Question

1. (3%) You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data are obtained: [S] V0 –Inhibitor +Inhibitor 0.0001 33 17 0.0002 50 29 0.0005 71 50 0.001 83 67 0.002 91 80 0.005 96 91 0.01 98 95 0.02 99 98 0.05 100 99 0.1 100 100 0.2 100 100

a) What is the Vmax in the absence of inhibitor?

b) What is the Km in the absence of inhibitor?

c) When [S] = 0.0004, what will V0 be in the absence of inhibitor?

d) When [S] = 0.0004, what will V0 be in the presence of inhibitor?

e) What kind of inhibitor is it likely to be?

2. (2%) An enzyme is inhibited by two distinct competitive inhibitors, IA and IB. IA is shown to bind with a Ki of 13 μM while IB binds with a Ki of 150 μM.

a) Which is the more potent of the two inhibitors?

b) The natural substrate for the enzyme has a Km value of 0.1 mM. How does the substrate binding affinity compare to each of the two competitive inhibitors?

3. (2%) Assuming they have equal affinity for the enzyme, why would a noncompetitive inhibitor be a more effective drug than a competitive inhibitor?

Answer 1

[S]	V0	V0 inhibitor	1/[S]	1/V0	1V0 inhibitor
0.0001	33	17	10000	0.030303	0.058823529
0.0002	50	29	5000	0.02	0.034482759
0.0005	71	50	2000	0.014085	0.02
0.001	83	67	1000	0.012048	0.014925373
0.002	91	80	500	0.010989	0.0125
0.005	96	91	200	0.010417	0.010989011
0.01	98	95	100	0.010204	0.010526316
0.02	99	98	50	0.010101	0.010204082
0.05	100	99	20	0.01	0.01010101
0.1	100	100	10	0.01	0.01
0.2	100	100	5	0.01	0.01

a) without inhibitor : we plot a graph between 1/[S] v/s 1/V

y = 2E-06x + 0.01

intercept = 1/ v max

Vmax = 1/ 0.01 = 100

Slope = Km / Vmax

Km = slope X Vmax = 2 X 10^-6 X 100 = 0.0002

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

V0 = 100 X (0.0004 /( 0.0002 + 0.0004) = 66.67

c) in presence of inhibitor

y = 5E-06x + 0.01

intercept = 1/ Vmax

Vmax = 1/ 0.01 = 100

Slope = Km / Vmax

Km = slope X Vmax = 5 X 10^-6 X 100 = 0.0005

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

V0 = 100 X (0.0004 /( 0.0005 + 0.0004) = 44.44

e) competitive inhibitor