The enzyme, phosphoglucomutase, catalyzes the interconversion
of glucose 1-phosphate and glucose 6-phosphate,

Glucose 1-phosphate image from custom entry tool Glucose 6-phosphate

If at chemical equilibrium at 25° C, 95% glucose 6-phospate is
present, calculate:

a. K’eq and ΔG°’ of the reaction forming glucose 6-phosphate;
b.ΔG’ of the reaction under cellular conditions where the glucose 1-phosphate concentration is 1.090 x 10-2 M and the concentration of glucose 6-phosphate is 1.395 x 10-4 M.

= 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant can be calculated as:

= 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant   can be calculated as:

0.01279816514  M

0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

particle   p is an electron and particle q is a neutron

the correct answer is:

a. formation of precipitation