Question 1 of 8: Nitrogen Requirement and Ethanol Yield in Fermentation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved Casio/Sharp 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 (renumbered Q1–Q8 continuously: Q1–Q5 = Part I, Q6–Q8 = Part II). Q1, Q2, Q3, and Q4 are calculation/stoichiometry questions; Q5, Q6, Q7, and Q8 are essay questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, maintenance-associated product formation, fermenter mass and energy balances; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial nutrition, transport mechanisms, cell-wall structure, pure-culture technique, sterilization methods; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant/animal tissue rheology and gross structure.
Question 1: Nitrogen Requirement and Ethanol Yield in Fermentation (20 marks)
Find. (a) minimum (NH4)2SO4 concentration; (b)(i) ethanol yield $Y_{PS}$; (ii) comparison to the thermodynamic maximum.
Approach. (a) Find biomass MW, extract the N mass fraction, scale to the target cell concentration, then convert moles of N to moles of the diammonium sulfate salt (2 N per formula unit). (b) Use $Y_{XS}$ and the ash correction to fix the biomass coefficient $c$ from a 1-mol-glucose basis, then close the C/H/O elemental balances for $d,e,f$; the electron (degree-of-reduction) balance is used only as an independent arithmetic check, not as a fifth equation.
(a) Biomass MW and nitrogen requirement. Ash-free MW of CH1.83N0.25O0.55:
$$\text{MW}=12(1)+1(1.83)+14(0.25)+16(0.55)=\boxed{26.13\ \text{g/mol}}.$$
Nitrogen mass fraction in the biomass $=14(0.25)/26.13=0.1339$. At $X=25$ g/L:
$$N_{\text{needed}}=25(0.1339)=3.349\ \text{g N/L} = 3.349/14=0.2392\ \text{mol N/L}.$$
Each mole of (NH4)2SO4 supplies 2 mol N, so
$$n_{(NH_4)_2SO_4}=0.2392/2=0.1196\ \text{mol/L}\ \Rightarrow\ \boxed{15.8\ \text{g/L}}\ \left(=0.1196\times132.1\right).$$
(b)(i) Ethanol yield. Basis: 1 mol glucose. Ash-free biomass MW $=12+1.8+16(0.5)+14(0.2)=24.6$. Since $Y_{XS}=0.05$ g total (incl. ash) biomass per g glucose, the ash-free organic mass produced is $0.05(180)(1-0.05)=8.55$ g, so
$$c=\frac{8.55}{24.6}=0.3476\ \text{mol biomass/mol glucose},\qquad b=0.2c=0.0695\ \text{mol NH}_3\text{/mol glucose}\ (\text{N-balance}).$$
The C, H, and O balances close the remaining three unknowns $d$ (CO2), $e$ (H2O), $f$ (ethanol):
$$\text{C: } d+2f=6-c,\qquad \text{H: } 2e+6f=12+3b-1.8c,\qquad \text{O: } 2d+e+f=6-0.5c.$$
Solving this $3\times3$ linear system: $d=1.896$, $e=0.156$, $\boxed{f=1.878\ \text{mol ethanol/mol glucose}}$.
$$Y_{PS}=\frac{f\,(\text{MW}_{EtOH})}{\text{MW}_{glc}}=\frac{1.878(46)}{180}=\boxed{0.480\ \text{g ethanol/g glucose}}.$$
Independent check (available-electron balance, $\gamma_{glc}=4$, $\gamma_{biomass}=4(1)+1.8-2(0.5)-3(0.2)=4.2$, $\gamma_{EtOH}=6$ per the given values): $6(4)=c(4.2)+f\cdot2(6)$, i.e. $24=0.3476(4.2)+1.878(12)=1.46+22.54=24.0$ ✓.
(b)(ii) Comparison with the thermodynamic maximum. The maximum possible ethanol yield comes from the no-growth fermentation stoichiometry $\text{C}_6\text{H}_{12}\text{O}_6\rightarrow 2\text{C}_2\text{H}_6\text{O}+2\text{CO}_2$ (the same reaction used in Q4(a)):
$$Y_{PS}^{max}=\frac{2(46)}{180}=\boxed{0.511\ \text{g/g}}.$$
$$\frac{Y_{PS}}{Y_{PS}^{max}}=\frac{0.480}{0.511}=\boxed{93.9\%}.$$
Zymomonas in this problem operates very close to the theoretical ceiling — only about 6% of the glucose carbon that could in principle become ethanol is instead diverted to biomass synthesis, consistent with the small (5%) biomass yield specified.
