Quick Princeton Review biology question- Allele frequency

[boxbrown]

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Hi all,

I had a quick question regarding a PR bio chapter problem. Here it is, with their explanation: "If 20% of the population is heterozygous for an allele Q and 10% homozygous, what will be the frequency of the allele in the population?

The answer is 20%. The explanation is: The total number of alleles is 100%. The frequency of the allele is .5*(20% heterozyygotes)+10% homozygotes= 20%.

Why doesn't the traditional Hardy-Weinberg procedure work? I.e, if we say that the homozygotes QQ=p^2 in Hardy-Weinberg law, then p^2=10%=.1 , therefore p=frequency of allele Q=square root of .1= .31 or 31% , not 20%.

Is it because the problem never stated that it was in Hardy-Weinberg equilibrium? 🙂

 

[ClemsonFirst]

Hit the nail right on the head. They never said it was at Hardy-Weinberg equilibrium. And by finding that the frequency at HW is 31% you know it is not at HW equilibrium.

 

[boxbrown]

Hit the nail right on the head. They never said it was at Hardy-Weinberg equilibrium. And by finding that the frequency at HW is 31% you know it is not at HW equilibrium.

Thanks a bunch Clemson. I only figured it out while typing out the post.