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%.
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Answer:
B.) The amounts of A, T, G and C in DNA varies from species to species and C.) In any ome species, the amount of adenine closely approximates that of thymine D.) In any one species, the amount of guanine closely approximates that of cytosine
Explanation:
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