Under which kinetic circumstance is classical (pure) noncompetitive inhibition most cleanly observed in steady-state enzyme studies?

Difficulty: Medium

Correct Answer: Rapid equilibrium conditions with equal affinity for E and ES

Explanation:


Introduction / Context:
Classical noncompetitive (the pure case) is a limiting scenario of mixed inhibition where the inhibitor binds free enzyme (E) and the enzyme–substrate complex (ES) with equal affinity. Experimentally, the cleanest expression of this behavior appears under rapid equilibrium assumptions.


Given Data / Assumptions:

  • Rapid establishment of binding equilibria relative to catalysis.
  • Ki for E equals Ki for ES (Ki,E = Ki,ES).
  • Steady-state conditions for initial-rate analysis.


Concept / Approach:
With rapid equilibrium and equal affinities, the apparent K m stays constant while V max decreases by 1/(1 + I/Ki). This yields Lineweaver–Burk lines intersecting on the x-axis (constant −1/Km) and increasing slopes with inhibitor.


Step-by-Step Solution:

Specify the defining equality: Ki,E = Ki,ES.Assume rapid equilibrium: binding adjusts quickly compared with catalysis.Predict kinetic outcome: Vmax decreases; Km unchanged.Thus, choose “Rapid equilibrium conditions with equal affinity for E and ES.”


Verification / Alternative check:
Global fitting of v vs [S] data to the pure noncompetitive model reproduces unchanged x-intercepts with decreased y-intercepts, matching the rapid-equilibrium prediction.


Why Other Options Are Wrong:

  • Slow/non-equilibrium: complicates interpretation; does not yield the textbook pattern cleanly.
  • Moderate equilibrium: vague and not defining the pure case.
  • Single-turnover: outside the steady-state initial-rate regime.


Common Pitfalls:
Confusing pure noncompetitive with mixed inhibition where Ki,E ≠ Ki,ES, which alters Km as well as Vmax.


Final Answer:
Rapid equilibrium conditions with equal affinity for E and ES

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