The problem. Ten quadrats are laid in a forest stand. Species A is found in 6 of them, with 3, 5, 2, 4, 6 and 4 individuals. Species B is found in 9 of them, with 1, 2, 1, 1, 1, 2, 1, 1 and 1 individuals. Calculate density, frequency and abundance for each, and comment on the distribution.
The three formulas. They differ only in what you divide by, which is exactly why they get muddled:
- Density = total individuals ÷ total quadrats laid
- Frequency (%) = (quadrats in which the species occurred ÷ total quadrats laid) × 100
- Abundance = total individuals ÷ quadrats in which it occurred
Species A. Total individuals = 3 + 5 + 2 + 4 + 6 + 4 = 24, present in 6 quadrats.
Density = 24 ÷ 10 = 2.40 per quadrat Frequency = (6 ÷ 10) × 100 = 60% Abundance = 24 ÷ 6 = 4.00
Species B. Total individuals = 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 = 11, present in 9 quadrats.
Density = 11 ÷ 10 = 1.10 per quadrat Frequency = (9 ÷ 10) × 100 = 90% Abundance = 11 ÷ 9 = 1.22
Now the interpretation, which is where the marks are. The abundance-to-frequency ratio is used to describe how individuals are spread:
Species A: 4.00 ÷ 60 = 0.0667 Species B: 1.22 ÷ 90 = 0.0136
On the ratio limits commonly taught — below 0.025 regular, 0.025 to 0.05 random, above 0.05 contagious — Species A is contagious, meaning clumped, and Species B is regular, meaning evenly spread.
Read that back against the raw data and it makes sense. Species A turns up in only six plots but in groups of four to six when it does. Species B is nearly everywhere but almost always as a single individual. Density alone would have hidden this: A looks twice as dense as B, yet B is the more widely distributed species. That sentence is the answer the examiner is looking for.
Check which ratio limits your own course uses before quoting them, since textbooks differ slightly.