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?

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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C) reflection (i also looked up the other choices just to make sure)