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Unformatted text preview: q = the frequency of the recessive allele (represented here by a) p2 = frequency of AA (homozygous dominant) 2pq = frequency of Aa (heterozygous) q2 = frequency of aa (homozygous recessive) • Your answer: 0.4 • Answer: • You should expect 160 to be homozygous dominant. • Calculations: • q2 for this population is 360/1000 = 0.36 • q = = 0.6 • p = 1  q = 1  0.6 = 0.4 • The homozygous dominant frequency = p2 = (0.4)(0.4) = 0.16. • Therefore, you can expect 16% of 1000, or 160 individuals, to be homozygous dominant....
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 Spring '08
 Reichler
 Evolution, Population Genetics, 16%, Zygosity

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