Question

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.

Answer 1

`K_{eq` = 19

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

ΔG° of the reaction under cellular conditions = 10817.46 J

Step-by-step 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
`K_{eq` can be calculated as:

`K_{eq`
`= ([glucose-6-phosphate])/([glucose-1-[phosphate])`

`K_{eq`
`= (0.95)/(0.05)`

`K_{eq` = 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
`K_{eq` can be calculated as:

`K_{eq`
`= ([glucose-6-phosphate])/([glucose-1-[phosphate])`

`K_(eq)= (1.395*10^(-4))/(1.090*10^(-2))`

`K_(eq) =` 0.01279816514 M

`K_(eq) =` 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J