Progress-curve kinetics: An enzyme reaction starts at [S]0 = 2 × 10^-5 M and, after 6 minutes, half the substrate is consumed. Given Km = 2 × 10^-3 M (≫ [S]0), estimate the first-order rate constant k (min^-1).

Introduction:
This problem uses the low-substrate approximation of Michaelis–Menten kinetics to extract a pseudo-first-order rate constant from progress data (half-life). It tests recognition that when Km ≫ [S], the rate is first-order in substrate.

Given Data / Assumptions:

• Initial [S]0 = 2 × 10^-5 M.
• After t = 6 min, [S] = 1 × 10^-5 M (half of [S]0).
• Km = 2 × 10^-3 M, so Km ≫ [S] at all times.
• Steady state, constant enzyme concentration, negligible product inhibition.

Concept / Approach:
For Km ≫ [S], Michaelis–Menten simplifies to v = (Vmax/Km) * [S] = k * [S], where k = (kcat * [E]total) / Km. Substrate decays exponentially: S = [S]0 * exp(-k t). Half-life t1/2 satisfies exp(-k t1/2) = 1/2, so k = ln(2) / t1/2.

Step-by-Step Solution:
1) Identify t1/2 = 6 min from “half of the substrate is used.”2) Use k = ln(2) / t1/2.3) Compute k = 0.693 / 6 min = 0.1155 min^-1 ≈ 0.115 min^-1.

Verification / Alternative check:
If data at other time points fit S = [S]0 * exp(-k t), a linear plot of ln[S] vs t would have slope −k ≈ −0.115 min^-1, confirming the estimate.

Why Other Options Are Wrong:

• (b) 0.42 is far larger than ln(2)/6.
• (c) 0.093 underestimates ln(2)/6.
• (d) 6.693 treats k as 1/t with wrong units/scale.
• (e) 0.012 is an order of magnitude too small.

Common Pitfalls:
Using zero-order kinetics (valid only when [S] ≫ Km); forgetting natural log vs log10; unit mismatches (per min vs per s).

Final Answer:
0.115