If the intracellular concentrations of a metabolite (M-OH) and its phosphorylate
ID: 1049617 • Letter: I
Question
If the intracellular concentrations of a metabolite (M-OH) and its phosphorylated form (M-OPO_3^2-) were 3.1 mM and 0.12 mM, respectively, and if the intracellular concentrations of ATP and ADP were 4.2 mM and 0.17 mM, respectively, what would be the numerical value of Delta G (in kcal per mol to the nearest hundredth) for the following reaction: M-OH + ATP doubleheadarrow M-OPO_3^2- + ADP + H^+? Assume a temperature of 25 degree C and a pH of 7.3. To solve this problem, you will need to know the standard free energies of hydrolysis of the phosphorylated metabolite and of ATP These value are -3.5 kcal/mol and -7.3 kcal/mol, respectively.Explanation / Answer
PH = 7.3
H+ = 10^-7.3 = 5.011 * 10^-8 M = 5.011 * 10^-5 mM
K = [H+] [ADP][M-OPO32-] / [MOH] [ ATP] = 5.011 * 10^-5 * 0.12 * 0.17 / 3.1*4.2 = 7.85*10^-8 mM
ATP+H2O.ADP+Pi+free energy
Go = -7.3 kal/mol
similarily for phosphorylelated metabolite
GO = -3.5 kcal/mol
total GO = -7.3-3.5 = -10.8 kcal/mol
G = GO + RT ln K = -10.8 + 8.314*298 / 4.2*1000 ln 7.85*10^-11 = -10.8- 13.725 kcal/mol = -24.53 kcal/mol