Cystic fibrosis is a genetic disorder. it is a recessive trait that is determined by two alleles. suppose two parents each carry at least one recessive allele, but neither of them has cystic fibrosis. which statement reflects the likelihood of a child of these parents having cystic fibrosis?
a. the child cannot have the disease because only the dominant alleles will be passed on.
b. the child will have the disease because only recessive alleles will be passed on.
c. the child may or may not have the disease because a recessive gene from each parent may be passed on.
d. the child may or may not have the disease because the disease does not depend on the alleles of the parents.

The correct answer is C. The child may or may not have the disease because a recessive gene from each parent may be passed on.

Cystic fibrosis is inherited as an autosomal recessive disorder i.e. it requires homozygous recessive condition in order to express its phenotype.

The heterozygous organisms act as a carrier of the diseases but do not have the disease.

According to the question, both the parents are heterozygous.

Let us assume C and c be the alleles for cystic fibrosis. Hence, CC will be non-carrier normal individual, Cc will be the carrier normal individual and cc will be the diseased person.

Now both the parents are Cc i.e. carrier or heterozygous.

Gametes produced by them will be C and c (in both)

Cross is shown below in an attached file.

It results in the formation of three genotypes CC, Cc, and cc in 1:2:1 respectively.

Hence, 1 out 4 can be the diseased offspring.

b

is correct hope this   also i learnd this in middle school