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macmilan leaning A suspension of purified mitochondria was prepared in a buffer

ID: 207028 • Letter: M

Question

macmilan leaning A suspension of purified mitochondria was prepared in a buffer at pH 6.4 and 25 °C. The pH of the matrix was determined to be 7.4, and the electrical charge difference across the inner mitochondrial membrane (IMM) was 0.19V (inside negative) Suppose that one mole of H* ions flow through the FoF1-ATP synthase (embedded in the IMM). What is the total energy available for ATP synthesis? Number k.J mol What percentage of the total energy calculated above is associated with: Number Number the differencein pH across the IMM. the differencein electrical charge across the IMM. %

Explanation / Answer

Ans. #I. Free energy change of a proton gradient, when proton moves from its higher to lower concertation, available to ATP synthase is given by-

            dG = -2.303 RT(dpH) + nF(dV)               - equation 1

            where,

R = (0.001987 kcal mol-1K-1 or 0.008314 kJ mol-1 K-1)

                        T = 37.00C = 300.15 K

                        (dpH) = (pH siteH+moves from, MM) – (pH siteH+moves to, IMS)

                                    Or, dpH = 7.4 – 6.4 = 1.0

                        (dV) = (V siteH+moves from, MM) – (pH siteH+moves to, IMS)

                                    Or, dV = 0.19V

Putting the values in equation 1-

dG’ = [-2.303 (0.008314 kJ mol-1 K-1 x 298.15 K) x 1.0] +

[1 x 96.485 kJ V-1 mol-1 x (-0.19V)]

Or, dG’ = (-5.71 kJ / mol) + (-18.33 kJ / mol)

Hence, dG = -24.04 kJ/ mol

#II. % energy associated with dpH = (energy of dpH / Total energy) x 100

                                                            = [-5.71 kJ mol-1 / (-24.04 kJ mol-1) x 100

                                                            = 23.75 %

#III. % energy associated with dV = (energy of dV / Total energy) x 100

                                                            = [-18.33 kJ mol-1 / (-24.04 kJ mol-1) x 100

                                                            = 76.25 %