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I got this question right, but I just wanted to know - for the parental genotypes, how do you decide between using XBXb with XbY and XBxb with XBY?

Like how do you know whether to give the homogametic female bird the recessive b rather than the dominant B?
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Report 1 year ago
Sex-linked conditions usually undergo X-linked recessive transmission, so the abnormal recessive allele (b in your example) is carried on the X chromosome and the Y chromosome acts as a "sleeping partner" (partner as in business partner NOT as in sexual partner) i.e. the Y chromosome has no effect on the phenotype.

So, the condition or disease is transmitted from an affected male XbY through his daughter (who will be a carrier XbXB) to his grandson XbY. If a girl has both parents who are affected (XbXb and XbY) OR if the father is a sufferer XbY AND the mother is a carrier (XbXB) AND the daughter has the genotype XbXb then she will be affected.

If you do a punnet square for each of these scenarios [over a few generations], you will see that the probability of a female suffering from the condition is 1/8 of that of a male.

An example is red-green colour blindness (protanopia) in which males are affected roughly 8 times more frequently than females. Similarly with haemophilia [a bleeding disorder due to lack of clotting factor VIII) which runs in the royal family, and affected mostly kings [not queens] in alternate thrones.

Specifially for your Q, in order to decide whether to use XBXB or XbXb: If the female is affected, she has to be XbXb, but if she is unaffected she wil be XBXB - you will be told if she is a carrier then XbXB.

Having said this, there are very rare situations where the transmission is sex-linked dominant e.g. Vitamin D resistant rickets, and more recently Y-linked conditions have also been described.


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