OCR GCSE Biology A: Field investigations, biodiversity and human impact on ecosystems (J247)

This study guide follows the OCR GCSE (9-1) Gateway Science Biology A specification (J247), Version 4.0 (August 2026), topic B6.1 “Monitoring and maintaining the environment”, statements B6.1a to B6.1d. Statements B6.1a to B6.1c are assessed in both Foundation and Higher Tier papers. Statement B6.1d is printed in bold in the specification, so it is Higher Tier only.

Use the OCR GCSE Biology hub and the printable checklist to track your progress. When you finish, work through the revision notes and the practice questions. For exam technique across the whole course, see exam preparation. If you are unsure where your gaps are, take one of the free 10-minute diagnostics.

What this unit covers

Statement	What you must be able to do	Tier
B6.1a	Explain how to carry out a field investigation into the distribution and abundance of organisms in a habitat, and how to find their numbers in a given area	Both
B6.1b	Describe positive and negative human interactions within ecosystems and explain their impact on biodiversity	Both
B6.1c	Explain some benefits and challenges of maintaining local and global biodiversity	Both
B6.1d	Evaluate evidence for the impact of environmental changes on the distribution of organisms, with reference to water and atmospheric gases	Higher Tier only

The specification also lists mathematical skills for this topic: calculating arithmetic means, plotting graphs with suitable scales, using percentiles, extracting information from charts, graphs and tables, and understanding the principles of sampling.

B6.1a: Field investigations

Distribution means where organisms are found. Abundance means how many there are. A field investigation measures one or both, usually to relate them to an abiotic factor such as light, soil moisture or distance from a path.

Why we sample

You cannot count every daisy in a field or every beetle in a wood. Instead you count in small, representative areas and scale up. The sample must be large enough and unbiased, otherwise the estimate is poor.

Random sampling

Random sampling gives every part of the habitat an equal chance of being chosen. It avoids bias, such as placing quadrats where plants look thickest.

1. Lay tape measures along two adjacent sides of the study area to form axes.
2. Generate pairs of random numbers, from a random number table or generator, as coordinates.
3. Place a quadrat at each coordinate and count.

Random sampling suits a fairly uniform habitat when you want the overall abundance.

Systematic sampling and transects

Systematic sampling places samples at regular intervals. A transect is a line across a habitat. You lay a tape measure along the line and place a quadrat at fixed intervals, for example every 2 m. Use a transect when conditions change gradually, such as from the sea towards the land, or from a woodland edge into the open. The distribution of each species can then be matched to the changing factor.

Equipment

Equipment	Used for
Quadrat (a square frame, often 0.5 m by 0.5 m, so 0.25 m² in area)	Counting plants or slow-moving animals in a known area; also estimating percentage cover
Pooter	Collecting small insects by suction without crushing them
Net (sweep net or pond net)	Catching flying insects in vegetation or organisms in water
Key	Identifying an organism from observable features
Tape measure	Laying out transects and measuring area

Percentage cover is useful when plants overlap or are hard to count as individuals, such as grass. Estimate the fraction of the quadrat that each species covers.

Scaling up from quadrats

The method is: find the mean number per quadrat, divide by the quadrat area to get the number per m², then multiply by the total area.

Worked example 1. A meadow measures 40 m by 25 m. A student uses twelve random quadrats, each 0.5 m by 0.5 m, and counts buttercups: 3, 5, 0, 4, 2, 6, 1, 3, 4, 2, 5, 3.

• Total area = 40 × 25 = 1000 m²
• Quadrat area = 0.5 × 0.5 = 0.25 m²
• Total count = 38, so mean = 38 ÷ 12 = 3.17 per quadrat
• Number per m² = 3.17 ÷ 0.25 = 12.67 per m²
• Estimate for the meadow = 12.67 × 1000 = 12 667, which is about 12 700 buttercups

This is an estimate. A larger sample, or more random placement, would usually make it more reliable.

Capture-recapture

Capture-recapture is used for animals that move. It estimates population size using this relationship:

estimated population = (number first marked × number in second sample) ÷ number of marked animals recaptured

Worked example 2. A student collects ground beetles with a pooter and marks 48 with a small dot of non-toxic paint on the underside. After a day, she collects 40 beetles and finds that 6 have a mark.

• Estimated population = (48 × 40) ÷ 6
• = 1920 ÷ 6 = 320 beetles

The method assumes that marking does not harm the animals or make them easier to see, that marked animals mix back in evenly, that no animals enter or leave the population, and that no large births or deaths occur between samples. If many marked animals were eaten, the estimate would be too high, because the recaptured number in the denominator would be too small.

Worked example 3: a transect

A student lays a 10 m transect from a hedge into the middle of a field and records percentage cover of dandelions and the light level.

Distance from hedge (m)	0	2	4	6	8	10
Dandelion cover (%)	2	6	15	28	36	38
Light (arbitrary units)	20	35	55	80	92	95

Dandelion cover is much lower near the hedge (2% compared with 38%), and it rises as light level rises. This suggests that dandelions are more abundant where light is higher. The data show a link, not proof that light is the cause, because other factors such as soil moisture and competition also change with distance from the hedge. A good answer states the pattern with figures, then gives the limit on the conclusion.

Percentiles and the symbols to use

A percentile tells you the value below which a given percentage of results fall. In the lengths of 20 limpet shells below, ordered from 21 mm to 40 mm in 1 mm steps, the 75th percentile is the value in position 0.75 × 20 = 15, which is 35 mm. So 75% of the shells are 35 mm or shorter. Use a clear, stated method.

Where an answer asks you to compare, use the symbols =, <, <<, > and >> correctly. For example, mean 12.7 per m² > mean 3.2 per m² describes a higher density, and << means much less than.

Common errors for B6.1a

• Forgetting to divide by the quadrat area, so the answer is per quadrat rather than per m².
• Placing quadrats by eye and calling the method random.
• Using a random method when the question describes a changing gradient, where a transect is the better choice.
• Saying that capture-recapture gives the exact population rather than an estimate.

B6.1b: Human interactions and biodiversity

Biodiversity is the variety of living organisms in an ecosystem or on Earth. High biodiversity usually makes an ecosystem more stable because there are more species to fill roles in food webs if one declines.

Negative interactions

Human activity	Impact on biodiversity
Land use: clearing forest for farming, building or roads	Habitat is destroyed or split into small patches, so populations shrink and some species disappear locally
Draining wetlands or ploughing meadows	Specialist plants and animals lose their habitat
Farming with one crop (monoculture)	Few species can live there and food sources are limited
Hunting and poaching	Numbers of the target species fall; if breeding cannot keep up, the species may become extinct
Pollution, such as acid rain from sulfur dioxide	Some plants and animals cannot survive the change in conditions

Positive interactions

Conservation can focus on individual species or on selected habitats.

• Individual species: legal protection from hunting, captive breeding and reintroduction, and seed banks that store seeds of rare plants.
• Selected habitats: nature reserves and protected areas, restoring hedgerows or ponds, and reducing pollution that damages a habitat.

Worked example 4: acid rain and germination

A student soaks 25 cress seeds in water of each pH and counts how many germinate.

pH	7	6	5	4	3
Seeds germinated	23	22	19	12	4
Percentage germinated	92	88	76	48	16

Percentage = germinated ÷ 25 × 100. At pH 3 the value is 4 ÷ 25 × 100 = 16%.

To plot this, put pH on the x-axis (the independent variable) and percentage germination on the y-axis, choose scales that use more than half the grid, label axes with units and join points sensibly. The graph shows that germination falls as acidity increases, and falls sharply below pH 5. The fall from 92% to 16% is a decrease of 76 percentage points. Acid rain could therefore reduce the plants that establish themselves in an affected habitat, which lowers biodiversity.

Indicator species

Lichens are used to monitor air pollution. Crusty lichens tolerate more sulfur dioxide, while bushy lichens are found only where the air is cleaner. Counting lichen types at several distances from a pollution source shows how the air quality changes.

B6.1c: Benefits and challenges of maintaining biodiversity

Benefits

• Stable ecosystems with more species are better able to cope with change.
• Many foods, materials and medicines come from wild species.
• Plants such as pollinated crops depend on insects and other wild animals.
• Wild areas give people recreation and income.

Challenges

• Gaining agreements: countries and groups have different needs. A country trying to develop its economy may be reluctant to limit logging, farming or fishing, and agreements across borders are difficult to negotiate and enforce.
• Monitoring conservation schemes: checking that a scheme works needs regular surveys, trained staff and money. Remote areas make monitoring hard, and poaching or illegal logging may go unnoticed.
• Local livelihoods: people who rely on land, hunting or fishing may lose income unless there are alternatives.

Ecotourism

Ecotourism brings visitors to see wildlife in a way designed to limit damage. It can pay for rangers and monitoring, give local people jobs, and give communities a reason to protect habitats rather than clear them. Its benefit depends on managing visitor numbers so that disturbance does not harm the wildlife it relies on.

B6.1d: Environmental change and distribution (Higher Tier only)

You must evaluate evidence, not only describe changes. Link the factor, the pattern in the data and the biology.

Water

Dissolved oxygen, temperature and pollution all affect where aquatic organisms live. Sewage and fertiliser run-off feed microorganisms, which use up oxygen as they respire. Animals needing high oxygen, such as mayfly nymphs, then disappear, while those that tolerate low oxygen, such as bloodworms, can remain. Warmer water also holds less dissolved oxygen.

Atmospheric gases

Rising carbon dioxide and its effect on temperature can move the range of species. Sulfur dioxide pollution removes sensitive lichens. Distribution is therefore shown by which species are present at each place.

Worked example 5: evaluating river evidence

	Above outflow	Below outflow
Dissolved oxygen (mg/dm³)	8.4	3.1
Mayfly nymphs in sample	46	12

• Oxygen fell by (8.4 − 3.1) ÷ 8.4 × 100 = 63%.
• Mayfly nymphs fell by (46 − 12) ÷ 46 × 100 = 74%.

Evaluation. The evidence supports the idea that the outflow reduced oxygen and so reduced mayflies, because both fall at the same place. However, one sample at each site is weak. Repeats on different days would show whether the result is typical, and other differences between sites, such as flow speed or temperature, are not controlled. A correlation does not prove cause. Better evidence would add samples at several distances downstream and show recovery further away.

When evaluating, comment on sample size, repeats, control of other variables, and whether the data cover enough time or places.

Official syllabus

Source: OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Topic B6.1 Monitoring and maintaining the environment, statements B6.1a to B6.1d.