p^2 +2pq +q^2 with A=p frequency a=q frequency Aa x Aa heterozygous cross Assume there is a heterozygous survival advantage, where an AA type egg has probability 0.8 of mating and an aa type egg has probability 0.2 of mating, but Aa type eggs have 1.0 probability of mating. Explain why the new ‘total’ frequency of all chromosomes in reproducing adults (i.e., in the next generation born) will not be 100% but instead will be F(p) = 0.8p^2+ 1.0 ∙2pq + 0.2q^2 This is called the balance equation. Find p1, the frequency of A in the adult population, if an AA type egg has probability 0.8 of mating and an aa type egg has probability 0.2 of mating, but Aa type eggs have 1.0 probability of mating. Remember that the frequency is the proportion out of the total and use the balance equation you derived above. Note I think finding p1 involves implicit differentiation. Frequency of A is described as variable p
p^2 +2pq +q^2 with A=p frequency a=q frequency Aa x Aa heterozygous cross
Assume there is a heterozygous survival advantage, where an AA type egg has
Find p1, the frequency of A in the adult population, if an AA type egg has probability 0.8 of mating and an aa type egg has probability 0.2 of mating, but Aa type eggs have 1.0 probability of mating. Remember that the frequency is the proportion out of the total and use the balance equation you derived above.
Note I think finding p1 involves implicit differentiation. Frequency of A is described as variable p.
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