Biochemical Reaction Networks
145
the present context we shall just use a value of 100 mmoles ATP (g DW)1
(corresponding to about 2.42
moles ATP (C mol)'1). Notice that in the current example the amount of C 02 formed in connection with
biomass formation in the first reaction is slightly different from what was used in Example 5.4, but this is a
consequence of a slightly different elemental composition of the biomass used in the present example.
The simple model can be presented in matrix form corresponding to Eq. (5.1) as:
V
^
1.12
-1.5
0
-1
^
r .
1
0
0
0
r a
0
1
-1
0
V
r e
0
0
1
0
V 2
F
0.12
0.5
0
0
V 3
r t
0
0
0
1
0
0.15
0.5
-0 .5
-0.333
,0 ,
- 2.42
0.5
0
-0.333,
(
2
)
As there are four fluxes and we have two constraints, one for NADH and one for ATP (the two lower
balances in Eq. (2)), the degrees of freedom
F= 2
and we therefore need to measure a minimum of 2 out of
the 6 non-zero rates. The other rates can easily be calculated once the four fluxes have been determined. If
more than 2 measurements are used the calculated values of v, to v4 will be least squares fitted values which
means that the constraints will not be exactly satisfied - but more accurate values of the fluxes will be
obtained in the real situation where the measured rates are contaminated by experimental errors. The use of
more than the minimum number of measured rates will be postponed until Section 5.4.
We now choose two measured non-zero rates out of the six available rates. The choice may seem arbitrary,
but it will shortly be shown that only some of the 15 possible combinations of 2 out of the 6 rates will lead
to solution of the problem. Let the choice be
rs
and
and in this case the matrix equation (5.24) becomes:
V
0
1
-1
0
N
r e
0
0
1
0
r c
0.12
0.5
0
0
(V
r s
0
0
0
1
V 2
F
-1.12
-1 .5
0
-1
V3
rx
1
0
0
0
, V4,
0
0.15
0.5
-0 .5
-0.333
, 0
J
v-2 .4 2
0.5
0
-0 .3 3 3 ,
And hereby we find using Eq. (5.25):
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