Aerobic Growth of S. cerevisiae on Ethanol — Stoichiometry from RQ: 04-BS-13, December 2019, Q6

Question 6: Aerobic Growth of S. cerevisiae on Ethanol — Stoichiometry from RQ (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

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Full combustion of ethanol ($C_2H_5OH+3O_2\rightarrow2CO_2+3H_2O$) has RQ $=2/3=0.667$. The given RQ $=0.66$ sits just below this ceiling, so the algebra below correctly forces a very small (but positive) biomass coefficient $c$ — i.e. this stoichiometry describes an almost fully oxidative (low-growth-yield) condition, not an error in the solution.

Given.

Quantity	Value
Substrate	ethanol, $C_2H_5OH$, MW = 46 g/mol
Biomass formula	$CH_{1.704}N_{0.149}O_{0.408}$, MW = 22.318 g/mol
Respiratory quotient, RQ $=d/a$	0.66

Find. (a) $a,b,c,d$ (and $e$); (b) $Y_{x/s}$, $Y_{x/O_2}$ (mass basis).

Approach. Basis 1 mol ethanol. Write C, H, N, O atom balances (4 equations) plus the RQ constraint $d=0.66a$ (5th equation) for the five unknowns $a,b,c,d,e$; solve the resulting linear system.

(a) Coefficients from the elemental balances + RQ. $$\text{C: } 2=c+d,\quad \text{N: } b=0.149c,\quad \text{H: } 6+3b=1.704c+2e,\quad \text{O: } 1+2a=0.408c+2d+e,\quad \text{RQ: } d=0.66a.$$ Solving this $5\times5$ linear system: $$\boxed{a=2.917},\qquad \boxed{b=0.01115},\qquad \boxed{c=0.0748},\qquad \boxed{d=1.925},\qquad e=2.953.$$ Check: $d/a=1.925/2.917=0.660$ ✓; C-balance $c+d=0.0748+1.925=2.00$ ✓.
Interpretation. Because the stated RQ (0.66) is so close to the full-combustion ceiling (0.667), almost all of the substrate carbon is respired to CO₂ ($d=1.925$ of the 2 available) and only a small remainder ($c=0.0748$) is diverted to biomass — this is the same "RQ-near-ceiling forces a tiny yield" behavior documented for other substrates more reduced than biomass (see the Practice Question 6 below for a substrate/RQ combination further from its own ceiling, giving a much larger, more typical biomass yield).
(b) Yield coefficients. $$Y_{x/s}=\frac{c\,(22.318)}{1(46)}=\frac{0.0748(22.318)}{46}=\boxed{0.0363\ \text{g biomass/g ethanol}},$$ $$Y_{x/O_2}=\frac{c\,(22.318)}{a\,(32)}=\frac{0.0748(22.318)}{2.917(32)}=\boxed{0.0179\ \text{g biomass/g O}_2}.$$

Coefficient	Value
$a$ (O₂)	2.917
$b$ (NH₃)	0.01115
$c$ (biomass)	0.0748
$d$ (CO₂)	1.925
$e$ (H₂O)	2.953
$Y_{x/s}$	0.0363 g/g
$Y_{x/O_2}$	0.0179 g/g

Topic: Aerobic fermentation stoichiometry fixed by a stated respiratory quotient.

Key relations: RQ $=$ mol CO₂ produced / mol O₂ consumed $=d/a$; full-combustion RQ (no growth) is the ceiling value that RQ approaches as biomass yield $c\to0$.

Why this works: as more of a reduced substrate's carbon and electrons are diverted into biomass rather than fully oxidized to CO₂/H₂O, the O₂:CO₂ stoichiometry shifts measurably away from the no-growth combustion ratio; a stated RQ therefore pins down exactly how much diversion (i.e. how much $c$) is occurring.

Common pitfall: treating a low computed biomass yield as an arithmetic mistake — whenever the given RQ sits close to the substrate's own full-combustion ceiling, a small $c$ is the mathematically forced (and physically correct) consequence, not an error.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (respiratory quotient and elemental-balance stoichiometry).

Question 6 of 9: Aerobic Growth of S. cerevisiae on Ethanol — Stoichiometry from RQ

Question 6: Aerobic Growth of S. cerevisiae on Ethanol — Stoichiometry from RQ (20 marks)