If the intracellular concentrations of creatine phosphate (CrP), creatine (Cr),
ID: 580867 • Letter: I
Question
If the intracellular concentrations of creatine phosphate (CrP), creatine (Cr), ADP, and ATP were 0.09 mM, 3.0 mM, 0.15 mM, and 4.0 mM, respectively, what would be the numerical value of G (in kcal per mol to the nearest hundredth) for the reaction in which ATP is the phosphoryl group donor to Cr to make CrP. Assume a temperature of 37 °C and a pH of 7.4. Consider the biochemical standard free energy change for ATP hydrolysis and CrP hydrolysis to be -7.3 kcal/mol and -10.3 kcal/mol, respectively. Consider also whether you need to include [H+].
Explanation / Answer
ATP + Cr <-----> CrP + ADP + H+
pH = -log[H+] ; [H+] = 10-7.4 = 3.98107*10-8 M = 3.98107*10-5 mM
G = Go + RT ln Keq
Keq = [CrP] [ADP] [H+] / [ATP] [Cr] = (0.09 * 0.15 * 3.98107*10-5) / (4 *3) = 4.4787*10-8
R = 1.9872*10-3 Kcal / mol K ; T = 37+273.15 = 310.15 K ; Go = -7.3 - (-10.3) = 3 Kcal/mol
G = 3 + 1.9872*10-3*310.15 ln (4.4787*10-8) = -7.4291 Kcal /mol