Question

Which of these is correct for mixed inhibition?

(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 had been asked this question during an interview for a job.

This interesting question is from Fundamentals topic in section Enzyme Kinetics Fundamentals of Enzyme Technology

Answer 1

Correct option is (d) Lowers to 0.67 Vmax and Km increases to 2Km

The explanation is: Mixed inhibition is said to occur when both EI and ESI complexes are formed. The rate equation is this case is given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{[I]}{K_i})+[S](1+\frac{[I]}{K_i’})}\)

Where

Km = Miachelis Menton constant, Ki = Dissociation constant for EI complex, [I] = Inhibitor concentration, S = Substrate concentration and Ki‘ = Dissociation constant for ESI complex

Now, if Ki = [I] = 0.5 Ki‘, then Vmax lowers to 0.67 and Km increases to 2Km.