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hardy weinberg equilibrium [Sep. 20th, 2008|09:47 am]
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[evathereal]
[Current Mood |argh]

I did this whole problem, and I keep on getting 2 extra people. But I add everything up and they all equal to one, so I don't know what I'm doing wrong. I'm in Evolutionary Ecology, and we're working on genetics problems, but they shouldn't be so bad 'cos I'm not in Genetics (this is a prereq for genetics).


Pretend there is a population of humans that exhibits a rare recessive allele for pink hair, and 12 individuals out of the entire population of 100,000 have pink hair. If the population is in Hardy-Weinberg Equilibrium, HOW MANY individuals are heterozygote CARRIERS of the allele, and HOW MANY individuals do not have the allele? What are the frequencies of p, q, p2, q2, and 2pq? SHOW YOUR WORK.

This is my work and answers:

pp = q2 = 12/100,000 = .00012

q = .011

p + q = 1
p + .011 = 1
p = .989

p2 = .9781

2 x .989 x .011 = .0218

Therefore --

p = .989
q = .011
p2 = .9781
q2 = .00012
2pq = .0218

# PP/100,000 = .9781
# PP = 97,810 people

# Pp/100,000 = .0218
# Pp = 2,180 people

checking:
97,810 + 2,180 + 12 = 100,002

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Comments:
[User Picture]From: casecob
2008-09-20 03:19 pm (UTC)
i mean what i said though, your answer is pretty close.

you have 2 extra people, from a population of 100,000.
0.002% error isn't that bad.
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[User Picture]From: evathereal
2008-09-20 03:25 pm (UTC)
*nods* I just wanted to be sure. I guess it's just for the homework purposes... I don't know if my prof would count off for rounding. He mentioned it in class once, but hopefully he isn't that strict... I'll just ask him.
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