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.
Distributive property, commutative property, and associative property (I believe. I tried my best but I’m not completely sure that I’m correct. Good luck!)
Answer:
35.235
Step-by-step explanation:
because 78.3*0.45 = 35.235
Answer:
2
Step-by-step explanation:
12-2=m(2+3)
m=10/5=2
m=2
Answer:
m<FLS=108 degrees
m<SLT=72 degrees
m<ALG=18 degrees
Step-by-step explanation:
(3x) + (4x +12) = 180
x = 24
4x + 12 = 108 = m<FLS
180 - 108 = 72 = m<SLT
SLG = 72 + 90 + x
x = 18 = m<ALG
