Question

The __________ inhibition gives the following rate equation.

\(V=\frac{V_{max} [S]}{K_m+[S](1+\frac{[S]}{K_s})}\)

(a) Substrate

(b) Mixed

(c) Non-competitive

(d) Competitive

This question was posed to me in a national level competition.

Query is from Fundamentals in division Enzyme Kinetics Fundamentals of Enzyme Technology

Answer 1

The correct answer is (a) Substrate

To explain: Substrate inhibition is a special case of uncompetitive inhibition which occurs in about 20% of all known enzymes. It is primarily caused by binding of parts of the substrate molecule to the subsites in the active site. At high substrate concentration, the resultant complex may become inactive causing reduction in the rate of reaction. Assuming the ESS complex may not form product, the rate equation may be given by \(V=\frac{V_{max} [S]}{K_m+[S](1+\frac{[S]}{K_s})}\)

Where,

Km = Miachelis Menton constant, Ks = Dissociation constant for substrate and S = Substrate concentration.