146
Chapter 5
V
'-1 .1 2
-1.5
0
-1
>
-1
( O
r,
V2
1
0
0
0
rx
-0.333rs +2.047
rx
V3
0.15
0.5
-0 .5
-0 .3 3 3
0
5.14r,
,VJ
2.42
0.5
0
- 0.333J
-0 .5 r, -4.19/-,
(4)
Now all four fluxes have been determined as linear combinations of the two measured rates. Calculated
values for the remaining non-zero rates are obtained by use of Eq. (5.26):
r o
f
0
1
-1
0N
| V
0.333/-, -3 .0 9 3 r,N
r e
0
0
1
0
V2
5.14/-,
r c
0.12
0.5
0
0
V3
-0.167/-, +1.143/-,
J
*
y
k 0
0
0
l
k V
,
-0.5/-,-4.19/-,
,
If we have measurements of
ra, r„ rc
and rg together with
r,
and rx we may now compare measured and
calculated values of the rates. In case measured and calculated rates agree to within the experimental error it
can be concluded that the simple metabolic network gives a good description of the anaerobic metabolism
of yeast at the environmental conditions of the experiment.
From Eq. (5) it is seen that
rc
and
rx
are proportional. Consequently the two rate measurements do not form
a basis on which the other four rates can be expanded. It is not immediately obvious why this combination
of two rates cannot be used - while the remaining 14 pairs of
r
can be used as is easily seen from Eq. (5).
Inspection of Eq. (4) explains why the system is not observable based on measurement of ethanol and
biomass. Both v, and v3 are proportional with
r„
and these two fluxes are therefore linearly dependent and if
we lack information about one of the two other fluxes. That measurements of
rK
and
rt
causes problems for
analysis of the system would have been difficult to identify before analysis, and it is typically only through
analysis of the matrix equation for different set of measurements that it can be concluded which set of
measurements ensures that the system is observable.
Finally some comments on manual checks of the correctness of the calculations
In each of the 4 reactions the carbon and redox balances close
In a correct formulation of T the sum of numbers in each column should therefore give zero for the
carbon containing species,
i.e.
the upper 6 rows.
In the final result the carbon balance should close,
i.e. ra + re + rc + rg
=
“ r,
In the final result the redox balance should close,
i.e.
5ra
+
6re
+ 4.667rg + 4.18/-,= (-1.667/-, -15.465/-,) + 30.840r,
+ ( - 2.333/-, -19.553/-,) + 4 .18r, = 4 (- /-,) + 0.002/-, * 4(- /-,)
From the result in Eq. (5) the stoichiometry of a black box model for the overall growth process can be
obtained. Let
ra
be zero in which case:
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