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Answer:
1/2 or 50%
Explanation:
This question involves a single gene coding for the number of digits per paw in cats or humans. The allele for polydactyly i.e six digits (P) is dominant over the allele for normal number of digits i.e 5 (p).
According to this question, If a man who is heterozygous (Pp) for this trait marries a woman with 5 fingers (pp), the following gametes will be produced by each parent:
Pp : P and p
pp : p and p
Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:
Pp, Pp, pp and pp
The offsprings with genotype Pp will have polydactyl, hence, the chances that they will have a child with polydactyly is 1/2 or 50%.
