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.
Answer:
Step-by-step explanation:
Solve for 
:
-Multiply both sides of the equation by the denominator 
:
-Subtract both sides by 
:
So, the value of is 
.
<span>CAE=95
GAE=90
CAG=95-90=5
ACG=5
CGA=180-(5+5)=170
CBA=12—170=85</span>
