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 is 1/6.

Step 1: Determine the inheritance pattern of cystic fibrosis

Cystic fibrosis is inherited in an autosomal recessive pattern, which means a child must inherit one defective gene from each parent to develop the disease.

Step 2: Identify Charles's genotype

Charles had a child with cystic fibrosis in his previous marriage, which means he must be a carrier of the cystic fibrosis allele (genotype: Cc), where 'C' is the normal allele and 'c' is the cystic fibrosis allele.

Step 3: Identify Elaine's genotype likelihood

Elaine's brother died of cystic fibrosis, which means her parents were both carriers (Cc x Cc). The possible genotypes for Elaine are: CC (not a carrier), Cc (carrier), and cc (affected), but since Elaine does not have cystic fibrosis, her genotype is either CC or Cc.

Step 4: Calculate Elaine's carrier probability

Using a Punnett square for Elaine's parents (Cc x Cc), the probability of Elaine being a carrier (Cc) is 2/3, since we exclude the 'cc' scenario (affected). Thus, 1 (affected) out of 4 possibilities, leaving 3, and 2 out of those 3 are carriers.

Step 5: Calculate the probability of the child being affected if both parents are carriers

There is a 1/4 chance of two carriers having a child with cystic fibrosis (Genotypes in this case: CC, Cc, Cc, cc).

Step 6: Combine the probabilities

The combined probability is the product of the probability that Elaine is a carrier (2/3), the probability that Charles is a carrier (certainty, 1), and the probability of two carriers having an affected child (1/4):

Step 7: Calculation and conclusion

Multiplying these probabilities gives \( (2/3) * 1 * (1/4) = 2/12 = 1/6 \). Therefore, the probability that Charles and Elaine will have a child with cystic fibrosis is 1/6.

Key Concepts

autosomal recessive inheritance

Cystic fibrosis is caused by a mutation in the CFTR gene, and it is inherited in an autosomal recessive pattern. This means that for a child to have cystic fibrosis, they must inherit two copies of the mutated gene, one from each parent. In this inheritance pattern, if both parents are carriers (i.e., they have one normal allele and one mutated allele, represented as Cc), they can potentially pass the disease onto their children.

• A child with two normal alleles (CC) is not affected.
• A child with one normal allele and one mutated allele (Cc) is a carrier but not affected.
• A child with two mutated alleles (cc) will have cystic fibrosis.

When neither parent shows symptoms of the disease, but both can pass the mutated gene to their offspring, this inheritance pattern highlights why understanding genetic risks is important for family planning.

Punnett square analysis

Punnett squares are a helpful tool in genetics for predicting the probability of offspring inheriting certain traits. By mapping out the alleles each parent can pass on, we can visualize the potential genetic combinations. For Charles and Elaine's situation, we use the Punnett square to see the likelihood of their child inheriting cystic fibrosis. Since cystic fibrosis follows an autosomal recessive pattern, both parents need to be carriers for there to be a chance of an affected child.
Assuming both parents are carriers (Cc), the Punnett square looks like this:

• Top row and left column: C and c (representing the alleles of each parent).
• Combining each parent's alleles: CC, Cc, Cc, cc.

From this, we see that there is a 1 in 4 chance (25%) that their child could inherit cystic fibrosis (cc), a 2 in 4 chance (50%) they will be carriers (Cc), and a 1 in 4 chance (25%) they will not inherit the mutated gene at all (CC).

carrier probability calculation

In the given scenario, we need to determine the likelihood of Elaine being a carrier for the cystic fibrosis gene. Since her brother had cystic fibrosis, we know both of her parents were carriers (Cc). Using a Punnett square for her parents:

• CC (not a carrier)
• Cc (carrier)
• Cc (carrier)
• cc (affected)

Since Elaine does not have cystic fibrosis, we exclude the 'cc' possibility, leaving us with three potential outcomes: CC, Cc, Cc. Thus, out of these three remaining scenarios, two of them indicate Elaine is a carrier (Cc). Therefore, the probability of Elaine being a carrier is 2 out of 3, or approximately 66.7%.

genotype identification

Identifying parent's genotypes helps assess the genetic risks for their offspring. Charles had a child with cystic fibrosis from a previous marriage, indicating he must be a carrier of the cystic fibrosis allele (Cc). Elaine's brother had cystic fibrosis, confirming her parents were both carriers (Cc). Since Elaine does not have cystic fibrosis, her genotype must be either CC or Cc.
Given the genotypes:

• Charles: Cc (carrier)
• Elaine: Can be either Cc (carrier) or CC (non-carrier)

We then combine the parent's genotypes to calculate the risk their child will have cystic fibrosis. If both parents are carriers (Cc x Cc), the Punnett square predicts a 1 in 4 chance (25%) their child could inherit cystic fibrosis.