Question

Two different enzymes are able to catalyze the same reaction, A → B. They both have the same Vmax but differ in their KM for the substrate A. For enzyme 1, the KM is 1.0 mM; for enzyme 2, the KM is 10 mM. When enzyme 1 was incubated with 0.1 mM A, it was observed that B was produced at a rate of 0.0020 mmoles/minute.

a) What is the value of the Vmax of these enzymes?

B) What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A?

c) What will be the rate of production of B when enzyme 1 is incubated with 1 M of A?

Answer 1

Answer a. 0.022 mmol/min;

b. 0.00022 mmol/min;

c. 0.022mmol/min

Explanation:

Use the following equation Michaelis and Menton equation for a, b and C     

Vo = Vmax [S] / (Km + [S])

a) Given Km = 1 mM, [S] = 0.1 mM, Vo = 0.002 mmoles/min, Vmax = ?

Using above equation

0.002 = Vmax * (0.1) / (1 + 0.1) on solving

Vmax = 0.022 mmoles/min

b)   Vmax = 0.022 mmoles/min, Km = 10 mM, [S] = 0.1 mM, Vo = ?

Vo = 0.022* (0.1)/(10 + 0.1) on solving

Vo = 0.00022 mmoles/min

c) [S] = 1000 mM, Vmax = 0.022 (answer from A), Km = 1, Vo = ?

Vo = 0.022 (1000)/(1 + 1000).....on solving

Vo = 0.022 mmoles/min

A key point to be noted is that Vo is the same as Vmax because [S] is so high that it approaches Vmax.