BCHM2972
2006
Lecture 6
CYTOCHROME C OXIDASE (aka COMPLEX IV)
Cytochrome c oxidase (Complex IV) catalyses the final reduction in the electron
transport chain. Overall, the reaction is O2 + 4 H+ + 4 e-  2 H2O (irreversible)and
involves 4 electrons (passed one at a time) from reduced cytochrome c to
cytochrome c oxidase to complete the four-electron reduction of one O2 molecule.
(It’s called cytochrome c oxidase because it oxidises cytochrome c!)
Complex IV is composed of multiple subunits and contains four redox centers. Two
redox centers, haem a and a pair of copper atoms known as the CuA center, are on
subunit II and accept electrons first from cytochrome c .(Remember that cytochrome
c can only carry one electron at a time so it shuffles back and forth, between Complex
III and IV, collecting and dropping off one electron at a time) Electrons are then
passed from haem a nd CuA to the other two redox centers , CuB and haem a3, which
are very closely associated and form the binuclear centre on subunit I. (Don’t confuse
haem a (which accepts the electrons from cytochrome c) with haem a3 in the
binuclear center (which accepts electrons from haem a and participates directly in the
reduction of oxygen). I don’t know why they can’t have better names!.
To start the cycle, 2 electrons are passed from cytochrome C to haem and-CuA. (so
then cytochrome c has become oxidised) These two electrons are then passed to the
oxidised binuclear center [Fe3+—Cu2+] to fully reduce it  [Fe2+ —Cu+]. At the same
time, an H+ from the mitochondrial matrix and a hydroxyl from the binuclear center
form the first water molecule produced in the electron transport chain. The hydroxyl
gets regenerated later in the cycle. Then a diatomic oxygen molecule binds, shared by
the Fe and Cu in the binuclear centre. This arrangement ensures that reactive oxygen
that will be produced as an intermediate in the reaction does not get released to cause
trouble!
The reduction of O2  O22- 2 H2O requires the nearly simultaneous
rearrangement of 4 electrons. However, the fully reduced [Fe2+—Cu+] binuclear
complex can readily donate only 3 electrons to its bound O2 before reaching its fully,
fully oxidized [Fe4+—Cu2+] state. (Don’t get confused, the [Fe3+—Cu2+] form of the
binuclear center at the start of the cycle is also considered to be fully oxidised!).
The 4th electron required to reduce O2  O22- is temporarily donated (with a proton)
by a the phenolic –OH on a tyrosine residue that is located close to the binuclear
center. Remember the other three electrons could be provided by the binuclear
complex: reduced Fe2+ loses 2 electrons Fe4+ and Cu + loses 1 electron  Cu2+. The
four electrons are given to oxygen. (at this point, only 2 electrons have entered the
complex). This redistribution of electrons yields an ‘oxyferryl’ complex.
A third electron (transferred from cytochrome c via haem a and CuA) and two protons
from the matrix complete the reduction of oxygen, releasing the second H2O.molecule
and regenerating the tyrosine (paying back the debt from earlier when tyrosine
participated in the electron rearrangement).
A fourth and final electron transfer (and proton from the matrix) yields the oxidized
[Fe3+ —Cu2+] complex, and regenerates the hydroxyl, thereby completing the
catalytic cycle.
Note that in the first H2O the O and one of the Hs came from the hydroxyl (OH)
which is later regenerated by one of the oxygen atoms from the incoming O2 while the
other the oxygen atom from the incoming O2 leaves as the second H2O molecule.
So, 4 protons have come from the matrix to form 2 H2O (one hangs out in the
hydroxyl group of the binuclear center for a while!) and 4 protons are meanwhile
pumped from the matrix to form the proton gradient.