In: Chemistry
Determine the Vmax and the Km of the enzyme in the absence and presence of each inhibitor. Calculate the KI for each inhibitor. Are the inhibitors all the same? Which is the best inhibitor and why? What is the catalytic efficiency and turnover number of the enzyme knowing that during the experiments its concentration was 2 nM? What happens to the catalytic efficiency and turnover number of the enzyme in the presence of each of the inhibitor?
The following velocity data were obtained for an enzymatic reaction in the absence and presence of three different inhibitors (A, B, and C):
Initial velocity (v) Initial velocity (v) Initial velocity (v) Initial Velocity (v)
[S] Control (no inhibitor) (+A at 6μM) (+B at 30μM) (+C at 4mM)
(mM) (nM/min) (nM/min) (nM/min) (nM/min)
0.200 16.67 6.25 5.56 10.00
0.250 20.00 7.69 6.67 11.11
0.333 24.98 10.00 8.33 12.50
0.500 33.33 14.29 11.11 14.29
1.00 50.00 25.00 16.67 16.67
2.00 66.67 40.00 22.22 18.18
2.50 71.40 45.45 23.81 18.52
3.33 76.92 52.63 25.64 18.87
4.00 80.00 57.14 26.67 19.00
5.00 83.33 62.50 27.77 19.23
Plot 1/[S] and 1/[Vo] for all the four data sets
x-intercept = -1/Km
y-intercept = 1/Vmax
enzyme turnover Kcat = Vmax/[Enzyme]
Graphs are shown below
1) without inhibitor
x-intercept = -1/Km = -0.0053
Km = 188.68 mM
y-intercept = 1/Vmax = 0.0561
Vmax = 17.82 nM/min
Kcat = 17.82/2 = 8.91
2) with +A 6 micromolar inhibitor
x-intercept = -1/Km = -0.0160
Km = 62.5 mM
y-intercept = 1/Vmax = 0.1482
Vmax = 6.75 nM/min
Kcat = 6.75/2 = 3.37
3) with +B 30 micromolar inhibitor
x-intercept = -1/Km = -0.0160
Km = 62.5 mM
y-intercept = 1/Vmax = 0.1682
Vmax = 5.94 nM/min
Kcat = 5.94/2 = 2.97
4) with +C 4 mM inhibitor
x-intercept = -1/Km = -0.0053
Km = 188.68 mM = 188.68
y-intercept = 1/Vmax = 0.0961
Vmax = 10.40 mM/min = 10.4 x 10^6 micromolar/min
Kcat = 10.4 x 10^6/2 = 5.2 x 10^6
the best efficiency and turnover number for the enzyme was seen in case of +C. With lowest turnover number seen in case of +B.
the relevant graphs are given below .