Question

what is the chance that two carriers of the cystic fibrosis allele will have a carrier child?
a 0%
b 50%
c 100%
d 25%

Explanation:

Step1: Determine the genotypes of the parents

Cystic fibrosis is an autosomal - recessive disorder. Let the normal allele be \(A\) and the mutant allele be \(a\). Carriers have the genotype \(Aa\).

Step2: Set up the Punnett square

Gametes of Parent 1	Gametes of Parent 2
\(A\)	\(AA\)	\(Aa\)
\(a\)	\(Aa\)	\(aa\)

The genotypes of the offspring are \(AA:Aa:aa = 1:2:1\)

Step3: Calculate the probability of having a carrier child

Carriers have the genotype \(Aa\). From the Punnett square, the probability of \(Aa\) is \(\frac{2}{4}=\frac{1}{2} = 50\%\)

Answer:

B. 50%