Question

If the intracellular concentrations of a metabolite (M-COO-) and its phosphorylated form (M-COOPO32-) were 3 mM and 0.12 mM, respectively, and if the intracellular concentrations of ATP and ADP were 3.9 mM and 0.14 mM, respectively, what would be the numerical value of \Delta Δ G (in kcal per mol to the nearest hundredth) for the following reaction: M-COO- + ATP <--> M-COOPO32- + ADP? Assume a temperature of 37 °C and a pH of 7.2. To solve this problem, you will need to know the standard free energies of hydrolysis of the phosphorylated metabolite and of ATP. These values are –10.3 kcal/mol and –7.3 kcal/mol, respectively.

Answer 1

The reaction is MCOO + ATP <----> MCOOPO32 + ADP

[M-COO-] = 3 * 10^-3M

[M-COOPO32-] = 0.12 * 10^-3M

[ATP] = 3.9 * 10^-3M

[ADP] = 0.14 * 10^-3M

Tempereature = 37degree.

= 37 + 273.15K

T= 310.15K

Delta G⁰ = -10.3Kcal/mol

To convert it into Joules, multiply by 4.184

Delta G⁰ = -43.095J/mol

At pH 7, the expression for delta G is

Delta G = Delta G⁰ +RT ln{ [ADP]  [M-COOPO32-] / [ATP] [M-COO-]}

= -43.0952 + 8.315 * 310.15 * ln {[0.14 * 10^-3M] [ 0.12 * 10^-3M] / [3 * 10^-3M] [ 3.9 * 10^-3M]

= -43.0952 + 2,578.897 * ln (0.144*10³)

ln (0.144*10³) =[ log (0.144*10³)] / 0.4343

= -8.846

Delta G = -43.0952 - 8.842 = - 51.941

Converting into kCal/mol by dividing with 4.184 J/mol

Delta G = - 12.414 KCal/mol