How does Hardy-Weinberg principle predict equilibrium?

[Uncommon Ancestor]

Hello, I'm new on these forums and I have a doubt I bet someone here could help me solving.

 

I'm a secondary-school student and whenI was doing a work that involves population genetics I found the Hardy-Weinberg equilibrium to be relevant as it constitutes one of the basis of population genetics. While I do totally understand the equations that show how we can know the allele frequencies if we know the frequency of the recessive character, I have troubles understanding why is it predicted that the result of the mixing will produce stable frequencies within the time.

Although most explanations on the subject I have consulted on the net take this point for granted, I have found two sources that do somehow explain it: one is Wikipedia and the another one is on Berkeley's webpage (look at the bottom of the post for links). However, I definitely have troubles understanding Wikipedia's so I will focus my question on Berkeley's webpage's one.

In the forementioned explanation, it states that "The condition for population stability (i.e. the genotype frequencies being the same from one generation to the next) is P·R=Q² ". I have been unable to find why is that assertion correct, so here comes my question. How do we know that?

Thanks in advace for any answer.

PD: I have checked a similar question to be already asked in these fora. Although I have found a post asking a similar question, the author's doubts weren't fully solved, so I have preferred to start a new topic.

EXPLANATIONS:

• From Wikipedia: https://en.wikipedia.org/wiki/Hardy%E2%80%93Weinberg_principle#Derivation
• From Berkeley's webpage: http://www.stat.berkeley.edu/~terry/Classes/s260.1998/Week5/week5/node1.html

[sauerkrautpie]

I am not good at genetical calculation but the equality is = p²+2pq+q²=1

[Uncommon Ancestor]

Thank you for the answer!

 

It seems that I haven not expressed myself clearly enough so I amb going to try to fix it.

 

I know that the Hardy-Weinberg equation that is applies for only one non-sexual related locus is p²+2pq+q²=1 and I have no troubles with that point. I understand where it comes from and I think I know how to use it in context. What Berkeley's webpage states is that P·R=Q²is an equation defining the conditions in which stability for allele frequencies are reached. It calculates that applying the Hardy-Weinberg equation for each possible case of breeding and then syntetizing the results on a furmula that calculates frequencies for each character (P being dominant homozigous, R being recessive and Q being the heterozigous), so it demonstrates that p²+2pq+q²=1 means that P·R=Q² so it predicts equilibrium. Summing up: the fact that the Hardy-Weinberg equation correlates with P·R=Q² signifies that it does actually predict stability so P·R=Q² must come from another source --not be a mere consequence of the first equation.

 

 

Both Hardy & Weinberg supposed a large population, mating ``random'' w.r.t. the trait under discussion (this does not mean choose your mate randomly), equal viability of offspring, ..., with other demographic issues ignored. Suppose that the proportions of AA, Aa, and aa genotypes in such a population are P, 2Q and R, respectively, with P+2Q+R=1. Consider what happens after one round of ``mating'':

Let the population frequencies in the new generation of AA, Aa, and aa be and , respectively. Then

The condition for population stability (i.e. the genotype frequencies being the same from one generation to the next) is , and , and satisfy this equation. Thus we reach a stable set of genotype frequencies in one round of ``random mating'' at a locus, regardless of the initial genotype frequencies.

 

The problem is that I don't understand why is that equation (P·R=Q²) related to stability.

 

 

 

If it is not clear enough, please tell me and I will try again. For english is not my mother tongue, any corrections are welcomed.

[Uncommon Ancestor]

Ok, now I have finally understood. It has took me days of Internet research, but now I finally do understand it.

 

If someone else finds himself in a similar situation, he can ask me. The aswer is a bit tricky, though.

[KipIngram]

Just out of curiosity, why do they write p^2 + 2pq + q^2 = 1 instead of (p+q)^2 = 1?

[Uncommon Ancestor]

It's because there happen to be three possible combinations involving 2 alleles (A and a) in a diploid population.

 

p is the frequency for A allele and q is the frequency for a.

 

The frequency of each allele is 1 minus the other one, but each organism has two alleles in its genome.

 

AA is dominant homozigote and its frequency is p^2

 

aa is recessive homozigote and its frequency is q^2

 

But there are also Aa members, heterozigotes, whose frequency is 2pq.

 

The fact that p^2+2pq+q^2=1 is deduced from the fact that (p+q)^2=1, 1 being the addition of all the frequencies in a population (of course, the calculus changes a little bit if it involves more than 2 alleles).

 

In fact, AA and Aa memebers are assumed to have an equal phenotype (physical appearence), so it is not possible to know whether are they homozigote or heterozigote without other techniques than physical checking.

 

Hardy-Weinberg can be used to estimate how many habitants in a population are Aa and how many are AA, knowing only how many of them are aa members. Even if so, as Hardy-Weinberg equilibrium assumes some ideal conditions that dont occur in actual populations (for instance, an infinite population or no evolution at all), our estimation can differ slighty (or not so slighty) from the real values.