Introduction to Population Genetics
Introduction to Population Genetics Biol 519
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Date Created: 09/17/15
Coarse Notes Population Genetics MUTATION INTRODUCTION TO MUTATION READING Hedrick pp 357 367 and 376 385 Mutation plays two key roles in evolution 1 It is an evolutionary force that changes gene frequencies 2 It is the ultimate source of all genetic variation 0 Note any complete theory of evolution must explain processes that create mutations Most mutations are rare Alleles are rarely incorrectly replicated Ballpark figures for spontaneous mutation rates Most mutations are deleterious Most common mutations involve loss of function Deleterious mutations are generally recessive Mutation is a destructive force Mutation is a creative force MUTATION AS A FORCE OF EVOLUTIONARY CHANGE Simplest case two alleles A and a 0 Reality consider an exon that is 100 bp long Approximation alleles refer to 2 classes of alleles e g Tnot T 2G C or A Will consider recurrent mutation How do allele frequencies change given recurrent mutation Rate of mutation probability a gene copy will mutate per generation III 1 Coarse Notes Population Genetics Let u 2 rate of mutation from A to a forward mutation rate v 2 rate of mutation from a to A backward mutation rate 0 Equations of genetic change 17 p1 u 1 pv App pupv1P A v 3 uvpv Observations 1 Since u v are small Ap due to mutation is typically small Mutation is a quotweakquot evolutionary force 2 Ap is linear inp Ie in contrast to selection no term like p1 p rate of creation of new mutations depends on both 0 the frequency of quotmutablequot copies e g p 0 the mutation rate per copy u 3 13 makes sense proportional to relative mutation rates at equilibrium the number of A39s converted to as number of as to As 4 Can rewrite above rate equation as A p u v p Shows that the rate of evolution towards equilibrium is proportional to 0 the total mutation rate 0 the deviation of the current allele frequency from the equilibrium 3 v u v 5 The rate of mutation does not depend on genetic variance No variance factor in the equation for Ap Mutation as an evolutionary force does not require pre existing variation 0 in fact it39s most effective as an evolutionary force when p 0 or p 1 0 How fast is equilibrium approached A I A p p1 u v 120 p Answer very slowly e g when u v 1076 it takes almost 350000 generations to evolve half way to the equilibrium f Q 0 5 111 2 Coarse Notes Population Genetics 0 Punch Line Mutation is an extremely weak force in changing allele frequencies MUTATION amp POPULATION SIZE 0 The randomness of mutation Have been assuming that population size is infinite Allows us to treat mutation as a deterministic force 0 Q But isn39t mutation random A Which allele copies mutate is decided by chance 0 In an infinite sized population mutations are always occurring if the mutation rate is positive so mutation is deterministic when viewed at the population level 0 In reality Consider a single locus determined by a 500 bp sequence of DNA Can have 4500 z 10300 alleles No real population will carry all these alleles Reason isn t mutation rather the constraint of finite population size 0 Finite Population Size and Mutation Selection interactions Some qualitative conclusions without the details which we ll consider soon Deleterious mutations Genetic diversity not affected much by population size as long as it s not too small Favorable mutations Adaptive evolution must wait for advantageous mutations to arise Larger populations provide more opportunities for advantageous mutations to appear each generation Is evolution faster in larger populations 0 Two Possibilities 1 Selection works mainly on existing variation III 3 Coarse Notes Population Genetics 2 Evolution is constrained by the appearance of unconditionally advantageous mutations Which prevails in nature is not known 0 Another possibility environmental change Formerly deleterious mutations become advantageous In this case evolution is limited by 0 the rate of relevant environmental changes 0 the qualities of deleterious mutations that are maintained MUTATION amp DRIFT The in nite alleles model 0 Q How much heterozygosity is maintained if mutation can produce numerous alleles at a locus Can mutation and drift alone explain observed levels of genetic variation in protein and DNA sequence data 0 Consider the infinite alleles model Assume No mutation is repeated ie every new mutant allele is novel 0 Will focus on ft Why Because as tbecomes large all homozygotes will have genes descended from a single ancestral mutant allele ie prob quotautozygosityquot gt 0 Thus f t will thus be a direct measure of homozygosity Intuitively expect f 75 1 since recurrent mutation inputs novel alleles but what balance is reached 0 f H1 2 Prneither allele mutated X Prdescended from one gene in the last generation PrIBD but not descended from the same gene in the last generation or f 1 u2 1 fl III 4 Coarse Notes Population Genetics A 1 0 As I gt 00 f t gt f 2 eventual probability of homozygosity or the 4Nu 1 average frequency of homozygotes A 1 The quantity 4Nu is often denoted by 9 so that f n 2 gives a good idea of homozygosity but how many different alleles are maintained in a population of size 2N Minimum number 1 Maximum 2N Define 1 J 2 quoteffective number of allelesquot in the population this is directly related to the amount of heterozygosity Rationale Consider a population in H W proportions with 3 equally frequent alleles Freq homozygotes 132 132 132 13 0 Then 1Freq of homozygotes 113 3 alleles Consider a pop whose alleles have unegual freqs 14 14 12 Freq homozygotes 1 22 142 1I42 58 gt 83 226 quotequally frequentquot of quoteffectivequot alleles At equilibrium effective number of alleles is 1 4Nu 1 0 1 Punchline Substantial genetic variability is possible with mutation if ZNu gt 1 ie at least one mutantlocus generation 0 Dynamics of Mutation amp Drift Q What is the rate of substitution of selectively neutral mutations 1 How frequently does a new neutral mutation arise Ans 2Nu where u is the mutation rate to selectively neutral alleles per generation III 5 Coarse Notes Population Genetics 2 How likely is this new mutant to be fixed Ans Since alleles are quotneutralquot lZN 3 How frequently will one neutral allele replace another Ans how often they arise x how often they fix 2 ZNu x 1 2N u per generation Conclude Rate of neutral substitution 2 neutral mutation rate 0 Note the substitution rate is independent of N Why While neutral mutations are more likely to fix in a smaller population the rate at which they arise 2Nu is small Just the opposite is true for larger pops The result is that substitution rate is independent of a population s size Property is consistent with a central observation motivating the quotmolecular clockquot hypothesis Property is also central to the quotneutral theoryquot of molecular evolution advanced by Kimura Problem Some observed substitution rates are constant per year rather than per generation III 6 Population Genetics HANDOUT 12 Sex linkage and HardyWeinberg Let s consider a diploid population with X Y sex determination females are XX males are XY We want to study evolution of a locus with two alleles on the X chromosome with no counterpart on the Y chromosome Some notation p ft 2 frequency of the A allele among X gametes in females in generation I pmt 2 frequency of the A allele among X gametes in males in generation t Under the H W assumptions the following offspring genotype frequencies are found Daughters Sons AA PAAt1pftpmt AY B1Yt1Pft Ad PMI1Pft1Pmt Pmt1Pft Y flyt11Pft aa PMt 1 1 pftl pmt Allele frequencies among the offspring computed from these genotype frequencies are l 0 p ft 1 5 p ft pmt average of allele frequencies in both parent sexes pmt 1 p ft the allele frequency among just the female parents Evolutionag gnamics Suppose we have the extreme case pm0 1 p f0 0 t 0 1 2 3 pt 1 0 05 025 pft 0 05 025 0375 1 0395 m 0 1 2 3 t Unlike cases we ve seen up until now the evolutionary paths oscillate towards an equilibrium What equilibrium is eventually reached It turns out that the frequency of the A allele becomes peq xpm0 2Apf0 in both sexes In the above case peq 1 3g 0 Z Coarse Notes Population Genetics NONRANDOM MATING amp GENETIC DRIFT NONRANDOM MATINGINBREEDING READING Hedrick pp 479 482 237 279 OWill distinguish two types of nonrandom mating 1 Assortative mating mating between individuals with similar phenotypes or among individuals that occur in a particular location 2 Inbreeding mating between related individuals Both types of nonrandom mating may have similar consequences since individuals with similar phenotypes often have similar genotypes It is often difficult to separate cause from effect Eg individuals with similar phenotypes may mate because a phenotypic assortative mating occurs b mating with relatives is preferred c matings are primarily based on proximity I Population subdivision The Wahlund Effect It turns out that population subdivision per 38 can effect the distribution of genotypes in the entire population Consider a locus with 2 alleles A and a and a collection of isolated subpopulations numbered 1 2 3 Let the frequencies of A and a in subpopulation i be pi and q Assuming random mating within each isolated subpopulation 39 FreQAA in subpopulation i pf FreqAd in subpopulation i 217in Freqaa in subpopulation i q Let 7 Av g pi average freq of A across all subpopulations Likewise let a Avgqi 1 1 7 What is the average frequency of each genotype over all the subpopulations II l Coarse Notes Population Genetics 0 Consider AA homozygotes first From Fun Facts VarX EX2 EX2 so EX2 VarX 2 So AYgp2 p2 p Varp Similarly for ad homozygotes Avgqi2 q2 2 Varq C72 Varp since Varq Var1 p Varp Finally for Ad heterozygotes AYg2pqi1 p2 q2 1 172 C72 2Varp 217C 2Varp A Thought Experiment 0 Suppose genotypes were randomly sampled from a population whose substructure was unknown The frequencies of A and a in sample would be 1 7 and q With random mating would then expect to find genotypes in proportions 2 2 AAAaaa 2p 2pqq But the genotype frequencies observed would be AA Aa ad 2 I 2 pq 1 72 Varp Z g 2Varp z Varp Ie would find an excess of homozygotes and a deficit of heterozygotes compared to expectations Why Simply because of population subdivision and in particular variance in allele frequencies across subpopulations Given across subpopulations differences in allele frequencies the apparent excess in homozygotes and deficit of heterozygotes from what is expected were the entire population to mate at random defines what is called the Wahlund E 39ect The Wahlund effect is a common cause of non conformity to Hardy Weinberg expectations in population samples I INBREEDING Will now consider the genetic consequences of mating between relatives inbreeding e g 11 2 Coarse Notes Population Genetics In 192039s Sewall Wright invented an ingenious E approach to tracking genotypes through pedigrees based on the probability that allele copies are identical by descent Will follow the French geneticist Malecot39s re working of Wright39s method here Q Q G D d Identity by descent IBD Two alleles are identical by descent if 1 both are descended from the same allele in a common ancestor or 2 one allele is descended from the other 0 Will mean IBD relative to a specific base population whose alleles are deemed to be not IBD Definition The inbreeding coefficient f 1 of an individual J is the probability that its two gene copies at a locus are identical by descent Once f I is known it s not hard to find the probabilities that J is AA Act or ad Consider a randomly chosen individual 0 With probability f 1 both gene copies in that individual are IBD Then both will be A if the allele they were copied from in the base population were A 0 But A occurs with frequency p in the base population so the probability of being an AA given both genes are IBD is p gt the probability of getting two A alleles that are IBD is f I p Likewise the probability of being ad with both genes are IBD is f I 1 p 0 With probability 1 f 1 the two genes in an individual will not be IBD Must have descended from different allele copies in the base population Assuming the copies are made independently then with probability p X p p2 the copied alleles are both A genotype AA etc 0 Putting this all together have BM 1 f p2 ffp 13 1 f12p1 p 11 3 Coarse Notes Population Genetics 12m1 1p2 f11 p 0 Great So now only need to determine f 1 Thanks to Wright this is very easy to do Eg Let39s find f in the pedigree above o O a on Need only concentrate on the central part of the pedigree a CD 0 f Probe E c X Probc E c X Probc E g 12 x 12 x 12 18 K o no So the expected genotype frequencies for this pedigree are R 782 1812 Rm 782p1 17 PM 781 p2 181 p ASIDE What is the average frequency of A among individuals with inbreeding coefficient f1 FreqA PAA 1139PA4 1 f1p2 fjp i391 f12P1 17 1 f1p2 17 p2f1p 1 f1p pr P Inbreeding does not affect allele frequencies on average but does affect the probabilities that 2 A s or 2 as co occur in an individual Computing Inbreeding Coefficients in general Say we want to find f in the pedigree at right Rupcs 1 Enumerate each loop 2 Each loop must a go through each individual no more than once b only change from up to down once II 4 Coarse Notes Population Genetics 3 Multiply by Z for each passage through an individual If the passage through an individual involves a change of direction updown l multiply by 1 f2 instead of 12 where f is the inbreeding coefficient for that individual 4 Add the probabilities of each loop 0 For the above example f 0 Some forbidden loops IGECBDEGI goes through G twice IJDBCEGIalready counted IGECBDEJI loop ECBDE already accounted for in fE Evolutionary Application Kin Selection 0 Probability that M individuals share an allele descended from a common ancestor is called the kinship coef cient or coefficient of consanguinity Kinship coefficient between individuals A and B is denoted F AB 0 What is Fa in the last example Clearly it is the inbreeding coefficient of their offspring I f 732 0 The connection between f and F The kinship coefficient of two individuals is equal to the inbreeding coefficient of their perhaps hypothetical offspring 11 5 Coarse Notes Population Genetics 0 e g Fmotherydaughter 14 assuming unrelated non inbred parents Kinship coefficients useful in studying evolution of social especially quotaltruisticquot traits In 1964 W D Hamilton proposed a rule of thumb to determine whether a rare allele will be favored by selection 0 A rare mutation which affects the fitness of its carrier and others will spread if br gt c where b quotbenefitquot 2 increase in fitness to recipients of action 0 2 quotcostquot 2 loss in fitness to actor r 2 quotdegree of relatednessquot can use above rules to find r r 2F Another view of f I 0 f I is the proportion by which heterozygosity is reduced relative to a random mating group with the same allele frequencies If f I 0 the population is in Hardy Weinberg proportions If f I 1 all individuals are homozygous In general Het 1 fHet0 where Het0 2p1 p 217g 0 Consider e g the Wahlund effect Overall frequencies of A and a are 17 and a If individuals from all subpopulations mate at random expect to find 21 a heterozygotes Would actually observe 217 2Varp heterozygotes Thus in this case 21 2Varp1 f217 In order for this to be true f Var 17 Suggested exercise show this Note f 0 even though there is no deficit of heterozygotes within subpopulations and no obvious inbreeding II 6 Coarse Notes Population Genetics THE COALESCENT INTRODUCTION TO THE COALESCENT READING Hedrick pp 452 462 Often want to use patterns of genetic variability to estimate parameters such as mutation and migration rates 4Nu 14Nu10 eg Under infinite alleles model saw that I where 1 is the expected equilibrium heterozygosity and 0 4Nu estimate 9 by replacing 11 with sample heterozygosity Hobserved and solving equation H0 SBYVB for 6 gestimaed observed This illustrates a prospective approach since estimate is based on a forward looking model Hobserved is also assumed representative of entire population s true heterozygosity Alternatively can develop estimates based on models that look backwards in time ie are retrospective and focus entirely on the set of samples alleles rather than entire population Called coalescent approaches Main assumption behind coalescent all alleles at a locus in a sample can be traced back to a single ancestral allele The coalescent lineage genealogy of sampled alleles traced back to their common ancestor mast creee nt Coarse Notes Population Genetics Tk 2 amount of time there are k distinct lineages each Tk is an independent random variable Are interested in T m the total time in all branches of the genealogy until the entire set of alleles coalesces ie can be traced back to a single common ancestor allele 0 For above example Tlot 4T4 3T3 2T2 Since the Tk s are random variables so is Tlot If u mutation rate generation then Edifferent alleles in sample 2 E mutations in genalogy from the common ancestor uETw Coalescent approach often used to estimate 0 4Nu based on the in nitesites model 0 assumes each allele is an infinitely long DNA or polypeptide sequence 0 every mutation occurs at a different site Note infinite sites model like infinite alleles model except in 1AM don t know how different the alleles are 0 Will see that S number of segregating sites ie variable sites can be used to estimate 9 since ES uETm 1 1 0 In fact will show thatETw 4N2 4N177 where n 121 1 2 3 n l number of alleles in sample So ES i Sobserved n SuggeStS3 gestimated 21 1 Logic behind formula for ETw 2 expected time to coalescence for a sample of n alleles Consider the probability of no coalescence in previous generation III 8 Coarse Notes Population Genetics 15I allele s ancestor in previous generation is one of 2N possible alleles 2nd allele has different ancestor in previous generation with probability 1 7 3 d allele has different ancestor from first two alleles with probability 1 7 nth allele has different ancestor from first n l alleles with probability 1 n So Pno coalescence in previous generation 2 Palleles have n distinct ancestors 1 2 n 1 1 1n prev1ous generation 2 1 7 1 7 b7 z17i 21v 21v 2N 2N 2N Finally Pat least one coalescence in previous generation 2 1 Pn distinct 1 2 1 1 2 1 ancestors i 2 L1WMm 2 2N 4N N E 2N 2N Implies time to first coalescence in a sample of n alleles T is geometrically distributed 2N Geometric distribution is well studied E g know that n n By similar argument i611 Know 71m n n 1YH2T2 so 4N 1 1 ETm ngm i E 4N1E Coalescence based derivation of equilibrium homozygosity 2 under 1AM Consider Ptwo alleles not IBD 1 f 0 Two alleles will be IBD if they coalesce before a mutation occurs on either lineage since Ptwo gene coalesce 1 2N per generation and Pmutation 2u per generation pIBD1i 1 12N2u 14Nu 19 Same result as before but coalescent approach far easier III 9 Coarse Notes Population Genetics Example Aguad et al 1989 yellow achaete scute region of D melanogaster examined 11 64 chromosomes found 9 polymorphic sites out of 2112 nucleotide sites 0 9mmde entire region 54 19 E i 1 i2 Implies about 3 effective alleles segregating 0per site 0region2112 9 X 104 per site 111 10
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