A) Fructose-6-phosphate is phosphorylated to fructose-(1-6)-bisphosphate as part of the glycolytic pathway. The phosphorylation of fructose-6-phosphate by phosphate is described by the following equation:
Fructose-6-phosphate + Pi ↔Fructose-(1-6)-bisphosphate
∆G0’ = +47.7 kJ/mol
Suppose that the phosphorylation of fructose-6-phosphate is coupled
to the hydrolysis of ATP
(-30.5 kJ/mol). What is the ∆G0’ for the coupled reaction?
A
-17.2 kJ/mol
B
+17.2 kJ/mol
C
+15.0 kJ/mol
D
-16.5 kJ/mol
E
+0.7 kJ/mol
B) Using the answer in question #5 to calculate the Keq what is the ratio of fructose-(1-6)- bisphosphate to fructose-6-phosphate at equilibrium for the reaction if the equilibrium concentration of [ATP] = 3mM and [ADP] = 1 mM?
A
0.00005
B
0.01
C
0.0033
D
0.001
E
0.003
A)G of coupled reaction is sum of G of individual reaction.So,G0'=+47.7+(-30.5)=+17.2 kJ/mol.
B)We know that one ATP is needed to convert one Fructose-6-phosphate into Fructose-(1,6)-bisphosphate.
So number of Fructose-1,6-bisphosphate formed is same as number of ATP reacted.
So,concentration of Fructose-6-phosphate is same as unreacted ATP and concentration of Fructose-(1,6)-bisphosphate is same as that of ADP formed.
At equilibrium,ratio of concentration (Fructose-6-phosphate /Fructose-(1,6)-bisphosphate) =3/1=3
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