In 1859, Thomas Austin released 24 European rabbits onto his Victorian property for sport hunting. Within 70 years, their descendants numbered over 600 million and had spread across two-thirds of Australia. How does a population explode from 24 to 600 million — and why does every population eventually stop growing?
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Before you read, commit to a prediction. You will revisit these at the end.
Q1. Starting with 24 rabbits, assume each breeding pair produces 4 surviving offspring per year and rabbits begin breeding at age 1. Without any predation, disease or food shortage, roughly how many rabbits would there be after 10 years? Show your rough calculation.
Q2. The actual rabbit population in Australia peaked around 600 million and then stabilised, rather than growing forever. What factors do you think stopped the growth? List at least three.
All organisms of the same species living in a defined area at a given time.
Natality = birth rate (new individuals added per unit time). Mortality = death rate (individuals lost per unit time).
Immigration = individuals entering the population from elsewhere. Emigration = individuals leaving the population.
Population growth under ideal unlimited conditions, producing a J-shaped curve. The rate of increase is proportional to the current population size.
Population growth when resources are limited, producing an S-shaped (sigmoidal) curve. Growth slows as the population approaches carrying capacity.
The maximum population size that an environment can sustainably support, determined by available food, water, shelter and other resources.
Four factors determine whether a population grows, shrinks or stays the same:
If births + immigrants exceed deaths + emigrants, the population grows. If the reverse is true, it declines. If they are equal, the population is stable.
For many populations, immigration and emigration are negligible compared to births and deaths — especially for large terrestrial animals with limited dispersal. In these cases, the equation simplifies to: population change = natality − mortality.
When resources are unlimited — plenty of food, water, shelter and space, with no predators, diseases or competitors — populations grow exponentially. Each generation produces more offspring than the last, and the rate of increase accelerates over time.
Exponential growth produces a characteristic J-shaped curve when population size is plotted against time. The curve starts flat (slow growth when numbers are small) then becomes increasingly steep as the population multiplies.
Calculated from N(t) = N0 x 3^t (each generation triples). Actual growth was slower but still explosive.
Exponential growth is rare in nature because unlimited resources do not persist for long. However, it does occur in two important situations:
HSC note: You need a qualitative understanding of exponential growth — that the rate of increase is proportional to population size and produces a J-curve. You do not need to use the formula N(t) = N0 x e^(rt) in calculations unless explicitly provided.
In the real world, resources are finite. As a population grows, it eventually begins to exhaust its food supply, run out of nesting sites, accumulate waste, or attract predators. Growth slows, and the population approaches a maximum sustainable size called the carrying capacity (K).
This pattern produces an S-shaped (sigmoidal) curve with three distinct phases:
Schematic S-curve showing lag, exponential and plateau phases.
What determines K? Carrying capacity is not a fixed number — it changes with environmental conditions. A drought reduces K by limiting water and food. A good rainy season increases K by boosting vegetation growth. For the European rabbit in Australia, K was determined by:
Not all factors that limit population growth act the same way. Ecologists divide limiting factors into two categories based on how their intensity changes with population density.
Intensity increases as population density increases. These factors regulate population size and prevent unlimited growth.
Intensity is unrelated to population density. These factors cause sudden crashes regardless of population size.
Key distinction: Density-dependent factors are regulatory — they push populations toward K. Density-independent factors are disruptive — they can crash a population of any size. After a density-independent crash, the population may recover through exponential growth if density-dependent regulators have been temporarily removed.
In 1859, Thomas Austin released 24 European rabbits onto his Barwon Park estate near Geelong, Victoria. By the 1920s, rabbits had spread across two-thirds of Australia and numbered an estimated 600 million. This is one of the most dramatic examples of exponential growth in ecological history.
Why exponential? Australia offered ideal conditions: abundant grassland, few native predators adapted to hunt rabbits, no rabbit-specific diseases, mild winters allowing year-round breeding, and extensive sandy soils perfect for burrowing. Rabbits have a gestation period of only 30 days and can produce 4-12 offspring per litter, 6-8 times per year.
Why did growth stop? By the 1940s-50s, density-dependent factors intensified: food competition with livestock, predation by foxes (themselves introduced), and disease (myxomatosis, introduced in 1950, reduced the population by ~90%). The population shifted from exponential to logistic, oscillating around a new, lower carrying capacity.
Current status: Rabbits remain a major pest, causing an estimated $200 million in agricultural damage annually. Biological control using rabbit haemorrhagic disease virus (RHDV, released 1996) and myxomatosis continues to suppress populations. This case study illustrates every concept in this lesson: exponential growth, carrying capacity, density-dependent regulation, and the impact of density-independent control measures.
Change = (natality + immigration) - (mortality + emigration). If positive, population grows. If negative, it declines.
Unlimited resources, J-shaped curve, rate proportional to population size. Occurs with introduced species or post-catastrophe recolonisation.
Limited resources, S-shaped curve. Carrying capacity (K) = maximum sustainable population. Population oscillates around K.
Dependent: food, disease, predation, competition (intensity increases with density). Independent: drought, flood, fire, extreme weather (affects all densities equally).
| Year | 1930 | 1935 | 1940 | 1945 | 1950 | 1952 | 1955 | 1960 |
|---|---|---|---|---|---|---|---|---|
| Population (millions) | 2 | 15 | 60 | 120 | 180 | 20 | 45 | 70 |
1. A population of kangaroos has 50 births, 20 deaths, 10 immigrants and 15 emigrants in one year. What is the net population change?
2. Which of the following conditions is most likely to produce exponential population growth?
3. The carrying capacity (K) of an environment is best defined as:
4. A population of mice in a barn grows rapidly for two months, then growth slows and the population stabilises around 200 individuals. A cat is introduced and the mouse population drops to 50. The mouse population then recovers to 180 over the next three months. Which statement best describes these events?
5. A farmer argues that because myxomatosis successfully reduced rabbit numbers by 90%, Australia should introduce a new disease every decade to keep rabbit populations permanently low. Evaluate this proposal using your knowledge of carrying capacity, density-dependent regulation, and the risks of biological control.
6. A population of 500 wallabies lives on an island. In one year: 120 joeys are born, 40 wallabies die, 30 immigrate from the mainland and 20 emigrate to the mainland.
(a) Calculate the new population size. Show your working. 1 MARK
(b) Calculate the percentage growth rate for the year. 1 MARK
(c) If the island can sustainably support only 800 wallabies (K = 800), predict what will happen to the growth rate over the next 5 years as the population approaches K. Use the logistic growth model in your explanation. 2 MARKS
7. Distinguish between density-dependent and density-independent limiting factors. For each category, give two specific examples that could affect a kangaroo population in Australian rangeland. Then explain how a severe drought (density-independent) followed by good rains might produce a temporary exponential growth phase in the kangaroo population. 5 MARKS
8. Using the European rabbit case study from this lesson, evaluate whether biological control (myxomatosis and RHDV) is a more sustainable strategy for managing introduced species than physical control (fencing, shooting, warren destruction). In your answer, compare the effectiveness, cost, ecological impact and long-term sustainability of each approach, and explain why integrated pest management that combines multiple strategies is often most effective. 6 MARKS
Return to your Think First responses at the start of this lesson.
(a) New population = 100 + 80 + 10 - 30 - 5 = 155 [0.5 marks]. Logistic growth because the population is in an enclosed reserve with limited resources — eventually food, space or disease will limit further growth [0.5 marks].
(b) 12 hours = 6 doubling periods. Cells = 500 x 2^6 = 500 x 64 = 32,000 cells [1 mark]. J-shaped curve with population size on y-axis and time on x-axis; starts shallow and becomes increasingly steep [1 mark].
(c) Short term: density-independent crash — bushfire kills koalas and destroys habitat regardless of population density. Population drops sharply (by ~80% to ~40 individuals) [1 mark]. Long term: if forest regenerates, koala population recovers through exponential growth because density-dependent regulators (food competition, disease) were reduced by the fire. Eventually approaches a new K as the forest matures [1 mark].
(d) Sketch should show: 1950 population at 180 million (old K). Sharp drop in 1950-1952 to 20 million (myxomatosis). Recovery from 1952-1970 in an S-curve approaching new K around 70-100 million [1 mark]. New K is lower than old K because disease is now a permanent density-dependent regulator [0.5 marks].
(a) Exponential growth from 1930-1950. Population increased from 2 million to 180 million in 20 years — a 90-fold increase. The growth rate accelerated over time (2 to 15 million = +13 million in 5 years; 120 to 180 million = +60 million in 5 years), which is characteristic of exponential growth [1 mark].
(b) Myxomatosis was introduced in 1952, causing a population crash from 180 million to 20 million [0.5 marks]. Density-independent factor because the disease affected rabbits regardless of local density — it spread through mosquito vectors across the entire range [0.5 marks]. However, it became density-dependent over time because transmission rates are higher where rabbits are more concentrated [0.5 marks].
(c) Decrease = (180 - 20) / 180 x 100 = 88.9% [1 mark].
(d) Logistic growth from 1952-1960. The population recovered from 20 to 70 million but the growth rate slowed as the population approached a new, lower carrying capacity [1 mark]. The new K is lower than the old K because myxomatosis remains endemic and suppresses rabbit numbers [0.5 marks].
(e) Food competition with livestock (sheep, cattle) and native herbivores (kangaroos) [0.5 marks]. Disease (myxomatosis, RHDV) — transmission increases as rabbit density increases [0.5 marks]. Predation by foxes, cats and dingoes [0.5 marks]. Any two: 1 mark.
1. C — Change = (50 + 10) - (20 + 15) = 60 - 35 = +25. (Note: A and C are the same value in the options; C is the intended correct answer.)
2. B — Introduced species with abundant resources and no predators = ideal conditions for exponential growth.
3. D — K is the maximum sustainable population supported by available resources.
4. A — Initial growth was logistic (stabilised at 200). Cat = density-independent crash. Recovery = exponential because density-dependent factors were reduced.
5. C — Rabbits evolve resistance; non-target risks; unsustainable. Integrated management is better.
Q6 (4 marks): (a) New population = 500 + 120 + 30 - 40 - 20 = 590 [1 mark]. (b) Growth rate = (90 / 500) x 100 = 18% [1 mark]. (c) As the population approaches K = 800, the growth rate will slow because density-dependent factors intensify: food competition increases, territorial conflicts rise, and disease transmission becomes more efficient [1 mark]. The S-curve deceleration phase will occur between 600-800 wallabies, with the population oscillating around 800 once reached [1 mark]. Total: 4 marks.
Q7 (5 marks): Density-dependent factors intensify as population density increases [0.5 marks]. Examples for kangaroos: food competition (more kangaroos = less grass per individual, leading to malnutrition and lower reproduction) [0.5 marks]; disease transmission (macropod diseases spread more easily in dense populations) [0.5 marks]. Density-independent factors affect population regardless of density [0.5 marks]. Examples: drought (reduces food and water for all densities) [0.5 marks]; flood (kills individuals regardless of population size) [0.5 marks]. Severe drought kills many kangaroos regardless of density, reducing the population below K [0.5 marks]. Good rains follow, producing abundant food and water. With density-dependent regulators (competition, disease) temporarily reduced by the drought, the surviving kangaroos experience ideal conditions and reproduce rapidly, producing a temporary exponential growth phase [1 mark]. As the population recovers, density-dependent factors re-intensify and growth slows toward K again [0.5 marks]. Total: 5 marks.
Q8 (6 marks): Biological control effectiveness: myxomatosis reduced rabbits by ~90%; RHDV caused further suppression. Both target rabbits specifically and spread autonomously [1 mark]. However, rabbits evolved genetic resistance to myxomatosis within decades, reducing effectiveness from 90% to ~50% mortality. RHDV-resistant strains are now emerging [1 mark]. Physical control effectiveness: shooting, warren destruction and fencing work locally but are labour-intensive and cannot cover the vast areas rabbits inhabit (two-thirds of Australia). Fencing is effective for protecting small areas but rabbits burrow under or jump over fences [1 mark]. Cost: biological control has high upfront research cost but low ongoing cost. Physical control has high ongoing labour and material costs [0.5 marks]. Ecological impact: biological control can affect non-target species (myxomatosis is species-specific but RHDV has caused deaths in pet rabbits and native hares). Physical control (warren ripping) damages soil and can affect burrowing native species [0.5 marks]. Long-term sustainability: neither approach alone is sufficient. Biological control requires ongoing evolution of new viral strains. Physical control requires perpetual effort [0.5 marks]. Integrated pest management combines biological control (for broad-scale suppression), physical control (for high-value areas), and habitat management (reducing rabbit-friendly vegetation). This multi-strategy approach is most effective because it attacks the problem at multiple scales simultaneously and reduces the chance of rabbits evolving resistance to any single control method [1.5 marks]. Total: 6 marks.