Covariance between relatives

Transcription

1 UNIVERSIDADE DE SÃO PAULO ESCOLA SUPERIOR DE AGRICULTURA LUIZ DE QUEIROZ DEPARTAMENTO DE GENÉTICA LGN5825 Genética e Melhoramento de Espécies Alógamas Covariance between relatives Prof. Roberto Fritsche-Neto roberto.neto@usp.br Piracicaba, March 23 rd, 2018

2 Definitions Coefficient of kinship (f) Probability that two gametes taken at random from two individuals are identical by descent (IBD, ) Expresses the degree of relatedness between individuals - coefficient of parentage Coefficient of relationship (r) It is the additive genetic relationship between individuals This is the twice the coefficient of kinship r = 2f It is also equal the inbreeding coefficient of their progeny Aditive covariance between relatives The covariance between the breeding values COV a(x,y) = r xy Va It can actually be due to additive genetic effects, as well as dominance and epistatic effects In general the contribution of dominance and epistatic effects to the genetic covariance is low

3 Calculating the inbreeding coefficient of X Lets consider one bi-allelic locus with two alleles A 1 and A 2 We assume that A, the common ancestor of P and Q, is not inbred, thus its genotype is A 1 A 2 The probability that X receives A 1 from A via P, is the probability that A passes A 1 to P multiplied by the probability that P passes A 1 to X This probability is 1/2 1/2 = 1/4 Now we need to know the probability that X receives A 1 from both P and Q 1/4 1/4 = 1/16 We now know the probability that A 1 is IBD in X X could also be IBD by receiving two copies of A 2 Probability IBD in X via either A 1 or A 2 is 1/16 + 1/16 = 2/16 = 1/8 P(IBD) = 1/2 3 = 1/2 n, where n is the number of common ancestral individuals However, if the parent A is inbred, the IBD increases and should be considered (1/2) n F A, where F A is the inbreeding coefficient of the common ancestor Thus, IBD is the sum the two probabilities: F X = (1/2) n + (1/2) n F A F X =(1/2) n (1+F A )

4 Calculating the inbreeding coefficient of X In more complex pedigrees, parents may be related to each other through more than one common ancestor, or from the same common ancestor, but through different paths The general formula is F X =(1/2) n (1+F A ) where n is the number of individuals in any path of relationship counting the parents of X and all individuals in the path which connects the parents to the common ancestor The summation is over all paths There are 2 common ancestors, A and E There are 2 possible paths

5 The coefficient of kinship Probability that two gametes taken at random (one from each individual carry alleles that are IBD The kinship (f) between two individuals is equal to the inbreeding coefficient of their progeny F X =f p1p2 where X is the progeny and p 1 and p 2 are the parents Basic rules to estimate f First: the f between P and Q is the mean of the four co-ancestries f BC = D E fac+ D E fad + D E fbc + D E fbd Second: the coefficient of kinship of an individual with itself f AA is the inbreeding coefficient of progeny that would be produced by self-mating f AA = ½(1+F A ) Third: the coefficient of kinship between parent and offspring f PA is the mean coefficient of kinship between A and both the parents of P, (A and B) f PA = ½(f AB +f AA )

6 Covariance between relatives f xy = ¼ [P(x i y i ) + P(x j y j ) + P(x i y j ) + P(x j y i )] u xy = [P(x i y i ; x j y j ) + P(x i y j ; x j y i )] simultaneous events (the same genotype - dominance) NON-INBRED RELATIVES Half-sibs F = 0, thus, F A = 0 F xy = ¼ [P(x y A i ) + P(x y A j ) + P(x A i y A j ) + P(x A j y A i )] If x and y are non-inbred, thus the last two parts are zero, because their parents are not inbred either f xy = ¼ [1/4 + 1/ ] f xy = 1/8 pollen pollen Since r = 2f A(A i A j ) r = 1/4 1/2 1/2 u xy = 0 Probability of transmit the genotype dominance effect x y

7 Full-sibs F = 0, thus, F A = F B = 0 Covariance between relatives f xy = ¼ [P(x y A i ) + P(x y A j ) + P(x y A k ) + P(x y A L ) + P(x A i y A j ) + P(x A j y A i ) + P(x A k y A L ) + P(x A L y A k )] Since A and B are non-inbred, P(A i A j ) = 0 f xy = ¼ [1/4 + 1/4+ 1/4 + 1/ ] f xy = 1/4 Since r = 2f r = ½ (½. ½). (½. ½) u xy = [P(x y A i ; x y A k ) + P(x y A i ; x y A L ) + P(x y A j ; x y A k ) + P(x y A j ; x y A k )] A(A i A j ) x B(A k A L ) y u xy = [(0.25 x 0.25) + (0.25 x 0.25) + (0.25 x 0.25) + (0.25 x 0.25)] u xy = 1/16 + 1/16 + 1/16 + 1/16 u xy = ¼

8 INBRED RELATIVES Half-sibs F p 0; P(A i A j ) = P(A k A L ) 0 Covariance between relatives f xy = ¼ [P(x y A i ) + P(x y A j ) + P(x A i y A j ) + P(x A j y A i )] these cases can be A i A j f xy = ¼ [1/4 + 1/4+ F(1/4) + F(1/4)] f xy = ¼ [1/2 + F(1/2)] f xy = 1/8[1 + F] pollen Since r = 2f r = ¼[1+F] A(A i A j ) 1/2 1/2 pollen u xy = 0 Probability of transmit the genotype dominance effect x y

9 Full-sibs F 0, thus, F A = F B = 0 Covariance between relatives f xy = ¼ [P(x y A i ) + P(x y A j ) + P(x y A k ) + P(x y A L ) + P(x A i y A j ) + P(x A j y A i ) + P(x A k y A L ) + P(x A L y A k )] Since A and B are non-inbred, P(A i A j ) = 0 f xy = ¼ [1/4 + 1/4+ 1/4 + 1/4 + F(1/4) + F(1/4) + F(1/4) + F(1/4)] f xy = ¼[1 +F] Since r = 2f A(A i A j ) r = ½[1 +F] B(A k A L ) (½. ½) + (½. ½ )F u xy = [P(x y A i ; x y A k ) + P(x y A i ; x y A L ) + P(x y A j ; x y A k ) + P(x y A j ; x y A k )] u xy = [(0.25 x 0.25) + (0.25 x 0.25) + (0.25 x 0.25) + (0.25 x 0.25)] u xy = 1/16 (1+ F) 2 + 1/16(1+ F) 2 + 1/16(1+ F) 2 + 1/16(1+ F) 2 u xy = ¼(1+ F) 2 x y

10 Covariance between relatives P(x y A i ) = P(x y A k ) + P(x A i y A j ) = ½. ½ + F(½. ½) = ¼ + ¼F = ¼(1 + F) P(x y A k ) = P(x y A k ) + P(x A k y A L ) = ½. ½ + F(½. ½) = ¼ + ¼F = ¼(1 + F) P(x y A i ; x y A k ) = P(x y A i ). P(x y A k ) = ¼(1+F). ¼(1+F) = 1/16(1+F) 2

11 Why r = 2f? x ij = u + α ix + α jx + S ijx y ij = u + α iy + α jy + S ijy E(x ij ) = E(y ij ) = u E(α i ) = i p i α i = 0 E(α j ) = j p j α j = 0 E(S ij ) = ij p i p j S ij = 0 E(α i,α j ) = E(α i ) E(α j ) = 0 E(α i,s ij ) = E(α i ) E(S ij ) = 0 E(α j,s ij ) = E(α i ) E(S ij ) = 0 COV(x ij, y ij ) = E[x ij -E(x ij )]. E[y ij -E(y ij )] = [u + α ix + α jx + S ijx u]. [u + α iy + α jy + S ijy u] = E(α ix,α iy ) + E(α ix,α jy ) + E(α ix,s ijy ) + E(α jx,α iy ) + E(α jx,α jy ) + + E(S ijx,s ijy ) E(α ix,s ijy ) = 0 Covariance between allele effect and genotype (all cases)

12 Why r = 2f? E(α ix,α iy ) = i p i α i P(x i y i )α i = i p i α i2 P(x i y i ) = 1/2Va. P(x i y i ) E(α ix,α jy ) = i p i α i P(x i y j )α i = i p i α i2 P(x i y j ) = 1/2Va. P(x i y j ) E(α jx,α iy ) = j p j α j P(x j y j )α j = j i p j α j2 P(x j y j ) = 1/2Va. P(x j y j ) E(α jx,α jy ) = j p j α j P(x j y i )α j = j i p j α j2 P(x j y i ) = 1/2Va.P(x j y i ) E(S ijx,s ijy ) = ij p i p j S ij [P(x j y i, x j y j ) + [P(x i y j, x j y i )]S ij = ij p i p j S ij2 [P(x j y i, x j y j ) + [P(x i y j, x j y i )] = Vd.[P(x j y i, x j y j ) + [P(x i y j, x j y i )] COV(x ij, y ij ) = E[x ij -E(x ij )]. E[y ij -E(y ij )] = [u + α ix + α jx + S ijx u]. [u + α iy + α jy + S ijy u] = E(α ix,α iy ) + E(α ix,α jy ) + E(α ix,s ijy ) + E(α jx,α iy ) + E(α jx,α jy ) + + E(S ijx,s ijy ) COV a (x ij, y ij ) = ½Va[P(x j y i ) + P(x j y j ) + P(x i y j ) + P(x j y i )] = ½Va[4fxy] = 2fVa COV d (x ij, y ij ) = ½Vd.[P(x j y i, x j y j ) + [P(x i y j, x j y i )] = Vd[u xy ] =u xy Vd COV g (x ij, y ij ) = 2.f xy.va + u xy Vd