- Polyols and oxygen yields and rates
Polyols.
For macroscopic analysis a simple process representation was obtained
by substitution of the overall set of excreted polyols (Röhr et
al., 1987) for an unique formula equivalent. The mean composition
used was [CH2.54O]n, n=3.72 since the average
composition of polyols showed minor variations among the different idiophase
stages (data not shown). This expression is a weighted average of the
elementary composition of all polyols involved, considering as weight
factor their own excretion rate. However, when the balance of intracellular
fluxes was done, individual polyols were considered. Macroscopic rates
were calculated by means equation (A.1) where f
stands for fluxes (in C-moles. g-1. h-1)
of medium exchanged compounds .
Oxygen.
Reductance balance was used to the determination of the oxygen yields
at the idiophase stages considered in the model: early, medium (A.2)
and late (A.3). In these equations f
stands for fluxes (in C-moles.g-1.h-1)
between cell and medium, while g stands for the reductance grade.
II. Determination of the
rate of GABA cycle
At the early idiophase
the NH3/NH4+ pool is considered at
steady state (free diffusion). This fact, together with the absence
of nitrogen at the culture medium allows us to write down the following
mass balance equation:
The balance equations
for the extracellular (A5), cytoplasmic (A6) and mitochondrial compartments
(A7) are the followings:
By summation of equations A5-A7 we obtain
(A.8), suitable for the R14 reaction.
Following the same procedure we can write
the corresponding equation for NH3 in each compartment:
From equations (A.9) to (A.11) we obtain
(A.12):
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(A.12)
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By substitution of A.12 in A.8 we finally
obtain:
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(A.13) |
III. Energy Balances
Assuming a steady state
the H+ balance equation
(1/3) R13 -
b R20
- R21 + 2 R22 = 0
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allows us to write equation A.14:
0.30133 + 2 R22 = R21
+ 1.1669 R20 |
(A.14)
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Analogously, the mitochondrial ATP balance
(1/3) R13 + 3 R15 +
2 R16 - (R20 + R21) = 0 |
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allows us to write equation A.15:
R21 = 2.7412 - R20
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(A.15)
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By substituting equation A.15 into A14
we obtain the uptake rate of mitochondrial Pi:
R22 = 0.08345 R20 +
1.21996 |
(A.16)
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From the mass balance for cytoplasm Pi,
we obtain A.17:
R19 - R22=(1/3)
R8 - [(1/6) R2+(1/10) R3+(1/5) R4+(1/6) R6+(1/3) R11+(1/6) R12]
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R19 - R22 = 0.7435 |
(A.17)
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By substitution in A.17 of A.16 we obtain
A.18:
R19 - 0.08345 R20 =
1.9635 |
(A.18)
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That together equation 6 (see Energy balance
section in the main text) we finally obtain A.19, that allow us the
determination of the H+-ATPase rates.
After evaluation of R19 and R20 it can
be determined the values for R21 and R22.
IV.
Energy and reductive balance relations |
The alternative respiration fraction can be envisaged as a consequence
of an imbalance between the energy and reductive mechanisms. Only if
the oxygen demand based in ATP is equal to the oxygen demand from the
reductive equivalents, all the respiration rate would be ATP associated.
Hence, for conservation of ATP equation A.20 must be fulfilled, ao,
ax, ap standing for ATP yields of respiration,
biomass and citric acid respectively; f
being the corresponding specific productivity respectively and m the
ATP cellular maintenance as defined in equation (6) in the text.
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(A.21) |
On the other hand, equation
A.21 must be satisfied by the reductive equivalents. The subindex i
refers to any product different but biomass that consume or produce
NADH, NADPH or FADH. aoH, axH
refers to yields in reduction equivalents for oxygen and biomass
respectively.
Accordingly the following
expression (A.22) is derived for the ratio of the alternative respiration
system rate, which equivalent to the previously presented for the alternative
respiration ratio [R17/(R15+R16+R17].
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(A.21) |
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