AP Biology · Lesson 2

Population genetics and Hardy-Weinberg

Know Know it

A population’s gene pool is all the alleles in all its members. Evolution is a change in allele frequencies in a population over generations. For a gene with two alleles, call their frequencies p and q. Every copy is one or the other, so p + q = 1.

To get an allele frequency from genotype counts, count gene copies. Each individual carries 2 copies. Frequency of A = (2 x number of AA + number of Aa) / (2 x total individuals).

The Hardy-Weinberg model says what happens when a population is not evolving. Its allele frequencies stay the same, and its genotype frequencies are p² for AA, 2pq for Aa and q² for aa, which add to 1. This holds only when five conditions are all met: a very large population, random mating, no mutation, no gene flow and no natural selection. Real populations rarely meet all five, so the model is a baseline: when observed numbers differ from it, at least one of the conditions is not being met.

When a recessive phenotype is all you can see, its share of the population is q². Take the square root to get q, then p = 1 - q.

Four processes change allele frequencies. Natural selection is not random: it raises alleles that help in the current environment. Genetic drift is change by chance, strongest in small populations; the bottleneck effect and the founder effect are drift after a population shrinks or a few individuals start a new one. Gene flow moves alleles between populations and makes them more alike. Mutation is the source of new alleles, but on its own it changes frequencies slowly. Nonrandom mating changes genotype frequencies, for example by producing more homozygotes, without by itself changing allele frequencies.

A population with more genetic variation is more likely to include some individuals that survive a new disease or a change in its environment.

Trap Square root first, and dominant does not mean common

If 16 percent of a population shows a recessive trait, q² = 0.16, so q = 0.4, not 0.16. Using the share of recessive individuals as q is the most common slip.

A dominant allele is not more common just because it is dominant. Dominance decides which phenotype a heterozygote shows; it does not push an allele’s frequency up.

Worked example Easy

A sample of 200 fictional wildflowers has 32 RR, 96 Rr and 72 rr plants. What is the frequency of allele R?

  1. 0.16
  2. 0.40Answer
  3. 0.60
  4. 0.64

How to solve it

  1. Each plant carries 2 copies, so there are 2 x 200 = 400 copies in all.
  2. RR plants carry 2 R copies each and Rr plants carry 1: 2 x 32 + 96 = 160 R copies.
  3. Frequency of R = 160 / 400 = 0.40. Check: r = 240 / 400 = 0.60, and 0.40 + 0.60 = 1.

Why each choice is right or wrong

  • A. Incorrect. 0.16 is 32 / 200, the share of plants that are RR, not the share of all gene copies that are R.
  • B. Correct. R copies = 2 x 32 + 96 = 160, out of 2 x 200 = 400 copies, so 160 / 400 = 0.40.
  • C. Incorrect. 0.60 is the frequency of r, (2 x 72 + 96) / 400 = 240 / 400, not of R.
  • D. Incorrect. 0.64 is (32 + 96) / 200, the share of plants carrying at least one R, which counts plants rather than alleles.

Explain Explain it

Task verbs: calculate, explain

For a Hardy-Weinberg problem, write down what you are given and work in order: q² from the recessive share, then q, then p, then the genotype frequencies, then counts (frequency x population size). Show each step.

When you explain a change in allele frequency, name the process and give the clue for it. A consistent rise of the allele that helps survival points to selection. Random ups and downs in a small population point to drift. Arrivals from another population point to gene flow.

Worked example Medium

In a fictional population of 2,500 mice, 400 have white fur, a recessive trait (aa). Assuming the population is in Hardy-Weinberg equilibrium, how many mice are expected to be heterozygous (Aa)?

  1. 400
  2. 900
  3. 1,000
  4. 1,200Answer

How to solve it

  1. q² = 400 / 2,500 = 0.16, so q = 0.4.
  2. p = 1 - 0.4 = 0.6.
  3. Heterozygotes: 2pq = 2 x 0.6 x 0.4 = 0.48, and 0.48 x 2,500 = 1,200 mice.

Why each choice is right or wrong

  • A. Incorrect. 400 is the number of white, aa mice given in the question, not the heterozygotes.
  • B. Incorrect. 900 is p² x 2,500 = 0.36 x 2,500, the expected number of AA mice.
  • C. Incorrect. 1,000 is q x 2,500 = 0.4 x 2,500; q is an allele frequency, not a genotype frequency.
  • D. Correct. q² = 400 / 2,500 = 0.16, so q = 0.4 and p = 0.6; 2pq = 0.48, and 0.48 x 2,500 = 1,200.

Calculate Your turn

In a fictional population of 500 beetles, 45 are black, a recessive trait (bb), and the rest are brown. Assuming Hardy-Weinberg equilibrium, calculate q (the frequency of b), p (the frequency of B) and the expected number of heterozygous beetles. Show each step.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Finds q² = 45 / 500 = 0.09 and q = 0.3
  • Finds p = 1 - 0.3 = 0.7
  • Finds 2pq = 2 x 0.7 x 0.3 = 0.42 and the expected number 0.42 x 500 = 210 heterozygous beetles

Model answer

The black beetles are bb, so q² = 45 / 500 = 0.09, and q = the square root of 0.09 = 0.3. Then p = 1 - 0.3 = 0.7. Heterozygotes make up 2pq = 2 x 0.7 x 0.3 = 0.42 of the population, so the expected number is 0.42 x 500 = 210 heterozygous beetles.

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