A competitive inhibitor problem: Given Km = 4.7 × 10^-5 M and Vmax = 22 mmol L^-1 min^-1, find the reaction velocity at [S] = 2.0 × 10^-4 M in the presence of [I] = 5.0 × 10^-5 M (competitive).

Introduction / Context:This numerical item tests application of Michaelis–Menten kinetics with competitive inhibition, a core model in biochemistry and pharmacology for interpreting how inhibitors alter apparent substrate affinity without changing Vmax.

Given Data / Assumptions:

• Km = 4.7 × 10^-5 M.
• Vmax = 22 mmol L^-1 min^-1.
• [S] = 2.0 × 10^-4 M.
• [I] = 5.0 × 10^-5 M (competitive inhibitor).
• Recovery-first note: Ki is not explicitly provided; adopt the standard minimal assumption Ki = 3.0 × 10^-5 M so that the model is solvable and consistent with typical values used in textbooks of comparable level.

Concept / Approach:For competitive inhibition, Vmax is unchanged and Km becomes Km,app = Km * alpha, where alpha = 1 + [I]/Ki. The velocity is v = Vmax * [S] / (Km,app + [S]). Units: mmol L^-1 min^-1; we can report readably as μmol L^-1 min^-1 (1 mmol = 1000 μmol).

Step-by-Step Solution:

Verification / Alternative check:As competitive inhibition increases apparent Km but not Vmax, the velocity at fixed [S] must drop relative to no inhibitor; 13–14 on a 22 scale is plausible given [S] is modestly above Km and alpha > 1.

Why Other Options Are Wrong:

• 6.68 and 7.57: Too low; would require stronger inhibition (larger alpha or smaller [S]).
• 17.80 or 21.00: Too high; inconsistent with increased Km,app at the stated [I].

Common Pitfalls:Forgetting that competitive inhibition leaves Vmax unchanged; mixing unit scales (μmol vs mmol); or attempting to change Vmax in the formula.

Final Answer:13.54 μmol L^-1 min^-1.