The Hardy-Weinberg Principle and Allele Frequencies · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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When gene pools stand still

Science · Genetics & Evolution · ages 17-18
Name ______________________   Date ____________
  1. Which of these does the model assume?

    • Populations stay tiny and isolated
    • Mating is random across the population
    • Selection favors one allele strongly in every generation
  2. Which pair of equations describes the Hardy-Weinberg model?

    • p plus q equals 1 with p squared plus 2pq plus q squared equals 1
    • p times q equals 1 with p squared minus 2pq minus q squared equals 1
    • p minus q equals 1 with 2p plus 2q equals 1
  3. A population in Hardy-Weinberg equilibrium keeps the same allele frequencies across generations.

    Circle one:   True   False

  4. How do you count copies of allele A from genotype counts?

    • Count every individual exactly once, regardless of genotype
    • Count only the heterozygotes and ignore the rest
    • Count each AA twice and each Aa once
  5. In 100 plants there are 24 AA, 36 Aa and 40 aa. What is the frequency of allele A?

    Answer: ______________

  6. A small inbred group mates among relatives. Which assumptions break?

    • Huge population size and random mating
    • No mutation and no gene flow
    • The equations themselves stop adding up
  7. A population's allele frequencies change across generations. What can this show?

    • The population has evolved at that gene
    • The population is in perfect equilibrium
    • The counting method must be wrong
  8. In 400 people, 64 are affected by a fully recessive condition. Assuming Hardy-Weinberg conditions, what is the carrier frequency?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

When gene pools stand still W1-mt_eJjwZ2qVAE-s1

  1. Mating is random across the population · Random mating, huge size, and no selection are all on the assumptions list.
  2. p plus q equals 1 with p squared plus 2pq plus q squared equals 1 · Allele frequencies sum to one, and the genotype mix follows from squaring that sum.
  3. True · Unchanging frequencies define equilibrium, and they mean no evolution is happening at that gene.
  4. Count each AA twice and each Aa once · Each AA individual carries two copies of A, each Aa carries one, and each aa carries none.
  5. 0.42 · A copies are 48 plus 36, which is 84 of 200 total alleles. The frequency of A is 84 divided by 200, or 0.42.
  6. Huge population size and random mating · Small groups drift by luck and relatives pair nonrandomly, breaking two assumptions at once.
  7. The population has evolved at that gene · A lasting change in allele frequency is evolution at that gene. It shows that the baseline is not holding, but it does not identify selection by itself.
  8. 0.48 · The affected frequency is q squared, so q squared is 64 divided by 400, or 0.16. Thus q is 0.4, p is 0.6, and 2pq is 0.48.
Worksheet · LightMySky