Classic hemophilia is transmitted as a recessive X-linked trait. What is the probability

Classic hemophilia is transmitted as a recessive X-linked trait. What is the probability of having a sick child from a marriage of a healthy woman with a man suffering from hemophilia?

Let’s designate the chromosome containing the gene that causes the development of hemophilia as Xg, then the chromosome with the gene that determines the well-being in relation to this disease will be XG.

A man with hemophilia has the Xg Y genotype. They produce two types of spermatozoa – Xg and Y. To their daughters, they transmit a chromosome containing a pathological gene – Xg, to sons – a hemophilia-neutral Y chromosome.

A healthy woman can be both homo- and heterozygous. Let’s consider both options.

1. A healthy woman with a genotype homozygous for the gene for normal blood coagulation – XGXG. It produces the same type of XG eggs.

The offspring of a married couple of a hemophilic man and such a woman will be represented by the following options:

heterozygote girls who are carriers of the pathological gene (XGXg) – 50%;

healthy boys (XG Y) – 50%.

The probability of having a sick child is 0%.

2. A healthy woman with a heterozygous genotype – XGXg. It produces two types of eggs – XG and Xg.

The offspring of a married couple of a hemophilic man and a heterozygous woman will include options:

girls with hemophilia (XgXg) – 25%;

heterozygote girls who are carriers of the pathological gene (XGXg) – 25%;

healthy boys (XG Y) – 25%;

boys with hemophilia (Xg Y) – 25%.

In this case, the probability of having a sick child is 50%.