Question

A Hemophiliac man marries a normal woman. Their offsprings will be?

Answer 1

Hint: Hemophilia is a sex-linked recessive disorder. The defective gene for this trait is located in the X-chromosome. This is a disease which causes failure of coagulation of blood due to defective blood protein factors present.

Complete answer:
Now since hemophilia is an X-linked recessive disease, the females who possess two XX chromosomes, can have a defective X chromosome for this trait yet exhibit normal phenotypes, i.e., they can become carrier (or heterozygous). But the males have only one X chromosome and so, if a male progeny inherits a defective X chromosome, then he is bound to become hemophilic. These types of hemophilia males often die before attaining maturity. Hemophilia females (homozygous for the defective trait) are also rare since the mother of such a child needs to be at least a carrier and the father should be hemophilic (unviable in the later stage of life).
Now, this means a normal female can be a carrier for this disease.
Thus, there are two possibilities while determining the nature of the offspring:
1. A hemophiliac man marries a normal woman who is phenotypically normal but is a carrier (heterozygous for the trait).
2. A hemophiliac man marries a normal woman (who does not carry any defective trait for the disease).
Thus, we shall discuss both the possibilities one by one.
A. A hemophiliac man marries a normal woman who is phenotypically normal but is a carrier (heterozygous for the trait):
Let us consider the defective X chromosome to be X^h and the normal X chromosome to be X.
Therefore, the genotype of the man is (22+X^hY) and the genotype of the carrier woman is (22+XX^h).
While making the cross, we will only consider the sex chromosomes.
Thus, we draw the Punnett square table:

FEMALE ↓ MALE →	Xh	Y
X	XhX (normal carrier female)	XY (normal male)
Xh	XhXh (hemophilia female)	XhY (hemophilic male)

Thus, we can see that phenotypically, there is 50% chances that the offspring born would be normal and 50% chances that the offspring born would be hemophilic.
And, if we consider genotypically, then 75% of the offspring will bear the defective X chromosomes.
B. A hemophiliac man marries a normal woman (who does not carry any defective trait for the disease):
Here the genotype of the man is (22+X^hY) and the genotype of the carrier woman is (22+XX).
Thus, we draw the Punnett square table:

FEMALE ↓ MALE →	Xh	Y
X	XXh (normal carrier female)	XY (normal male)
X	XXh (normal carrier female)	XY (normal male)

Thus, in this case we can see that phenotypically, there is a 0% chance that the offspring born would be hemophilic.
And, if we consider genotypically, then 50% of the offspring will bear the defective X chromosome.
And, one other notable fact is that the males born from this possible couple will never be hemophilic, but the females born will always be the carrier.

Note:
One point to be remembered is that one has to consider both the possibilities while considering cases of a normal female since such a kind of normal female can also be a carrier and a carrier carries the defective trait but does not express it phenotypically.