Question

If the intracellular concentrations of a metabolite (M-OH) and its phosphorylated form (M-OPO32-) were 2.8 mM and 0.1 mM, respectively, and if the intracellular concentrations of ATP and ADP were 4 mM and 0.16 mM, respectively, what would be the numerical value of [\Delta] ΔG (in kcal per mol to the nearest hundredth) for the following reaction: M-OH + ATP <--> M-OPO32- + ADP + H+? Assume a temperature of 37 °C and a pH of 7.4. 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 –3.5 kcal/mol and –7.3 kcal/mol, respectively.

Answer 1

We know,

dG = dGo + RTlnK

where,

dGo = -7.3 - (-3.5) = -3.8 kcal/mol [calculated from standard free energies]

with 1 kcal/mol = 4.184 kJ/mol

dGo = -3.8 x 4.184 = -16 kJ/mol

also,

R = 8.315 J/K.mol

T = 37oC + 273 = 310 K

pH = -log[H+] = 7.4

So, [H+] = 3.98 x 10^-8 M

K = [M-OPO3^2-][ADP][H+]/[M-OH][ATP]

Feeding the given values,

K = (0.1)(0.16)(3.98 x 10^-8)/(2.8)(4) = 5.7 x 10^-11

Feeding all the values into the first equation, (1 kJ/mol = 1000 J/mol)

dG = -16 + (8.314 x 310/1000)ln(5.7 x 10^-11) = -76.794 kJ/mol

So,

numerical value of dG would be,

dG = -76.794 kJ/mol/4.184 = -18.354 kcal/mol