Question

Imagine that inside a cell the [creatine phosphate] = 3.5 mM, [creatine] = 0.6 mM, and [Pi] = 4.3 mM. Furthermore, imagine that the pH inside the cell is 7.4. Given that ΔG°′ for creatine phosphate hydrolysis is -10.3 kcal/mol, what is the ΔG for creatine phosphate hydrolysis inside the above-described cell, which is at 37 °C? Report your answer in kcal/mol to the nearest tenths. Hint: you might wish to consult the chapter in our text that discusses thermodynamics.

Answer 1

Hydrolysis of creatin phosphate

Creatine phosphate + ADP + H⁺ Creatine + ATP : G^o = -10.3 Kcal/mol

G = G^o + RT ln ([products] / [reaction])

R = 1.98 x 10^-3 Kcal K^-1 mol^-1

T = 37 + 273 = 310K

G = -10.3 + 2.303 x 1.98 x 10^-3 x 310 log ([0.6] /[3.5] [4.3])

G = -7.7 Kcal/mol