Heritability Practicals

Heritability Estimation: 

Heritability Estimation Dept. of AGB Veterinary College, Hebbal, Bangalore

Estimation of heritability by Half sib correlation method.: 

Estimation of heritability by Half sib correlation method. Half sibs are of two types Paternal Half sib Maternal half sib The sibs resemble each other because they have common genes. Therefore the resemblance between the sibs is used as the basis to estimate the covariance, which contains a fraction of additive genetic variance.

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S1 S3 S2 D1 D2 D3 D1 D2 D3 D1 D2 D3 Half sibs of S1 Half sibs of S2 Half sibs of S3

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Covariance in half sibs is estimated as Correlation (t) between them HS of S1 HS of S2 HS of S3 V P =σ 2 s+σ 2 e

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Variation in this half sibs is due to half the effect of genes coming from both the parents and as well the environmental variation. If we consider only the half of the variance caused due to sire and even in that there is variation caused due to environment within the progeny. Hence the formula is σ 2 S = COV HS = ¼ VA+ 1/16 VAA……… σ 2 w = σ 2 p – COV HS σ 2 p = Total phenotypic variance.

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The correlation between sibs is estimated as In case of half sibs the Genetics variance in the sire component is σ 2 S = COV HS = ¼ V A + 1/16 V AA ……… Environmental variance component: σ 2 e (or) σ 2 w = σ 2 p – COV HS V P =σ 2 s+σ 2 e

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t = t = σ 2 s σ 2 s+σ 2 e

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Sires 1 2 i Total Y 11 Y 21 Y i1 Y 12 Y 22 Y i2 Y 13 Y 23 Y ij Yi=sire total Y1 Y2 ∑Yi X=Y1+Y2+Yi n i =No.of observation/sire n 1 n 2 n i N Equal number of progenies per sire group

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Equal number of progenies per sire group This method is used when each of a number of sires are mated to a number of females allotted at random and one offspring is produced by each dam. The observations Y ij on the j th offspring of the i th sire is expressed by the following statistical model. Yij =μ +S i +e ij

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y ij =μ +Si+e ij y ij = is the record of j th offspring of the i th sire µ = Population mean S i = Effect of i th sire e ij = environmental deviation attributed to individuals within the sire groups.

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∑(S i ) 2 = σ 2 s The size of the variance component (σ 2 s) is due to the fact that the sire group differs and indicates the difference between the progeny of different sires. This variance of the sire group means is equal to Cov HS ∑( e ij ) 2 = σ 2 e The variance component σ 2 e indicates the difference between the progeny of the same sire group within the sire component. t= σ 2 s+σ 2 e σ 2 s

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The analysis of variance is done to partition the phenotypic variance σ 2 p into σ 2 s and σ 2 e. MSs = K σ 2 s+σ 2 e. The mean square (MSs) between the sire group is an estimate of the variance component within the half sib group.

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The mean square within the group (MSe) is the variance of the individual observation about their group mean and is equal to σ 2 e. Therefore σ 2 s+σ 2 e= σ 2 p σ 2 s =MSs-MSe K σ 2 s σ 2 s+σ 2 e t =

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1. Calculation of correction factor = 2. Calculation of total sum of squares Individual observation total sum of square minus correction factor. ∑ Yij 2 - C.F that is( Y11 2 + Y12 2 ………… Y33 2 )- C.F Steps involved in the calculation of Heritability by ANOVA

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- C.F 3.Calculation between sires sum of square (Bss) 4.Calculation of within sires between progenies sum of squares (Wss) Wss =T.S.S- Bss, that is step2-3

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ANOVA Source of variation d.f Sum of squares Mean squares Expected mean squares Between sires S-1 Bss MSs=Bss/S-1 K σ 2 s+σ 2 e Within sires between progenies N-S Wss Mse=Wss/N-S σ 2 e Total N-1 TSS

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5. Calculation of σ 2 s = MSs-MSe K Vp=σ 2 s+σ 2 e 6. Intra class correlation (t)= σ 2 s σ 2 s+σ 2 e 7. Calculation of heritability, h 2 = 4 X t

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Example: There were three sires each mated to six dams and one progeny of each dam was recorded for their first litter size which is given below. Calculate the heritability by half sib correlation method and give the inference. Sires 1 2 3 Total 10 5 9 7 8 11 9 7 8 9 7 9 7 6 8 5 8 5 Total 47 41 50 X=138 n i =No.of observation/sire 6 6 6 N=18

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1. Calculation of correction factor = N (X) 2 CF = (138) 2 18 1 2 3 10 7 9 9 7 5 5 8 7 7 6 8 9 11 8 9 8 5 47 41 50 Steps in the calculation of ANOVA Correction factor = 1058 ∑X= 138

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2. Calculation of total sum of squares (∑ Yij 2 - C.F) that is( Y11 2 + Y12 2 ………… Y33 2 )- C.F 1 2 3 10 7 9 9 7 5 5 8 7 7 6 8 9 11 8 9 8 5 47 41 50 ∑ Yij 2 = 10 2 +7 2 +9 2 +…….8 2 +5 2 TSS= 1108-1058 =50

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3.Calculation between sires sum of square (Bss) - C.F - 1058 = 1065-1058 = 7 BSS = 7

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4.Calculation of within sires between progenies sum of squares (Wss) or Ess Wss =T.S.S- Bss, that is step 2-3 TSS= 1108-1058 =50 BSS = 7 Ess= 50 -7 =43

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Source of variation d.f Sum of squares Mean squares Expected mean squares Between sires S-1 3-1=2 Bss =7 MSs=Bss/S-1 =7/2=3.5 K σ 2 s+σ 2 e Within sires between progenies N-S 18-3=15 Wss =43 Mse=Wss/N-S =43/15=2.87 σ 2 e Total N-1 18-1=17 TSS=50 ANOVA

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σ 2 s =MSs-MSe K MSs= 3.5 MSe= 2.87 K= 6 σ 2 s = 3.5-2.87 6 = 0.105 5. Calculation of variance due to sire

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Vp= σ 2 s+ σ 2 e =0.105 + 2.87 =2.975 6. Intra class correlation ‘t’= σ 2 s σ 2 s+σ 2 e t = 7. Heritability= 4 x t = 4 x 0.035 = 0.14

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Inference: The calculated heritability value of litter size at birth is 0.14. It means only 14 % of the genetic superiority of the individuals selected for litter size will be transmitted to the next generation.