Question 1 of 10: Ethanol Production by Yeast — Actual vs. Thermodynamic-Maximum Yield
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved Casio/Sharp calculator allowed). Format: Part I offers 6 questions (any 3 constitute a complete answer, 20 marks each) and Part II offers 4 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 10 are solved below for completeness. Most questions require an essay-format answer; Q1–Q4 and Q7 are calculation questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/viral morphology, physiology and growth control; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant/animal tissue structure.
Question 1: Ethanol Production by Yeast — Actual vs. Thermodynamic-Maximum Yield (20 marks)
Biomass yield from glucose, $Y_{XS}$ (dry wt., incl. ash)
0.11 g/g
Ash content of biomass
5%
Biomass formula (ash-free)
CH1.8O0.5N0.2, MW = 24.6 g/cmol
MW glucose / ethanol
180 / 46 g/mol
Atomic masses
C = 12, H = 1, N = 14, O = 16
Find. (a) Ethanol yield $Y_{P/S}$ (g ethanol/g glucose); (b) that yield as a percentage of the stoichiometric (thermodynamic) maximum.
Approach. $Y_{XS}$ and the ash correction fix the biomass coefficient $c$ (converting dry biomass mass to cmol of the ash-free formula); the nitrogen balance then gives $b$, and the remaining C, H, O atom balances (three equations) solve simultaneously for the three unknowns $d$, $e$, $f$. The ethanol coefficient $f$ converts directly to a mass yield, which is then compared against the classical Gay-Lussac stoichiometric maximum (glucose → 2 ethanol + 2 CO2, no biomass).
Biomass coefficient $c$ from the yield and ash correction. Per mole of glucose (180 g), dry biomass formed $=Y_{XS}(180)=0.11(180)=19.8$ g. Only 95% of this is the ash-free CH1.8O0.5N0.2 material (the rest is inert ash, carrying no C/H/N/O balance obligation): ash-free mass $=19.8(0.95)=18.81$ g, so
$$c=\frac{18.81}{24.6}=\boxed{0.7646\ \text{cmol biomass/mol glucose}}.$$
Nitrogen balance → $b$. NH3 is the only N source and biomass the only N sink:
$$b = 0.2c = 0.2(0.7646) = 0.1529\ \text{mol NH}_3\text{/mol glucose}.$$
Carbon, hydrogen and oxygen balances → $d,e,f$. Three simultaneous linear equations in the three remaining unknowns:
$$\text{C: } c+d+2f=6, \qquad \text{H: } 1.8c+2e+6f=12+3b, \qquad \text{O: } 0.5c+2d+e+f=6.$$
Solving (matrix elimination) gives
$$d=1.7706,\quad e=0.3441,\quad f=\boxed{1.7324\ \text{mol ethanol/mol glucose}}.$$
Part (a): mass yield of ethanol.
$$Y_{P/S}=\frac{f\cdot\text{MW}_{\text{ethanol}}}{\text{MW}_{\text{glucose}}}=\frac{1.7324(46)}{180}=\boxed{0.443\ \text{g ethanol/g glucose}}.$$
Part (b): thermodynamic (stoichiometric) maximum. With no biomass or CO2-only side reaction diverting carbon, the classical anaerobic fermentation equation $\text{C}_6\text{H}_{12}\text{O}_6\rightarrow2\,\text{C}_2\text{H}_6\text{O}+2\,\text{CO}_2$ gives the absolute ceiling:
$$Y_{P/S}^{\max}=\frac{2(46)}{180}=0.511\ \text{g/g}.$$
The actual yield is
$$\frac{0.443}{0.511}\times100=\boxed{86.6\%}\ \text{of the thermodynamic maximum}.$$
The 13.4% shortfall is exactly the carbon/electron flux diverted from ethanol into biomass (and its associated CO2) to support cell growth — ethanol formation and growth are competing sinks for the same substrate.
