Why do scientists use hardy weinberg

When a population satisfies conditions 1—5, Observation 1 ensures that allele frequencies will remain unchanged in every succeeding generation that satisfies those conditions. Applying condition 6 and the crucial H-W insight, in each generation, after, 6 the genotype frequencies are p 2 , 2pq, and q 2. This is the genius of the H-W principle : after one generation of fully random mating, both the genotype and allele frequencies are fixed until one of conditions 1—6 is violated.

To summarize, 1 allele frequencies can always be computed from the genotype frequencies in the same generation if all genotypes can be identified , but not vice versa; and 2 if the population is in H-W equilibrium, genotype frequencies in the current or the next generation can be computed from the current allele frequencies. These two equations are widely used in biology teaching, but all too often they are used as a mathematics exercise that does not promote understanding.

Equation 1 is true for any monogenic Mendelian trait because there are only two outcomes. Squaring Equation 1 yields. So, Equation 2 simply follows mathematically from Equation 1. There is no assumption about random mating and no other biological assumption in the step from Equation 1 to Equation 2.

Interestingly, using only these formulae, we can determine whether H-W eq exists in a single generation of a population by determining whether the genotype distribution matches that predicted from the allele distribution, but this requires that both the allele frequencies and the genotype frequencies are known. Hardy provided an ingenious way to determine whether H-W eq exists in a single generation, given only the genotype frequencies. For a short account of Hardy's proof in modern language accessible to advanced students, along with several other proofs of the H-W principle, see Baldwin, The following textbook problems are built on the assumption that, if a population is in H-W eq which is often a dubious assumption , then it is possible to calculate the allele frequencies from the frequency of the homozygous recessives which can be found by observation.

The data below demonstrate the frequency of tasters and non-tasters in an isolated population at H-W eq. The allele for non-tasters is recessive. How many of the tasters in the population are heterozygous for tasting? An acceptable answer would be any number in the range of —, depending on how the students rounded the variables in the H-W equation.

This is a standard H-W eq problem. This yields the number of heterozygotes as 0. You sampled individuals and determined that could detect the bitter taste of PTC and 65 could not. Calculate the following frequencies. The frequency of the recessive allele. The frequency of the dominant allele. The frequency of the heterozygous individuals.

Both problems are focused on making calculations that students can do without understanding what H-W eq is. In Problem 1, H-W eq is explicitly assumed so that the problem is technically correct. But it doesn't say what chemical was being tasted presumably PTC , so it doesn't ask students whether H-W eq conditions could be met for this trait.

And students can solve Problem 2 only by assuming H-W eq, which is not justified. The ability to taste PTC phenylthiocarbamide , a bitter substance that cannot be tasted by some individuals, is frequently used in H-W eq problems, likely because it is assumed to be selected neither for nor against, given that PTC does not occur in nature.

Thus, the student is expected to deduce or more likely assume that the H-W conditions apply. Teachers and textbooks, however, rarely make this reasoning explicit, leaving students with the misperception that understanding PTC tasting is just a game or a puzzle that likely seems unimportant to them because it doesn't relate to their daily lives.

In fact, recent research has shown that the ability to taste PTC is strongly correlated with the ability to taste other bitter substances that do occur naturally, many of which are toxins. Thus, it seems likely that the ability to taste bitter substances such as PTC is positively selected for. Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals SS have normal blood cells that are easily infected with the malarial parasite.

Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait ss have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect.

However, individuals with the heterozygous condition Ss have some sickling of red blood cells, but generally not enough to cause mortality. Thus, heterozygotes tend to survive better than either of the homozygous conditions. The solution above, which assumes H-W eq and that natural selection is not occurring with regard to this gene, contradicts the statement of the problem, which notes selective pressures for one and against another of two blood-cell phenotypes.

Justify your answer. Because the members of the population that contract sickle-cell because they are homozygous recessive will likely die before reproducing, the frequency of alleles in the population is not stable.

There is natural selection taking place. Although this problem instructs students to use the H-W equations, again the known effects of natural selection at this locus mean that H-W eq is impossible. The problem therefore asks for what is, in fact, a meaningless calculation. Then it asks students to answer a question that demonstrates that the computation was meaningless but does not ask them to recognize that it was meaningless!

The H-W eq-based frequencies are irrelevant. A second issue that arises in problems about sickle-cell anemia is that two opposing selective pressures are at work — a positive selection for heterozygosity and a negative selection against affected homozygotes. Such a situation can produce balanced polymorphism equilibrium but not H-W eq, because the calculation to keep the genotype constant requires further parameters. Percentage of variants raw numbers are shown in B deviating from HWE due to D heterozygote deficiency or E heterozygote excess in each population.

Another factor that could affect detection of HetExc variants, was the degree to which HWE assumptions were satisfied in each population. Consequently, these populations also had the lowest proportion of HetExc variants, 0. Figure 2. Only variants with 0. This initial analysis has been performed on variants that remained following the gnomAD sequence quality filtering process and might therefore be assumed to be real.

However, variant databases are known to contain errors that could give a significant HetExc signal. To explore this we developed a set of more stringent filters. In particular, variant properties that could produce a false positive HetExc signal were investigated. At this stage, only variants that had an excess of heterozygotes in all populations, and were statistically significant in at least one population were classified as HetExc.

Therefore, HetExc of variants located in segmental duplications might be a result of genotyping errors. Figure 3. Impact of tandem repeats, segmental duplications and allele balance on the probability of variant deviation from Hardy-Weinberg Equilibrium HWE due to heterozygote excess HetExc.

For each variant these statistics are aggregated into a single metric that represents cumulative percentage of Variant Carriers with Normal 0.

Secondly, to investigate the correlation between HetExc and Allele Balance AB , which is a known indicator of systematic genotyping errors Muyas et al. In gnomAD, variant AB data is stored as a number of variant carriers converted to percentages here in 20 AB groupings from 0 to 1, 0. For an average variant, the majority of variant carriers Only 2. These HetExc variants were then compared with a group of variants that survived the same filtering process, but did not have an excess of heterozygotes HetExc-.

The HetExc- and HetExc groups consisted of 39, and variants 50, and if counted in seven ethnic populations separately, Figure 4A in 11, and genes, respectively.

Figure 4. B Proportions of missense, synonymous and other protein coding variants in HetExc and HetExc- datasets. C ClinVar status e. To determine which of the HetExc candidate recessive disease causing variants were already known, their clinical significance in the disease variant database ClinVar; Landrum et al.

HetExc variant enrichment in known AR genes adds evidence that some of the selected variants might deviate from HWE due to natural selection and could have some disease association. Since gnomAD v3 mostly consisted of individuals that were not present in gnomAD v2. However, our findings that HetDef is a major cause of deviations from HWE in all populations is contrary to previous studies Chen et al. However, previous studies focused on error detection in older and smaller datasets, some of which were corrected in gnomAD.

Graffelman et al. We observed a higher rate of HetExc variants in these regions, as well as those that had low allele balance, which correlates with previous work Graffelman et al. Chen et al. However, this approach resulted in the exclusion of rare variants that were analyzed in this study and might be more affected by the Wahlund effect i.

Moreover, some of the HetExc variants detected in previous studies were marked as non-pass quality or were no longer HetExc in gnomAD, possibly due to differences in variant filtering and genotype calling procedures. For example, c.

Another example, the BRSK2 variant c. Therefore, a higher rate of HetDef variants in our study could be explained by a larger population size and a different variant dataset, as well as improvements in variant filtering and genotype calling procedures. Analysis of HetExc variants Supplementary Table S1 , selected as recessive disease causing candidates, led to somewhat contradictory results, which should be interpreted with caution.

Moreover, despite applying our extensive filtering strategies, many of the HetExc variants might still be deviating from HWE due to genotype errors or by chance due to insufficient population size. However, the c. Therefore, the difference between the number of homozygote in gnomAD v2. Nevertheless, the presence of known pathogenic and heterozygote advantageous variants such as HBB c. Especially, the CHD6 gene variant c. Although CHD6 is not yet linked with any disease, it is known to act as transcriptional repressor of different viruses including influenza and papiloma virus Alfonso et al.

Similarly, natural selection and nonrandom mating disrupt the Hardy-Weinberg equilibrium because they result in changes in gene frequencies.

This occurs because certain alleles help or harm the reproductive success of the organisms that carry them. Another factor that can upset this equilibrium is genetic drift, which occurs when allele frequencies grow higher or lower by chance and typically takes place in small populations.

Gene flow, which occurs when breeding between two populations transfers new alleles into a population, can also alter the Hardy-Weinberg equilibrium. Because all of these disruptive forces commonly occur in nature, the Hardy-Weinberg equilibrium rarely applies in reality.

The Hardy-Weinberg equation is expressed as:. In the equation, p 2 represents the frequency of the homozygous genotype AA, q 2 represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. If the p and q allele frequencies are known, then the frequencies of the three genotypes may be calculated using the Hardy-Weinberg equation. In population genetics studies, the Hardy-Weinberg equation can be used to measure whether the observed genotype frequencies in a population differ from the frequencies predicted by the equation.