Sickle cell anemia is caused by a recessive allele. Roughly 1 out of every 500 african americans is afflicted with sickle cell anemia. How would you use the hardy weinberg equation to calculate the % of heterozygotes for sickle cell allele (.002=q^2)?

1 Answer

The percent of heterozygous individuals in African American population, for sickle cell allele, is roughly 8.54%.

Explanation:

The Hardy-Weinberg equation is:

p^2 + 2pq + q^2 = 1" " and also " "p+q=1

  • where p^2 is the percentage of homozygous dominant phenotype
  • where 2pq is the percentage of heterozygous dominant phenotype
  • where q^2 is the percentage of homozygous recessive phenotype

One double recessive afflicted individual means there are two individuals among 1000. Thus,

q^2=0.002

q = 0.0447

Then

p + q = 1

p = 1 - q

p = 0.9553

p^2 = 0.9126

2pq = 2(0.9553)(0.0447) = 0.0854

Thus % of heterozygous individuals in the population is = 8.54%

Double Check:

0.9126 + 0.0854 + 0.002 = 1