1. The following kinetic data were obtained for an enzyme-catalyzed reaction:
[S]
(mM)
Velocity (rate) M/min
0.1
3.33
0.2
5.00
0.5
7.14
0.8
8.00
1.0
8.33
2.0
9.09
a. Plot the velocity (rate) of this reaction as a function of substrate concentration; draw the horizontal maximum rate line.b. Complete a Lineweaver-Burk plot of these data, determine KM and Vmax.
c. The enzyme concentration was 1.0 x 10-6 M. Calculate the turnover number and explain the physical significance of this number.2. (Extra Credit) The following kinetic data were obtained for an enzyme-catalyzed reaction in the presence and absence of inhibitor Y:
[S]
(mM)
Velocity (rate) M/min Velocity (rate) M/min
No Inhibitor Y
With Inhibitor Y
0.2
5.0
3.0
0.4
7.5
5.0
0.8
10.0
7.5
1.0
10.7
8.3
2.0
12.5
10.7
4.0
13.6
12.5
a. Plot the velocity (rate) of this reaction as a function of substrate concentration for both the uninhibited and inhibited reaction; draw the horizontal maximum rate lines.b. Outline the specific underlying reason why the inhibited rates of reaction are slower by discussing how Y interacts with E or with the ES complex.
c. Do Lineweaver-Burk plots of these two sets of data, determine KM and Vmax for each,
d. For inhibited reactions, KM is increased by a factor of (1 + [I]/Ki), where [I] is the inhibitor concentration and Ki is the inhibitor dissociation constant. Assuming the inhibitor concentration [I] was 0.2 mM in these experiments, uses the Lineweaver-Burk slope data and calculated KM 's for uninhibited and inhibited reactions to determine the Ki inhibitor dissociation equilibrium constant.