The enzyme ribose‑5‑phosphate isomerase catalyzes the conversion between ribose‑5‑phosphate (R5P) and ribulose‑5‑phosphate (Ru5P) through an enediolate intermediate. In the Calvin cycle, Ru5P is used to replenish ribulose‑1,5‑bisphosphate, a substrate for rubisco. For the conversion of R5P to Ru5P, if Δ G ° ′ = 0.460 kJ / mol ΔG°′=0.460 kJ/mol and Δ G = 3.30 kJ / mol ΔG=3.30 kJ/mol , calculate the ratio of Ru5P to R5P at 298 K 298 K . [ Ru 5 P ] [ R 5 P ] = [Ru5P][R5P]= Which of the statements is true? This reaction is favorable, and it is not likely regulated. This reaction is favorable, and it is likely regulated. This reaction is not favorable, and it is not likely regulated. This reaction is not favorable, and it is likely regulated.

This reaction is favorable, and is likely regulated.

Explanation:

The equation to calculate delta G (dG) of a reaction is dG = dGo' + RTln [initial P]/[initial R]. You could use just dG = RTln [initial P]/[initial R] if (and that's a big IF) dGo' is zero, meaning that the reaction is at equilibrium when we have equal amounts of [P] and [R] (which is rarely the case). What really matters is the ratio of Q ([initial P]/[initial R]) to Keq ([P at equilibrium]/[R at equilibrium]), meaning how far off are we from equilibrium.

If Q=Keq, we are already at equilibrium (EQ).

If Q<Keq, we are not yet at EQ, having relatively more [R], or less [P] than under EQ conditions. This means the reaction will move forward to produce more P until EQ is achieved (dG is therefore NEGATIVE).

If Q>Keq, we are also off EQ, but we have relatively more [P], or less [R] than under EQ conditions. This means the reaction will move backwards to produce more R until EQ conditions are achieved (dG is therefore POSITIVE).

Try to understand these equations below (they say what I tried to describe in words)

dGo' = -RTlnKeq (under "standard conditions", i.e. we try to figure out how a reaction "behaves" if we start out with the same molar concentrations of R and P)

dG = dGo' + RTlnQ Q=[initial P]/[initial R]   or

dG = -RTlnKeq + RTlnQ or

dG = RTlnQ - RTlnKeq   or

dG = RTln Q/Keq

i think so it’s will be b yes i believe its