Question

Example 1: Two different enzymes are able to catalyze the same reaction, A → B. They both have the same Vmax, but differ their Km 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 the enzymes? 0.022 b) What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A? 0.00022 c) What will be the rate of production of B when enzyme 1 is incubated with 1 M (i.e., 1000 mM) A? 0.022

Answer 1

Answer: (a) = 0.022mmoles/minute; (b) = 0.00022mmoles/minute; (c) = 0.022mmoles/minute

Step-by-step explanation:

The Michaelis- Menten equation is used to show the relationship between substrate concentration and reaction rate.

The equation is Vo = Vmax[S]/(Km + [S]

where Vo is the initial velocity, Vmax is the maximum velocity, [S] is the substrate concentration, Km is the Michaelis constant.

(a) For enzyme 1; Vo = 0.002mmoles/minute, Vmax = ?, [S] = 0.1mM, Km =1.0mM; making Vmax subject of the formula;

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

Vmax = 0.002*(1.0 + 0.1)/ 0.1

Vmax = 0.022mmoles/minute (Vmax for Enzyme 1&2)

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

Vo = 0.022 * 0.1/(10+0.1) = 0.00022mmoles/minute

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

Vo = 0.022*1000/(1.0+1000) = 0.022mmoles/minute