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
Sep 29, 2016
The percent of heterozygous individuals in African American population, for sickle cell allele, is roughly
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
#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
Double Check:
#0.9126 + 0.0854 + 0.002 = 1#
