The answer is #1. Hope this helps
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
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You use division
over 2 or 4 or 3 or 5 or 7 and so on
in this case 2 is good
64 = 2 * 2 * 2 * 2 * 2 * 2 ( use division) = 2^6
16 = 2 * 2 * 2 * 2 = 2^4
number reminder
64 0
32 0
16 0
8 0
4 0
2 0
1 1
so 64 = 2^6 (number of zeros) no reminder
1.2 8 x 15 percent = 8 x .15 = 1.2
