Selection Coefficients and Predicting Allele Frequency Change
Giving each genotype a relative fitness turns selection into arithmetic: the frequency after one generation follows from the frequencies before and the fitnesses. Working the change out generation by generation shows how slowly a rare recessive allele is removed.
What a learner can do afterwards
- Calculates allele frequency after one generation from genotype frequencies and fitnesses.
- Explains why selection against a recessive allele slows as it becomes rare and uses q squared to show why affected homozygotes become scarce.
- Distinguishes relative fitness from survival or family size taken alone.
1 · Read
Relative fitness, w, is a genotype's expected reproductive contribution divided by the best genotype's contribution in the same population. Values are usually scaled so the best is 1, but only the ratios matter, so any common scale gives the same next frequency. On the scale where the best is 1, the selection coefficient s = 1 - w measures how far a genotype falls short. A fitness of 0.8 gives s = 0.2.
Mean fitness is the sum of each genotype frequency multiplied by its fitness. For frequencies 0.25, 0.50, and 0.25 with fitnesses 1.0, 0.8, and 0.6, the contributions are 0.25, 0.40, and 0.15, so mean fitness is 0.80.
For a diploid population, first multiply each genotype frequency by its fitness and add the products; that sum is the mean fitness. Then divide each weighted frequency by the mean fitness to normalize it. The new A frequency is the normalized AA frequency plus half of the normalized Aa frequency.
A recessive harmful allele is hidden from selection in heterozygotes only when those heterozygotes have no fitness disadvantage. If q is the allele frequency, random pairing predicts recessive homozygotes at q squared. As q falls, q squared falls faster, so fewer homozygotes expose the allele and removal slows.
A tenfold drop in q gives a hundredfold drop in q².
The same idea applies to a lethal recessive with no heterozygote effect: most copies remain in carriers while selection removes affected homozygotes. If carriers instead pay even a small fitness cost, selection acts on every carrier too. Because carriers hold most copies of a rare allele, removal then does not stall as the allele becomes rare, the way it does when only homozygotes are exposed.
In a haploid population, A and a each start at 0.5. Each A individual leaves 4 offspring and each a individual leaves 1. Dividing by the best count gives fitness 1 for A and 0.25 for a. Mean fitness is 0.5 times 1 plus 0.5 times 0.25, or 0.625, so the next A frequency is 0.5 divided by 0.625, or 0.8. Raw counts give the same answer, 2 divided by 2.5, because only the ratio matters.
Weight genotype frequencies by relative fitness, divide by mean fitness, and remember why a rare recessive can hide in carriers.
Original lesson, assessment, and flashcard text authored for this topic in the Open Source Taxonomy project.
The two rewritten readings were written for this dataset; the videos are Khan Academy embeds; publisher terms are unknown. The reading on selection arithmetic and extra questions x1 to x5 were written for this topic in the Open Source Taxonomy project.
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