For a Michaelis–Menten enzyme, what fraction of Vmax is observed when the substrate concentration [S] equals 2 × Km?

Introduction / Context:Understanding how velocity depends on [S] relative to Km is central to predicting reaction rates and designing assays. This conceptual question checks whether you can manipulate the Michaelis–Menten expression without arithmetic complexity.

Given Data / Assumptions:

• v = Vmax * [S] / (Km + [S]).
• We are given [S] = 2 Km.

Concept / Approach:Substitute [S] = 2 Km directly into the equation and simplify as ratios; no absolute numbers are needed.

Step-by-Step Solution:

Verification / Alternative check:At [S] = Km, v = 0.5 Vmax; increasing [S] to 2 Km should raise v above 0.5 but still below Vmax, consistent with 0.66.

Why Other Options Are Wrong:

• 0.09 and 0.33: Too low for [S] > Km.
• 0.91: Approaches saturation; would require [S] ≫ Km.
• 0.50: Occurs at [S] = Km, not 2 Km.

Common Pitfalls:Adding rather than dividing; or assuming linearity at all [S].

Final Answer:0.66.