Final answer:
To calculate the probability that a child of a phenotypically normal individual and a CF carrier will develop cystic fibrosis, one must know the genotype of the normal individual. If the normal individual is a carrier, the chance is 25%. Otherwise, if they have two normal alleles, the chance is 0%.
Step-by-step explanation:
If a phenotypically normal individual who has a sibling with cystic fibrosis marries a known carrier of the CF mutation, the probability that their child will develop cystic fibrosis depends on the genotype of the phenotypically normal individual. Since cystic fibrosis is an autosomal recessive disorder, an individual must inherit two recessive alleles to express the disease. Carriers of the CF mutation, who are phenotypically normal, possess one dominant (normal) and one recessive (mutant) allele. If we assume the phenotypically normal individual is a carrier due to having a sibling with the condition (which suggests that both of their parents would be carriers), the individual would most likely have a genotype of Ff (where 'F' is the normal allele and 'f' is the mutant allele).
When a carrier of cystic fibrosis (Ff) mates with another carrier (Ff), the Punnett square shows that their offspring have a 25% chance of having cystic fibrosis (ff), a 50% chance of being a carrier (Ff), and a 25% chance of being phenotypically normal and not a carrier (FF). However, if the phenotypically normal individual is not a carrier (FF), the chance that their child with a known carrier (Ff) will have cystic fibrosis is 0%, and the chance of being a carrier is 50% (Ff). Without knowing the exact genotype of the phenotypically normal individual, we cannot calculate the exact probability.