Solved by a verified expert:A locus that affects susceptibility to a degenerative disease has two alleles, A and a. In a population, 16 people have genotype AA, 92 have genotype Aa, and 12 have genotype aa. Is this population evolving? Explain.I found this answer online: there are 120 individuals in the population, so there are 240 alleles, of these, there are 124 “A” alleles (32 from the 16 “AA” individuals and 92 from the 92 “Aa” individuals) thus, the frequency of the “A” allele is p=124/240=.52; hence the frequency of the ‘a’ allele is q=.48. Based on the hardy-Weinberg equation, if the population were not evolving, the frequency of genotype “AA” should be p2=.52x.25=.27; the frequency of genotype “aa” should be q2=.48x.48=.23. In a population of 120 individuals, these expected genotype frequencies lead s to predict that there would be 32 “AA” individuals (.27x.120), 60 “Aa” individuals (.51×120), and 28 “aa” individuals ( .23×120). The actual numbers for this population deviate from these expectations (fewer homozygotes and more heterozygotes than expected. This indicates that the population is not in hardy Weinberg equilibrium and hence may be evolving at this locus.I’m confused about how “Based on the hardy-Weinberg equation, if the population were not evolving, the frequency of genotype “AA” should be p2=.52x.25=.27; the frequency of genotype “aa” should be q2=.48x.48=.23.”. How were 0.27 and 0.23 found? Why are these used?
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