ncentration of 3 X 10 M. Find the time required to convert 70% of substrate when Vmax = 2.167 M/min. PBL 1: An enzyme with a Km = 1 X 10² M was assayed to convert the substrate with initial co Km So In (So-S) Vmax t = Vmax S = So - xSo
ncentration of 3 X 10 M. Find the time required to convert 70% of substrate when Vmax = 2.167 M/min. PBL 1: An enzyme with a Km = 1 X 10² M was assayed to convert the substrate with initial co Km So In (So-S) Vmax t = Vmax S = So - xSo
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Chapter20: Enzymes
Section: Chapter Questions
Problem 20.45E
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Question
![PBL 1: An enzyme with a Km = 1 X 102 M was assayed to convert the substrate with
initial co
ncentration of 3 X 10$M. Find the time required to convert 70% of substrate when Vmax = 2.167
M/min.
Km
In
So
1
- (So – S)
t =
Vmax
S' Vmax
S = So - xS,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F372ed87c-03f1-4ddf-a754-fda92e9474ec%2F74a76e6a-b990-4b8d-9b4d-c35beed9f2df%2Fzkbuz7t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:PBL 1: An enzyme with a Km = 1 X 102 M was assayed to convert the substrate with
initial co
ncentration of 3 X 10$M. Find the time required to convert 70% of substrate when Vmax = 2.167
M/min.
Km
In
So
1
- (So – S)
t =
Vmax
S' Vmax
S = So - xS,
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