Final answer:
The conversion of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P) has a standard free energy change of 1.7 kJ/mol, indicating it is non-spontaneous under standard conditions. However, under typical cellular conditions with concentrations of 120 μM G6P and 28 μM F6P, the reaction may proceed spontaneously if the actual free energy change (ΔG) calculated using these concentrations is negative.
Step-by-step explanation:
The reaction of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P) is an essential step in the glycolysis pathway. The standard free energy change (ΔG°) for this reaction is 1.7 kJ/mol. To address the questions:
- Under standard thermodynamic conditions (where ΔG° = 1.7 kJ/mol), the reaction is non-spontaneous because a positive ΔG° value indicates that the reaction requires an input of energy to proceed.
- However, in a living cell the concentrations of G6P and F6P are not 1 M. To determine the direction of spontaneity under these conditions, the actual free energy change (ΔG) must be calculated using the formula ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient, R is the gas constant (8.314 J/K·mol), and T is the temperature in Kelvin (310.15 K for 37°C). Substituting the given concentrations into Q (as [F6P]/[G6P]), and with ΔG° as 1.7 kJ/mol, we calculate the actual ΔG for this reaction. If ΔG is negative under these cellular conditions, the forward reaction (G6P to F6P) is spontaneous.
If the calculated ΔG is negative, it suggests that under cellular conditions, contrary to the standard conditions, the conversion of G6P to F6P occurs spontaneously.