Hardy-Weinberg Equilibrium Help

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jewish_scientist
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Hardy-Weinberg Equilibrium Help

Postby jewish_scientist » Thu Feb 15, 2018 4:22 pm UTC

p = f(A) = the probability that a randomly selected zygote in a population will have allele A
q = f(B) = the probability that a randomly selected zygote in a population will have allele B
1 = p + q
12 = (p + q)2 = p2 + 2pq + q2

f(A)2 = f(AA) = the probability that a randomly selected organism in a population with have the genotype AA
f(B)2 = f(BB) = the probability that a randomly selected organism in a population with have the genotype AA
f(AB) = the probability that a randomly selected organism in a population with have the genotype AA
1 = f(AA) + f(AB) + f(BB)

p2 + 2pq + q2 = f(AA) + f(AB) + f(BB)
f(A)2 + 2pq + f(B)2 = f(A)2 + f(AB) + f(B)2
2pq = f(AB)

The only things I did to derive the above equations is square both sides of an equation, substitute equivalent variables, and subtract identical amounts from both sides of an equation. Mathematics says that doing any of these things to an equation CANNOT change the validity* of an equation.

I was just told that iff a population is at Hardy-Weinberg Equilibrium, then all of the above is true. That means that there are populations where the above is not true. In other words, values exist that are counterexamples to equations above, which in turn means that these mathematical truths are false. Because a mathematical truth cannot be false, all populations must be in Hardy-Weinberg Equilibrium. The whole point of the Hardy-Weinberg Equilibrium is that an evolving population will not be in an Hardy-Weinberg Equilibrium.

What am I missing?

*What I mean by this is that iff the original equation is true, then the derived equation is true.
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Re: Hardy-Weinberg Equilibrium Help

Postby doogly » Thu Feb 15, 2018 4:34 pm UTC

Not a biologist, but from a probability perspective, the f(AA) = f(A)^2 assumes that getting the A allele from each parent are entirely independent events, otherwise you'd need to worry about A|A and A|B being different. The independence of these sounds like it requires the equilibrium.
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Re: Hardy-Weinberg Equilibrium Help

Postby elasto » Thu Feb 15, 2018 9:00 pm UTC

I'm not a biologist either but f(A)2 = f(AA) does seem like the key to it.

If we imagine the population consists entirely of an organism of genotype AB: f(A) = 1, so f(A)2 = 1, yet f(AA) = 0

So that would seem to be an example of a population not at Hardy-Weinberg Equilibrium.

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Re: Hardy-Weinberg Equilibrium Help

Postby gmalivuk » Thu Feb 15, 2018 9:09 pm UTC

jewish_scientist wrote:p = f(A) = the probability that a randomly selected zygote in a population will have allele A
q = f(B) = the probability that a randomly selected zygote in a population will have allele B
Zygotes are fertilized so they have two alleles. You're talking about gametes or chromosomes.

doogly wrote:The independence of these sounds like it requires the equilibrium.
Yeah, obviously p2 + 2pq + q2 = f(AA) + f(AB) + f(BB) = 1, because they're probabilities, but you can only match those up term-by-term if the probabilities are all independent.

f(AA) = f(A from mother AND A from father) = f(A from father) * f(A from mother GIVEN A from father) = f(A from mother) * f(A from father GIVEN A from mother).
You need them to be independent in order to say f(A from mother GIVEN A from father) = f(A from mother).
You furthermore need them NOT to be sex-linked in order to say f(A from mother) = f(A from father) = p.
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qetzal
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Re: Hardy-Weinberg Equilibrium Help

Postby qetzal » Fri Feb 16, 2018 4:10 am UTC

Just imagine a situation where AB heterozygotes are lethal.

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Re: Hardy-Weinberg Equilibrium Help

Postby jewish_scientist » Fri Feb 16, 2018 1:49 pm UTC

elasto wrote:If we imagine the population consists entirely of an organism of genotype AB: f(A) = 1, so f(A)2 = 1, yet f(AA) = 0

f(A) = the probability that a randomly selected gamete* produced by a member of the population has allele A =/= the probability that a randomly selected member of a population has allele A

qetzal wrote:Just imagine a situation where AB heterozygotes are lethal.

Spoiler:
Lethality is not important, fertility is. This difference is important because all dead things do not reproduce, but not all living things do reproduce. Therefor, we can model your scenario as if AB results in living organisms that never reproduce.

p = 0.5
q = 0.5
f(AA) = 0.25
f(AB) = 0.5
f(BB) = 0.25

In the next generation, these values will not change.


*Thank you gmalivuk for the correction.
Last edited by jewish_scientist on Fri Feb 16, 2018 7:04 pm UTC, edited 1 time in total.
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Re: Hardy-Weinberg Equilibrium Help

Postby gmalivuk » Fri Feb 16, 2018 3:32 pm UTC

Elasto's numbers were wrong (thanks to your mistake in the OP) but the point remains:

If everyone is AB, then f(A) is 0.5 and f(A)^2 is 0.25, but f(AA) is 0.

The lethality point is also valid, as it's basically the reverse situation, where nobody is AB instead of everybody. If half are AA and half are BB then f(A)=f(B)=0.5 and f(A)*f(B)=0.25 but f(AB)=0.

(Of course you can't model it as a completely different population where f(AB)=.5, because 0 and 0.5 are, as it turns out, different numbers.)
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Re: Hardy-Weinberg Equilibrium Help

Postby jewish_scientist » Fri Feb 16, 2018 7:37 pm UTC

Strictly speaking, f(A) refers to the probabilities in the current generation and f(AA) refers to the probabilities in the next generation. This is because f(A) * f(A) is the chance two gametes carrying the allele A will be used to create an descendant. If every organism is AB, then f(AA) = 0.25 since that is the chance a member of the next generation will be AA. Similarly, if all organisms with genotype AB immediately die after being conceived, f(AB) =/= 0 since they were conceived.

When the gene is sex linked, then you model it as if there are 3 alleles, which means you end up having to do (p + q + r)2. It gets further complicated when you start trying to figure out what values to give p, q, and r.
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Re: Hardy-Weinberg Equilibrium Help

Postby doogly » Fri Feb 16, 2018 7:47 pm UTC

jewish_scientist wrote:Strictly speaking, f(A) refers to the probabilities in the current generation and f(AA) refers to the probabilities in the next generation. This is because f(A) * f(A) is the chance two gametes carrying the allele A will be used to create an descendant. If every organism is AB, then f(AA) = 0.25 since that is the chance a member of the next generation will be AA. Similarly, if all organisms with genotype AB immediately die after being conceived, f(AB) =/= 0 since they were conceived.

I mean, you could interpret things that way, and then as you say it is a vacuous result. So that is not how it is meant to be interpreted.

Instead, the interpretation is as you originally and rightly (well with a typo) stated:
f(AB) = the probability that a randomly selected organism in a population with have the genotype AB

If you are randomly selecting among members of a population, it is meaningful and indicates the equilibrium works out that way. The reason why it is a meaningful result about equilibrium in a population is because you take the probability as selecting from among the population. If instead you interpret the sample space out of which you do probability to not be the population, but to instead include unviable zygotes and the like, then you are correct to observe that there is not a meaningful theorem.
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Re: Hardy-Weinberg Equilibrium Help

Postby gmalivuk » Fri Feb 16, 2018 8:09 pm UTC

You can derive proportions for subsequent generations, but the actual equation is one of equilibrium, meaning subsequent generations don't have different proportions than the current generation.

The seven assumptions underlying Hardy–Weinberg equilibrium are as follows:

organisms are diploid
only sexual reproduction occurs
generations are nonoverlapping
mating is random
population size is infinitely large
allele frequencies are equal in the sexes
there is no migration, mutation or selection

If there is selection (such as lethality for hetero- or homozygotes), you can't derive the equilibrium equation.

(And as was pointed out in the very first reply, you can't assume things like f(AA) = f(A)2 unless those probabilities are independent, which they are if mating is random but which they need not be in general.)
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Re: Hardy-Weinberg Equilibrium Help

Postby Eebster the Great » Mon Feb 19, 2018 12:37 am UTC

Right, and selection is a pretty straightforward example. We can imagine a population initially at equilibrium with f(A) = f(B) = 0.5. Then we can imagine a change in the environment that makes the BB genotype 100% lethal in infancy, while the AA and AB genotypes are unaffected. For a while, we will have f(A) > f(B) > 0 and f(BB) < [f(B)]2. Why the last inequality? Because while newborns may have the expected proportion, f gives the fraction in the overall population, and newborns represent only a small part of it. No adults will have BB, so those early deaths skew the total. After an infinite time, f(B)=0, restoring equilibrium.


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