Question

question 8
2 points
two parents who have no history of cystic fibrosis and do not, themselves, have cystic fibrosis have a child who has cystic fibrosis. how is this possible?
both parents were homozygous recessive
one parent was homozygous dominant and the other was homozygous recessive
one of the parents was a carrier
both of the parents were carriers

Explanation:

Step1: Understand cystic fibrosis inheritance

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

Step2: Analyze each option

• Option A: If both parents were homozygous recessive (\(aa\)), they would have cystic fibrosis themselves. But the problem states parents do not have cystic fibrosis. So this option is wrong.
• Option B: If one parent was \(AA\) (homozygous dominant) and the other was \(aa\) (homozygous recessive), all their children would be \(Aa\) (carriers) and not have cystic fibrosis (\(aa\)). So this option is wrong.
• Option C: If only one parent was a carrier (\(Aa\)) and the other was \(AA\), children would be \(AA:Aa = 1:1\). No child would have cystic fibrosis (\(aa\)). So this option is wrong.
• Option D: If both parents were carriers (\(Aa\)), using Punnett square:

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

The probability of having a child with \(aa\) (cystic fibrosis) is \(\frac{1}{4}\). Since parents (\(Aa\)) do not have cystic fibrosis (because of dominant \(A\) allele), this option is correct.

Answer:

D. Both of the parents were carriers