What is the impact of the lower value Vmax on the affinity for enzyme for substrate? And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why?
What is the impact of the lower value Vmax on the affinity for enzyme for substrate? And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why?
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
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What is the impact of the lower value Vmax on the affinity for enzyme for substrate?
And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why?
![Michaelis Menten Plot.ds (Compatibility Mode] - Microsoft Excel
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1 [Ethanol)
Vo
Calc. Vo delta^2
0.007
0.06
0.0327 0.000745
0.35
3
0.015
0.11
0.057444 0.002762
4.
0.031
0.16
0.087264
0.00529
0.3
0.068
0.21
0.118696 0.008336
0.1
0.23
0.131384 0.009725
0.25
0.2
0.28
0.148218 0.017366
0.2
0.3
0.29
0.154831 0.018271
• Vo
0.4
0.28
0.158364 0.014795
0.15
Calc. Vo
10
0.077292
11
Km
0.029391
0.1
12
Vmax
0.17
0.05
13
14
16
0.1
0.2
03
0.4
0.5
17
18
Fig. 11. Michaelis-Menten Plot using a value of 0.17 for Vmax.
Notice: 1) that a new set of calculated values in column C has been generated and
2) that by using these new data a new best fit Michaelis-Menten plot (red curve)
has been generated.
Fig. 11 shows that the rates for the new enzyme-catalyzed reaction (red curve) are
significantly lower than those originally obtained (blue symbols).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ef5f725-b08b-4f71-a80c-f37e47d07785%2F5bf624a6-df3c-40cd-b3f1-ac4eb51bf95d%2Fo3jgn49_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Michaelis Menten Plot.ds (Compatibility Mode] - Microsoft Excel
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F24
A
B
D
G
H
M
1 [Ethanol)
Vo
Calc. Vo delta^2
0.007
0.06
0.0327 0.000745
0.35
3
0.015
0.11
0.057444 0.002762
4.
0.031
0.16
0.087264
0.00529
0.3
0.068
0.21
0.118696 0.008336
0.1
0.23
0.131384 0.009725
0.25
0.2
0.28
0.148218 0.017366
0.2
0.3
0.29
0.154831 0.018271
• Vo
0.4
0.28
0.158364 0.014795
0.15
Calc. Vo
10
0.077292
11
Km
0.029391
0.1
12
Vmax
0.17
0.05
13
14
16
0.1
0.2
03
0.4
0.5
17
18
Fig. 11. Michaelis-Menten Plot using a value of 0.17 for Vmax.
Notice: 1) that a new set of calculated values in column C has been generated and
2) that by using these new data a new best fit Michaelis-Menten plot (red curve)
has been generated.
Fig. 11 shows that the rates for the new enzyme-catalyzed reaction (red curve) are
significantly lower than those originally obtained (blue symbols).
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