Question 4 of 8: Oxygen Demand With vs. Without Growth; Maximum Biomass Yield From Glucose vs. Ethanol
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: Part I offers 5 questions (any 3 constitute a complete answer, 20 marks each) and Part II offers 3 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 8 numbered questions are solved below for completeness. Q2–Q5 and Q7 are calculation/derivation questions; Q1, Q6, and Q8 are essay questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter mass and energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/eukaryotic cell structure, fungi, protozoa/algae, Gram-stain cell envelope; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant tissue structure and cereal grain morphology.
Question 4: Oxygen Demand With vs. Without Growth; Maximum Biomass Yield From Glucose vs. Ethanol (20 marks)
Find. (a) O2 consumed per mole glucose with growth vs. without growth (full combustion); (b) maximum possible mass yield of biomass from ethanol vs. from glucose.
Approach. Convert the given mass yield to a molar biomass coefficient $c$ via the ash-free biomass molecular weight, close the C/H/N/O elemental balances for $a,b,d,e$, and cross-check $a$ with the degree-of-reduction (electron) balance; for part (b), recognise that the maximum theoretical yield is capped by whichever of the carbon balance or the electron balance is more restrictive, which depends on whether the substrate is more or less reduced than the biomass.
Convert $Y_{xs}$ to a molar biomass coefficient. Biomass (ash-free) MW $=12+1.84(1)+0.55(16)+0.2(14)=25.44\ \mathrm{g/Cmol}$. Per mole of glucose (180 g), total dry biomass produced $=0.5(180)=90$ g, of which $90(0.95)=85.5$ g is ash-free organic matter:
$$c=\frac{85.5}{25.44}=3.361\ \mathrm{Cmol\ biomass/mol\ glucose}.$$
Close the elemental balances for $\mathrm{C_6H_{12}O_6}+aO_2+bNH_3\to cX+dCO_2+eH_2O$. C: $d=6-c=2.639$. N: $b=0.2c=0.672$. H: $12+3b=1.84c+2e \Rightarrow e=3.916$. O: $6+2a=0.55c+2d+e$, solving:
$$\boxed{a=2.521\ \mathrm{mol\ O_2/mol\ glucose\ (with\ growth)}}.$$
Cross-check via the degree-of-reduction (electron) balance. Available electrons: substrate supplies $\gamma_S\cdot(\text{C atoms})=4.0(6)=24$ per mole glucose; these are split between biomass ($\gamma_Bc=4.14(3.361)=13.91$) and O2 ($4a$, since O2 accepts 4 electrons per mole): $24=13.91+4a\Rightarrow a=2.52$, matching step 2 — a reliable audit of the elemental-balance arithmetic.
Compare with the no-growth (pure respiration) case. Without growth, all glucose carbon and electrons are fully oxidised: $\mathrm{C_6H_{12}O_6}+6O_2\to6CO_2+6H_2O$, i.e. $a_{no\text{-}growth}=6$ mol O2/mol glucose. Compared with $a_{growth}=2.52$:
$$\boxed{\text{growth reduces O}_2\text{ demand by }\frac{6-2.52}{6}\times100\%\approx58\%}.$$
Diverting a large share of the substrate's electrons into biomass (rather than fully oxidising them to CO2/H2O) means far fewer electrons remain to be dumped onto O2 — growth is energetically "cheaper" in O2 terms than pure maintenance respiration on the same substrate.
Part (b): maximum theoretical yield — identify the binding constraint. The theoretical maximum biomass yield is capped either by carbon availability ($c\le n_C$, the substrate's carbon count) or by electron availability ($\gamma_Sn_C=\gamma_Bc$, i.e. $a=0$) — whichever binds first. Since glucose is less reduced than biomass ($\gamma_S^{glu}=4.0<\gamma_B=4.14$), converting all of its carbon to biomass would need more electrons than glucose supplies, so the electron balance binds:
$$c_{max}^{glu}=\frac{\gamma_S^{glu}n_C}{\gamma_B}=\frac{4.0(6)}{4.14}=5.797\ \mathrm{Cmol/mol\ glucose}\quad(a=0,\text{ verified}).$$
Ethanol is more reduced than biomass ($\gamma_S^{eth}=6.0>\gamma_B=4.14$), so converting all its carbon to biomass leaves a genuine electron surplus that must be dumped on O2 — here the carbon balance binds instead:
$$c_{max}^{eth}=n_C^{eth}=2\ \mathrm{Cmol/mol\ ethanol},\qquad a=\frac{\gamma_S^{eth}n_C-\gamma_Bc_{max}^{eth}}{4}=\frac{12-8.28}{4}=0.93\ \mathrm{mol\ O_2/mol\ ethanol}\;(>0,\text{ consistent}).$$
Convert both to a mass basis and compare.
$$Y_{max}^{glu}=\frac{5.797(25.44)}{180}\Big/0.95=0.862\ \mathrm{g/g},\qquad Y_{max}^{eth}=\frac{2(25.44)}{46}\Big/0.95=1.164\ \mathrm{g/g}.$$
$$\boxed{\frac{Y_{max}^{eth}}{Y_{max}^{glu}}=1.35\ \Rightarrow\ \text{ethanol's maximum possible biomass yield is} \approx35\%\text{ higher than glucose's}.}$$
Note on the shortcut: applying the electron-balance formula $c_{max}=w\gamma_S/\gamma_B$ (the one the given $w$ values invite) to ethanol too would give $2(6.0)/4.14=2.90$ Cmol biomass per mol ethanol, i.e. $1.69$ g/g, about twice glucose's yield. That is not physically attainable: ethanol has only $w=2$ carbon atoms, and 2.90 Cmol of biomass would need more carbon than the substrate supplies. The carbon balance therefore caps the ethanol yield at 1.16 g/g. Either way ethanol gives the higher maximum yield per gram of substrate, because it is the more reduced substrate.
