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And how is the energy gained from the lowering of the "energy level" of the electron used to generate the chemiosmotic gradient?

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    $\begingroup$ I think this is a completely reasonable question, why the close votes? Can someone at least comment on what the problem is, and give the OP a chance to edit, before voting to close? $\endgroup$
    – Roland
    Commented Feb 25, 2017 at 8:57

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All of the members of the Electron Transport System (ETS) are organized in "complexes", clusters of enzymes and other proteins. These complexes are in proximity to one another such that electrons can be passed between them. The different cytochromes, and other electron handlers, have different affinities (attractions) for electrons. Those at the beginning of the ETS have less affinity for electrons than those at the end. The reason the electrons move through the ETS is due to that greater attraction of electrons by the subsequent complexes in the pathway. These complexes are positioned with respect to one another so that no "short circuits" occur; the electrons skip no steps down the ETS. Finally, the "low energy" electron from the last complex is mopped up by that great electron-hog oxygen.

The energy lost by the electron as it passes between carriers is not directly harvested to pump protons. The importance of this "loss of energy" is that it makes the transfer of the electrons a spontaneous process insuring their passage down the ETS. Typical entry into the ETS involves NADH donating a hydrogen ion to the matrix and a pair of electrons to NADH dehydrogenase. The electrons are eventually transferred to the mobile coenzyme Q. This now attracts a pair of hydrogen ions from the mitochondrial matrix which then induces the operation of a proton pump. This pumps hydrogen ions into the inner-membrane space creating the electrochemical gradient.

So, in general the arrival of electrons, with the attendant hydrogen ions, changes the conformation (shape) of pumps which translocate the protons to build up the gradient. The electrons are ultimately passed on to oxygen, forming water with the help of some spare hydrogen ions. This removal of the electrons allows the ETS members to be receptive to another load of electrons from the NADH.

Reference 2nd paragraph: http://www.physiologymodels.info/metabolism/ETSOXPHOS/ets.htm This is a good site to read more and view some diagrams, especially focus on the second panel, "Complex I".

Reference 1st paragraph: Molecular Biology of the Cell; Alperts, Watson, et.al.

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  • $\begingroup$ Thanks for the really detailed answer - the link was very interesting; it's just that I'm in the dark as to how these electrons are transferred between complexes? It seems counter intuitive that they can just hop between molecules when all other electron transfers in biochemistry require some kind of enzyme-mediated redox reaction. $\endgroup$
    – ooakley
    Commented Feb 25, 2017 at 8:18
  • $\begingroup$ @user30109. These are oxidation-reduction reactions. And some of the members of the complexes are appropriately called enzymes. $\endgroup$
    – bpedit
    Commented Feb 25, 2017 at 16:02
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Good question. The common picture of the electron transport chain as a sequence of molecular machines that pass along a "high energy electron" is biochemically rather misleading I think. What is really going on is just a series of energy-releasing (exothermic) redox reactions. The respiratory complexes are enzymes that catalyze these reactions and couple the energy released in each reaction to proton pumping against a gradient.

As an example, lets consider the reaction carried out by Complex I, where NADH is oxidized:

NADH + H$^+$ + CoQ $\iff$ NAD$^+$ + CoQH$_2$

This reaction transfers a hydride ion (H$^-$) carrying two electrons from NADH, which is accepted by CoQ. Since CoQ is a much better electron acceptor than NAD$^+$, this reaction is highly favorable with a $\Delta G$ of about -85 kJ. So quite a bit of energy is released, and part of this energy is captured by Complex I to pump four protons across the inner memberane.

Note that electrons are not "traveling" on their own through Complex I somehow. The electrons are bound to molecules that participate in a redox reaction. And it doesn't make much sense to say that a specific electron's "energy level" is decreased. Rather, the compounds on the right hand side have a lower free energy $G$ in total than those on the left hand side, and therefore the reaction overall releases energy (the free energy difference $\Delta G$ is negative). Also, if you look at the chemical structures of CoQ and CoQH2 --- and you absolutely should look at structures in biochemistry, the names alone are not very helpful --- you will find that the electrons involved are actually delocalized in CoQH2, so there's no way to figure out which electron goes where.

The same reasoning goes for the other respiratory complexes. Complex III oxidizes CoQH$_2$ back to CoQ by coupling it to reduction of cytochrome C,

CoQH$_2$ + 2 Ferricytochrome-C $\iff$ CoQ + 2 Ferrocytochrome-C + 2 H$^+$

This reaction is also favorable, and again the energy released is used to pump protons. Finally, Complex IV oxidizes cytochrome C back and transfers electrons to O$_2$, also an energy-releasing redox reaction. The reaction mechanisms are much more complicated of course, involving various chemical groups bound to the enzymes, but this is the net result.

So the "respiratory chain" is not some conveyor belt for electrons; it is a sequence of coupled redox reactions. The net result of the Complex I + III + IV reactions is that NADH has lost electrons and oxygen has gained electrons, but it's not necessarily the same electrons, and it doesn't make sense to speak of "energy level" of electrons in this context. To understand the energetics, we must look at all the compounds and reactions involved.

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  • $\begingroup$ @tomd, as I interpreted the question, the OP was aware of the chemoosmotic principle but was unclear about the energetics that provide the energy for proton pumping. $\endgroup$
    – Roland
    Commented Feb 25, 2017 at 21:28

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