An enzyme has a Km of 2 mM. At a substrate concentration of 2 mM, what fraction of Vmax is the reaction velocity?
AOne quarter (25%)
BOne half (50%)
CTwo thirds (67%)
DFully saturated (100%)
By definition, Km is the substrate concentration at which v = Vmax/2. Substituting [S] = Km into the Michaelis-Menten equation: v = Vmax × Km / (Km + Km) = Vmax/2. This is the most direct conceptual test of what Km means.
Question 2 True / False
A lower Km value usually means the enzyme has a higher binding affinity for its substrate.
TTrue
FFalse
Answer: False
Km reflects substrate affinity only under the rapid-equilibrium assumption (where kcat << k-1). In the general steady-state derivation, Km = (k-1 + kcat) / k1, so it incorporates the catalytic rate as well. An enzyme with fast catalysis (high kcat) can have a high Km despite tight binding. Km is an operational parameter, not a pure binding constant.
Question 3 Short Answer
Why does the Lineweaver-Burk (double reciprocal) plot transform the Michaelis-Menten curve into a straight line, and what kinetic parameters can be read directly from it?
Think about your answer, then reveal below.
Model answer: Taking the reciprocal of both sides of v = Vmax[S]/(Km + [S]) gives 1/v = (Km/Vmax)(1/[S]) + 1/Vmax, which is linear in 1/[S]. The y-intercept equals 1/Vmax and the x-intercept equals -1/Km, allowing both parameters to be extracted graphically.
The hyperbolic Michaelis-Menten equation is hard to fit by eye; linearizing it via double reciprocal allows Km and Vmax to be estimated from a straight-line plot. Understanding this transform connects enzyme kinetics to the broader skill of linearizing non-linear equations, a technique used throughout quantitative biology.