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.
We are given :
The ratio of orange juice to pineapple juice in tropical treat punch = 4:3 or 4/3.
Number of oz of orange juice = 64 oz.
Let us assume number of oz pineapple juice does he need = p.
We can setup an proportion:
64 : p = 4 : 3
On cross multiplication we get
64 × 3 = 4 × p
192 = 4p
Dividing both sides by 4, we get
p = 48.
<h3>Therefore, he needs 48 oz of pineapple juice.</h3>
If it is to match 3+1 which is 4 the value to satisfy X is -1
Answer:
dependent
Step-by-step explanation:
