DP Biology HL C4.1 Populations and Communities

DP Biology HL C4.1 Populations and Communities - Full notes

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DP Biology HL | C4.1 Populations and Communities

DP BIOLOGY · THEME C: INTERACTION AND INTERDEPENDENCE

C4.1

Populations and Communities

Full notes for HL · IB-style

Contents

1	Populations and how to estimate their size
2	Quadrat sampling for sessile organisms
3	Capture-mark-release-recapture for motile organisms
4	Population growth and carrying capacity
5	Communities and interactions between species
6	Competition, predation and herbivory
7	Mutualism, parasitism and keystone species
8	Populations and communities defined
9	Estimating plant populations with quadrats
10	Capture–mark–release–recapture
11	Population growth curves
12	Carrying capacity and limiting factors
13	Modelling population growth
14	Intraspecific and interspecific competition
15	Predator–prey relationships
16	Herbivory, parasitism and mutualism
17	Allelopathy and antibiotic secretion
18	Keystone species and community structure
19	Chi-squared test for association
20	Mixed review: worked problems
21	Additional exam-style practice
22	Human population growth
23	Common misconceptions and links
24	Exam-style question with markscheme
25	Check your understanding
26	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define population and community
estimate population size using random quadrat sampling and calculate standard deviation and confidence
estimate population size using capture-mark-release-recapture and the Lincoln index
explain the sigmoid population growth curve and carrying capacity
explain density-dependent and density-independent factors
describe types of interspecific relationships: competition, predation, herbivory, mutualism, parasitism
explain predator-prey cycles and the role of keystone species
test for association between species using the chi-squared test

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do interactions between organisms affect the size of populations? How do species in a community depend on each other?

How this topic is examined

C4.1 is heavily examined through calculations: population estimates from quadrats and the Lincoln index, and chi-squared tests for association. Graph questions include sigmoid curves and predator-prey cycles.

Show every step of calculations, state assumptions and limitations of sampling methods, and use 'correlation' carefully.

Command term	What the examiner wants
State / Identify	a brief answer with no explanation needed
Outline	a brief account or summary
Describe	a detailed account, including trends or features, without reasons
Explain	a detailed account including causes, reasons or mechanisms
Distinguish / Compare and contrast	give the differences (and similarities), point by point
Calculate	a numerical answer with working and units
Deduce / Suggest	reach a conclusion from data; propose a reasonable explanation
Evaluate / Discuss	weigh strengths and limitations and reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the biology shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Populations and how to estimate their size

Count a sample, then scale up.

Think about it

How could you estimate the number of daisies on a football field, or the number of fish in a lake, without counting every one?

A population is a group of organisms of the same species living in the same area at the same time, which can interbreed. A community is all the populations of different species living and interacting in an area.

Counting every individual in a population is usually impossible, so ecologists use sampling: they count a sample and use it to estimate the total. Random sampling avoids bias. The method depends on whether the organism is sessile (does not move, such as plants or barnacles) or motile (moves, such as animals).

2 Quadrat sampling for sessile organisms

Count organisms in random squares.

A quadrat is a square frame of known area (often 1 m²). Quadrats are placed at random positions, chosen using random number tables or generators to give coordinates on a grid; this prevents the investigator unconsciously choosing typical or interesting spots. In each quadrat, the number of individuals of the species is counted. The mean number per quadrat is calculated and multiplied by the number of quadrats that would fit in the whole area.

Population estimate = mean number per quadrat × (total area ÷ area of one quadrat)

Standard deviation measures the spread of the counts around the mean. A large standard deviation means that the organisms are clumped or that the sample is too small, so the estimate is less reliable. Taking more quadrats improves reliability.

WORKED EXAMPLE 1 Quadrat estimate

Ten 1 m² quadrats in a 500 m² meadow contained 4, 7, 2, 5, 6, 3, 8, 5, 4, 6 dandelions. Estimate the population.

Solution

Mean = (4 + 7 + 2 + 5 + 6 + 3 + 8 + 5 + 4 + 6) ÷ 10 = 50 ÷ 10 = 5.0 per m²M1

Population = 5.0 × 500 = 2 500 dandelionsA1

Standard deviation is about 1.8, so counts vary moderatelyA1

Quick check

1. Why must quadrats be placed randomly?

2. What does a large standard deviation indicate?

Answers: 1. to avoid bias 2. data widely spread: clumped distribution or unreliable estimate

3 Capture-mark-release-recapture for motile organisms

Mark some, release them, and see how many turn up again.

For motile animals, the capture-mark-release-recapture method is used. A sample is captured, counted (n₁), marked harmlessly, and released. After enough time for them to mix back into the population, a second sample is captured (n₂), and the number of marked individuals in it is counted (n₃). The population size is estimated by the Lincoln index:

Population size = (n₁ × n₂) ÷ n₃

Assumptions: marked individuals mix randomly with the population; marks are not lost and do not affect survival or behaviour; there is no significant migration, birth or death between samples; and all individuals are equally likely to be captured.

WORKED EXAMPLE 2 Lincoln index

40 woodlice were captured, marked and released. A week later, 50 were captured, of which 8 were marked. Estimate the population.

Solution

Population = (40 × 50) ÷ 8M1

= 250 woodliceA1

Common mistake

Using the method on organisms that do not mix, or marking them in a way that makes them more visible to predators. Both violate the assumptions and bias the estimate.

Quick check

1. State the Lincoln index formula.

2. Give two assumptions of the method.

Answers: 1. N = n₁ × n₂ ÷ n₃ 2. random mixing; no change in population; marks not lost or harmful

4 Population growth and carrying capacity

Populations grow fast at first, then level off.

When a population colonizes a new area with plenty of resources, it shows a typical sigmoid (S-shaped) growth curve. In the exponential phase, resources are plentiful and there are few limiting factors, so the birth rate greatly exceeds the death rate and numbers rise rapidly. In the transitional phase, resources become limiting and competition increases; the birth rate falls and the death rate rises, so growth slows. In the plateau phase, birth rate equals death rate and the population stays near the carrying capacity: the maximum population size that the environment can support sustainably.

Factors that limit population growth are density-dependent if their effect increases as the population becomes more crowded: competition for food and space, predation, disease and accumulation of toxic wastes. They regulate populations by negative feedback. Density-independent factors affect the population regardless of its density: extreme weather, natural disasters and fires.

Figure 1. A sigmoid population growth curve levelling off at the carrying capacity (about 1 000).

Quick check

1. Define carrying capacity.

2. Give two density-dependent factors.

Answers: 1. maximum population size the environment can sustain 2. competition, predation, disease

5 Communities and interactions between species

No species lives alone.

Within a community, species interact in several ways. These interspecific relationships affect population sizes and distributions. They include competition, predation, herbivory, mutualism and parasitism, along with pathogenicity. Interactions within a species, such as competition for mates, are intraspecific.

Relationship	Effect on species A	Effect on species B	Example
Competition	−	−	red and grey squirrels
Predation	+ (predator)	− (prey)	lynx and snowshoe hare
Herbivory	+ (herbivore)	− (plant)	caterpillars and leaves
Mutualism	+	+	corals and zooxanthellae
Parasitism	+ (parasite)	− (host)	tapeworm in humans

6 Competition, predation and herbivory

Resources and predators regulate populations.

Interspecific competition occurs when two species need the same limited resource. It reduces the population of both, and can exclude one species from an area. The introduced grey squirrel in Britain outcompetes the native red squirrel for food (and carries a virus that kills red squirrels), causing the red squirrel's decline.

Predation links the populations of predator and prey in cycles. When prey are abundant, predators have more food and their numbers increase; more predators eat more prey, so prey numbers fall; with less food, predator numbers then fall; with fewer predators, prey recover. Predator peaks lag behind prey peaks. The classic example is the lynx and snowshoe hare in Canada, recorded from fur-trading records over about a century.

Herbivory is feeding on plants. Plants respond with physical and chemical defences (B4.2).

Figure 2. Predator-prey cycles: lynx numbers lag behind hare numbers.

Quick check

1. Why do predator peaks lag behind prey peaks?

2. Give an example of interspecific competition.

Answers: 1. predators need time to reproduce after food increases 2. grey and red squirrels

7 Mutualism, parasitism and keystone species

Some relationships help both partners; some species hold communities together.

Mutualism is a relationship in which both species benefit. Examples: root nodules in legumes contain Rhizobium bacteria that fix nitrogen for the plant and receive sugars; mycorrhizal fungi help orchids and many trees absorb minerals in exchange for sugars; zooxanthellae (photosynthetic algae) live inside coral polyps, providing sugars and receiving shelter and CO₂. Coral bleaching occurs when stressed corals expel their zooxanthellae.

Parasitism benefits the parasite at the expense of the host: tapeworms, head lice, mistletoe.

A keystone species has a disproportionately large effect on its community relative to its abundance; removing it causes major changes. Sea otters eat sea urchins; without otters, urchin populations explode and destroy kelp forests. Wolves in Yellowstone regulate elk numbers, allowing vegetation to recover along rivers.

7.1 Testing for association: the chi-squared test

To test whether two species tend to occur together (association), ecologists record presence or absence in many quadrats and use a chi-squared (χ²) test. The null hypothesis is that the two species are distributed independently. Expected frequencies are calculated as (row total × column total) ÷ grand total. Then χ² = Σ (observed − expected)² ÷ expected, with 1 degree of freedom for a 2 × 2 table. If χ² is greater than the critical value (3.84 at p = 0.05), the null hypothesis is rejected: there is a significant association.

	Species B present	Species B absent	Total
Species A present	30	10	40
Species A absent	15	45	60
Total	45	55	100

WORKED EXAMPLE 3 Chi-squared test for association

Using the table, test whether species A and B are associated.

Solution

Expected (A present, B present) = 40 × 45 ÷ 100 = 18; (A present, B absent) = 22; (A absent, B present) = 27; (A absent, B absent) = 33M1

χ² = (30−18)²/18 + (10−22)²/22 + (15−27)²/27 + (45−33)²/33 = 8.0 + 6.5 + 5.3 + 4.4 = 24.2M1 A1

24.2 > 3.84 (df = 1, p = 0.05): reject the null hypothesis; significant positive associationR1

Quick check

1. Define keystone species.

2. What is the critical χ² value for 1 degree of freedom at p = 0.05?

Answers: 1. a species with a disproportionately large effect on its community 2. 3.84

8 Populations and communities defined

Levels of ecology.

A population is a group of organisms of the same species living in the same area at the same time, able to interbreed. A community is all the populations of different species living and interacting in an area. An ecosystem includes the community and its abiotic environment.

WORKED EXAMPLE 4 Definitions (basic)

Distinguish a population from a community.

Solution

Population: one speciesR1

Community: all species in an areaR1

9 Estimating plant populations with quadrats

Random sampling.

Place quadrats at random coordinates (using random numbers on a grid), count individuals or estimate percentage cover, then calculate mean density and scale up to the whole area.

WORKED EXAMPLE 5 Estimate (standard)

Ten 1 m2 quadrats contain a total of 34 daisies. Estimate the population in a 200 m2 lawn.

Solution

Mean density 3.4 m−2M1

3.4 × 200 = 680 daisiesA1

Quick check

1. Why should many quadrats be used?

Answers: 1. to reduce the effect of chance and give a more reliable mean

10 Capture–mark–release–recapture

The Lincoln index.

For mobile animals: capture and mark a sample (M), release, recapture a second sample (N) and count the marked individuals (R). Population estimate = M × N/R. Assumptions: marks are not lost and do not affect survival; marked animals mix randomly; no births, deaths or migration between samples.

WORKED EXAMPLE 6 Lincoln index (standard)

40 snails are marked. Later, 50 are caught, of which 8 are marked. Estimate the population.

Solution

40 × 50/8M1

= 250A1

Quick check

1. State two assumptions of the capture–recapture method.

2. 60 beetles are marked; 45 of a second sample of 72 are unmarked. Estimate the population.

Answers: 1. marks not lost; random mixing; closed population 2. 27 marked: 60 × 72/27 = 160

11 Population growth curves

Exponential and logistic.

With unlimited resources, populations grow exponentially (J-curve). In reality, resources become limiting, and the curve becomes sigmoid (S-shaped): exponential phase, transitional phase, and plateau at the carrying capacity, where birth rate equals death rate.

Figure 3. Sigmoid growth reaching a carrying capacity of about 400.

WORKED EXAMPLE 7 Phases (exam level)

Explain why population growth slows in the transitional phase.

Solution

Resources (food, space) become limitingR1

Competition, predation and disease increase; death rate risesR1

12 Carrying capacity and limiting factors

Density-dependent and independent.

Carrying capacity is the maximum population an environment can sustain. Density-dependent factors (competition, predation, disease) have a greater effect at higher density. Density-independent factors (floods, fires, temperature) affect populations regardless of density.

WORKED EXAMPLE 8 Classifying (standard)

Classify disease and a hard frost as density-dependent or independent.

Solution

Disease: density-dependent (spreads faster at high density)R1

Frost: density-independentR1

13 Modelling population growth

Using growth rates.

Population change = (births + immigration) − (deaths + emigration). Percentage growth rate = (change/initial) × 100 per unit time. Exponential growth can be modelled by N = N0(1 + r)t.

WORKED EXAMPLE 9 Growth rate (standard)

A population grows from 1200 to 1350 in one year. Calculate the percentage growth rate.

Solution

150/1200 × 100M1

= 12.5% per yearA1

Quick check

1. At 12.5% per year, estimate the population after 3 more years.

Answers: 1. 1350 × 1.1253 = 1922

14 Intraspecific and interspecific competition

Within and between species.

Intraspecific competition occurs between members of the same species for the same resources; it regulates population size. Interspecific competition occurs between species with overlapping niches; it can lead to competitive exclusion or resource partitioning.

WORKED EXAMPLE 10 Red and grey squirrels (exam level)

Explain the decline of red squirrels in Britain after grey squirrels were introduced.

Solution

Interspecific competition for food (greys digest acorns better)R1

Greys carry squirrelpox, which kills redsR1

15 Predator–prey relationships

Oscillating populations.

Predator and prey populations often oscillate: as prey increase, predators increase (more food); predation reduces prey; predators then decline; prey recover. The snowshoe hare and lynx cycles in Canada are a classic example.

Figure 4. Predator–prey cycles: predator peaks lag behind prey peaks.

WORKED EXAMPLE 11 Lag (standard)

Explain why predator peaks occur after prey peaks.

Solution

Predator numbers increase only after food becomes plentifulR1

Reproduction takes timeR1

16 Herbivory, parasitism and mutualism

Other interactions.

Herbivores eat plants; parasites live on or in hosts and harm them; mutualists benefit each other (mycorrhizae, nitrogen-fixing bacteria in root nodules, zooxanthellae in corals). Pathogens are parasites that cause disease.

Quick check

1. Give an example of mutualism involving nitrogen.

Answers: 1. Rhizobium bacteria in legume root nodules

17 Allelopathy and antibiotic secretion

Chemical competition.

Some plants release chemicals that inhibit the growth of nearby plants (allelopathy), such as black walnut (juglone). Some fungi and bacteria secrete antibiotics that kill competitors, as Penicillium does.

WORKED EXAMPLE 12 Allelopathy (standard)

Define allelopathy.

Solution

Release of chemicals by a plant that inhibit other plants' growthA1

18 Keystone species and community structure

Big effects.

Keystone species have disproportionately large effects on community structure. Removing them can cause major changes: removing sea otters allowed sea urchins to destroy kelp forests.

Quick check

1. Explain why sea otters are keystone species.

Answers: 1. they control sea urchin populations, protecting kelp forests

19 Chi-squared test for association

Do two species occur together?

The χ2 test determines whether two species are associated, using presence/absence data from quadrats in a contingency table. Calculate expected frequencies, χ2 = Σ(O − E)2/E, and compare with the critical value (3.84 at p = 0.05 with 1 degree of freedom).

	B present	B absent	Total
A present	22	8	30
A absent	6	14	20
Total	28	22	50

WORKED EXAMPLE 13 Chi-squared (exam level)

Test whether species A and B are associated.

Solution

Expected: 16.8, 13.2, 11.2, 8.8M1

χ2 = 1.61 + 2.05 + 2.41 + 3.07 = 9.15A1

9.15 > 3.84: significant associationR1

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 14 Mixed problem (exam level)

A capture–recapture estimate of a fish population is 500, but some marks fell off. Explain the effect on the estimate.

Solution

Fewer marked fish recaptured (R too low)R1

Population overestimatedR1

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 15 Paper 1 question (standard)

Which is density-independent? A. disease B. competition C. predation D. drought

Solution

DA1

WORKED EXAMPLE 16 Paper 2 question (exam level)

Outline the phases of a sigmoid population growth curve.

Solution

Exponential: plentiful resourcesR1

Transitional: resources limiting; growth slowsR1

Plateau at carrying capacityR1

22 Human population growth

Exponential so far.

The human population grew exponentially over the last two centuries, reaching 8 billion in 2022, due to improved medicine, agriculture and sanitation. Growth rates are now falling as birth rates decline in many countries.

Quick check

1. Suggest two factors that have reduced human death rates.

Answers: 1. vaccination and antibiotics; improved sanitation and food supply

23 Common misconceptions and links

Students often think…	Actually…
population and community mean the same	a population is one species; a community is all species
predators keep prey numbers constant	they produce linked cycles
a significant χ² proves one species causes the other	it shows association, not cause

How C4.1 connects to the rest of the course

Topic	Connection
B4.2 Niches (HL)	competitive exclusion
C4.2 Energy and matter	food chains
A4.2 Conservation	keystone species
B4.1 Adaptation (HL)	transects and quadrats
D4.2 Sustainability	sustainable harvesting

24 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Figure 5. Lynx and hare populations.

(a) Describe the relationship between hare and lynx populations. [2]

(b) Explain the pattern. [3]

(d) State two assumptions of this method. [2]

(e) Explain the shape of a sigmoid growth curve. [3]

Markscheme

Part	Answer	Marks
(a)	both cycle with a period of about 10 years	A1
	lynx peaks lag behind hare peaks	A1
(b)	more hares: more food, lynx increase	R1
	more lynx: more predation, hares fall	R1
	fewer hares: lynx starve, fall; hares recover	R1
(c)	60 × 45 ÷ 9	M1
	= 300	A1
(d)	random mixing; no births/deaths/migration; marks not lost or harmful (any two)	A2
(e)	exponential: plentiful resources	R1
	transitional: limiting factors increase	R1
	plateau at carrying capacity: births = deaths	R1

Examiner's comments. In (b), the best answers explained both halves of the cycle and the lag.

25 Check your understanding

Section A: multiple choice

1. A community is

A. all the populations in an area B. one species C. the abiotic environment D. a food chain

2. In the Lincoln index, n₃ is

A. the first sample B. the second sample C. the number of marked individuals recaptured D. the population

3. Carrying capacity is

A. the birth rate B. the maximum population an environment can sustain C. the growth rate D. the death rate

4. Which is density-independent?

A. disease B. competition C. predation D. a flood

5. Mutualism benefits

A. one species only B. both species C. neither species D. only the host

6. Sea otters are an example of

A. a keystone species B. a parasite C. a producer D. an invasive species

Section B: structured

7. Describe how to estimate the population of a plant species using quadrats. [4]

8. Explain predator-prey cycles. [3]

9. Explain why the chi-squared test cannot show cause. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Random coordinates; known-size quadrats; count; mean; scale up to area; repeat for reliability.

8. Prey up, predators up (lag); prey down; predators down; prey recover.

9. Association may result from shared preference for an abiotic factor; correlation is not causation.

26 Summary and topic map

Idea	Remember
Population vs community	one species vs all species
Quadrats	random sampling; mean × area; standard deviation
Lincoln index	n₁ × n₂ ÷ n₃; assumptions
Sigmoid curve	exponential, transitional, plateau; carrying capacity
Limiting factors	density-dependent vs independent
Interactions	competition, predation, herbivory, mutualism, parasitism
Keystone species; χ²	disproportionate effect; association testing

Before the exam, make sure you…

•show every step of population calculations
•state sampling assumptions and limitations
•explain cycles with lag
•interpret χ² against the critical value

Populations and Communities on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 6. Topic map for Populations and Communities.

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