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:
C.4 is the answer
Explanation:
By the end of meiosis, the resulting reproductive cells, or gametes, each have 23 genetically unique chromosomes. The overall process of meiosis produces four daughter cells from one single parent cell. Each daughter cell is haploid because it has half the number of chromosomes as the original parent cell.-
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Answer:
creo que es una población pequeña
Explanation:
Answer:
D. Which cell organelles carry out the process of photosyinthesis and cellular respiration?
I don't think the answer is A as that seems a bit drastic, I think it's C.