The following metabolite concentrations were measured in a perfused rat heart at

Question

The following metabolite concentrations were measured in a perfused rat heart at 308 K

Metabolite

Concentration

   fructose 6-phosphate

   60 µm

   fructose bisphosphate

   9 µm

   ATP

   5.3 µm

   ADP

   1.1 µm

   AMP

   95 µm

The enzyme phosphofructokinase catalyses the reaction

fructose 6-phosphate + ATP <=> fructose bisphosphate + ADP               0’= -17.7 kJ/mol

and the enzyme adenylate kinase catalyses the reaction

2ADP <=> ATP + AMP               0’= +2.1 kJ/mol

Are these enzyme catalyzed reactions at equilibrium in perfused rat heart? If not, what are the values of 0’ for the reactions? Comment on results.

Metabolite

Concentration

   fructose 6-phosphate

   60 µm

   fructose bisphosphate

   9 µm

   ATP

   5.3 µm

   ADP

   1.1 µm

   AMP

   95 µm

Explanation / Answer

Ans. F-6-P + ATP ß-------------> F1,6-bisP + ADP                   ; dG0’ = - 17.7 kJ/mol

Equilibrium constant, K = ( [F1,6-bisP] [ADP] ) / ([F-6-P] [ATP])

Or, K = [ (9 x 10-6 M) (1.1 x 10-6 M) ] / [(60 x 10-6 M) (5.3 x 10-6 M)]

            = (9 x 1.1) / (60 x 5.3)

            = 3.11x 10-2

Hence, equilibrium constant, K = 3.11x 10-2

Now,

Using the equation dG = dG0’ + RT lnK              - equation 1

                        Where, dG = calculated/ experimental free energy change = ?

                                                dG0’ = standard/ theoretical free energy change = -17.7 kJ/mol

                                                R = 0.008314 kJ mol-1K-1

                                                T = 308 K

Putting the values in equation 1

            dG = (-17.7 kJ/mol) + (0.008314 kJ mol-1K-1) (308 K) ln (3.11x 10-2)

            or, dG = (-17.7 kJ/mol) +2.56 kJ mol-1 x (- 3.47)

            or, dG = -17.7 kJ/mol - 8.8832 kJ/mol = - 26.5832 kJ/mol

Hence, experimental free energy change = - 26.5832 kJ/mol

Conclusion, since dG < dG0 i.e. experimental free energy is less than (or, NEGATIVE) when compared to standard free energy change, the reaction spontaneously proceeds in forward direction. That is, the reaction facors formation of F1,6-bisP.

If dG = dG0’, the reaction would be at equilibrium.

Part 2: Equilibrium constant for adenylate cyclase reaction,

            K = ([ATP] [AMP]) / [ADP]2 = (5.3 x 10-6 M) (95 x 10-6 M)] / (1.1 x 10-6 M)2

            Hence, K = 416.115

Putting the values in equation 1-

            dG = (2.1 kJ/mol) + (0.008314 kJ mol-1K-1) (308 K) ln (416.15)

            or, dG = (2.1 kJ/mol) + 15.49 kJ mol-1 = 17.59 kJ mol-1

Hence, dG = 17.59 kJ mol-1

Since dG > dG0’, the reaction is non-spontaneous.

Since, dG is NOT equal to dG0’ the reaction is also NOT at equilibrium.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.