Hemophilia is another example of a X-linked disease caused when a recessive allele (Xh) is expressed. If a normal male reproduces with a heterozygous normal female, what are the expected genotypes and phenotypes? Will any of their daughters develop hemophilia?

Genotypes: XX(75%), XhX(25%)
Phenotypes: normal
Their daughters won't get it because they would need to be homozygous for it and since the father is normal, it's not possible for them to inherit this from each parents and be homozygous (draw out a punnet square if ur confused). Hope this helps'

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aliasger2709 aliasger2709 · Answer 2 · 2022-04-06T10:27:47+02:00

X-linked diseases are caused by the genes present on the X chromosome of the germinal cell. The genotype is 1:2:1 and the phenotype is 3:1.

What is Hemophilia?

Hemophilia is an X- linked disorder that cannot clot blood and results in excessive bleeding after an injury. It is due to the absence of the blood coagulating factor VIII.

The genotype male parent = [tex]\rm X^{h}Y\\[/tex]

The genotype female parent = [tex]\rm X^{h}X^{H}[/tex]

The cross between the normal male and heterozygous normal female is shown in the image attached below.

From the cross the genotype: carrier : normal : diseased = 1:2:1

The phenotype will be: normal: diseased = 3:1

No, none of the daughters will develop hemophilia as they are not purebred and that is not possible as both the parent needs to be purebred for the inheritance.

Therefore, the genotype is 1:2:1 and the phenotype is 3:1.

Learn more about X-linked diseases here:

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