The shape of the lesson
By the end of the lesson, learners can
- Ecosystems, niches and three levels of biodiversity: Biodiversity is more than a list of species.
- Random quadrats and Simpson's index: Choose the wild meadow and press Place 10 random quadrats.
- Mark, release, recapture: Press 1.
- Three levels: ecosystems and habitats, species and abundance, genetic variation.
- Random coordinates on a grid remove sampling bias.
- D = 1 − Σ(n/N)2: near 1 high diversity, near 0 low.
- Lincoln: N = n1 × n2 ÷ m2.
One picture every idea returns to
An ecosystem is a community of organisms and its physical environment, interacting; a niche is the role of a species in its ecosystem, including what it feeds on, where it lives and how it interacts with other organisms and the environment. Biodiversity can be assessed at three levels: the number and range of ecosystems and habitats, the number of species and their relative abundance, and the genetic variation within each species. Samples must be random to avoid bias. Frame quadrats placed at random coordinates give density, frequency or percentage cover of plants and slow-moving animals, and Simpson's index of diversity, D = 1 − Σ(n/N)2, combines richness and evenness on a scale of 0 to 1. Mobile animals are estimated by mark-release-recapture with the Lincoln index, N = (n1 × n2) ÷ m2.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Count a sample, estimate the whole.. Nobody can count every plant in a meadow or every beetle in a wood. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Ecosystems, niches and three levels of biodiversity | Biodiversity is more than a list of species. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Random quadrats and Simpson's index | Choose the wild meadow and press Place 10 random quadrats. Project tab 4 of the lab (Biodiversity lab, quadrats and Simpson's index) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Mark, release, recapture | Press 1. Project tab 6 of the lab (Biodiversity lab, mark-release-recapture) and let a student drive it while the class predicts. | Corners |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Mark, release, recapture, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Square each n/N before adding, then subtract the total from 1. | Adding the n/N values first always gives 1, and forgetting the final subtraction gives a number that falls as diversity rises. |
| Put the recaptured marked number m 2 on the bottom. | If marked animals are more likely to be caught a second time (trap-happy), m 2 is too large and the population is underestimated. |
Sampling and measuring biodiversity: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Define the terms ecosystem and niche.
Explain the three different levels at which the biodiversity of an area can be assessed.
Describe how you would use random sampling with frame quadrats to estimate the percentage cover of a moss on a lawn, and explain why sampling must be random.
Random quadrats in a hedge bottom contain: species A 24, species B 12, species C 8, species D 6. Calculate Simpson's index of diversity, D = 1 − Σ(n/N)2, to two decimal places.
In a pond, 45 water boatmen were caught, marked and released. Two days later 60 were caught, of which 9 were marked. Use the Lincoln index, N = (n1 × n2) ÷ m2, to estimate the population, and state two assumptions of the method.
Habitat X has D = 0.21 and habitat Y has D = 0.78. Explain what these values show about the two habitats.
Answer key and marking notes
- Ecosystem: a relatively self-contained, interacting community of organisms and the abiotic environment in which they live.
- Niche: the role of a species within an ecosystem, including how it gains energy and nutrients, where it lives and its interactions with other organisms and the environment.
- Ecosystem / habitat diversity: the number and range of different ecosystems and habitats in the area.
- Species diversity: the number of different species (richness) and their relative abundance (evenness).
- Genetic diversity: the variation within each species, the number of different alleles of each gene in the population.
- Lay two tape measures at right angles along two edges of the lawn to form a grid.
- Use random numbers (calculator or table) to generate coordinates and place the quadrat there; repeat for many quadrats (for example 20, or until the running mean levels off).
- In each quadrat estimate the percentage of the area covered by moss (for example using a 100-square gridded quadrat); calculate the mean.
- Random placement avoids bias from the investigator choosing sites, so the sample is representative of the whole lawn.
- N = 24 + 12 + 8 + 6 = 50.
- (n/N)2: 0.482 = 0.2304; 0.242 = 0.0576; 0.162 = 0.0256; 0.122 = 0.0144; sum = 0.328.
- D = 1 − 0.328 = 0.67.
- N = (45 × 60) ÷ 9 = 2700 ÷ 9 = 300.
- Assumption: marks are not lost and do not affect survival or the chance of being recaptured.
- Assumption: no births, deaths, immigration or emigration between samples, and the marked animals mix randomly with the population.
- Y has a high D, close to 1: high biodiversity, with many species and individuals spread evenly among them.
- X has a low D: low biodiversity, dominated by one or a few species, so it is likely to be less stable and more easily affected by a change in conditions.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which is the best definition of an ecosystem?
- Aa community of organisms interacting with each other and with their abiotic environment
- Ball the organisms of one species living in one area at the same time
- Cthe place where an organism normally lives
- Dthe role of a species within its community
Which measurement assesses biodiversity at the genetic level?
- Athe number of different alleles of a gene in a population of one species
- Bthe number of different species found in an area
- Cthe number of different habitats in a nature reserve
- Dthe number of individuals of each species in a quadrat
A student places quadrats only where plants are easy to see. What is the main problem with this?
- Athe sample is biased and may not represent the whole area
- Bthe quadrats will contain too many individuals to count
- CSimpson's index cannot be calculated from such data
- Dthe results will be less precise but still unbiased
In a mark-release-recapture study, 50 snails are marked and released. In a second sample of 40, 8 are marked. Estimate the population using N = (n1 × n2) ÷ m2.
- A250
- B2000
- C400
- D160
Which method is most suitable for estimating the abundance of grass, where separate plants cannot be counted?
- Apercentage cover in random frame quadrats
- Bdensity in random frame quadrats
- Cmark-release-recapture with the Lincoln index
- Dcounting individuals along a line transect
Use the table of counts from random quadrats at site P to calculate Simpson's index of diversity, D = 1 − Σ(n/N)2.
- A0.42
- B0.58
- C0.75
- D0.25
Using the same table, which statement compares sites P and Q correctly?
- AQ has higher diversity because its individuals are spread more evenly among the same four species
- BP has higher diversity because it has more grass plants
- Cboth sites have equal diversity because both have four species and 40 individuals
- DQ has higher diversity because it has more species than P
In a mark-release-recapture study of voles, some marked voles learn that traps contain food and enter them more often. What effect does this have on the estimate?
- Am2 is too high, so the population is underestimated
- Bm2 is too high, so the population is overestimated
- Cn1 is too low, so the population is overestimated
- Dthere is no effect, because the formula corrects for it
Quiz key
An ecosystem is the community plus its physical (abiotic) environment, all interacting. One species in an area is a population; the place where an organism lives is its habitat; and its role is its niche.
Genetic diversity is the variation within a species, measured for example as the number of alleles of a gene. Counting species and the individuals of each is the species level (richness and relative abundance), and counting habitats is the ecosystem or habitat level.
Choosing sites deliberately introduces bias, so the sample will over-represent conspicuous plants. Random coordinates avoid this. Simpson's index can still be calculated, but it will be wrong; and the problem is bias, not precision, so the data are not unbiased.
N = (50 × 40) ÷ 8 = 2000 ÷ 8 = 250. 2000 forgets to divide by m2; 400 is 50 × 8; 160 is 40 × 8 ÷ 2, a muddled rearrangement.
Percentage cover estimates the area of the quadrat occupied by the species, ideal when individual plants cannot be distinguished. Density needs countable individuals; mark-release-recapture is for mobile animals; and a line transect records presence along a gradient, not abundance.
N = 30 + 4 + 3 + 3 = 40. (n/N)2 = 0.5625 + 0.01 + 0.005625 + 0.005625 = 0.58375, so D = 1 − 0.584 = 0.42. 0.58 forgets to subtract from 1. 0.75 is the value for site Q, read from the wrong column. 0.25 comes from 1 − 30/40, using only the grass proportion without squaring or adding the other species.
Both sites have four species and N = 40, but at P grass is dominant, while Q is much more even; Simpson's index rewards evenness, so Q has the higher D. More grass means lower, not higher, diversity. Equal richness does not mean equal diversity, and Q does not have more species.
Trap-happy marked voles inflate m2, the number of marked animals in the second sample. Because m2 is the divisor in N = n1n2 ÷ m2, the estimate is too small. n1 is fixed at the first capture, and the formula assumes an equal chance of capture, so it cannot correct for this.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.