Cystic fibrosis is a genetic disorder caused by a recessive allele. if two parents each carry an allele for the disorder but have normal phenotypes, indicate what percentage of their children will be phenotypically normal and what percentage will have cystic fibrosis.

75% normal phenotype, 25% cystic fibrosis.

Explanation:

Since cystic fibrosis is caused by a recessive allele, then it needs both affected alleles to develop. If the parents have normal phenotyps but they carry one allele for the disorder, then their genotypes will be Cc for both (they are heterozygous).

C is the dominant allele that codes for the trait, and c is the affected allele.

They only have one copy of the affected allele, thereby they will not develop cystic fibrosis.

The next step is to find out the genotypes of the gametes.

Gametes are sex cells, sperm or eggs produced by the parents. Those cells only have one allele of the gene, that means the gametes produced by them can be either C or c. Those gametes are used in the punnett square as shown in the picture.

There we can see 50% of the offspring is Cc, 25% is cc and 25% is CC.

As it was stated before, individuals who are cc will develop cystic fribrosis. Then, 25% of them will have it. And 75% of them (25% of CC + 50% of Cc) will not.

well, i believe that the answer is 2. : 3