Using the Michaelis–Menten framework, which expression correctly represents the initial rate in pure noncompetitive inhibition as a function of substrate [S], inhibitor I, and constants?

Difficulty: Medium

Correct Answer: Vmax * S / ((Km + S) * (1 + I/Ki))

Explanation:


Introduction / Context:
Rate laws under inhibition can be obtained by modifying the Michaelis–Menten expression with factors that reflect inhibitor binding. In pure noncompetitive inhibition, the inhibitor reduces the effective V max without changing K m.


Given Data / Assumptions:

  • Pure noncompetitive: Ki for E and ES are equal, so only V max is scaled.
  • Initial-rate regime with constant [E] total and steady-state assumptions.
  • Symbols: Vmax, Km, inhibitor concentration I, inhibition constant Ki.


Concept / Approach:
The effect is Vmax,app = Vmax / (1 + I/Ki) while Km,app = Km. Substitute these into v = (Vmax,app * S) / (Km,app + S).


Step-by-Step Solution:

Start with v = (Vmax * S) / (Km + S).Apply pure noncompetitive scaling: Vmax → Vmax/(1 + I/Ki).Keep Km unchanged.Obtain v = Vmax * S / ((Km + S) * (1 + I/Ki)).


Verification / Alternative check:
Lineweaver–Burk form gives 1/v = (1 + I/Ki) * ((Km + S) / (Vmax * S)), showing increased slope with unchanged x-intercept, consistent with decreased V max and constant K m.


Why Other Options Are Wrong:

  • Forms that alter Km but not Vmax correspond to competitive or mixed inhibition, not pure noncompetitive.
  • rmax * S / Km and Vmax / (1 + Km/S) are not valid Michaelis–Menten forms for inhibited systems.


Common Pitfalls:
Confusing symbols (Vmax vs rmax) or placing the inhibition factor in the numerator instead of the denominator.


Final Answer:
Vmax * S / ((Km + S) * (1 + I/Ki))

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