Answer:
50% of their children are likely to be carriers of cystic fibrosis
Explanation:
Since the normal allele "F" will be the dominant allele while the mutated CFR allele "f" will be the recessive allele, <u>the gene (pair of alleles) of the person that is a CFR carrier will be "Ff" while that of the normal person who isn't a carrier will be "FF"</u>. The attachment shows the crossing between the two parents. From the illustration in the attachment, for every 4 children given birth to, 2 of them will likely be normal, "FF", (not a carrier and doesn't have cystic fibrosis) while 2 others will likely be carriers of cystic fibrosis (Ff). Hence, 50% of their children are likely to be carriers of cystic fibrosis.
