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biology

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The frequency of the sickle cell disease in parts of Central Africa is as high as 10 percent compared to a frequency of .5 percent in the US.

Use the percentages for the Central Africa population and the US population to calculate the frequency of the heterozygous and homozygous dominant genotypes in each of these areas.

  • biology - ,

    how do i do this?

  • biology - ,

    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.

  • biology - ,

    thanks! so the heterozygous and homozygous dominant genotypes are p^2 and 2pq?

  • biology - ,

    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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