Hemophilia (a blood clotting disorder) is determined by an X-linked recessive ge

Question

Hemophilia (a blood clotting disorder) is determined by an X-linked recessive gene (alleles are H and h). Brachydactyly (abnormally short digits) is caused by an autosomal gene (allele is B). A man who has hemophilia and normal digits is married to a woman who displays brachydactyly.   A genetic test indicated that she is a carrier for hemophilia. The man’s mother and father were phenotypically normal. The woman’s mother displayed brachydactyly (but was otherwise normal), and her father was completely normal for these traits. What is the probability that their first child will suffer from both disorders?

Explanation / Answer

It is given that man is hemophilic with normal digits while female is carrier for hemophilia and suffers from Brachydactyly. With reference to hemophilia the genotypes will be XhY and XhXH for man and woman. A cross between them is likely to give offsprings with genotypes XhXh, XhXH, XhY and XHY. This means that there is 50% or 1/2 chance that their offspring will be born with hemophilia. For brachydactyly we know that it is an autosomal dominant disease and the trait for the disease does not skip generation and is continuously passed. Since one parent has normal digits and other suffers from brachydactyly therefore there is a chance that 50% or 1/2 of the offsprings will be affected by this disorder.

Finding the probabilty that their first child will suffer from both diseases we have,

1/2 X 1/2 = 1/4

This means there is a 25% chance that the first child born to this couple will suffer from both the disorders.

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