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:
26
Step-by-step explanation:
welll i have no clue just guess i cant even read the question
Answer:
b. f(n) = 2n^2 + 1
Step-by-step explanation:
Thanks for explaining the nth thing to me. It all makes sense now :'D
I'll be honest I think you're at least a grade level above me in math.
It's b. because
f(n) = 2(1)^2 + 1
2(1) + 1
2 + 1 = 3 (first term)
f(n) = 2(2)^2 + 1
2(4) + 1
8 + 1 = 9 (second term) ...and so on. :)
No. 61 is an odd number not divisible
