Two different enzymes are able to catalyze the same reaction, but differ their K

Question

Two different enzymes are able to catalyze the same reaction, but differ their Kn the substrate A. For enzyme I, the Kin is I ,0 niM; for enzym When enzyme I was incubated with 0.1 mM A it was observed that B was pro 0. AB. They both have the same V ma duced at a rate of 0020 mmoles/minute. a) What is the value of the Vmax of the 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 producti of B when enzyme l is incubated with 1 M (ie, l 000 mM) A?

Explanation / Answer

Answer:

a. 0.022 mmol/min;

b. 0.0022 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 = 0.0022 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.

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