Question 7 of 10: E. coli Batch Culture — Specific Growth Rate and Biomass Yield

Question 7: E. coli Batch Culture — Specific Growth Rate and Biomass 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.

Find. (a) $\mu(t)$; (b) $\mu_{\max}$; (c) $Y_{XS}$ over the batch, and whether it is constant.

Approach. The instantaneous specific growth rate is $\mu=\dfrac{1}{x}\dfrac{dx}{dt}=\dfrac{d(\ln x)}{dt}$, estimated at each interior point from a central difference of $\ln x$ against the neighbouring time points (and a one-sided difference at the two ends). $\mu_{\max}$ is read off as the (constant) slope of the log-linear exponential-growth region. $Y_{XS}$ is computed as $-\Delta x/\Delta s$ over successive intervals and compared for constancy.

1. Compute $\mu_i$ at every time point. Using $\mu_i\approx\dfrac{\ln x_{i+1}-\ln x_{i-1}}{t_{i+1}-t_{i-1}}$ (central difference; forward/backward difference at the endpoints) on the tabulated $x(t)$ gives a lag phase (small $\mu\approx0.15$–$0.19\ \text{h}^{-1}$ at $t\le0.33$ h), a transition point ($\mu\approx1.0\ \text{h}^{-1}$ at $t=0.5$ h, where the central difference already straddles the onset of growth), a clean exponential plateau ($\mu\approx1.49$–$1.52\ \text{h}^{-1}$ for $0.75\le t\le3.0$ h), and a sharp decline to zero/negative as the culture exhausts glucose and enters stationary phase ($t\ge3.1$ h).
2. Part (a): plot $\mu$ vs. $t$ (Fig. 7.1). The plateau in the middle of the plot is the true exponential (balanced) growth phase; the low values at $t\le0.33$ h reflect the lag phase (cells adapting, not yet growing exponentially), and the collapse after $t\approx3.1$ h reflects substrate exhaustion ($s\to0$) forcing $\mu\to0$.
3. Part (b): $\mu_{\max}$ from a linear regression of $\ln x$ vs. $t$ over the exponential plateau ($t=0.75$ to $3.0$ h, 7 points). A least-squares fit of $\ln x=\mu t+\text{const.}$ over this window gives slope $$\boxed{\mu_{\max}=1.50\ \text{h}^{-1}}$$ with residuals below 0.01 in $\ln x$ at every point — an excellent straight-line fit confirming this is genuine exponential growth, not curve-fitting noise.
4. Part (c): observed yield $Y_{XS}=-\Delta x/\Delta s$ over successive intervals. Computing $Y_{XS}$ interval-by-interval (skipping the lag-phase intervals $t=0$–$0.5$ h, where $\Delta s$ is at or below measurement resolution — it is zero between 0.33 and 0.5 h — and the post-exhaustion interval $t=3.5$–$3.7$ h, where $\Delta s=0$, so the ratio is undefined) gives values scattered between $0.42$ and $0.53$ kg/kg, with no clear trend up or down as the culture progresses through exponential growth and into the transition to stationary phase.
5. Overall yield over the whole batch. $$Y_{XS}=\frac{x_{\text{final}}-x_0}{s_0-s_{\text{final}}}=\frac{11.6-0.20}{25.0-0.0}=\boxed{0.456\ \text{kg cells/kg glucose}}.$$ Given the $0.42$–$0.53$ kg/kg scatter across individual intervals, $Y_{XS}\approx0.45$–$0.50$ kg/kg is approximately constant through most of the batch (consistent with a single fixed true-growth-yield coefficient, as assumed in simple unstructured kinetic models); the scatter reflects reading/interpolation error in the tabulated data rather than a real, systematic drift in yield.