Question

Mutations in CFTR are recessive. If two people that are genetic carriers of Cystic Fibrosis (CTFR/ctfr) have children together, what percentage of children would be homozygous for the normal (non-cystic fibrosis) allele?

a) 0%
b) 25%
c) 50%
d) 75%

Answer 1

The percentage of children who would be homozygous for the normal CFTR allele when both parents are carriers is 25%, as determined by Mendelian genetics and a Punnett square analysis.

Step-by-step explanation:

The question concerns the inheritance pattern of cystic fibrosis, which is an autosomal recessive disorder. In the scenario provided, both parents are genetic carriers of cystic fibrosis, meaning they each have one normal CFTR allele (F) and one mutated CFTR allele (f). According to Mendelian genetics, when two carriers (Ff) have children, there is a 25% chance the children will be homozygous recessive (ff) with cystic fibrosis, a 50% chance they will be carriers (Ff) like their parents, and a 25% chance they will be homozygous dominant (FF) with normal alleles.

Therefore, the percentage of children that would be homozygous for the normal (non-cystic fibrosis) allele is 25%. This is found by combining the probability of one normal allele from the mother with the probability of one normal allele from the father in a classical Punnett square cross where the parental genotypes are Ff x Ff.