Discussion
For a simple enzyme that follows Michaelis-Menten kinetics, kinetic data was collected in the absence (dotted line, -I), or presence (solid line, +I) of an uncompetitive inhibitor (I). Which one of the following Eadie-Hofstee plots best describes the expected result?
(\(v_{o}\) = initial velocity, [S] = free substrate concentration)
- A.
![Eadie-Hofstee plot of v0 against v0/[S]: the solid +I line has the same y-intercept as the dotted -I line but a larger x-intercept](/uploads/question-images/gate-xl-2026/q40-a.png)
- B.
![Eadie-Hofstee plot of v0 against v0/[S]: the solid +I line lies below and roughly parallel to the dotted -I line, with smaller intercepts on both axes](/uploads/question-images/gate-xl-2026/q40-b.png)
- C.
![Eadie-Hofstee plot of v0 against v0/[S]: the solid +I line starts lower on the y-axis than the dotted -I line, is shallower and crosses it before reaching a larger x-intercept](/uploads/question-images/gate-xl-2026/q40-c.png)
- D.
![Eadie-Hofstee plot of v0 against v0/[S]: the solid +I line starts lower on the y-axis than the dotted -I line and meets it at the same x-intercept](/uploads/question-images/gate-xl-2026/q40-d.png)
Answer: D
The expected Eadie-Hofstee plot is (D): the two lines meet on the x-axis, and the inhibited line is lower and less steep.
An Eadie-Hofstee plot graphs \(v_{0}\) against \(v_{0}/[S]\):
\[ v_{0} = V_{max} - K_{m}\frac{v_{0}}{[S]} \]
The y-intercept is \(V_{max}\), the slope is \(-K_{m}\) and the x-intercept is \(V_{max}/K_{m}\). An uncompetitive inhibitor binds only the enzyme-substrate complex and divides both \(V_{max}\) and \(K_{m}\) by the same factor. So the y-intercept falls, the line becomes less steep, and the x-intercept \(V_{max}/K_{m}\) stays where it was.
Plot (B) has parallel lines (same slope, so unchanged \(K_{m}\)), the pattern of a pure non-competitive inhibitor. Plot (A) keeps the y-intercept, as a competitive inhibitor would, but moves the x-intercept the wrong way. Plot (C) changes the x-intercept, so \(V_{max}/K_{m}\) is not conserved.
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