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24 February, 02:15

Both members of a couple are heterozygous for a recessive mutation that causes tay-sachs disease (normal = a, mutant =

a. if they have six children, what is the probability of them having two affected children and four unaffected ones?

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  1. 24 February, 02:39
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    Tay-sachs disease - aa/Aa

    Normal - AA

    Since both parents are heterozygous

    P1 = Aa x Aa

    by doing punett square

    Possible genotypes: AA, Aa, Aa, aa

    75% chance that they will have affected children.

    25% chance that they will have unaffected children if the disease is a recessive.
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