C

N in N index in biomass

Degree of reduction of atoms

+ 3 for NH+ or NH3 as N source 0 for N2 as N source

EXAMPLE 7a

Calculation of the stoichiometry of example 1a using y values

In Example 1a, five equations were solved to calculate the full macrochemical equation. Using the y values of Table 3 we can first make the balance of degree of reduction. For the electron donor oxalate yD = 1 per carbon or 2 per mole oxalate; for biomass yX = 4.2 and for the electron acceptor O2yA = — 4 (Table 3). For all the other chemicals (N source, H+, H2O, HCO—) y = 0 by definition. The y balance is now

Because f = —5.815 we obtain c = — 1.857 directly.

From the C balance we then obtain e = + 10.63. From the N balance a = —0.20, from the charge balance b = —0.8, and from

+ 5 for NO— or HNO3 as N source the O or H balance we finally find d = — 5.42. This is, as expected, the same result as before in Example 1a.

EXAMPLE 7b

Calculation of the Gibbs energy of reaction in example 1b using DGe values

Using the now-available full macrochemical stoichiometry, it is possible to calculate the Gibbs energy of reaction using the Gibbs energy balance.

For each chemical compound the Gibbs energy contribution follows from the product of its number of electrons and its DGe number. For example (using Table 3), the Gibbs energy contribution for oxalate, O2, and biomass in the growth reference system follows as oxalate = 2 x 1 x 52.522 = +105.04 kJ O2 = —4 x (—78.719) = 314.876 kJ biomass = 4.2 x 33.840 = 142.128 kJ

For all other reactants the Gibbs energy contribution in the growth reference system is zero. For the Gibbs energy of the macrochemical-reaction equation we obtain then from the available full stoichiometry:

— 5.815(105.04) — 1.857(314.876) + 142.128 = — 1053 kJ

This is the same as obtained before, but now the calculation has only three terms.

Energetics of Redox Couples, Catabolic Redox Reactions, RET, and Energetic Regularities

It was pointed out earlier that for each microbial growth system a catabolic redox reaction is needed where an electron donor couple reacts with an electron acceptor couple. For the generation of maintenance energy the catabolic reaction is also required. For example, in aerobic growth on glucose the electron donor couple is glucose/HCO— (glucose is oxidized to HCO—) and the electron acceptor couple is O2/H2O (O2 is reduced to H2O). However, for anaerobic growth on glucose, where the catabolic reaction is the conversion of glucose into ethanol, the electron donor couple is glucose/HCO3 and the electron acceptor is the HCO—/ ethanol couple. To be able to quickly calculate the catabolic energy production, it is relevant to define the DGe, DHe, and y values of redox couples i/j. These can be calculated from y, DGe, and DHe values (Table 3) using equations 5a-5c.

(DGe) couple couple y couple = yi yj

Equations 5a-5c have the following properties:

• If we invert the redox couple, for example, i/j into j/ i, we are in fact inverting the redox half reaction of the redox couple. The value of y changes sign, which is logical because produced electrons become consumed electrons. The value of DGe and DHe does not change, which is logical because the energetics of the reaction do not change.

• If the redox couple i/j contains a biological reference compound (e.g., j = HCO3, H2O, H +, NO3, SO4—) then for compound j the value of y, DGe, and DHe is zero. Equations 5a-5c then show that the value of y, DGe, and DHe of the redox couple i/j becomes equal to the tabulated values of the i component in Table 3. For example, if the redox couple is an organic compound/HCO—, then the y, DGe-, and DHe value of the organic compound follow directly from Table 3.

• If the redox couple does not contain a reference chemical then equations 5a-5c must be used to calculate y, DGe, and DHe. Table 5 shows some examples.

It can be seen from Table 5 that electron donor couples are characterized by positive y values and electron-acceptor couples by negative y values (which is logical). As stated earlier, in each microbial growth system there functions a catabolic reaction between an electron donor and an electron acceptor, which generates the required (for anabolism and maintenance) Gibbs energy. It is noted (equation 6c) that in a full catabolic redox reaction, the redox couple with the highest DGe value must be the electron donor; the redox couple with the lowest DGe value is the acceptor (Example 8a).

EXAMPLE 8a

Recognizing electron donor and acceptor

Consider the following catabolic reaction: C6H12O6 + 6O2 r 6HCO3 + 6H+. The two redox couples are C6H12O6/HCOj" and O2/H2O. According to Table 3, the glucose couple has DGe = + 39.744 kJ/e-mol and the O2 couple has DGe = —78.719 kJ/ e-mol. Clearly glucose is the electron donor and O2 is the acceptor. Consider now the catabolic reaction C6H12O6 + 2H2O r 2HCO— + 2H+ + 2C2H5OH. This is the catabolic reaction in the ethanol fermentation. The redox couples are C6H12O6/HCO— and 