Assume that the free energy change \((\Delta G)\) associated with the movement of 1 mole of protons from the outside to the inside of a bacterial cell is \(-23 \mathrm{kJ} / \mathrm{mol}\) and \(3 \mathrm{H}^{+}\) must cross the bacterial plasma membrane per ATP formed by the bacterial \(\mathrm{F}_{1} \mathrm{F}_{0}-\mathrm{ATP}\) synthase. ATP synthesis thus takes place by the coupled process: $$3 \mathrm{H}_{\mathrm{out}}^{+}+\mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \rightleftharpoons 3 \mathrm{H}_{\mathrm{in}}^{+}+\mathrm{ATP}+\mathrm{H}_{2} \mathrm{O}$$ a. If the overall free energy change \(\left(\Delta G_{\text {overall }}\right)\) associated with ATP synthesis in these cells by the coupled process is \(-21 \mathrm{kJ} / \mathrm{mol}\), what is the equilibrium constant \(\left(K_{\mathrm{eq}}\right)\) for the process? b. What is \(\Delta G_{\text {synthesis }},\) the free energy change for ATP synthesis, in these bacteria under these conditions? c. The standard free energy change for ATP hydrolysis ( \(\Delta G^{\text {o' }}\) hydrolysis) is \(-30.5 \mathrm{kJ} /\) mol. If \(\left[\mathrm{P}_{\mathrm{i}}\right]=2 \mathrm{m} M\) in these bacterial cells, what is the \([\mathrm{ATP}] /[\mathrm{ADP}]\) ratio in these cells?

Step by step solution

01

Free Energy and Equilibrium Constant

First, one should recall the formula for the equilibrium constant using the Gibbs free energy: \(K_{eq} = e^{\frac{-\Delta G_{overall}}{RT}}\), where \(R = 8.314 \times 10^{-3} \mathrm{kJ} / \mathrm{mol.K}\) and \(T = 298 K\) (standard room temperature). Here, \(\Delta G_{overall} = -21 \mathrm{kJ/mol}\). Convert \(\Delta G_{overall}\) to J: \(\Delta G_{overall} = -21000\). Then, substitute the values into the equation: \(K_{eq} = e^{\frac{-(-21000)}{(8.314 \times 10^{-3} \times 298)}} = e^{848.45}\). This indicates that the reaction strongly favours the formation of ATP for this \(\Delta G_{overall}\).

02

Free Energy Change for ATP Synthesis

Given that 3 protons must cross the membrane for each ATP formed and each proton contributes \(-23 \mathrm{kJ/mol}\) of free energy, determine \(\Delta G_{synthesis}\) by calculating the total amount of energy contributed by the H+ ions: \(3 \times -23 \mathrm{kJ/mol} = -69 \mathrm{kJ/mol}\). \(\Delta G_{synthesis}\) is equal to \(\Delta G_{overall}\) - total energy contributed by the protons: \(-21 \mathrm{kJ/mol} - (-69 \mathrm{kJ/mol}) = 48 \mathrm{kJ/mol}\). Thus, \(\Delta G_{synthesis}\) = 48 \mathrm{kJ/mol}\).

03

ATP to ADP Ratio Calculation

Finally, to calculate the \([\mathrm{ATP}] /[\mathrm{ADP}]\) ratio one could use the relationship with the standard free energy of ATP hydrolysis, \(\Delta G^{\circ '}\) (hydrolysis) = -30.5 kJ mol^-1. Applying the expression: \(\Delta G^{\circ '}\) = -RT ln \([\mathrm{ATP}]/[\mathrm{ADP}][\mathrm{Pi}]\). Here, it is given that \([\mathrm{Pi}]\) = 2 mM (0.002 mol/L). Rearranging, you get \([\mathrm{ATP}]/[\mathrm{ADP}]\) = \(e^{\frac{-\Delta G^{\circ '}}{RT[\mathrm{Pi}]}}\). Substituting values: \([\mathrm{ATP}]/[\mathrm{ADP}]\) = \(e^{\frac{-( - 30.5 \times 10^3)}{(8.314 \times T \times 0.002)}}\). This gives the ATP/ADP ratio in these cells.

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