Biochemistry 440
Fall, 2008
Problem Set 6 Answers
1. What is the ∆G of the hydrolysis of ATP in an E. coli cell given that [ATP] is 8 mM, [ADP]
is 1 mM and the [Pi] is 16 mM.
= -30 kJ/mol + 2.3 RT log (1 x 10-3 x 16 x 10-3 / 8 x 10-3)
= -30 kJ/mol + 5 x log 2.0 x 10-3
= -30 kJ/mol + 5 kJ/mol x -2.7
= -43 kJ/mol
2. The catabolism of glucose is an important source of energy for all cells. It begins with the
following transformation which is the first step of the glycolytic pathway:
Glucose --> Glucose 6-phosphate
Theoretically, the cell could phosphorylate glucose directly with inorganic phosphate (Pi) like
so:
Glucose + Pi --> Glucose 6-phosphate + H2O ∆G'0 = +10 kJ/mol
a. Calculate the equilibrium constant for this reaction at 25˚C.
∆G’˚ = -2.3RT log K’eq
10 = -5 * log K’eq
-2 = log K’eq
0.01 = K’eq
b. Is it favorable under standard conditions?
No, because ΔG'0 > 0.
c. If in a typical cell, [glucose] is 5 mM and [Pi] is 1 mM, what would be the concentration of
glucose-6 phosphate at equilibrium if phosphorylation occurred by the reaction above?
K'eq = [G 6-P]/[Glucose]*[Pi]
0.01 = [G 6-P]/5 x 10-3 * 1 x 10-3
0.01 * 5 X 10-6 = [G 6-P]
5 x 10-8 or 50 nM = [G 6-P].
d. Does this direct phosphorylation of glucose represent a reasonable route for the catabolism of
glucose? Explain.
No, 50 nM glucose 6-phosphate is very low.
e. The cell actually accomplishes the phosphorylation of glucose by coupling it to the
hydrolysis of ATP in a reaction catalyzed by the enzyme hexokinase:
!G=!G'0+2.3RT log
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