GATE BT 2022 • Q50
MSQ • 2 Marks
Section 2: General Biotechnology
Biomolecules, enzyme kinetics
Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
Which of the following plot(s) is(are) CORRECT for an enzyme that obeys Michaelis-Menten kinetics in the presence of a competitive inhibitor?
Options:
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(A) 1/v vs 1/[S] shows identical Y-intercept (1/Vmax) with increased slope (apparent Km / Vmax)
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(B) v vs [S] reaches the same Vmax at saturating substrate concentration [S]
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(C) Eadie-Hofstee plot (v vs v/[S]) shows unchanged Y-intercept with more negative slope
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(D) Lineweaver-Burk plot shows parallel lines with decreased Vmax
Official Answer Key:
A,B,C
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2022 Q.50)**
**1. Governing Concept**:
- **Topic**: Enzyme Kinetics & Diagnostic Kinetic Plots
- For an enzyme obeying Michaelis-Menten kinetics:
1. **Direct Plot ($v_0$ vs $[S]$)**: Hyperbolic curve passing through the origin, asymptotically approaching $V_{\max}$.
2. **Lineweaver-Burk Plot ($1/v_0$ vs $1/[S]$)**: Straight line with y-intercept $= 1/V_{\max}$, x-intercept $= -1/K_m$, slope $= K_m/V_{\max}$.
3. **Eadie-Hofstee Plot ($v_0$ vs $v_0/[S]$)**: Straight line with y-intercept $= V_{\max}$, slope $= -K_m$.
4. **Hanes-Woolf Plot ($[S]/v_0$ vs $[S]$)**: Straight line with y-intercept $= K_m/V_{\max}$, slope $= 1/V_{\max}$.
**2. Option Analysis**:
- **Option (A) is CORRECT**: Correctly depicts the hyperbolic $v_0$ versus $[S]$ curve reaching $V_{\max}/2$ at $[S] = K_m$.
- **Option (B) is INCORRECT**: Inverts the canonical Lineweaver-Burk intercepts.
- **Option (C) is INCORRECT**: Shows sigmoidal cooperativity (allosteric Hill kinetics), which violates hyperbolic Michaelis-Menten kinetics.
- **Option (D) is CORRECT**: Accurately shows the linear reciprocal Lineweaver-Burk plot ($1/v_0$ vs $1/[S]$).
**3. Final Answer**:
- **Correct Options**: **A, D**
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