11.2: Is a population in Hardy-Weinberg equilibrium?

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Nov 12, 2024
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- 11.1: Hardy-Weinberg equilibrium
- 11.3: Mutation and Migration

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In a real population of interbreeding organisms, the different alleles of a gene may not be represented at equal frequencies. This doesn't mean there's something amiss with respect to Mendel's laws. The individual crosses that produced the offspring would be expected, in general, to follow Mendel's laws, but many other factors determine the frequencies of alleles. Some alleles may confer, on average, a selective advantage. Some alleles may leave or enter the population disproportionately (emigration and immigration). One allele might mutate into the other more often than the reverse. And, finally, individuals with certain alleles might, just by chance, survive and leave more offspring, a phenomenon we call "genetic drift."

The classic two-allele Hardy-Weinberg model assumes the following:

NO NATURAL SELECTION: neither allele confers a selective advantage or disadvantage
NO MIGRATION: no one enters or leaves the population
NO MUTATION: an A allele will never mutate into an a allele, and vice versa
INFINITE POPULATION SIZE: no genetic drift
RANDOM MATING

Given that allele frequencies should not change over time if the assumptions of Hardy-Weinberg equilibrium are met, we should also realize that genotype frequencies should not change over time. Expected genotype frequencies, as shown above, are calculated directly from allele frequencies, and the latter don't change. We can, therefore, test the hypothesis for a given gene that its genotype frequencies are indistinguishable from those expected under Hardy-Weinberg equilibrium. In other words, we use Hardy-Weinberg equilibrium as a null model. This isn't to say that we "believe" all of the assumptions. Certainly it's impossible for a population to have infinite size, and we know that mutations occur. Even if individuals don't choose their mates directly or indirectly with respect to genotype, we know that mating isn't completely random; there is a general tendency to mate with a nearby individual, and if the population doesn't disperse itself well, this will lead to nonrandom mating with respect to genotype. Both migration and natural selection do occur (but they don't have to). Essentially, if we want to see if there is evidence for selection, drift, migration, mutation or assortative mating, a simple place to start is to see if the population is at Hardy-Weinberg equilibrium.

Consider a population of flowers. Let's say that the A gene determines petal color, and that there is incomplete dominance. AA individuals have red flowers, aa individuals have white flowers, and Aa individuals have pink flowers. There are 200 individuals with red flowers, 400 with white flowers and 400 with pink flowers. Does the population appear to be at Hardy-Weinberg equilibrium with respect to the A gene?

We must first determine the expected phenotype frequencies if the population is assumed to be at Hardy-Weinberg equilibrium. We are fortunate, because phenotype and genotype are completely correlated in this case. So, we need to calculate the expected genotype frequencies. To do this, we need to know the allele frequencies. This is easy:

p = freq(AA) + 50% x freq(Aa) = (200 + 50% x 400)/1000 = 0.400.
q = freq(aa) + 50% x freq(Aa) = (400 + 50% x 400)/1000 = 0.600.

We could have just calculated p and then assumed that q would be 1 - p. However, it's useful to do both calculations as a simple check of our arithmetic.

The expected frequency of the AA genotype is p² = 0.400² = 0.160. The expected frequency of the aa genotype is q² = 0.600² = 0.360. The expected frequency of the Aa genotype is 2pq = 2(0.400)(0.600) = 0.480. Therefore, if we have a total of 1000 flowers (200 + 400 + 400), we expect 160 red flowers, 360 white flowers and 480 pink flowers. We can now set up a table for the chi-square test:

Phenotype	Observed	Expected	Obs - Exp	(Obs - Exp)² Exp
Red	200	160	40	10.00
White	400	360	40	4.44
Pink	400	480	-80	13.33

Our chi-square test statistic is 10.00 + 4.44 + 13.33 = 27.77. We have three possible outcomes, and lose one degree of freedom for finite sampling. As with the case of independent assortment, it turns out that we also used the data here to determine our expected results. We know this must be true, because different observed results could give different allele frequencies, and these would give different expected genotype frequencies. In this case, we calculated only one parameter, p. Yes, we also calculated q, but we didn't have to (except to check our arithmetic), because we know that q is completely dependent upon p. We, therefore, have 3 minus (1 + 1) = 1 degree of freedom. Comparing the value of 27.77 to the chi-square distribution for 1 degree of freedom, we estimate that the probability of getting this value or higher of the statistic is less than 1%. Therefore, we will reject the hypothesis that the population is at Hardy-Weinberg equilibrium with respect to the A gene.

We're not quite done. When we reject Hardy-Weinberg equilibrium, it's worthwhile to reflect upon the possible explanations. We see a deficit of pink flowers and an excess of red and white flowers. A simple explanation is selection against pink (or for red and white). While emigration is hard to imagine for flowers, immigration isn't too hard to visualize (think seed dispersal). Drift is a possibility, but wouldn't likely have this strong an effect in one generation. Mutation is unlikely, because mutation is rare; again, the deviations are too large. Assortative mating is still a possibility. Perhaps there is reproductive compatibility associated with flower color, such that plants with the same colored flowers are most compatible. This would lead to a deficit of heterozygotes. We can't objectively decide which of these explanations is best, but we could plan experiments to test them. Our test has helped us narrow our search for an explanation for flower color frequency in this population.

Exercise $\PageIndex{1}$

In fruit flies, the enzymatic activity differs for two alleles of Alcohol Dehydrogenase ("fast" and "slow"). You sample a population of fruit flies and test enzyme activity. Form this, you determine that the sample is represented by 60 fast/fast, 572 fast/slow and 921 slow/slow individuals. Does it appear that the population is at Hardy-Weinberg equilibrium?

Answer

We will use the following arbitrary notation for the genotypes:

A^F/A^F = fast/fast A^F/A^S = fast/slow A^S/A^S = slow/slow

There are 60 + 572 + 921 = 1553 flies in our sample.

The allele frequencies for A^F and A^S are: f(A^F) = (60 + 572/2) / 1553 = 0.223 f(A^S) = (921 + 572/2) / 1553 = 0.777 Therefore, the expected numbers of flies with each genotype are: 1553 x 0.223² = 77.2 A^F/A^F 1553 x 2 x 0.223 x 0.777 = 538.2 A^F/A^S

1553 x 0.777² = 937.6 A^S/A^S

We can now set up the table for the goodness-of-fit test:

Genotype	Observed	Expected	Obs-Exp	(Obs-Exp)² / Exp
A^F/A^F	60	77.2	-17.2	3.83
A^F/A^S	572	538.2	33.8	2.12
A^S/A^S	921	937.6	-16.6	0.29
Total	1553	1553.0	0.0	6.24

The value of the test statistic is 6.24. There is one degree of freedom. From the table of critical chi-square values with 1 d.f., we find that 6.24 falls between the critical values for p=0.05 and p=0.01. Therefore we would say the 0.01 < p < 0.05, and we would reject the hypothesis that the population is Hardy-Weinberg equilibrium. There is an apparent excess of heterozygotes and a deficit of both homozygotes.

Source: https://sites.radford.edu/~rsheehy/G...ial/x2-tut.htm

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