What is the probability of a child being born with cystic fibrosis if both parents are heterozygous?

In genetics, cystic fibrosis is an autosomal recessive disorder, which means that a child must inherit two copies of the mutated gene (one from each parent) to be affected by the disease. When both parents are heterozygous for the cystic fibrosis allele, they carry one normal allele (let's denote it as "N") and one mutated allele (denote it as "c"). The possible allele combinations for their offspring can be determined using a Punnett square.

Each parent can contribute either the normal or mutated allele, resulting in the following possible combinations for their children:

1. NN (normal)

2. Nc (carrier, not affected)

3. Nc (carrier, not affected)

4. cc (affected by cystic fibrosis)

From these combinations, there's one instance of a child being affected by cystic fibrosis (cc), and three instances where the child would either be normal (NN) or a carrier (Nc). Thus, the probability that a child will be born with cystic fibrosis – requiring the cc combination – is 1 out of 4, or 25%.

This understanding of inheritance patterns for recessive genetic disorders is crucial in genetics and helps in assessing the risk for future generations.