1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)
1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)
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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![Q1: the following table gives the results of enzyme-catalyzed reaction.
The enzyme concentration [E] = 0.00012 mmol/L
Use the information of substrate concentration [S] and rate Vo for both experiments, with
inhibitor and without inhibitor to answer the following questions
(NOTE: [E] =
0.00012 mmol/L)
[S], mmol/L
0.710
0.400
0.310
0.098
0.066
0.040
Experiment 11
without inhibitor
VO, mmol/L-min
0.200
0.180
0.160
0.120
0.100
0.070
Experiment 2
with inhibitor
[1] = 0.033 mmol/L
VO, mmol/L-min
0.180
0.150
0.110
0.070
0.050
0.040
1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo
2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo
3- Calculate Kcat for the enzyme (without inhibition)
4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the
type of the Inhibition (competitive, non-competitive or uncompetitive)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e616cea-d678-417c-9781-9b502ffb47e0%2F86b3e828-91c1-4bef-9c47-883a9c29a341%2Fd84fu4w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1: the following table gives the results of enzyme-catalyzed reaction.
The enzyme concentration [E] = 0.00012 mmol/L
Use the information of substrate concentration [S] and rate Vo for both experiments, with
inhibitor and without inhibitor to answer the following questions
(NOTE: [E] =
0.00012 mmol/L)
[S], mmol/L
0.710
0.400
0.310
0.098
0.066
0.040
Experiment 11
without inhibitor
VO, mmol/L-min
0.200
0.180
0.160
0.120
0.100
0.070
Experiment 2
with inhibitor
[1] = 0.033 mmol/L
VO, mmol/L-min
0.180
0.150
0.110
0.070
0.050
0.040
1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo
2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo
3- Calculate Kcat for the enzyme (without inhibition)
4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the
type of the Inhibition (competitive, non-competitive or uncompetitive)
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