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:
The answer is C
Explanation:
Nutrients (Nitrogen, Phosphorus, and Potassium) They are the third most frequent stressor in impaired rivers and streams, and the fourth greatest stressor in impaired estuaries. The three primary nutrients in manure are nitrogen, phosphorus, and potassium.
Answer:
More than one of the above
Explanation:
I strongly recommend sticking with the prescribed dosage of a drug.
A drug works by binding itself to the receptor site of a cell or tissue by non-covalent interactions.
Repeated doses of the same drug however may make the drug start behaving as an inverse agonist by blocking (instead of binding) the receptor site of the cell thus inducing a reduced response instead of an increased response to the drug.
