Question

In a population of snails, 64% are found to have a coil pattern determined by the dominant allele "C." Calculate the frequency of the recessive snails in the population, as well as the frequencies of the heterozygous and homozygous dominant snails. Also, what is the frequency of the coil "C" allele, and the recessive noncoil pattern allele "c"?

Answer 1

Answer and Explanation:

Data:

• 64% of snails in the population are found to have a coil pattern determined by the dominant allele "C".

This 64% of the population is composed of individuals with genotype CC and Cc. This percentage equals a frequency of 0.64.

36% of the population is composed of homozygote individuals expressing the recessive trait, cc. This percentage equals a genotypic frequency of 0.36.

If CC=p² and cc=q², then:

cc=q²=0,36

√cc=√0.36=0.6

c=0.6

0.6 is the allelic frequency for the recessive trait.

Considering that the sum of frequencies of the dominant allele plus the recessive allele equals one, then C + c = 1.

If we know the frequency of the recessive allele c, but not the frequency of the dominant allele, we can obtain it by clearing the equation. This is

C + c = 1

C + 0.6 = 1

1 - 0.6 = C

1 - 0.6 = 0.4

So the frequency of the dominant allele, C is 0.4.

The genotypic frequency of the heterozygous is 2pq=2Cc=2x0.4x0.6= 0.48

So the genotyic frequency for the heterozygous is 0.48

So, up to now, we have:

• 64% of the population is composed of individuals with genotype CC and Cc. This percentage equals a frequency of 0.64.

• 36% of the population is composed of homozygote individuals for the recessive trait, cc. This percentage equals the genotypic frequency for the recessive trait, 0.36.

• The allelic frequency for the recessive allele is 0.6

• The allelic frequency for the dominant alelle is 0.4

• The dominant genotypic frequency is 0.16

• The heterozygous genotypic frequency is 0.48

• The proportion of individuals expressing the dominant trait is 0.16 + 0.48 = 0.64 = 64%