Phosphoglucomutase catalyzes the reaction of glucose 6-phosphate (G6P) to fructose 6-phosphate (F6P). You are starting the reaction in a test tube with the 0.8M substrate (G6P), and you let the reaction reach equilibrium. The product (F6P) concentration at equilibrium is 0.6M. There are no intermediates in this reaction and no products at the beginning. The Keq for this reaction is:.

0.75

Explanation:

Mathematically, the equilibrium constant Keq is the concentration of product divided by the concentration of the reactant.

And since the product is fructose 6-phosphate (F6P) while the reactant is glucose 6-phosphate (G6P):

Keq = [F6P] / [G6P]

Keq = 0.6 / 0.8

Keq = 0.75 (since Keq is almost equal to 1, it means the amount of F6P and G6P in the reaction is almost the same)

Thus, the equilibrium constant Keq for this reaction is 0.75

b is the correct answer

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the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be