Question

The rate equation for _______________ inhibition is given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{[P]}{K_p})+[S]}\).

(a) substrate

(b) non-competitive

(c) product

(d) competitive

The question was posed to me in class test.

Question is from Fundamentals in chapter Enzyme Kinetics Fundamentals of Enzyme Technology

Answer 1

Right choice is (c) product

Explanation: Product inhibition is case of competitive inhibition, where the substrate and the inhibitor compete for the active site of the enzyme. The inhibitors in this case bear structural similarity to the substrate. This causes substantial loss of productivity and hence the rate equation can be given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{[P]}{K_p})+[S]}\)

Where,

Km = Miachelis Menton constant, Kp = Dissociation constant for product, P = Product concentration and S = Substrate concentration.