Within the classic Michaelis–Menten framework, which step is taken to be the slow, rate-limiting “substrate-consuming” step?
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AConversion of ES to product (E + P), i.e., the substrate-consuming step
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BRelease of product from the active site after chemistry is complete
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CFormation of the enzyme–substrate (ES) complex from E + S
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DNone of these
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EDiffusional encounter between E and S in bulk solution
Answer
Correct Answer: Conversion of ES to product (E + P), i.e., the substrate-consuming step
Explanation
Introduction / Context:The Michaelis–Menten model assumes a particular hierarchy of rates to obtain a tractable rate law. Knowing which step is taken as slow clarifies why Vmax relates to kcat and why Km relates to binding and catalytic parameters.
Given Data / Assumptions:
- Simple mechanism: E + S ⇌ ES → E + P.
- Either rapid equilibrium (Michaelis–Menten) or steady-state (Briggs–Haldane) approximations are invoked.
Concept / Approach:In the classic picture, chemistry—conversion of ES to E + P—is the slow, rate-limiting step. ES formation and breakdown back to E + S are faster processes that establish a quasi-steady distribution, enabling derivation of v = (Vmax * [S])/(Km + [S]).
Step-by-Step Solution:
Identify the “substrate-consuming” step: ES → E + P, which removes substrate from the pool.Associate this step’s rate constant with kcat (turnover number).Under this assumption, Vmax = kcat * [E]_total and Km reflects a composite of rate constants.Verification / Alternative check:When product formation is artificially accelerated or slowed (mutations, pH changes), Vmax shifts accordingly, consistent with chemistry being rate limiting.
Why Other Options Are Wrong:
- Product release can be rate-limiting in some enzymes but is not the generic assumption of the classic model.
- ES formation is not assumed to be the slow step in Michaelis–Menten derivations.
- “None” and pure diffusion control are outside the classic simplifying assumption.
Common Pitfalls:Equating steady-state with slow binding, or assuming product release always limits rate.
Final Answer:Conversion of ES to product (E + P), i.e., the substrate-consuming step.