A cholinesterase enzyme was tested in the absence and presence of an inhibitor, McBlocker. The data were transformed and plotted as a Lineweaver-Burk plot, where: The Lineweaver-Burk plot showed that both control and McBlocker lines had similar y-intercepts. However, the line for McBlocker had a steeper slope and a less negative x-intercept than the control. Which one of the following interpretations of these results is MOST appropriate? McBlocker:

Question

BlackTom AI · Accepted Answer

Lineweaver–Burk linear form: 1/v = (Km/Vmax)(1/[S]) + 1/Vmax, so the y-intercept equals 1/Vmax, the slope equals Km/Vmax, and the x-intercept equals −1/Km.

1) Relevant concepts:
- Y-intercept (1/Vmax) reports Vmax. If y-intercepts are the same for control and inhibitor, Vmax is unchanged.
- Slope is Km/Vmax. If the slope increases while Vmax is unchanged, Km must increase.
- X-intercept is −1/Km. A “less negative” x-intercept (closer to zero) corresponds to a larger Km (because −1/Km magnitude decreases as Km increases).

2) Step-by-step reasoning:
- The problem states both lines have similar y-intercepts → 1/Vmax unchanged → Vmax is unchanged in the presence of McBlocker.
- The McBlocker line has a steeper slope → Km/Vmax is larger. With Vmax unchanged, Km must be larger.
- The McBlocker line’s x-intercept is less negative → −1/Km is closer to zero → Km is larger. Both slope and x-intercept observations consistently indicate an increased Km with unchanged Vmax.

3) Numerical illustration (intermediate steps):
- Assume control: Vmax = 1 so 1/Vmax = 1 (y-intercept); slope = Km/Vmax = 1 → Km = 1; x-intercept = −1/Km = −1.
- With inhibitor: y-intercept still 1 (Vmax still 1); slope = 2 → Km = slope×Vmax = 2; x-intercept = −1/2 = −0.5 (less negative). This matches the described plot behavior.

4) Why this indicates competitive inhibition:
- Competitive inhibitors increase apparent Km (need more substrate to reach same velocity) but do not affect Vmax (sufficient substrate outcompetes inhibitor). That signature is exactly “Km increases, Vmax unchanged.”

5) Why other choices are wrong:
- Noncompetitive inhibition: Vmax decreases (y-intercept changes) while Km stays the same — not observed here.
- Mixed inhibition: both Km and Vmax change — here Vmax is unchanged.
- Uncompetitive inhibition: both Km and Vmax decrease, producing parallel LB lines — not the case.

Conclusion: The correct interpretation is that McBlocker increases Km but does not change Vmax, consistent with competitive inhibition, 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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