Biochemical Reaction Networks
141
For
Y
.„ = 0.167 and
Yx
= 0.355 one obtains
Ysx =
0.286 and for ys0 = 0.044 and
Yx
= 0.513 one obtains Tsx =
0.177. In both cases the calculated values of F„ are close to the experimentally determined values of the
table in Example 3.5. The difference is due to round-off errors.
We shall next consider the ATP balance:
- av, + (2v2 + 0. Iv, )P/0 +
f.3v2
+ 0.5v3 = 0
Inserting v, and v2 from Eqs. (10) and (11) and v3
=
Y^
(-rs) one obtains:
2 P/O
-1.048(1 -
Y„)
+ jj).0714 + ~ j y „
a ~
0.952(1 -T j-1 .4 2 9 T „
(14)
05)
and consequently:
For
D=
0.30 h'1
(ß = 0.50) : a = p 168 P/O + 0.887
(16)
For
D =
0.40 h 1
(ß = 0.36) :
a = 0.496 P/O + 1.517
(17)
It hardly matters if ß is 0.36 or 0.5. If the value ß = 0.5 is used in (17) then a = 0.4960 P/O + 1.545, i.e., the
influence
of P/O is the same while the ATP generated by substrate level phosphorylation is about 2%
higher. Likewise the stoichiometric coefficient of C02 in reaction (1) has only a small influence on the
relation between a and P/O - and no influence at all on T„. If a stoichiometric coefficient of 0.12 is used
instead of 0.10 Eq. (16) is changed to a = 1.168 P/O + 0.877.
While the result of the sensitivity analysis for the influence of the C02 coefficient in reaction ( 1
) and of ß in
reaction (2) shows that these parameters have virtually no influence on the result there is a dramatic
influence of the specific growth rate rx =
D
on the relation between
a
and P/O. When
D
increases above the
critical dilution rate, i.e., a changes from Eq. (8) to Eq. (16) or Eq. (17) the influence of the ATP generation
from the oxidative phosphorylation decreases significantly - at least if the effective P/O ratio is the same as
is generally accepted in the literature. Inserting P/O = 1.25 in the three expressions yields:
D
(h ')
0.15
0.30
0.40
a (moles ATP (C-mole biomassf1
2.42
2.34
2.14
In Example 5.5 we will show that
Y ^TP
for anaerobic conditions is about 1.80 moles ATP (C-mole
biomass)'1
for all dilution rates. The higher value for a found in the above table may be explained by a
less effective energy generation in the oxidative phosphorylation, i.e., the P/O ratio is lower than the
assumed value of 1.25. If we on the other hand assume that TxATP = a is the same at anaerobic and aerobic
conditions, we can calculate the P/O ratio from Eq. (16) and find it to be about 0.78. This may be too
low, and probably the P/O ratio is somewhere between this value and 1.25. We might also speculate that
the P/O ratio is fixed at e.g. 1.5 and that T^ajp = a is higher for aerobic growth than for anaerobic growth.
This may be due to a higher cost for biomass synthesis (different macromolecular composition)
previous page 165 Bioreaction Engineering Principles, Second Edition  read online next page 167 Bioreaction Engineering Principles, Second Edition  read online Home Toggle text on/off