population genetics question
Started by klm0101
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Edit: ...
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Ok, I think I got this now.
Heterozygous: 2pq = 20% = .2
Homozygous: p^2 = 10% = .1
The total frequency of allele Q is simply: pq + p^2 = .2/2 + .1 = 20%😀 Tell me if you still don't understand.
***It just simply means that the entire heterozygous population contains 50% of the allele Q. Analogously, 2pq / 2 = pq accounts for a half of that heterozygous population. Make sense???
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Explained a little easier
p^2 + 2pq + q^2 (equation) (more info-this is from 1-2-1 punnett sqaure [hardy weinberg eq])
.7 + .2 + .1 (fill in percentages)
.2 + .1 (isolate variable)
.2/2 + .1 (narrow to selected allele)
=0.20 = 20%
p^2 + 2pq + q^2 (equation) (more info-this is from 1-2-1 punnett sqaure [hardy weinberg eq])
.7 + .2 + .1 (fill in percentages)
.2 + .1 (isolate variable)
.2/2 + .1 (narrow to selected allele)
=0.20 = 20%
mathematically,
p^2 + pq = p(p+q) = p
p^2 + pq = p(p+q) = p
shouldn't just finding the square root of the homozygote frequency be enough to find the allele frequency of p? Unless this population isn't in Hardy-Weinberg equilibrium?
That is the general misconception that I ran into also 🙂 ... You have to distinguish the difference between the homozygous and heterozygous alleles. In homozygous p2 or q2 alone does not account for its total frequency. So when you use the sqr of homozygous frequency, you only pick out the probability of 100% homozygous pp allele and doesn't include the allele p in pq of the heterozygous one. Make sense? 🙂
So p^2 + 1/2 (2pq) will give you the frequency of the "p" allele right?
If the question stem states a dominant/recessive disorder is expressed in 80%/20% of the population, that means 0.80 = p^2 or 0.20 = q^2 respectively, correct?
If the question stem states a dominant/recessive disorder is expressed in 80%/20% of the population, that means 0.80 = p^2 or 0.20 = q^2 respectively, correct?
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I do understand, but let's take another example
Suppose the homozygote dominant frequency is .64, the heterozygote is .32, and the homozygote recessive is .04. To find the p allele, you can do 1 of two things:
1) just square root the homozygote dominant frequency, so sqrt(.64) = .8
OR
2) do your method. p^2 + pq = .64 + .16 = .8
Either way, you should get the same answer. I'm just not getting why you can't do the same thing with the ratios posted in the OP's question.
Suppose the homozygote dominant frequency is .64, the heterozygote is .32, and the homozygote recessive is .04. To find the p allele, you can do 1 of two things:
1) just square root the homozygote dominant frequency, so sqrt(.64) = .8
OR
2) do your method. p^2 + pq = .64 + .16 = .8
Either way, you should get the same answer. I'm just not getting why you can't do the same thing with the ratios posted in the OP's question.
the terms p^2, 2pq, and q^2 are estimates of what fraction of a population in HW equilibrium are homozygous dominant, heterozygous, and homozygous recessive respectively. p and q ARE the allele frequencies in said population. either there's some info missing from the question, the answer is wrong, or im dumb.
oh. i think the point is that it doesn't say the population is in HW equilibrium.
edit: haa ok so, given 100 people: 10% are homozygous = 10 people = 20 q alleles.
20% heterozygous = 20 people = 20 q alleles, 20 p alleles.
so 100 people = 200 alleles total
40 q alleles / 200 alleles = 20%
edit: haa ok so, given 100 people: 10% are homozygous = 10 people = 20 q alleles.
20% heterozygous = 20 people = 20 q alleles, 20 p alleles.
so 100 people = 200 alleles total
40 q alleles / 200 alleles = 20%
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Sorry to bring this up again 😀
The percentages here are relative. So when you have a 80% out of 64% is much less than 80% out of 100%, right? Therefore,
1) This is 80% out of 64% homozygous alleles
2) This is 80% out of 100% total of p and q.
They are not equal! Make sense?
