Final answer:
According to the Punnett square, two carrier parents have a 25% chance of having a child with cystic fibrosis.
The probability that children of parents each carrying a different mutation of the CFTR gene will have cystic fibrosis is 25%, as both parents must pass on their faulty CFTR allele for the child to be affected with the disease.
Step-by-step explanation:
According to the Punnett square, two parents that are carriers (Ff) of the cystic fibrosis gene have a 25% chance of having a child with cystic fibrosis (ff). The affected child must inherit two recessive alleles. The carrier parents are not affected.
The probability that children of parents each carrying a different mutation of the CFTR gene will have cystic fibrosis is 25%, as both parents must pass on their faulty CFTR allele for the child to be affected with the disease.
Cystic fibrosis (CF) is an autosomal recessive disorder, which means that a child must inherit two copies of the faulty CFTR gene, one from each parent, to develop the disease. When one parent carries the R117H mutation and the other carries the G542X mutation in one of their CFTR alleles, there is a 25% chance that a child will inherit both mutated alleles (one from each parent) and therefore have cystic fibrosis. Each parent must pass on the affected CFTR allele for a child to display symptoms of CF. Each parent has one affected and one unaffected allele, so their genotypes are Rf and Gf respectively, where 'R' and 'G' represent the flawed alleles and 'f' represents the normal allele. The possible allele combinations for their offspring are Rf, Gf, rf (normal), and RG (cystic fibrosis). Only the RG combination results in cystic fibrosis, and since it's only one of the four possibilities, the chance is 25%.