Question

What happens to Vmax and Km in case of uncompetitive inhibition when I = Ki‘?

(a) Lowers to 0.5 Vmax and 0.5 Km

(b) Vmax is unchanged and Km increases 2Km

(c) Lowers to 0.5 Vmax and Km remains unchanged

(d) Lowers to 0.67 Vmax and Km increases to 2Km

I got this question in an international level competition.

I want to ask this question from Fundamentals in portion Enzyme Kinetics Fundamentals of Enzyme Technology

Answer 1

Correct choice is (a) Lowers to 0.5 Vmax and 0.5 Km

The best explanation: Uncompetitive inhibition occurs during multi-substrate reaction where the inhibitor is competitive to one substrate and uncompetitive to the other. The inhibition cannot be overcome as Vmax and Km are equally reduced. The rate equation is given by, \(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i’}}\right ) [S]}{\left (\frac{K_m}{1+\frac{[I]}{K_i’}}\right )+[S]}\).

When I = Ki‘ in this equation, the Vmax and Km is reduced to half.