Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles, Elaine, nor their parents have cystic fibrosis.)

Verified step by step guidance

1. Determine the genetic status of Charles and Elaine. Since Charles had a child with cystic fibrosis (CF) and CF is a recessive disorder, Charles must be a carrier of the CF gene (Ff). Similarly, since Elaine's brother had CF and neither of her parents had the disease, it is likely that both of her parents were carriers, making Elaine likely to be a carrier as well (Ff).
2. Set up a Punnett square with both Charles and Elaine as carriers (Ff x Ff). This will help visualize the possible genetic outcomes for their children.
3. Fill in the Punnett square. The possible genotypes of the offspring will be: FF (normal), Ff (carrier), Ff (carrier), and ff (affected by CF).
4. Calculate the probability of each genotype. Since there is one ff combination out of four possible outcomes, the probability that Charles and Elaine will have a child with CF is 1/4 or 25%.
5. Explain the results to the couple, emphasizing that there is a 25% chance with each pregnancy that their child will have cystic fibrosis, a 50% chance the child will be a carrier, and a 25% chance the child will neither have CF nor be a carrier.