Glucose-1-phosphate is converted to Glucose-6 phosphate by an enzyme. The equilibrium constant in biochemical standard conditions is 19. Answer the following four questions regarding this enzyme-catalyzed reaction. K! = 19 R 8.314 AG-RTln K' AG' = AG' + RTlnQ' Ran J mol K Glucose 1- PO²- →→ Glucose - 6 - PO
Glucose-1-phosphate is converted to Glucose-6 phosphate by an enzyme. The equilibrium constant in biochemical standard conditions is 19. Answer the following four questions regarding this enzyme-catalyzed reaction. K! = 19 R 8.314 AG-RTln K' AG' = AG' + RTlnQ' Ran J mol K Glucose 1- PO²- →→ Glucose - 6 - PO
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
Related questions
Question
there are A-D questions to this picture set up.
A) What enzyme catalyzes this reaction?
B) What is Delta G, please answer in Joules, K=19
C) If concentration of Glucose-1_Phosphate is 48.82 uM at equalibrium, what is the concentration of Glucose-6-phosphate in uM?
D) If the reaction is NOT at equalibrium, what is delta G at 25C if the concentration of Glucose-1-phosphate is 15.04 uM and concentration of Glucose -6-phosphate is 1.62 mM? please answer in Joules and in significant figures. *note, 10^3uM in 1 mM
Thank you!!
![Glucose-1-phosphate is converted to Glucose-6 phosphate by an enzyme. The equilibrium constant in
biochemical standard conditions is 19. Answer the following four questions regarding this enzyme-catalyzed
reaction.
K!
= 19
R 8.314
AG-RTln K'
AG' = AG' + RTlnQ'
Ran
J
mol K
Glucose 1- PO²- →→ Glucose - 6 - PO](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fb4eccc-7198-4ff0-ba32-4ca9cbdd36ee%2Faeafa3dd-44c8-4700-a55d-406fb8d5b702%2Fobg7pbi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Glucose-1-phosphate is converted to Glucose-6 phosphate by an enzyme. The equilibrium constant in
biochemical standard conditions is 19. Answer the following four questions regarding this enzyme-catalyzed
reaction.
K!
= 19
R 8.314
AG-RTln K'
AG' = AG' + RTlnQ'
Ran
J
mol K
Glucose 1- PO²- →→ Glucose - 6 - PO
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