An enzyme catalyzed a substrate with the rates shown in the tables below as a fu
ID: 55685 • Letter: A
Question
An enzyme catalyzed a substrate with the rates shown in the tables below as a function of several inhibitor concentrations. Answer the following data. Note: The reactions were taken place in a 96 well plate with each set of concentrations in a separate well.
a. Is the DATA good? Explain.
b. Are there enough data points to acceptably determine Km, Vmax and KI? Explain.
c. Would you prefer to have Vmax or kcat? Explain.
d. Can you find kcat from the data? If not, what other information do you need?
e. Use Hanes-Woolf or Eadie-Hofstee to determine Km, Vmax and KI. Explain if the results are acceptable and justify your answer.
f. Use the Lineweaver-Burke linearization method to calculate Km, Vmax, and KI. Are the results acceptable? Explain and if they aren’t give the most likely reason they are not.
g. What type of inhibition is occurring? Justify your answer.
h. For the uninhibited data ([I] = 0), what will the substrate concentration be after 20 minutes of reaction in a batch reactor if you start with the same initial concentration of substrate?
RxnW - [I] = 0 (uM)
Sample
[S] - mM
Wells
v (mOD/min)
Rx01
0.100
B1
B2
2.414
2.483
Rx02
0.150
C1
C2
3.330
2.735
Rx03
0.200
D1
D2
3.729
3.709
Rx04
0.300
E1
E2
4.609
4.054
Rx05
0.400
F1
F2
5.128
4.970
RxnW - [I] = 0.5 (uM)
Sample
S- mM
Wells
V(mOD/min)
Rx01
0.100
B3
B4
1.528
1.422
Rx02
0.150
C3
C4
1.828
1.922
Rx03
0.200
D3
D4
2.426
2.190
Rx04
0.300
E3
E4
2.962
3.282
Rx05
0.400
F3
F4
3.674
3.786
RxnW - [I] = 1.0 (uM)
Sample
S- mM
Wells
V(mOD/min)
Rx01
0.100
B5
B6
1.001
0.992
Rx02
0.150
C5
C6
1.367
1.448
Rx03
0.200
D5
D6
1.819
1.790
Rx04
0.300
E5
E6
2.338
2.359
Rx05
0.400
F5
F6
2.660
2.608
Sample
[S] - mM
Wells
v (mOD/min)
Rx01
0.100
B1
B2
2.414
2.483
Rx02
0.150
C1
C2
3.330
2.735
Rx03
0.200
D1
D2
3.729
3.709
Rx04
0.300
E1
E2
4.609
4.054
Rx05
0.400
F1
F2
5.128
4.970
Explanation / Answer
a.
The given data is a good, but it needs simple modification of the v (mOD/min) at all three inhibitor concentration with the mean values.
Form the data, the graph is as follows:
All these values shows parabolic axis, so there enough data points to acceptably determine Km, Vmax and KI.
b.
I would prefer to the Vmax, because we should first see the Vmax from the graph to calculate Kcat.
f.
To draw Lineweaver-Burke we should change the data into:
The LB pot is:
Form the graph, the
The in the absence of inhibitor:
The in the 0.5 uM inhibitor:
The in the 0.1 uM inhibitor:
The Km = 1/Vmax
g.
The inhibition that is occurring is competitive