It’s 2 using the rule -b/2a —> 8/4=2
Its the second answer, you're welcome my dude
1) 1 hour and 15 minutes
2) 4:07
3) 30 minutes
4) 8:35
9) 1 hour and 53 minutes
10) 4:17
11) 2 hours and 52 minutes
12) 3:35
13) 9:55
14) 8:10
15) 1 hour and 45 minutes
16) 3:05
17) 6:48
18) 9:45
19) 1 hour and 25 minutes
20) 2 hours and 45 minutes
Brainliest would be appreciated :)
Answer:
42
Step-by-step explanation:
uih
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%
