The Hardy-Weinberg Principle andestimating allele frequencies inpopulationsIntroductionThe genetic composition of a population consists of three components when we confineourselves to the study of variation at a single locus:1. The number of alleles at a locus.2. The frequency of alleles at the locus.3. The frequency of genotypes at the locus.It may not be immediately obvious why we need both (2) and (3) to describe the geneticcomposition of a population, so let me illustrate w ith two hypothetical populations:A1A1A1A2A2A2Population 1 50 0 50Population 2 25 50 25It’s easy to see that the frequency of A1is 0.5 in both populations,1but the genotypefrequencies are very different. In point of fact, we don’t need both genotype and allelefrequencies. We can always calculate allele frequencies from genotype frequencies, but wecan’t do the reverse unless . . .1p1= 2(50)/200 = 0.5, p2= (2(25) + 50)/200 = 0.5.c 2001-2004 Kent E. HolsingerDerivation of the Hardy-Weinberg principleWe saw last time using the data from Zoarces viviparus that we can describe empirically andalgebraically how genotype frequencies in one generation are related to genotype frequenciesin the next. Let’s explore that a bit further. To do so we’re going to use a technique that isbroadly useful in population genetics, i.e., we’re going to construct a mating table.Offsrping genotypeMating Frequency* A1A1A1A2A2A2A1A1× A1A1x2111 0 0A1A2x11x1212†120A2A2x11x220 1 0A1A2× A1A1x12x1112120A1A2x212141214A1A2x12x2201212A2A2× A1A1x22x110 1 0A1A2x22x1201212A2A2x2220 0Believe it or not, in constructing this table we’ve already made three assumptions aboutthe transmission of genetic variation from one generation to the next:Assumption #1 Genotype frequencies are the same in males and females, e.g., x11is thefrequency of the A1A1genotype in both males and females.*Assumption #2 Genotypes mate at random with respect to their genotype at this partic-ular locus.*Assumption #3 Meiosis is fair. More spec ifically, we assume that there is no segregationdistortion, no gamete competition, no differences in the developmental ability of eggs,or the fertilization ability of sperm.†Now that we have this table we can use it to calculate the frequency of each genotype innewly formed zygotes, provided that we’re willing to make three additional assumptions:Assumption #4 There is no input of new genetic material, i.e., gametes are producedwithout mutation, and all offspring are produced from the union of gametes withinthis population.2Assumption #5 The population is of infinite size so that the actual frequency of matingsis equal to their expected frequency and the actual frequency of offspring from eachmating is equal to the Mendelian expectations.Assumption #6 All matings produce the same number of offspring, on average.Taking these three assumptions together allows us to conclude that the frequency of a par-ticular genotype in the pool of newly formed zygotes isX(frequency of mating)(frequency of genotype produce from mating) .Sofreq.(A1A1in zygotes) = x211+12x11x12+12x12x11+14x212= x211+ x11x12+14x212= (x11+ x12/2)2= p2freq.(A1A2in zygotes) = 2pqfreq.(A2A2in zygotes) = q2Those frequencies probably look pretty familiar to you. They are, of course, the familiarHardy-Weinberg proportions. But we’re not done yet. In order to say that these proportionswill also be the genotype proportions of adults in the progeny generation, we have to maketwo more assumptions:Assumption #7 Generations do not overlap.Assumption #8 There are no differences among genotypes in the probability of survival.The Hardy-Weinberg principleAfter a single generation in which all eight of the above assumptions are satisfiedfreq.(A1A1in zygotes) = p2(1)freq.(A1A2in zygotes) = 2pq (2)freq.(A2A2in zygotes) = q2(3)3It’s vital to understand the logic here.1. If Assumptions #1–#8 are true, then equations 1–3 must be true.2. If genotypes are in Hardy-Weinberg proportions, one or more of Assumptions #1–#8may still be violated.3. If genotypes are not in Hardy-Weinberg proportions, one or more of Assumptions #1–#8 must be false.Point (3) is why Hardy-Weinberg is so important. There isn’t a population of anythinganywhere in the world that satisfies all 8 assumptions, even for a single generation.2Butall possible evolutionary forces within populations cause a violation of at least one of theseassumptions. Departures from Hardy-Weinberg are one way in which we can detect thoseforces and estimate their magnitude.3Estimating allele frequenciesBefore we can determine whether genotypes in a population are in Hardy-Weinberg propor-tions, we need to be able to estimate the frequency of both genotypes and alleles. This iseasy when you can identify all of the alleles within genotypes, but suppose that we’re tryingto estimate allele frequencies in the ABO blood group system in humans. Then we have asituation that looks like this:Phenotype A AB B OGenotype(s) aa ao ab bb bo ooNo. in sample nANABNBNONow we can’t directly count the number of a, b, and o alleles. What do we do? Well, if weknew pa, pb, and po, we could figure out how many individuals with the A phenotype havethe aa genotype and how many have the ao phenotype, namelyNaa= nA p2ap2a+ 2papo!Nao= nA 2papop2a+ 2papo!.2There may be some that come reasonably close, but none that fulfill them exactly.3Actually, there’s a ninth assumption that I didn’t mention. Everything I said here depends on theassumption that the locus we’re dealing with is autosomal.4Obviously we could do the same thing for the B phenotype:Nbb= nB p2bp2b+ 2pbpo!Nbo= nB 2pbpop2b+ 2pbpo!.Notice that Nab= NABand Noo= NO(lowercase subscripts refer to genotypes, uppercaseto phenotypes). If we knew all this, then we could calculate pa, pb, and pofrompa=2Naa+ Nao+ Nab2Npb=2Nbb+ Nbo+ Nab2Npo=2Noo+ Nao+ Nbo2N,where N is the total sample size.Surprisingly enough we can actually estimate the allele frequencies by using this trick.Just take a guess at the allele f requencies. Any guess will do. Then calculate Naa, Nao, Nbb,Nbo, Nab, and Nooas described in the preceding paragraph.4That’s the Expectation part ofwhat’s called the EM algorithm. Now take the values for Naa, Nao, Nbb, Nbo, Nab, and Noothat you’ve calculated and use them to calculate new values for the allele frequencies. That’sthe Maximization part of the EM algorithm. Chances are your new values for pa, pb, andpowon’t match your initial guesses, but5if you take these