PBL 1: An enzyme with a Km = 1 X 102 M was assayed to convert the substrate withinitial concentration of 3 X 10$M. Find the time required to convert 70% of substrate when Vmax = 2.167M/min.KmInSo1- (So – S)t =VmaxS' VmaxS = So - xS,

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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

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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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,

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