What is the probability that her child will have cystic fibrosis?

1/32-to work this problem draw out the family lineage, calculate the risk of having the cf gene at each branch and multiply each risk. the parent of this woman through whom she is related to her first cousin has a 1/2 chance of carrying the mutant cf, this parental sib has a 1/2 chance of carrying the allele and this persons child (the first cousin) also has a 1/2 chance of carrying the mutant allele. cumulative risk = 1/2 x 1/2 x 1/2 = 1/8 that the first cousin has a mutant allele. then the risk to have an affected child is 1 x 1/2 x 1/8 x 1/2 = 1/32.