Posted by Sarah on Wednesday, February 27, 2013 at 8:22pm.
how do i do this?
Sickle cell anemia is autosomal dominant.
Let p=sickle cell and q=normal cell.
p+q=1
(p+q)^2=p^2+2pq+q^2=1
p^2= homozygous dominant
q^2=sickle cell
2pq=heterozygous
For Central Africa
0.1=q^2=sickle cell
q=0.316
p=1-0.316=0.684
p^2= homozygous dominant=(0.684)^2=0.468
q^2=sickle cell=(0.316)^2=0.0999
2pq=heterozygous=2(0.684)(0.316)=0.432
p^2+2pq+q^2=0.468+0.432+0.0999=1
Repeat for U.S.
thanks! so the heterozygous and homozygous dominant genotypes are p^2 and 2pq?
Yes, but when you do theses Hardy-Weinberg problems you are suppose to let the recessive gene equal p and the autosomal dominant gene equal q. When I started doing this for you I accidentally switched it, but the numbers are correct. The post should read
Let p=sickle cell and q=normal cell.
p+q=1
(p+q)^2=p^2+2pq+q^2=1
q^2= homozygous dominant
p^2=sickle cell
2pq=heterozygous
For Central Africa
0.1=p^2=sickle cell
p=0.316
q=1-0.316=0.684
q^2= homozygous dominant=(0.684)^2=0.468
p^2=sickle cell=(0.316)^2=0.0999
2pq=heterozygous=2(0.684)(0.316)=0.432
p^2+2pq+q^2=0.0999 +0.432+0.468=1
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