Cellulomonas SCP from Glucose — Actual vs. Maximum Theoretical Yield: 04-BS-13, December 2019, Q5

Question 5: Cellulomonas SCP from Glucose — Actual vs. Maximum Theoretical Yield (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.

Given.

Quantity	Value
Substrate	glucose, $C_6H_{12}O_6$, MW = 180 g/mol, $w=6$ carbon atoms
Biomass formula (ash-free)	$CH_{1.56}N_{0.16}O_{0.54}$, MW = 24.44 g/mol
Given reaction	1 mol glucose $\rightarrow$ 6 mol biomass (all substrate carbon converted)
Ash content	5% of total dry cell mass
Degree of reduction, glucose $\gamma_s$	4

Find. Actual biomass yield (mass and molar) vs. the maximum theoretically possible yield.

Approach. Under anaerobic conditions with no external electron acceptor and no reduced product other than biomass (the by-product H₂O carries no available electrons), every available electron in the substrate must end up in the biomass — so the maximum possible biomass yield is set by the electron (degree-of-reduction) balance, cross-checked against the carbon balance (maximum one biomass-carbon per substrate carbon, since "all carbon" is converted).

Actual yield, from the given reaction. Basis 1 mol glucose. Molar yield $=6$ mol biomass/mol glucose. Mass yield (ash-free, organic biomass only): $$Y_{x/s}^{\text{organic}}=\frac{6(24.44)}{180}=\boxed{0.815\ \text{g/g}};$$ including the 5% ash (total dry cell mass $=$ organic mass$/(1-0.05)$): $$Y_{x/s}^{\text{total}}=\frac{0.815}{0.95}=\boxed{0.858\ \text{g total biomass/g glucose}}.$$
Maximum yield from the carbon balance. Since the biomass formula has exactly 1 carbon atom per formula unit and "all carbon in the substrate is converted," the carbon-balance ceiling is simply $c_{max}=w=6$ mol biomass/mol glucose — i.e. the given reaction already sits at the carbon-balance maximum.
Maximum yield from the electron balance. Degree of reduction of the biomass (per carbon, referenced to CO₂/H₂O/NH₃ at $\gamma=0$): $$\gamma_{biomass}=4(1)+1(1.56)-2(0.54)-3(0.16)=4+1.56-1.08-0.48=\boxed{4.00}.$$ Total available electrons supplied by 1 mol glucose $=\gamma_s\cdot w=4(6)=24$. All of them must appear in biomass (anaerobic, no other electron sink), so $$c_{max}=\frac{\gamma_s\,w}{\gamma_{biomass}}=\frac{24}{4.00}=\boxed{6.00\ \text{mol biomass/mol glucose}}.$$
Comparison. Both independent bounds — the carbon balance ($c_{max}=6$) and the electron balance ($c_{max}=6.00$) — coincide exactly, and the actual given yield is also 6 mol/mol. Hence $$\frac{Y_{x/s}^{\text{actual}}}{Y_{x/s}^{\text{max}}}=\frac{6}{6.00}=\boxed{100\%}\quad\text{(both in molar and in mass terms, since the same ash factor applies to both).}$$

Quantity	Result
Biomass degree of reduction $\gamma_{biomass}$	4.00 (identical to glucose's $\gamma_s=4$)
Actual molar yield	6 mol biomass/mol glucose
Max. molar yield (carbon-balance)	6 mol/mol
Max. molar yield (electron-balance)	6.00 mol/mol
Actual mass yield (organic / total incl. ash)	0.815 g/g / 0.858 g/g
Actual/maximum ratio	100% — the culture already operates at the theoretical ceiling

Topic: Carbon-balance vs. electron (degree-of-reduction)-balance limits on anaerobic biomass yield.

Key relations: degree of reduction $\gamma=4c+h-2o-3n$ per formula unit; anaerobic electron-balance ceiling $c_{max}=\gamma_s w/\gamma_{biomass}$; carbon-balance ceiling $c_{max}=w/(\text{carbons per biomass formula unit})$.

Why this works: anaerobically, with no O₂ and no other reduced product besides biomass, conservation of available electrons is exactly as binding as conservation of carbon atoms — here the substrate and biomass happen to share the same degree of reduction (4.00), so the two independent bounds coincide and the stated reaction is already 100% efficient by both measures, a deliberately constructed teaching case.

Common pitfall: assuming 100% carbon conversion automatically implies the electron balance is also satisfied — in general $\gamma_s\neq\gamma_{biomass}$, and whichever balance is more restrictive (smaller $c_{max}$) is the one that actually limits the yield (see Q6's very different outcome for a case where the two bounds do not coincide).

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (degree of reduction, theoretical maximum yield).

Question 5 of 9: Cellulomonas SCP from Glucose — Actual vs. Maximum Theoretical Yield

Question 5: Cellulomonas SCP from Glucose — Actual vs. Maximum Theoretical Yield (20 marks)