Figure 3.3 in SKD 3rd Edition shows the effect of initial substrate concentration on the initial reaction rate (v) for a constant total enzyme concentration ([E0]). Now consider a plot of initial reaction rate (v) vs. total enzyme concentration ([E0]) for constant initial substrate concentration ([S0]). Which of the following describes the shape that would be obtained for such a plot?

Question

BlackTom AI · Accepted Answer

Michaelis–Menten kinetics relates the initial rate v to enzyme and substrate through the enzyme–substrate complex: v = kcat[ES]. Two key limits give the shape of v versus total enzyme concentration [E0] when initial substrate [S0] is held constant. Relevant concepts - v = kcat[ES]; the concentration of ES depends on [E0] and free substrate [S]. - Under the usual MM steady-state (when [E0] << [S0]): [ES] ≈ [E0][S0]/(Km + [S0]), so v ∝ [E0]. - If [E0] becomes very large relative to [S0], substrate becomes limiting and the maximum possible [ES] is [S0], so v approaches kcat[S0] (a plateau). Step-by-step reasoning with intermediate steps 1. General MM relation (steady-state approximation, assuming free S ≈ S0 when enzyme is small): [ES] ≈ [E0][S0] / (Km + [S0]) therefore v ≈ kcat [E0] [S0] / (Km + [S0]). This is linear in [E0] with slope kcat[S0]/(Km + [S0]) — so for low [E0] the plot is a straight line through v = 0 at [E0] = 0. 2. Consider the opposite extreme: if [E0] >> [S0], most enzyme molecules cannot be bound because there are not enough substrate molecules. The maximum possible [ES] ≤ [S0]. Thus v_max_limited_by_substrate = kcat [ES]_max ≈ kcat [S0]. Increasing [E0] further cannot increase [ES] beyond [S0], so v approaches this constant value (a plateau). Why the correct answer is correct - The described curve (linear rise at low [E0], then slower increase and plateau at high [E0]) follows directly from the two regimes above: proportionality when enzyme is limiting, and substrate-limited saturation when enzyme is in excess. Why the other choices are incorrect - “Horizontal line; [E0] has no effect on v.” contradicts the proportional relationship at low [E0]; v clearly increases with added enzyme until substrate limits it. - “Straight line with positive slope … v increases linearly with [E0].” is only true when [E0] is small relative to [S0]; it ignores the substrate-limiting plateau at high [E0]. - “None of the above.” is incorrect because one option correctly describes the two-regime behavior. Conclusion The correct description is: for low [E0] the plot is linear through v = 0, but at high [E0] the rate increases more slowly and approaches a maximum plateau — which matches the provided answer.

类似问题

Which of the following statements about V0, Vmax, [S] and Km values is not true?

The Vmax of an enzyme is:

Is the following statement true or false?'There is an inverse relationship between the Km and an enzyme’s affinity for its substrate'.

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