Carrying Capacity, Limiting Factors and Growth Curves
Red squirrels were once found in woodlands all over Ireland. Today, in many counties, greys have taken over. Why can a woodland support only a certain number of squirrels, and why does one species sometimes lose out to another? In this section you will learn how the environment sets a ceiling on a population, and how to describe the way populations grow towards it.
Carrying capacity and limiting factors
A population cannot grow for ever. Every habitat has a limited supply of food, water, space and shelter, so there is a maximum number of individuals it can support year after year.
Anything that holds a population back is a limiting factor. Limiting factors are sorted into two groups, depending on whether their effect changes with how crowded the population is.
| Density-dependent | Density-independent | |
|---|---|---|
| Effect | Gets stronger as population density rises | Acts the same whatever the density |
| Type | Mostly biotic | Mostly abiotic |
| Examples | Food supply, competition, predation, disease, parasites | Severe frost, flooding, storms, fire, drought |
Abiotic factors such as water, oxygen, space, temperature and soil can also limit a population, for example by setting how much food a habitat can grow. Take red deer in Killarney National Park. In a crowded herd, there is less grazing per animal and disease spreads more easily, so food and disease are density-dependent. A harsh winter storm kills deer whether the herd is large or small, so it is density-independent.
Define the term carrying capacity* of an environment. Distinguish between a density-dependent and a density-independent limiting factor, giving one example of each for a…
Define the term carrying capacity of an environment. Distinguish between a density-dependent and a density-independent limiting factor, giving one example of each for a population of red deer in Killarney National Park.
J-curves and S-curves
If a few individuals arrive in a habitat with plenty of resources, they breed freely and the population grows by a constant percentage each time period. This is exponential growth, and it gives a J-shaped curve.
When resources begin to run out, growth slows and the population levels off around K. This gives an S-shaped (sigmoid) curve.
The S-curve has three phases:
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Lag phase: few individuals, slow start while the population adjusts to its new habitat.
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Exponential phase: rapid increase, resources plentiful.
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Stationary phase: births equal deaths, and the population fluctuates around K.
A pure J-curve cannot last. It is typical of a short burst, such as bacteria in fresh broth or an invasive species arriving in a new habitat. Over a long period, natural populations show S-shaped growth, because limiting factors always catch up.
A population of bacteria grows in a flask of nutrient broth. Name the shape of the growth curve (S-shaped or J-shaped) shown by a population that grows exponentially…
A population of bacteria grows in a flask of nutrient broth. Name the shape of the growth curve (S-shaped or J-shaped) shown by a population that grows exponentially with no limits, and name the shape shown by a population that levels off. State which of the two curves is more commonly seen in natural populations over a long period.
The number of rabbits on an island off the Cork coast is recorded each year. | Year | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | Rabbits | 20 | 40 | 80 | 160 | 320…
The number of rabbits on an island off the Cork coast is recorded each year.
| Year | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Rabbits | 20 | 40 | 80 | 160 | 320 |
- a) Name the type of growth shown.
- b) Calculate the percentage increase in population from year 1 to year 2.
- c) State two factors that would eventually limit this growth.
Zebra mussels, an invasive species, colonise a stretch of the River Shannon. In the first year there are 1,000 mussels per square metre, and the population increases by…
Zebra mussels, an invasive species, colonise a stretch of the River Shannon. In the first year there are 1,000 mussels per square metre, and the population increases by 60% each year when resources are plentiful.
- a) Calculate the density after 3 years of unrestricted growth.
- b) Explain why this J-shaped growth cannot continue indefinitely.
- c) Explain what is meant by a density-dependent factor and give two examples relevant to this population.
Working with population data
Exam questions give you a table or graph and ask for a handful of standard things.
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Carrying capacity: the value at which the population levels off, so read the plateau.
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Average rate of increase: $\frac{\text{change in population}}{\text{change in time}}$.
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Phases: find where the slope changes.
A farmer's field shows the same idea. Total gain is the number of sheep multiplied by the gain per sheep. With 10 sheep: $10 \times 8 = 80$ kg. With 30: $30 \times 5 = 150$ kg. With 60: $60 \times 1 = 60$ kg. Total gain rises, then falls: beyond a certain stocking level, competition for grass reduces what each animal gets. At 60 sheep each animal gains only 1 kg, and at 80 some lose weight, so the field is at or beyond its carrying capacity. The best stocking level is near the peak (around 30 sheep), where the total is highest and each animal still does well. Carrying capacity can be raised by reseeding, adding fertiliser, or providing supplementary feed.
A population of 50 grey squirrels is introduced to a woodland in County Wicklow. The population size is recorded every two years: | Year | 0 | 2 | 4 | 6 | 8 | 10 | 12 |…
A population of 50 grey squirrels is introduced to a woodland in County Wicklow. The population size is recorded every two years:
| Year | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Squirrels | 50 | 90 | 210 | 420 | 560 | 600 | 602 |
- a) Describe the shape of the curve and identify the phases of growth (lag, exponential, stationary).
- b) Estimate the carrying capacity of the woodland.
- c) Calculate the average rate of increase in squirrels per year between year 4 and year 6.
- d) Suggest two limiting factors responsible for the levelling off.
A farmer in County Meath stocks a field with sheep. When 10 sheep are kept, each gains 8 kg per season. When 30 sheep are kept, each gains 5 kg per season. When 60 sheep…
A farmer in County Meath stocks a field with sheep. When 10 sheep are kept, each gains 8 kg per season. When 30 sheep are kept, each gains 5 kg per season. When 60 sheep are kept, each gains 1 kg per season, and when 80 sheep are kept some lose weight.
- a) Calculate the total weight gain of the flock at 10, 30 and 60 sheep.
- b) Explain these results in terms of carrying capacity and competition.
- c) Suggest, with a reason, a sensible stocking level for the farmer and how the carrying capacity might be raised.
The logistic equation
The S-curve can be described mathematically by the logistic equation:
$$\frac{dN}{dt} = rN\left(\frac{K-N}{K}\right)$$-
$N$ is the population size, $K$ the carrying capacity.
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$r$ is the intrinsic rate of increase, the growth rate per individual when resources are unlimited.
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$\frac{K-N}{K}$ is the fraction of K still available. It is close to 1 when $N$ is small (almost no resistance) and close to 0 when $N$ nears K. When $N = K$, the bracket is 0, so $\frac{dN}{dt} = 0$ and growth stops.
Growth is fastest at $N = K/2$ (here 2,500), where the product $N(K-N)$ is largest: there are plenty of breeders and plenty of room. This is the idea behind a sustainable yield: harvest no more than the population's growth per year, ideally near $K/2$, so the stock replaces itself. Harvesting at or near K gives little surplus, and overfishing far below $K/2$ means few breeders to replace what is taken.
The logistic growth of a population is described by $\frac{dN}{dt}=rN\left(\frac{K-N}{K}\right)$, where $N$ is population size, $r$ is the intrinsic rate of increase and…
The logistic growth of a population is described by $\frac{dN}{dt}=rN\left(\frac{K-N}{K}\right)$, where $N$ is population size, $r$ is the intrinsic rate of increase and $K$ is the carrying capacity. A population of fish in a lake has $r = 0.4$ per year and $K = 5,000$.
- a) Calculate $\frac{dN}{dt}$ when $N = 500$, $N = 2,500$ and $N = 4,900$.
- b) State at which population size growth is fastest and explain why.
- c) Explain the biological meaning of $\frac{K-N}{K}$ and what happens to growth when $N = K$.
- d) A fishing quota is to be set to give a sustainable yield. Using your results, recommend a population size to maintain and justify your answer.
Predators, competitors and moving targets
Real populations rarely sit neatly at a fixed K. Carrying capacity changes with the seasons, the food supply and other species.
Predator–prey cycles. This is a classic textbook example (an Irish equivalent would be foxes and rabbits on grassland or hedgerow). In a boreal forest, snowshoe hare numbers rise when food is plentiful. Lynx, which eat hares, then increase about two years later, because they need time to find food and reproduce. More lynx reduce hares, and then the lynx starve and fall in turn. Predation is density-dependent: the more hares there are, the easier they are to catch, and the more predators the food supports. The result is an oscillating curve, not a smooth S-curve.
Hence K for the hares is not fixed: it depends on vegetation quality, which varies with weather, and on how many lynx are present.
Competition and disease. Red and grey squirrels compete for the same food, but greys use it more efficiently and can feed on unripe seeds, so greys win. Greys also carry the squirrelpox virus, to which they are immune, but which is usually fatal to reds. A disease therefore acts differently on the two species: for the greys it is almost harmless, for the reds it is a major, density-dependent mortality factor. The effective carrying capacity for reds falls, and the population declines.
Management measures include:
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Grey squirrel control (trapping or culling in red squirrel areas), which removes the competitor and the source of virus.
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Habitat protection and planting conifers or mixed woodland where reds do better than greys, which raises the carrying capacity for reds.
A population of lynx and a population of snowshoe hare (a predator and its prey) are monitored in a boreal forest. The hare population peaks, and the lynx population…
A population of lynx and a population of snowshoe hare (a predator and its prey) are monitored in a boreal forest. The hare population peaks, and the lynx population peaks about two years later, followed by a crash in hares and then in lynx.
- a) Explain how predation acts as a density-dependent limiting factor for the hares.
- b) Explain the time lag between the two peaks.
- c) Discuss whether the carrying capacity of the hares is a fixed number, with reference to food supply and predation.
- d) Describe how this pattern differs from a simple S-shaped curve.
Red squirrels in Ireland are declining where grey squirrels have been introduced. Both species compete for food and the grey squirrel also carries the squirrelpox virus,…
Red squirrels in Ireland are declining where grey squirrels have been introduced. Both species compete for food and the grey squirrel also carries the squirrelpox virus, to which it is immune.
- a) Explain, using the concept of carrying capacity, why the red squirrel population falls in areas shared with greys.
- b) Identify two limiting factors affecting the red squirrel and classify each as density-dependent or density-independent.
- c) Explain why a disease may act differently on the two species.
- d) Propose two management measures to protect red squirrels and justify them ecologically.
Back to the start: grey squirrels can use food better and carry a virus that kills reds, which lowers the carrying capacity for red squirrels where the two species share a wood.
Summary
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Carrying capacity (K) is the maximum population an environment can support indefinitely.
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Density-dependent factors (food, competition, predation, disease) act more strongly as density rises. Density-independent factors (frost, flood, fire) do not depend on density.
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Unlimited growth gives a J-curve. Limited resources give an S-curve with lag, exponential and stationary phases, and this is the usual pattern over a long period.
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Read K from the plateau, and find a rate using change in population ÷ change in time.
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The logistic equation $\frac{dN}{dt} = rN\left(\frac{K-N}{K}\right)$ gives zero growth at $N = K$ and fastest growth at $K/2$. This is the basis of sustainable yield.
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Carrying capacity is not fixed. It shifts with food supply, predators, competitors and disease.