The conversion of of fructose-6-phosphate (F6P) to fructose-1.,6-bisphosphate (F

Question

The conversion of of fructose-6-phosphate (F6P) to fructose-1.,6-bisphosphate (F1,6BP) is an important and highly regulated step in glycolysis. The concentrations of F6P F1,6BP, ATP and ADP were measured in muscle tissue at 25 C at pH-7.0. The concentrations and chemical reaction are given in the table below AG 4.4 kcal Concentration (mM) F6P F1,6BP 0.012 ADP 1.2 (2pts) Calculate the Keq for this reaction (Ipt). Where does the equilibrium lie? With the reactants (F6P) or products (F1,6P)? (2pts) Calculate the free energy change (AG') for the reaction in muscle tissues under these conditions (Ipt) Is this reaction exergonic or endergonic?

Explanation / Answer

Ans. #I. Using the equation dG0’ = - RT lnKeq               - equation 1

Where, dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = equilibrium constant under standard condition

R = (0.001987 kcal mol-1K-1 or 0.008315 kJ mol-1 K-1)

Putting the values in equation 1-

            -4.4 kcal mol-1 = - (0.001987 kcal mol-1K-) x 298.15 K x ln Keq

            Or, 2.303 log Keq = 4.4 kcal mol-1 / 0.59242405 kcal mol-1

            Or, log Keq = 7.42711 / 2.303 = 3.2250

            Or, Keq = antilog 3.225

            Hence, Keq = 1678.70

Therefore, Keq for the reaction = 1678.70

Note 1: It is NOT mentioned how you have to calculate Keq. If the Keq is to be calculated using the concentrations of chemical species (as it would be more appropriate for the step #III) Q can be taken equivalent to Keq under experimental conditions.

#II. Since the Keq is greater than 1.0, the equilibrium lies to the far right, towards the product (F-1,6-bisP).

Note 2: If the Keq is to be calculated using the concentrations of chemical species (as it would be more appropriate for the step #III), the value of calculated Keq’ = 0.01348.

Since Keq’ is less than 1, the equilibrium shifts to the left (towards, F-6-P).

#III. Balanced reaction-       F-6-P + ATP --------> F-1,6-bisP + ADP

# All concentrations must be in terms of Molarity.

Equilibrium constant under experimental condition,

Keq’ = [F-1,6-bisP] [ADP] / ([F-6-P] [ATP])

            Or, Keq’ = (0.012 x 10-3) (1.2 x 10-3) / [(0.089 x 10-3) (12.0 x 10-3)]

            Hence, Keq’ = 0.01348

Using the equation dG’ = dG0’ + RT lnKeq’                   - equation 2            

Where, dG’ = calculated/ experimental free energy change

dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq’ = Equilibrium constant under experimental condition,

Putting the values in equation 2-

            dG’= (-4.4 kcal mol-1) + (0.001987 kcal mol-1K-) x 298.15 K x ln 0.01348

            dG’ = (-4.4 kcal mol-1) + (0.001987 kcal mol-1K-) x 298.15 K x (-4.3065481735)

            Or, dG’ = (-4.4 kcal mol-1) – 2.551 kcal mol-1

            Hence, dG’ = -6.951 kcal mol-1

Hence, calculated dG’ = -6.951 kcal mol-1

#IV. The original reaction (F-6-P + ATP <-----> F-1,6-bisP + ADP) consumes ATP, so it is an ENDERGONIC reaction.

However, note that, dG’ < dG0’, i.e. dG’ is more negative than dG0’, the reaction (in muscles) under experimental conditions is exergonic because the reaction drives in backward direction with hydrolysis of F-1,6-bisP into F-6-P and release of some energy- which is used to form ATP from ADP.

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