43. An enzyme has a Vmax of 50 mol product formed (minute X mg protein)-1 and a
Km of 10 M for the substrate. When a reaction mixture contains the enzyme and
5
M substrate, which of the following percentage of the maximum velocity will be
closest to the initial reaction rate?
a. 5%
b. 15%
c. 33%
d. 50%
Michaelis-Menten Kinetics Basics
The Michaelis-Menten equation gives the initial enzyme velocity as:
Given:
- \( V_{max} = 50 \, \text{mol product/min/mg protein} \)
- \( K_m = 10 \, \mu M \)
- \( [S] = 5 \, \mu M \)
Substituting the values:
The percentage of \( V_{max} \) achieved is:
Conceptual Understanding
Enzyme kinetics follows a hyperbolic saturation curve. The velocity (\(v\)) approaches \(V_{max}\) asymptotically with increasing substrate concentration. At \([S] = K_m\), the velocity equals half the maximum (\(v = 0.5 V_{max}\)).
When \([S] = 0.5 K_m\):
This confirms the fractional velocity directly without substitution.
Option-by-Option Analysis
- a. 5%: Too low. Would occur if \([S] \ll K_m\), such as 0.5 μM. Underestimates by about sixfold.
- b. 15%: Underestimated rate corresponding to \([S] \approx 1.76 \, \mu M\) (≈ 0.18 \(K_m\)). Incorrect for 5 μM substrate.
- c. 33%: Correct. At \([S] = 0.5 K_m\), fractional velocity equals \(1/3\) or 33.33% \(V_{max}\).
- d. 50%: Occurs at \([S] = K_m = 10 \, \mu M\). Overestimates actual fraction.
Final Answer: Option (c) – 33%
