Question

Biological Thermodynamics

A. Calculate the ΔG°’ for the reaction of ATP with Glucose to form Glucose-6-phosphate and ADP.

ΔG°’ = -16.7 kJ/mol

B. Calculate the equilibrium constant for this reaction under standard conditions. (T = 25 °C; R = 8.314 J KA1 molA1; pH=7.0)

Keq = 852.8

C. If the equilibrium concentration of ATP is 3 mM and the equilibrium concentration of glucose is 1 mM, calculate the equilibrium concentrations of both Glucose-6-phosphate and ADP.

D. Repeat this calculation assuming that the concentrations given in part C are instead initial concentrations. You need to calculate the equilibrium concentrations of both Glucose-6-phosphate and ADP.

[glucose-6-phosphate] = 1 mM

[ADP] = 1 mM

Answer 1

                        ATP   + Glucose ------> Glucose 6 Phosphate + ADP

initially   

At equilib     3 mM        1mM                  x                x

dG = -16.7 kJ/mol and Keq = 852.8

Keq = [Glucose 6 phophate] [ADP]/[ATP][Glucose]

852.8 = (x)*(x)/(1mM)*(3mM)

X = 50.58 mM = [gliucose 6 phosphate] = [ADP]

                 ATP       + Glucose ------> Glucose 6 Phosphate + ADP

initially         3 mM          1mM             0                0

At equilib      (3-x)           (1-x)                   x                                      x

Keq = [Glucose 6 phophate] [ADP]/[ATP][Glucose]

852.8 = (x)*(x) / ((3-x)*(1-x))

x = 3.0053 mM = [glucose 6 Phosphate] = [ADP]