In humans alkaptonuria is a metabolic disorder in whichaffected persons produce black urine. Alkapotonuria results from anallele (a) that is recessive to the allele for normal metabolism (A). Sally has a normal metabolism, but her brother hasalcaptonuria. Sally's father has alcoptonuria, and her mother hasnormal metabolism.

a) Give the genotype of Sally, her mother, father and herbrother.

b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria?

c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?

a) The genotype of her father = "aa".

The genotype of her mother = "Aa".

The genotype of Sally is "Aa".

b) 1/2

c) 1/2

Explanation:

a) Sally's father has alkaptonuria. So, the genotype of her father is "aa". Her mother is normal. Since her brother is affected by the disease, he should be homozygous recessive (aa). This means that the mother also has one copy of "a" allele and is heterozygous dominant (Aa). A cross between Aa (mother) and aa (father) would give 50% affected progeny with "Aa" genotype. So, the genotype of Sally is "Aa".

b) The genotype of her mother = Aa

The genotype of her father = aa

Aa x aa = 1/2 Aa (affected) : 1/2 aa (normal)

So, if Sally's parents have another child, there is a 1/2 probability that this child will have alkaptonuria.

c) The genotype of Sally = Aa

Genotype of the man Sally marries = aa (since he is affected with alkaptonuria)

Aa x aa = 1/2 Aa (affected) : 1/2 aa (normal)

There is a 1/2 probability that their child will have alkaptonuria.