A man who is heterozygous for the disease haemophilia marries a woman who is double recessive for haemophilia. What percentage of their offspring would have...

A man who is heterozygous for the disease haemophilia marries a woman who is double recessive for haemophilia. What percentage of their offspring would have the disease?

Answer Details

Haemophilia is an X-linked recessive genetic disorder. This means that the gene responsible for haemophilia is located on the X chromosome, and individuals with two copies of the defective gene (one inherited from each parent) will have the disease. The man in this scenario is heterozygous, meaning he has one normal X chromosome and one X chromosome with the defective gene. The woman is double recessive, meaning she has two copies of the defective gene. When the man and woman have offspring, each child will inherit one of their father's X chromosomes and one of their mother's X chromosomes. The possible combinations of X chromosomes the children could inherit are: - Normal X chromosome from the father and normal X chromosome from the mother - Normal X chromosome from the father and defective X chromosome from the mother - Defective X chromosome from the father and normal X chromosome from the mother - Defective X chromosome from the father and defective X chromosome from the mother Of these possible combinations, only the last one (defective X chromosome from both parents) will result in the child having haemophilia. The probability of a child inheriting a defective X chromosome from the father is 50%, since the father is heterozygous. The probability of a child inheriting a defective X chromosome from the mother is 100%, since the mother is double recessive. Therefore, the probability of a child inheriting a defective X chromosome from both parents and having haemophilia is 50% x 100% = 50%. So the correct answer is: 50% of their offspring would have the disease.