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3 Non-random mating, Inbreeding and Population Structure.
Jewelweed, Impatiens capensis , is a common woodland flower in the Eastern US. You may have seen the swollen seed pods that explosively pop when you touch time, which is the source of an alternative name for this plant, “touch-me-not”. Another interesting thing is that it produces two kinds of flowers. Many of the flowers are orange and conspicuous. Those flowers have various features to attract pollinators, primarily nectar that is secreted at the base of the curved spur in the back of the flower. Bees and hummingbirds visit the flower to feed on the nectar and in the process transfer some of the pollen from one flower to another. Most plant species produce both pollen and ovules in the same flower so there is a potential for self-fertilization to occur. That can happen either when pollen is transferred to the stigma of the same flower, or when pollen is transferred between flowers on the same plant. The showy flowers have evolved various mechanisms to limit self-fertilization within a flower. They typically produce pollen for one day, after which the anther falls off to reveal the stigma. Therefore pollen receipt (female function) is separated in time from pollen donation (male function). Within flower selfing is minimal because pollen and stigma are not functional at the same time. In addition to the showy orange flowers, the plant produces another set of flowers that never even open. Those tiny green cleistogamous (meaning “closed mating”) flowers look more like buds. Pollen is shed inside the unopened flower directly onto its stigma, so all of the seeds are produced by self-fertilization. Presumably the cleistogamous flowers have a reproductive assurance advantage, since they are able to produce seeds even in the absence of pollinators. Thus Impatiens shows two extremes of mating: 100% self-fertilization in the tiny cleistogamous flowers, and normal crossing between plants in the showy (“ chasmogamous ”) flowers.
3.1 What are the effects of nonrandom mating on allele and genotype frequencies? The Hardy Weinberg Equilibrium was defined for the case of random mating. However it is rare that individuals in a population truly mate at random 1
. Very often populations are spatially structured and individuals are more likely to mate with others that are nearby than with individuals from farther away. Often social structure will limit mating opportunities on others within the group. And some species, like jewelweed, routinely self-fertilize. Self-fertilization: Lets start with the extreme case of self-fertilization. How do allele and genotype frequencies change after one generation of self-fertilization? As before we'll consider a single locus with two alleles, A and a. We'll use upper case letters for the genotype frequencies in that population. Let P be the frequency of AA homozygotes, H be the frequency of heterozygotes and Q be the frequency of aa homozygotes. Because those are all of the possibilities, P+H+Q =1. After selfing, all of the AA homozygotes produce only AA offspring and all of the aa homozygotes produce only aa offspring. However, selfing within the heterozygotes produces all three genotypes of offspring: 1/4 will be AA, 1/2 will be Aa and 1/4 will be aa. Genotype Initial Frequency Frequency after selfing AA P P + 1/4 H Aa H 1/2 H aa Q Q + 1/4 H The result is that the frequency of the homozygous genotypes increases with each generation of selfing, and the frequency of heterozygotes decreases. In particular, the change in heterozygosity with selfing is: ! H ' = 1 2 H That shows that the frequency of heterozygotes decreases by half each generation. Eventually, after many generations of complete selfing the frequency of heterozygotes will decline to zero. Effect of selfing on allele frequency: You can always calculate the allele frequency as ! p = P + 1 2 H. So after selfing, and substituting the new values for P and H we get (^1) It is important to remember that we are talking about random mating with respect to a particular locus. Birds may choose mates non-randomly with respect to plumage, but since unlinked loci are inherited independently, they may very well be mating at random with respect to a certain enzyme locus.
D. Stratton 2008- 2011 4 The obligate self-fertilization in the tiny cleistogamous flowers will result in a steady increase in the frequency of homozygous genotype but it will have no effect on the allele frequencies in the population. What happens with the showy flowers? Do their seeds fit the expectations of random mating? Observed and predicted genotypes for chasmogamous flowers: Imagine that you collect a sample of 200 seeds that were all produced by the showy chasmogamous flowers. Fertilization of those flowers requires visits by pollinators so those seeds should represent matings between various individuals. You then determine the genotype of each seed to test whether they fit the Hardy-Weinberg expectations. Below are results for one enzyme locus, isocitrate dehydrogenase or IDH. The two alleles are labeled “1” and “2”. IDH Genotype Observed number Predicted number under HWE 11 104 92. 12 64 87. 0 22 32 20. 5 (Data are simplified, but are based on real information in Stewart ( 1994 ) for I. pallida ) What are the allele frequencies in this population? p=(frequency of allele “1”)= _____________ q=(frequency of allele “2”)=_____________ Under HWE, we expect the proportions of the three genotypes to be p 2 , 2pq and q 2
. The observed allele frequencies are p=0.68 and q=0.32, so the expected genotype frequencies are p^2 =0.46, 2pq=0.44, and q^2 =0.10 for genotypes 11, 12, and 22 respectively. That means that in a sample of 200 individuals, we expect 92.5 “11”, 87.0 “12” and 20.5 “22” genotypes. For this sample the observed number of seeds of each genotype does not quite match the predicted numbers. The data show an excess of the two homozygotes and a deficit of heterozygotes compared to the Hardy-Weinberg expectations. Is that difference real or could it just be a random deviation in this sample? 3.2 Testing for departures from HWE If we took repeated samples of seeds from this population we would likely find slight variation in the numbers of each genotype just by chance. So it is not surprising that the numbers don’t exactly match the predicted numbers. The question is whether our results are within the normal range of variation or not. To decide whether it is unusual or not we need to do a statistical test, in this case a chi-square ( !
2 ) test. This test calculates the sum of squared deviations from the expected values and compares that sum to the value you would expect from random chance alone. The ! " 2 value is calculated as followed:
( Oi # Ei )^2 i Ei
where Oi is the observed number of genotype i, and Ei is the expected number of that genotype. IDH Genotype Observed number Expected number !
( O " E )^2
E
Sum 200 200 13. Notice that the expected values are not integers. The observed numbers will always be counts of individuals but the expected values can be, and usually are, decimal numbers. The observed and expected values will always add up to the same total number of individuals, however. For our example, !
2
2
2
2
By itself, the !
2 value doesn’t mean much. Even if the population actually fit the assumptions for HWE we would expect slight deviations in the numbers of genotypes in our particular sample of individuals just by random chance. Fortunately statisticians have calculated the probability of observing particular ! "^2 values by chance alone. That probability depends on the degrees of freedom : the number of independent data points that we have. For this test, the degrees of freedom is the number of genotypic classes (3) minus the number of parameters estimated from our data (1, for the allele frequency p) minus 1. That leaves 1 degree of freedom. 2 This figure shows the probability of observing a chi square value at least as big as this by chance alone. By convention, statisticians usually say a result is “statistically significant” if there is less than 5% of chance of it occurring by chance alone. That 5% threshold corresponds to a ! "^2 value of 3.84: there is only a 5% chance of observing a chi-square value at least as big as 3.84. (^2) You might wonder why we don’t also have to subtract another degree of freedom for the allele frequency q. The answer is because once p is known, we also know q since it is just 1-p.
3.4 FST is a measure of spatial structure Although it would be nice to just define a population as the collection of individuals that are mating together randomly, the boundaries of a population are often very indistinct. A plant that is growing on one side of a meadow is much more likely to pollinate others growing nearby, so in that sense two sides of a meadow may be genetically separated. But when the plants are scattered evenly across that meadow, where would you draw the boundary? In addition, there may be behavioral patterns that cause individuals to mate non-randomly. If individuals live in family structured groups (e.g. wolves) they may be likely to mate with relatives. The result is that populations are often genetically structured, at a variety of scales. Behavioral traits may result in individuals that are inbred with respect to the local subpopulation. And spatial separation may result in sub-populations that are "inbred" with respect to the population as a whole. In the latter case, inbreeding arises because individuals are on average more related to others within their own subpopulation than they are to others in more distant sites. If mating takes place within a subpopulation then they are mating with closer relatives than they would with a truly random sample of mates from the population as a whole. In section 3.1 we used the deficit of heterozygotes to quantify inbreeding as !
F = 1 "
Ho He
Here we can use a similar approach to measure that level of inbreeding due to population structure, this time by comparing the heterozygosity within subpopulations relative to what we would expect if there were random mating over the total population.^3 ! FST = 1 " Average Expected Heterozygosity within Subpopulations Expected Heterozygosity of the Total Population or ! FST = 1 " H (^) s HT The heterozygosity (H) is the expected proportion of heterozygous, 2 pq. If individuals mate at random over the entire population, then the expected total heterozygosity is simply ! HT = 2 p ( 1 " p ). On the other hand, if there is spatial structure and individuals mate within subpopulations then the frequency of heterozygotes will depend on the local allele frequency ( p i) in each subpopulation: ! Hi = 2 pi ( 1 " pi ) for subpopulation i. If there are a total of k (^3) Sewall Wright developed this index of population structure, which he called the "fixation index" or F. That fixation index or inbreeding coefficient can be partitioned into the various levels of population structure. FIS is the fixation index of an i ndividual with respect to its local s ubpopulation and FST is the average fixation index of s ubpopulations relative to the t otal population. It can be applied at any spatial scale. Sometimes people use FST to measure the differentiation among distinct populations in a large region.
subpopulations, then the average expected heterozygosity within subpopulations is: ! H (^) S = 1 k 2 pi ( 1 " pi ) i = 1 k
Example: Imagine that there are two nearby patches of 200 plants each. Mating takes place only among plants from the same local area, but within those patches mating is completely random. The two subpopulations in the two areas have slightly different allele frequencies: In the first patch the frequency of allele A is p = 0.8 and in second the allele frequency is p = 0.5. But in both sites mating is random so the genotype frequencies exactly match the expected frequencies under HWE. In the first patch the heterozygosity is 64/200 = 0.32 and in the second the heterozygosity is 100/200=0.5. Now, imagine that when we sampled the population we didn't keep track of where each plant came from. All we had was a sample of 400 plants that we assumed was from a single large population. The data we collected showed that there were a total of 178 AA homozygotes, 164 Aa heterozygotes and 58 aa homozygotes. What is the frequency of allele A in the combined population? p =
What are the expected genotype frequencies in that combined sample?
The allele frequency in the combined population is 0.65, so the expected heterozygosity is HT= 20.650.35 = 0.455. The average expected heterozygosity in the two subpopulations is HS= (0.32+0.5)/2 = 0.41 so ! FST = 1 " H (^) s HT = 1 "
= 0. To understand the properties of FST, we can look at some extreme conditions. If individuals are mating completely at random over the entire population, then there will be no local variation in allele frequency and each of the subpopulations will have the same expected heterozygosity as the total population. In that case Fst will be 0: there is no differentiation among subpopulations. At the other extreme, each subpopulation may be Genotype Allele Frequency AA Aa aa pA Subpopulation 1 128 64 8 0. Expected 128 64 8 Subpopulation 2 50 100 50 0. Expected 50 100 50 Combined Population 178 164 58 Expected
Answers p 3. After selfing, the genotype frequencies will be: ! p^2 +^12 p (1" p ) ,
p ( 1 " p ) , and ! (1" p )^2 +^12 p (1" p ). Calculate the allele frequency as ! p = P +^12 H so ! p ' = p 2
1 2 p^ (1^ "^ p )^ + 1 2 p^ (^1 "^ p ) = p^2 + p ( 1 " p ) = p^2 + p " p^2 = p p 4. p=(104+(64/2)) / 200 = 0.68; q= (32 + (64/2)) / 200=0. p 6. IDH Genotype Observed number Expected number ! ( O " E )^2 E 11 91 81.92 1. 12 74 92.16 3. 22 35 25.92 3. sum 200 200 7. p 8. The allele frequency in the combined population is p=0. The expected numbers for each genotype are 169, 82, and 49 p 9. Site allele frequency pi Expected Heterozygosity Hi 1 0.4 0. 2 0.2 0. 3 0.35 0. 4 0.7 0. 5 0.6 0. HS = 0.431. HT = 0.495. FST = 1 - 0.431/0.495 = 0.
