p g p g. This implies that for any single locus, the

Transcription

1 SOLUTIONS TO EXERCISES FOR CHAPTER 3 Agronomy 65 Saisical Geneics W. E. Nyquis March 004 Exercise 3.. a. Random maing means ha he arens or maes are geneically or genoyically indeenden of one G i G j = G i G j. anoher, i.e., ( ) ( ) ( ) b. In a single oulaion comosed of unisexual individuals, he frequencies of he genoyes are usually he same in he wo sexes. Assuming ha he frequencies are he same, he frequency of any homozygoe is ( g ) i and he frequency of any heerozygoe is ( i) ( j) g g. This imlies ha for any single locus, he genoyic frequencies are Hardy-Weinberg frequencies. If he frequencies are differen in he wo sexes, hen he frequency of any homozygoe is ( i ) ( i ) ( g i ) ( g j ) + ( g j ) ( g i ). g g and he frequency of any heerozygoe is On he oher hand, for a single oulaion comosed of bisexual individuals, i is wih cerainy ha he frequencies of he genoyes are he same in he sexes because every individual funcions equally as boh a male and a female. Hence, he frequency of any homozygoe is ( g ) i and he frequency of any heerozygoe is ( gi) ( g j). c. In a cross beween wo oulaions he frequencies of he genoyes are usually differen. Hence, he frequency of any homozygoe in he cross is ( i ) ( i ) ( g i ) ( g j ) ( g j ) ( g i ) roorions. g g and he frequency of any heerozygoe is +. This imlies ha for any single locus, he genoyic frequencies are no in H-W d. If an individual is defined as one comosed only of he loci reresened in G, hen he definiion of random maing ha any individual of one sex has an equal chance of maing wih any oher individual of he oosie sex is equivalen. For bisexual individuals his includes self-maing. On he oher hand, if an individual is defined as one comosed of more loci han are included in G han i is no roer o define random maing wih resec o G by saying ha every individual has an equal chance of maing wih any oher individual of he oosie sex. This is rue because he individuals wih any given G i may no be maing a random, i.e., here may be osiive henoyic or geneic assoraive maing wih resec o loci no included in G. For examle, if G denoes a single loci for seed color or blood ye as described in my noes (see. 3.8), random maing may occur wih resec o he genoyes a his locus bu ye wih resec o he individuals (comosed of loci for oher rais or all remaining ones in he organism) here may be osiive henoyic assoraive maing such as flowering ime or inelligience. Exercise 3.. The random maing heorem is imoran in ha one can obain he genoyic srucure of he nex generaion under random maing by using only he gameic frequencies of he revious generaion. Thus, one can avoid he laborious rocedure of randomly airing he arens, considering heir individual gameics ouus, deermining he genoyes of heir offsring and heir frequencies, and hen allying over all ossible genoyic maings.

2 Exercise 3.3. The allelic frequencies are [see equaion (.0)] = (0.0) = = 0.3 A a = (0.0) = = 0.7 The genoyic frequencies afer random maing are [equaion (3.35)] AA = (0.3) = 0.09 Aa = (0.3)(0.7) = 0.4 aa = (0.7) = 0.49 Exercise 3.4. Poulaion : Allelic and Hardy-Weinberg genoyic frequencies are A = ( 0.0) = = 0.3 a = ( 0.0) = = 0.7 AA = A = ( 0.3) = 0.09; Aa = ( 0.3)( 0.7) = 0.4; aa = a = 0.49 Poulaion : Allelic and Hardy-Weinberg genoyic frequencies are A = = 0.3 a = = 0.7 AA = A = ( 0.3) = 0.09; Aa = ( 0.3)( 0.7) = 0.4; aa = a = 0.49 Poulaion 3: Allelic and Hardy-Weinberg genoyic frequencies are A = ( 0.60) = 0.3 a = ( 0.60) = 0.7 AA = A = ( 0.3) = 0.09; Aa = ( 0.3)( 0.7) = 0.4; aa = a = 0.49 Exercise 3.5. A random-maing oulaion showing 5% recessives has he following ercenage of heerozygoes: q = 0.05, so q= 0.05 and = 0.05 ( )( ) = (0.7764)(0.36) = = frequency of heerozygoes or 34.7% heerozygoes Exercise 3.6. Poulaion : A 0.0 q; aa 0.98 q ; q 0.98 q= = q= = 0.00 q = (0.00)(0.9899) = Poulaion : aa = 0.0 q= 0.0 = 0.44 = 0.44 = q = (0.8586)(0.44) = 0.48

3 3 Exercise 3.7. ( ) lim = = lim ( ) =, 0 0 namely, frequency of heerozygoes when is very small. An alernaive way o solve his roblem is o aly L Hosial s Rule (given in any calculus book), which says ha if boh he numeraor and denominaor vanish as he variable aroaches he limi 0, so ha he fracion assumes he indeerminae form 0/0, hen he limi of he original fracion is equal o he limi of he raio of he derivaives of he numeraor and denominaor, namely, ( ) d ( ) d ( ) + ( ) + ( ) ( 0) + ( 0) lim = = lim = lim = lim = = 0 0 d d 0 0 where we aly he derivaive of a roduc o he numeraor, namely, d( uv) dv du = u + v dx dx dx Exercise 3.8. Firs, assuming random maing, we solve his exercise by aking a more brue force, definiional aroach as follows: We lis only hose maings (7 ou of 9 ossible maings) ha give heerozygoes. A A A A A A offsring offsring offsring frororom Male Female Maing Condiional robabiliy A A A A A A aren aren frequency A A A A A A female female female AA AA AA AA AA AA q q q AA AA ( q ) AA AA AA AA AA AA 4 qq q q q 4 3 q q 3 q q 3 q q 3 q q q q Summing he column oals, we have q + q + q = q = oal frequency of heerozygoes in he random-maing oulaion, as i should be. frequency of heerozygous offsring from heerozygous female aren q Hence, = = frequency of heerozygous female aren q

4 An alernaive, more direc way o solve his exercise is he following: Male allele Random Nonrandom A A Female maing maing Offsring Offsring aren frequency frequency AA AA AA AA AA AA AA AA q Q AA q P Q 4 From he above i is easily seen ha regardless of wheher he male allele is A or A one-half of he offsring from he female arenal genoye A A is A A. Finally we resen even a hird nonabulaing, urely robabiliy way o solve he exercise. In se noaion, le A = female aren A A, B = offsring A A, B = offsring A A, B 3 = offsring A A,. P( female aren AA and offsring AA ) P( female aren is AA offsring is AA ) = P( offsring is AA) P( A B ) ( ) ( ) ( ) ( ) P A P B A q P A B = = = = P( B ) ( ) P B q Or, for nonrandom-maing oulaion ( ) Q P( A B ) = = Q Exercise 3.9. a. The frequencies of he hree alleles in he oulaion are A A A 3 AA 6 AA AA Toal Or, alernaively for A only: P( A) = P( AA) P( A AA) + P( AA3) P( A AA3) = + = 3 = 6 3 4

5 b. Afer random maing he genoyic frequencies from acual allying is P G G A A A A A A A A A A A A ( i j) ( AA )( AA ) ( ) = ( AA 4 3)( AA 3) ( 3) = = ( AA)( AA) ( ) = = = = = = = = ( AA )( AA ) ( AA )( A A ) ( AA )( A A ) 5 Mulilying he maing frequencies by he condiional robabiliy for each offsring genoye gives he following frequencies of he offsring genoyes: P A A = = 9 = ; P A A = = 4= ; P A A = = 30 = 5 ; ( ) ( ) ( 3) ( ) ( ) ( ) = + + = = ; = = = ; = = ; P A A P A A P A A c. Yes, he offsring genoyic frequencies are in Hardy-Weinberg equilibrium. If he male and female gameic arrays are equal, Hardy-Weinberg.roorions are obained immediaely in he firs generaion afer random maing. P A A = = ; P A A = = ; P A A = 5 = 5 ; ( ) ( ) ( 3) ( ) ( ) ( ) = = ; = = ; = = ; 44 P A A P A A P A A Exercise 3.0. Three-fourhs of he AO individuals would be execed o be A O, and one-fourhs of he AO individuals would be execed o be A O. The execed roorions of A A, A A, and A A among AA individuals would be A A A A A A 9/6 6/6 /6 Exercise 3.. The maximum roorions of heerozygoes in a random-maing oulaion are m,, and for wo, hree, and m alleles, resecively. See equaion (3.59). 3 m Exercise 3.. Le A be he frequency of he recessive allele. A( A) 8 = A 8A = ( A) = A 0 A = A = = see equaion (3.35) 5

6 ( ) A 98 = A 98 A = ( A) = A 00 A = A = = = Exercise 3.3. (3, 655) ,988 ˆ A = = = , ˆ a = = Execed number (E) O E χ AA ˆ (4,345) = 3, Aa q ˆˆ(4,345) = aa ˆ q (4, 345) = < χ ν =, α = 0.05 = 3.84, nonsignifican Degrees of freedom equal he number of cells minus minus he number of indeenden arameers which are esimaed from he daa and are used o calculae he execed numbers. In his case, he degrees of freedom equal one. Exercise 3.4. Two generaions of random maing are required o reach Hardy-Weinberg roorions when he allelic frequencies a one or more loci are differen in he wo sexes. In firs generaion of random maing, he genoyic frequencies are no in H-W roorions, bu he allelic frequencies are equalized beween he wo sexes due o indeenden assormen of he sex chromosomes in gamee formaion in every individual or for each allele. There are equal frequencies of males and females for every genoye. Then in he second generaion of random maing wih equal allelic frequencies in boh males and females, H-W roorions are esablished. Exercise 3.5. The allelic frequencies in he wo oulaions are A = = 0.5 A = = 0.4 oulaion : oulaion : a = = 0.5 a = = 0.6 In he firs generaion of he cross: Females 0.4 A 0.6 a Males 0.5 A 0.0 AA 0.30 Aa 0.5 a 0.0 Aa 0.30 aa 0.0 AA Aa aa

7 In he second generaion of he cross afer random maing: A 0.55 a 0.45 A 0.05 AA Aa 0.55 a Aa aa 0.05 AA Aa aa Equilibrium roorions would be achieved afer wo generaions of random maing or one generaion afer he cross iself. Exercise 3.6. a. The allelic frequencies in each of he wo oulaions are: Poulaion X X X X X X = + = + ( ) ( ) X X X X X = + = + Poulaion Y Y Y Y Y Y = + ( ) = + Y Y Y Y Y = + ( ) = + b. The genoyic srucure in he cross oulaion is AA AA AA X Y X Y + X Y X Y The allelic frequencies in he cross oulaion are [equaion (3.00)] X Y c + = X Y c + = ( )( ) ( )( ) ( )( ) ( )( ) c. If individuals of he cross generaion mae a random, Hardy-Weinberg equilibrium will be obained in he nex generaion, because wih equal gameic frequencies on he male and female sides and random maing Hardy-Weinberg roorions are obained. The genoyic srucure of he cross generaion afer random maing is AA AA AA c c c c ( ) ( )( ) ( ) X Y X Y X Y X Y = = = d. The numerical values for he allelic frequencies in each of he wo oulaions are Poulaion X X = = 0.45 X = = 0.55

8 Poulaion Y Y = = 0.60 Y = = 0.40 The genoyic srucure for he cross oulaion is 8 Poulaion Y A A Poulaion X A A A A A 0.45 (0.45)(0.60) (0.45)(0.40) = 0.7 = A A A A A 0.55 (0.55)(0.60) (0.55)(0.40) = 0.33 = The allelic frequencies in he cross oulaion are c = = 0.55 c = = The genoyic srucure of he cross oulaion afer random maing is AA AA A A c c c c ( ) ( )( ) ( ) ( ) ( )( ) ( ) = 0.55 = = = = = Exercise 3.7. No, a sable allelic srucure a every locus does no imly a sable mulilocus gameic srucure or a sable genoyic srucure from generaion o generaion. If he oulaion is no in linkage equilibrium, he gameic srucure will change from generaion o generaion. Exercise 3.8. a. Le i = frequency of ih allele a A locus, and r j = frequency of jh allele a B locus. Wih linkage equilibrium he frequency of A i B j gamee is i r j. Using he random-maing heorem, we know ha he frequency of any genoye is he roduc of he gameic frequency, i.e., ( ) ( )( ) A B = r r = rr i i j j i j i j i i j j Thus, for any fixed combinaion i and i, we sum over all combinaions of j and j

9 mb mb mb mb ( AA i i BjBj ) = i i rr j j j= j = j= j = mb mb = i i rr j j j= j = = i i () = ii for every i and i This roves ha H-W roorions exis for he A locus in ha every genoye a he A locus is he roduc of he corresonding allelic frequencies. In like manner we can rove ha H-W roorions exis for he B locus. We can, of course, double sum wih resec o i and i, in addiion o he double sum wih resec o j and j, ma ma mb mb = ii rjrj i= i = j= j = and, in like manner o ha above, esablish H-W roorions for boh he A and B loci. b. Using he random-maing heorem, we know ha he frequency of any genoye is he roduc of he gameic frequencies, i.e., ( i i j j ) = ( i j + ij )( i j + i j ) A A B B r r Furhermore o obain he frequency of any genoye, say AA i i, we mus sum over all genoyes a he B locus mb mb mb mb A A B B r r j= j = j= j = ( i i j j ) = ( i j + ij )( i j + i j ) m B m 9 B ( r i j ij ) ( i rj ij ) = + + j= j = mb mb mb mb = i rj + ij i rj + i j j j j j = = = = = i() + 0 i () + 0 [see equaions (3.6A) and (3.6B)] = i i which roves ha H-W roorions or random maing exiss a he A locus. Likewise we can show ha H-W roorions or random maing exiss a he B locus. Exercise 3.9. Yes, wih mulile alleles linkage equilibrium may exis for some combinaion of i and j in ha A ib = j A i B, bu for oher combinaions of i and j, j AiB j A i B. All ha is required for he exisence of j linkage disequilibrium in he oulaion as a whole is o have A ib j A i B for a leas one combinaion of i j and j. Exercise 3.0. a. From equaion (3.63) he frequency of he gamee is = A B + = ( 0.4)( 0.7) = 0.35

10 The frequencies of all gamees are B b 0 A = 0.35 Ab = a ab = 0.35 ab = The frequencies of he following genoyes in he oulaion are = = 0.5 ( ) ( )( ) ( ab ) ( )( ) Ab ( ) ( )( ) = = = = ab b. When he loci are linked ( ρ = ) 4, he frequency of he gamees from he oulaion are obained direcly as follows: Genoye Frequency Ab ab ab / (0.35) (0.35) = 0.5 /Ab (0.35) (0.05) = /ab (0.35) (0.35) = /ab (0.35) (0.5) = Ab/Ab (0.05) (0.05) = Ab/aB (0.05) (0.35) = Ab/ab (0.05) (0.5) = ab/ab (0.35) (0.35) = 0.5 ab/ab (0.35) (0.5) = ab/ab (0.5) (0.5) = Toal By use of equaion (3.8), we obain each gameic frequency direcly as a funcion of he gameic frequency of he revious generaion: = ρ0 + ρa B = 0.75( 0.35) + 0.5( 0.4)( 0.7) = Ab = ρ0ab + ρa b = 0.75( 0.05) + 0.5( 0.4)( 0.3) = ab = ρ0ab + ρa B = 0.75( 0.35) + 0.5( 0.6)( 0.7) = ab = ρ0ab + ρa b = 0.75( 0.5) + 0.5( 0.6)( 0.3) = 0.35

11 c..afer random maing he linkage disequilibriuor gamee by definiion is [see equaion (3.6)] = A B = ( 0.4)( 0.7) = The linkage disequilbrium calculaed from ha of he revious generaion by equaion (3.85) is = ρ0 = 0.75( 0.07) = Exercise 3.. a. Condiional robabiliy Genoye Frequency Ab ab ab 4 ab ab ab 4 Toal b. ( ) ( )( ) ( )( ) 0 f = 9 = 9 = 9 = ab c. ( ) = Ab = ab = ab = = 0.5 f = = ab d. ( ) ( )( ) e. 9 ( ) 0 = = = 0.0 [see equaion (3.94)] 0 = 0.0 = ρ0 0 = ( 0.8) = ( ) log 0.8 = log( 0.5) ( ) = ( ) = = generaions required for he linkage disequilibrium o decrease halfway o equilibrium 9 0

12 Exercise 3.. The roof ha mb ma ij j= i= and (3.6B). For he firs equaliy we have mb ij = 0 j= mb ( ij i j ) = 0 j= ij i j = 0 j j i i() = 0 i i = 0 0= 0 Likewise for he second equaliy we have ma ij = 0 i= ma ( ij i j ) = 0 i= ij j i = 0 i i j j() = 0 j j = 0 0= 0 = 0 and ij = 0 for m A and m B greaer han wo is derived in equaions (3.6A) Exercise 3.3. a. The Hardy-Weinberg condiion for an auosomal locus is ha condiion when every genoyic frequency a any single locus in a single oulaion is equal o he roduc of he corresonding allelic frequencies. I is achieved afer one generaion of random maing when he male and female gameic arrays are equal. b. Linkage disequilibrium exiss for any oulaion when he frequency of any kind of gamee differs from he roduc of he frequencies of he corresonding genes. For examle, linkage disequilibrium exiss a wo loci if he linkage disequilibrium value AiB = j i j A i B differs from zero. j Linkage disequilibrium is brough abou in he iniial oulaion or amongs he founder individuals from selecion, or migraion, or hybridzaion, or by chance in a small oulaion, when he aricular genes are associaed more or less frequenly han he roduc of he frequencies of he corresonding genes. No, associaion of genes on he same chromosome is no essenial for linkage disequilibrium; i simly deends uon how hey are associaed in he individuals in he iniial oulaion. The only role of chromosomal associaion uon linkage disequilibrium is is influence uon he loss of linkage disequilibriurom generaion o generaion. The rae of loss is ρ. Linkage disequilibrium affecs he frequencies of all 0 ossible wo-locus genoyes from generaion o generaion unil. c. Hardy-Weinberg is a one-locus genoyic henomenon whereas linkage equilibrium is a wo or more locus gameic and genoyic henomenon.

13 3 Exercise 3.4. Probably he simles aroach is o consruc a able, incororaing he resricions ha he sum of he disequilibrium values, weighing each cell equally in every row and column, equals zero and he furher resricion ha all gameic frequencies mus be zero or greaer (i.e., no negaive gameic frequencies can exis). Focusing on he cell wih he lowes equilibriurequency (cell for A B ), he maximum absolue disequilibrium would occur for = 0.00 or = 0.0. Hence, we resen wo ables: ) when he gameic frequency = 0.00, and ) when he gameic frequency = 0.0. ) When he gameic frequency = 0.00 : B B Toal A B A B.. = = 0.4. = 0.3 A = 0.00 = 0.06 = 0.30 = 0.06 A B A B.. = = = 0.7 A = 0.0 = 0.06 = 0.50 = 0.06 Toal. = 0.. = 0.8

14 ) When he gameic frequency = 0.0 : 4 B B Toal A B A B.. = = 0.4. = 0.3 A = 0.0 = 0.4 = 0.0 = 0.4 A B A B.. = = = 0.7 A = 0.00 = 0.4 = 0.70 = 0.4 Toal. = 0.. = 0.8 The absolue linkage disequibrium value is he greaes when = 0.0, namely, = 0.4. The linkage disequilibrium values for he oher gamees are: = 0.4, = 0.4, and = 0.4. = A B where min A = 0.3, B = , so More formally, one can wrie ( ) When = 0.00, = 0.06 ; and when = 0.0, = 0.4. Thus, he maximum absolue linkage disequilibrium exiss when = 0.0. All of he gameic frequencies and linkage disequilibrium values for ha case of = 0.0 are given above. Exercise 3.5. Way : A ib= j A i B + j = i j A i B + ρ j 0 i j0 see equaions (3.63) and (3.86) Way : A 0 ( ib= j ρ i j 0) i j0 see equaion (3.95B) In he firs way, we add he linkage disequilibrium in generaion o he gameic frequency wih indeendence of allelic frequencies. In he second way, we subrac he loss in linkage disequilibrium over generaions from he iniial gameic frequency.

15 Exercise = 0 A B [see equaion (3.6)] i j i j i j = 0.8 (0.3)(0.8) = = 0.04 i j ρ0 i j0 = 0 0 (0.9) i j i j0 = = + i j0 Ai Bj i j0 0 = (0.3)(0.8) + (0.9) (0.04) [see equaion (3.63)] 5 Exercise 3.7. [3.] The fracion of he linkage disequilibrium los in each generaion is ρ [see equaion (3.9)], and ha los afer generaions is ρ0 [see equaion (3.93)]. Exercise 3.8. The loss in linkage disequilibrium afer he firs generaion of random maing is ρ A B 0. The amoun of loss in he second generaion is amoun of loss in he hird generaion is and so on for he h generaion. Hence, he oal loss is oal loss = ρ 0 + ρρ0 0 + ρρ0 0 + L+ ρρ0 0 = ρ+ ρρ0 + ρρ0 + L+ ρρ0 0 ( ) The quaniy ( ρ+ ρρ0 + ρρ0 + + ρρ0 ) common raio r is ρ 0, and he number of erms n is, so ρ of ha reained afer he firs generaion, namely, ρ ( ) ρ0 i j0. The ρ of ha reained afer he second generaion, namely, ( 0 i j0) i j ρ ρ, L is he sum of a geomeric series where he firs erm a is ρ, he A more direc soluion is he following. We desire o rove ha n a ar ρ0 = S = r Saring wih ρ0 ( ρ ) ( ρ0 ) ρ ρρ0 0 S = = = ρ0 ρ, we mulile boh he numeraor and denominaor by ( ) ( )( ) ( ) ( ) ρ ρ ρ ρ ρ ρ0 = = ρ0 ρ0 ρ ρ 0 Leing ρ0 = a, ρ0 = r, and = n, we have ( ρ0) ( ρ n 0) ρ0 a ar = which is he sum of a geomeric series. ρ0 r In his aricular geomeric series a and r are funcionally relaed, i.e., a= r..

16 6 Exercise 3.9. I would exec inbreeding o slow he rae of aroach o linkage equilibrium in a large inbred oulaion. Inbreeding leads o an increased robabiliy of union of like gamees. Conversely, inbreeding leads o a decrease in he robabiliy of union of unlike gamees, and hus a decrease in he robabiliy of recombinaion beween unlike gamees. Since he recombinaion beween unlike gamees is wha leads o linkage equilibrium, inbreeding would reduce he rae of aroach o linkage equilibrium. Exercise a. See Secion 3.4.5, o The iniial linkage disequilibria in males and females for linked loci are m 0 = = 0.8 (0.3)(0.8) = 0.04 f 0 m i f j i j = = 0.6 (0.3)(0.8) = 0.0 i j A A i B B j The linkage disequilibria in males and females for linked loci in generaion are m m = ρ + = () ( ) = 0.03 ( ) m ( 0 ) f f f ρ0 0 = + = (0.9) ( ) = 0.07 The linkage disequilibria in males and females for linked loci in generaion are m m = ρ + = () ( ) = ( ) m ( ) 0 f f f ρ0 = + = (0.9) ( ) = The general exressions for linkage disequilibrium in he males and females as a funcion of iniial linkage disequilibrium are m m f f = ρ = ρ f f f f ( ) ρ0 ( ) f f f ( 0 ) 0 ( 0 ) m m m m ρ0 m m m = ρ0 ρ ρ 0 = ρ ρ ρ 0 m ρ0 + ρ0 f ρ0 + ρ = ρ = ρ0 0 m f = ρ0 ρ0 0 = ρ0 ρ0 0 = + = + f f f f ( ) ρ0 ( ) f f f ( 0 ) 0 ( 0 0 ) m m m m 3 ρ0 m m m = ρ0 ρ 0 ρ0 0 + ρ ρ0 0 = ρ ρ 0 ρ0 0 + ρ ρ0 m ρ0 + ρ0 f ρ0 + ρ = ρ 0 0 ρ0 0 = ρ0 ρ0 0 m f = ρ0 ρ0 0 = ρ0 ρ0 0 = + = + m m ( ) f f ( ) = ρ ρ = ρ ρ

17 By he use of hese general exressions he linkage disequilibrium values in generaion 3 are m m ( ) = ρ ρ = ()(0.95) (0.03) = = f f ( ) = 3 = ρ0 ρ0 0 = (0.9)(0.95) (0.03) = We calculae he mean linkage disequilibria for generaion,, and 3: = ρ0 0 = 0.95(0.03) = = ρ0 0 = (0.95) (0.03) = = ρ0 0 = (0.95) (0.03) = which agrees wih he mean of he disequilibrium values for he males and females: + = = = = = = b. For loci on searae chromosomes (so recombinaion does occur) we have m m = ρ + = 0.5 ( ) = 0.03 = 0.05 ( ) ( ) m ( 0 ) ( ) f f f 0.03 ρ0 0 = + = 0.5 ( ) = = 0.05 The linkage disequilibrium values become he same for boh sexes. The easies way o calculae he linkage disequilibrium is o calculae he mean of he iniial linkage disequilibria and o aly a formula analogous o (3.86). ρ ( ) 8 = = (0.03) = = I is dissiaed in he same manner as occurs in a single random-maing oulaion. AiB ρ j 0 i j0 = Exercise 3.3. Equaion (3.0) embodies he condiions when eiher linkage equilibrium or linkage disequilibrium exiss in he gamees roduced from he cross oulaion iself. The equaion is reeaed here. c 0 4( 0 )( = ρ ρ ) + i j i j + i j Ai A i Bj B j I is observed ha here are wo major erms on he righ-hand side of he equaion, erm and erm. Linkage equilibrium exiss in he gamees of he cross oulaion when equaion (3.0) is equal o zero. This occurs when boh erms equal zero, or when hese wo erms are equal in magniude bu oosie in sign. The firs erm is equal o zero when and are boh equal o zero or when hey are equal in magniude AiBj i j AiBj i j bu oosie in sign, i.e., =. The second erm is equal o zero when ρ = or Ai Ai Bj Bj = or =. 0

18 8 Linkage disequilibrium exiss in he gamees of he cross oulaion when equaion (3.0) is no equal o zero. This occurs when one or boh erms are no equal o zero and do no cancel. This occurs when ) m f and/or are no equal o zero, i.e., one or boh oulaions are in linkage disequilibrium, and do no AiBj i j AiBj i j cancel, i.e.,, and ) erm is no equal o zero when he genes are linked ( ) ( A A ) i i, and ρ 0 >, B j is no equivalen in boh oulaions A i is no equivalen in boh oulaions Bj B j. Exercise 3.3. The exression for linkage disequilibrium in a cross is equaion (3.0): c 0 4( 0 )( = ρ ρ ) + i j i j + i j Ai A i Bj B j If ha exression sums o zero for every combinaion, we have linkage equilibrium a loci A and B. There can be secific cancelling effec wihin he firs major erm and/or beween he wo major erms. If he firs major erm sums o zero for all combinaions of i and j and here is indeendence (no linkage) or he same frequency of every A i allele exiss in he wo oulaions or he same frequency of every B j allele exiss in he wo oulaion, here is linkage equilibrium. If any one of he hree facors is zero, he second erm equals zero. Exercise In Examle 3., he observed dearures are all due o linkage disequilibrium. For examle, in generaion he frequency of any genoye is a funcion of linkage disequilibrium, i.e., for /, we have ( ) = (, ) = ( A B +, ) = ( ) + 8 = 3 ( 8 ) = 9 64 Likewise for he double heerozygoes, /ab and Ab/aB, he sum of he frequencies of couling and reulsion hase, heerozygoes is ( ) + Ab ( ) = (, ab, ) + ( Ab, ab, ) ab ab = ( A B +, )( a b + ab, ) + ( A b + Ab, )( a B + ab, ) = = 3 ( 8) + ( 8) = ( 64 ) + ( 64 ) = + = The difference beween he frequencies of he couling and reulsion hase, double heerozygoes is 9 ( ) Ab ( ) = = ab ab This difference in genoyic frequencies beween couling and reulsion hase, double heerozygoes is an equivalen measure of linkage disequilibrium [see equaion (3.66)]; he difference in generaion is wice he linkage disequilibrium in generaion as we have designaed he generaions [see equaions (3.64) and (3.7)], namely, = ( 8) =. 4

19 9 In Examle 3.4, he frequency of all nine genoyes (couling and reulsion added ogeher) is equal o he roduc of he marginal frequencies of he corresonding single-locus genoyic values. This is rue because in his examle boh arenal oulaions were in linkage equilibrium. I is his fac ha makes he following rue (along wih random maing): ( ) = ( AA ) c ( BB ) c c = A B ( ) = A B -- he roduc A imes B imlies linkage equilibrium, and and he roduc A B imes A B imlies random maing = ( ) This las exression above imlies ha he linkage disequilibrium is zero because ( ) = ( A B + ) where = 0. This is also rue for he sum of he couling and reulsion double heerozygoes ( ) + Ab ( ) = = 0.70 which is equal o he roduc of he wo single-locus ab ab frequencies, Aa Bb = 0.340( 0.50) = 0.70, imlying linkage equilibrium. Now why do he frequencies of he couling and reulsion double heerozygoes differ? I is simly because he frequency of he four kinds of gamees in he wo arenal oulaions are differen, ( ) Ab ab ( ) = ab ab + ab Ab ab + ab Ab = 0.4( 0.05) + 0.( 0.45) 0.8( 0.05) + 0.8( 0.45) = ( ) = If he corresonding frequencies of he four kinds of gamees were he same in he wo arenal oulaions, hen he difference beween he frequencies of he couling and reulsion double heerozygoes would be zero. For examle, if he f oulaion would have he same gameic frequencies as he m oulaion, hen we would have 0.4(0.) + 0.(0.4) [0.8(0.8) + 0.8(0.8)] = [ ] = 0. Exercise a. The linkage disequilibriuor he gamee from he m oulaion is m m c c = A ( B ) = 0.4 ( 0.8)( 0.55) = = 0.0 The linkage disequilibriuor he gamee from he f oulaion is f f c c = A( B) = 0.45 ( 0.8)( 0.55) = = 0.0 Noice ha he linkage disequilibrium value of 0.0 for he m oulaion and ha of 0.0 for he f oulaion are oosie in sign. The mean of he wo values is = + = An alernaive way o calculae is by use of he mean of he gameic frequencies: = + ( 0.8)( 0.55) = = 0.005

20 0 b. The corresonding values for he oher hree gamees are: m m c c = = = = 0.08 Ab ( ) ( )( ) Ab Ab A b f Ab = = Ab = + = = m m c c ab = ab a B = = = 0.07 f ab = = ab = = = m m c c ab = ab a b = = = 0.03 f ab = = ( 0.04) 0.0 ab = = = ab ( ) ( )( ) ab ( ) ( )( ) Ab ab ab c. The mean of he gameic frequencies from oulaion m and oulaion f are = = = = = = 0.5 = = The linkage disequilibrium is [see equaion (3.64)] = ab Ab ab = 0.435( 0.085) 0.365( 0.5) = = No, i is no equal o 0.00, he value given in he Exercise iself; i is one half of ha value. I is he same as ha in (a) and (b). Why? I am no oo clear on he ossible reason, bu ha measure of 0.00 is he resul of union of individual gamees from he m oulaion and from he f oulaion. There is no averaging effec o reduce he deviaion. When we define linkage equilibriuor each gamee from he m and f oulaions relaive o he cross iself, we observe ha hey are oosie in sign and show less deviaion (like a mean). I would sugges ha he reason for he relaion is ha we are using he mean of he gameic frequencies from he wo oulaions, i.e., = + ( 0.8)( 0.55) = = The original measure of 0.00 is he union of gamees from he m and f oulaion and here is no averaging involved. Exercise Similariies beween wo-locus and hree-locus disequilibria are: () Two-locus and hree-locus linkage disequilibria are defined in he same way, namely, as a dearure of heir gameic frequencies from ha of indeendence of he corresonding genes = equaion (3.6) i j i j Ai Bj C = C A B C i j k i j k i j k equaion (3.)

21 () Gameic frequencies [and hence heir linkage disequilibria, see equaion (3.63)] for boh can be exressed in generaion in erms of comlemenary gameic grou robabiliies 0 equaion (3.8) i j = ρ i j + ρ A i Bj 00 0 equaion (3.4) i jck = ρ i jck + ρ C k i j + ρ0 ABC + ρb AC i j k k i k (3) The higher-order, hree-locus linkage disequilibrium measure or he adjused hree-locus linkage disequilibrium value behaves like he wo-locus linkage disequilibrium measure, i.e., δ 00 vs. 0 equaion (3.9) vs. equaion (3.86) i jck = ρ δ i jck i j = ρ i j δ 00 0 vs. 0 0 equaion (3.9) vs. equaion (3.86) i jck = ρ δ i jc k i j = ρ i j We regard he hree-locus linkage disequilibrium as less imoran because he adjused hree-locus linkage disequilibrium value is dissiaed more raidly han he wo-locus linkage disequilibium value, i.e., rae of loss is 3 ρ / [see equaion (3.5)] which is 50% greaer han ha for wo loci for an equivalen recombinaion value. Exercise We assume he following allelic frequencies: A = 0.6, a = 0.4 B = 0.7, b = 0.3 C = 0.5, c = 0.5 Then he equilibrium gameic frequencies are: C = 0.6( 0.7)( 0.5) = 0. c = 0.6( 0.7)( 0.5) = 0. AbC = 0.6( 0.3)( 0.5) = 0.09 Abc = 0.6( 0.3)( 0.5) = 0.09 abc = 0.4( 0.7)( 0.5) = 0.4 abc = 0.4( 0.7)( 0.5) = 0.4 abc = 0.4( 0.3)( 0.5) = 0.06 abc = = 0.06 ( )( ) We creae he hree-locus linkage disequilibrium by adding or subracing an arbirary amoun (0.0) from he hree-locus gameic equilibrium values. Since i is a hree-way able, we consruc wo wo-way ables one for gamees which carry he C allele and one which carries he c allele. Gamees wih C allele Gamees wih c allele B b B b A C AbC AC A c Abc Ac = = =(0.6)(0.5) = = =(0.6)(0.5) a abc abc ac a abc abc ac = = =(0.4)(0.5) = = =(0.4)(0.5) BC bc C Bc bc c =(0.7)(0.5) =(0.3)(0.5) =(0.7)(0.5) =(0.3)(0.5)

22 From he marginal frequencies of he wo above ables i is observed ha linkage equilibrium exiss for loci A and C and for loci B and C. By adding he corresonding cells in he wo above ables one obains he gameic frequencies for loci A and B which is also in linkage equilibrium: gamee = = 0.4 = (0.6)(0.7) gamee Ab = = 0.8 = (0.6)(0.3) gamee ab = = 0.8 = (0.4)(0.7) gamee ab = = 0. = (0.4)(0.3) The hree-locus gameic frequencies, of course, exhibi linkage disequilibrium: C = abc = Abc = abc = 0.0 AbC = abc = c = abc = 0.0 Exercise Offsring Maing ye Male Female Male Female Frequency A a AA Aa aa A AA Aa q aa a AA q Aa q q q aa q q Male: : () + q + q () + q q A 3 3 = + q+ q+ q = ( + q) + q( + q) = + q = ( + q) = a : q + q () + q q + q q () 3 = q + q + q + q = q( + q) + q ( + q) = q + q = q( + q) = q

23 3 Female: 3 AA : () + q = + q = ( + q ) = Aa : q + q () + q () + q q = q + q + q + q = q( + q+ + q) = q 3 aa : q q + q q () = q + q = q ( + q) = q Exercise The ercenage of males showing he rai is 0.05 = 0.45 or 45% Tha roorion is he same for a nonrandom-maing oulaion as for a random-maing oulaion. Exercise The robabiliy ha he woman is a carrier is one-half, or P ( woman is carrier) = If she is a carrier, he robabiliy ha she will have an affeced son is one-half, or P( son is affeced woman is carrier ) =, or P ( son is affeced) = P( woman is carrier) P( son is affeced woman is carrier) = 4 Exercise a. q+ q = q( + q) = q= ( q) q= q q b. c. d. e. 3 q + q = q ( q+ ) = ( ) ( q) = [ ( )] = [ + ] = ( ) q = ( + q)( q) = q q q = + q q = ( q ) or + q + q 4q = q q + = ( q ) f. + q q = = = q = q q q q q q