Biology · Introductory biology · Worked example

Estimate a growth rate from two counts

Gause’s curve for Paramecium caudatum gives $20.4$ cells on day 1 and $137.2$ on day 2. Treat that day as exponential growth and find $r$ . Why is it lower than the $2.309 per day$ that Gause fitted for the uncrowded culture?

Write the growth over one day

Over $t = 1 day$ , exponential growth from $N_{1} = 20.4$ to $N_{2} = 137.2$ means $137.2 = 20.4 e^{r \cdot 1}$ .

137.2 = 20.4 e^{r}

Divide by the starting count

Dividing both sides by $20.4$ leaves the factor by which the culture grew, about $6.73$ .

\frac{137.2}{20.4} = e^{r}

Take the natural $log$

The natural $log$ undoes $e$ , so $r$ is the $log$ of the growth factor. Writing $r$ on the left keeps the next row checkable.

r = ln \frac{137.2}{20.4}

Evaluate

The culture grew at about $1.906$ per cell per day over that day.

r \approx 1.906

Compare with the uncrowded rate

Uncrowded, $20.4$ cells growing at $2.309 per day$ would reach $20.4 e^{2.309} \approx 205$ cells by day 2, not $137$ . Crowding is the difference. The free share of the tube, $1 - N / 375$ , fell from $0.946$ to $0.634$ during that day, so each cell reproduced more slowly as the day went on.

20.4 e^{2.309} \approx 205.31

Know what the number means

An $r$ found from two counts is an average over the interval, not the population’s $r$ . Only counts taken while $N$ is far below $K$ estimate the uncrowded rate, so ecologists fit the whole curve, as Gause did.

Result

$r \approx 1.906 per day$ . It is lower than $2.309$ because crowding was already slowing the culture, and the free share of the tube fell from about $0.95$ to $0.63$ that day.

Your turn

1. One like the example

A bacterial culture grows from $2, 000$ to $32, 000$ cells in $2 hours$ . Assuming exponential growth, find $r$ and the doubling time.

Show the answer and explanation

$r \approx 1.39 per hour$ , and the culture doubles every $30 minutes$ .

$32, 000 = 2, 000 e^{2 r}$ , so $16 = e^{2 r}$ , $ln 16 = 2 r$ and $r = ln 16 / 2 \approx 1.386$ . Since $16 = 2^{4}$ , the culture doubled four times in $2 hours$ , so the doubling time $ln 2 / r$ is $0.5 hours$ .

\begin{array}{l} 32000 = 2000 e^{2 r} \\ 16 = e^{2 r} \\ ln 16 = 2 r \\ r = \frac{ln 16}{2} \\ r \approx 1.39 \end{array}

2. A variation

Gause’s curve gives $319.0$ cells on day 3 and $369.0$ on day 4. Find $r$ over that day, and compare it with the day 1 to day 2 value.

Show the answer and explanation

$r \approx 0.146 per day$ , about a thirteenth of the $1.906$ found for day 1 to day 2.

$r = ln (369 / 319) \approx 0.146$ . By day 3 only about $15 %$ of the tube was free, so crowding had cut each cell’s growth far below the uncrowded $2.309$ .

ln \frac{369}{319} \approx 0.146

3. Somewhere new

A population counted once a year grows by $5 %$ each year. What is its continuous growth rate $r$ , and how long does it take to double?

Show the answer and explanation

$r \approx 0.0488 per year$ , and it doubles in about $14.2 years$ .

Each year multiplies $N$ by $1.05$ , so $e^{r} = 1.05$ and $r = ln 1.05 \approx 0.0488$ . The doubling time is $ln 2 / ln 1.05$ $\approx 14.21 years$ . Using $0.05$ for $r$ would give $13.86 years$ , a common slip.

\begin{array}{l} ln 1.05 \approx 0.0488 \\ \frac{ln 2}{ln 1.05} \approx 14.21 \end{array}

Keep exploring

The population growth tool opens with the logistic model from Gause’s $20.4$ cells on day 1, $r = 2.309 per day$ and $K = 375$ . It reaches about $137.5$ cells on day 2. The tool computes the curve again from the constants as printed, so it differs slightly from the $137.2$ in Gause’s table, which the lesson quotes.

Return to the concept →

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 6 key terms Open the population growth tool Solve for r in Math Open worked example on a board Population equations in Biology Reference

Your existing work stays on this device. Examples open as editable copies.

Write the growth over one day

Divide by the starting count

Take the natural loglog

Evaluate

Compare with the uncrowded rate

Know what the number means

Result

Your turn

1. One like the example

r≈1.39 per hourr≈1.39 per hour, and the culture doubles every 30 minutes30 minutes.

2. A variation

r≈0.146 per dayr≈0.146 per day, about a thirteenth of the 1.9061.906 found for day 1 to day 2.

3. Somewhere new

r≈0.0488 per yearr≈0.0488 per year, and it doubles in about 14.2 years14.2 years.

Keep exploring

Study it, then try it

Take the natural $log$

$r \approx 1.39 per hour$ , and the culture doubles every $30 minutes$ .

$r \approx 0.146 per day$ , about a thirteenth of the $1.906$ found for day 1 to day 2.

$r \approx 0.0488 per year$ , and it doubles in about $14.2 years$ .