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?

Question

Litzy Bird

Biology

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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Answers (1)

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Grayson Mitchell · Answer 1

Key:

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.

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?

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?