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 think the answer is C
sorry if i am wrong
Step-by-step explanation:
I ain’t givin u no answer do yo work
f^3+11g-4hf 3 +11g−4hf, cubed, plus, 11, g, minus, 4, h when f=3f=3f, equals, 3, g=2g=2g, equals, 2 and h=7h=7h, equals, 7.
Alik [6]
Given:
The expression is:
The given values are 
.
To find:
The value of the given expression for the given values.
Solution:
We have,
After substituting , we get
Therefore, the value of the given expression is 21.
