = (Vmax * [S]) / (Km + [S] + [S]^2 / Ki). Given Vmax = 10 µmol/(min.mg), Km = 1 mM, and Ki = 4 mM, the substrate concentration [S] at which the reaction velocity achieves its maximum value is ________ mM.
Options:
(A) Curve showing substrate inhibition with a maximum at intermediate [S] followed by a decline at high [S]
(C) Sigmoidal curve characteristic of positive cooperativity
(D) Linear increase in velocity with increasing substrate concentration
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2012 Q.46)**
**1. Governing Concept**:
- **Topic**: Enzyme Kinetics & Substrate Inhibition (Haldane Model)
- The given rate equation is the canonical Haldane substrate inhibition model:
$v = rac{V_{\max} [S]}{K_m + [S] + rac{[S]^2}{K_i}}$
- At low substrate concentrations ($[S] \ll \sqrt{K_m K_i}$): The $[S]^2/K_i$ term is negligible, and the velocity increases with $[S]$ similarly to standard Michaelis-Menten kinetics.
- At the optimal substrate concentration $[S]_{ ext{opt}} = \sqrt{K_m K_i}$: The reaction velocity reaches its maximum peak.
- At high substrate concentrations ($[S] \gg \sqrt{K_m K_i}$): The $[S]^2/K_i$ term dominates the denominator, causing the velocity to decline progressively toward zero due to unproductive dead-end complex formation ($E S_2$).
**2. Option Analysis**:
- **Option (A) is CORRECT**: Accurately depicts the characteristic bell-shaped substrate inhibition curve (rising to a peak then decreasing at high $[S]$).
- **Options (B), (C), and (D) are INCORRECT**: Do not account for the inhibitory decrease in rate at elevated substrate concentrations.
**3. Final Answer**:
- **Correct Choice**: **A**
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