Question

cystic fibrosis (cf) is an inherited disease that affects the respiratory and digestive systems, among others. it is a recessive disease, meaning that offspring must receive a recessive allele from each parent in order to have the disease (genotype - cc). a person may not have the disorder but can carry the allele for the disease (genotype - cc). if a person does not have the disease and does not carry the allele for the disease, their genotype would be cc. if both parents do not have the disease, but both are carriers, what are the chances that they will have a child born with cystic fibrosis? work the punnett square and then choose the correct answer based on your results.

Explanation:

Step1: Determine Parent Genotypes

Both parents are carriers (not having the disease but carry the allele), so their genotype is \( Cc \).

Step2: Set Up Punnett Square

The alleles from each parent: Parent 1 ( \( Cc \)) can contribute \( C \) or \( c \); Parent 2 ( \( Cc \)) can contribute \( C \) or \( c \). The Punnett Square is:

	\( C \)	\( c \)
\( c \)	\( Cc \)	\( cc \)

Step3: Analyze Offspring Genotypes

• \( CC \): 1/4 (no disease, no carrier)
• \( Cc \): 2/4 (no disease, carrier)
• \( cc \): 1/4 (has cystic fibrosis)

Step4: Calculate Probability for \( cc \)

The probability of having a child with genotype \( cc \) (cystic fibrosis) is \( \frac{1}{4} \) or 25%.

Answer:

The chance of having a child with cystic fibrosis is \( \frac{1}{4} \) (or 25%).