# Hardy-Weinberg Equilibrium Test Result for Merged Genotypes

I was curious if two populations are in Hardy-Weinberg Equilibrium (HWE), and if they are merged, then what happens? To find out, I considered populations from the 1000 genome project data. For example, at first, I considered the genotypes of the American (AMR) population from a specific SNPs and checked whether they were in HWE and found that they were. Then I selected the East Asian (EAS) population and found that the genotypes were also in HWE. However, when I merged the genotypes from both populations and checked HWE, I found that it was not in HWE. Surprisingly, when I added 50% of genotypes from each population, they were in HWE. So, I did it for different combinations, for example, 80% genotypes from the AMR population and 20% from the EAS population and vice versa, and found that they were also in HWE. I used the chi-squared goodness of fit test and the Haldane Exact test. The allele and genotype frequencies for each combination are different from the initial populations. What conclusion can I draw from my findings? Should I check to see if there are any assumptions about the test that are violated?

• Welcome to Biology StackExchange! If I understand your question correctly: Imagine a limiting situation with two alleles and two populations, one of which only has one allele present ($p_1 = 1, q_1=0$) while the other has only the other allele present ($p_2=0, q_2=1$). Even if both of these populations are in HWE, when you merge them, they won't be ($p = 1/2, q=1/2$ but there are no heterozygotes). Commented Dec 3, 2023 at 13:10
• Here are the allele frequencies and genotype frequencies: G T GG GT TT 0.408 0.592 0.184 0.447 0.369 → AMR Population || 0.651 0.349 0.427 0.448 0.125 → EAS Population || 0.550 0.450 0.327 0.447 0.226 → Combined all || 0.550 0.450 0.327 0.446 0.227 → 50% each combined || 0.471 0.529 0.247 0.447 0.306 → 80% AMR and 20% EAS || 0.614 0.386 0.390 0.448 0.162 → 80% EAS and 20% AMR @Domen Commented Dec 3, 2023 at 21:02
• your initial question was: I was curious if two populations are in Hardy-Weinberg Equilibrium (HWE), and if they are merged, then what happens? I have given you an example which tells you that merging allele frequencies from two populations in HWE will generally result in a population not being in HWE. Commented Dec 4, 2023 at 6:20
• Thank you very much! I get your point. My question is: how can I draw conclusions? Since different percentages of merging of genotypes result in HWE, except for all of them combined. Can I say something from the allele and genotype frequencies? Or do I need to take any further steps to reach the conclusions? Commented Dec 4, 2023 at 7:45
• @statm I suggest that you look into the idea of statistical power. More extreme deviations from HWE will naturally be easier to detect by your test; "in equilibrium" is not a binary after all, every population is somewhere in a continuum of disequilibrium. I'd also suggest consulting prior attempts to test for HWE in stratified populations. I think that you can safely reach the trivial conclusion that well mixed populations are closer to HWE than stratified populations, but no more. Commented Dec 4, 2023 at 19:47

As @Domen points out, there is no expectation that adding two populations that are each in Hardy-Weinberg equilibrium will initially create a new population in HWE.

OP's procedure of combining different proportions of the initial populations, then checking for HWE, is essentially like asking how much of one population can be added to another before HWE is lost. This vaguely gets at the question of how different the two populations are to begin with, but doesn't provide any information on whether the assumptions underlying HWE are met.

A better function to ask about the initial difference between populations, and that can be calculated from the initial allele frequencies, is the fixation index, FST.

Incidentally, if the assumptions of HWE are still met after the two populations are merged, then after one generation of mating HWE will be reestablished across the whole population.

Well, the absence of migration is one of the assumptions of Hardy-Weinberg law. By merging the two populations you are basically simulating migration, violating the assumption, and thus (as @Domen and @Darlingtonia correctly pointed out) there is no reason why you should expect that the two populations are in HWE (https://www.nature.com/scitable/knowledge/library/the-hardy-weinberg-principle-13235724/). The fact that you get different results when mixing different proportions may be due to different factors, such as (but not limited to):

1. Lower number of individuals obtained when mixing a fraction of each population may influence the power of the chi-square test. Seeing the numbers used in the chi-square test may help to understand.
2. The difference in frequencies between the two populations is small enough that only a total mixing brings to a significant deviation from equilibrium.
3. When you mix a large proportion of one population and a small proportion of the other, you likely get a final population which is sufficiently close to the first one, and thus does not significantly deviate from HWE.