In a family where both parents are heterozygous carriers for cystic fibrosis (genotype Ff), what is the probability that their daughter Irene is a carrier (heterozygous) for the disease? Express your answer as a percentage.

• A. 25%
• B. 50%
• C. 100%
• D. 75%

Answer: (B)

Verified step by step guidance

1. Identify the genotypes of the parents. Both parents are heterozygous carriers for cystic fibrosis, so their genotypes are Ff and Ff.
2. Set up a Punnett square to determine the possible genotypes of their offspring. The alleles from each parent (F and f) combine to form the following genotypes: FF, Ff, fF, and ff.
3. List the possible genotypes of the children: FF (homozygous dominant, not a carrier), Ff (heterozygous carrier), fF (heterozygous carrier), and ff (homozygous recessive, affected by cystic fibrosis). Note that Ff and fF are functionally the same genotype (heterozygous carrier).
4. Calculate the probability of each genotype by counting the number of occurrences in the Punnett square. There is 1 FF, 2 Ff (or fF), and 1 ff out of 4 total possibilities.
5. Determine the probability that Irene is a carrier (heterozygous) by dividing the number of heterozygous genotypes by the total number of genotypes: \(\frac{2}{4} = 0.5\), which corresponds to 50%.