114
Chapter 4
the substrate level is much higher than the
Km
of the enzyme.
If cop(=cpo) = 0, the processes are energetically uncoupled and oxidation will not drive phosphorylation.
The ratio between
vp
and v0 is the operational P/O ratio:
v„
c BaAa + cmA a
( c„„
_
P_
_
po
o
____
PP
P
_ [
PP
v„
c
+ cn„A„
Q
OO
0
Op
p
^ po f ^ p p ^ 0
A + copic ooAp
ic pp! c J n {Ap /A 0^ { c l p ! Cgoc J >l
(CootcVA/AJ +
l
z(Ap/A0)+q
qz{Ap /A0)
+ 1
(
10
)
where
fr
V
V
Coo }
and
q
=
0
)
'
(
11
)
q
describes the degree of coupling between oxidation and phosphorylation. Since it can be proved that the
energy dissipation
D
is nonnegative if and only if
coocPP ~ cl
P>°>it follows that 0<^<1 when c^O , i.e.
when oxidation drives phosphorylation. The case
q~*
1 is interesting. Here v/v„—>z. If no other processes
scavenge the proton gradient set up by the oxidation process, the ratio between the overall rates can be
shown to be v/v0
=
z = njnp.
The numbers
n0
and
np
are not theoretically given, but most researchers agree
that the ratio P/O is between 2 and 3 unless, e.g. the inward flow of protons is used to support transport of
substrates or other cations by active transport processes.
The ratio between output and input energy is the thermodynamic efficiency *1 of the energy coupling system
1
=
(
\
z{Ap/AB) + q ^
f + q
qz(Ap/A0) + l
1 qf +
1
(
12)
where/ =
z(A/A„)(<
0, since
Ap
is negative). The maximum efficiency is obtained for
q f + 2 f + q = 0,
or
f
\
(1 - V T V )
(13)
( i + v ^ V ) 2
(14)
Experimental values for the force ratio/ obtained for living microorganisms are fairly constant at -0.7 to -
0.8 (Roels, 1983). Thus with a coupling coefficient of
q
= 0.95 it follows from Eq. (10) that
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