180
Chapter 5
result with respect to changes in kx. Also relate the present results with the discussion in Example 5.5 in
which exactly the same data were used in the end. In Example 5.5 one reached the conclusion that
y
is
more likely to be 1.8 than the value found here. What is the reason for this different conclusion?
Problem
5.2 Analysis of a PFL deletion mutant of
Lactococcus lactis.
We shall study a
Lactococcus lactis
deletion mutant in which the gene encoding pyruvate formate lyase
(PFL) has been knocked out. The metabolic capability of the strain is exactly as shown in Fig. 2.4 and
2.6B except that HCOOH cannot be produced. The biomass composition is X = CFLO
0
sN
0
25
and biomass
is produced according to
v
2
:
- 1.1 CH20 + 0.1 CO, + X + 0.075 NADH - 2.48 ATP = 0
The other network reactions are:
v,
glucose —* pyruvate
v3:
pyruvate—»AcCo A (by PDH)
v4:
pyruvate—»lactate
v5:
AcCo A—»ethanol (e)
v6:
AcCo A—»acetate (HAc).
a.
Assume that the net rates of production of pyruvate, AcCoA, NADH and ATP are zero.
Calculate the six fluxes as well as the remaining non-zero production rates in terms of the
measured rates rx = p and
rs
= relhanol. List the three pairs of two rate measurements for which the
system is non-observable. Give a physiological reason for the negative value of
rHAc = v„.
b.
Repeat the calculations of question a., but assume that rpynlvate =
rPYR ф
0.
c.
Why does this not help to make
rHAc
positive?
d.
Finally consider a
pdh(-)
strain. Here the functioning of PFL is unimpaired while PDH does not
function. Repeat the calculation of question a. for this strain. Explain why
rHAc
is positive.
Comment : A
pfl(-)
strain will not grow anaerobically unless it is fed with a little HAc besides
glucose (which is of course the main carbon and energy substrate.). This has been proved
experimentally by Henriksen and Nilsson (2001) who also found that the
pfl(-)
strain fed properly
will have an almost two-fold increase of гШас compared to the wild type strain. Why is that so?
Problem 5.3
The maximum theoretical yield of lysine on glucose
In Problem 4.2 it was shown that a growth yield of 6/7 for L-lysine on glucose (Example 3.9) could not
be excluded for thermodynamic reasons. In the present problem it will be shown that the explanation for
the lower theoretical yield (3/4) - see references to Problem 4.2 - is to be found in the need for NADPH
and not NADH in the synthesis of lysine from glucose. This will help the reader to appreciate that the
crude definition of redox equivalents as “H
2
“ cannot always be used.
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