Problem 6 - Campbell Biology

Question

Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? (a) All three children are of normal phenotype. (b) One or more of the three children have the disease. (c) All three children have the disease. (d) At least one child is phenotypically normal. (Note: It will help to remember that the probabilities of all possible outcomes always add up to 1.)

Step-by-Step Solution

Answer

a) 42.19%, b) 57.81%, c) 1.56%, d) 98.44%

Step 1 - Determine Probability per Child

Both parents are carriers, meaning they have one normal allele (N) and one recessive allele (n). The genotypes possible for their children are: NN, Nn, and nn. The probabilities for these are: NN = 25%, Nn = 50%, and nn = 25%. A child has a 75% chance of being phenotypically normal (NN or Nn) and a 25% chance of having the disease (nn).

Step 2 - Calculate Probability for All Three Children Normal (Part a)

To find the probability that all three children are normal, multiply the probability for one child by itself three times: \[ (0.75)^3 = 0.421875 \] Thus, the probability that all three children have a normal phenotype is approximately 42.19%.

Step 3 - Calculate Probability for One or More Children have the Disease (Part b)

The probability that at least one child has the disease is the complement of the event that all three children are normal. Use the previous result: \[ 1 - (0.75)^3 = 1 - 0.421875 = 0.578125 \] Thus, the probability that one or more children will have the disease is approximately 57.81%.

Step 4 - Calculate Probability All Three Children have the Disease (Part c)

To find the probability that all three children have the disease, multiply the probability for one child having the disease by itself three times: \[ (0.25)^3 = 0.015625 \] Thus, the probability that all three children have the disease is approximately 1.56%.

Step 5 - Calculate Probability At Least One Child is Phenotypically Normal (Part d)

The probability that at least one child is phenotypically normal is the complement of all three children having the disease. Use the previous result: \[ 1 - (0.25)^3 = 1 - 0.015625 = 0.984375 \] Thus, the probability that at least one child is phenotypically normal is approximately 98.44%.

Key Concepts

inheritance patterns

Inheritance patterns describe how genetic traits are handed down from parents to offspring. In this exercise, we focus on autosomal recessive inheritance, which is relevant for diseases like hemochromatosis.
In autosomal recessive patterns, both parents must carry one defective copy of the gene for an offspring to express the disease. Individuals have two copies of most genes, one from each parent.
These genetic combinations can follow simple rules:

• If both genes are normal (NN), the person is healthy.
• If one gene is normal and the other is defective (Nn), the person is a carrier but does not show disease symptoms.
• If both genes are defective (nn), the person will express the disease.

Understanding these rules helps us predict the likelihood of different genetic outcomes in children.

recessive alleles

A recessive allele is a gene variant that does not produce a trait unless paired with another recessive allele. Unlike dominant alleles, which can mask the presence of recessive alleles, recessive traits only show up when an individual has two copies of the recessive allele.
In our hemochromatosis example, 'n' represents the recessive allele causing the disease. If a child inherits 'n' from both parents, they will have hemochromatosis. If they inherit 'N' from at least one parent, they will not have the disease. This understanding is crucial for predicting phenotypic outcomes.
Knowing whether an allele is recessive helps predict the combination of genes possible in offspring and enables accurate calculation of phenotypic probabilities.

Punnett square calculations

A Punnett square is a grid used to predict the genetic outcome of a cross between two individuals. It maps the possible combinations of parental genes and helps visualize inheritance patterns.
For our exercise with hemochromatosis, both parents are carriers, with genotypes Nn. The Punnett square for two Nn carriers would show:

• 25% chance (NN)
• 50% chance (Nn)
• 25% chance (nn)

These probabilities help us calculate the likelihood of different phenotypes in their children.
Probabilities derived from a Punnett square enable precise genetic predictions and are foundational to genetics problem-solving.

phenotypic probability

Phenotypic probability refers to the likelihood of an individual displaying a particular trait based on their genetic makeup. In the context of our exercise, we are looking at the probability that children will show a normal or diseased phenotype.
For example, a single child of two carriers has a 75% chance of having a normal phenotype (NN or Nn) and a 25% chance of having a diseased phenotype (nn). By extending this to three children, we use multiplication to find probabilities for all children:

• All normal: \(0.75^3 = 0.421875\)
• One or more diseased: \(1 - 0.421875 = 0.578125\)
• All diseased: \(0.25^3 = 0.015625\)
• At least one normal: \(1 - 0.25^3 = 0.984375\)

Understanding phenotypic probabilities allows us to predict the occurrence of traits over multiple generations.