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:
I am currently doing that one as well right know and I think that you need two octahedrons, and 8 tetrahedrons
Step-by-step explanation:
28 each book 155-15=140 and then 140÷5=28
Answer:
34.5
Step-by-step explanation:
divide 145 by 4.2
Step-by-step explanation:
W=-6, x=1.2, and z=-6/7
(W²x-3)÷10-z
we substitute
((-6²)(1.2)-3)÷10-(-6/7)
((-36)(1.2)-3) ÷10-(-6/7)
(-43.2-3) ÷10(6/7)
(-46.2)÷60/7
-46.2÷60/7
-46.2*7/60
-46.2/1*7/60
-323.4/60
-5.39
