Lesson 2 of 5 / Population genetics
Build the Hardy-Weinberg model
Why is the heterozygote frequency 2pq?
I can connect random gamete combinations to p squared, 2pq and q squared.
I can explain the Hardy-Weinberg model and its assumptions.Syllabus
Syllabus 9477, Core 4(m). Explain the Hardy-Weinberg model and its assumptions.
I can calculate two-allele frequencies and genotype/phenotype proportions.Syllabus
Syllabus 9477, Core 4(n). Calculate two-allele frequencies and genotype/phenotype proportions.
About 6 min
Make a guess. You are not marked.
In a population, alleles A and a each have frequency 0.5. Under Hardy-Weinberg, what fraction are heterozygous?
Show the answer
Half, because 2pq = 0.5.
An Aa individual can arise as A from the father and a from the mother, or the reverse, hence the 2.
The key idea
Random combination of gametes gives AA = p squared, Aa = 2pq and aa = q squared, when the model applies.
Explore the idea
Two routes make a heterozygote
Aa = pq + qp = 48.0%. Moving the slider changes allele frequencies; it keeps the random-mating model in place.
Explanation
The Hardy-Weinberg model is a reference for inheritance at a locus in a population. With two alleles, let their frequencies be p and q, so p + q = 1. Under the model, allele frequencies remain stable between generations and genotype frequencies follow a predictable distribution.
Imagine drawing an allele independently from each of two gamete pools with the same allele frequencies. The probability of drawing A from both is p x p, giving p squared for AA. Drawing a from both gives q squared for aa.
A heterozygote can form in two ways: A from the first gamete and a from the second, or a from the first and A from the second. These mutually exclusive possibilities each have probability pq, so their total is 2pq. They are two routes to one genotype, not two different genotypes.
The three proportions sum to 1 because they include every possible genotype: p squared + 2pq + q squared = 1. Multiply a proportion by the population size to obtain an expected number of individuals. Expected numbers describe the model; actual samples can vary by chance.
If A is completely dominant, the A phenotype includes AA and Aa. Its expected frequency is p squared + 2pq, also equal to 1 - q squared. With codominance, the heterozygote has a distinguishable phenotype, so there are three phenotype classes. Match the genotype-to-phenotype rule given in the question.
Step by step
- 1
Start with allele frequencies
Define p and q before substituting numbers.
- 2
Combine independent gametes
Multiply frequencies for each combination and add the two heterozygote routes.
- 3
Translate to the requested quantity
Keep genotype proportions, phenotype proportions and expected counts distinct.
Worked example
Work through the evidence
For p = 0.6 and q = 0.4, predict genotype numbers in 500 individuals under Hardy-Weinberg equilibrium.
One way to explain it
AA: 0.6 squared x 500 = 180. Aa: 2 x 0.6 x 0.4 x 500 = 240. aa: 0.4 squared x 500 = 80. The expected counts sum to 500.
Why this answer works
- Square each allele frequency only for its homozygote.
- Include both routes to the heterozygote.
- Convert proportions to counts only at the last step.
Watch out for this
A student says: "2pq represents two different kinds of heterozygote." What is wrong with this?
Show the answer
Aa and aA are the same genotype in this autosomal model. The two products count the two possible parental origins of the alleles.
Check your understanding
When p = q = 0.5, which expected genotype distribution follows?
Show the answer
25% AA, 50% Aa, 25% aa
The homozygotes each have frequency 0.25 and the two heterozygote routes total 0.50.
Quick recall
Why is the heterozygote frequency 2pq?