The incidence of phenylketonuria (PKU), an autosomal recessive genetic disorder, is approximately 1 in 4,500 in Ireland. Assuming Hardy-Weinberg equilibrium, what is the probability that an Irish female, who is phenotypically normal with no family history of PKU, and an Irish male, who like his parents does not have PKU, but does have a sister with PKU, will have a child with PKU?

Probability: 0%

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.

the two answers are a, the population of birds with large beaks will increase on the island, and c, the trait for large beaks will be passed on to future generations through natural selection. the reason for that is because if the birds that have beaks developed for eating small seeds lose all of their food, they will go hungry and eventually starve because they physically cannot eat the large seeds, therefore making the population of large beaked birds higher. also, if the trait of large beaks are dominant on the island over small beaks, then over time the large beaked birds will reproduce more and more, while the smaller beaked birds will reproduce less and less because they can't survive with the food provided. that leads to the trait of small beaks disappearing and the trait of large beaks taking over the island.