Measuring biodiversity: sampling methods and Simpson's index – Introduction to Biodiversity (Senior 4 Biology) | Ganzaa

Introduction to Biodiversity

Lessons

Senior 4 · Biology

Introduction to Biodiversity

3 lessons

Lesson 3 of 3

Measuring biodiversity: sampling methods and Simpson's index

25 min

Which stream is healthier?

A class wants to compare two streams that drain the same hill: one flows out of a forest, the other runs past a village and fields. They cannot count every insect larva on the stream bed, so they must sample, and then turn their counts into one number that captures diversity, so that the two streams can be compared fairly.

In S3 you estimated population size with quadrats and with capture–recapture. At A-level you must also choose the right method for each organism and habitat, decide how many samples are enough, judge whether a method's assumptions hold, and calculate Simpson's index of diversity.

Definition

Random sampling: choosing sample positions so that every point in the area has an equal chance of being chosen, usually from random-number coordinates on a grid. It avoids bias from the person choosing.

Systematic sampling: taking samples at regular intervals, usually along a transect (a line laid across a habitat), to show how organisms change along an environmental gradient.

Abundance: how many of a species there are, measured as density (individuals per unit area), percentage cover (the percentage of the ground covered by the species, seen from above) or frequency (the percentage of quadrats in which the species occurs).

Simpson's index of diversity, $D$ : a number from 0 to just under 1 that combines species richness and evenness; the higher it is, the more diverse the community.

Choosing a method

Method	How it works	Use it for
Quadrat, placed at random	Two tapes at right angles form axes; random numbers give coordinates; at each, count individuals or estimate percentage cover inside the frame	Plants and slow-moving animals in a fairly uniform area
Line transect	Record every organism that touches the tape at fixed intervals	A quick picture of how species change along a gradient
Belt transect	Place quadrats at fixed intervals along the tape (or continuously, end to end)	Measuring abundance, not just presence, along a gradient such as forest into grassland, or lakeshore to dry ground
Sweep net	Swing a strong cloth net through vegetation a fixed number of times	Insects and spiders in grass and low shrubs
Kick sampling	Hold a net on the stream bed downstream of your feet and disturb the stones for a fixed time, so that animals wash into the net	Invertebrates (insect larvae, worms, snails) in streams
Mist net	A fine, almost invisible net between poles; trained and licensed ringers remove each bird quickly	Small birds and bats, often with ringing for capture–recapture
Pitfall trap	A container sunk level with the soil, with a cover against rain; checked often	Beetles, spiders and other ground animals active at night
Capture–recapture	Catch, mark and release a sample; later catch a second sample and count the marked ones	Population size of mobile animals

To make two samples comparable, keep the effort constant: the same quadrat size, the same number of sweeps, the same kicking time, at the same time of day and in similar weather.

How big, and how many?

Quadrat size. Too small, and most quadrats are empty; too large, and counting takes too long. One way to choose: record the number of species in quadrats of increasing size; when larger quadrats add almost no new species, that size is enough. Typical sizes are $0.25\,\mathrm{m^2}$ for short grassland and up to $100\,\mathrm{m^2}$ for trees.
Number of samples. Plot the running mean (the mean of all samples so far) against the number of samples. While it still jumps about, take more; once it levels off, extra samples add little (Figure 2).

Worked example 1: from quadrat counts to a population

Twelve $0.25\,\mathrm{m^2}$ quadrats were placed at random on a 30 m by 20 m school lawn, and the wood sorrel (Oxalis) plants in each were counted (illustrative data): 4, 7, 2, 5, 6, 3, 5, 4, 6, 5, 4, 5. Estimate the population on the lawn, and say whether 12 quadrats were enough.

Mean per quadrat: $\bar{x} = \frac{\sum x}{n} = \frac{56}{12} = 4.67$ .
Density: $\frac{4.67}{0.25\,\mathrm{m^2}} = 18.7$ plants per $\mathrm{m^2}$ .
Area of lawn: $30 \times 20 = 600\,\mathrm{m^2}$ . Population $\approx 18.7 \times 600 = 11\,200$ plants (exactly $\frac{56}{12} \times 4 \times 600 = 11\,200$ ).
Figure 2 shows that from the fourth quadrat on the running mean stays within 5 % of the final value, so 12 random quadrats were enough for this lawn.

Capture–recapture at A-level

If $n_{1}$ animals are caught, marked and released, and a later sample of $n_{2}$ contains $m_{2}$ marked animals, the Lincoln index estimates the population $N$ :

N = \frac{n_1 \times n_2}{m_2}

The estimate is only as good as its assumptions: the marks are not lost and do not affect survival or the chance of recapture; marked animals mix fully with the rest; and there are no births, deaths, immigration or emigration between the two samples. At A-level you must say which way a broken assumption moves the estimate.

Worked example 2: butterflies at a forest edge

On day 1, 48 butterflies of one species were caught in a forest clearing with hand nets, each was given a small dot of ink on the underside of a hind wing, and all were released. On day 3, 55 were caught, of which 12 were marked (illustrative data). (a) Estimate the population. (b) Birds catch more of the marked butterflies because the dots make them easier to see. State and explain the effect on the estimate.

(a) $N = \frac{48 \times 55}{12} = \frac{2640}{12} = 220$ butterflies.
(b) Fewer marked butterflies survive to be recaptured, so $m_{2}$ is too small. Because $m_{2}$ is the denominator, a smaller $m_{2}$ gives a larger $N$ : the population is overestimated. That is why marks go on the underside of the wing, where they are hidden when the butterfly rests.

Simpson's index of diversity

Richness alone ignores evenness (Lesson 1). Simpson's index uses both. Imagine picking two individuals at random from a community: the index is the probability that they belong to different species.

D = 1 - \sum \left( \frac{n}{N} \right)^{2}

where $n$ is the number of individuals of each species and $N$ is the total number of individuals of all species. $D = 0$ means one species only; values close to 1 mean many species, evenly spread. Some books use $D = 1 - \frac{\sum n(n-1)}{N(N-1)}$ , which gives a slightly higher value in small samples; use the form your examination gives you, and the same form for every site you compare.

Worked example 3: two streams compared

Kick samples of the same duration were taken from a stream leaving a forest (A) and from a stream below a village (B) (illustrative data).

Animal	Stream A	Stream B
Mayfly nymphs	18	0
Stonefly nymphs	9	0
Caddisfly larvae	12	2
Freshwater shrimps	6	0
Flatworms	3	0
Leeches	0	3
Sludge worms	0	14
Bloodworms (midge larvae)	2	31
Total, $N$	50	50

Stream A: $\sum \left( \frac{n}{N} \right)^{2} = 0.36^2 + 0.18^2 + 0.24^2 + 0.12^2 + 0.06^2 + 0.04^2 = 0.1296 + 0.0324 + 0.0576 + 0.0144 + 0.0036 + 0.0016 = 0.2392$ , so $D_{A} = 1 - 0.2392 = 0.761$ .
Stream B: ${0.62}^{2} + {0.28}^{2} + {0.06}^{2} + {0.04}^{2} = 0.3844 + 0.0784 + 0.0036 + 0.0016 = 0.4680$ , so $D_{B} = 1 - 0.4680 = 0.532$ .
Stream A has more species (6 against 4) and a more even spread, so it is more diverse.
Interpretation: mayfly and stonefly nymphs need well-oxygenated water and are absent from B, while bloodworms and sludge worms, which tolerate low oxygen, dominate it. This pattern suggests organic pollution in B. A lower $D$ alone does not prove pollution: you would also measure dissolved oxygen and repeat the samples.

Miss Estruch Biology: "How to Sample in Ecology". Watch how random coordinates are generated for quadrats, and note when the video says to use quadrats and when to use mark–release–recapture.

Warning

Safety and good practice in fieldwork

Work in groups, near an adult, and only in streams that are shallow and slow on the day; never enter fast or flooded water. Wear boots, avoid skin contact with still or slow fresh water, which can carry bilharzia, and wash your hands with soap afterwards.
Handle animals as little as possible and return them where you found them. Mist-netting is done only by trained, licensed people.
Get permission before working in a protected area or on someone's land.

Tip

Exam technique for Simpson's index. Lay out a table with columns $n$ , $\frac{n}{N}$ and $\left( \frac{n}{N} \right)^2$ ; sum the last column, then subtract from 1. Give $D$ to 2 or 3 significant figures. When comparing sites, refer to both richness and evenness, and name a factor that could explain the difference.

✦

Practice: sampling and Simpson's index

Explain why quadrats are placed at random coordinates rather than where the student thinks is typical.
State which method you would use to sample: (a) beetles active on the forest floor at night; (b) grasshoppers in a meadow; (c) the change in plants from a lakeshore to dry ground.
Ten $1\,\mathrm{m^2}$ quadrats contained a total of 36 plants of a species, which was present in 7 of them. Calculate its density and its frequency.
In a capture–recapture study of land snails, 60 were marked, and a second sample of 45 contained 9 marked snails. Estimate the population.
In question 4, some marks were lost before the second sample. Explain how this affects the estimate.
A grassland has three species with 70, 20 and 10 individuals. Calculate Simpson's index of diversity.
Two sites have the same species richness, but $D = 0.31$ at site X and $D = 0.72$ at site Y. Explain what this tells you, and suggest one human activity that could cause the difference.

Show answersHide answers

Choosing positions by eye introduces bias (people pick "interesting" or easy spots). Random coordinates give every point an equal chance, so the sample represents the whole area and statistical tests are valid.
(a) Pitfall traps; (b) sweep net (with capture–recapture if a population size is needed); (c) a belt transect.
Density $= \frac{36}{10 \times 1\,\mathrm{m^2}} = 3.6$ per $\mathrm{m^2}$ ; frequency $= \frac{7}{10} \times 100 = 70\,\%$ .
$N = \frac{60 \times 45}{9} = 300$ snails.
Some marked snails were counted as unmarked, so $m_{2}$ is too small and $N$ is overestimated.
$N = 100$ ; $\sum \left( \frac{n}{N} \right)^2 = 0.49 + 0.04 + 0.01 = 0.54$ ; $D = 1 - 0.54 = 0.46$ .
Site X is dominated by one or a few species (low evenness), while at Y individuals are spread more evenly. Possible causes at X: heavy grazing, fertiliser (which favours a few fast-growing species) or trampling.

Quick check

You want to show how plant species change from a forest into a grassland. Which method fits best?

Key points

Sample at random for uniform areas and along transects for gradients; match the method (quadrat, sweep net, kick sample, mist net, pitfall trap) to the organism, and keep the effort the same at every site.
Choose quadrat size from a species–area test and the number of samples from a running-mean plot.
Lincoln index $N = \frac{n_1 \times n_2}{m_2}$ ; lost or harmful marks reduce $m_{2}$ and overestimate $N$ .
Simpson's index $D = 1 - \sum \left( \frac{n}{N} \right)^2$ combines richness and evenness; compare sites with the same formula and explain differences with measured factors.

1Which stream is healthier?

2Choosing a method

3How big, and how many?

4Capture–recapture at A-level

5Simpson's index of diversity

Which stream is healthier?

Choosing a method

How big, and how many?

Capture–recapture at A-level

Simpson's index of diversity