Answer:
(a) 1/2
(b) Possible Genotype = [tex]X^HX^H and X^HX^h[/tex]
Possible phenotype = non-haemophilic
Explanation:
Let the allele for haemophilia be h, the alternate form would be H.
Genotype of non-haemophilic man would be [tex]X^HY[/tex]
Genotype of non haemophilic woman could be either [tex]X^HX^H or X^HX^h[/tex]
If they have an haemophilic son, it means the genotype of the woman is [tex]X^HX^h[/tex]
Crossing the two:
[tex]X^HY[/tex] x [tex]X^HX^h[/tex] = [tex]X^HX^H, X^HX^h, X^HY, X^hY[/tex]
(a) One out of the two sons is affected. Hence the probability of their next son having haemophilia is 1/2.
(b) the possible genotypes for their daughters are [tex]X^HX^H and X^HX^h[/tex] and both of them have no haemophilia.
