Question 8 of 10: Mean Generation Time for a Ternary-Fission Microorganism

Question 8: Mean Generation Time for a Ternary-Fission Microorganism (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. The mean time between successive cell divisions, $t_d$, for an organism that triples (not doubles) its population at each division.

Approach. Biomass still grows exponentially, $m=m_0e^{\mu t}$, regardless of how many daughters a division produces; find $\mu$ from the slope of $\ln m$ vs. $t$ as usual. What changes is the relationship between $\mu$ and the generation time: each division multiplies the population (and biomass) by a factor of 3 rather than the usual factor of 2 for binary fission, so $\mu$ and $t_d$ are linked by $\ln 3$, not $\ln 2$.

1. Specific growth rate from the data. Successive point-to-point slopes of $\ln(\text{mass})$ ($\Delta t=0.5$ h throughout) are $0.811$, $0.855$, $0.782$, $0.811\ \text{h}^{-1}$ — consistent within data-rounding, confirming clean exponential growth. A least-squares regression of $\ln m$ vs. $t$ over all 5 points gives $$\mu=\boxed{0.815\ \text{h}^{-1}}$$ (intercept $\ln m_0=-2.30$, i.e. $m_0=0.100$ g/L, matching the $t=0$ data point exactly).
2. Relate $\mu$ to the generation time for ternary (not binary) fission. After one generation time $t_d$, the population/biomass has multiplied by 3 (three daughters replace the parent): $$m(t_d)=3\,m_0=m_0e^{\mu t_d}\;\Rightarrow\;\mu t_d=\ln 3\;\Rightarrow\;t_d=\frac{\ln 3}{\mu}.$$ (For ordinary binary fission this same derivation gives the familiar $t_d=\ln2/\mu$; the factor changes from $\ln2\approx0.693$ to $\ln3\approx1.099$ here because each division event triples, rather than doubles, the population.)
3. Numerical answer. $$t_d=\frac{\ln 3}{0.815}=\frac{1.099}{0.815}=\boxed{1.35\ \text{h}}\ (\approx81\ \text{minutes}).$$