![Q. 43 An enzyme follows Michaelis-Menten kinetics with substrate S. The fraction of the maximum velocity ( 𝑽max ) will be observed with the substrate concentration [ S ] = 𝟒𝑲𝐦 is ____ (correct to one decimal place). ( 𝑲𝐦 is Michaelis-Menten constant)](https://www.letstalkacademy.com/wp-content/uploads/2026/01/slide_43-2.png)
Q. 43 An enzyme follows Michaelis-Menten kinetics with substrate S. The fraction of the maximum
velocity ( 𝑽max ) will be observed with the substrate concentration [ S ] = 𝟒𝑲𝐦 is
____ (correct to one decimal place). ( 𝑲𝐦 is Michaelis-Menten constant)
The Michaelis-Menten equation describes enzyme kinetics where velocity v = Vmax[S]/(Km+[S]). For [S] = 4Km, this yields a specific fraction of maximum velocity Vmax. The correct answer to Q.43 is 0.8 (or 80%).
Correct Answer Calculation
Substitute [S] = 4Km into the Michaelis-Menten equation:
v = (Vmax ⋅ 4Km)/(Km + 4Km) = (4Vmax Km)/(5Km) = (4/5)Vmax = 0.8Vmax
Thus, the fraction is 0.8, correct to one decimal place. At [S] = Km, v = 0.5 Vmax by definition; higher [S] approaches 1.0 asymptotically.
Key Concepts in Michaelis-Menten Kinetics
Km represents substrate concentration at half Vmax, reflecting enzyme-substrate affinity. The equation assumes steady-state conditions with no inhibitors.
- Low [S] << Km: v ≈ (Vmax / Km) [S] (first-order kinetics).
- [S] = Km: v = 0.5 Vmax (standard reference).
- High [S] >> Km: v ≈ Vmax (zero-order kinetics).
Common Fraction Values Table
| [S]/Km Ratio | v/Vmax Fraction | Decimal (1 place) | Percentage |
|---|---|---|---|
| 1 | 1/2 | 0.5 | 50% |
| 2 | 2/3 | 0.7 | 67% |
| 4 | 4/5 | 0.8 | 80% |
| 9 | 9/10 | 0.9 | 90% |
| 19 | 19/20 | 0.95 | 95% |
Exam Tips for Enzyme Kinetics Questions
- Practice Lineweaver-Burk plots (1/v vs 1/[S]) to visualize: intercept 1/Vmax, slope Km/Vmax.
- Relate to real enzymes like hexokinase (low Km, high affinity).
This solves GATE/NEET-style biotech queries precisely for molecular biology students.
