When succinate is oxidized to fumarate in the citric acid cycle, electrons are t

Question

When succinate is oxidized to fumarate in the citric acid cycle, electrons are transferred to bound
FAD to produce FADH2 in complex II instead of NADH. Electrons then travel through Complex II to QH2,
similar to those harvested from fatty acid oxidation transferred to FAD+ bound to acyl-Co dehydrogenase.

a. For the reaction, succinate + FAD(ComplexII) --> fumarate + FADH2 (Complex II), calculate the delta G '

b. Calculate the delta G ' for oxidation of succinate to fumarate using NAD+ as the electron acceptor instead.

c. Based on your calculations, can you explain why a different electron acceptor must be used in this case other than the more commonly observed NAD+?

Explanation / Answer

A) What I am doing in the equations below is making it so I can add the two equations so the hydrogens and electrons cancel out. This allows me to just switch the sign of E and add the values together getting an Etotal value.

Succ -> Fumerate + 2H+ + 2e-     E= - 0.031V

NAD+ + 2H+ + 2e- -> NADH          E = - 0.315 V

Adding the two gives Etotal = - 0.346 V showing it's very very negative for delta E.

delta G = - nF Etotal where n is the number of electrons moving (2 in this case) and F is the faraday constant 96485 J/V*mol. Delta G for this turned out to be 66.7 KJ/mol, a very large positive gibbs free value which shows this will not happen readily.

Doing this same thing with succinate but now with FAD.

Succ -> Fumerate + 2H+ + 2e-     E= - 0.031V

FAD + 2H+ + 2e- -> FADH2           E= - 0.040 V

Adding these values gives E = - 0.071 V

Delta G = 13.7 KJ/mol, a less positive value than NAD+ produced. This means it is a better choice for the body to use FAD because the barrier to overcome is much smaller. 66.7 KJ/mol vs 13.7 KJ/mo <-- smaller barrier

For future calculations if you need to multiply the equation through by 2 or 3 to make terms cancel, do not multiply the E value by 2 or 3. E values are an "intrinsic" property just like density. If you have double the amount of water it does not mean you have double the density of water. It does not depend on how much you have, it is a set value for the reaction and only needs to switch signs if you swap products and reactants.

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