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UNIT 07 / REVIEW

Hardy-Weinberg

For two alleles, p+q=1. Under Hardy-Weinberg assumptions, genotype frequencies are p², 2pq, and q².

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01 / Hardy-Weinberg

Hardy-Weinberg

The equilibrium model assumes a very large population, random mating, no selection, negligible mutation, and no migration. It is a comparison model for asking whether observed genotype frequencies match expected values. Allele frequency counts copies: heterozygotes contribute one copy of each allele. In a completely dominant system, the recessive phenotype identifies the recessive genotype, but the dominant phenotype combines two genotypes. A mismatch with expectations does not alone reveal which assumption failed.

MAKE IT CONCRETE

If the recessive phenotype frequency is 0.09 and equilibrium applies, q=0.30, p=0.70, and the heterozygote frequency is 2pq=0.42. The square root is doing actual work here.

DON’T FALL FOR THIS

Do not plug the dominant phenotype frequency straight into p². Dominant-looking individuals may be homozygous or heterozygous.

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Original review notes informed by standard biology principles. Check the official AP Biology framework ↗ and OpenStax Biology 2e ↗ for further reading.