Question 7 of 10: Hairy-Root Culture Kinetics — Specific Growth Rate, Sugar Uptake and 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 7: Hairy-Root Culture Kinetics — Specific Growth Rate, Sugar Uptake and Yield (20 marks)
Given. The 12-point biomass (g/L) and residual-sugar (g/L) time course above, $t=0$ to 55 d.
Find. $\mu(t)$ and $q_S(t)$ over each 5-day interval; identify any interval of constant $\mu$; the overall observed $Y_{XS}$.
Approach. Both specific rates are computed interval-by-interval from the discrete data: $\mu\approx\ln(X_{i+1}/X_i)/\Delta t$ (log-mean growth rate over the interval) and $q_S\approx-(\Delta S/\Delta t)/\bar X$ (sugar consumed per unit time per unit average biomass), each plotted against the interval midpoint. The overall yield is simply total biomass formed divided by total sugar consumed across the whole run.
(a) Specific growth rate per interval.
$$\mu_i=\frac{\ln(X_{i+1}/X_i)}{t_{i+1}-t_i}.$$
Evaluated for all 11 intervals (full table plotted below): $\mu$ falls from $0.223\ \text{d}^{-1}$ (0–5 d) to $0.154\ \text{d}^{-1}$ (5–10 d) to $0.055\ \text{d}^{-1}$ (10–15 d), and continues declining, with minor noise, down to $0.003\ \text{d}^{-1}$ by 50–55 d.
(b) Specific sugar-uptake rate per interval.
$$q_{S,i}=\frac{S_i-S_{i+1}}{(t_{i+1}-t_i)\,\bar X_i},\qquad \bar X_i=\tfrac12(X_i+X_{i+1}).$$
$q_S$ tracks $\mu$ closely in shape, falling from $0.40\ \text{g sugar g}^{-1}\text{d}^{-1}$ (0–5 d) to $0.009\ \text{g g}^{-1}\text{d}^{-1}$ by 50–55 d.
(a) When is $\mu$ constant? Strictly, it is not: $\mu$ is highest in the very first interval and declines monotonically (with minor scatter) over the whole 55-day run — there is no window in which successive $\mu_i$ values are equal within the data's noise. This is a genuine biological result, not a data-reading failure: differentiated hairy-root tissue grows as a branching, increasingly diffusion-limited mat (nutrient and O2 penetration into the root interior becomes progressively harder as biomass accumulates), so hairy-root cultures characteristically show a continuously decaying specific growth rate rather than the sharp, constant-$\mu$ exponential phase seen in well-mixed single-cell suspension cultures. The closest approach to "constant" is the initial 0–10 d window, where $\mu$ is highest and changing most slowly in relative terms.
Specific growth rate μ(t) and specific sugar-uptake rate qₛ(t), computed interval-by-interval from the batch data. Both decline continuously from the start — there is no window of constant μ in this hairy-root culture.