Question

Polydactyly is expressed when an individual has extra fingers and/or toes. Assume that a man with six fingers on each hand and six toes on each foot marries a woman with a normal number of digits. Having extra digits is caused by a dominant allele. The couple has a son with normal hands and feet, but the couple's second child has extra digits. What is the probability that their third child will have polydactyly?

Answer 1

The polydactyly is caused by autosomal dominant inheritance way. Let the dominant allele causing the polydactyly be 'P' and the wild type normal allele be 'p'. As the male parent is having polydactyly, he must be having atleast one 'P' allele as one of his offspring has a normal phenotype. Hence the genotype of the male parent is Pp and female parent is pp. A cross between them will result in the following offsprings:

	P	p
p	Pp Polydactyly	pp Normal digits
p	Pp Polydactyly	pp Normal digits

Hence there is a 50% probability of polydactyly offspring being produced.

Hence the probability that their third child will be polydactyly is 50% or 1/2.

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