Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
I don’t understand how to do part b of question 5. The other picture is my work of part a. How would changing the enzyme concentration affect Km and Vmax. The answer given to us is Vo= 0.796mM/s. I just don’t understand how to get to that answer.
![Shift
Slope
km
X (2 uM --
0.002 mm)
6v -
V.
max
[S] (mM)
V (M/s)
0-009
0.860
0.01
1.307
Slope =
1-367
0.860
2.651198 x/0-3
D.01
0.004
5.302396 x10-3
y=mx+b
5.302396 x10-@M
Km
Vmax [s]
Vmax
Km
Vo
Vmax
CS]
Vmax
2.451198
1.367
0.01
Vmax
6.49999
V max
Kcat =
Vmax=
2.0000MM/S
km
Vmax
Vmax
Krat
[E]
m.Vmax
km=
2.0000
Krat
Km
= 2.651198 xr03.2
0.002
-5.30 237e x/0-3
appreoched
catalytic perfection
has
= 1600 30 xo-3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3591fe4b-0e6f-406c-a068-d60024e5a170%2F92a2319c-3e03-4699-afac-b8589166d740%2F4eyjfrc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Shift
Slope
km
X (2 uM --
0.002 mm)
6v -
V.
max
[S] (mM)
V (M/s)
0-009
0.860
0.01
1.307
Slope =
1-367
0.860
2.651198 x/0-3
D.01
0.004
5.302396 x10-3
y=mx+b
5.302396 x10-@M
Km
Vmax [s]
Vmax
Km
Vo
Vmax
CS]
Vmax
2.451198
1.367
0.01
Vmax
6.49999
V max
Kcat =
Vmax=
2.0000MM/S
km
Vmax
Vmax
Krat
[E]
m.Vmax
km=
2.0000
Krat
Km
= 2.651198 xr03.2
0.002
-5.30 237e x/0-3
appreoched
catalytic perfection
has
= 1600 30 xo-3
![4. For an enzyme that has a Km of 25 mM, Vmax of 50 mM/s and kcat of 250 s1, how long does a
single reaction take?
Hkct= tine of simgle
Singe
A. 25 s
B. 250 ms
C. 50 ms
D. 4 ms
E. 4s
5. A kinetics experiment was performed on enzyme X (2 µM), and gave the data shown below.
[S] (uM)
V. (mM/s)
4.00
0.860
10.0
1.307
A. Does enzyme X approach catalytic perfection? Justify your answer.
B. What will be the reaction rate (Vo) with 1.5 uM enzyme X and 6.0 uM substrate?
6. Why does an enzyme's activity decrease as the temperature rises above its optimal
temperature? Why does activity decrease below the optimal temperature?
7. Which of the following statements is true about a competitive enzyme inhibitor?
A. It has no effect on Km
It binds to the enzyme:substrate complex and prevents it from reacting further
C. It decreases the Vmax
D. It causes the enzyme to refold into an inactive conformation
В.
P Type here to search
DI
ム×
F1
F2
F3
F4
F5
F6
@
2$
%
近
%23](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3591fe4b-0e6f-406c-a068-d60024e5a170%2F92a2319c-3e03-4699-afac-b8589166d740%2F2ccsexj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. For an enzyme that has a Km of 25 mM, Vmax of 50 mM/s and kcat of 250 s1, how long does a
single reaction take?
Hkct= tine of simgle
Singe
A. 25 s
B. 250 ms
C. 50 ms
D. 4 ms
E. 4s
5. A kinetics experiment was performed on enzyme X (2 µM), and gave the data shown below.
[S] (uM)
V. (mM/s)
4.00
0.860
10.0
1.307
A. Does enzyme X approach catalytic perfection? Justify your answer.
B. What will be the reaction rate (Vo) with 1.5 uM enzyme X and 6.0 uM substrate?
6. Why does an enzyme's activity decrease as the temperature rises above its optimal
temperature? Why does activity decrease below the optimal temperature?
7. Which of the following statements is true about a competitive enzyme inhibitor?
A. It has no effect on Km
It binds to the enzyme:substrate complex and prevents it from reacting further
C. It decreases the Vmax
D. It causes the enzyme to refold into an inactive conformation
В.
P Type here to search
DI
ム×
F1
F2
F3
F4
F5
F6
@
2$
%
近
%23
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