Q.44. An enzyme obeying Michaelis-Menten kinetics shows a reaction velocity (v) of 𝟏𝟎𝝁 𝐦𝐨𝐥/𝐦𝐢𝐧
when the substrate concentration [𝐒] equals its 𝐊𝐌. The maximal velocity 𝐕max for this enzyme is ____
𝝁𝐦𝐨𝐥/𝐦𝐢𝐧 (correct to integer number).
( 𝐊𝐌 is Michaelis-Menten constant)

The Michaelis-Menten equation describes enzyme kinetics for the reaction velocity v = ^V_max [S] / (K_m + [S]), where V_max is the maximum velocity, [S] is substrate concentration, and K_m is the Michaelis constant. In this problem, v equals 10 μmol/min when [S] = K_m, directly leading to the correct V_max value.

Core Principle

When substrate concentration [S] equals K_m, the reaction velocity reaches half of V_max, as substituting [S] = K_m into the equation yields v = V_max/2. Here, given v = 10 μmol/min at [S] = K_m, V_max must be twice that value.

Correct Answer

V_max = 20 μmol/min.
Given v = 10 = V_max/2, solving for V_max gives V_max = 20 μmol/min (integer as required).

Equation Derivation

Start with the Michaelis-Menten equation:
v = V_max [S] / (K_m + [S])
Set [S] = K_m:
v = V_max K_m / (K_m + K_m) = V_max K_m / (2 K_m) = V_max/2
Thus, V_max = 2v = 2 × 10 = 20 μmol/min.

Common Options Explained

• 10 μmol/min: Incorrect; this is v at [S] = K_m, not V_max (V_max is always higher).
• 20 μmol/min: Correct, as v = V_max/2 at K_m.
• 5 μmol/min: Incorrect; this would imply v = V_max/4, occurring at lower [S] ≈ K_m/3.
• 40 μmol/min or higher: Incorrect; overestimates, as [S] = K_m gives exactly half V_max, not a quarter or less.

Exam Tips

Plot v vs [S] shows a hyperbola approaching V_max asymptotically; K_m is [S] at v = 0.5 V_max. For MCQs, recall this hallmark property to quickly identify V_max without full calculations. Practice Lineweaver-Burk plots (1/v vs 1/[S]) where x-intercept = -1/K_m and y-intercept = 1/V_max.