The enzyme aldolase catalyzes the following reaction: Fructose 1,6-bisphosphate

Question

The enzyme aldolase catalyzes the following reaction:

Fructose 1,6-bisphosphate dihydroxyacetone phosphate + glyceraldehyde 3-phosphate For this reaction, Gº =+23.8 kJ/mol (+ 5.7 kcal/mol).

(a)        Calculate the change in free energy G for this reaction under typical intracel-lular conditions using the following concentrations: fructose 1,6-bisphosphate, 0.15 mM; dihydroxyacetone phosphate, 4.3 X 106 M; and glyceraldehyde 3-phosphate, 9.6 105 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.

For question a should i cange the mM to M?

Explanation / Answer

The reactions involded:

a)

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

Apply

dG = dG° + RT*ln(Q)

substitute 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

substitute

dG = dG° + RT*ln(Q)

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

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

b)

this is negative, so this is favoured inside a cell, but not outside of it!

Q3.

For question a should i cange the mM to M?

YES! always change to M, via 10^-3

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