Genetics question help

[x-linked]

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if a male hemophiliac (XhY) is crossed with a female carrier of both color-blindness and hemophilia (XcXh), what is the probability that a female child will be phenotypically normal?

 

[coodoo]

if a male hemophiliac (XhY) is crossed with a female carrier of both color-blindness and hemophilia (XcXh), what is the probability that a female child will be phenotypically normal?

answer = 50 percent?

 

[x-linked]

answer = 50 percent?

yes, but how to do it 😕

 

[sinned]

in sex linked genes,
the female child needs to have both genes of a single disorder in order to get a phenotypic disorder. the male child needs to only have a X chromosome with the disorder and he will get a phenotypic disorder.

so do a cross, and you will get XcXh, XhXh, XcY, XhY as your generations...well for the females, you have XcXh, you need both of the same disorders to be on both of the chromosomes, in this case its different, so it has NO disorders...on the other hand XhXh has both same disorders on both of the chromosomes, so it will be hemophilic...so out of 2 possible females only one has a disorder that is hemophilic...so 50%

 

[coodoo]

notice that the question states female children and not all children.

 

[x-linked]

notice that the question states female children and not all children.

can you explain? 😀 😀 😀 THANK YOU

 

[coodoo]

yeah no problem. As the other poster already alluded to, the cross b/w the parents produces 4 children. Of these 4 children, two are females. Of these 2 FEMALES, one of them produces a phenotypic trait. Hence, it is 50 percent

 

[x-linked]

thank you, sinned and coodoo ! ! ! !