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#