You observe a population of three-toed frogs that lives in the cloud forests of Panama. In this population, you observe the following genotype counts at a locus controlling coloration: 411 (AA), 800 (Aa), and 911 (aa). This population is not at Hardy-Weinberg Equilibrium (HWE). For this population to be in HWE, what would the frequency of homozygous AA frogs have to be

0.1460

Explanation:

Total population size= 411 + 911 + 800 = 2122

Let's assume that the population is in Hardy Weinberg Equilibrium.

Under these conditions, frequency of recessive allele (a)= 911 x 2 + 800 / 2122 x 2   = 2622/4244= 0.6178 .

This is because each homozygous recessive genotype has two copies of the recessive allele while each heterozygous dominant genotype has one copy of the recessive allele.  

For this population to be under Hardy Weinberg Equilibrium, frequency of dominant allele (A) = 1- 0.6178= 0.3822

Frequency of homozygous dominant genotype (AA) = 0.1460

good luck

explanation:

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