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

Principles Of Pharmacology Med Assist
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Chapter5: Calculating Adult Dosages: Oral And Parenteral Forms
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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 A 59% b Answered: Determine the X + a bartleby.com Get live help whenever you need from online tutors! Try bartleby tutor today → bartleby IS SEARCH & ASK AN EXPERT CHAT Vx MATH SOLVER 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)
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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
Transcribed Image Text:10:26 AM Wed Mar 9 A 59% b Answered: Determine the X + a bartleby.com Get live help whenever you need from online tutors! Try bartleby tutor today → Q SEARCH & ASK AN EXPERT 보 CHAT Vx MATH SOLVER 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
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