OxfordAQA International AS and A-level Biology: Populations, communities and ecological succession (9610) – Revision Notes

These notes condense section 3.3.1 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, which sits in Unit 3 (International A-level only). For full explanations use the study guide. Then test yourself with the practice questions. Course pages: OxfordAQA A-level Biology hub, printable checklist and the free diagnostics.

3.3.1.1 Definitions to learn word for word

Word	Meaning
Population	Every organism of a single species living in one habitat
Community	Populations of different species in the same place
Ecosystem	A community together with the non-living parts of its surroundings
Habitat	The place where an organism lives
Niche	The role of a species in its habitat, set by its adaptations to biotic and abiotic conditions
Biotic factor	Effect of other living organisms: food, competitors, predators, parasites
Abiotic factor	Non-living: temperature, light, pH, water, salinity, mineral ions

Must-know distinction: habitat is where, niche is how. Two species can share a habitat and differ in niche.

3.3.1.2 What changes population size

Cause	Type	Effect	Typical wording
Abiotic factor	Non-living	Changes survival, growth or breeding	“Low temperature slows enzyme-controlled reactions, so fewer offspring survive”
Intraspecific competition	Same species	Stabilises the population around a level the environment can support	“Each individual gets a smaller share of a limited resource”
Interspecific competition	Different species	The better competitor rises and the weaker falls or disappears	“Niches overlap, so one species is out-competed for light”
Predation	Predator kills prey	Cyclical change, with the predator peak after the prey peak	“More prey, then more food, then more predators after a delay”

Method in steps for a predator-prey graph

1. Name the peak and trough years for each species.
2. State the lag between prey peak and predator peak.
3. Link the two with food supply.
4. Quote at least one figure.
5. Say what else could cause the change (weather, disease, other food).

Worked reminder. Aphids rise from 480 to 1200 per plant between week 2 and week 6. Percentage change = (1200 − 480) / 480 × 100 = +150%. The starting figure is the divisor.

Analysing data: correlation and cause

• A positive correlation means both variables rise together. A negative correlation means one rises as the other falls.
• A correlation tells you that two variables move together. On its own it cannot tell you that one produces the other.
• A third factor can drive both (a confounding variable). Along a transect, light, soil moisture and soil depth often change together.
• Support for a causal link: a plausible mechanism, a consistent result at many sites, and an experiment in which only one factor is changed.
• Random sampling and a suitable sample size make data representative.

Statistics in this section

Statistic	Formula	Use
Mean	sum of values / n	Typical value
Standard error	standard deviation / √n	How precise a mean is
95% confidence limits	mean ± about 2 × standard error	If limits of two means do not overlap, the difference is unlikely to be chance
Spearman rank	rₛ = 1 − 6Σd² / (n(n² − 1))	Tests correlation between two variables whose values you can rank

Method for Spearman rank in steps

1. Rank each variable separately, 1 for the smallest value.
2. Find d, the difference between each pair of ranks, and square it.
3. Add the d² values to get Σd².
4. Substitute into the formula. The answer lies between −1 and +1.
5. Look up the critical value for your n at p = 0.05. If rₛ, ignoring its sign, is bigger, the correlation is significant. The sign gives the direction.

Worked reminder. Six quadrats give Σd² = 10. rₛ = 1 − (6 × 10) / (6 × 35) = 1 − 60/210 = 0.714. If the critical value for n = 6 at p = 0.05 is 0.886, then 0.714 is smaller, so the correlation is not significant.

Worked reminder. Five counts of snails per tile: 12, 15, 9, 14, 10. Mean = 60 / 5 = 12. Standard deviation = 2.55, standard error = 2.55 / √5 = 1.14, so the 95% limits are about 9.7 to 14.3.

Required practical 9: competition and seedling growth

Item	Notes
Aim	Investigate the effect of competition on seedling growth
Independent variable	Seedling density (intraspecific) or proportion of two species (interspecific)
Dependent variable	Mean dry mass per seedling, or shoot height, after a fixed time
Control variables	Species and seed batch, compost, tray area, sowing depth, watering, light, temperature, harvest time
Replicates	At least three trays per treatment, to calculate a mean and spot anomalies
Dry mass	Dry the shoots until the mass stops changing, then weigh them
Conclusion	Mean mass per seedling falls as density rises, because of competition for light, water and mineral ions

Why dry mass is better than fresh mass: it removes the variable water content. Why tray area must be the same: density is the number of seedlings per unit area.

Worked reminder. Mean mass per seedling falls from 48 mg at 8 per tray to 30 mg at 32 per tray. Percentage decrease = (48 − 30) / 48 × 100 = 37.5%. Total mass per tray rises from 384 mg to 960 mg (8 × 48 and 32 × 30), so competition lowers the mass per plant, not the total.

Evaluation points: edge effects (more light at the tray edge), one harvest time only, small number of replicates, seedlings of different size at the start.

3.3.1.3 Primary succession

Primary succession starts on newly formed or exposed land that has no soil, such as bare rock, new sand or volcanic ash. Pioneer species arrive first and a climax community forms at the end.

Stage	Typical organisms	What changes
Pioneer	Lichens, mosses, cyanobacteria	Rock weathers slightly, dead remains add organic matter
Early soil	Small herbaceous plants, grasses	Soil holds more water and mineral ions, environment less hostile
Intermediate	Shrubs, small trees	Taller species shade and out-compete the herbs
Climax	Trees (woodland in many regions)	Relatively stable community

Key idea in your own words: the species at one stage alter the environment in ways that favour newcomers with different adaptations, and the newcomers may then make conditions worse for the earlier species.

Biodiversity and succession. Early on, harsh conditions allow few species. As soil develops, more niches appear and species number rises. Near the climax it may level off or fall, because dense canopy reduces light and a few dominant species out-compete others.

Worked reminder. A dune system has 4 plant species per 100 m² on the young dune and 26 on the older, vegetated dune. Increase = (26 − 4) / 4 × 100 = +550%. The older dune has deeper soil and more organic matter.

Conservation and managing succession

Conservation of habitats frequently involves managing succession, which holds a habitat at one stage.

Habitat	Natural direction	Management	Effect
Species-rich grassland	Scrub, then woodland	Grazing, mowing, scrub removal	Keeps the grassland stage open
Woodland floor	Shade increases	Coppicing	Light reaches ground plants
Heathland	Invaded by trees	Clearing saplings	Keeps heathland species

Costs: lost income, ongoing labour. Benefits: protects species that depend on the earlier stage. The unit introduction stresses that sustainability depends on managing the conflict between human needs and conservation.

Quick self-test

1. Define a population.
2. Define an ecosystem.
3. Distinguish a habitat from a niche.
4. Aphid numbers rise from 480 to 1200 per plant. Calculate the percentage increase.
5. Calculate the 95% confidence limits (about mean ± 2 × standard error) for the snail counts 12, 15, 9, 14 and 10.
6. A Spearman rank test on 6 pairs gives Σd² = 10. Calculate rₛ and state whether it is significant if the critical value is 0.886.
7. Explain why the predator peak follows the prey peak.
8. Why is intraspecific competition said to stabilise a population?
9. Grass cover falls from 38% to 12% after shrubs invade. Calculate the percentage decrease.
10. Explain why plant species number may fall near the climax.
11. Give two reasons why dry mass is a better measure than height in required practical 9.
12. Explain why grazing can conserve a grassland.

Answers

1. Every organism of a single species living in one habitat.
2. A community together with the non-living parts of its surroundings.
3. Habitat is the place where it lives. Niche is the role it plays, set by its adaptations to biotic and abiotic conditions.
4. (1200 − 480) / 480 × 100 = 150%.
5. Mean 12, standard deviation 2.55, standard error 1.14, so about 9.7 to 14.3.
6. rₛ = 1 − 60/210 = 0.714. This is below 0.886, so not significant.
7. Prey increase gives predators more food, but predators need time to breed, so they increase after a delay.
8. As numbers rise each individual gets less of a limited resource, so births fall or deaths rise and the population levels off.
9. (38 − 12) / 38 × 100 = 68.4%.
10. Dense canopy lets little light reach the ground, so light-demanding plants are out-competed and a few shade-tolerant species remain.
11. It ignores differences in water content, and it measures total growth, not just one dimension.
12. Grazing removes shrub seedlings, so succession does not move on to scrub and woodland, and the light-demanding grassland plants survive.

Where marks are usually lost

• Writing “ecosystem” when the answer needs “community”, or leaving out the non-living part.
• Giving a niche as a place.
• Explaining a predator-prey peak without saying that the predator rises after a delay.
• Using the end figure as the divisor in a percentage change.
• Describing a correlation as proof of cause.
• Giving the sign of rₛ without comparing its size with a critical value.
• Leaving out tray area or replicates in a design for the seedling practical.
• Stating that succession simply “adds” species, when the key idea is that earlier species change the environment.
• Writing that biodiversity always increases through succession.
• Naming a management method without saying which stage it holds the habitat at.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), specification Version 5.1, for International A-level exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.1 and required practical 9 in Unit 3.