Can anyone help me with this question??

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.

a- all cells come from pre-existing cells