Alkaptonuria is an infrequent autosomal recessive condition. It is first noticed in newborns when the urine in their diapers turns black upon exposure to air. The condition is caused by the defective transport of the amino acid phenylalanine through the intestinal walls during digestion. About 4 people per 1000 are carriers of alkaptonuria.
Sara and James had never heard of alkaptonuria and were shocked to discover that their first child had the condition. Sara's sister Mary and her husband, Frank, are planning to have a family and are concerned about the possibility of alkaptonuria in one of their children.
The four adults (Sara, James, Mary, and Frank) seek information from a neighbor who is a retired physician. After discussing their family histories, the neighbor says, "I never took genetics, but I know from my many years in practice that Sara and James are both carriers of this recessive condition. Since their first child had the condition, there is a very low chance that the next child will also have it, because the odds of having two children with a recessive condition are very low. Mary and Frank have no chance of having a child with alkaptonuria because Frank has no family history of the condition." The two couples each have babies and both babies have alkaptonuria.


What are the genotypes of the four adults?

Verified step by step guidance

1. Step 1: Understand the inheritance pattern of alkaptonuria. It is an autosomal recessive condition, meaning that an individual must inherit two copies of the defective allele (one from each parent) to express the condition. Carriers have one defective allele and one normal allele, and they do not show symptoms.
2. Step 2: Analyze the information about Sara and James. Their first child has alkaptonuria, which means the child inherited one defective allele from each parent. Therefore, both Sara and James must be carriers (heterozygous) for the condition. Their genotype is represented as Aa, where 'A' is the normal allele and 'a' is the defective allele.
3. Step 3: Consider Mary and Frank's situation. Despite the neighbor's claim that Frank has no family history of the condition, their child also has alkaptonuria. This indicates that both Mary and Frank must be carriers of the defective allele. Like Sara and James, their genotypes are also Aa.
4. Step 4: Use a Punnett square to confirm the inheritance probabilities. For each couple (Sara and James, Mary and Frank), a Punnett square can be constructed using their genotypes (Aa x Aa). This will show the probabilities of their children inheriting different combinations of alleles: AA (normal), Aa (carrier), or aa (affected).
5. Step 5: Summarize the genotypes of the four adults based on the analysis. Sara and James are both carriers (Aa), and Mary and Frank are also carriers (Aa). This explains why both couples had children with alkaptonuria, as the condition arises when a child inherits the 'aa' genotype.

Key Concepts

Autosomal Recessive Inheritance

Autosomal recessive inheritance occurs when a trait or condition is expressed only when an individual has two copies of the recessive allele, one inherited from each parent. In the case of alkaptonuria, both parents must be carriers (heterozygous) for their child to inherit the condition. If both parents are carriers, there is a 25% chance with each pregnancy that their child will be affected.

Carrier Status

A carrier is an individual who possesses one copy of a recessive allele for a genetic condition but does not exhibit symptoms of the condition themselves. In the context of alkaptonuria, both Sara and James are carriers, meaning they each have one normal allele and one defective allele. This carrier status is crucial for understanding the risk of passing the condition to their offspring.

Genotype Determination

Genotype determination involves identifying the specific alleles present in an individual. For the four adults in the scenario, Sara and James must have the genotype 'Aa' (where 'A' is the normal allele and 'a' is the allele for alkaptonuria), while Mary and Frank's genotypes need to be inferred based on their family history. Since Frank has no family history of the condition, he is likely 'AA', while Mary’s genotype could be 'AA' or 'Aa', depending on her family background.