Michaelis–Menten mechanism: What is the rate-determining step under the usual assumption k2 ≪ k−1?

Introduction:
Classic Michaelis–Menten kinetics assumes a mechanism E + S ⇌ ES → E + P with the chemical conversion (k2) slower than ES dissociation (k−1). Under this common condition, the product formation step is rate-determining.

Given Data / Assumptions:

• Steady-state approximation: d[ES]/dt ≈ 0 after a brief pre-steady-state period.
• k2 ≪ k−1, which makes chemistry slower than ES breakdown to E + S.
• No product inhibition and single-substrate reaction.

Concept / Approach:
When k2 is the smallest forward rate constant, the overall turnover is gated by ES → E + P. Thus Vmax = kcat * [E]total where kcat ≈ k2 under this mechanism.

Step-by-Step Solution:
1) Mechanism: E + S ⇌ ES (k1, k−1) followed by ES → E + P (k2).2) If k2 is slow, each catalytic cycle’s duration is dominated by the chemistry step.3) Therefore, the product formation step determines the maximal rate; increasing [S] saturates ES but cannot exceed the limit set by k2.

Verification / Alternative check:
At saturation, V = Vmax = k2 * [E]total. Experimentally, turnover numbers (kcat) measured at saturating substrate reflect the rate-determining chemical step, corroborating that ES → E + P is limiting when k2 is smallest.

Why Other Options Are Wrong:

• (a) ES formation is usually fast under typical MM assumptions.
• (b) ES dissociation to E + S is faster than chemistry in the stated regime.
• (d) Only one step (chemistry) is rate-determining here.
• (e) Inhibitor binding is not part of the uninhibited MM mechanism.

Common Pitfalls:
Confusing pre-steady-state binding steps with overall steady-state turnover; assuming ES formation is rate-limiting even at high [S].

Final Answer:
Product formation step (ES → E + P).