POPULATION GENETICS Quantitative traits and evaluation of phenotypic variance. Heritability of quantitative traits Ing. Radovan KASARDA, PhD. Estimation of Heritability • Method „ parent – offspring“ h2OP – – – – similarity (kinship) between parents and offspring according to Wright (1920) relationship coefficient is 0,5 comparison of relatives between groups h2 is twice of rXY resp. regress coefficient bYX.. Principle of calculation: Pair X Y 1 production of mother production of daughter ΣX ΣY X2 Y2 X.Y ΣX2 ΣY2 ΣX.Y n To calculate rXY resp. bYX estimate semi-values A, B, C. AX 2 ( X ) 2 variance of X n B Y 2 ( Y ) 2 variance of Y n X Y C XY n covariance between X and Y Estimation of Heritability • Method „ parent – offspring“ h2OP than: rXY C XY A.B ( X2 Y2 ) C bYX A h 2 OP 2.rXY h 2 OP 2.bYX In case data are divided by fathers, calculate partial Ai, Bi, Ci separatelly for each father and sumarize it. Estimation of Heritability • Method „ parent – offspring“ h2OP With use of ANOVA: Father Variance σ2 Covariance ni ni .σ2i ni.covXY 1 710 77 20 14200 1540 2 520 154 30 15600 4620 3 800 100 30 24000 3000 4 400 20 20 8000 400 1 640 128 40 25600 5120 2 700 70 10 7000 700 4 530 78 20 10600 1560 2 480 35 20 9600 700 3 550 90 30 16500 2700 4 640 64 30 19200 1920 Σ 250 150300 22260 bOP covOP P2 22260 2 0,148; hOP 2.0,148 0,296 150300 Method „half-sibs“ - ANOVA • half-sibs have one common parent (father) – use of genetic similarity between half-sibs (0,25) – universal method – Conditions of use: • • • • groups of half-sibs in same conditions (min. 3 in group) group based by father (min. 3 fathers) (p) in group min. 3 measurements (ni) i.e. min. 9 measurements in total (n) ANOVA – initial informations Father i Daughter j Prod. kg Yij 1 1 5 25 20 400 σ2 2 2 g 30 1 30 900 2 20 400 p=3 σ2e 30 Σiyij = 80 1008,33 2133,33 900 900 1 30 900 2 25 625 3 Σ Σiyij= 55 Y2ij 3 σ2g 3 3 σ2e Yi.2/ni Yi. σ2e ni=3 n=9 35 Σiyij = 90 Y..= 225 2700 Y2i.=5841,66 1225 Σi Σj Y2ij= 6275 ANOVA – One Way σ2 S DF MS S g Yi.2 / ni Y..2 / n fg=p -1 MSg=Sg:fg Se fe=n - p MSe=Se:fe 2 2 Y Y ij i. / ni i g2 MS g MSe ni e2 MSe j ri 2 g 2 p h2=4.ri σ2P =σ2g + σ2e = 84,259; S F MS σ2 216,666 2 108,333 12,037 433,33 6 72,222 72,222 ri = 0,143; h2 = 0,57 ANOVA – „full-sibs“ • in populations of multiparous animals (pigs, poultry,..) – high genetic similarity of full-sibs (0,5), whose should under equal conditions show equal phenotype – difficult for segmentation • in group – variability of environment • between groups – variability depended on diverse genetics of mothers • between classes – variability deperded on diverse genetics of fathers • σ2P =σ2F + σ2M + σ2E • two-factorial ANOVA is used – i = father; j = mother; k = progeny ANOVA – initial informations Father i 1 Mother j 1 Daughter k Trait Yijk Yij. Y2i../ni Yi.. Yij.2/nij Y2ijk 1 2 3 2 1 2 3 3 1 2 3 Σ p mi nj ΣY... ΣY2i../n Σi Σj Y2ij/nij Σi Σj Σk Y2ijk i ANOVA – two factorial S DF MS S F Yi.2 / ni Y...2 / n fF=p -1 MSF=SF:fF SM i Se i Yij2. / nij Yi..2 / ni j 2 kij j k i Y 2 ij. j / nij F2 k1.MSF k2 .MSM (k2 k1 ).MS E k1.k3 MSM MSE k1 fM=m - p MSM=SM:fM M2 fe=n - m MSe=Se:fe e2 MSe i Y σ2 k1 – average # of daughter per mother k2 – average # of mother per father k3 – average # of daugh. per father ANOVA – two factorial 2 FM h 2 2 P h 4 2 F 2 F 2 F 2 P h 4 2 M 2 M 2 P 2 M ANOVA: Half-sibs • Exercise: Calculate heritability p=8 Yi. Yi.2/ni Y2ij 1 75,94 288,3430 288,6938 2 75,50 285,0112 285,7600 3 79,33 314,6611 315,2639 4 77,98 304,0430 304,6091 5 73,65 271,2151 271,7600 6 76,38 291,6943 292,1326 7 75,34 283,8047 284,1924 8 75,33 283,7295 284,1279 ni=20 ΣY..= ΣY2i. /ni= Σi Σj Y2ij= ANOVA: Full-sibs • Exercise: p =10 m = 130 MSF=391,3 MSM=37,50 MSE=6,40 n = 1955 k1=15,00 k2=15,40 k3=193,00 • Calculate: σ2 F= h2 F= σ2M= h2M= σ2E= h2FM= σ2P= Non-parametric estimates of heritability • very good elaborated • used if less information is known – low number of observation, ranking evaluation • could resuld in non-realistic estimates • only informative results of h2 • analyses of influence of genotypes on traits – Estimates by Young – estimates using Spearman correlation coeficients – estimates based on realised heritability Non-parametric estimates of heritability – Estimates by Young: D X X h 2 (X M D M X ).2 above-average mothers below-average mothers above-average daughters XM XM XD XD below-average daughters – Estimates using Spearmann rank-order correlation coefficient: h2 = r S rS 1 6 d 2 n(n 2 1) mother rank 5,5 5,5 10 daughter rank 9 8 4 d 3,5 2,5 6 d – absolute difference between mother and daughter ranking • Calculate heritability: Non-parametric estimates of heritability • estimates based on realised heritability – by division of pop. on better (+) resp. worser (-) half h 2 O O P P Y Y Y Y O – offspring; P - parents – according to realized selection Gr h d 2 h 2 - realized genetic gain - selection difference (d=i.s) XO X XPX - difference of selected offspring from average of off. generation - difference of selected parents from average of parental generation