Question

If the intracellular concentrations of a metabolite (M-OH) and its phosphorylated form (M-OPO32-) were 3 mM and 0.08 mM, respectively, and if the intracellular concentrations of ATP and ADP were 3.8 mM and 0.17 mM, respectively, what would be the numerical value of \Delta Δ G (in kcal per mol to the nearest hundredth) for the following reaction: M-OH + ATP <--> M-OPO32- + ADP + H+? Assume a temperature of 37 °C and a pH of 7.4. To solve this problem, you will need to know the standard free energies of hydrolysis of the phosphorylated metabolite and of ATP. These values are –3.4 kcal/mol and –7.3 kcal/mol, respectively.

Answer 1

solution:

MOH + ATP -----> MOPO3^2- + ADP

Given,

            Standard free energy change of ATP hydrolysis = -7.3 kcal/ mol

            Standard free energy change of MOPO3² hydrolysis = -3.4 kcal/ mol

ATP is on left side of equation, whereas MOPO3^2- on the right side of the equation. So, overall standard free energy change of the reaction is given by-

            dG⁰ (overall) = dG⁰ (ATP) - dG⁰ (MOPO3^2-)

                                    = (- 7.3 kcal/ mol) – (3.4 kcal/ mol)

                                    = -3.9 kcal/ mol

                                    = -16.3176 kJ/ mol

The value of overall dG^o = -3.9 kcal/ mol means that out of total 7.3 kcal/ mol energy released from ATP hydrolysis during the reaction, only 3.9 kcal/ mol is released to surrounding. The rest energy 3.4 kcal/ mol is still conserved in (MOPO3^2-), which is released if the compound undergoes hydrolysis.

Using the equation dG = dG^0’ + RT lnK             - equation 1

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

                                                dG^0’ = standard/ theoretical free energy change                                                            R = 0.0083146 kJ mol^-1K^-1

                                                T = temperature in kelvin

                                                K = equilibrium constant under given condition.

Now, equilibrium constant of the overall reaction is given by (all concentrations are converted to Molar concertation)-

            K = ([MOPO3^2-] [ADP]) / ([MOH] [ATP])

            Or, K = [(0.08 x 10^-3 ) (0.17 x 10^-3)] / [(3 x 10^-3 ) (3.8 x 10^-3)]

            Or, K = 0.0136 / 11.4 = 0.001192

            Hence, K = 0.001192

Putting the values of K’, overall dG⁰ and others in equation 1

            dG = (- 16.3176 kJ mol^-1) + (0.0083146 kJ mol^-1 K^-1) x 310K x (ln 0.001192)

            or, dG⁰ = (-16.3176 kJ mol^-1) + (2.49438 kJ mol^-1) x (- 6.73)

            Or, dG⁰ = ( -16.3176 kJ mol^-1) + (- 17.35222 kJ mol^-1)

            Or, dG⁰ = -33.669 kJ mol^-1

Therefore, dG⁰ of given reaction under given condition =  -33.669 kJ mol^-1