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anonymous

  • one year ago

If a women who is heterozygote for color-blindness married to a color-blind man with normal vision,the probability that her daughter is color blind is: a.0% b.25% c.50% d.75% e.100%

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  1. Rushwr
    • one year ago
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    it's going to be 50% answer C. Colourblindness is recessive. Parental genotypes are going to be \[X ^{C}X ^{c} \] from mother and \[X ^{c}Y\] from mother. Here there are @ sons and @ daughters. Nut out of the 2 daughters one will be colour blind and the other one will be heterozygote. Hence we take it as 50%

  2. anonymous
    • one year ago
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    @rushwr It is not just because it is recessive. The gene(s) involved are only on the X chromosome. Males only have 1 X chromosome so if they have faulty gene(s) they will always pass this along to their daughter. Females have 2 X chromosomes and now the dom/res gene thing comes into play.

  3. Rushwr
    • one year ago
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    yeah that is not what I meant ! Isn't it a fact that all the sex linked disorders involves X chromosomes only.

  4. anonymous
    • one year ago
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    @Rushwr There are some Y linked diseases. Most are not terrible because those usually die out as all men with it would be affected. You still need to know both if it is dom/res and the relationship with the sex chromosomes. If it was a dominant mutation then all the daughters of the couple would have it, despite the heterozygocity of the mother. Since it is now, only 50% will.

  5. anonymous
    • one year ago
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    thanks @Rushwr @mrdoldum

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