Question

a healthy, non - carrier man has a baby with a woman with cystic fibrosis. what is the chance that their baby will have cystic fibrosis? 100% chance 75% chance 50% chance 25% chance 0% chance

Explanation:

Step1: Understand cystic fibrosis inheritance

Cystic fibrosis is an autosomal - recessive disorder. Let the normal allele be represented by 'A' and the mutant allele by 'a'. The woman with cystic fibrosis has a genotype of 'aa' since it is a recessive disorder. The healthy non - carrier man has a genotype of 'AA'.

Step2: Determine possible genotypes of offspring

When 'AA' (man) and 'aa' (woman) have children, all of their offspring will have the genotype 'Aa' through the cross:
\[

$$\begin{array}{|c|c|c|} \hline & a&a\\ \hline A&Aa&Aa\\ \hline A&Aa&Aa\\ \hline \end{array}$$

\]
Since having the 'aa' genotype is required to have cystic fibrosis and none of the offspring have the 'aa' genotype, the probability of the baby having cystic fibrosis is 0.

Answer:

E. 0% chance