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B
X0Y
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C
aa
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D
Aa
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Solution
The correct option is DAa As the child from the first parents is diseased that means the parents are the carrier of the disease. Individual 2 is diseased and the other parent is normal according to pedigree. The normal parent must be a carrier in order to express the disease in the next generation. Now the marriage between diseased and normal parent results in all normal children that mean mother is homozygous dominant. So, the genotype of the individual II (3) is heterozygous. Thus, the correct answer is D.
Since 3 one is carrier so its genotype will be Aa.