Problem 14 - Campbell Biology

Question

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.)

Step-by-Step Solution

Answer

The probability that Charles and Elaine will have a baby with cystic fibrosis is \(\frac{1}{6}\).

Step 1: Understand the Genetics of Cystic Fibrosis

Cystic fibrosis (CF) is a recessive genetic disorder. This means that a child must inherit two copies of the defective gene, one from each parent, to have the disease. If the child inherits only one defective gene, they will be a carrier but will not have the disease.

Step 2: Determine Charles' Carrier Status

Charles has a child from his first marriage with CF, indicating that he must be a carrier (genotype Cc, where C is the normal allele and c is the CF allele). This is because the child with cystic fibrosis must have inherited a recessive allele from Charles.

Step 3: Determine Elaine's Carrier Status Probability

Elaine's brother had CF, which means both of her parents were carriers (Cc). Therefore, each child they had, including Elaine, had a 1/4 chance of having CF (cc), a 1/2 chance of being a carrier (Cc), and a 1/4 chance of having two normal alleles (CC). Since Elaine does not have CF, there is a 2/3 probability that she is a carrier (Cc).

Step 4: Assess the Probability of Both Being Carriers

The probability that Charles is a carrier is 1 (100%) because he has already fathered a child with CF. The probability that Elaine is a carrier is 2/3 (as calculated above). So, the combined probability that both are carriers is: \(\text{P(Carrier)} = 1 \times \frac{2}{3} = \frac{2}{3}\).

Step 5: Calculate the Probability of Their Child Having CF

If both Charles and Elaine are carriers (Cc), the probability of their child having CF (cc) is 1/4 (since each carrier parent has a 1/2 chance of passing the CF gene). So, the probability is: \(\text{P(CF) if both are carriers} = \frac{1}{4}\).

Step 6: Combine the Probabilities

Combine the probabilities calculated in steps 4 and 5 to find the overall probability of their child having CF: \(\text{Overall Probability} = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}\).

Key Concepts

Cystic Fibrosis

Cystic fibrosis (CF) is a genetic disorder that affects the lungs, pancreas, and other organs. It leads to the production of thick, sticky mucus that can clog airways and trap bacteria. This often results in severe respiratory and digestive problems.
CF is caused by mutations in the CFTR gene. The most common mutation is called ΔF508. People with CF must inherit two copies of the defective gene, one from each parent. This genetic pattern classifies CF as a recessive genetic disorder.
It's important to know that not everyone with a single CF gene mutation will develop the disease. They are known as 'carriers'. They can pass the defective gene to their offspring, but they do not exhibit the symptoms themselves.

Recessive Genetic Disorders

Recessive genetic disorders require an individual to inherit two copies of the defective gene to show symptoms. Only if both parents are carriers can they pass the disorder to their children.
Here's a breakdown of what can happen if both parents carry one defective gene each:

• There's a 25% (1 in 4) chance that the child will inherit both defective genes, resulting in the disorder.
• There's a 50% (1 in 2) chance that the child will inherit one defective gene and be a carrier without symptoms.
• There's a 25% chance that the child will inherit two normal genes.

Understanding this pattern helps in predicting the risks of passing a recessive genetic disorder, like cystic fibrosis, to offspring.

Carrier Probability

Carrier probability helps determine the chances a person has of being a carrier for a certain genetic disorder. In the context of cystic fibrosis:
Charles already has a child with CF from his first marriage. This confirms he is a carrier. We represent carriers with the genotype Cc (C for the normal gene and c for the CF gene).
Elaine's brother had CF, which means both of her parents were likely carriers (Cc). Elaine could have inherited one of three genotypes: CC (not a carrier), Cc (carrier), or cc (affected). Since Elaine does not have CF, the possibilities narrow down to CC or Cc.
Statistically speaking, the probability she is a carrier is 2/3 or about 67%. Assuming both are carriers, the probability of their child having CF is calculated by combining the probability of both parents being carriers with the probability of two carriers having a child with CF:
\[\text{P(child has CF) = P(both carriers) \times P(CF child from two carriers)} = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} \] So, there is a 1 in 6 chance Charles and Elaine will have a child with CF.