The standard free energy change, ΔGº‘ , for the phosphorylation of glucose to glucose-6-phosphate (glucose + Pi yields glucose-6-P) is +3.3 kcal/mol. The standard free energy change, ΔGº‘ , for the hydrolysis of ATP (ATP yields ADP + Pi) is -7.4 kcal/mol. What is the cellular free energy change, ΔG, if the concentration of glucose is 0.005 M, the concentration of ATP is 0.005M, the concentration of glucose-6-phosphate is 0.001M and the concentration of ADP is 0.001M? Cellular temperature = 37 C Gas constant = 0.00198 kilocalories/mol K Temperature = C + Kelvin (C + 273.15)
The standard free energy change, ΔGº‘ , for the phosphorylation of glucose to glucose-6-phosphate (glucose + Pi yields glucose-6-P) is +3.3 kcal/mol. The standard free energy change, ΔGº‘ , for the hydrolysis of ATP (ATP yields ADP + Pi) is -7.4 kcal/mol. What is the cellular free energy change, ΔG, if the concentration of glucose is 0.005 M, the concentration of ATP is 0.005M, the concentration of glucose-6-phosphate is 0.001M and the concentration of ADP is 0.001M? Cellular temperature = 37 C Gas constant = 0.00198 kilocalories/mol K Temperature = C + Kelvin (C + 273.15)
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
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The standard free energy change, ΔGº‘ , for the phosphorylation of glucose to glucose-6-phosphate (glucose + Pi yields glucose-6-P) is +3.3 kcal/mol.
The standard free energy change, ΔGº‘ , for the hydrolysis of ATP (ATP yields ADP + Pi) is -7.4 kcal/mol.
What is the cellular free energy change, ΔG, if the concentration of glucose is 0.005 M, the concentration of ATP is 0.005M, the concentration of glucose-6-phosphate is 0.001M and the concentration of ADP is 0.001M?
Cellular temperature = 37 C
Gas constant = 0.00198 kilocalories/mol K
Temperature = C + Kelvin (C + 273.15)
the answer is supposed to be -6.1 kcal/mol but I am not sure how to get there - please show all steps
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