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%.
