Question

Why does the son of a carrier mother and a normal father suffer from haemophilia whereas the son of a haemophiliac father and normal mother would not?

Answer 1

Hint: Haemophilia is a sex-linked recessive disorder which is also known as bleeder's disease. This disease affects the X- chromosome of sex chromosome which leads to failure of the blood clot and in severe cases it proves to be fatal.

Complete answer:
Haemophilia was discovered by John Otto in 1803 and it is also known as bleeder's disease because due to deficiency or failure of blood clotting factor the blood fails to coagulate and lead to continuous loss of blood from the body. It is a sex-linked disease which affects the X chromosome, therefore when women get affected from this disease they became the carrier or when male get effected from this disease they will become the sufferer because in female XX chromosome and male XY chromosome are sex chromosome. In case of carrier mother, the chromosome would be XhX and for normal father, chromosome would be XY when cross takes place four pair of chromosome result out in form of ${X^h}$X, XX, ${X^h}$Y, XY, which show that one boy and one girl is normal whereas one girl is a carrier and one boy is haemophilia, but in case of haemophilia father and normal mother the chromosome of father would be ${X^h}$Y and that of mother would be XX and after crossing the chromosome of offspring would be ${X^h}$X, ${X^h}$X, XY, XY, which shows that only daughter is sufferer whereas son would be normal. In the case of the female when both sex chromosome is haemophilic then the offspring will die before its birth.

Note: The haemophilic disease is of mainly three types Haemophilia A, Haemophilia B and Haemophilia C. Haemophilia A cause by failure of clotting factor VIII and Haemophilia B cause by failure clotting factor IX whereas Haemophilia C occurs in less than 1% of the patient by the failure of clotting factor XI.