Haemophilia is a recessive X-linked trait. A man who does not have hemophilia marries a woman who is a carrier. what is the probablity, compared to all possibilities, they will have?
Hemophilia is inherited as a sex-linked gene found on the X chromosome. Let us use the symbol X to represent a person who does not have hemophilia(normal) and X- to represent an X chromosome with the hemophilia gene. Since women have two X chromosomes, they can have the following genotypes: XX(normal) XX- (female carrier but still normal because the normal gene is dominant) and X-X- ( female with hemophilia). Males are hemizygous for the X chromosome because they are XY. Therefore, there are only two possible genotypes for males: XY(normal male) and X-Y (male with hemophilia). In the problem, the male is normal XY and the female is a carrier X X-. Therefore, when they produce gametes by meiotic cell division, the male with make an X sperm and a Y sperm. The female will make an X egg and an X- egg. The genotypic results are as follows-
XX (normal female)
X X- (carrier female)
XY (normal male)
The phenotypes are: 25% normal female, 25% normal male, 25% carrier female and 25 % male hemophiliac.
Sex linkage or sex-linked inheritance is the transmission of characters and their determining genes along with sex determining genes which are born on the sex chromosomes and, therefore, are inherited together from one generation to the next. Hemophilia is popular example of sex linked inheritance in human beings. It is a recessive character and is, therefore, masked in the heterozygous condition. Individuals suffering with this disease lack a factor responsible for clotting of blood. Consequently, even a minor cut may cause prolonged bleeding leading to death. Since it is recessive character, a lady may carry the disease and would transmit the disease to 50% of her sons, even if the father is normal. 100% female will be normal and 50% male will be affected.