AP Biology · Unit 7 · Topic 7.5
Hardy-Weinberg describes no real population. It is the prediction you make when nothing is happening, so that a departure from it means something.
Learning objective 7.5.A Describe the conditions under which allele and genotype frequencies will change in populations.
Each population below is deliberately built wrong — all heterozygotes, or none at all, or two populations just merged. Nothing is done to any of them except one generation of random mating. Watch two things separately: the bars, and the number underneath them.
Hardy–Weinberg proportions are a statement about a single generation: the genotype frequencies are p², 2pq and q². As the panel above shows, one round of random mating produces them from any starting point at all — no conditions required beyond the mating being random.
Hardy–Weinberg equilibrium is a statement about successive generations: the allele frequencies do not change. That is what needs the five conditions, and it is what the next tab is about.
This is also why the heterozygote is the term that never appears alone in a phenotype count: a heterozygote looks like a dominant homozygote, so p² and 2pq are measured together as one number and only q² can be counted directly.
Every run starts at the same allele frequency. The first button is the control — all five conditions met. Each of the others breaks exactly one of them and changes nothing else. The solid line is the allele frequency; the red dashed line is the heterozygote frequency, and the wide gray band under it is what 2pq predicts. When the two agree, the red line sits on the band.
| Condition | What breaks it | What happens | Do allele frequencies move? |
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| Across the replicate runs | Value |
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The single run drawn in the panel above is one of these. It is worth clicking back to it with this table in mind: on its own it looks like a trend, and it is not one.
Only the recessive phenotype tells you a genotype directly — anyone showing it must carry two recessive alleles. So every Hardy–Weinberg problem starts by finding q², taking its square root, and working outwards. The square below is the whole equation: its area is one, its sides are p and q.
| Scenario | q² | q | p | p² | 2pq | Carriers per affected |
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This is the Topic 7.2 result reached by a different route. There, selection could not remove a recessive allele because most copies of it were hidden in heterozygotes; here, the same fact falls out of the algebra, because 2pq shrinks far more slowly than q² does as q gets small.
Answer these in your own words before you open the answers.
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