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Best unbiased linear Prediction: Sire and Animal models

Raphael Mrode

Training in quantitative genetics and genomics

30th May to 10th June 2016

ILRI , Nairobi

2

BLUP

• The MME of provided a framework to obtain • BLUE of k'b, given that k'b is estimable • BLUP of the vector of random effects. • The properties of BLUP are more or less incorporated in the

name: – Best - means it maximises the correlation between true (a) and

predicted breeding value (â) or minimises prediction error variance (PEV) (var(a - â)).

– Linear - predictors are linear functions of observations.

– Unbiased - estimation of realized values for a random variable such as animal breeding values and of estimable functions of fixed effects are unbiased (E(a = â)). Accounts for selection if all data on which selection has been based is included

– Prediction - involves prediction of true breeding value.

3

Numerator relationship matrix

• In lecture 1, we estimated the fixed effect solutions and predicted genetic merit of 3 sires.

• Assumed that the sires were unrelated. • Usually animals or sires tended to be related and the genetic

relationship among these animals is incorporated • The genetic covariance among individuals is comprised of three

components: – the additive genetic variance

– the dominance variance and

– the epistatic variance.

• This lecture will address only the additive genetic relationship

• Use of additive genetic relationship matrix usually increases the accuracies of evaluations and should help account for previous selection decisions if all pedigrees are utilised

4

Numerator relationship matrix

• The numerator relationship matrix (A) describes the additive genetic relationship among individuals

• The additive genetic relationship between animals i and j is twice the probability two genes taken at random from i and j are of identical by descent.

• It is equal to twice the coancestry or the coefficient of kingship

• The matrix A is symmetric and its

– diagonal element for animal i (aii) is equal to 1 + Fi, with Fi is the inbreeding coefficient

– off-diagonal element, aij equals the numerator of the coefficient of relationship

• When multiplied with genetic variance (σ2u) it is equal to the covariance of

breeding values. Thus var(ui) = aiiσ2

u = (1 + Fi)σ2

u.

5

Recursive method for computing A

• Henderson (1976) described method for calculating the matrix A

• Pedigree are coded 1 to n and ordered such that parents precede their progeny.

If both parents (s and d) of animal i are known

aji = aij = 0.5(ajs + ajd ) ;j = 1 to i-1

aii = 1 + 0.5(asd)

If only one parent s is known and assumed unrelated to the mate

aji = aij = 0.5(ajs) ;j = 1 to i-1

aii = 1

If both parents are unknown and are assumed unrelated

aji = aij = 0 ;j = 1 to i-1

aii = 1

6

Example pedigree

• Calf Sire Dam • --------------------------- • 3 1 2 • 4 1 unknown • 5 4 3 • 6 5 2

• a11 = 1 + 0 = 1 • a12 = 0.5(0+0) = 0 = a12 • a22 = 1 + 0 = 1 • a13 = 0.5(a11+a12) = 0.5(1.0 + 0 ) = 0.5 = a31 • a23 = 0.5(a12+a22) = 0.5(0 + 1.0) = 0.5 = a32

7

Recursive method for computing A

1 2 3 4 5 6 ---------------------------------------------------- 1 1.00 0.0 0.50 0.50 0.50 0.25 2 0.00 1.0 0.50 0.00 0.25 0.625 3 0.50 0.50 1.00 0.25 0.625 0.563 4 0.50 0.00 0.25 1.00 0.625 0.313 5 0.50 0.25 0.625 0.625 1.125 0.688 6 0.25 0.625 0.563 0.313 0.688 1.125 • a66 = 1 + 0.5(a52) = 1 + 0.5(0.25) =1.125 • From the above calculation the inbreeding coefficient for calf 6 is

0.125

8

Decomposing the matrix A

A = TDT' where T is a lower triangular matrix. Non-zero elements of T, say tij, is the coefficient of relationship between

animals i and j, if i and j are direct relatives It can easily be computed applying the following rules: For the ith animal tii = 1 If both parents (s and d) are known tij = 0.5(tsj + tdj) If only one parent (s) is known tij = 0.5(tsj) If neither parents is known tij = 0

9

Decomposing the matrix A

The Mendelian sampling (m) for an animal i

mi = ui - 0.5(us + ud)

var(mi) = var(ui) - var(0.5us + 0.5ud)

= var(ui) - var(0.5us) - var(0.5ud) + 2cov(0.5us,0.5ud)

= 1 + Fi)σ2

u - 0.25assσ2

u - 0.25addσ2u - 0.5asdσ2

u

var(mi)/σ2u = dii = (1 + Fi) - 0.25ass - 0.25add - 0.5asd

Since Fi = 0.5asd

dii = 1 - 0.25(1 + Fs) - 0.25(1 + Fd)

dii = 0.5 - 0.25(Fs + Fd)

10

Decomposing the matrix A

D is a diagonal matrix of the covariance matrix for Mendelian sampling. It is calculated as:

if both parents of animal i are known, the

dii = 0.5 - 0.25(Fs + Fd)

If only one parent (s) is known diagonal element is

dii = 0.75 - 0.25(Fs)

and if no parent is known

dii = 1

11

Matrix T for example pedigree

The matrix T is

1 2 3 4 5 6

-------------------------------------

1 1.0 0.0 0.0 0.0 0.0 0.0

2 0.0 1.0 0.0 0.0 0.0 0.0

3 0.5 0.5 1.0 0.0 0.0 0.0

4 0.5 0.0 0.0 1.0 0.0 0.0

5 0.5 0.25 0.5 0.5 1.0 0.0

6 0.25 0.625 0.25 0.25 0.5 1.0

12

Matrix D for example pedigree

• D = diag(1.0, 1.0, 0.5, 0.75, 0.5, 0.469)

• For instance, animal 4 has only the sire known which is not inbred, therefore

• d44 = 0.75 - 0 = 0.75

• and

• d66 = 0.5 - 0.25(0.125 + 0) = 0.469

• because both parents are known and the sire has inbreeding coefficient of 0.125

13

Deriving rules for A inverse

A-1 = (T-1)'D-1T-1

I - M = T-1 , where M is a matrix that traces genes from parents to offspring

• D-1 = Diag(1,1,2,1.333,2,2.133)

1.00.5-0.00.00.5-0.0

0.01.00.5-0.5-0.00.0

0.00.01.00.00.0.5-

0.00.00.01.00.0.5-

0.00.00.00.01.00.0

0.00.00.00.00.01.0

=

0.00.50.00.00.50.0

0.00.00.50.50.00.0

0.00.00.00.00.0.

0.00.00.00.00.0.5

0.00.00.00.00.00.0

0.00.00.00.00.00.0

-

100000

010000

001000

000100

000010

000001

0

5

05

5

14

Deriving rules for A inverse

• Therefore A-1 = (T-1)'D-1T-1 is:

•

• 1 2 3 4 5 6

• |------------------------------------------------

• 1 | 1.833 0.5 -1.0 -0.667 0.0 0.0

• 2 | 0.5 2.033 -1.0 0.0 0.533 -1.067

• 3 | -1.0 -1.0 2.50 0.5 -1.0 0.0

• 4 | -0.667 0.0 0.5 1.833 -1.0 0.0

• 5 | 0.0 0.533 -1.0 -1.0 2.533 -1.067

• 6 | 0.0 -1.067 0.0 0.0 -1.067 2.133

15

Rules for the inverse of A (ignoring inbreeding

Henderson (1976) used equation to develope rules for A-1

D-1 = 2 if both parents are known

4/3 if one parent is known

1 if no parent is known

If di = diagonal element of D-1 for animal i.

di = 4/(2 + no of parents unknown)

16

Rules for the inverse of A (ignoring inbreeding

A-1 = 0 If both parents of the ith animal are known, add di to the (i,i) element -di/2 to the (s,i), (i,s), (d,i) and (i,d) elements di/4 to the (s,s), (s,d), (d,s) and (d,d) elements If only one parent (s) of the ith animal is known, add di to the (i,i) element -di/2 to the (s,i) and (i,s) elements di/4 to the (s,s) element Neither parents of the ith animal are known, add di to the (i,i) element

17

Rules for the inverse of A

• The same rules can be used when accounting for inbreeding but with the elements of D-1 computed accounting for the inbreeding of parents. Verify this using the pedigree above

18

Sire model

• Using of the performance of progeny to evaluate the genetic merits of their sires is referred to as the sire model.

• For instance, the genetic merits of bulls can be predicted on the basis of the milk production

• Given that the covariance between the sire and his progeny is 0.5, this method predicts the probable transmitting ability (PTA) of the sires which is half the breeding values of the sires.

19

Sire model

• MME with the relationship matrix incorporated are

• with α = σ2e/σ2

s = 4-h2/h2

=

+

yZ

yX

a

b

AZZXZ

ZXXX

1 ˆ

ˆ

20

Example: Sire model

• Same data as in lecture1 , a sire model with A among the 3 sires incorporated.

• Assume that α = σ2e/σ2

s = 100/200 and the data of recoding is as follows:

• --------------------------------------------------------

• Cow Herd Calving class Sire Test day milk yield (kg)

• -----------------------------------------------------

• 8 1 2 5 36.2

• 9 1 1 6 25.8

• 10 1 2 7 31.5

• 11 1 1 5 42.0

• 12 1 2 6 12.3

• 13 2 2 7 28.5

• 14 2 1 5 10.6

• 15 2 2 6 23.4

• 16 2 2 7 22.4

• 17 2 1 5 14.8

• -------------------------------------

• The sires 1,2, 3 were recorded as 5,6,7 respectively and cows 1 to 10 recoded from 8 to 17

21

Example : Sire model

• Bull Sire dam

• 5 1 0

• 6 2 3

• 7 2 4

• --------------------------------------------

• Using the rules outlined earlier, the A-1 for the above pedigree is

2001010

200110

1.3330000.667

symmetric1.500.50

1.50.50

20

1.333

A1

22

Example: Sire Model

• The matrix X is as defined in lecture 1.

• Z is now set up to included the 4 animals in the pedigree

• The addition of A-1 α to Z’Z gives

3000000

0300000

0040000

0000000

0000000

0000000

0000000

ZZ

01001001000000

00100100100000

10010010010000

Z and

4000.500.50

04000.50.50

004.6670000.333

0.5000.7500.250

00.5000.750.250

0.50.500.250.251.000

000.3330000.667

AZZ1

23

Example : Sire Model

• The MME are:

82.4

61.5

103.6

0

0

0

0

154.3

93.2

99.7

147.8

a

a

a

a

a

a

a

b

b

b

b

4.0000.500.503021

4.0000.50.502112

4.6670000.3331322

0.7500.2500000

0.750.2500000

sym.1.0000000

0.6670000

6033

422

50

5

7

6

5

4

3

2

1

4

3

2

1

24

Example: Sire Model

• Fixed effects

• Herds

• 1 11.239

• 2 0.000

• Calving class

• 1 17.606

• 2 20.071

• Random sire effects

• 1 0.915

• 2 -0.915

• 3 -3.155

• 4 2.240

• 5 1.830

• 6 -5.190

• 7 2.903

• ---------------------------------

25

Accuracy and prediction error variance

• The accuracy (r) of predictions is the correlation between true and predicted breeding values.

• Dairy cattle evaluations, the accuracy of evaluations is usually expressed in terms of reliability, which is r2.

• Calculation for r or r2 require the diagonal elements of the inverse of the MME.

• If the coefficient matrix of the MME is represented as C:

26

Accuracy and prediction error variance

• Prediction error variance (PEV) = var(a-â) = C22σ2e

• PEV could be regarded as the fraction of additive genetic variance not accounted for by the prediction. Therefore

• PEV = C22σ2e = (1 - r2)σ2

a or PEV = (1 - r2)σ2s for a sire

model

• with r2 = squared correlation between the true and estimated breeding values

2221

12111

2221

1211

CC

CCC

CC

CCC inverse dgeneralise a and

27

Accuracy and prediction error variance

• Thus for animal i

• diσ2

e = (1 - r2)σ2a

• where di is the ith diagonal element of the C22

•

• di σ2

e/σ2a = 1 - r2

• r2 = 1 - diα

• The standard error of prediction (SEP) is

• SEP = var(a - â)

• = diσ2

e for animal i

• Note that r2 = 1 - ( SEP2 /σ2a ). ASReml gives SEP and not r2, so compute r2 from SEP

• Diagonal elements from the 3 by 3 block for the 3 sires in example were 0.975, 0.970 and 1.045. The corresponding reliabilities for the 3 sires therefore equals were 0.55, 0.57 and 0.54 respectively.

28

Uses of the PEV

• Note:

• PEV = (1 – r2)σ2a = α / (ne + α)σ2

a

• Therefore ne can be computed as:

•

• ne = [PEV-1 – σs-2]σe

2 with a sire model

• ne = [PEV-1 – σA-2]σE

2 with an animal model

v ar A v ar ˆ A PEV

The effective number of progeny or records (ne) also related to PEV:

v ar ˆ A A2PEV

29

Individual Animal model

• In the previous example, the genetic merit of only the sires of cows were predicted.

• The main advantage: number of equations are reduced since only sires are evaluated compared to all animals in the data set.

• Disadvantages: genetic merit of the mate (dam of progeny) is not accounted for. It is assumed that all mates are of similar genetic merit and this can result in bias in the predicted breeding values if there is preferential mating.

• Therefore an evaluation that predicts the genetic merit of all animals in the data set would overcome this problem. This is called the individual animal model

• Uses the usual MME to evaluate all animals with records and all their relatives in the pedigree. However the α term with an animal model, α = σ2

e/σ2a = 1-

h2/h2

30

Individual Animal model

Calves Pens Sire Dam WWG (kg) --------------------------------------------------- 4 Pen1 1 unknown 4.5 5 Pen2 3 2 2.9 6 Pen2 1 2 3.9 7 Pen1 4 5 3.5 8 Pen1 3 6 5.0 -------------------------------------------------- Assume = σ2

a=20 and σ2e= 40

and α= σ2e/σ2

a= 40/20 =2

31

Incidence matrices

0.55.39.39.25.4';

y

00110

11001=X

;

10000000

01000000

00100000

00010000

00001000

= Z

00110000

11001000=ZX

32

Matrices for the MME

• Z’X = transpose of X’Z

)1,1,1,1,1,0,0,0(

0.55.39.39.25.4000 ' of Tranpose

8.6

0.13'

diag

ZZ'

yZ

yX

33

The A-1 for the example data

2.0000.0001.000-0.0000.0001.000-0.0000.000

0.0002.0000.0001.000-1.000-0.0000.0000.000

1.000-0.0002.5000.0000.0000.5001.000-1.000-

0.0001.000-0.0002.5000.5001.000-1.000-0.000

0.0001.000-0.0000.5001.8330.0000.0000.667-

1.000-0.0000.5001.000-0.0002.0000.5000.000

0.0000.0001.000-1.000-0.0000.5002.0000.500

0.0000.0001.000-0.0000.667-0.0000.5001.833

= A1-

34

Solutions to the MME

Effects Solutions Pens 1 4.358 2 3.404 Animals 1 0.098 2 -0.019 3 -0.041 4 -0.009 5 -0.186 6 0.177 7 -0.249 8 0.183

35

Fixed effect solutions

From first row of MME

(X’X)b = X’y - (X’Z)â

b = (X’X)-1X’(y - Z â)

- Solution for calves in pen 1 is

b1 = [(4.5 + 3.5 + 5.0) - (-0.009 + -0.249 + 0.183)]/ 3 = 4.358

ij j

ijijidiagay /)ˆ(b̂

36

Understanding animal EBV

From second row of MME

(Z’Z+A-1α)â = Z’y - (Z’X)b

(Z’Z+A-1α)â = Z’(y - Xb)

(Z’Z+A-1α)â = (Z’Z)YD

with YD = ZZ-1 Z’(y – Xb)

(Z’Z+ uiiα)âi = αuip(âs + âd) + (Z’Z)YD + αΣkuim(âanim - 0.5âm)

(Z’Z+uiiα)âi = αupar(PA) + (Z’Z)YD + 0.5αΣkuprog(2âanim - âm)

37

Understanding animal EBV

EBV is made of contributions from: Parents average (PA), its record and progeny (PC) EBV = n1PA + n2YD + n3 PC Where n1 + n2 + n3 = 1 Numerator of: n1 = 2α, 4/3α or 1α if both,1 or no parents is known n2 = number of records n3 = ½α or ⅓α when mate is known or not Denominator = sum of numerators for n1 ,n2 ,n3 YD = record of animal corrected for fixed effects

38

Understanding animal EBV

Animal 8 as an example:

EBV8 = n1(PA) + n2(y8 – b1)

where

n1 = 2 λ /5=4/5 and n2 = 1/5

EBV8 = n1(EBV3+EBV6)/2) + n2 (5.0 -4.358)

EBV8 = n1(0.068) + n2 (5.0 -4.358) = 0.183

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