Question

Can you answer the following questions with work, please?A) The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reactions was allowed to come to equilibrium. If the keq' is 0.50 what is the Delta G (degree prime) in kJ/mol?B) The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Under cellular conditions (37*C), the glucose-6-phosphate is 6.6 uM and the fructose-6-phosphate is 1.3 uM. If the keq' is 0.50 what is the Delta G in kJ/mol? (Use the Delta G (degree prime) from the previous question)

Answer 1

a) dG0= 1.573 kJ/mol

b)  dG = -2.614 kJ/mol

 Explanation:

a)

Write down the values given in the question,

Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reactions was allowed to come to equilibrium the keq is 0.50

To calculate the delta value ,

dG = -RT ln (Keq)

Putting values:

dG0 = -8.314*273*ln(0.5)

dG0= 1.573 kJ/mol

b)

Glucose-6-phosphate is 6.6 uM and the fructose-6-phosphate is 1.3 uM. If the keq' is 0.50.

To calculate the delta value,

dG = dG0 + RT*ln(Q)

Putting values:

dG = 1573 + 8.314*310*ln(1.3/6.6)

Solving we get:

dG = -2.614 kJ/mol