In cats a form of polydactyly is inherited as an autosomal, complete dominant trait. In Key West, Florida 16% of the cats have a normal number of digits. What is the frequency of the polydactyly allele in that population? Show your work.

The answer is 0.6.

If polydactyly is an autosomal dominant trait, that means that dominant allele A is responsible for the disease. Both dominant homozygous (genotype: AA) and heterozygous (Aa) cats have the disease. So, recessive homozygous cats (aa) have a number of digits.

To calculate the frequency of the polydactyly allele (A) in the population the Hardy-Weinberg principle can be used:

p² + 2pq + q² = 1 and p + q = 1

where p and q are the frequencies of the alleles, and p², q² and 2pq are the frequencies of the genotypes.

If 16% of the cats have a normal number of digits (aa), then the frequency of recessive homozygous genotype q² is:

q² = 0.16

⇒ q = √0.16

⇒ q = 0.4

The frequency of recessive allele is 0.4

If the frequency of recessive allele is known, the frequency of dominant (polydactyly) allele p is:

p + q = 1

⇒ p = 1 - q

⇒ p = 1 - 0.4

⇒ p = 0.6

Thus, the frequency of the polydactyly allele in that population is 0.6 or presented in percentage 60%.