Please explain the formula (process) used to get the Red undelived numbers, Thanks!! example each Row, please. ISI 1/IS1
Please explain the formula (process) used to get the Red undelived numbers, Thanks!! example each Row, please. ISI 1/IS1
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
Related questions
Question
![4. Graphing the results from kinetics experiments with enzyme inhibitors
The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the
presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [ET] is the same in
each experiment.
[S]
(mM)
1
2
4
8
12
1
2
4
8
12
[S] 1/[S]
(mm)
a. Determine Vmax and Km for the enzyme.
b. Determine the type of inhibition and the Kifor each inhibitor.
v (1)
M-1 μmol/mL.se
Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a.
and b.
1000
500
250
125
83.3
Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding v, we
will calculate 1/[S] and 1/v.
Please explain the formula (process) used to get the Red undelined numbers,
example
each Row, please.
Thanks!!
C
12
20
29
35
40
1/v
(1)
v (μmol/mL sec
12
20
29
35
40
-400
Condition
No inhibitor
→
20 x 104
1/v (1)
mL sec/μmol
10 x 104
8.33x104
5.00x104
3.45x104
2.86x104
2.50x104
5 mM inhibitor (2)
5 mM inhibitor (3)
0
Plots of 1/v vs. 1/[S] indicate straight lines given by
(1) 1/v = 63.2(1/[S] ) + 1.95 x 104
(2) 1/v = 211.8(1/[S]) +2.00 x 104
(3) 1/v = 137.2(1/[S)) + 4.38 x 104
In general,
(2)
v (μmol/mL sec
4.3
8
14
21
26
1/v (2)
v (3)
μmol/mL sec mL sec/μmol μmol/mL-se
1 K
v (2)
400 800
1/[S]
m
4.3
8
14
X
21
26
1 1
V [S] V
max
2.33x105
1.25x105
7.14×104
4.76x104
3.85x104
max
Thus, the y-intercept is equal to 1/Vmax and the x-intercept is equal to -1/Km.
(a)
Vmax (umol/mL-sec)
51.2
Vmax,app
49.9
22.8
(3)
v (µmol/mL·sec
5.5
1200
9
13
16
18
Km (MM)
3.2
Km,app
10.6
3.1
с
5.5
9
13
16
18
inhibitor 3
inhibitor 2
no inhibitor
1/v (3)
mL-sec/μmol
1.82×105
1.11×105
7.69x104
6.25x104
5.56x104
(b) Inhibitor (2) increases the apparent Km of the enzyme without affecting Vmax. This is characteristic of a
competitive inhibitor. In this case Ki is calculated as follows](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeadb183-1175-48ee-be1d-be4cbd36e1e9%2F86ad7030-65d9-4ae5-b1bd-bdf2f636c249%2Fs64aaso_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. Graphing the results from kinetics experiments with enzyme inhibitors
The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the
presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [ET] is the same in
each experiment.
[S]
(mM)
1
2
4
8
12
1
2
4
8
12
[S] 1/[S]
(mm)
a. Determine Vmax and Km for the enzyme.
b. Determine the type of inhibition and the Kifor each inhibitor.
v (1)
M-1 μmol/mL.se
Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a.
and b.
1000
500
250
125
83.3
Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding v, we
will calculate 1/[S] and 1/v.
Please explain the formula (process) used to get the Red undelined numbers,
example
each Row, please.
Thanks!!
C
12
20
29
35
40
1/v
(1)
v (μmol/mL sec
12
20
29
35
40
-400
Condition
No inhibitor
→
20 x 104
1/v (1)
mL sec/μmol
10 x 104
8.33x104
5.00x104
3.45x104
2.86x104
2.50x104
5 mM inhibitor (2)
5 mM inhibitor (3)
0
Plots of 1/v vs. 1/[S] indicate straight lines given by
(1) 1/v = 63.2(1/[S] ) + 1.95 x 104
(2) 1/v = 211.8(1/[S]) +2.00 x 104
(3) 1/v = 137.2(1/[S)) + 4.38 x 104
In general,
(2)
v (μmol/mL sec
4.3
8
14
21
26
1/v (2)
v (3)
μmol/mL sec mL sec/μmol μmol/mL-se
1 K
v (2)
400 800
1/[S]
m
4.3
8
14
X
21
26
1 1
V [S] V
max
2.33x105
1.25x105
7.14×104
4.76x104
3.85x104
max
Thus, the y-intercept is equal to 1/Vmax and the x-intercept is equal to -1/Km.
(a)
Vmax (umol/mL-sec)
51.2
Vmax,app
49.9
22.8
(3)
v (µmol/mL·sec
5.5
1200
9
13
16
18
Km (MM)
3.2
Km,app
10.6
3.1
с
5.5
9
13
16
18
inhibitor 3
inhibitor 2
no inhibitor
1/v (3)
mL-sec/μmol
1.82×105
1.11×105
7.69x104
6.25x104
5.56x104
(b) Inhibitor (2) increases the apparent Km of the enzyme without affecting Vmax. This is characteristic of a
competitive inhibitor. In this case Ki is calculated as follows
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