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38. 1,3-Bisphosphoglycerate is used in Glycolysis to generate ATP in the second half of the Glycolytic Cycle. The Go for hydrolysis of Bisphosphoglycerate is -49.4 kJ/mol. What would be the equilibrium constant for the reaction at human body temperature. (HINT: Remember the hydrolysis energy for ATP, how much extra energy would be given in transfer of this phosphate to ADP, and that there are two hydrolysis steps involved in this calculation!) a. 4134.0 b. 2055.8 c. 1530.2 d. 2.0 x 10E6

Explanation / Answer

Ans. Given, standard enthalpy of hydrolysis-

ATP + H2O --------> ADP + Pi                   ; dG0 = -30.5 kJ/mol             - Rxn 1

1.3-BisPGA + H2O -----> ATP + 3-PGA     ; dG0 = -49.4 kJ/mol             - Rxn 2

# To determine the Keq for “Formation of ATP from ADP and 1,3-BisPGA”-

The desired reaction can be written as the sum of “reverse of Rxn 1” and reaction 2 as follow-

            ADP + Pi -----------> ATP + H2O               ; dG0 = + 30.5 kJ/mol           - Rxn 1’

    (+) 1,3-BisPGA + H2O ---> ATP + 3-PGA       ; dG0 = - 49.4 kJ/mol            - Rxn 2

    1,3-BisPGA + ADP + PI --> 2 ATP + 3-PGA; dG0net = - 18.9 kJ/mol      - Rxn 3

Body temperature = 37.00C = 310.15 K

# 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 given condition

R = 0.0083146 kJ mol-1 K-1

Putting the values in equation 3-

            -18.9 kJ/mol = - (0.0083146 kJ mol-1 K-1) x 310.15K ln Keq

            Or, -18.9 kJ mol-1 / (- 2.57877319 kJ mol-1) = ln Keq

            Or, 7.329= 2.303 log Keq

            Or, log Keq = 7.3291 / 2.303

            Or, Keq = antilog (3.1824)

            Hence, Keq = 1521.95

Hence, equilibrium constant for the specified reaction (Rxn 3), Keq = 1530.2

Correct option. c. 1530.2 (nearest value).

The deviation from given value might have arisen due to accounting different number of significant figures.

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