uncompetitive
Summary
An uncompetitive inhibitor binds only to the enzyme–substrate complex, not to free enzyme, so it cannot be outcompeted by more substrate. It lowers both Vmax and apparent Km by the same factor, which gives parallel lines on a Lineweaver–Burk plot.
Detail
Meaning: The inhibitor binds the ES complex at a site separate from the active site and traps it in a form that cannot make product. Removing ES lowers the usable enzyme, so Vmax falls. Removing ES also pulls E + S toward ES (Le Châtelier), so apparent substrate affinity rises and apparent Km falls.
Equation: 1/v = (Km/Vmax)(1/[S]) + α′/Vmax, with α′ = 1 + [I]/Ki′; v is initial rate (M/s), Vmax is maximum rate (M/s), Km is the Michaelis constant (M), [S] and [I] are substrate and inhibitor concentrations (M), Ki′ is the dissociation constant of the inhibitor from ES (M), and α′ is unitless. Apparent Vmax and Km are each the original value divided by α′.
On the MCAT: A passage gives rate data or a Lineweaver–Burk plot with and without inhibitor. Uncompetitive inhibition shows parallel lines: the slope (Km/Vmax) is unchanged, the y-intercept rises, and the x-intercept (−1/Km) moves to a more negative value. On a Michaelis–Menten curve, the plateau is lower and half-maximal rate occurs at a lower [S].
Common confusion: Students mix it up with noncompetitive inhibition, where Km stays the same and Vmax falls (lines meet on the x-axis). Uncompetitive lowers both Vmax and Km. Adding substrate does not overcome it. More substrate makes more ES, which gives the inhibitor more to bind.
Sources
- Lehninger Principles of Biochemistry
- Biochemistry (Berg, Tymoczko, Stryer)
- Fundamentals of Biochemistry (Voet and Voet)
Written for the MCAT and checked before it was published. For study reference only. Spotted an error? Let us know.