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IrinaVladis [17]
3 years ago
13

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

Biology
1 answer:
Andre45 [30]3 years ago
6 0
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.


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