In a hypothetical population of jackalopes living in the Southwestern US, individuals homozygous dominant for locus...

In a hypothetical population of jackalopes living in the Southwestern US, individuals homozygous dominant for locus E have very large ears, heterozygous (Ee) individuals have medium sized ears and homozygous recessive (ee) individuals have very small ears. The population is composed of 500 individuals with the following phenotypes: 245 individuals with big ears, 210 individuals with medium sized ears, and 45 individuals with very small ears. Assuming p = the frequency of allele E and q = the fequency of allele e, fill in the blanks below.

Freq. of EE =

Freq. of Ee =

Freq. of ee =

p =

q =

Expert Solution

According to Hardy - Weinberg equation,

The frequency of EE, is p² = 245/500 or 0.49 or p=0.7

frequency of ee, q² = 45/500= 0.09 or q= 0.3

And of Ee becomes, p² + 2pq + q² =1,

Thus, pq =1- (0.49+0.03)/2

Or freqeucny of Ee = 0.74

(Check that p+q=1 according to Hardy Weinberg law)

gladiator answered 2 years ago

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