Cell Stoichiometry

General equation of aerobic growth#

The general equation of aerobic growth balances a carbon source ($C_wH_xO_yN_z$) and a nitrogen source ($H_gO_hN_i$) against oxygen, forming biomass ($CH_\alpha O_\beta N_\delta$, per mole of carbon — a "C-mole" basis), carbon dioxide, and water:

$$C_wH_xO_yN_z + aO_2 + bH_gO_hN_i \rightarrow cCH_\alpha O_\beta N_\delta + dCO_2 + eH_2O$$

• $w, x, y, z$ — subscripts of the carbon source's molecular formula
• $a$ — stoichiometric coefficient of oxygen (mol)
• $b$ — stoichiometric coefficient of the nitrogen source (mol)
• $g, h, i$ — subscripts of the nitrogen source's molecular formula
• $c$ — stoichiometric coefficient of biomass (mol)
• $\alpha, \beta, \delta$ — subscripts of the biomass formula, on a C-mole basis (one carbon atom per formula unit)
• $d$ — stoichiometric coefficient of carbon dioxide (mol)
• $e$ — stoichiometric coefficient of water (mol)

There are five unknown coefficients ($a, b, c, d, e$) but only four elemental balances (C, H, O, N), so the system is one equation short. That missing equation has to come from somewhere else — a measured respiratory quotient, a measured biomass yield, or an available electron balance (both covered below).

Worked example: aerobic growth on glycine#

Question. A culture grows aerobically on glycine ($C_2H_5NO_2$) as its sole carbon and nitrogen source, with ammonia ($NH_3$) as a supplementary nitrogen source and an assumed cell formula of $CH_{1.8}O_{0.5}N_{0.2}$. If $1\ \text{mol}$ of biomass forms per mole of glycine consumed, find the remaining stoichiometric coefficients.

Solution.

1. Write the equation with the known subscripts. Glycine gives $w=2, x=5, y=2, z=1$; ammonia gives $g=3, h=0, i=1$; biomass is $\alpha=1.8, \beta=0.5, \delta=0.2$; and $c = 1$ is given: $C_2H_5NO_2 + aO_2 + bNH_3 \rightarrow 1\,CH_{1.8}O_{0.5}N_{0.2} + dCO_2 + eH_2O$.
2. Carbon balance: $2 = c + d = 1 + d \Rightarrow d = 1$.
3. Nitrogen balance: $1 + b = 0.2c = 0.2(1) = 0.2 \Rightarrow b = -0.8$. $b$ comes out negative — glycine alone already supplies more nitrogen than $1\ \text{mol}$ of biomass needs, so $0.8\ \text{mol}$ of $NH_3$ is actually a product, not a reactant.
4. Hydrogen balance: $5 + 3b = 1.8c + 2e$. Substituting $b = -0.8$: $5 - 2.4 = 1.8 + 2e \Rightarrow 2.6 = 1.8 + 2e \Rightarrow e = 0.4$.
5. Oxygen balance: $2 + 2a = 0.5c + 2d + e = 0.5 + 2 + 0.4 = 2.9 \Rightarrow 2a = 0.9 \Rightarrow a = 0.45$.
6. Check every balance against the final numbers (C: $2 = 1+1$; H: $5 = 1.8 + 0.8 + 2.4$; O: $2.9 = 0.5+2+0.4$; N: $1 = 0.2+0.8$) — all hold.

Answer: $C_2H_5NO_2 + 0.45\,O_2 \rightarrow CH_{1.8}O_{0.5}N_{0.2} + CO_2 + 0.4\,H_2O + 0.8\,NH_3$

Respiratory quotient (RQ)#

Respiratory quotient (RQ) is the ratio of the rate of carbon dioxide produced to the rate of oxygen consumed during aerobic metabolism.

$$RQ = \frac{\text{no. of moles of } CO_2 \text{ produced}}{\text{no. of moles of } O_2 \text{ consumed}}$$

• $RQ$ — respiratory quotient (ratio of CO$_2$ produced to O$_2$ consumed — indicates how reduced or oxidized the substrate is: a reduced substrate is rich in hydrogen relative to oxygen (fats, hydrocarbons — still holds a lot of energy/available electrons, so needs a lot of $O_2$ to fully combust), while an oxidized substrate already has more oxygen bonded in and fewer available electrons left to give up (organic acids, and $CO_2$ at the fully-oxidized extreme) — see degree of reduction, below, for the precise measure of this), dimensionless

Physical significance. $RQ$ is a direct readout of a culture's metabolic state, and it's cheap to measure in real time — off-gas analyzers on a fermenter track $CO_2$ and $O_2$ continuously without needing a liquid sample. That's why it's used industrially as a live process-monitoring signal: a shift in $RQ$ can flag a change in metabolic mode (oxidative vs. fermentative), a substrate switch, or an approaching oxygen limitation, often before anything else in the run shows it.

Scenario	Typical $RQ$	Why
Aerobic growth on a carbohydrate (e.g. glucose → biomass + $CO_2$)	$\approx 1$	little rearrangement needed between substrate and $CO_2$
Growth on fats or hydrocarbons as substrate	$< 1$	the reduced substrate needs extra $O_2$ per $CO_2$
Growth on organic acids or amino acids as substrate	$> 1$	an already-oxidized substrate needs less $O_2$ per $CO_2$
Excretion of an oxidized organic-acid product (citric, oxalic acid) from a carbohydrate	$< 1$	forming the product itself consumes $O_2$ while releasing little or no $CO_2$ (e.g. $C_6H_{12}O_6 + 1.5O_2 \rightarrow C_6H_8O_7 + 2H_2O$, zero $CO_2$) — this dilutes RQ down, not up
Lipogenesis: converting carbohydrate into neutral lipid (fat synthesis, not fat oxidation)	$> 1$, up to $\approx 1.2\text{–}1.3$	synthesizing a highly reduced product needs extra NADPH, supplied by decarboxylating pathways (pentose phosphate pathway, malic enzyme) that release $CO_2$ without consuming matching $O_2$
Overflow metabolism on a carbohydrate (byproduct secreted instead of full respiration, e.g. acetate or aerobic ethanol overflow)	slightly $> 1$	the reduced byproduct branch needs little or no $O_2$ while still releasing $CO_2$ via decarboxylation
Pure anaerobic fermentation (e.g. to ethanol)	undefined / $\rightarrow \infty$	no $O_2$ is consumed at all

Don't confuse substrate identity with product identity. A reduced substrate (fat fed to the cell) needs more $O_2$ per $CO_2$ → $RQ<1$. But synthesizing that same kind of reduced molecule as a product from a carbohydrate feed has the opposite effect on $RQ$ — it pushes $RQ$ above 1, because building reducing power for the product releases $CO_2$ without matching $O_2$ use. Which one applies depends on whether the reduced compound is what's being fed in or what's being made.

$RQ$ can also shift during growth on a fixed substrate, not just from substrate choice — if part of the carbon is diverted into a byproduct ("overflow metabolism") instead of going to biomass or full oxidation, that carbon carries electrons away with it instead of needing $O_2$ to dispose of them, so $O_2$ consumption drops relative to $CO_2$ evolution and $RQ$ creeps above $1$ even on a carbohydrate substrate. Two common real examples:

• Acetate overflow (E. coli): at high glycolytic flux, pyruvate is produced faster than the TCA cycle/respiratory chain can process it, so the excess is excreted as acetate (a mildly oxidized, growth-inhibiting byproduct) instead of being fully respired.
• Crabtree effect (S. cerevisiae): even with plenty of $O_2$, yeast growing fast on excess glucose ferments part of it to ethanol instead of fully respiring it ("aerobic fermentation"). Ethanol is a reduced byproduct, not oxidized, but $RQ$ still rises — diverting carbon away from full respiration cuts $O_2$ use more than it cuts $CO_2$ evolution.

The same direction applies to lipogenesis: converting carbohydrate into a highly-reduced storage product like fat pushes $RQ$ above $1$ (often measured $1.1$–$1.3$ in indirect calorimetry) — synthesizing a reduced product needs extra NADPH, generated by decarboxylating pathways (pentose phosphate pathway) that release $CO_2$ without a matching $O_2$ step, the same mechanism as ethanol overflow above. Excreting an oxidized organic-acid product (citric acid, oxalic acid) instead consumes $O_2$ to build the extra oxygen into the product while contributing no matching $CO_2$ (that carbon exits in the acid, not as $CO_2$) — this pulls $RQ$ below $1$, the opposite direction from a first glance at "oxidized = less $O_2$ needed" might suggest.

Direction of $\gamma_P$ vs. substrate	Example product	Effect on $RQ$
More reduced ($\gamma_P > \gamma_S$)	ethanol, lipid	pushes $RQ$ above 1
More oxidized ($\gamma_P < \gamma_S$)	citric acid, oxalic acid	pushes $RQ$ below 1

Worked example: interpreting an elevated RQ during aerobic growth#

Question. An aerobic culture converts glucose to biomass and $CO_2$. The measured $RQ = 1.05$. What does this indicate?

Solution.

1. Glucose is a carbohydrate, so the baseline expectation for straightforward oxidative growth (biomass + $CO_2$ only) is $RQ \approx 1$.
2. $1.05$ is slightly above that baseline, not below — ruling out explanations that push $RQ$ down: growth on a reduced substrate (fat fed in) or excretion of an oxidized organic-acid product (citric, oxalic acid — consumes $O_2$ to build in oxygen, contributes no matching $CO_2$). Also ruling out pure anaerobic fermentation (no $O_2$ term at all, so $RQ$ would be undefined/very large, not a small number near 1).
3. A value above $1$ on a carbohydrate substrate points to some carbon exiting as a reduced byproduct (ethanol-type overflow metabolism, or net lipogenesis) instead of being fully respired — that pathway releases $CO_2$ (via decarboxylation, e.g. pyruvate decarboxylase or the pentose phosphate pathway) without a matching $O_2$-consuming step, nudging $CO_2$ output up relative to $O_2$ consumed.
4. Nitrogen fixation doesn't touch the carbon/$O_2$ balance at all, so it's unrelated to any $RQ$ shift.

Answer: Significant overflow metabolism or lipogenesis — carbon exiting as a reduced byproduct, not an oxidized one. (Note: $RQ \approx 1$ is a rounded rule of thumb for carbohydrate substrates, not an exact value — reading a specific measured $RQ$ this way is a reasonable exam-level call, but a rigorous diagnosis would need the culture's actual yield to compute the true expected baseline before judging a deviation.)

Available electrons#

Available electrons are the number of electrons a substance can transfer to oxygen on combustion to $CO_2$, $H_2O$, and $N_2$. Available electrons can also be considered the valency of each element in the substance.

Element	Valency (available e⁻ per atom)
$C$	$+4$
$H$	$+1$
$O$	$-2$
$P$	$+5$
$S$	$+6$
$N$	variable — depends on the compound

$N$ has no fixed valency — found the same way as any oxidation state, by setting the sum of valencies in the compound to zero (or to its charge, for an ion):

• $-3$ in $NH_3$ or an organic amine
• $0$ in $N_2$
• $+5$ in $NO_3^-$

Worked example: available electrons in glycine#

Question. Find the number of available electrons in glycine ($C_2H_5NO_2$), treating its nitrogen as amino nitrogen (same reference state as $NH_3$).

Solution.

1. Assign valencies: $C = +4$, $H = +1$, $O = -2$, $N = -3$ (amino nitrogen).
2. Multiply each valency by the number of atoms of that element: $C: 2 \times 4 = 8$; $H: 5 \times 1 = 5$; $O: 2 \times (-2) = -4$; $N: 1 \times (-3) = -3$.
3. Sum: $8 + 5 - 4 - 3 = 6$.

Answer: Glycine has $6$ available electrons per mole.

Degree of reduction ($\gamma$)#

Degree of reduction ($\gamma$) is the number of available electrons in a substance, normalized per carbon atom.

$$\gamma = \frac{\text{no. of available electrons}}{\text{no. of carbon atoms}}$$

• $\gamma$ — degree of reduction (available electrons per carbon atom — measures how reduced/energy-rich a substance is; higher $\gamma$ needs more $O_2$ to fully oxidize), dimensionless

Quick reference — valency (available e⁻ per atom):

Element / species	Valency
$C$	$+4$
$H$	$+1$
$O$	$-2$
$P$	$+5$
$S$	$+6$
$N$ — amine / $NH_3$ / $NH_2^-$ group	$-3$
$N$ — $N_2$	$0$
$N$ — $NO_2^-$ (nitrite)	$+3$
$N$ — $NO_3^-$ (nitrate)	$+5$

Biomass nitrogen is always at the $-3$ (amine) level regardless of the nitrogen source fed — see the common confusion note below.

Worked example: degree of reduction of glycine#

Question. Find the degree of reduction of glycine.

Solution.

1. Available electrons in glycine $= 6$ (from the worked example above).
2. Glycine has $2$ carbon atoms.
3. $\gamma = 6/2 = 3$.

Answer: $\gamma_{\text{glycine}} = 3$.

Substrate	Formula	$\gamma$
Carbon dioxide	$CO_2$	$0$
Glycine	$C_2H_5NO_2$	$3$
Glucose	$C_6H_{12}O_6$	$4$
Acetic acid	$C_2H_4O_2$	$4$
Methanol	$CH_4O$	$6$
Ethanol	$C_2H_6O$	$6$
Methane	$CH_4$	$8$
Typical biomass	$CH_{1.8}O_{0.5}N_{0.2}$	$\approx 4.2$

A higher $\gamma$ means the substrate is more reduced (more available electrons per carbon) — more oxygen is needed to fully oxidize it. $CO_2$ sits at $\gamma = 0$ because it is the reference state for carbon — every other compound's degree of reduction is measured relative to it. $CH_4$, with no oxygen and the maximum possible hydrogen, is the highest-$\gamma$ single-carbon compound there is.

Conservation of available electrons#

Available electrons are conserved in a bioreaction, the same way atoms are: every electron entering in the carbon source leaves again — in biomass, in extracellular product, or transferred to $O_2$ as the electron acceptor. This principle supplies the fifth equation the general growth equation is missing (see above). Rather than memorizing a symbolic equation for it, apply the conservation idea directly each time:

electrons in $=$ electrons in biomass $+$ electrons in product $+$ electrons to $O_2$ $+$ electrons spent reducing $N$

• Electrons carried by a compound = its degree of reduction $\gamma$ (electrons per carbon atom) × the moles of carbon it contributes — on a C-mole basis, that's just $\gamma \times$ (C-moles present); for a named molecule like a product, it's $\gamma \times$ (carbon atoms per molecule) × (moles of that molecule).
• One mole of $O_2$ = $4$ electron equivalents, from $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ — divide the electrons left over after biomass and product by $4$ to get moles of $O_2$.
• Reducing nitrogen costs electrons too: biomass nitrogen is always incorporated at the amine level ($-3$ valency, same as $NH_3$). If the feed nitrogen source is more oxidized than that (e.g. $NO_3^-$, $+5$), each mole of nitrogen atom reduced down to $-3$ before assimilation costs $8$ electrons out of the substrate's budget, before anything reaches biomass, product, or $O_2$. This term is $0$ whenever the nitrogen source is already at the amine level ($NH_3$ or an amine).

Worked example: electron balance with nitrate as the nitrogen source#

Question. A culture grows aerobically on glucose ($C_6H_{12}O_6$) using nitrate ($NO_3^-$) as its sole nitrogen source, forming biomass $CH_{1.8}O_{0.5}N_{0.2}$. If $0.5\ \text{mol}$ of biomass forms per mole of glucose consumed, find the moles of $O_2$ consumed per mole of glucose.

Solution.

1. Electrons entering with $1\ \text{mol}$ glucose: $\gamma_S \times (\text{C atoms}) = 4 \times 6 = 24$ (from the degree-of-reduction table above).
2. Degree of reduction of biomass: $\gamma_X = 4(1) + 1.8 - 2(0.5) - 3(0.2) = 4.2$ — using amine-nitrogen valency, because that's the oxidation state nitrogen sits at once it's inside the cell, regardless of the feed.
3. Electrons landing in biomass: $\gamma_X \times (\text{mol biomass}) = 4.2 \times 0.5 = 2.1$.
4. Electrons spent reducing nitrogen: the biomass formula has $0.2\ \text{mol N}$ per C-mole, and $0.5\ \text{mol}$ biomass forms, so $0.2 \times 0.5 = 0.1\ \text{mol N}$ reduced, costing $8 \times 0.1 = 0.8$ electrons.
5. Electrons left for $O_2$: $24 - 2.1 - 0.8 = 21.1$.
6. Moles of $O_2$: $21.1 / 4 = 5.275$.

Answer: $5.275\ \text{mol}\ O_2$ per mole of glucose. Re-running the same balance with $NH_3$ instead (no nitrogen-reduction cost) gives $a = (24-2.1)/4 = 5.475\ \text{mol}$ — more $O_2$, not less. $\gamma_X$ was $4.2$ in both cases; what changed is that nitrate reduction diverted $0.8$ of the substrate's available electrons away from $O_2$, so less $O_2$ was needed to balance the rest.

Working per C-mole of substrate#

Cell yield is often reported as $Y_{X/S}$ — C-mol biomass per C-mol substrate, not mole of biomass per mole of substrate. On that basis, one C-mole of substrate carries exactly $\gamma_S$ electrons (no multiplying by carbon count needed — that's what "per C-mole" already means), and $Y_{X/S}$ is directly the moles of biomass to plug into "electrons in biomass" above.

Worked example: oxygen requirement from cell yield and degrees of reduction#

Question. Yeast grows aerobically on glucose with ammonia as the nitrogen source. The cell yield $Y_{X/S} = 0.50$ C-mol biomass / C-mol glucose. The degree of reduction of glucose is $4.00$ and of biomass is $4.20$. Find the oxygen requirement $a$, in mol $O_2$ per C-mol glucose.

Solution.

1. Electrons in, per C-mole of glucose: $\gamma_S = 4.00$.
2. $NH_3$ is the nitrogen source, so nothing is spent reducing nitrogen.
3. Electrons landing in biomass: $\gamma_X \times Y_{X/S} = 4.20 \times 0.50 = 2.10$.
4. Electrons left for $O_2$: $4.00 - 2.10 = 1.90$.
5. Moles of $O_2$: $1.90 / 4 = 0.475$.

Answer: $a = 0.475\ \text{mol}\ O_2$ per C-mol glucose.

Electron yields ($\eta$)#

Electron yield ($\eta$) is the fraction of the substrate's available electrons that end up at a given destination — biomass, $O_2$, or product — instead of the absolute count used elsewhere in the electron balance.

$$\eta_X = \frac{\gamma_X c}{\gamma_S w} \qquad\qquad \eta_{O_2} = \frac{4a}{\gamma_S w}$$

• $\eta_X$ — fraction of substrate electrons transferred to biomass, dimensionless
• $\eta_{O_2}$ — fraction of substrate electrons transferred to $O_2$, dimensionless
• $\gamma_S w$ — total available electrons entering with the substrate (denominator for every $\eta$)

With no product and no nitrogen-reduction cost, dividing the electron balance $\gamma_S w = \gamma_X c + 4a$ through by $\gamma_S w$ gives $\eta_X + \eta_{O_2} = 1.0$ — every electron that comes in lands in biomass or $O_2$, so the two fractions must add to 1 when there's no third destination.

A more reduced substrate does not mean a higher $\eta_X$. Highly reduced substrates like methane ($\gamma=8$) have a lower growth efficiency per available electron than moderately-reduced substrates like carbohydrates ($\gamma\approx4$–$4.5$) — methanotrophs waste proportionally more of the substrate's electrons as heat/to $O_2$. So $\eta_X$ is lower and $\eta_{O_2}$ is higher for methane, the opposite of what "more electrons available" might suggest.

Worked example: electron yields for growth on glycine#

Question. Using the glycine worked example above ($\gamma_S=3$, $\gamma_X=4.2$, $c=1$ mol biomass per mol glycine, $w=2$ carbon atoms), find $\eta_X$ and $\eta_{O_2}$.

Solution.

1. Total substrate electrons: $\gamma_S w = 3 \times 2 = 6$.
2. Electrons to biomass: $\gamma_X c = 4.2 \times 1 = 4.2$, so $\eta_X = 4.2/6 = 0.70$.
3. Remaining electrons to $O_2$: $6 - 4.2 = 1.8$, so $\eta_{O_2} = 1.8/6 = 0.30$.
4. Check: $\eta_X + \eta_{O_2} = 0.70 + 0.30 = 1.0$ ✓ (no product, and glycine's own nitrogen already sits at amine level, so no nitrogen-reduction term).

Answer: $\eta_X = 0.70$, $\eta_{O_2} = 0.30$ — 70% of glycine's available electrons are captured in biomass, 30% are spent reducing $O_2$.

General equation with product formation#

When a cell also excretes an extracellular product, the general growth equation gains a sixth term:

$$C_wH_xO_yN_z + aO_2 + bH_gO_hN_i \rightarrow cCH_\alpha O_\beta N_\delta + fC_jH_kO_lN_m + dCO_2 + eH_2O$$

• $f$ — stoichiometric coefficient of the product (mol)
• $j, k, l, m$ — subscripts of the product's molecular formula — written as the real molecule (e.g. $C_2H_6O$ for ethanol), not on a C-mole basis the way biomass is

There are now six unknowns ($a, b, c, d, e, f$) but still only four elemental balances, so two more relations are needed — typically two measured yields, or one measured yield plus the electron balance below.

Electron balance with product formation#

A product excreted from the cell is a third place electrons can land, alongside biomass and $O_2$ — its electron content is $\gamma_P \times j$ per mole, where $j$ is carbon atoms per product molecule (needed because, unlike biomass, a product is written as the real molecule, not on a C-mole basis).

Worked example: fermentation of glucose to ethanol#

Question. Saccharomyces cerevisiae ferments glucose anaerobically to ethanol and carbon dioxide, with no biomass formed and no oxygen consumed: $C_6H_{12}O_6 \rightarrow f\,C_2H_5OH + d\,CO_2$. Find $f$ and $d$, and confirm the result conserves available electrons.

Solution.

1. Carbon balance: $6 = 2f + d$.
2. No biomass and no $O_2$ means every electron from the substrate has to land in the product instead: electrons in = electrons in product.
3. Electrons in $1\ \text{mol}$ glucose: $\gamma_S \times 6 = 4 \times 6 = 24$. Electrons per mole of ethanol: $\gamma_P \times j = 6 \times 2 = 12$ (from the degree-of-reduction table above).
4. $24 = 12f \Rightarrow f = 2$.
5. Substitute into the carbon balance: $6 = 2(2) + d \Rightarrow d = 2$.

Answer: $C_6H_{12}O_6 \rightarrow 2\,C_2H_5OH + 2\,CO_2$ — the classic fermentation equation. All $24$ of glucose's available electrons land in ethanol; none go to $CO_2$ ($\gamma = 0$) or $O_2$ (none consumed).

Yield coefficients#

Yield coefficients measure how much of the substrate consumed ends up in biomass or in product, most often reported on a mass basis — grams formed per gram of substrate consumed. This is a different convention from the C-mole-basis $Y_{X/S}$ used in the electron balance above; both are in everyday use, so always check which one a number is quoting.

$$Y_{X/S} = \frac{\Delta X}{\Delta S} \qquad\qquad Y_{P/S} = \frac{\Delta P}{\Delta S}$$

• $Y_{X/S}$ — biomass yield on substrate (fraction of substrate mass converted into biomass), mass basis, in grams of biomass per gram of substrate consumed ($\text{g/g}$)
• $Y_{P/S}$ — product yield on substrate (fraction of substrate mass converted into product), mass basis, in grams of product per gram of substrate consumed ($\text{g/g}$)
• $\Delta X$, $\Delta P$, $\Delta S$ — biomass formed, product formed, and substrate consumed, in grams ($\text{g}$)

From a balanced equation, $Y_{X/S} = cMW_X/MW_S$ and $Y_{P/S} = fMW_P/MW_S$, where $MW_X$, $MW_P$, $MW_S$ are the formula weights of biomass, product, and substrate.

Mass yield can exceed $1.0\ \text{g/g}$. $Y_{P/S}$ only charges substrate mass in the denominator — it doesn't charge the mass of $O_2$ (or any other co-reactant) that gets chemically incorporated into the product. If the product picks up oxygen atoms during synthesis, $MW_P$ can exceed $MW_S$ per mole converted, pushing $Y_{P/S}$ above $1.0\ \text{g/g}$ without violating any conservation law — the extra mass came from $O_2$, not $S$.

Worked example: mass-basis $Y_{X/S}$ for growth on glycine#

Question. Using the glycine worked example above ($c = 1$), find the mass-basis $Y_{X/S}$.

Solution.

1. Formula weight of biomass $CH_{1.8}O_{0.5}N_{0.2}$: $MW_X = 12.011 + 1.8(1.008) + 0.5(15.999) + 0.2(14.007) = 24.63\ \text{g/mol}$.
2. Formula weight of glycine $C_2H_5NO_2$: $MW_S = 2(12.011) + 5(1.008) + 14.007 + 2(15.999) = 75.07\ \text{g/mol}$.
3. $Y_{X/S} = cMW_X/MW_S = (1)(24.63)/75.07 = 0.328\ \text{g/g}$.

Answer: $Y_{X/S} \approx 0.33\ \text{g biomass/g glycine}$.

Maximum theoretical yield#

Maximum theoretical yield is the ceiling on $Y_{X/S}$ or $Y_{P/S}$: the value reached if every available electron in the substrate landed in biomass (or in product) with none spent on $O_2$.

Maximum biomass yield ($Y_{X/S}^{max}$)#

If no electrons are spent on $O_2$ or product, all of the substrate's electrons must land in biomass instead — that's the ceiling, $c_{max}$ (moles of biomass per mole of substrate at the theoretical maximum).

Worked example: maximum $Y_{X/S}$ for growth on glycine#

Question. Find the theoretical maximum biomass yield on glycine, mass basis.

Solution.

1. Electrons in $1\ \text{mol}$ glycine: $\gamma_S \times (\text{C atoms}) = 3 \times 2 = 6$ (from the degree-of-reduction worked example above).
2. With no $O_2$ and no product, all $6$ electrons must land in biomass: $c_{max} = 6/\gamma_X = 6/4.2 = 1.43\ \text{mol biomass/mol glycine}$.
3. $Y_{X/S}^{max} = c_{max}MW_X/MW_S = (1.43)(24.63)/75.07 = 0.469\ \text{g/g}$.

Answer: $Y_{X/S}^{max} \approx 0.47\ \text{g/g}$ — well above the $0.33\ \text{g/g}$ actually achieved in the worked example above, because that reaction also spent electrons on $O_2$ ($a = 0.45\ \text{mol}$). This is a theoretical ceiling, not a target real aerobic growth reaches.

Maximum product yield ($Y_{P/S}^{max}$)#

Same idea, aimed at product instead: if no electrons are spent on $O_2$ or biomass, all of the substrate's electrons must land in product — the ceiling, $f_{max}$ (moles of product per mole of substrate at the theoretical maximum).

Worked example: maximum $Y_{P/S}$ for glucose fermented to ethanol#

Question. Find $Y_{P/S}^{max}$ for ethanol from glucose, and compare it to the fermentation worked example above.

Solution.

1. That fermentation already has $a = 0$ and $c = 0$, so it sits exactly at the ceiling: $f_{max} = f = 2$.
2. $MW_P$ (ethanol, $C_2H_6O$) $= 2(12.011) + 6(1.008) + 15.999 = 46.07\ \text{g/mol}$.
3. $Y_{P/S}^{max} = f_{max}MW_P/MW_S = (2)(46.07)/180.16 = 0.511\ \text{g/g}$.

Answer: $Y_{P/S}^{max} \approx 0.51\ \text{g ethanol/g glucose}$ — the textbook figure for anaerobic yeast fermentation, and here it's also the actual yield, since the reaction never spends electrons on growth or $O_2$.

Worked example: maximum $Y_{P/S}$ for homolactic fermentation#

Question. Find $Y_{P/S}^{max}$ (mass basis) for glucose fermented anaerobically to lactic acid ($C_3H_6O_3$) with no biomass or byproducts.

Solution.

1. Balanced equation: $C_6H_{12}O_6 \rightarrow 2\,C_3H_6O_3$ — glucose and $2\times$ lactic acid have the same total atom count, so this is a straight isomerization with no $O_2$, $CO_2$, or $H_2O$ involved.
2. Check degree of reduction: glucose $\gamma_S = 4$; lactic acid $\gamma_P = [4(3)+6-2(3)]/3 = 12/3 = 4$ — identical, confirming the conversion is redox-neutral (no electrons need an $O_2$ or byproduct sink).
3. $MW_{glucose} = 180.16\ \text{g/mol}$; $2 \times MW_{\text{lactic acid}} = 2(90.08) = 180.16\ \text{g/mol}$ — exact match.
4. $Y_{P/S}^{max} = 180.16/180.16 = 1.0\ \text{g/g}$.

Answer: $Y_{P/S}^{max} = 1.0\ \text{g lactic acid/g glucose}$ exactly — when substrate and product share the same degree of reduction, the conversion is mass-neutral and the theoretical yield is $1.0\ \text{g/g}$ with no computation needed beyond checking $\gamma_S = \gamma_P$.

Worked example: maximum molar $Y_{P/S}$ for sucrose to citric acid#

Question. Find $Y_{P/S}^{max}$ (mol citric acid/mol sucrose) for sucrose ($C_{12}H_{22}O_{11}$) converted to citric acid ($C_6H_8O_7$), based strictly on electron conservation.

Solution.

1. Available electrons in sucrose: $4(12) + 1(22) - 2(11) = 48$.
2. Available electrons in citric acid: $4(6) + 1(8) - 2(7) = 18$.
3. At the ceiling ($a=0$, $c=0$), all electrons land in product: $\gamma_S w = \gamma_P j f \Rightarrow 48 = 18f \Rightarrow f_{max} = 2.67$.

Answer: $Y_{P/S}^{max} = 2.67\ \text{mol citric acid/mol sucrose}$ — molar yields skip the $MW$ conversion the mass-basis examples above need, since the electron balance already gives you the mole ratio directly.

Theoretical oxygen demand (ThOD)#

Theoretical oxygen demand (ThOD) is the $O_2$ needed to fully oxidize a substance to $CO_2$, $H_2O$, and $N_2$ — the electron-balance ceiling on $O_2$ consumption, reached when $c = 0$ and $f = 0$ (no biomass, no product; every electron goes to $O_2$).

$$\text{ThOD} = \frac{\text{available electrons per mole}}{4}$$

• $\text{ThOD}$ — theoretical oxygen demand ($O_2$ needed to fully combust the substance — the ceiling on $O_2$ consumption, reached when no electrons go to biomass or product), in moles of $O_2$ per mole of substance ($\text{mol } O_2/\text{mol}$); multiply by $32\ \text{g/mol}$ and divide by the molar mass for a mass basis ($\text{g } O_2/\text{g}$)

The converse matters just as much: any biomass or extracellular product formed — even a partially-oxidized one like an organic acid — carries some of the substrate's electrons away with it, so the actual $O_2$ consumed is always less than ThOD. Complete oxidation to $CO_2$ is the only case that reaches the ThOD ceiling.

Worked example: ThOD of glycine#

Question. Find the theoretical oxygen demand of glycine, in grams of $O_2$ per gram of glycine.

Solution.

1. Available electrons in glycine $= 6$ (from the worked example above).
2. $\text{ThOD} = 6/4 = 1.5\ \text{mol } O_2/\text{mol glycine}$.
3. Convert to mass: $1.5 \times 32/75.07 = 0.639\ \text{g } O_2/\text{g glycine}$.

Answer: $\text{ThOD} \approx 0.64\ \text{g } O_2/\text{g glycine}$.

$Y_{X/S}^{max}$, $Y_{P/S}^{max}$, and ThOD are all the same conservation idea (electrons in = electrons in biomass + electrons in product + electrons to $O_2$), just with a different pair of destinations zeroed out — all electrons to biomass, all to product, or all to $O_2$.

Elemental composition of biomass#

A biomass (dry cell) formula written on a C-mole basis — $CH_\alpha O_\beta N_\delta S_\epsilon P_\zeta$ — gives each element's mass contribution to dry biomass as (subscript × atomic weight), divided by the total formula weight.

$$\%\,\text{element}_i = \frac{n_i M_i}{\sum_j n_j M_j} \times 100$$

• $\%\,\text{element}_i$ — mass contribution of element $i$ to dry biomass (what fraction of the cell's dry weight that element accounts for), percent ($\%$)
• $n_i$ — subscript (atom count) of element $i$ in the biomass formula, dimensionless
• $M_i$ — atomic weight of element $i$, in grams per mole ($\text{g/mol}$)
• $\sum_j n_j M_j$ — sum of (subscript × atomic weight) over every element in the formula, in grams per mole ($\text{g/mol}$) — the total C-mole formula weight

Worked example: carbon contribution to yeast dry biomass#

Question. The elemental composition of the dry biomass of a yeast species is $CH_{1.6}O_{0.4}N_{0.2}S_{0.0024}P_{0.017}$. Calculate the contribution of carbon to dry biomass, to two decimal places.

Solution.

1. Multiply each subscript by its atomic weight ($C = 12.011$, $H = 1.008$, $O = 15.999$, $N = 14.007$, $S = 32.06$, $P = 30.974$):
   • $C: 1 \times 12.011 = 12.011$
   • $H: 1.6 \times 1.008 = 1.613$
   • $O: 0.4 \times 15.999 = 6.400$
   • $N: 0.2 \times 14.007 = 2.801$
   • $S: 0.0024 \times 32.06 = 0.077$
   • $P: 0.017 \times 30.974 = 0.527$
2. Sum all contributions to get the total formula weight: $12.011 + 1.613 + 6.400 + 2.801 + 0.077 + 0.527 = 23.43$.
3. Divide carbon's contribution by the total and convert to a percentage: $12.011 / 23.43 \times 100 = 51.27\%$.

Answer: Carbon contributes $51.27\%$ of dry biomass.

Key points#

• The general growth equation has five unknown coefficients but only four elemental balances (C, H, O, N) — a fifth relation (measured $RQ$, measured yield, or an electron balance) is always needed to solve it.
• If a nitrogen-containing carbon source supplies more nitrogen than the biomass formed needs, the nitrogen-source coefficient $b$ comes out negative — meaning $NH_3$ is a product, not a reactant.
• Respiratory quotient $RQ = \dfrac{\text{mol } CO_2}{\text{mol } O_2}$: $\approx 1$ for carbohydrates, $< 1$ for reduced substrates (fats, hydrocarbons), $> 1$ for already-oxidized substrates (organic acids, amino acids).
• Available electrons = the sum of (valency × atom count) for every element in a formula; $C=+4$, $H=+1$, $O=-2$, $P=+5$, $S=+6$ are fixed, but $N$'s valency depends on the compound it's found in.
• Degree of reduction $\gamma$ = available electrons ÷ number of carbon atoms — it measures how reduced a substance is per carbon, and is what lets an electron balance replace an elemental balance when solving the general growth equation. $CO_2$ is the zero-point ($\gamma=0$); $CH_4$ is the highest single-carbon value ($\gamma=8$).
• One mole of $O_2$ = $4$ electron equivalents — the conversion used to turn "electrons left over" into moles of $O_2$ in any electron balance, and it follows directly from $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$.
• $\gamma_X$ (biomass) doesn't depend on the nitrogen source — it's fixed by biomass's elemental composition, using amine-nitrogen ($-3$) valency, because that's the level nitrogen sits at once it's inside the cell either way. Feeding $NO_3^-$ instead of $NH_3$ changes the electron balance instead (an extra $8$ electrons per N atom to reduce nitrate down to the amine level), which shows up as lower $O_2$ demand, not a lower $\gamma_X$.
• A biomass elemental formula is written per C-mole (one carbon atom per formula unit), so mass-percent composition is just each element's (subscript × atomic weight) over the formula's total weight.
• Product formation adds a sixth term to the general equation ($f\,C_jH_kO_lN_m$) and a matching term to the electron balance ($\gamma_P j f$) — the product's electron total needs the $\times j$ because, unlike biomass, it isn't written on a C-mole basis.
• Maximum theoretical yield ($Y_{X/S}^{max}$, $Y_{P/S}^{max}$) is the electron balance with $O_2$ zeroed out — all substrate electrons land in biomass or product instead. Real aerobic cultures always fall below it, since some electrons are always spent on $O_2$.
• Theoretical oxygen demand (ThOD) is the same electron balance run the other way: $c = f = 0$, every electron goes to $O_2$ — the $O_2$ needed to fully combust the substance.

Common confusion#

Changing the nitrogen source vs. changing $\gamma_X$ — it's tempting to think that feeding $NO_3^-$ instead of $NH_3$ "uses up" some of nitrogen's reducing power and lowers biomass's degree of reduction. It doesn't: $\gamma_X$ is fixed by what's actually in the cell (amine nitrogen, $-3$ valency) and never changes with the feed. The nitrate/ammonia choice only changes the electron balance around growth — nitrate has to be reduced by $8\ e^-$ per N atom before it can be assimilated, and those electrons come out of the carbon source's budget instead of going to $O_2$.

Proof, for $CH_{1.8}O_{0.5}N_{0.2}$ biomass: $\gamma_X = 4(1) + 1(1.8) - 2(0.5) - 3(0.2) = 4 + 1.8 - 1 - 0.6 = 4.2$.

Nothing in this line references the feed — the $-3$ is amine-nitrogen valency, fixed because that's the oxidation state nitrogen has once it's in the cell, not the state it arrived in. So $\gamma_X = 4.2$ whether the nitrogen source was $NH_3$ or $NO_3^-$.

C-mole $Y_{X/S}$ vs. mass-basis $Y_{X/S}$ — the same symbol is used for two different numbers. The C-mole version (C-mol biomass/C-mol substrate, used in the electron balance) and the mass-basis version (g biomass/g substrate, used everywhere else) will not match numerically, because they carry different formula weights in the conversion. For the glycine example above, $c = 1\ \text{mol biomass/mol glycine}$ is already a molar ratio (not C-mole, since glycine has $2$ carbons), while $Y_{X/S} = 0.33\ \text{g/g}$ is the same result on a mass basis — always check which basis a reported $Y_{X/S}$ is on before using it in a calculation.

"Nitrogen contributes zero available electrons" — false. Nitrogen's valency is variable (it depends on the compound), which gets mistaken for zero. In biomass specifically it sits at the amine level, $-3$, a real nonzero (and negative) contribution — it's the $-3\delta$ term in every $\gamma_X$ calculation above, not an absent one. $N_2$ is the one case where nitrogen's valency actually is $0$, but that's a special case, not the general rule.