Question

The enzyme, phosphoglucomutase, catalyzes the interconversion

Answer 1

Answer:

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

$K_{eq$ $= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}$

$K_{eq$$= \frac{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:

$K_{eq$ $= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}$

$K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}$

$K_{eq} =$ 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