Answer:
heterozygous (Aa) are 0.085 = or 8.54%
Step-by-step explanation:
The heterozygous individuals are carriers of the sickle cell trait. They have a genotype of Aa and are represented by the 2pq term
in the H-W equilibrium equations.
According to the question 0.2% of the population is affected with sickle cell anemia, thus q^2
= 0.2% = 0.002 in decimal. So, q =
sqr(q^2)
or sqr(0.002) =
0.04472
and p + q = 1, thus p = 1 – q = 1 – 0.04472 = 0.96
Thus, A allele has a frequency of 0.96 and the a allele has a frequency of 0.04472. Therefore, the
percentage of the population that is heterozygous (Aa) and are carriers is = 2pq = 2× 0.04472× 0.96 =0.085 = or 8.54%
Answer:
all me know is u make lot of dollar
Step-by-step explanation:
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B, look at picture for explanation
Please let me know if this helped or rate this the brainlist! Thanks
Answer:
The answer to your question is True. Hope it helps and have a good day!
The answer is C. 1/4
* the work may not make sense but it’s the correct answer ( I think)
