2 The first step in glycolysis is the formation of glucose-6-phosphate from glu

Question

2 The first step in glycolysis is the formation of glucose-6-phosphate from glu glucose + Pi glucose-6-phosphate + H2O (a) Calculate the equilibrium ratio at 25 °C of glucose-6-phosphate to glucose if the phosphate concentration is 2.5 mmol L-1 . Species glucose glucose-6-phosphate -1325.00 H20 Pi AGo/kJ mol-1 -436.42 -157.28 -1058.56 (b) If the phosphate ion is transferred from ATP, what is the cquilitbrium ntio of glucose-6-phosphate to glucose? The in vivo concentrations of ATP and ADP are, respectively, 5 mmol L- and 0.5 mmol L-1 and the standard free energy change of hydrolysis of ATP in the biochemical convention is -37.64 kJ mol-1.

Explanation / Answer

Ans. #2.a. dG0’ of reaction = Sum of dfG0’ of products - Sum of dfG0’ of reactants

                                                = (-1325.00 – 157.25) kJ mol-1 - (-436.00 – 1058.56) kJ mol-1

                                                = -1482.25 kJ mol-1 + 1494.56 kJ mol-1

                                                = + 12.31 kJ mol-1

Therefore, dG0’ of reaction = 12.31 kJ mol-1

# Given, [Pi] = 2.5 mmol L-1 2.5 mM = 0.0005 M

Equilibrium constant for the reaction, k = [G-6-P] / ([G] [Pi])

            Or, k = [G-6-P] / ([G] x 0.0025)

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

                        Where,

                        dG0 = Standard Gibb's free energy

                        R = Universal gas constant = 0.0083146 kJ mol-1 K-1

T = Temperature in kelvin

k = Equilibrium constant (Keq)

Putting the values in equation 1-

            dG0 = -RT ln { [G-6-P] / ([G] x 0.0025) } = -RT { (ln [G-6-P] / [G]) – ln 0.0025}

            Or. 12.31 kJ mol-1 = - (0.0083146 kJ mol-1 K-1 x 298 K) { ln [G-6-P] / [G]) + 5.9914}

            Or, 12.31 kJ mol-1 / (- 2.4777508 kJ mol-1) = ln [G-6-P] / [G]) + 5.9914

            Or, - 4.968 – 5.9914 = 2.303 log [G-6-P] / [G]

            Or, -10.9594 / 2.303 = log [G-6-P] / [G]

            Or, [G-6-P] / [G] = antilog (- 4.75875)

            Hence, [G-6-P] / ([G]) = 0.000017428 = 1.7428 x 10-5

#b. Balanced reaction:      G + ATP <--------> G-6-P + ADP

Equilibrium constant, k = [G-6-P] [ADP] / ([G] [ATP])

Or, k = [G-6-P] [0.5] / ([G] [5])

Or, k = 0.10 [G-6-P] / [G]

Putting the values in equation 1-

            Or. -37.64 kJ mol-1 = -(0.0083146 kJ mol-1 K-1 x 298 K) ln ( 0.10[G-6-P] / [G])

            Or, -37.64 kJ mol-1 / (- 2.4777508 kJ mol-1) = (ln [G-6-P] / [G]) + ln 0.1

            Or, 15.1912 + 2.303 = (ln [G-6-P] / [G]

            Or, 17.4942 = 2.303 log [G-6-P] / [G]

            Or, 17.4942 / 2.303 = log [G-6-P] / [G]

            Or, [G-6-P] / [G] = antilog (7.5962)

            Hence, [G-6-P] / [G] = 3.946 x 108

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