Question

The enzyme aldolase catalyzes the following reaction:

Fructose 1,6-bisphosphate ↔ dihydroxyacetone phosphate + glyceraldehyde 3-phosphate For this reaction, ΔGº′ =+23.8 kJ/mol (+ 5.7 kcal/mol).
a. Calculate the change in free energy ΔG for this reaction under typical intracellular conditions using the following concentrations: fructose 1,6-bisphosphate, 0.15 mM; dihydroxyacetone phosphate, 4.3 X 10−6 M; and glyceraldehyde 3-phosphate, 9.6 10−5 M. Assume that the temperature is 25ºC.
b. Explain why the aldolase reaction occurs in cells in the direction written despite the fact that it has a positive free-energy change under standard conditions.

Answer 1

The reactions involved were:-

Step-by-step explanation:

a) Fructose 1,6-bisphosphate ↔ dihydroxyacetone phosphate + glyceraldehyde 3-phosphate dG° = 23.8 kJ/mol

Apply to the:-

dG = dG° + RT*ln(Q)

Replace known values, R = 8.31 J / molK, T = 25 ° C = 298 K

Q = [dihydroxyacetone phosphate][glyceraldehyde 3-phosphate] /[Fructose 1,6-bisphosphate]

Q = (4.3*10^-6)(9.6*10^-5) / (0.15*10^-3) = 0.000002752

Replace it:-

dG = dG° + RT*ln(Q)

dG = 23800 + 8.314*298 * ln(0.000002752)

dG = -7920.8068 J/mol = -7.92 kJ/mol

(b) It was negative, so this is preferred inside a cell, and not outside it!