Question

Using 1.0 nM enzyme X for the enzyme kinetics experiments, you obtained the reaction rates at different concentrations of substrate (shown in the following table). Assume that the enzyme X follows a typical Michaelis-Menten kinetics.

[substrate] (μM) Reaction rate (μmol/min)

2 7

5 17

10 29

20 44

50 67

100 80

200 89

500 95

1000 98

a) Estimate the Vmax and Km values of the enzyme in this reaction using double reciprocal plot.

b) If 5.0 nM of the enzyme X is used in the experiment, what will be the Vmax and Km of the enzyme X?.

Answer 1

a) So I have taken reciprocal of both data series and plotted the following,

1/[S] (micromolar-1)     1/[Vo] (min/micromolar)

0.5                                         0.143

0.2                                         0.059

0.1                                         0.034

0.05                                       0.023

0.02                                       0.015

0.01                                       0.0125

0.005                                     0.0112

0.002                                     0.0105

0.001                                     0.0102

plot 1/[S] on x-axis and 1/[Vo] on the y-axis

Take intercept on both axis,

x-axid intercept = -1/Km = 0.005

So, Km = -200

y-axid intercept = 1/Vmax = 0.09

So, Vmax = 11.11 micromolar/min

b) Vmax is directly proporational to enzyme ocncentration

So, when we increase the concentration of enzyme from 1.0 nM to 5.0 mM, that is 5-times of the original values,

the Vmax = 5 x 11.11 = 55.55 micromolar/min

Km however does not change with the enzyme concentration and remains the same.