For a couple with a known heterozygous gene for cystic fibrosis, what should the nurse tell them about their baby’s probability of having the disease?

In the context of cystic fibrosis, the disease is inherited in an autosomal recessive manner. This means that an individual must inherit two copies of the faulty gene—one from each parent—to express the disease.

For a couple where one parent is heterozygous (carries one copy of the faulty gene, also known as a carrier) and the other parent is also heterozygous, the possibilities for their offspring’s genotype can be analyzed using a Punnett square. The combinations of the parent genotypes would lead to the following potential outcomes for their child:

• 25% chance of being homozygous recessive (inheriting the disease),

• 50% chance of being heterozygous (a carrier, but not affected by the disease),

• 25% chance of being homozygous dominant (neither a carrier nor affected).

Based on this, there is a 25% chance that the baby will inherit the cystic fibrosis gene from both parents and therefore have cystic fibrosis. The categorization "25% or less chance" reflects the actuality that while there is a direct chance of 25%, it is not more than that. Thus, the correct probability that the couple should be informed about