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)?

Question

Madeline Andrews · Accepted Answer

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

The formula for Hardy-Weinberg is: #p^2 + 2pq + q^2 = 1" "# and also #" "p+q=1# One double recessive afflicted individual means there are two individuals among #1000#. Thus, #q^2=0.002# #q = 0.0447# Next #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%# Verify Again: #0.9126 + 0.0854 + 0.002 = 1#

Nicholas Ahn · Answer

undefined To calculate the percentage of heterozygotes for the sickle cell allele using the Hardy-Weinberg equation, you can use the equation:

$$2pq = 2 \times 0.02 \times 0.98 = 0.0392$$

So, approximately 3.92% of African Americans would be heterozygous carriers of the sickle cell allele.