25%
Explanation:
Let's assume that the allele "c" causes cystic fibrosis in homozygous conditions. The genotype of the carrier parent would be "Cc". A cross between two carrier parents, each with a genotype "Cc" would produce progeny in the following phenotype ratio=
Cc x Cc= 3/4 normal (1/4 CC: 1/2 Cc) : 1/4 affected (1/4 cc).
Therefore, the probability that one of their children will have CF disease is= 25%
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