Question

What will the progeny be if a woman carrier for haemophilia marries a normal man?

Hint:

In X-linked recessive traits like haemophilia, the disease is transmitted from a carrier mother to her sons (Criss-cross inheritance).

Answer 1

Step 1: Haemophilia is an X-linked recessive disorder. The genotype of the carrier woman is $X^H X^h$ and the normal man is $X^H Y$.
Step 2: The possible gametes are: Woman ($X^H, X^h$) and Man ($X^H, Y$).
Step 3: The Punnett square results in: $X^H X^H$ (Normal daughter), $X^H Y$ (Normal son), $X^H X^h$ (Carrier daughter), and $X^h Y$ (Haemophilic son).
Step 4: Conclusion: Half of the male offspring will suffer from the disease, while all female offspring will have normal vision/phenotype (though half will be carriers).