Mitochondria and Chloroplast Bioenergetics

(Terminology Consistent With Lehninger, Nelson, and Cox, Biochemistry , 1993)

Suggestions and Corrections to

Gale Rhodes
Chemistry Department
University of Southern Maine

Revised 2006/07/14

This handout shows how to calculate the free energy available from A) redox reactions; B) concentration gradients; C) voltage gradients; and D) proton or other ion gradients, which have two components, concentration and voltage. I have taken pains in all sections to employ sign conventions consistently, so take note of how each process is defined, and how the definition determines the sign of the free-energy change.

A. Free-energy change during a redox reaction

Consider the oxidation of ubiquinone by cytochrome c. How much free energy is available from this process?

Half reactions	E0'(V)
a) UQ (ox) + 2e- + 2H+ --> UQH2 (red)	0.04
b) Cytochrome c-Fe3+ (ox) + e- --> Cytochrome c-Fe2+ (red)	0.23

Obtain balanced equation for the process described: reverse a) and add 2 x b):

rev-a):	UQH2 ---> UQ + 2 e- + 2 H+
2 x b):	2 Cytochrome c-Fe3+ + 2 e- ---> 2 Cytochrome c-Fe2+
Sum:	UQH2 + 2 Cytochrome c-Fe3+ ---> UQ + 2 H+ + 2 Cytochrome c-Fe2+

DE⁰' = E⁰'_red - E⁰'_ox = 0.23 V - 0.04 V = 0.19 V.

DG⁰' = -nFDE ⁰' = -(2)(96.48 kJ/V-mol)(0.19V) = -36.7 kJ/mol

This process is spontaneous under standard conditions (K_eq > 1).

The actual, or cellular, DG depends on the cellular ratios of oxidants and reductants:

DG = DG ⁰' + RT ln ([UQ][H⁺]²[Cyt c-Fe²⁺]²)/([UQH₂][Cyt c-Fe³⁺]²)_cell

NOTE: In calculations using this equation, and whenever you have concentration terms inside a logarithm term, remember that the proper terms are really activities and not concentrations. The activity of the solute is its actual concentration divided by its standard concentration, so activities are unitless (which is good, because it's very hard to attach a physical meaning to units like ln[mol/L] ). Click here to learn how to compute activies for solutes, gases, hydrogen ions, and water or other solvents.

B. Free-energy change during solute movement across a concentration gradient

Consider the movement of protons from the cytoplasm into the matrix of the mitochondrion:

H⁺_out <==> H⁺_in

How much free energy is available from the movement of protons down the concentration gradient created by electron transport?

DG = DG ⁰' + RT lnQ = DG ⁰' + RT ln([H⁺_in]/[H⁺_out])

DG ⁰' = 0 because K_eq for the process is 1.0 (DG ⁰' = - RT ln K_eq),

so DG = RT ln([H⁺_in]/[H⁺_out])

To express DG in terms of the pH gradient (rather than the concentration gradient), change ln to log (that is, log₁₀) and expand the log term:

DG = 2.303 RT log([H⁺_in]/[H⁺_out])
DG = 2.303 RT (log[H⁺_in] - log[H⁺_out]) = -2.303 RT (pH_in - pH_out)
or

DG = - 2.303 RT DpH ===> (NOTE: DpH = pH_in - pH_out )

If proton pumping maintains a pH gradient of 1.4 units (lower outside ), then

DpH = + 1.4 and

DG = - 2.303 (8.315 x 10^-3 kJ/mol-K)(298K)(1.4) = - 7.99 kJ/mol

This is the free-energy change attributable to the concentration gradient. This would be the free energy available from a gradient of a nonionic solute, or from an ionic gradient if the movement of other ions maintained equal voltage on both sides of the membranes, as is true in chloroplasts. The movement of other ions across the thylakoid membrane maintains electrical neutrality across the membrane, despite light-driven proton pumping into the thylakoid lumen. In particular, as H⁺ moves from the stroma into the lumen, Mg²⁺ moves out of the lumen into the stroma. So in chloroplasts, the proton gradient is simply a concentration gradient, and there is no accompanying voltage gradient.

C. Free-energy change during solute movement across a voltage gradient

In mitochondria, electron transport drives proton pumping from the matrix into the intermembrane space. There is no compensating movement of other charged ions, so pumping creates both a concentration gradient and a voltage gradient, the latter resulting from the excess of proton charges outside the inner mitochondrial membrane. This voltage component makes the proton gradient an even more powerful energy source. Here's how to calculate the contribution of voltage to the energy available from such a gradient.

Define the membrane voltage gradient, or membrane potential, as

Dy_m = y_in - y_out.

How much free energy is available from the movement of protons down the voltage gradient created by electron transport? NOTE: Same process, as before: H⁺_out <==> H⁺_in.

D G = -nF Dy⁰' + nFDy_m , and Dy⁰' = 0 (membrane potential = 0 under standard conditions.)

so

DG = nF Dy_m

(NOTE: Dy_m = y_in - y_out .)

If proton pumping maintains a voltage gradient of 0.14 V, positive outside, then Dy_m is negative, as defined here.

Dy_m = y_in - y_out = - 0.14 V.

DG = (1)(96.48 kJ/V-mol)(- 0.14 V) = - 13.5 kJ/mol

This is the free-energy change attributable to the voltage gradient.

D. Proton-motive force: free energy change during movement across a proton gradient

"Proton-motive force" (Dp) is a Dy- or DE-like term (DE is electromotive force ) that combines the concentration and voltage effects of a proton gradient such that

DG = - nFDp⁰' + nFDp, and Dp⁰' = 0 (proton motive force = 0 under std conds),

so

DG = nF Dp

DG can also be expressed as the sum of the DpH and Dy_m contributions:

DG = - 2.303 RT DpH + nFDy_m

so nFDp = - 2.303 RT DpH + nFDy_m

or

Dp = (-2.303 RT DpH)/nF + Dy_m

This way of expressing the proton-motive force was probably adopted because of its elegant resemblance to the Nernst equation, but the clearest expression of the energy available from a proton gradient is probably

DG = - 2.303 RT DpH + nFDy_m

Remember (this is a recording) that DpH = pH_in - pH_out and Dy_m = y_in - y_out because we started out by considering the movement of protons from the cytoplasm to the matrix:

H⁺_out <==> H⁺_in

The total free energy available from the movement of 1 mole of protons from the cytoplasm to the matrix under cellular conditions (DpH = 1.4, Dy = 0.14 V) is the sum of the free energy changes calculated in sections B and C:

DG = - 2.303 RT DpH + nFDy = -7.99 kJ/mol - 13.5 kJ/mol
DG = -21.5 kJ/mol

E. Mitochondrial proton gradient as a source of energy for ATP synthesis

Estimated consumption of the proton gradient by ATP synthesis is about 3 moles protons per mole ATP. If DG = 50 kJ/mol for ATP synthesis in mitochondria, then by Hess's law, DG = 50 + 3(- 21.5) = - 3.4 kJ/mol, and the process of synthesis of ATP at the expense of the proton gradient is spontaneous under mitochondrial conditions.

Estimated proton pumping associated with electron transport is 10 protons per electron pair passed from NADH to O₂. In addition to the 3 protons consumed per ATP synthesized, one proton is spent in transporting ATP to the cytoplasm, for a total of 4 H⁺ per cytoplasmic ATP. So the yield of cytoplasmic ATP per electron pair is

(1 ATP/4 H⁺)/(10 H⁺/electron pair) = 2.5 ATP/electron pair.

Only 6 protons are pumped for each electron pair passed from FADH₂ to O₂, so ATP yield is

(1 ATP/4 H⁺)/(6 H⁺/electron pair) = 1.5 ATP/electron pair.

These are the ATP yields most commonly quoted by careful textbook authors.

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