Question

Hemophilia is an X-linked recessive trait. If a non-hemophiliac man mates with a non-hemophiliac woman and they have a hemophiliac son, (a) what is the probability that their next son will have hemophilia? (b) What phenotypes and genotypes are possible for their daughters?

Answer 1

(a) 1/2

(b) Possible Genotype =
`X^HX^H and X^HX^h`

Possible phenotype = non-haemophilic

Step-by-step explanation:

Let the allele for haemophilia be h, the alternate form would be H.

Genotype of non-haemophilic man would be
`X^HY`

Genotype of non haemophilic woman could be either
`X^HX^H or X^HX^h`

If they have an haemophilic son, it means the genotype of the woman is
`X^HX^h`

Crossing the two:

`X^HY` x
`X^HX^h` =
`X^HX^H, X^HX^h, X^HY, X^hY`

(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
`X^HX^H and X^HX^h` and both of them have no haemophilia.