20. Applications of the Genetic-Drift Model

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1 0. Applicaions of he Geneic-Drif Model 1) Deermining he probabiliy of forming any paricular combinaion of genoypes in he nex generaion: Example: If he parenal allele frequencies are p 0 = 0.35 and q 0 = 0.65, wha is he probabiliy ha a random sample of 4 progeny will consis of hree heerozygoes and a homozygoe: A 1 A, A 1 A, A 1 A, A A? Soluion: The probabiliy of sampling n 1 = 3 A 1 alleles in a sample of = 8 is given by he binomial erm C p 3 q 5 = 8! ( 0. 35) ( 0. 65) ! 5! = ( )(. )(. ) =.. Thus he probabiliy of sampling his paricular combinaion of progeny genoypes is 7.9% Relaive frequency Allele frequency (p) ) Deermining he probabiliy of fixaion in he nex generaion: Example for small populaion size (exac binomial probabiliies): Given an iniial populaion of N = 4 wih p 0 = 0. and q 0 = 0.8, wha is he probabiliy of one allele being los due o drif? Soluion: From he binomial disribuion, find he probabiliies of sampling n 1 of gamees, where n 1 is equal o 0 or 8 (he condiions for loss of he A 1 and A alleles, respecively, due o sampling variaion). 0 8 Pr( 0 of 8) = C p q = ( 1)( 1)( ) = Pr( 8 of 8) = C p q = ( 1)( )( 1) = The sum of hese probabiliies is 0.168, indicaing ha here is a 16.8% probabiliy of fixaion in he nex generaion. Noe ha, since here is only one way o sample eigh alleles of he same kind, he N oal probabiliy of fixaion in general reduces o p + q.

2 0. Applicaions of drif model Relaive frequency Allele frequency (p) Example for larger populaion size (normal approximaion): Given an iniial populaion of N = 15 wih p 0 = 0. and q 0 = 0.8, wha is he probabiliy of one allele being los due o drif? Soluion: Deermine he variance expeced due o drif, and model he expeced variaion in allele frequencies wih a normal disribuion cenered on he iniial frequency and having he expeced variance. From his disribuion, deermine he probabiliies of randomly drawing an allele frequency 0 or 1. z z 0 0 L U p q ( 0. )( 08. ) = = = ( 15) p0q0 = = = = -. = = = -. = z L and z U are he lower and upper values of he sandard-normal disribuion ha correspond o allele frequencies of 0 and 1, respecively. The probabiliy of randomly drawing a value z L, he lef-ailed probabiliy from he sandard-normal disribuion, is virually zero. The corresponding righ-ailed probabiliy of randomly drawing a value z U is The probabiliy of fixaion is he sum of hese, indicaing ha, on average, in only 3 imes ou of a 1000 would fixaion occur in he nex generaion.

3 0. Applicaions of drif model 3 N = 30, p = Relaive frequency Allele frequency (q)

4 0. Applicaions of drif model 4 3) Deermining wheher an observed change in allele frequency is consisen wih he model of geneic drif: Example: We have observed a change from q 0 = 0.50 o q 1 = 0.45 in a populaion of size N = Is his amoun of change consisen wih a model of geneic drif in he absence of oher effecs (selecion, gene flow, ec.)? Soluion: Express he observed change in relaion o he variance expeced due o drif. Because he variance is in squared unis, he square roo of he variance (= sandard error) is used insead. The raio of he change o he sandard error is a z-saisic, disribued as a sandard-normal disribuion. = q1 - q0 = p q ( 05. )( 05. ) = = = ( 5000) 0 0 p0q0 = = z = = = The z value is he number of sandard-deviaion unis of he observed change. The probabiliy of observing such a large magniude of change is virually zero, and hus is unlikely under he geneicdrif model. This resul suggess ha some oher facor(s) have probably conribued o he change. Example: We have observed he same change from (q 0 = 0.50 o q 50 = 0.45) over a period of 50 generaions. Is his amoun of change consisen wih a model of geneic drif in he absence of oher effecs? = q - q = , ( 5000),50 50 = p q ( 1- ( 1- ) ) = ( 05. )( 05. )( 1- ( 1- ) ) = 001. = z = = = The probabiliy of randomly observing a value 1.4 sandard deviaions from he mean is 0.16 (- ailed), assuming ha such changes are normally disribued. In oher words, approximaely 16% of he ime a change of 0.05 or greaer could be observed by chance alone, so ha he observaion is consisen wih (= no significanly differen from) he drif model.

5 0. Applicaions of drif model 5 4) Pariioning variaion due o drif ino wihin-populaion and among-populaion componens: Example: Two populaions, each of size N = 50, change independenly due o drif over he same number of generaions (). The iniial frequencies in boh populaions were p 0 = q 0 = 0.5. In populaion X, he frequencies afer generaions are p = 0.4, q = 0.6. In populaion Y, he frequencies afer generaions are p = 0.55, q = (a) By how much has variabiliy been reduced wihin populaions X and Y due o drif? Soluion: Because he variabiliy due o drif is variance in allele frequencies raher han oal geneic variance (heerozygosiy, =pq, which is he variance in genoype frequencies), we mus esimae allelic variance before and afer drif. The reducion in variance due o drif is he difference beween hem. Before drif: Recall ha he general definiion of a variance is x = E x - [ E x ] ( ) ( ), where he expeced value E(x) is he weighed mean of x. For he allele frequency q, we can consruc he following able ha expresses he allele frequency for each genoype. The variance we wan is a funcion of he mean q, weighed by genoype frequencies. Noe he dependence on Hardy-Weinberg assumpions in esimaion of he genoype frequencies. Genoypes: A 1 A 1 A 1 A A A Genoype frequencies: q: [ ] w( q) = Eq ( ) Eq ( ), where var w(q) is he wihin-populaion variance. ) = ( 0. 5)( 0) + ( 050. )( 05. ) + ( 05. )( 1) = 05. ) = ( 05. )( 0) + ( 050. )( 05. ) + ( 0. 5)( 1) = ( q ) = (0.5) = 0.15 w Noe ha 0.15 is he maximum allelic variance ha is possible wihin a populaion.

6 0. Applicaions of drif model 6 Afer drif, populaion X: p = 0.4, q = 0.6. Genoypes: A 1 A 1 A 1 A A A Genoype frequencies: q: ) = ( 016. )( 0) + ( 0. 48)( 05. ) + ( 0. 36)( 1) = ) = ( 016. )( 0) + ( 0. 48)( 05. ) + ( 036. )( 1) = ( q ) = 0.48 (0.6) = 0.10 w The wihin-populaion variance of populaion X has been reduced from 0.15 o Afer drif, populaion Y: p = 0.55, q = Genoypes: A 1 A 1 A 1 A A A Genoype frequencies: q: ) = ( )( 0) + ( )( 05. ) + ( 003. )( 1) = ) = ( )( 0) + ( )( 05. ) + ( 003. )( 1) = ( q ) = 0.37 (0.451) = 0.14 w The wihin-populaion variance of populaion Y has been reduced from 0.15 o (b) By how much has variabiliy beween populaions X and Y been increased due o drif? Soluion: The beween-populaion variance is based on he mean q, weighed by is frequencies wihin populaions raher han wihin genoypes. Populaion: X Y Populaion frequencies: = = q: ) = ( 05. )( 0. 60) + ( 05. )( 045. ) = 055. ) = ( 05. )( 060. ) + ( 05. )( 0. 45) = ( q) = = 0.81 (0.55) = b, The beween-populaion variance has been increased by drif from 0 (because he populaions iniially were idenical) o

7 0. Applicaions of drif model 7 5) Esimaion of he number of generaions needed o obain a given amoun of divergence among populaions: Example: The wo populaions above, each of size N = 50, changed independenly due o drif over he same number of generaions (). Esimae he number of generaions required o obain he observed amoun of beween-populaion variance. 1 Soluion: The expeced variance in change across generaions is, p0q0[ 1 ( 1 N ) ] = - -. Given he observed beween-populaion variance, he iniial allele frequencies, and he populaion size, his equaion can be solved for : 1 [ 1 1 ], = p0q0 - ( - N ) Ø = ( 05. )( 05. ) 1-1- º ŒŒ Ł ( 50) ł [ ] = ( 05. ) 1- ( 099. ) = 1- ( 0. 99) (. ) = -. = ln( 0. 99) = ln( ) ln( ) = = - =. 4 ln( 0. 99) Thus he observed amoun of differeniaion beween he wo populaions could be expeced o occur, on average, in less han 3 generaions. ø ß œœ In general, q δ ln 1 p0q0 =. 1 ln 1 Recall ha he expeced ne change of (over numerous subpopulaion replicaes) is zero: E( ) = 0. Thus, if p 0 and q 0 are no known, heir bes esimaes are he mean allele frequencies among he wo (or more) populaions a ime.