Question

A population of sea anemones has 64 individuals with “bumpy tentacles” and 36 individuals with “smooth tentacles.” This characteristic is controlled by one locus that has two alleles. The “bumpy” allele is recessive. Assuming that the population is in Hardy-Weinberg equilibrium, what are the allele frequencies?

Answer 1

Answer:

The a allele has a frequency of 0.8

The A allele has a frequency of 0.2

Explanation:

Lets say the allele for smooth tentacles is T and the allele for bumpy tentacles is t.  We know that the bumpy allele is recessive, so that means they must be tt. Smooth tentacles could either be Tt or TT.

tt = 64

TT + Tt = 36

The frequency of tt = bumpy tentacles / the whole population

= 0.64

We use the following equations

For genotype frequencies:

p² + 2pq + q² = 1, where p² is the homozygous dominant genotype, and q² is the homozygous recessive genotype.

For allele frequencies:

p + q = 1, where p is the dominant allele and q is the recessive allele

We know that q² = 0.64

Therefore, q = $\sqrt{0.64}$ = 0.8.

p + q = 1

p + 0.8 = 1

therefore, p = 0.2

The a allele has a frequency of 0.8

The A allele has a frequency of 0.2