Q.50 An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate
of 20 μM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is
_____ μM/min .

The value of V_max for the given enzyme is 40 μM/min.

Given data and concept

Substrate concentration, [S] = 4 × 10⁻⁴ M
Rate of reaction at this substrate concentration, v = 20 μM/min
Michaelis constant, K_m = 2 × 10⁻⁴ M

For a simple one‑substrate enzyme, the Michaelis–Menten equation is

v = ^{V_max [S]}_{Km + [S]}

where v is reaction velocity, V_max is maximum velocity, [S] is substrate concentration and K_m is the Michaelis constant.

Step‑by‑step numerical solution

Insert the known values (keep units consistent; both [S] and K_m are in molar):

20 = V_max × ^{4 × 10−4}₂