Question

Electron transfer translocates protons from the mitochondrial matrix to the external medium, establishing a pH gradient across the inner membrane (outside more acidic than inside). The tendency of protons to diffuse back into the matrix is the driving force for ATP synthesis by ATP synthase. During oxidative phosphorylation by a suspension of mitochondria in a medium of pH 7.4, the pH of the matrix has been measured as 7.7.

(a) Calculate [H+ ] in the external medium and in the matrix under these conditions.

(b) What is the outside-to-inside ratio of [H+ ]? Comment on the energy inherent in this concentration difference (just the energy from the concentration gradient, not the electrochemical potential.

(c) Calculate the number of protons in a respiring liver mitochondrion, assuming its inner matrix compartment is a sphere of diameter 1.5 microns. (volume of a sphere is 4/3 π r^3)

(d) From these data, is the concentration gradient sufficient to generate ATP from ADP and Pi? If not, suggest how the necessary energy for synthesis of ATP from ADP and phosphate arises.

Answer 1

a) Medium pH 7.4

[H+] = 10^-pH = 10^-7.4 = 3.98x10^-8 M

Matrix pH 7.7

[H+] = 10^-7.7 = 1.995x10^-8 M

(b) Ratio of Outside/inside = 3.98x10⁻⁸ / 1.99x10⁻⁸ = 2

Equation for free energy of the concentration difference across the membrane is:

ΔG= RT ln (C2)/(C1)

ΔG = 8.3144 * 298 ln (1.99/3.98)

ΔG = -1,717.40 J/mol

This value represents the energy in the concentration difference between the matrixand the external medium.

c) We can calculate the volume of the inner matrix compartment of the mitochondrionas; Vm= 4/3πr3

where; r = 0.75 μm.

Vm = 4/3 * 3.14 * 0.75 = 1.77 μm³

Matrix [H+] = 1.995x10⁻⁸ mol/L

(1.995x10⁻⁸mol)/(1x10¹⁵ μm³) x (6.24x10²³ molecules/mol) X 1.77 μm³= 22 protons

d) The energy available from the concentration difference can be calculated from thefollowing equation which is derived from the free energy change for a concentrationgradient as above:

ΔGt= RT ln (C2)/(C1)

for protons at 25°C: ln (C2)/(C1) = 2.3(log [H+]_P− log [H+]_N = 2.3 (pHN– pHP) = 2.3 ΔpH

therefore: 2.3(0.3)(2.48kJ/mol) = 1.7 kJ/mol of energy available forthe synthesis of 1 mol of ATP. Clearly this is not enough energy for ATP synthesis.

Hope this helps.