in response to this question, could you explain why in Step 3, R became 0.500 and why in Step 4, Vmax is sudden multiplied by 10^-3? It's unclear in the response
in response to this question, could you explain why in Step 3, R became 0.500 and why in Step 4, Vmax is sudden multiplied by 10^-3? It's unclear in the response
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
in response to this question, could you explain why in Step 3, R became 0.500 and why in Step 4, Vmax is sudden multiplied by 10^-3? It's unclear in the response
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Science » Chemistry » Q&A Library » Determine the values of KM and Vmax. For the Vmax obtained, calculate the turnover num...
Question
• Determine the values of Ky and Vmax. For the V,
obtained, calculate the turnover number assuming
тах*
таx
that 1X 104mol of enzyme were used.
Substrate concentration (mol L²)
Velocity (mM min 1)
2.500
0.588
1.000
0.500
0.714
0.417
0.526
0.370
0.250
0.256
Expert Solution
Step 1
The graph formed will be as follows:
Graph of Velocity v/s Substrate
concentration
0.7
0.6
0.5
0.4
0.3
0.2
Velocity (R)"
Transcribed Image Text:10:25 AM Wed Mar 9
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Science » Chemistry » Q&A Library » Determine the values of KM and Vmax. For the Vmax obtained, calculate the turnover num...
Question
• Determine the values of Ky and Vmax. For the V,
obtained, calculate the turnover number assuming
тах*
таx
that 1X 104mol of enzyme were used.
Substrate concentration (mol L²)
Velocity (mM min 1)
2.500
0.588
1.000
0.500
0.714
0.417
0.526
0.370
0.250
0.256
Expert Solution
Step 1
The graph formed will be as follows:
Graph of Velocity v/s Substrate
concentration
0.7
0.6
0.5
0.4
0.3
0.2
Velocity (R)
![10:26 AM Wed Mar 9
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Step 3
Putting the value of vmax in equation (1):
V.
max
[S]
R =
[S]+Km
0.666 mM min×(1×10³) mM
(1×10³) mM +Km
0.500 mM min-
-3
K = 0.332×10 mM
%3D
m
Step 4
Turn over number can be calculated as follows:
V.
max
= k,[E,]
(3)
....
where E, is the enzyme concentration, k2 is the turn over number, vmax is the maximum rate. Assuming
that all of the enzyme has bind, k2 can be calculated by putting the values in equation (3) as follows:
0.666 ×10³M min = k, ×1×10ª M
-1
k, = 6.66 min](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb922951d-2773-4842-822d-60d5ada7ed63%2F4d4bc96e-aea4-461e-b080-6885b08f5c91%2Fakx07d_processed.png&w=3840&q=75)
Transcribed Image Text:10:26 AM Wed Mar 9
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보 CHAT
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Step 3
Putting the value of vmax in equation (1):
V.
max
[S]
R =
[S]+Km
0.666 mM min×(1×10³) mM
(1×10³) mM +Km
0.500 mM min-
-3
K = 0.332×10 mM
%3D
m
Step 4
Turn over number can be calculated as follows:
V.
max
= k,[E,]
(3)
....
where E, is the enzyme concentration, k2 is the turn over number, vmax is the maximum rate. Assuming
that all of the enzyme has bind, k2 can be calculated by putting the values in equation (3) as follows:
0.666 ×10³M min = k, ×1×10ª M
-1
k, = 6.66 min
Expert Solution
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