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destroyer 221

Discussion in 'DAT Discussions' started by arpitpatel86, Jun 3, 2008.

  1. arpitpatel86

    2+ Year Member

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    Sickle cell anemia is an autosomal recessive trait. if two individuals are heterozygotes, what is the risk of having an affected child from a single mating
    a.50%
    b.10%
    c.0%
    d. 25%

    I thougth it be 50% but the answer is 25% and the solution said: AO A is the heterozygote. Let us set up our Punnett square. Kepp in mind that only A O A O has the disease, the heterozygotes are only carriers


    AO
    A
    AO
    AOAO
    AOA
    A
    AOA
    AA
    thus only 25% would be affect....

    so yea this makes sense but the problem i have is that they did it as if two females had a chld together *not possible i know* but they set up the punnet square pretty much like XO X vs XO X...(xO being a carrier).....why did they do this instead of having XO X for female carrier and XO Y for a male carrier....and i dont think a male can be a carrier neways its either he has the disease or doesnt.....and if that's the case



    XO
    Y
    XO
    XOXO (infected)
    XOY(infected)
    X
    XOX
    XY

    so that be 50%

    i just dont understand the way they did it bc if the male is heterozygote....doesnt that make him have the disease since he doesnt have the other X chromosome like the famale does?
     
    #1 arpitpatel86, Jun 3, 2008
    Last edited: Jun 3, 2008
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  3. Mamona

    5+ Year Member

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    Be careful autosomal and sex-link traits are different and I think you are confusing them together.
     
  4. Orgodox

    7+ Year Member

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    Yea this has nothing to do with X or Y or male or female it says autosomal so equal chances of both and not on sex chromosome. Dont set up the whole punnet sqaure takes too long for this problem.
    Its simple... the mother is heterozygous meaning she has one defected copy and one good one. she must give the defected copy which is a 1/2 chance. The same goes for the father he also has a 1/2 chance of giving the defected copy. Since the 2 events are independent multiply 1/2 X 1/2 and that 1/4 chance the child will get both defected copies and express the recessive trait
     

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