Expanded thermodynamic true yield prediction model: adjustments and limitations

Abstract

Bacterial yield prediction is critical for bioprocess optimization and modeling of natural biological systems. In previous work, an expanded thermodynamic true yield prediction model was developed through incorporating carbon balance and nitrogen balance along with electron balance and energy balance. In the present work, the application of the expanded model is demonstrated in multiple growth situations (aerobic heterotrophs, anoxic, anaerobic heterotrophs, and autolithotrophs). Two adjustments are presented that enable improved prediction when additional information regarding the environmental conditions (pH) or degradation pathway (requirement for oxygenase- or oxidase-catalyzed reactions) is known. A large data set of reported yields is presented and considered for suitability in model validation. Significant uncertainties of literature-reported yield values are described. Evaluation of the model with experimental yield values shows good predictive ability. However, the wide range in reported yields and the variability introduced into the prediction by uncertainty in model parameters, limits comprehensive validation. Our results suggest that the uncertainty of the experimental data used for validation limits further improvement of thermodynamic prediction models.

Appendices

Appendix A: Estimation of the usable energy released during oxygenase reaction

Oxygenase reactions are utilized by organisms to create more biodegradable forms of substrates. Some examples are when alkanes are hydroxylated, alkenes are converted into the corresponding epoxides (Peters and Witholt 1994), CO is oxidized to CO₂ (Timkovich and Thrasher 1988), ammonia is oxidized to hydroxylamine (Arp et al. 2002), and some aromatic compounds and cyclic alkanes are hydroxylated (Enroth et al. 1998; Kauppi et al. 1998). Based on the enzyme participating in the reaction, oxygenase reactions can be divided into two groups: monooxygenase-catalyzed reactions and dioxygenase-catalyzed reactions. Different conceptual models and energy calculation are needed for them individually.

Monooxygenase reaction

Monooxygenase enzymes catalyze the NADPH- (or NADH-) and oxygen- dependent oxidation of a wide range of chemicals. (Fosdike et al. 2005; Demirdogen and Adali, 2005) When 1 mole of molecular oxygen is reduced by a monooxygenase enzyme, 4 electron equivalents of electrons are needed: 2 e⁻ equivalents are from the oxidation of substrate and the other 2 e⁻ equivalents are invested by NADPH (or NADH). In order to keep NADH at a constant level within the cell, 2 e⁻ equivalents of electrons released during the further oxidation of substrate are used to regenerate the NADH from NAD⁺. Thus, in total, 4 moles of electrons released from the substrate are utilized for each mole of substrate transformed by the monooxygenation reaction. The process of oxygenase reaction can be simplified as Fig. 4.

Fig. 4

Conceptual model of monooxygenase reaction (During one monooxygenase reaction, the reduction of 1 mole of molecular oxygen, O₂, is directly coupled with the oxidation of substrate and NADH but only 1 mol oxygen atom, [O], is inserted into the substrate. In total, 4 e⁻ equivalents of electrons are required where 2 e⁻ eq. are from substrate and 2 e⁻ eq. are from NADH. But 2 e⁻ eq. released from the further oxidation of substrate have to recoup to NAD⁺ in order to keep the concentration of NADH in microorganisms constant.)

Because the oxidation half reactions (of NADH and substrate) and the reduction half reaction (of oxygen as co-substrate) are coupled directly, the electrons involved do not pass through other coenzymes such as Flavoprotein or Cytochrome c. Thus, the electron flow can not enhance the hydrogen gradient between the inner and outer membrane that is the mechanism of ATP formation. Energy produced during monoxygenase reactions is dissipated rather than stored; thus it is unavailable for cell synthesis. However, a small amount of energy is stored in the product, C_aH_bO_c+1 (the energy carried per electron in the substrate is increased after inserting oxygen). This energy is symbolized as ΔG _monooxy in the unit of kJ/mol or ΔG _e-monooxy in kJ/e⁻ eq. Therefore, during 1 mole of monooxygenase reactions, the electrons accepted by O₂ should be g(i)  = 4 e⁻ equivalents and the corresponding energy released is ΔG(i) = ΔG _monooxy or g(i) × ΔG _e-monooxy kJ.

If several monooxygenase reactions occur in the degradation pathway and the direct products of each time are known, we can calculate $\Delta G_{e\text{-}{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}}$ to replace $\Delta G_{e\text{-}{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}}$ directly (e.g., we use $\Delta G_{e\text{-}{\rm CH}_{4}{\rm O}}$ to replace $\Delta G_{e\text{-}{\rm CH}_{4}}$ in the methane example). Further, we modify the degree of reductance for the substrate to $\gamma_{{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}}-\frac{4}{a}$ where a is the carbon number in the substrate molecule. If the ΔG _monooxy (in kJ/mol monooxygen reaction) is not known, it can be estimated based on the group Gibbs formation energy estimation (Mavrovouniotis 1990, 1991). Appendix A provides details and an example of this type of estimation. When applying the expanded model, the oxygen involved in the monooxygenase reaction is considered a sink for electrons (an electron acceptor). We term this as EA(O₂− monooxy) and the corresponding energy released is ΔG _e-EA(O ₂−monooxy) = ΔG _monooxy/4 kJ/e⁻ eq, where 4 is the amount of electrons obtained by 1 mol monooxygen reaction. The electrons flowing into this EA are t _monooxy × 4 e⁻ eq/mol-C, where t _monooxy refers to the number of oxygeanse reactions during the degradation of 1 mol-C substrate.

The most common monooxygenase reaction happens on functional group of –CH₃. Here we use this common case as an example to explain the estimation method based on Mavrovouniotis (1991).

Generally, 1 mol oxygen atoms are added into the substrate and group –CH₃ becomes group –CH₂OH after one mol times of monooxygenase reaction. Correspondingly, the substrate (denoted as C_aH_bO_cN ⁿ⁺_d ) changes to C_aH_bO_c+1N ⁿ⁺_d . The half reactions of C_aH_bO_cN ⁿ⁺_d and C_aH_bO_c+1N ⁿ⁺_d to CO₂ are shown as Eqs. (3) and (4), respectively. The reductance degree of carbon in the substrate reduces from γ_s to (γ_s − 2/a), but the coefficients of carbonate species, α1, α2 and α3, do not change as long as pH does not change.

$$ \begin{array}{l} {\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}+(3a-c){\rm H}_{2}{\rm O}= (a\times \alpha 1){\rm H}_{2}{\rm CO}_{3}+(a\times \alpha 2) {\rm HCO}_{3}^{-} +(a\times \alpha 3){\rm CO}_{3}^{2-}\\ \quad +d{\rm NH}_{4}^{+}+(a\times (6-2\alpha 1-\alpha 2)+b-4d-2c){\rm H}^{+} +(\gamma_{s}\times a)e^{-} \end{array} $$

(3)

$$ \begin{array}{l} {\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}{\rm N}_{\rm d}^{n+}+(3a-c-1){\rm H}_{2}{\rm O}= (a\times \alpha 1){\rm H}_{2}{\rm CO}_{3}+(a\times \alpha 2){\rm HCO}_{3}^{-}+ (a\times \alpha 3){\rm CO}_{3}^{2-}\\ \quad +d{\rm NH}_{4}^{+}+(a\times (6-2\alpha 1-\alpha 2)+b-4d-2(c+1)) {\rm H}^{+}+(\gamma_{s}\times a)e^{-} \end{array} $$

(4)

Therefore, ΔG _monooxy can be calculated as Eq. (5), where ΔG _f(H₂O) is −237.18 kJ/mol (Benjamin 2002) and the change of standard Gibbs energy of formation between ${\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}$ and ${\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}$ can be estimated based on their molecular structure, $\Delta G_{\rm f}({\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}-{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}{\rm N}_{\rm d}^{n+})=7.9-(-29.3)-1.7=35.5\,\hbox{kcal/mol}$ , i.e., 148.39 kJ/mol, where 7.9, −29.3 and 1.7 kcal/mol is the estimated Gibbs energy of formation of group –CH₃, –OH and –CH₂, respectively. (Mavrovouniotis 1991)

$$ \begin{array}{l} \Delta G_{\rm monooxy}=\Delta G_{{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}{\rm N}_{\rm d}^{n+}}- \Delta G_{{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}}=\Delta G_{\rm f}({\rm C}_{\rm a}{\rm H}_{\rm b} {\rm O}_{\rm c}{\rm N}_{\rm d}^{n+}-{\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}{\rm N}_{\rm d}^{n+})\\ \quad +\Delta G_{\rm f}({\rm H}_{2}{\rm O})-2{\rm RT} \hbox{ln} [{\rm H}^{+}] \end{array} $$

(5)

Thus, ΔG _monooxy = −88.79−2 × (−39.87) = −9.05 kJ/mol monooxygenase reaction at pH 7 or ΔG _e-EA(O₂−monooxy) = ΔG _monooxy/4 = −2.263 kJ/e⁻ eq. Then, the energy of C_aH_bO_c+1N ⁿ⁺_d carried per electron can be estimated as Eq. (6):

$$ \Delta G_{\rm e} =\frac{\Delta G_{\rm e}^{0}\times \gamma_{\rm s} \times a-9.05}{\gamma_{\rm s} \times a-2} $$

(6)

where, Δ Ge⁰ is the energy carried per electron in $\hbox{C}_{\rm a}\hbox{H}_{\rm b}\hbox{O}_{\rm c}\hbox{N}_{\rm d}^{n+}$. Some estimation examples are shown in Table 10 and the results suggest this estimation method won’t introduce distinct error to $\Delta G_{\rm e}({\rm C}_{\rm a}{\rm H}_{\rm b}{\rm O}_{\rm c+1}{\rm N}_{\rm d}^{n+})$.

Table 10 The estimation of ΔG _e of C_aH_bO_c+1 (at pH 7)

Dioxygenase reaction

Dioxygenase-catalyzed reactions always happen on the double bond either in aromatic compounds or in alkenes. It is very similar to a monooxygenase reaction. The major difference is both oxygen atoms of O₂ are inserted into the substrate after reaction, shown in Fig. 5. After one mole of substrate is converted via the dioxygenase reaction, the substrate loses 4 moles of electrons and two –H bonds in the substrate are replaced by two –OH bonds. The gained energy, termed as ΔG _dioxy in the unit of kJ/mol dioxygenase reaction, is that stored in C_aH_bO_c+2. For example, a dioxygenase reaction happens on the benzene ring. Same as monooxygenase reaction discussed above, the energy stored is estimated as Eu. (7).

$$ \begin{array}{l} \Delta G_{\rm dioxy} =\Delta G({\rm C}_{\rm a} {\rm H}_{\rm b} {\rm O}_{\rm c+2} )-\Delta G({\rm C}_{\rm a} {\rm H}_{\rm b} {\rm O}_{\rm c} )=(\Delta G_{\rm f} ({\rm C}_{\rm a} {\rm H}_{\rm b} {\rm O}_{\rm c+2} ) \\ -\Delta G_{\rm f} ({\rm C}_{\rm a} {\rm H}_{\rm b} {\rm O}_{\rm c} ))+2\Delta G_{\rm f} ({\rm H}_2 {\rm O})-4\Delta G_{\rm f} ({\rm H}^{+})=265.98-474.36 \\ -4\Delta G_{\rm f} ({\rm H}^{+})=-208.38-\Delta G_{\rm f} ({\rm H}^{+}) \\ \end{array} $$

(7)

At pH = 7, ΔG _dioxy = −48.9 kJ/mol dioxygenase reaction or ΔG _e-EA(O₂−dioxy) = −48.9/4 = −12.23 kJ/e⁻ eq.

Fig. 5

Simplified process model of dioxygenase reaction (During one dioxygenase reaction, the reduction of one mole molecular oxygen, O₂, and the oxidation of substrate are coupled together. In total, the 4 e⁻ equivalents of electrons needed for O₂ reduction are from the oxidation of substrate. But in contrast to monooxygeanse reactions, two moles of oxygen atoms, [O], are inserted into the substrate.)

Generally, the stored energy during oxygenase reaction is very small compared with the energy released from the reduction of the common electron acceptor, oxygen (ΔG _e-EA(O ₂) = −78.685 kJ/e⁻ eq). Therefore, this mini energy modification, while providing explicit consideration of the effect of the inserted oxygen, generally does not alter the yield estimation for heterotrophs significantly. Thus, it is feasible to ignore it and only consider the importance of the 4 electron equivalents lost per oxygenase reaction. However, for autotrophic systems, the stored energy due to the oxygenase reaction has to be considered since the overall energy generation is also very low.

Appendix B: Application of the expanded thermodynamic true yield prediction model—methanotrophic growth example

In order to show the calculation process step by step, an example is considered. We consider methanotrophic bacteria growing on the media containing methane and nitrate under aerobic condition at pH 7 and temperature 35°C. Based on this information, the different roles during degradation are identified in Table 11. In this example, methane acts as sole CS and sole ED; oxygen acts as the terminal EA associated with energy generation, EA(1), and nitrate acts as the nitrogen source. Nitrate is also considered a secondary EA, EA(2), because it is an additional sink for electrons. Correspondingly, the products of the CS and ED are CO₂ and new biomass; the products of EA(1) and EA(2) are O(−2) and N(−3), respectively; and the product of the NS is N(−3). Therefore, the carbon balance (Eq. (1)) requires that the fraction of carbon in $\hbox{CO}_{2}\,\, (f_{{\rm CO}_2})$ and in new biomass (f _cell) sum to 1.

$$ f_{{\rm CO}_{2}}+f_{\rm Cell}=1 $$

(8)

The electron balance (Eq. (9)) requires the electrons donated during oxidation of methane to CO₂ and biomass equal the electrons accepted by O₂ and nitrate.

$$ f_{{\rm CO}_2} \times (\gamma_{\rm ED} -\gamma_{{\rm CO}_2 } )+f_{\rm cell} \times (\gamma_{\rm ED} -\gamma_{\rm X} )=g(1)+g(2)+g(3) $$

(9)

The electrons associated with the oxidation of methane (to CO₂ and biomass) are calculated by considering the fraction of carbon going to $\hbox{CO}_{2}(f_{{\rm CO}_2})$ and the difference in reductance degree between methane and $\hbox{CO}_{2}(\gamma_{\rm ED} -\gamma_{{\rm CO}_2})$ and separately considering the fraction of carbon going to cells (f _cell) and the difference in reductance degree between methane and cells (γ_ED −γ_X). For the right side of Eq. (2), although the reduction product of O₂ can be expressed as O(−2) generally, evidence shows that the first step of the aerobic degradation of methane is catalyzed by a monooxygenase enzyme. (Colby et al. 1977) The monooxygenase reaction has special energy consumption which is discussed below. Consequently, we split the electrons accepted by oxygen into two parts, the electrons accepted by oxygen during the oxygenase reaction, g(3), and the electrons accepted by oxygen as a terminal electron acceptor for energy generation, g(1). The electrons accepted by nitrate are considered g(2).

Table 11 Compounds and relevant roles in methanotrophic example system

Generally, the nitrogen balance can be ignored as long as nitrogen is not the limiting element of bacterial growth. For example, in this case, ignoring the nitrogen balance does not affect the whole model since the nitrogen balance is expressed as nitrogen in NS is equal to the nitrogen incorporated into new biomass. But when nitrogen limits bacteria growth, it has to be considered more carefully. For example, the nitrogen balance is very important for nitrifiers (where ammonia is the electron donor) and denitrifiers (where nitrate is the electron acceptor) where nitrogen is used in catabolism and anabolism. Nitrogen balance is less critical for aerobic heterotrophs using dissolved ammonia or nitrate only as a nitrogen source for anabolism.

After defining the carbon balance, nitrogen balance and electron balance for the example, the next step is to analyze the energy transfer during metabolism. Metabolism is a complex, multi-faceted process involving numerous reaction steps. However, for thermodynamic modeling of yield prediction, Fig. 6 provides sufficient detail. The catabolic process in our example includes four half reactions: (1) from methane to CO₂ (see Fig. 6a); (2) from O₂ to O(−2) during energy generation; (3) from O₂ to O(−2) during the oxygenase reaction (Fig. 6b); (4) from nitrate to ammonia (Fig. 6c). The anabolic process is simplified as two half reactions by assuming acetate is the critical intermediate: (1) from methane to acetate; (2) from acetate to new cell.

Fig. 6

Simplified metabolism process and the energy consumption of each step in the methanotrophic system with O₂ as electron acceptor and nitrate as nitrogen source (During catabolism, the methane is oxidized into carbon dioxide, oxygen is reduced to the valence of −2 and nitrate is reduced to ammonia; during anabolism, methane is transformed to acetate and then new cells are synthesized based on acetate and ammonia.)

With the two systematic adjustments that we propose to improve the predictive ability of the expanded thermodynamic model, the energy computation of the target system can be calcluated. Based on the model formulation, the energy balance is written as Eq. (10).

$$ K\times \sum_{i=1}^4 {\Delta G(i)} +\frac{E_{\rm syn} (1)}{K^{\rm m}}+\frac{E_{\rm syn} (2)}{K}=0 $$

(10)

The sum of the ΔG(i) includes ΔG(1): the energy released during the oxidation process of CH₄ to $\hbox{CO}_{2}.\,\, \Delta G(1)= f_{{\rm CO}_2} \times \Delta G_{{\rm CH}_4}$, where $\Delta G_{{\rm CH}_4}$ is the Gibbs energy change of CH₄ + 3H₂ O = 0.183H₂ CO₃ + 0.817HCO ⁻₃  + 8.818H⁺ + 8e ⁻, $\Delta G_{{\rm CH}_4} =-198.827\,\hbox{kJ/mol-C}$ and $f_{{\rm CO}_2}$ is the fraction of the carbon in methane that is oxidized to CO₂;   ΔG(2): the energy released during the reduction process of O₂ to O(−2) excluding oxygenase reaction. $\Delta G(2)=g(1)\times \Delta G_{e\text{-}{\rm O}_2}$, where, $\Delta G_{e\text{-}{\rm O}_2}$ is the Gibbs energy change of $\frac{1}{4}{\rm O}_2 +{\rm H}^{+}=\frac{1}{2}{\rm H}_2{\rm O}, \Delta G_{e-{\rm O}_2} =-78.685\,\hbox{kJ/e}^{-}\,\hbox{eq.}$ and g(1) is the equivalents of electrons accepted by oxygen during all reductive reactions except the oxygenation reaction; ΔG(3): the energy released during the reduction process of O₂ to O(−2) in oxygenase reaction only. ΔG(3) = g(3) × ΔG _e-monooxy, where g(3) = 4 e⁻ eq./mol substrate and ΔG _e-monooxy = −2.263 kJ/e⁻ eq.; and ΔG(4): the energy released during the reduction process of $\hbox{NO}_{3}^{-}$ to NH₃ during cell synthesis. ΔG(4) = g(2) × ΔG _e-nitrate, where ΔG _e-nitrate is the Gibbs energy change of the half reaction of $\frac{1}{8}{\rm NO}_{3}^{-}+\frac{9}{8}{\rm H}^{+}+e^{-}=\frac{1}{8}{\rm NH}_{3}+\frac{3}{8}{\rm H}_{2}{\rm O}$, ΔG _e-nitrate = −33.54 kJ/e⁻ eq. and g(2) is again the equivalents of electrons gained by nitrate, $g(2)=f_{\rm cell} \times (\gamma_{{\rm NH}_3} -\gamma_{{\rm NO}_3^{-}})=8\times f_{\rm cell}$.

The energy associated with synthesis in the model is described with two terms. E _syn (1): the energy released during the process of transferring f _cell mol methane to acetate. $E_{\rm syn} (1)=f_{\rm cell} \times (\Delta G_{{\rm CH}_4} -\Delta G_{\rm acetate})$. ΔG _acetate is the Gibbs energy change of the half reaction of acetate, $\frac{1}{2}{\rm C}_2 {\rm H}_3 {\rm O}_2^{-}+2{\rm H}_2 O=0.183{\rm H}_2 {\rm CO}_3 +0.817{\rm HCO}_3^{-}+4.317{\rm H}^{+}+4e^{-}$, ΔG _acetate = −106.302 kJ/mol-C. E _syn (2): the energy needed for cell synthesis based on acetate and ammonia. $E_{\rm syn} (2)=f_{\rm cell} \times \frac{\Delta G_{\rm ATP} \times {\rm MW}_{\rm cell}}{Y_{\rm ATP} \times 0.9}=f_{\rm cell} \times \frac{30.53\times 26.4}{10.5\times 0.9}=85.29f_{\rm cell}$ .

The energy efficiency term, K, is taken as 0.41 (see Xiao and VanBriesen (2006) for details of K estimation). m = + 1 if the sign of E _syn (1) is positive (the corresponding half reaction consumes energy), otherwise, m = −1 (energy is released).

Consequently, the model becomes the three equations, Eqs. (8), (9) and (10). Then the estimation of the true yield in this case can be obtained by solving these equations simultaneously, f _cell = 0.51 mol-C cell/mol-C CH₄. Comparing with the experimental true yield reported by Heijnen and Roels (1981), 0.506 mol-C cell/mol-C CH₄ (calculated from observed yield, maintenance and growth rate), the estimation error is low (+ 1%).