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Hardy-Weinberg Equilibrium - DressageQueen - 08-25-2012
I am studying for the biology CLEP and I came across the Hardy-Weinberg equilibrium. I don't fully understand it. I understand where the numbers go and what the equation is but on some of the questions I have come across on practice tests, my knowledge was useless. Can someone please explain it to me? For example (this is a real practice test question)
A given trait occurs in two alternative types, M and m, in a population at Hardy-Weinberg equilibrium. If 49 percent of the population has only type M alleles, what percentage of the population is expected to be heterozygous for the trait?
A) 9%
B) 14%
C) 21%
D) 42%
E) 51%
I guessed and got the question right (correct answer is D) but I don't understand how you are supposed to figure it out with only one number. P^2+2pq+q^2=1 so p^2 would equal 49 but where do you go from there? I have told that there are a few questions about this on the CLEP so I want to fully understand it. Thank You!!!!!
Hardy-Weinberg Equilibrium - clep3705 - 08-25-2012
Demo Question
Hardy-Weinberg practice
Hardy-Weinberg Equilibrium - tylisasweet - 08-26-2012
There are two equations when dealing with Hardy-Weinberg Equilibrium. p^2+2pq+q^2=1 is one, and the other is p+q=1 (that is, the percentage present for each allele must equal 1).
As you said, p^2 = .49 which implies that p = .7 (since p can't be a negative number). Therefore, .7 + q = 1 ... which means q = .3
So what's the percentage for heterozygous?
2pq = 2(.7)(.3) = .42 = 42%
Between this and clep3705's post, I hope you understand.
Hardy-Weinberg Equilibrium - DressageQueen - 08-26-2012
Awesome!! I get it! THANK YOU both!!!!!
Hardy-Weinberg Equilibrium - OE800_85 - 08-27-2012
I guess I'm late to the party.
Hardy-Weinberg Principle | Biology | Khan Academy
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