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13 April, 06:58

If a woman and her husband, who are both carriers, have three children, what is the probability that all three children have the normal phenotype? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/16)

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  1. 13 April, 08:32
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    Answer: 1/64 would be normal

    Explanation:

    Let the woman phenotype be Yy, since a carrier is usually heterozygous;

    Also let the husband phenotype be Yy

    Then Yy seperate to yield two alleles "Y" and "y"

    So, the cross of the two alleles:

    "Y", "y" X "Y", "y"

    F1 will be YY, Yy, Yy and yy

    From the crossing,

    - the normal phenotype are YY,

    -the carriers are Yy, Yy

    - while yy is recessive.

    Since, the probability that one of their offspring is normal is 1/4; for three children, then 1/4 x 1/4 x 1/4 = 1/64

    Thus, probability that all three children are normal is 1/64
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