Tom and Lesley are married and have a child, Sally, who has Cystic Fibrosis. Lesley has just found out she is pregnant, so what is the probability that the baby will have the condition?

Cystic Fibrosis is a recessive disease, therefore for Sally to have the condition she must have a homozygous recessive genotype, with one recessive allele of the gene from the mother and one from the father. This means both Tom and Lesley are carriers, as neither of them has the condition (don't have the phenotype) but the both have the recessive allele (a heterozygous phenotype).

Now that we know that both Tom and Lesley are carriers with heterozygous phenotypes, we can use a punnet square to determine the possible genotypic outcomes for the baby that Lesley is pregnant with, and therefore the possible phenotypes. Using the punnet square, we will find that the only genotype that will lead to the phenotype of having Cystic Fibrosis is homozygous recessive, so two recessive alleles. This will only occur in 1/4 possible combinations of the alleles, therefore the probability of the baby having Cystic Fibrosis is 0.25 or 25%.