Q.50 Which of the following plot(s) is(are) CORRECT for an enzyme that obeys
Michaelis-Menten kinetics, assuming [𝑆] ≪ 𝐾_m?
[𝑆] is the concentration of the substrate, 𝐾_m is the Michaelis constant, and v0 is
the initial reaction velocity.

Introduction

Enzyme kinetics is a key topic in CSIR NET, GATE Biotechnology and Biochemistry exams.
A commonly tested concept is how enzyme reaction velocity behaves when the substrate concentration [S] is far lower than the Michaelis constant Km ([S] ≪ Km).

Under these circumstances, the Michaelis–Menten equation simplifies to a linear relationship between initial reaction velocity (v₀) and substrate concentration ([S]). Let’s analyze the given plots to find the correct one.

Correct Answer

✔ Option (A) only

Explanation of Each Option

(A) Linear Increase of v₀ with [S] — ✔ Correct

When [S] ≪ Km, the Michaelis–Menten equation:

v₀ = (Vmax[S])/(Km + [S])

approximates to:

v₀ ≈ (Vmax/Km)[S]

This shows a linear proportionality between reaction rate and substrate concentration.
Therefore, Plot A correctly represents this relationship.

(B) Hyperbolic Saturation Curve — ✘ Incorrect

This curve represents the entire Michaelis–Menten response at rising [S], where v₀ approaches Vmax.
However, the question restricts the case to [S] ≪ Km, where saturation does not occur.
Hence, Plot B is not applicable.

(C) Exponential-like Increase — ✘ Incorrect

This graph implies accelerating rate increase, which does not occur in low substrate conditions.
Here the reaction is first-order and linear, not exponential.
Thus, Plot C is incorrect.

(D) Decline of [S] Over Time — ✘ Incorrect

This plot shows substrate concentration decreasing over time, which is unrelated to the required comparison of v₀ vs [S].
Therefore, Plot D does not address the question.

Conclusion

For enzymes following Michaelis–Menten kinetics under conditions where [S] ≪ Km, reaction velocity shows:

v₀ ∝ [S]

Thus, the only correct graph representing this behavior is:

✔ Option (A)