Please help. Got this question wrong on my test but I keep redoing it and gettin
ID: 64573 • Letter: P
Question
Please help. Got this question wrong on my test but I keep redoing it and getting the same answer. IIl. Enzyme Kinetics. Acting a range finding study. 1 mL of a ution of enzyme concentrate was used in each assay An inhibitor was present in some trials at 10 M. eciprocal plot (Lineweaver-Burke) method, 1:10 dil tube. Using double r determine Km and Vmas of the enzyme. Determine the type of inhibitor. What is the estimated K, of the inhibitor? (30 points) 08 12.5 13.3 14.1 0 13312 11.1 12 2 14.2 16.6 20.0 .DE 25.0 33.3 01 16.6 18.2 20.0 02 222 11.1 11.8 12.5 13.3 14.3 15.4 04 , 012 50.0-1-02 | 25 0-04 100.0 28.6 Key: completion of the above table will gain 6 points; the Lineweaver-Burk plots for enzymatic kinetie with inhibitor and without inhibitor will gain 8 points each (including the correct extension of the regression line to crossover at the x-axis, estimation of Km and Vnat, respectively); identification of the type of inhibitor will gain 3 points; and the correct calculation of the K, values will gain 5 points (wrong units lose 2 point). will gain S points ( in)lttExplanation / Answer
[S] in M
1/[S]
(M-1)
V0
(n mol/min)
No inhibitor
present
1/V0
No inhibitor
present
V0 (n mol/min)
inhibitor present
(10 M)
1/V0
inhibitor present
(10 M)
10
0.1
12.5
0.08
11.1
0.09009009
13.3
0.07518797
12.2
0.081967213
14.1
0.070921986
14.2
0.070422535
15.3
0.065359477
16.6
0.060240964
16.6
0.060240964
11.1
0.09009009
20
0.05
18.2
0.054945055
11.8
0.084745763
25
0.04
20
0.05
12.5
0.08
33.3
0.03003003
22.2
0.045045045
13.3
0.07518797
50
0.02
25
0.04
14.3
0.06993007
100
0.01
28.6
0.034965035
15.4
0.064935065
The Vmax for No inhibitor is calculated as follows:
y = 0.5009x + 0.03
Vmax = 1/ 0.03
= 33.33 n mol/min.
The Km for No inhibitor:
y = 0.5009x + 0.03
Km = 0.03/ 0.5009
= 0.059 n mol.
The Vmax in presence of inhibitor is calculated as follows:
y = 0.4984x + 0.06
Vmax = 1/ 0.06
= 16.66 n mol/min.
The Km for No inhibitor:
y = 0.4984x + 0.06
Km = 0.06/ 0.4984
= 0.120 n mol.
Thus, the kind of inhibitor used in the experiment is uncompetitive inhibitor.
[S] in M
1/[S]
(M-1)
V0
(n mol/min)
No inhibitor
present
1/V0
No inhibitor
present
V0 (n mol/min)
inhibitor present
(10 M)
1/V0
inhibitor present
(10 M)
10
0.1
12.5
0.08
-11.1
0.09009009
13.3
0.07518797
-12.2
0.081967213
14.1
0.070921986
-14.2
0.070422535
15.3
0.065359477
-16.6
0.060240964
16.6
0.060240964
11.1
0.09009009
20
0.05
18.2
0.054945055
11.8
0.084745763
25
0.04
20
0.05
12.5
0.08
33.3
0.03003003
22.2
0.045045045
13.3
0.07518797
50
0.02
25
0.04
14.3
0.06993007
100
0.01
28.6
0.034965035
15.4
0.064935065