Answer:
It controls opening and closing movements of the stomata, but that isn't one of the answer choices.
Explanation:
If you're answers consist of:
<span>
a. mutation
b. genetic drift
c. inheritance of acquired characteristics
d. natural selection
Then you're answer is D.) Natural Selection
Hope this helped!! :)</span>
Answer:
6,25%
Explanation:
Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:
Parents: Aa X Aa
Children: AA(A x A), Aa(A x a), Aa (a x A) and aa(a x a)
Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:
A x A = 1/4 = 25%
A x a = 1/4 = 25%
a x A = 1/4 = 25%
a x a = 1/4 = 25%
The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:
1/4 × 1/4 = 1/16 = 0.0625 = 6.25%
It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.
Answer:Prophase
Explanation: Prophase is when chromosomes are made and the nuclear membrane breaks it down and later into the appearance of spindle fibers.
Answer:
This poisonis an example of an aboitic factor
