Chemostat Data – Yield Coefficient and Monod Parameters: 23-Chem-B4 Biochemical Engineering, May 2013, Q1

Question 1: Chemostat Data – Yield Coefficient and Monod Parameters (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.

D (h^-1)	x (g/L)	S (mM)	Regime
0.05	0.248	0.067	steady growth
0.5	0.208	1.667	steady growth
5	0	10	washout

Find. Y_X/S (g biomass/mol glucose), μ_max (h^-1), K_S (mM).

Approach. The third row (D = 5 h^-1, x = 0) is a washout point — at D ≥ μ_max the culture cannot sustain itself, biomass drops to zero, and the exit substrate equals the feed value, so it directly reveals S_in. With S_in known, Y_X/S follows from a biomass/substrate balance on each steady row, and μ_max/K_S follow from the fact that a chemostat at steady state always has μ = D, fit through the Monod equation.

Fig. 1 — Lineweaver–Burk linearisation of the two steady-state rows (D = μ).

Recover S_in from the washout row. At D = 5 h^-1 the biomass is completely washed out (x = 0), so no substrate is consumed and the measured exit concentration equals the feed concentration: $$S_{in}=S(D=5\ \text{h}^{-1})=\boxed{10\ \text{mM glucose}}$$
Glucose yield coefficient. With sterile feed (X_in = 0), a steady-state biomass balance on each non-washout row gives Y_X/S = x /(S_in − S), in g per mM; converting mM → mol (1 mM = 10^-3 mol/L) gives g per mole: $$Y_{X/S}\Big|_{D=0.05}=\frac{0.248}{(10-0.067)\times10^{-3}}=\frac{0.248}{9.933\times10^{-3}}=24.97\ \text{g/mol}$$ $$Y_{X/S}\Big|_{D=0.5}=\frac{0.208}{(10-1.667)\times10^{-3}}=\frac{0.208}{8.333\times10^{-3}}=24.96\ \text{g/mol}$$ The two rows agree to within 0.1%, confirming a single yield coefficient describes both operating points: $$Y_{X/S}=\boxed{24.96\ \text{g biomass / mol glucose}\ (\approx 25.0)}$$
Monod parameters from the two steady rows. At chemostat steady state the specific growth rate equals the dilution rate, μ = D = μ_maxS/(K_S+S). Linearising as 1/D = 1/μ_max + (K_S/μ_max)(1/S) and fitting the line through the two points (1/S, 1/D) = (14.93, 20.0) h and (0.600, 2.00) h: $$\text{slope}=\frac{K_S}{\mu_{max}}=\frac{20.0-2.00}{14.93-0.600}=1.257\ \text{h}\cdot\text{mM},\qquad \text{intercept}=\frac{1}{\mu_{max}}=1.246\ \text{h}$$ $$\mu_{max}=\frac{1}{1.246}=\boxed{0.802\ \text{h}^{-1}}\qquad K_S=1.257\times0.802=\boxed{1.01\ \text{mM}}$$ Substituting back into D = μ_maxS/(K_S+S) for both rows reproduces D = 0.0500 and 0.500 h^-1 exactly, confirming the fit.

Quantity	Value
S_in (recovered from washout)	10 mM
Y_X/S	24.96 g biomass / mol glucose
μ_max	0.802 h^-1
K_S	1.01 mM

Topic: Chemostat (continuous stirred-tank bioreactor) steady-state analysis: washout, yield coefficients, and Monod kinetic parameter estimation.

Key relations: steady state μ=D; sterile-feed yield Y_X/S=X/(S_in−S); Monod μ=μ_maxS/(K_S+S); Lineweaver–Burk linearisation 1/D=1/μ_max+(K_S/μ_max)(1/S); washout criterion D≥μ_max⇒X=0, S=S_in.

Why this works: A chemostat's defining property is that dilution rate IS the control variable for specific growth rate — the organism cannot grow faster than it is diluted at steady state, so scanning D and reading off the resulting (x, S) pairs is exactly equivalent to scanning μ and reading off Monod's rate-vs-substrate curve. The washout point is not "wasted" data: it is the one row that measures S_in directly, without needing it as a separately-given quantity.

Common pitfall: Forgetting to convert mM to mol/L before computing g/mol (a factor-of-1000 error), and forgetting that Y_X/S requires S_in, which is not stated explicitly in the table — it must be read off the washout row.

Source: Shuler & Kargi, Bioprocess Engineering, 2nd ed., Ch. 6 (chemostat theory, washout, Monod parameter estimation).

Question 1 of 6: Chemostat Data – Yield Coefficient and Monod Parameters

Question 1: Chemostat Data – Yield Coefficient and Monod Parameters (20 marks)