SUPPORTING INFORMATION
A modified metabolic model for mixed culture fermentation with energy conserving
electron bifurcation reaction and metabolite transport energy
Fang Zhang1,2*, Yan Zhang1*, Man Chen2, Mark C. M. van Loosdrecht3, Raymond J.
Zeng1,2**
1
School of Earth and Space Sciences, University of Science and Technology of China,
Hefei, Anhui 230026, People’s Republic of China
2
Department of Chemistry, University of Science and Technology of China, Hefei,
Anhui 230026, People’s Republic of China
3
Department of Biotechnology, Delft University of Technology, Julianalaan 67, 2628
BC Delft, The Netherlands
*
These authors contributed equally to this work.
**
Correspondence concerning this article should be addressed to Raymond Zeng at
rzeng@ustc.edu.cn. Tel: +86 551 63600203; Fax: +86 551 63601592
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1. The metabolic reaction of pyruvate, acetate, ethanol, lactate, biomass and ATP
1.1 Pyruvate (C3H3O3-) formed form glucose: the pyruvate is produced from
glucose via Embden-Meyerhof-Parnas (EMP) pathway.
C6 H 12O6 + 2 NAD + + 2 ADP + 2Pi → 2C3 H 3O3 + 2 NADH + 2H + + 2 ATP + 2H 2 O
(Eq.S1)
1.2 Acetate (C2H3O2-) formed from pyruvate: the acetate is formed directly from
acetyl-CoA via phosphotransacetylase and acetate kinase with 1 ATP generated, as
shown in Eq. 4. The acetyl-CoA and reduced ferredoxin (Fdred) are produced from
pyruvate.
C3 H 3 O3 + Fdox + ADP + Pi → C 2 H 3 O2 + CO2 + H + + Fdred + ATP
(Eq.S2)
1.3 Ethanol (C2H6O) formed from pyruvate: ethanol is produced from the reduction
of acetyl-CoA by NADH with acetaldehyde as the intermediate.
C3 H 3O3 + Fdox + 2 NADH + H + → C2 H 6 O + CO2 + Fdred + 2 NAD +
(Eq.S3)
1.4 Lactate (C3H5O3-) formed from pyruvate: for the higher energy yield, most
microorganisms produce lactate from glucose via homofermentation (Eq.S4)
(Lengeler et al. 1999).
C3 H 3O3 + NADH + H + → C3 H 5O3 + NAD +
(Eq.S4)
1.5 Biomass (CH1.8O0.5N0.2) formed form glucose: the biomass production is
simplified to directly relate to glucose, with 12% C-mol transferred to CO2 (Gommers
et al. 1988), ATP consumed for biomass yield is 10.5 g_biomass/mol_ATP (Russell
and Cook 1995), and NADH is generated for the redox balance. Since on glucose
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there is a net production of NADH in the anabolism we do not need to consider
reverse electron flow for producing NADH. The stoichiometry is shown in Eq.S5:
C6 H 12O6 + 12.37 ATP + 0.912 NAD + + 12.58H 2 O + 1.056 NH 3 →
5.28CH1.8 O0.5 N 0.2 + 12.37 ADP + 12.37 Pi + 0.912 NADH + 0.72CO2 + 1.968H +
(Eq.S5)
1.6 The ATP synthesis reaction: the stoichiometry of ATP synthesis is shown in
Eq.S6:
H + + ADP + Pi → ATP + H 2 O
(Eq.S6)
2. The lumped metabolic model
The lumped metabolic model and the algebraic form are shown in Figure S1 and
Figure S2, respectively, what is used to our new structured model of anaerobic
glucose fermentation of mixed culture. The degrees of freedom are 6 in the MCF
model (V1, V2, V3, V5, V6, V7), and the equations are shown as below.

Metabolite production:
 Glucose consumption: V1;
 Acetate production: V2;
 Butyrate production: V3. While the V4 is not considered in our model.
 Ethanol production: V5;
 Lactate production: V6;
 Propionate production: V7;
 Hydrogen production: in the form of VH2, is equal to V8, V9 or V10 under
different hydrogen pressure. VH2 = 2*V2 + 2* V3 – V7 + 0.912 * V12;
 Biomass production: V12. V12 =5.28 * (V1 – V3 -2*(V2 + V5 + V6 + V7));
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 Carbon dioxide production: in the form of VCO2. VCO2 = V2 + 2* V3 + V5 +
0.72* V12/5.28;

Metabolite transporting:
 Glucose transporting: t1 = V1;
 Acetate transporting: t2 = V2;
 Butyrate transporting: t3 = V3;
 Ethanol transporting: t4 = V5;
 Lactate transporting: t5 = V6;
 Propionate transporting: t6 = V7;
 Carbon dioxide transporting: t7 = V CO2 = V2 + 2* V3 + V5 + 0.72* V12/5.28;
 Water transporting: t8= 2* V1 + V3 + 5/3 * V7 - 12.50 * V12;
 Hydrogen transporting: t9 = VH2 = 2*V2 + 2* V3 – V7 + 0.912 * V12;
 Ammonia transporting: t10= t1*1.056
 Biomass transporting: t11 = V12 =5.28 * (V1 – V3 -2*(V2 + V5 + V6 + V7));
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Figure S1 lumped metabolic model of anaerobic glucose fermentation
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Figure S2 the algebraic form of the metabolic network in the model
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Figure S3 the evaluation scheme of the metabolic model
3. The optimization calculation of glucose metabolites
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The transport of metabolites requires energy involved in the pmf relevant to
membrane-bound ATPase and the transport rate coefficient of metabolites, as shown
in Figure S4. The calculation of transport energy is shown in Eq. S6, in which, pmf
energy is 1/4 ATP, Vi is the producing rate of metabolic acid, mol/(L·h); Diff AH is the
transport rate coefficient, L/(molx·h); CAH,e and CAH,i are the concentrations of
undissociated organic acid, mol/L; X is the concentration of biomass, equal to
(CGlucose * Yxs), molx/L. CGlucose is the glucose concentration, mol/L; Yxs is the yield of
biomass production from glucose, mol/mol. The initial biomass yield (Yxs) is 0.5
mol/mol_glucose in Figure S3. In Eq.S7, the metabolic acid production rate Vi
(mol/(L·h)) is derived from vPi * CGlucose * D, in which, vPi is the factor of metabolic
acid from glucose, mol/mol_Glucose; D is dilution rate, 1/h. To other neutral metabolites,
such as ethanol, the transportation across cytomembrane consumes no energy.
ATPtr =
1
(Vi + Diff AH × (C AH ,e - C AH ,i ) × X )
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(Eq.S7)
Figure S4 metabolic acid transporting through cytomembrane
Herein, butyrate is demonstrated to be an example to describe the detail
calculation. To the equilibrium reaction of butyrate, as shown in Eq. S8, the value of
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HBu/VHBu is (H+/(H+ + 10-pKa,HBu)), in which, HBu is the undissociation concentration
of butyrate, mol/L; VHBu is the total butyrate yield from glucose, mol/L; H+ is the
proton concentration under pH condition (equal to 10
-pH
), mol/L. VHBu is derived
from vPBu * CGlucose * D, in which vPBu is butyrate yield from glucose, mol/mol;
CGlucose is glucose concentration, mol/L; D is dilution rate, 1/h. Therefore, CHBu,e in
Eq1.S1 is equal to (H+/(H+ + 10-pKa,HBu)) * vPBu * CGlucose * D. For the intracellular pH
value is 7, and the maximum dissociated butyrate is 10 mM, the intracellular
undissociated butyrate is omitted in the initial optimization calculation, but, would be
considered in the evaluation of the metabolic network, as shown in Eq.S9.
HBu = H + + Bu -
(Eq.S8)
ATPnet = ATPyield - ATPTr
(Eq.S9)
The net energy of butyrate (in the form of ATP, molTP/ (L·h)) is (3* VHBu –
1/4*(VBu + DiffBu * (H+/(H+ + 10-pKa,HBu)) * VHBu * X /D)), in which 3 is ATP yield
per butyrate from glucose; 1/4 is ATP consumed for proton motive force. Therefore,
the ATP net of butyrate yield is shorten as f_Bu * vP. And f_Bu is (3 – 1/4 – 1/4 *
DiffBu * (H+/(H+ + 10-pKa,HBu)) * X /D) * CGlucose * D. The other metabolites, such as
acetate, propionate, lactate and ethanol, are all deduced in the same form of f_Bu.
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f_ac = 2-1/4 - 1/4 * DiffBu * (H+/(H+ + 10-pKa,HAc)) * X /D) * CGlucose * D;
f_Bu = (3 – 1/4 – 1/4 * DiffBu * (H+/(H+ + 10-pKa,HBu)) * X /D) * CGlucose * D.
f_Pro = 5/3-1/4 - 1/4 * Diffpro * (H+/(H+ + 10-pKa,Hpro)) * X /D) * CGlucose * D;
f_Lac = 1-1/4 - 1/4 * DiffLac * (H+/(H+ + 10-pKa,HLac)) * X /D) * CGlucose * D;
f_EtOH = 1 * CGlucose * D;
f = [f_Ac, f_Bu, f_Pro, f_Lac, f_EtOH]; ﹒﹒﹒﹒﹒﹒(Block 1)
As known, the function LINPROG is utilized to realize minimized calculation,
so, all equations in Block 1 would multiply -1. To linear optimization function,
LINPROG (f, A, b, Aeq, beq, lb, ub) in MATLAB 7.8.0 (2009a) is shown in Block 2.
Linprog: min fT﹒X
s.t.:
﹒﹒﹒﹒﹒﹒(Block 2)
Aeq ﹒X = beq;
A ﹒X ≦ b;
Lb ≦X ≦ ub;
In which,
 f is the coefficient matrix, f = [f_Ac, f_Bu, f_Pro, f_Lac, f_EtOH];
 X is the factor, X = [vP_Ac, vP_Bu, vP_Pro, vP_Lac, vP_EtOH];
 According to mass balance in the form of pyruvate (C3), Aeq is equation
matrix, Aeq = [1, 2, 1, 1, 1]; beq is constant value as 2 * (1- Yxs/5.28);
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 According to redox balance of NADH, A is inequation matrix, A = [-2 -2 1 0 0;
0 -3 -2 -2 -2; 0 3 2 2 2]; b = [0; (-2 * (1-Yxs/5.28) - 0.912 * Yxs/5.28) * Hs;
4 * (1-Yxs/5.28) + 0.912 * Yxs/5.28]. 5.28 is the biomass yield per mole
glucose, as shown in Eq.S5 and Figure S2. Hs is the stage value of hydrogen
production, 0, 0.5, 1 according to stage 1, stage 2 and stage 3, respectively.
The initial Hs value is 1, and would be re-evaluated as shown in Figure S3
under environmental hydrogen partial pressure.
 According to mass balance in the form of pyruvate (C3), the range of
metabolites is demonstrated, lb = [0; 0; 0; 0; 0]; ub = [2 * (1-Yxs/5.28), (1Yxs/5.28), 2 * (1-Yxs/5.28), 2 * (1-Yxs/5.28), 2 * (1-Yxs/5.28)].
 Vm = LINPROG (f, A, b, Aeq, beq, lb, ub).
And, the Vm is the optimization result, evaluated via Figure S3.
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References
Gommers PJF, Schie BJv, Dijken JPv, Kuenen JG. 1988. Biochemical limits to
microbial growth yields: An analysis of mixed substrate utilization.
Biotechnology and Bioengineering 32(1):86-94.
Lengeler JW, Drews G, Schlegel HG. 1999. Biology of the Prokaryotes: Georg
Thieme Verlag.
Russell J, Cook G. 1995. Energetics of bacterial growth: balance of anabolic and
catabolic reactions. Microbiol. Rev. 59(1):48-62.
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