Question

An enzyme with a Km of 5mM has a reaction rate of 100 mmol/min at substrate concentration of 0.25 mmol. What is the maximum reaction rate that this enzyme can achieve when its saturated with substrate?

(a) 2100

(b) 1500

(c) 1900

(d) 9000

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This interesting question is from Kinetics of Enzyme Catalysed Reaction in division Enzyme Kinetics of Enzyme Technology

Answer 1

Right choice is (a) 2100

Easiest explanation: Miachelis Menten equation is given by \(V_0=\frac{V_{max} [S]}{K_m+[S]}\)

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

\(V_{max} = \frac{100 (5+0.25)}{0.25}\)

\(V_{max} = \frac{525}{0.25}\)

\(V_{max}\) = 2100 mmol/min.