Lesson 3 of 5 / Population genetics
Work backwards from a recessive phenotype
If 9% show a recessive phenotype, are 9% of the alleles recessive?
I can infer allele, genotype and phenotype frequencies from suitable equilibrium data.
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 an equilibrium population, 9% show a recessive trait. What is the frequency of the recessive allele?
Show the answer
0.3, the square root of 0.09.
The recessive phenotype is aa, with frequency q squared. So q is the square root of 0.09.
The key idea
Under Hardy-Weinberg equilibrium and complete dominance, the recessive phenotype frequency is q squared; take its square root to find q.
Explore the idea
Start with q squared, then work back
q squared = 0.09; q = 0.300; p = 0.700
In 10,000 individuals, the model predicts about 4,200 heterozygotes. The Aa bar is the carrier group in this simple recessive model.
Assumes a two-allele autosomal locus, complete dominance and Hardy-Weinberg equilibrium. Displayed percentages and counts are rounded.
Explanation
A fully expressed recessive phenotype identifies aa when one autosomal locus with complete dominance controls the character. If the population is also assumed to be in Hardy-Weinberg equilibrium, its aa frequency can be treated as q squared. Both the inheritance rule and the population model matter.
Convert a percentage to a proportion first. For 9%, q squared = 0.09, so q = 0.3 and p = 0.7. The heterozygote frequency is then 2 x 0.7 x 0.3 = 0.42. It is neither 0.09 nor 0.3.
In a population of 10,000, these proportions predict 900 aa, 4,200 Aa and 4,900 AA individuals. When the recessive allele is associated with a recessive condition, carriers are the heterozygotes. An individual with the dominant phenotype may be AA or Aa.
If asked for the proportion of carriers among individuals showing the dominant phenotype, change the denominator. In this example it is 4,200 / 9,100, about 46.2%, rather than 42% of the whole population. Read whether the question refers to everyone or to a selected group.
Without the equilibrium assumption, the recessive phenotype still identifies aa in this simple model, but it does not determine how many of the remaining individuals are AA or Aa. Do not silently use q = square root of the aa frequency for every observed population.
Step by step
- 1
Check the model
Confirm two alleles, autosomal inheritance, complete dominance and equilibrium.
- 2
Work from q squared to q
Take the positive square root, then calculate p = 1 - q.
- 3
Answer the requested count or proportion
Calculate the relevant genotype frequency and use the correct population denominator.
Worked example
Work through the evidence
In an equilibrium population of 2,500, 100 individuals show a completely recessive phenotype. Predict the number of heterozygotes.
One way to explain it
q squared = 100/2,500 = 0.04. Therefore q = 0.2, p = 0.8 and 2pq = 0.32. The expected heterozygote count is 0.32 x 2,500 = 800.
Why this answer works
- Use the recessive proportion, not the count 100, in the equation.
- Find q before calculating heterozygotes.
- State that 800 is a model prediction.
Watch out for this
A student says: "The dominant phenotype frequency is p." What is wrong with this?
Show the answer
p is the A allele frequency. Under complete dominance the A phenotype includes two genotypes, AA and Aa, giving p squared + 2pq under equilibrium.
Check your understanding
A recessive phenotype has frequency 0.16 in an equilibrium population. What is the expected heterozygote frequency?
Show the answer
0.48
q = 0.4, p = 0.6, and 2pq = 2 x 0.6 x 0.4 = 0.48.
Exam-style questions
Check your understanding
In an equilibrium population, 1 in 2,500 people has a recessive condition. About what fraction of people are carriers?
Show the answer
About 1 in 25
q squared = 1/2,500, so q = 1/50 = 0.02, p = 0.98, and 2pq = 2 x 0.98 x 0.02, about 0.04 or 1 in 25.
Quick recall
If 9% show a recessive phenotype, are 9% of the alleles recessive?