Cell Stoichiometry

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

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.)

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.

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.

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.

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.

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 $O_2$.

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).

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

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.

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 $O_2$.

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 $\gamma_S = \gamma_P$.

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.

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 $O_2$.

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.