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2 October, 13:37

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.

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  1. 2 October, 13:42
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    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%.
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