Final answer:
The probability that their child will not have either disease is 12.5%.To determine the probability of their child not having polydactyly and phenylketonuria, we consider the independent inheritance of these traits. The calculated probability is 12.5%, as the traits follow autosomal dominant and recessive patterns.
Step-by-step explanation:
To determine the probability that their child will not have either disease, we need to consider the inheritance patterns of polydactyly and phenylketonuria.
Polydactyly is an autosomal dominant trait, which means that if one parent has the trait, there is a 50% chance that their child will have it as well. Since the mother is heterozygous for polydactyly (Pp), there is a 50% chance that the child will have polydactyly and a 50% chance that the child will not have it.
Phenylketonuria is an autosomal recessive trait. The father does not have the trait, so he is most likely homozygous for the dominant allele (PP). The mother is a carrier for phenylketonuria, which means she is heterozygous (Pp). There is a 25% chance that the child will inherit two recessive alleles (pp) and have phenylketonuria, a 50% chance that the child will be a carrier (Pp), and a 25% chance that the child will not have the trait (PP).
Since the probabilities for polydactyly and phenylketonuria are independent, we can multiply the probabilities together:
- Probability of not having polydactyly: 50%
- Probability of not having phenylketonuria: 25%
Therefore, the probability that their child will not have either disease is 50% * 25% = 12.5%.