Greetings, I am working on a problem that states: In a certainpopulation, 16% of

Question

Greetings,

I am working on a problem that states: In a certainpopulation, 16% of the individuals have thalassemia major. Whatpercentage of this population is heterozygous for the thalassemiagene? The answer is 48%.

I understand the p+q=1 and p^2+2pq+q^2=1 equations.

I was thinking: If q=.16, p=.84.

Then, rounding up a little bit to make calculation easier (since Iam not allowed to use calculators), I then inputted:(.9^2)+2pq+(.16^2)=1

Then I was confused. I thought to solve for pq, but wasn't surehow. So, I used a calculator just to see if I would get the rightanswer and got something else that was wrong. So, I am wonderinghow to solve this problem?

Explanation / Answer

16% of the population has a certain disease, which I assume is anautosomal recessive disease. This means that in order for thedisease to express itself in the population, in must be ahomozygote recessive. So the frequency given is that of theGENOTYPE, q^2. When we assume HWE, then (q^2)=0.16 or q=0.4,which is the frequency of the ALLELE. Now we have all our info. We know p+q=1, so when q=0.4,p=0.6. The frequency of the heterozygote in the population isgiven by 2pq. 2pq 2(0.4)(0.6) 0.48 So the frequency of the heterozygote to express itself in an HWEpopulation is 48%. I hope this helps.

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