Hybridoma Batch Culture — Specific Growth Rate and Doubling Time: 04-BS-13, May 2016, Q8

Question 7: Hybridoma Batch Culture — Specific Growth Rate and Doubling Time (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. Batch cell-concentration time course (table above); exponential growth is expected to appear as a straight line on a plot of $\ln X$ vs. $t$.

Find. (a) specific growth rate $\mu$ during the growth phase; (b) doubling time $t_d$.

Approach. Plot $\ln X$ against $t$, identify the growth phase (rising, before the culture peaks and declines), and take $\mu$ as the slope of a least-squares regression line through that phase — not a two-point estimate, which is far more sensitive to measurement scatter in any one reading. The doubling time follows directly from $\mu$.

Figure 6. Semi-log plot of the hybridoma batch data. The culture peaks at $t=3.5$ d ($X=4.02\times10^6$ cells/mL) and declines thereafter (death phase, red points, excluded from the growth-rate regression).

Identify the growth phase. $X$ rises monotonically from $t=0$ to the peak at $t=3.5$ d ($4.02\times10^6$ cells/mL), then falls at $t=4.0$ and $t=4.5$ d (death phase, nutrient depletion/waste accumulation) — these last two points are excluded from the growth-rate fit.
Least-squares regression of $\ln X$ on $t$ over the growth phase ($t=0$ to $3.5$ d, 9 points). $$\mu=\text{slope}=\frac{\sum(t_i-\bar t)(\ln X_i-\overline{\ln X})}{\sum(t_i-\bar t)^2}=\boxed{0.645\ \text{d}^{-1}}\qquad(R^2=0.993),$$ confirming a genuinely exponential (straight-line, on the semi-log plot) growth phase across the full 3.5-day window despite realistic scatter in the raw counts.
Part (b): doubling time. $$t_d=\frac{\ln2}{\mu}=\frac{0.693}{0.645}=\boxed{1.07\ \text{days}\ (\approx25.8\ \text{h})}.$$

Quantity	Result
Growth phase	$t=0$ to 3.5 d (9 points, $R^2=0.993$)
(a) Specific growth rate, $\mu$	0.645 d⁻¹
(b) Doubling time, $t_d$	1.07 days (≈25.8 h)

Topic: Batch-culture growth kinetics — specific growth rate from a semi-log regression, and its relation to doubling time.

Key relations: exponential growth $X=X_0e^{\mu t}\Rightarrow\ln X=\ln X_0+\mu t$ (linear in $t$); doubling time $t_d=\ln2/\mu$.

Why this works: regressing over every point in the identified exponential window (rather than picking any single pair of points) averages out the inevitable measurement scatter in individual cell counts, giving both a more defensible $\mu$ and, via $R^2$, an explicit check that the assumed exponential model actually fits the data.

Common pitfall: computing $\mu$ from just the first and last growth-phase points (e.g. $t=0$ and $t=3.5$) — this two-point shortcut is highly sensitive to noise in exactly those two readings and discards the information in the other seven; always regress the whole identified window.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (batch growth kinetics, specific growth rate determination).

Question 8 of 9: Hybridoma Batch Culture — Specific Growth Rate and Doubling Time

Question 7: Hybridoma Batch Culture — Specific Growth Rate and Doubling Time (20 marks)