Population genetics
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A sample contains 20 AA, 60 Aa and 20 aa individuals. What is the frequency of A?
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0.50
(2 x 20 + 60) / (2 x 100) = 100/200 = 0.50.
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When p = q = 0.5, which expected genotype distribution follows?
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25% AA, 50% Aa, 25% aa
The homozygotes each have frequency 0.25 and the two heterozygote routes total 0.50.
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A recessive phenotype has frequency 0.16 in an equilibrium population. What is the expected heterozygote frequency?
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0.48
q = 0.4, p = 0.6, and 2pq = 2 x 0.6 x 0.4 = 0.48.
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In an equilibrium population, 1 in 2,500 people has a recessive condition. About what fraction of people are carriers?
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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.
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Which change can alter genotype frequencies without immediately changing allele frequencies?
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Individuals preferentially mate with others of the same genotype.
Non-random combinations can redistribute the same alleles among genotypes.
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Which is NOT an assumption of the Hardy-Weinberg model?
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The dominant allele is more common
The model makes no assumption about which allele is commoner; it works for any p and q.
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Which finding most directly supports heterozygote advantage at a locus?
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Aa individuals leave more surviving offspring than both AA and aa in the same environment.
Higher heterozygote reproductive success can maintain both alleles.
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