IB DP Biology: Populations and communities: sampling, growth, interactions (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology section C4.1 (Populations and communities, outcomes C4.1.1 to C4.1.18) from the IB Biology guide, first assessment 2025. The content is the same for Standard Level and Higher Level, as C4.1 has no additional higher level content. For full explanations read the study guide, then try the practice set. Hub: IB DP Biology. Tick off outcomes on the checklist, and find gaps with the free diagnostics.

Key definitions

Term	Meaning
Population	Individuals of one species in an area, normally interbreeding
Reproductive isolation	Little or no interbreeding; separates populations
Sampling error	Difference between the estimate and the true population size
Sessile	Fixed in place (plants, barnacles, mussels)
Carrying capacity	Largest population the environment can support indefinitely
Density-dependent factor	Effect grows stronger as density rises
Density-independent factor	Effect is the same whatever the density, such as a fire or drought
Community	All populations in an ecosystem, including plants, animals, fungi and bacteria
Endemic species	Native to the area
Invasive species	Introduced species that spreads and harms endemic species

Sampling methods (C4.1.2 to C4.1.4)

Why estimate: counting everything is too slow, too costly or impossible. Why random: avoids bias in site choice, using random coordinates.

Method	Use for	Calculation
Random quadrats	Sessile organisms	(mean per quadrat ÷ quadrat area) × total area
Capture, mark, release, recapture	Motile animals	Lincoln index: M × N ÷ R

Meaning of letters: M = number marked first time; N = total number caught second time; R = marked individuals in the second sample.

Method in steps (quadrats):

1. Choose random coordinates with tapes along two edges.
2. Place each quadrat, count the individuals.
3. Find the mean, then the standard deviation of the mean number per quadrat.
4. Scale up to the whole area.

Standard deviation: small means the individuals are fairly evenly spread; large means patchy or clumped.

Assumptions for the Lincoln index:

• Full mixing of marked and unmarked animals.
• No births, deaths, immigration or emigration between samples.
• Marks do not harm the animal or alter its chance of recapture, and do not wear off.
• Marked animals are not caught more or less easily than others.

Worked reminder: 25 snails marked; second sample 30 with 6 marked. Estimate = 25 × 30 ÷ 6 = 125.

Effect of a broken assumption: if marks wear off, R falls and the estimate is too high. If marked animals are harder to catch (“trap-shy”), R is again too low and the estimate too high. If marked animals die or leave, the estimate is also too high.

Sampling error: a different set of random quadrats gives a slightly different estimate. More quadrats narrow the gap but never remove it.

Regulation of population size (C4.1.5 to C4.1.6)

• Limiting resources: food, water, light, space, nesting sites, mineral nutrients.
• Density-independent factors make numbers fluctuate.
• Density-dependent factors give negative feedback that pushes numbers back towards carrying capacity: competition for resources, predation (easier to find prey when dense), and spread of pathogens and pests.

Growth curves (C4.1.7 to C4.1.8)

Phase	What happens	Why
Exponential	Fast, accelerating increase	Resources plentiful, births exceed deaths, more parents each generation
Transition	Growth slows	Competition and density-dependent factors build
Plateau	Level near carrying capacity	Births roughly equal deaths

• Test for exponential growth: plot population on a logarithmic vertical axis against time on a normal horizontal axis. A straight line means a constant percentage increase.
• No lag phase is expected in the guide’s sigmoid model.
• It is an idealised model. Real populations overshoot, oscillate or crash.
• Worked reminder: counts of 3, 9, 27, 81 per day are tripling each day. Their log₁₀ values (0.48, 0.95, 1.43, 1.91) rise by 0.48 each day, so the log graph is straight.
• Practical organisms: yeast or duckweed.
• Case study: reindeer on St Matthew Island, Alaska, grew from about 29 (1944) to roughly 6000 (1963), then crashed after overgrazing of lichen: overshoot shows the model is a simplification.

Must-know distinctions

• Population (one species) versus community (all species in the ecosystem).
• Quadrats (sessile organisms) versus capture, mark, release, recapture (motile animals).
• Density-dependent (negative feedback, regulates) versus density-independent (disturbs numbers).
• Intraspecific (within one species) versus interspecific (between species).
• Parasite (lives on or in a host and feeds from it) versus pathogen (causes disease).
• Experiment (manipulates a variable, can show cause) versus observation (records existing patterns, shows association).
• Top-down (predators limit) versus bottom-up (resources limit).
• Endemic (native) versus invasive (introduced and spreading).

Intraspecific relationships (C4.1.9)

• Competition: same needs, so competition for food, mates, territory, light, nesting sites. Example: red deer stags competing for mates.
• Cooperation: gains for group members. Examples: wolf packs hunting, meerkat sentries.

Interspecific relationships (C4.1.10 to C4.1.13)

Relationship	Effect	Example
Herbivory	Animal eats plants	Caterpillar on leaves
Predation	Predator kills prey	Lynx and hare
Interspecific competition	Both lose through shared resource	Grey and red squirrels
Mutualism	Both gain	Coral and zooxanthellae
Parasitism	Parasite gains, host harmed, lives on or in host	Tapeworm
Pathogenicity	Microbe or virus causes disease	Plasmodium in malaria

Mutualism, both sides:

• Legume root nodules: Rhizobium fixes nitrogen for the plant; the plant supplies carbohydrate and a low-oxygen home.
• Orchid mycorrhizae: fungus supplies carbon and minerals to the seedling; fungus gains a habitat and, in many species, carbon later.
• Coral zooxanthellae: algae supply sugars; coral supplies shelter, carbon dioxide and nitrogenous waste.

Invasive example: the grey squirrel digests acorns more efficiently than the red squirrel and also carries squirrelpox. A local example is equally valid.

Testing for competition (C4.1.14 to C4.1.15)

Competition is indicated but not proven when one species does better without the other. Three approaches: laboratory experiment, field observation by random sampling, and field manipulation by removing one species. Experiments manipulate a variable; observations record what exists.

Chi-squared for association, steps:

1. Null hypothesis: the two species are distributed independently.
2. Tabulate presence and absence at sites (2 × 2).
3. Expected = row total × column total ÷ grand total.
4. χ² = Σ (O − E)² ÷ E.
5. Degrees of freedom = (rows − 1)(columns − 1) = 1.
6. If χ² is more than the critical value (3.84 at 5%), reject the null.
7. If both present is below expected, the association is negative.

Worked reminder: 100 sites, A at 40, B at 30. Expected both present = 40 × 30 ÷ 100 = 12.

Predator and prey, control, chemicals (C4.1.16 to C4.1.18)

• Predator-prey: delayed negative feedback. Case study: snowshoe hare and Canada lynx cycles of roughly ten years, lynx peaks lagging hare peaks. Food supply for hares matters too.
• Top-down: predators limit herbivores. Bottom-up: nutrients limit producers and everything above. Both occur; one usually dominates.
• Allelopathy: plant releases chemicals deterring neighbours, such as juglone from black walnut.
• Antibiotic secretion: microbes kill rivals, such as penicillin from Penicillium.

Quick self-test

1. State why reproductive isolation matters for the definition of a population.
2. 36 insects marked; 30 caught later, 9 marked. Estimate the population.
3. Four random 0.5 m² quadrats hold 14, 10, 12 and 16 plants. Estimate the number in 200 m².
4. A survey has 100 sites. Species A is at 40 and B at 30. Find the expected number with both.
5. State the degrees of freedom for a 2 × 2 chi-squared table.
6. What does a straight line on a logarithmic vertical axis show?
7. Give one density-dependent and one density-independent factor.
8. State one benefit to each partner in coral and zooxanthellae.
9. Give one similarity between allelopathy and antibiotic secretion.
10. Why does a negative association not prove competition?
11. A population grows from 80 to 120. Find the percentage increase.
12. Contrast top-down and bottom-up control.

Answers

1. It separates one population from another; members of one population normally breed with each other.
2. 36 × 30 ÷ 9 = 120.
3. Mean = 13; density = 26 per m²; 26 × 200 = 5200.
4. 40 × 30 ÷ 100 = 12.
5. 1.
6. Exponential growth: constant percentage increase per time interval.
7. Disease spreading in crowded hosts (dependent); fire or drought (independent).
8. Coral gains sugars; zooxanthellae gain shelter and carbon dioxide.
9. Both release a chemical into the environment to deter competitors.
10. Another factor, such as different habitat preferences, could explain the pattern.
11. 40 ÷ 80 × 100 = 50%.
12. Top-down: consumers limit lower trophic levels. Bottom-up: resources limit producers and everything above.

Where marks are usually lost

• Swapping M, N and R in the Lincoln index.
• Omitting units or the scaling step from quadrat density to total area.
• Listing assumptions without saying what goes wrong when one fails.
• Calling a chi-squared result proof of competition.
• Describing a plateau as “growth stops” without saying births roughly equal deaths.
• Giving “competition” as the only density-dependent factor, with no predation or disease.
• Naming a mutualism but giving a benefit to one partner only.
• Using a model curve as if it were real data.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Section C4.1, Populations and communities.