Phosphoglycerate mutase catalyzes a reaction that occurs in both glycolysis and

Question

Phosphoglycerate mutase catalyzes a reaction that occurs in both glycolysis and gluconeogenesis. a. The Delta G degree' for the phosphoglycerate mutate reaction is 4.4 kJ/mol (in the glycolysis direction). What is the equilibrium constant for this reaction at 25 degree C (298K)? (R = 8.314 J/mol K, T = 298K) b. In typical cellular conditions, Delta G for this reaction is 0.83 kJ/mol. If the concentration of 2-phosphoglycerate is 0.028 mM, what is the concentration of 3-phosphoglycerate? (Use T = 298K.)

Explanation / Answer

Ans. #A. 3-phosphoglycerate --PGA mutase -> 2-phosphoglycerate ; dG0’=+4.4kJ/mol

At equilibrium,

            Standard Gibb’s free energy, dG0’ = -RT ln K     - equation 1

                        Where,

                        dG0’ = Standard Gibb's free energy

                        R = Universal gas constant = 0.008314 kJ mol-1 K-1

T = Temperature in kelvin

K = Equilibrium constant

Putting the values in equation 1-

            4.4 kJ mol-1 = - (0.0083146 kJ mol-1 K-1) x 298K x (ln K)

            Or, 4.4 kJ mol-1 = - 2.477572 kJ mol-1 x (ln K)

            Or, 4.4 / 2.477572 = - ln K

            Or, ln K = - 1.7759

            Or, 2.303 log K = - 1.7759

            Or, log K = - 1.7759 / 2.303 = - 0.7711

            Or, K = antilog (- 0.7711) = 0.169

Therefore, the equilibrium constant of the forward reaction = 0.169

Ans. #B. Using the equation dG = dG0’ + RT lnK                      - equation 2

                        Where, dG = experimental free energy change

                                                dG0’ = standard/ theoretical free energy change                                                            R = 0.008314 kJ mol-1K-1

                                                T = temperature in kelvin

                                                K = equilibrium constant under given condition.

Putting the values in equation 2-

            0.83 kJ mol-1 = (0.0083146 kJ mol-1 K-1) x 298K x (ln K)

            Or, 0.83 kJ mol-1 = 2.477572 kJ mol-1 x (ln K)

            Or, ln K = 0.83 kJ mol-1 / 2.477572 kJ mol- = 0.3350

            Or, log K = 0.3350 / 2.303 = 0.1456

            Or, K = antilog (0.1456) = 1.398

            Hence, K = 1.398     - under experimental (cellular) conditions.

Now,

Equilibrium constant under cellular condition, K = [2-PGA] / [3-PGA]

            Or, 1.398 = 0.028 mM / [3-PGA]

            Or, [3-PGA] = 0.028 mM / 1.398 = 0.0200 mM

Thus, [3-PGA] = 0.0200 mM         - under cellular conditions.

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