Question

Hemophilia is a recessive sex-linked trait (Xh) caused by a defective gene. The blood of individuals with this condition does not clot properly. Without injections of synthetic clotting factors, hemophiliacs are at risk of dying due to excessive bleeding. Make a Punnett square to show a cross between an affected male and a female who is heterozygous.

Give the expected genotype and phenotype.
How likely are they to have a sick child?

Answer 1

See the answer and explanation below.

Step-by-step explanation:

Hemophilia is sex linked. Traits that are sex linked are found on the sex chromosomes. For a man, the sex chromosome if XY and for a woma, the sex chromosome if XX. The Y chromosome in man is hypothesized not to carry any gene and hence, a man only need one allele of sex-linked traits in order to be affected while a woman would need two. Also, it means that a man cannot be heterozygous.

For hemophilia:

An affected male will have the genotype:
`X^hY`

Heterozygous female will have the genotype:
`X^HX^h`

`X^hY` x
`X^HX^h`

progeny:
`X^HX^h, X^hX^h, X^HY, X^hY` (See the attached image for the Punnet's square).

Genotype and phenotype

`X^HX^h` = Physically normal

`X^hX^h` = Hemophilic

`X^HY` = Physically normal

`X^hY` = Hemophilic

From the resulting progeny, 2 out of 4 are hemophilic. Hence the likelihood of having a sick child is 2/4 which is equal to 1/2 or 50%.