Investigating Ecosystems

This lesson covers the practical methods used to investigate ecosystems, including quadrats, transects, kick sampling, mark-release-recapture, and statistical analysis, as required by the Edexcel A-Level Biology specification (9BI0), Topic 10 -- Ecosystems.

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Why Sample?

It is usually impossible to count every organism in an ecosystem. Instead, ecologists take samples and use the data to estimate population size, distribution, and abundance. Good sampling must be:

• Representative -- the sample should reflect the whole area
• Random -- to avoid bias (using random number generators for coordinates)
• Sufficient in size -- more samples = more reliable data

Common Misconception: Students often think random sampling means "putting quadrats anywhere." True random sampling requires random number generators (or random number tables) to generate coordinates. Choosing locations by eye introduces unconscious bias (e.g. you might avoid muddy areas or be drawn to areas with more flowers).

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Choosing the Right Sampling Method

flowchart TB
    A["What do you\nwant to investigate?"] -->|"Distribution across\nan environmental gradient"| B["Use a\nTRANSECT"]
    A -->|"Abundance / density\nof organisms in\na uniform area"| C["Use random\nQUADRATS"]
    A -->|"Population size\nof mobile animals"| D["Use MARK-RELEASE-\nRECAPTURE"]
    A -->|"Freshwater\ninvertebrate\ncommunity"| E["Use KICK\nSAMPLING"]
    B --> B1["Line transect:\nquick survey"]
    B --> B2["Belt transect:\nmore detailed"]
    B --> B3["Interrupted belt:\ncompromise"]
    C --> C1["Record: species frequency,\ndensity, or % cover"]
    D --> D1["Lincoln Index:\nN = Mn/m"]
    E --> E1["Assess water quality\nusing indicator species"]

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Quadrats

A quadrat is a square frame (commonly 0.5 m x 0.5 m or 1 m x 1 m) placed on the ground to define a sample area. Organisms within the quadrat are counted or their cover is estimated.

Types of Data Collected

Measure	Description	When to Use
Species frequency	The number of quadrats in which a species is present, expressed as a percentage	Quick assessment of distribution; less useful for abundance
Species density	The number of individuals of a species per unit area (e.g. per m^2)	Countable organisms (e.g. individual daisy plants)
Percentage cover	The percentage of the quadrat area covered by a species	Plants that are difficult to count individually (e.g. grass, moss); total can exceed 100% if species overlap vertically
ACFOR scale	Abundance estimated on a subjective scale: Abundant, Common, Frequent, Occasional, Rare	Quick preliminary survey; subjective and less reliable

Random Quadrat Placement

To avoid sampling bias:

1. Lay out two tape measures at right angles along two edges of the study area (forming x and y axes)
2. Use a random number generator to produce pairs of coordinates
3. Place the quadrat at each pair of coordinates
4. Record the species and their abundance within each quadrat

How Many Quadrats?

A common exam question asks about the number of quadrats needed. More quadrats give more reliable results, but there is a point of diminishing returns. A species area curve can help determine the minimum quadrat number: plot the cumulative number of species found against the number of quadrats. When the curve levels off, you have sampled enough quadrats.

Worked Example: Estimating Population Size

Question: A student places 20 random quadrats (each 0.5 m x 0.5 m) in a field measuring 100 m x 50 m. The mean number of daisy plants per quadrat is 4.2. Estimate the total daisy population in the field.

Answer:

$\text{Area of one quadrat} = 0.5 \times 0.5 = 0.25 \text{ m}^2$Area of one quadrat=0.5×0.5=0.25 m2

$\text{Mean density} = \frac{4.2}{0.25} = 16.8 \text{ daisies per m}^2$Mean density=0.254.2​=16.8 daisies per m2

$\text{Total area of field} = 100 \times 50 = 5{,}000 \text{ m}^2$Total area of field=100×50=5,000 m2

$\text{Estimated total population} = 16.8 \times 5{,}000 = 84{,}000 \text{ daisies}$Estimated total population=16.8×5,000=84,000 daisies

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Transects

Transects are used to investigate how species distribution changes across an area, particularly where there is an environmental gradient (e.g. from a shoreline to inland, or from a path into a woodland).

Types of Transect

Type	Description	Advantage	Disadvantage
Line transect	A tape measure is stretched across the area. Species touching the line are recorded at regular intervals.	Quick; covers a long distance	Less detailed; misses species between points
Belt transect	A strip of fixed width (e.g. 0.5 m) is laid along the transect line. Quadrats placed contiguously along the strip.	Very detailed data	Very time-consuming
Interrupted belt transect	Quadrats placed at regular intervals along the transect (not contiguously).	Good compromise of detail and efficiency	May miss changes between quadrats

When to Use a Transect

Transects are used when you expect species distribution to change along a gradient, for example:

• Along a sand dune system (from embryo dunes to mature dunes) -- links to succession lesson
• Across a rocky shore (from low tide to high tide mark)
• From the edge to the centre of a woodland (light gradient)
• Along a river from upstream to downstream (pollution gradient)

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Kick Sampling (Freshwater Ecosystems)

Kick sampling is used to sample aquatic invertebrates in streams and rivers:

1. Hold a net downstream of your position
2. Kick the riverbed substrate upstream of the net for a set time (e.g. 30 seconds)
3. Disturbed invertebrates are carried by the current into the net
4. Empty the net contents into a white tray
5. Identify and count the organisms

Standardising the Method

To make results comparable between sites, you must standardise: the kick duration, the net size and mesh, the area disturbed, and the water depth. At each site, take multiple replicate samples and calculate a mean.

Biotic Index: Indicator Species

Kick sampling can be used to assess water quality because different invertebrate species have different tolerances to pollution:

Indicator Species	Pollution Tolerance	Water Quality Indicated	Biotic Index Score
Mayfly nymphs (Ephemeroptera)	Very low (intolerant)	Clean, well-oxygenated water	High (e.g. 10)
Stonefly nymphs (Plecoptera)	Very low	Clean water	High (e.g. 10)
Freshwater shrimps (Gammarus)	Moderate	Moderate water quality	Medium (e.g. 6)
Water louse (Asellus)	Moderate to high	Fairly polluted	Medium (e.g. 4)
Bloodworms (Chironomus larvae)	High (tolerant)	Polluted, low-oxygen water	Low (e.g. 2)
Rat-tailed maggots (Eristalis)	Very high	Very polluted, very low oxygen	Low (e.g. 1)

A biotic index assigns a score to each indicator species found. The total score indicates water quality: high scores = clean water; low scores = polluted water.

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Mark-Release-Recapture (Lincoln Index)

The mark-release-recapture method is used to estimate the population size of mobile animals (e.g. snails, woodlice, butterflies).

Method

1. Capture a sample of animals from the population
2. Count them -- this is the first sample (M)
3. Mark each individual (e.g. with a small spot of non-toxic paint)
4. Release the marked individuals back into the habitat
5. Allow time for the marked individuals to mix back into the population
6. Recapture a second sample
7. Count the total number in the second sample (n) and the number that are marked (m)

The Lincoln Index Formula

$N = \frac{M \times n}{m}$N=mM×n​

Where:

• N = estimated total population size
• M = number marked and released in the first sample
• n = total number captured in the second sample
• m = number of marked individuals recaptured in the second sample

Worked Example 1: Basic Lincoln Index

Question: A student captures 40 woodlice, marks them, and releases them. A week later, they capture 50 woodlice, of which 8 are marked. Estimate the population size.

Answer:

$N = \frac{M \times n}{m} = \frac{40 \times 50}{8} = \frac{2{,}000}{8} = 250 \text{ woodlice}$N=mM×n​=840×50​=82,000​=250 woodlice

Worked Example 2: Effect of Violated Assumptions

Question: In the above example, if 10 of the marked woodlice had been eaten by predators (because the paint made them more visible), how would this affect the estimate?

Answer:

If marked individuals are preferentially predated, fewer marked individuals survive to be recaptured. So m (recaptured marked individuals) would be smaller (e.g. 4 instead of 8).

$N = \frac{40 \times 50}{4} = 500$N=440×50​=500

The estimate would be too high (an overestimate). The true population is 250, but the violated assumption produces an estimate of 500.

Assumptions of Mark-Release-Recapture

Assumption	Explanation	If Violated...
No births, deaths, immigration, or emigration between sampling events	Population size remains constant	If deaths occur, N is overestimated; if births occur, N is underestimated
Marks do not affect survival	Marked individuals are not more likely to be predated	If marks increase predation, N is overestimated
Marks are permanent and visible	Marks do not rub off or become invisible	If marks rub off, N is overestimated (fewer recaptured)
Mixing is complete	Marked individuals mix randomly back into the population	If mixing is incomplete, results are unreliable
Equal probability of capture	Marked and unmarked individuals are equally likely to be caught	If marked animals become trap-shy, N is overestimated

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Simpson's Index of Diversity

Simpson's Index of Diversity (D) measures biodiversity, accounting for both species richness (number of species) and evenness (how evenly individuals are distributed among species):

$D = 1 - \sum\left(\frac{n}{N}\right)^2$D=1−∑(Nn​)2

Where:

• n = number of individuals of a particular species
• N = total number of individuals of all species
• The sum (\sum) is calculated across all species

D ranges from 0 to 1:

• Values close to 1 = high diversity (many species, evenly distributed)
• Values close to 0 = low diversity (dominated by one species)

Worked Example: Simpson's Index

Question: A student samples a meadow and records the following data:

Species	Number of Individuals (n)
Daisy	15
Buttercup	12
Clover	8
Dandelion	5
Total (N)	40

Calculate Simpson's Index of Diversity.

Answer:

Species	n	n/N	(n/N)^2
Daisy	15	0.375	0.1406
Buttercup	12	0.300	0.0900
Clover	8	0.200	0.0400
Dandelion	5	0.125	0.0156
		Sum	0.2862

$D = 1 - 0.2862 = 0.714$D=1−0.2862=0.714

D = 0.714 indicates moderately high diversity. If one species dominated overwhelmingly (e.g. 37 daisies and 1 of each other species), D would be much lower (closer to 0).

Interpreting Simpson's Index

D Value	Interpretation
0.0 -- 0.3	Low diversity (community dominated by one or few species)
0.3 -- 0.6	Moderate diversity
0.6 -- 0.8	High diversity
0.8 -- 1.0	Very high diversity (many species, evenly distributed)

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Reliability and Validity in Ecological Investigations

Concept	How to Achieve It
Reliability	Use a large sample size; repeat the investigation; calculate means and standard deviations
Validity	Control variables (standardise quadrat size, kick duration, etc.); use random sampling to avoid bias
Accuracy	Calibrate equipment; use appropriate identification keys; train observers

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Summary

• Ecologists use sampling techniques because it is impractical to count every organism.
• Quadrats measure species frequency, density, and percentage cover; placed randomly or along transects.
• Transects (line, belt, interrupted) investigate changes in distribution along environmental gradients.
• Kick sampling assesses freshwater invertebrate communities and water quality using indicator species and biotic indices.
• Mark-release-recapture (Lincoln index) estimates population size of mobile animals: N = Mn/m. Always state and evaluate the assumptions.
• Simpson's Index (D = 1 - sum of (n/N)^2) measures biodiversity, accounting for species richness and evenness.
• Good ecological investigations require random sampling, sufficient sample size, and controlled variables to ensure reliability and validity.

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A-Level Deep Dive: Investigating Ecosystems

Investigating ecosystems is the empirical and quantitative core of Topic 5 — the methodological lesson that turns ecological hypothesis into defensible measurement. The Edexcel 9BI0 treatment rests on four methodological pillars: (i) sampling design — random vs systematic, the choice between quadrats (sessile / slow-moving organisms) and transects (gradients) and mark-release-recapture (mobile animals); (ii) abiotic measurement — light, soil moisture, soil pH, temperature, dissolved O₂ — with explicit attention to precision, calibration and replication; (iii) sample size determination — the species-area / running-mean / cumulative-species curve as evidence that effort has been sufficient; and (iv) statistical analysis — Spearman's rank for monotonic correlation along a gradient, Student's t-test for comparing means between two sites, χ² for testing distributional independence. This deep dive ties each to the relevant Core Practical (CP10, CP11, CP12), works a Lincoln–Petersen calculation under a violated assumption, and equips candidates with the mark-scheme literacy needed for Paper 2 Section B fieldwork questions.

Spec mapping

The Edexcel 9BI0 specification places investigating ecosystems in Topic 5: On the Wild Side — Photosynthesis, Energy and Ecosystems, on Paper 2 (Energy, Exercise and Coordination). This is the fieldwork-and-statistics lesson of Topic 5 — the methodological hub in which CP10 (random sampling to investigate a habitat), CP11 (transect / quadrat / mark-release-recapture sampling along an environmental gradient) and CP12 (factors affecting decomposition rate) all converge. Statements concern: random vs systematic sampling and the situations that select each; the use of frame quadrats (frequency, density, percentage cover) and point quadrats (tussocky / layered vegetation); belt and line transects for gradient surveys; mark-release-recapture by the Lincoln–Petersen estimator and its assumptions; abiotic measurement with appropriate sensors (light meter, soil-moisture probe, pH meter, thermometer, dissolved-oxygen meter); the species–area (running-mean / cumulative-species) curve for determining sample-size adequacy; and the statistical tests appropriate to ecological data — Spearman's rank for correlation, the t-test for comparison of two means, χ² for goodness-of-fit and contingency — refer to the official Pearson Edexcel 9BI0 specification document for exact wording. Synoptic links radiate to lesson 6 — Population Dynamics (the theory the methods estimate), lesson 5 — Ecological Succession (transects across seral stages), lesson 1 — Ecosystems and Communities (the units sampled), lesson 2 — Energy Transfer in Ecosystems (biomass and productivity measurement) and Topic 4 — Biodiversity (Simpson's index and Lincoln estimates underpin diversity comparisons).

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Worked example with full mark scheme

Question (8 marks):

A student investigated the population of common limpets (Patella vulgata) on a 200 m × 8 m rocky-shore study site. In the first sampling session 80 limpets were captured, marked with a small dot of non-toxic enamel paint on the apex of the shell, and released. Forty-eight hours later 100 limpets were re-sampled and 16 of these carried a mark.

(a) Calculate the estimated population size using the Lincoln–Petersen estimator and express the limpet density per m². (2)

(b) During the 48 h interval, oystercatchers (Haematopus ostralegus) preferentially predated marked limpets because the white paint made them visually conspicuous. Predict, with reasoning, the direction of the resulting bias in the population estimate, and recalculate the estimate if predation reduced the recaptured-marked count from a true 16 to an observed 10. (3)

(c) The student wishes to test whether limpet density differs significantly between an exposed headland and a sheltered cove. Identify the appropriate statistical test, justify the choice over the alternatives, and state the null hypothesis. (3)

Solution with mark scheme:

(a) M1 (AO2.1) — apply N̂ = (M × C) / R = (80 × 100) / 16 = 8000 / 16 = 500 limpets. A1 (AO2.1) — site area = 200 × 8 = 1600 m², so density = 500 / 1600 ≈ 0.31 limpets m⁻² (accept 0.3 limpets m⁻² to 1 s.f.).

(b) M1 (AO3.1a) — preferential predation of marked individuals reduces R (the recaptured-marked count) below the value it would take under random mortality, so the ratio R / C falls. M1 (AO3.2a) — because N̂ = MC/R, an artificially small R inflates the denominator's effect and produces an overestimate of population size — the violation always biases N̂ upward when marks raise mortality or detectability. A1 (AO2.1) — recalculation: N̂ = (80 × 100) / 10 = 800 limpets, compared with the true 500 — a 60% overestimate, illustrating the sensitivity of the Lincoln estimator to violations of the equal-survival assumption.

(c) M1 (AO1.1) — the appropriate test is Student's t-test (specifically the unpaired / two-sample t-test) because the question compares two means (mean limpet density at headland vs cove) drawn from continuous quantitative data. M1 (AO3.1a) — Spearman's rank is rejected because the question asks about a difference, not a monotonic correlation; χ² is rejected because the data are continuous densities, not counts in discrete categories. A1 (AO1.2) — null hypothesis (H₀): there is no significant difference between the mean limpet densities at the exposed headland and the sheltered cove; any observed difference is due to sampling variation. The alternative hypothesis (H₁) is that the means differ significantly. The decision rule compares the calculated t to the tabulated critical t at p = 0.05 and df = n₁ + n₂ − 2.

Total: 8 marks.

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Specimen question modelled on the Edexcel 9BI0 paper format