ES → ES' v = k,[ES] ES' → E+ P v = k,[ES'] Show that the rate of formation of product has the same form as that shown in egn 17F.16, but with vmar and K, given by kk[E]o V max = k.(K+k,) KM = K,(k, +k. ) P17F.6 For many enzymes, the mechanism of action involves the formation of two intermediates: E+S→ ES v = k,[E][S] ES → E +S v = k, [ES]
ES → ES' v = k,[ES] ES' → E+ P v = k,[ES'] Show that the rate of formation of product has the same form as that shown in egn 17F.16, but with vmar and K, given by kk[E]o V max = k.(K+k,) KM = K,(k, +k. ) P17F.6 For many enzymes, the mechanism of action involves the formation of two intermediates: E+S→ ES v = k,[E][S] ES → E +S v = k, [ES]
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
![ES → ES'
v = k,[ES]
ES' → E+ P
v = k,[ES']
Show that the rate of formation of product has the same form as that shown in
egn 17F.16, but with vmar and K, given by
kk[E]o
V max =
k.(K+k,)
KM =
K,(k, +k. )
P17F.6 For many enzymes, the mechanism of action involves the formation of
two intermediates:
E+S→ ES
v = k,[E][S]
ES → E +S
v = k, [ES]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ea53192-722b-473b-bcc9-12317a79ebc7%2F62ddb7fb-b57a-4d1b-8f9e-b1bb69e0b873%2Frg0fkc.png&w=3840&q=75)
Transcribed Image Text:ES → ES'
v = k,[ES]
ES' → E+ P
v = k,[ES']
Show that the rate of formation of product has the same form as that shown in
egn 17F.16, but with vmar and K, given by
kk[E]o
V max =
k.(K+k,)
KM =
K,(k, +k. )
P17F.6 For many enzymes, the mechanism of action involves the formation of
two intermediates:
E+S→ ES
v = k,[E][S]
ES → E +S
v = k, [ES]
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