A) Is this reaction ( in picture provided) in equilibrium? B) If it is not then ,what is ∆G' at 25°C if the concentration of Glucose-1-phosphate is 15.04µM and the concentration of Glucose-6-phosphate is 1.62 mM?  Answer in Joules. Round to the correct number of significant figures.

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
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A) Is this reaction ( in picture provided) in equilibrium?

B) If it is not then ,what is ∆G' at 25°C if the concentration of Glucose-1-phosphate is 15.04µM and the concentration of Glucose-6-phosphate is 1.62 mM?  Answer in Joules. Round to the correct number of significant figures.  

(There are 103 µM in 1mM.)

Thank you so Much!!!

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!
Ran
= 19
J
R 8.314-
mol K
AG° = -RTlnK'
AG' = AG + RTlnQ'
Glucose-1- PO² → Glucose - 6 -
PO²-
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! Ran = 19 J R 8.314- mol K AG° = -RTlnK' AG' = AG + RTlnQ' Glucose-1- PO² → Glucose - 6 - PO²-
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!
Ran
= 19
J
R 8.314-
mol K
AG° = -RTlnK'
AG' = AG + RTlnQ'
Glucose-1- PO² → Glucose - 6 -
PO²-
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! Ran = 19 J R 8.314- mol K AG° = -RTlnK' AG' = AG + RTlnQ' Glucose-1- PO² → Glucose - 6 - PO²-
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You have been the only one who has been able to this.  It has three other parts as well,

A) Which Enzyme Catalyzes this reaction?

choices are in image provided.

B) What is ∆G°' for this reaction?  Answer in Joules.  K' = 19

C)

If the concentration of Glucose-1-phosphate is 48.82 µM at equilibrium, what is the concentration of Glucose-6-phosphate in µM? 
 
D)

If the reaction is not at equilibrium, what is ∆G' at 25°C if the concentration of Glucose-1-phosphate is 15.04µM and the concentration of Glucose-6-phosphate is 1.62 mM?  Answer in Joules.  Pay attention to units.  Round to the correct number of significant figures.  

There are 103 µM in 1mM.

 

Thank you and you are the winner for Genius of the day!!

 
 
 

 

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!
Ran
= 19
J
R 8.314-
mol K
AG° = -RTlnK'
AG' = AG + RTlnQ'
Glucose-1- PO² → Glucose - 6 -
PO²-
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! Ran = 19 J R 8.314- mol K AG° = -RTlnK' AG' = AG + RTlnQ' Glucose-1- PO² → Glucose - 6 - PO²-
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