For the random two substrate binding reaction shown here: ΚΑ R+A RA + B + B KB 11 RB + A ΚΑ Jak KB RAB You determine that aKA = 500 nM, KB = 20 nM and aKB = 250 nM. What is the values for a and KA? a = 2, KA = 250 nM a = 0.08, KA = 6250 nM = a =25, KA 10 nM Oα = 12.5, K₁ = 40 nM

For the random two substrate binding reaction shown here: ΚΑ R+A RA + B + B KB 11 RB + A ΚΑ Jak KB RAB You determine that aKA = 500 nM, KB = 20 nM and aKB = 250 nM. What is the values for a and KA? a = 2, KA = 250 nM a = 0.08, KA = 6250 nM = a =25, KA 10 nM Oα = 12.5, K₁ = 40 nM

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter23: Enzymes

Section: Chapter Questions

Problem 23.4P: Compare the activation energy in uncatalyzed reactions and in enzyme-catalyzed reactions.

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Question

For the random two substrate binding reaction shown here:
ΚΑ
R+A
RA
+ B
+ B
KB
11
RB + A
ΚΑ
Jak
KB
RAB
You determine that aKA = 500 nM, KB = 20 nM and aKB = 250 nM.
What is the values for a and KA?
a = 2, KA = 250 nM
a = 0.08, KA = 6250 nM
=
a =25, KA 10 nM
Oα = 12.5, K₁ = 40 nM

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