Quantitative Genetic Models Least Squares Genetic Model. Hardy-Weinberg (1908) Principle. change of allele & genotype frequency over generations

Transcription

1 Quanttatve Genetc Models Least Squares Genetc Model Hardy-Wenberg (1908) Prncple partton of effects P = G + E + G E P s phenotypc effect G s genetc effect E s envronmental effect G E s nteracton effect partton of varance components V ar(p ) = V ar(g) + Cov(G, E) + V ar(e) gnorng G E V ar(p ) = V ar(g) + V ar(e) further gnorng Cov(G, E) change of allele & genotype frequency over generatons deal assumptons: 1 large (nfnte) populaton random matng (no nbreedng or random drft) 3 no natural or artfcal selecton 4 no mutaton 5 closed to mgraton 6 dscrete generatons sexual reproducton 1:1 gamete rato 7 separate sexes c Z Zeng & S Yandell 4.1 February 8, 001 c Z Zeng & S Yandell 4. February 8, 001 Sexual Reproducton to H-W Proportons genotype female male p 1f p 1m b p f p m genotype b bb frequency p 1f p 1m p 1f p m + p f p 1m p f p m autosomal locus: same for male, female p 1 = p + p b / = p 1f p 1m + [p 1f p m + p f p 1m ]/ = [p 1f (p 1m + p m ) + p 1m (p 1f + p f )]/ = [p 1f + p 1m ]/ next and subsequent generatons genotype b bb frequency p 1 p 1 p p Hardy-Wenberg Equlbrum only 1 generaton to remove sex dfferences only generatons to stablze proportons random matng can stablze proportons subected to selecton n 1- generatons Hardy-Wenberg Dsequlbrum b p + D p p b D p = p 1 b p b p D p b + D p b = p p = p 1 p b = p 1 1 c Z Zeng & S Yandell 4.3 February 8, 001 c Z Zeng & S Yandell 4.4 February 8, 001

2 Test of Hardy-Wenberg Dsequlbrum ch-square test X = (obs-exp) /exp X = D [ N p log lkelhood test G = obs log( exp/obs ) ( G = [N log 1 + D ) p ( +N bb log 1 D ) p ] b + N b + N ] bb p p b p b ( ) D + N b log p p b Hardy-Wenberg Equlbrum gametes b total p p(1 p) p 1 = p b (1 p)p (1 p) p = 1 p total p 1 = p p = 1 p 1 genotype b bb frequency f = (p 1 ) f b = p 1 p f bb = (p ) Genetc Varance Components for a Locus Genotype value common Fsher G 11 m + a µ + a 1 + d 11 b G 1 = G 1 m + d µ + a 1 + a + d 1 bb G m a µ + a + d µ = g f g G g = p p G = m+(p 1 p )a+p 1 p d, c Z Zeng & S Yandell 4.5 February 8, 001 c Z Zeng & S Yandell 4.6 February 8, 001 Addtve and Domnance Varance Components Mather-Jnks Partton G 11 = m + a, G 1 = m + d, G = m a σg = f g (G g µ) = p p G µ g, = p 1 a + p 1 p d + p ( a) (µ m) = p 1 p [a + (p p 1 )d] + 4p 1 p d = σ a + σ d addtve varance component σ a = p 1p [a + (p p 1 )d] domnant varance component σ d = 4p 1 p d Fsher s Least Square Partton of Genetc Varance Components G = µ + a + a + d = µ + a 1 N 1 () + a N () + d N 1, N = number copes of allele (1=, =b) parameters µ = Ḡ = p p G, a = a = Ḡ Ḡ p G µ = Ḡ Ḡ d = G a a µ = G Ḡ Ḡ + Ḡ whch satsfy the condtons p a = p p d = p d = p d = 0, c Z Zeng & S Yandell 4.7 February 8, 001 c Z Zeng & S Yandell 4.8 February 8, 001

3 LS Reference Mean and Addtve Effects µ = p 1 G 11 + p 1 p G 1 + p G a 1 = p [p 1 (G 11 G 1 ) + p (G 1 G )] a = p 1 [p 1 (G 11 G 1 ) + p (G 1 G )] relaton to Mather-Jnks µ = m + a(p 1 p ) + dp 1p a 1 = p [a + d(p p 1 )] a = p 1 [a + d(p p 1 )] a 1 a = a + d(p p 1 ) LS Domnance Effects d 11 = G 11 µ a 1 = p [G 1 G 11 G ] d 1 = d 1 = G 1 µ a 1 a = p 1 p [G 1 G 11 G ] d = G µ a = p 1 [G 1 G 11 G ] c Z Zeng & S Yandell 4.9 February 8, 001 LS genetc varance components σg = p p (G µ) = p p (a + a + d ),, = p a + p p d = σa + σ d Addtve Varance σa = p a = [p 1a 1 + p a ] = p 1 p [p 1 (G 11 G 1 ) + p (G 1 G )] = p 1 p [p 1 (a d) + p (a + d)] = p 1 p [a + (p p 1 )d] Domnance Varance σd = p p d = p 1 d 11 + p 1p d 1 + p d, = p 1 p [G 1 G 11 G ] = 4p 1 p d c Z Zeng & S Yandell 4.10 February 8, 001 Hardy-Wenberg Dsequlbrum genotype b bb frequency P P b P bb H-W equlbrum p p p b p b wth p = P + 1 P b and p b = P bb + 1 P b dsequlbrum coeffcents P = p + D P b = p p b D b P bb = p b + D bb constrants of gene and genotype frequences P + P b + P bb = 1 D = 0 p + p b / = p D = D b = D bb D measures H-W dsequlbrum for locus P = p + D P b = p p b D P bb = p b + D c Z Zeng & S Yandell 4.11 February 8, 001 Lnkage dsequlbrum Let us consder the dstrbuton of gametes for a two-allele and two-locus model: Gamete A Ab a ab Frequency P A P Ab P a P ab L.-E. p A p p A p b p a p p a p b Defne dsequlbrum coeffcents as: P A = p A p + D A P Ab = p A p b + D Ab P a = p a p + D a P ab = p a p b + D ab constrants of gene and genotype frequences D A + D Ab + D a + D ab = 0 p A = P A + P Ab = p A + D A + D Ab = D A = D Ab and D ab = D a, c Z Zeng & S Yandell 4.1 February 8, 001

4 In general, Thus D = D = 0. The lnkage dsequlbrum n these populatons are determned by the gametes of F 1. F 1 produces four gametes n the followng proporton: P A = p A p + D A P Ab = p A p b D A P a = p a p D A P ab = p a p b + D A Correlaton coeffcent between two loc: γ = D A pa p p a p b Relatonshp between lnkage dsequlbrum and recombnaton frequency: For both backcross and F populatons, segregatng gametes are produced from F 1 Gamete A Ab a ab Frequency 1 r r r 1 r r = recombnaton frequency (0 < r < 0.5). p A = p = p a = p b = 1 D A = P A p A p = 1 r 1 4 = 4 1 (1 r) thus γ = D A (1 r)/4 = = 1 r pa p p a p b 1/16 c Z Zeng & S Yandell 4.13 February 8, 001 c Z Zeng & S Yandell 4.14 February 8, 001 Lnkage dsequlbrum b A p A p + D A p A p b D A p A a p ap D A p ap b + D A p a = 1 p A p p b = 1 p 1 correlaton between loc X = 1(A), 0(a); Y = 1(), 0(b) E(X) = p A 1 + p a 0 = p A V ar(x) = p A (1 p A ) + p a (0 p A ) = p A p a Cov(X, Y ) = P xy (x E(X))(y E(Y )) x,y = (p A p + D A ) (1 p A )(1 p ) + (p A p b D A ) (1 p A )(0 p ) + = D A Cov(X, Y ) γ = Corr(X, Y ) = V ar(x)v ar(y ) D = A pa p a p p b Lnkage dsequlbrum and partton of genetc varance Consder two loc, no epstass locus A:, ; locus : k, l least square genetc model G k l = µ + a + a k + a + a l + d + dk l k = gamete from parent 1 (superscrpt) l = gamete from parent (subscrpt) genotype frequency Pl k = frequency of G k l n populaton Hardy-Wenberg equlbrum: Pl k = P k P l lnkage dsequlbrum: P k = p p k + D k constrants p a = p d = p d = D k = D k = 0 k c Z Zeng & S Yandell 4.15 February 8, 001 c Z Zeng & S Yandell 4.16 February 8, 001

5 mean of the populaton effects G k l = µ + a + a k + a + a l + d + dk l constrants p a = p d = p d = D k = D k = 0 k Ḡ = Pl k Gk l = P k P l G k l,k,l,k,l = (p p k + D k )(p p l + D l )G k l,k,l = p p k p p l G k l = µ = Ḡ,k,l lnkage dsequlbrum does not affect mean (not true wth epstass) varance of the populaton σg = P k P l (G k l µ),k,l = (p p k + D k )(p p l + D l )(G k l µ),k,l = p (a ) + k p k (a k ) + p a + l p l a l + D k a a k + D l a a l,k,l + p p d + p k p l d k l + D k D l d dk l, k,l,k,l lnkage dsequlbrum varance contans covarances between addtve effects a and a k, a and a l between domnance effects d and dk l H-W equlbrum no covarance between a and d nbreedng yelds some covarance c Z Zeng & S Yandell 4.17 February 8, 001 c Z Zeng & S Yandell 4.18 February 8, 001 Genetc model for C and F genotype effects at loc, = 1,,, m genotype b b b effect a d a P 1 populaton fxed wth alleles P populaton fxed wth allele b gnore epstass phenotypc value of ndvdual k P 1 populaton: y k = µ + a + e k, e k N(0, σe ) =1 mean and varance of trat n P 1 : µ P1 = µ + a, =1 σ P 1 = σ e examne trat dstrbuton famlarze wth varous crosses underlyng model for QTL get ready for epstass Smlarly for P : y k = µ a +e k, µ P = µ a, =1 =1 F 1 = P 1 P populaton: y k = µ+ d +e k, µ F1 = µ+ d, =1 =1 σ P = σ e σ F 1 = σ e c Z Zeng & S Yandell 4.19 February 8, 001 c Z Zeng & S Yandell 4.0 February 8, 001

6 ackcross 1 = P 1 F 1 F = F 1 F 1 Populaton y k = µ + [x k a + (1 x k )d ] + e k =1 = µ + d + x k (a d ) + e k =1 =1 { 1 f a allele n k s F x k = 1 gamete 0 f a b allele E(x k ) = 1/, V ar(x k ) = 1/4 Cov(x k, x k ) = (1 r )/4 mean and varance σ 1 µ 1 = µ + (a + d )/ =1 m = (a d ) + 4 =1 smlarly for = P F 1 (1 r )(a d )(a d ) + σe 4 c Z Zeng & S Yandell 4.1 February 8, 001 y k = µ + [x k1 x k a + (1 x k1 )(1 x k )( a ) =1 +[x k1 (1 x k ) + (1 x k1 )x k ]d ] + e k = µ a + [x k (a + d ) x k1 x k d ] + e k =1 =1 x k1 = { 1 f allele n frst gamete from F 1 0 f b allele E(x k1 ) = 1, V ar(x k1) = 1 4 and smlarly for x k Cov(x k1, x k1 ) = 1 4 (1 r ) Under the assumpton of Hardy-Wenberg equlbrum, x k1 and x k are ndependent. Wth c Z Zeng & S Yandell 4. February 8, 001 these condtons, the mean and varance of ndvduals n F populaton become µ F = µ + 1 d =1 σf = 1 a + 1 d = (1 r )a a =1 + 1 (1 r ) d d + σe 4. frst term: addtve varance wthn loc second: domnance varance wthn loc thrd: addtve covarance between loc fourth: domnance covarance between loc Epstass secton 4.6 of Zeng s notes postponed to chapter 8 Multple Interval Mappng c Z Zeng & S Yandell 4.3 February 8, 001 c Z Zeng & S Yandell 4.4 February 8, 001