Question

Which of the following plot(s) is(are) CORRECT for an enzyme obeying Michaelis–Menten kinetics, assuming that [S] $\ll K_m$?

• A. A
• B. B
• C. C
• D. D

Hint:

If substrate concentration is much lower than $K_m$, the enzyme operates in the first-order region where $v_0$ increases linearly with [S].

Answer 1

The Correct Option is A, D

Step 1: Michaelis–Menten behaviour when [S] $\ll K_m$
When substrate concentration is very low compared to $K_m$, the Michaelis–Menten equation simplifies to:
\[ v_0 = \frac{V_{\max}}{K_m} [S] \] This means that the reaction rate is directly proportional to substrate concentration. Hence the plot must be a straight line.
Step 2: Examination of each plot
(A) Shows a straight-line increase of $v_0$ with [S]. This is exactly what is expected. Hence correct.
(B) Sigmoidal behaviour indicates cooperative binding and allosteric enzymes, not Michaelis–Menten enzymes. Hence incorrect.
(C) Exponential increase is not predicted by enzyme kinetics at low [S]. Hence incorrect.
(D) This plot shows substrate concentration decreasing with time. Any enzymatic reaction consumes substrate, so such a plot is always correct regardless of $K_m$. Hence correct.
Thus the correct answers are A and D.