Question

Given an enzyme with Km of 0.25mM, at what substrate concentration will the velocity of the enzyme reach 1/2 of the Vmax? (Vmax = 100 mmol/s)

(a) 0.5 mM

(b) 0.25 mM

(c) 0.75 mM

(d) 25 mM

This question was posed to me in quiz.

This intriguing question originated from Kinetics of Enzyme Catalysed Reaction in section Enzyme Kinetics of Enzyme Technology

Answer 1

The correct answer is (b) 0.25 mM

To explain I would say: \(V_0 = \frac{1}{2} V_{max}\)

\(V_0 = \frac{1}{2}*100\)

V0 = 50 mmol/s

\(V_0 = \frac{V_{max} [S]}{K_m+[S]}\)

50 = \(\frac{100 [S]}{0.25+[S]}\)

\(\frac{1}{2} = \frac{[S]}{0.25+[S]}\)

0.25+[S]=2[S]

[S]=0.25 mM.