Answer:
3 and 4 are both Aa
8 and 9 cannot be h0m0zygous.
Explanation:
Individual 3 and 4 are non-shaded, which means they are either AA or Aa.
They have three children, 9, 10, and 11. 9 and 11 are also non-shaded, meaning they are either AA or Aa.
However, 10 is shaded, which means he is aa.
For 10 to be aa, he must have gotten one a allele from each parent. Therefore, 3 and 4 must both be heterozygous Aa.
8 and 9 are both non-shaded, which means they are either AA or Aa.
They have three children, 13 is non-shaded, so either AA or Aa.
12 and 14 are both shaded, which means they are both aa. They much each have inherited an a allele from each parent.
This means it is impossible for individuals 8 and 9 to be h0m0zygous.
