Question

The following data was obtained for an enzyme in the absence of an inhibitor, and in the presence of two different inhibitors. The concentration of each inhibitor was 10 mM. The total concentration of enzyme was the same for each experiment.

[S] {mM}	without inhibitor v, {umol/(ml*s)}	with inhibitor A v, {umol/(ml*s)}	With inhibitor B v, {umol/(ml*s)}
0.0	0.0	0.0	0.0
1.0	3.6	3.2	2.6
2.0	6.3	5.3	4.5
4.0	10.0	7.8	7.1
8.0	14.3	10.1	10.2
12.0	16.7	11.3	11.9

Determine Km and Vmax for the uninhibited and inhibited enzyme. Make a Lineweaver- Burk Plot showing enzyme activity in the presence and absence of inhibitor. Graph should contain labels. Provide tables containing data analysis. USE EXCEL

Determine the type of inhibition for the two experiments that contain inhibitor. Provide an explanation for your answer.

Answer 1

The data is shown below

The graphs are plotted and linear fitting was performed. The obtained parameters are also shown with the graph.

Intercept = 1/Vmax. Then Vmax = 1/intercept.

For Without inhibitor, Vmax = 1/0.0402 = 24.88 V

For inhibitor A, Vmax = 1/0.0677 = 14.77 V

For inhibitor B, Vmax = 1/0.0577 = 17.33 V

Slope = km/Vmax => km = Slope x Vmax

For Without inhibitor, km = 0.2375 x 24.88 = 5.909

For inhibitor A, km = 0.2441 x 14.77 = 3.6054

For inhibitor B, km = 0.3275 x 17.33 = 5.6756

VA and VB are competitive inhibitor. The reason is that the two curves intersect at a point and the slopes are different.