AP Biology · Lesson 2

Population growth and community interactions

Know Know it

With unlimited resources, a population grows exponentially: dN/dt = rmax x N, where N is the population size and rmax is the maximum per capita growth rate. The number added each year rises as the population grows, giving a J-shaped curve.

Real resources run out. The carrying capacity, K, is the largest population the environment can support over time. In logistic growth, dN/dt = rmax x N x (K - N) / K. When N is small, (K - N) / K is close to 1 and growth looks exponential. As N nears K, the factor nears 0 and growth slows to a stop, giving an S-shaped curve. Growth is fastest when N is half of K.

Density-dependent factors hit crowded populations harder: competition for food and space, disease, predation and the buildup of wastes. They lower birth rates or raise death rates as density rises, which holds a population near K. Density-independent factors, such as frost, floods and fire, kill the same share whatever the density.

Species in a community interact. In competition (-/-) both lose; by competitive exclusion, two species cannot share the very same limited niche for long, so one dies out locally or they split resources (resource partitioning). In predation, herbivory and parasitism (+/-) one gains and the other loses. In mutualism (+/+) both gain, and in commensalism (+/0) one gains and the other is unaffected. Predator and prey numbers often rise and fall in cycles, with the predator peak following the prey peak.

A change in one population can ripple through a food web. A keystone species has an effect on its community far larger than its numbers suggest. Removing a top predator can let its prey multiply and overeat the producers, a trophic cascade.

Trap Use (K - N) / K, not N / K

In the logistic equation the brake is (K - N) / K, the share of capacity still unused. With K = 1,000 and N = 800, it is 200 / 1,000 = 0.2, not 800 / 1,000 = 0.8. The closer N is to K, the smaller the factor and the slower the growth.

Worked example Easy

A fictional population of 800 rabbits on a large, empty island grows with rmax = 0.2 per year and no limits on resources. Using the exponential model, at this size about how many rabbits per year is the population adding?

  1. 960
  2. 4,000
  3. 160Answer
  4. 80

How to solve it

  1. With no limits, use exponential growth: dN/dt = rmax x N.
  2. dN/dt = 0.2 x 800 = 160.
  3. At this size the population is adding about 160 rabbits per year.

Why each choice is right or wrong

  • A. Incorrect. 960 is 800 + 160, a population total, not the number of rabbits being added per year.
  • B. Incorrect. 4,000 comes from dividing 800 by 0.2; the equation multiplies rmax by N instead.
  • C. Correct. dN/dt = rmax x N = 0.2 x 800 = 160 rabbits added in a year.
  • D. Incorrect. 80 uses a rate of 0.1 per year, half of the given rmax of 0.2.

Explain Explain it

Task verbs: calculate, predict

For a growth problem, first decide whether the population is limited. If a carrying capacity is given, use the logistic equation and compute (K - N) / K before you multiply. Show each number.

To predict how a change spreads through a community, follow each feeding link one step at a time and say why each population rises or falls.

Worked example Medium

A fictional deer population of 400 lives where the carrying capacity is 1,000 deer and rmax = 0.5 per year. Using the logistic model, how many deer does the population add per year at this size?

  1. 80
  2. 200
  3. 300
  4. 120Answer

How to solve it

  1. (K - N) / K = (1,000 - 400) / 1,000 = 0.6.
  2. rmax x N = 0.5 x 400 = 200.
  3. dN/dt = 200 x 0.6 = 120 deer per year.

Why each choice is right or wrong

  • A. Incorrect. 80 uses N / K = 400 / 1,000 = 0.4 as the factor instead of (K - N) / K = 0.6.
  • B. Incorrect. 200 is rmax x N with no carrying capacity, the exponential answer, which ignores K.
  • C. Incorrect. 300 is rmax x (K - N) = 0.5 x 600, which leaves out the population size and the division by K.
  • D. Correct. 0.5 x 400 x (1,000 - 400) / 1,000 = 200 x 0.6 = 120 deer per year.

Predict Your turn

In a fictional grassland, foxes eat rabbits and rabbits eat grass. Hunters remove most of the foxes. Predict how the rabbit population and then the amount of grass will change over the next few years, and explain each prediction using the feeding links.

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Compare with a model answer

A full answer does this

  • Predicts that the rabbit population increases
  • Explains that with fewer foxes, fewer rabbits are eaten, so more survive and reproduce
  • Predicts that the amount of grass decreases because the larger rabbit population eats more of it

Model answer

The rabbit population will increase. With most of the foxes gone, far fewer rabbits are killed, so more of them survive and reproduce. The amount of grass will then decrease, because the larger rabbit population eats more grass than before.

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