Answer: Probability: 0%
Step-by-step explanation:
Hardy-Weinberg establishes that the genetic composition of a population remains in equilibrium as long as no natural selection or other factors act and no mutations occur. This means that the children of this couple will inherit the genes of the parents.
The woman is phenotypically normal with no family history of PKU, this mean that probably there are no affected (recessive) alleles in her family. In an autosomal recessive disorder, both copies of the gene in each cell are recessive. Therefore, her genotype will be AA with A being the dominant and normal allele. As for the Irish man, his sister does have PKU, which means her genotype is aa. So, she must have received a defective allele from each parent. Then the parents' genotypes are both Aa (heterozygous, only one allele affected as they are healthy). So, we know that the Irish man does not have the disease (he can't be aa), but he could be Aa (having inherited a normal allele from one parent, and an affected one from the other parent) or AA (having inherited both normal alleles from each parent).
Then we have two possible punnetts square to do: AA (the Irish woman) and AA (he Irish man); or AA (woman) and Aa (man).
In the attached figure both punnetts squares are shown. The first punnett square (AA x AA) shows that 100% of the offspring will be AA (this means, the phenotypic ratio is also 100% not affected and non-carrier). The second punnett square (AA x Aa) shows that 50% will be AA (healthy and non-carrier) and 50% will be Aa (heterozygous, healthy and carrier) But since it is an autosomal recessive disease, a single affected allele is not enough to develop the disease. So in neither case is there a chance of having a child with PKU.