Populations and Communities · Paper 2 — Free Biology SL Practice Questions

A researcher cultures yeast (Saccharomyces cerevisiae) in a chemostat and records the following data. Time / hours — Population / cells $\text{mL}^{-1}$ 0 — 50 2 — 100 4 — 200 6 — 400 | 800 | 10 — 1600 The exponential growth equation is $N_t = N_0 e^{rt}$, where $r$ is the intrinsic rate of increase and $t$ is time in hours. The doubling time is given by $t_d = \dfrac{\ln 2}{r}$.

A population of Escherichia coli is grown in a nutrient-rich broth in a closed flask. The initial population is $1000$ cells. The population doubles every hour during the first $6$ hours, reaching $64\,000$ cells at $t = 6\,\text{h}$. After $12$ hours the population stabilises at approximately $1 \times 10^{9}$ cells. $$N_t = N_0\,e^{rt} \qquad t_d = \frac{\ln 2}{r}$$

A population of Paramecium caudatum was cultured in a closed flask with a constant volume of nutrient medium. The population size was recorded every 24 hours. The culture medium was not replenished and waste products accumulated over time. Time (hours) — 0 — 24 — 48 — 72 — 96 — 120 — 144 — 168 — 192 — 216 — 240 Population (cells $\text{mL}^{-1}$) — 10 — 20 — 40 — 80 — 155 — 280 — 420 — 510 — 550 — 560 The following equations are provided: $$\frac{dN}{dt} = rN \quad \text{(exponential growth)}$$ $$\frac{dN}{dt} = rN\!\left(1 - \frac{N}{K}\right) \quad \text{(logistic growth)}$$