Lesson 1 of 5 / Population genetics
Count alleles, not just organisms
How can 200 organisms contain 400 copies of one gene?
In this lesson: Calculate allele frequencies directly from diploid genotype counts.
About 5 min
The key ideaAt a diploid autosomal locus, each individual contributes two alleles: p = (2AA + Aa) / (2N).
Here AA, Aa and aa represent genotype counts. N is the number of diploid individuals; p is the frequency of A and q is the frequency of a.
Explore the idea
Turn individuals into allele copies
p = 260 / 400 = 0.65. q = 140 / 400 = 0.35. The numbers below AA, Aa and aa count individuals. Each Aa individual contributes one copy to each colour.
Explanation
An individual has a genotype at a locus. A population has a distribution of genotypes and a pool of alleles. Frequency means a proportion, so always identify what you are counting and what belongs in the denominator.
For a diploid autosomal locus with alleles A and a, every AA individual contributes two A copies, every Aa individual contributes one A and one a, and every aa individual contributes two a copies. N individuals therefore contribute 2N copies at that locus.
Calculate p by dividing the number of A copies by 2N. Calculate q similarly for a, or use q = 1 - p when these are the only two alleles. This direct count does not assume Hardy-Weinberg equilibrium. You already know the genotypes and can count their copies.
The frequency of genotype Aa is the number of heterozygous individuals divided by N. It is not the frequency of allele a, which uses copies divided by 2N. Likewise, a dominant phenotype can include AA and Aa, so phenotype counts alone may not reveal the exact allele counts.
Step by step
- 1
Count the individuals
Add AA, Aa and aa to obtain N.
- 2
Count each allele
A copies = 2 x AA + Aa; a copies = 2 x aa + Aa.
- 3
Divide and check
Divide each allele count by 2N. The two frequencies must add to 1.
Worked example
Work through the evidence
A sample contains 90 AA, 80 Aa and 30 aa individuals. Calculate p and q.
One way to explain it
N = 200, so there are 400 allele copies. A copies = (2 x 90) + 80 = 260, giving p = 260/400 = 0.65. The remaining 140 copies are a, so q = 0.35.
Why this answer works
- Each heterozygote contributes to both allele counts.
- The denominator is 400 copies, not 200 individuals.
- 0.65 + 0.35 = 1.
Is this true? "If 30 of 200 individuals are aa, the frequency of allele a is 0.15."
0.15 is the observed aa genotype frequency. The a allele also occurs in every Aa individual. Count all a copies, or use a stated equilibrium model when genotypes are unknown.