Exercises Day 1
Part 1: EBV accuracy
Exercise 1
Effect of using relatives’ information on selection accuracy
For single trait prediction of breeding value, write out the P-Matrix and the G-vector for the
following cases:
1. One own performance record (X1)
P-matrix:
var(X1) = 2P is the phenotypic variance
G-vector:
Cov(X1, A)
= Cov(A+E, A) = Cov(A, A) + cov(E,A) = 2A + 0 = 2A
b = P-1G = (2P )-1 2A = h2
accurcay2 = b’G/2A = h2. 2A/ 2A = h2. Hence, accuracy = equal to h.
2. Information known on own performance and performance of sire (1 record each)
Information sources:

X1 = own performance
X2 = performance of sire
variance and covariance of information sources:
cov( X1 , X 2 ) 
X 
 var( X1 )
var  1   P  

var( X 2 ) 
 X2 
 cov( X 2 , X1 )
var(X1) = 2P is the phenotypic variance
var(X2) = 2P is the phenotypic variance
Cov(X1,X2)
= Cov(A+E, As +Es)
= Cov(A, As) + cov(A,Es) + cov(E,As) +cov(E,Es)
= ½2A+ 0 + 0 + 0.

covariance between information sources and the animal’s breeding value
X

 cov( X1 , A) 
cov( 1 , A   G  

 X2 
 cov( X 2 , A) 
Cov(X1, A)
= 2A
Cov(X2, A)
= Cov(As +Es , A) = Cov(As, A) + cov(Es,A) = ½2A + 0 .
such that index weights obtained by regression = covariance/variance:
Exercise Day1
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  P2
 b1 
1

P
G

1 2
 
 b2 
 2 A
1
 A2    A2   1
 

 P2   1 2  A2   1 2 h2
1
2
1
1
h2   h2 
 

1   1 2 h2 
2
3. Information known on own performance and an EBV of the sire (accuracy = 0.9)

variance and covariance of information sources:
cov( X1 , X 2 ) 
X 
 var( X1 )
var  1   P  

var( X 2 ) 
 X2 
 cov( X 2 , X1 )
var(X1) = 2P is the phenotypic variance
var(X2) = r22A where r is accuracy and r2 is reliability
Cov(X1,X2)

= ½ r2. 2A
covariance between information sources and the animal’s breeding value
X

 cov( X1 , A) 
cov( 1 , A   G  

 X2 
 cov( X 2 , A) 
Cov(X1, A)
= 2A
Cov(X2, A)
= ½ r2 2A
4. Information known on own performance and an EBV of the sire (acc=0.9) and dam (acc=0.5)
P is a 3 by 3 matrix, G a 3 by 1 vector
5. Information on own performance and 25 half sibs
Information sources:

X1 = own performance
X2 = mean performance of n half sibs
variance and covariance of information sources:
cov( X1 , X 2 ) 
X 
 var( X1 )
var  1   P  

X
cov(
X
,
X
)
var( X 2 ) 
2
1
 2

var(X1) = 2P is the phenotypic variance
var(X2) = tHS 2P + ((1-rHS)/n) 2P where tHS = 1/4 h2
Cov(X1,X2) = tHS

is the intra class correlation
covariance between information sources and the animal’s breeding value
X

 cov( X1 , A) 
cov( 1 , A   G  

 X2 
 cov( X 2 , A) 
Cov(X1, A)
= 2A
Exercise Day1
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Cov(X2, A)
= ½ 2A
such that index weights obtained by regression = covariance/variance:
1
1
  P2
   A2   1
r
r P2
 b1 
  h2 
1

  P G  2
 
2 1
2
2
 b2 
 r P [r  (1  r ) / n] P   2  A   r [r  (1  r ) / n]   1 2 h 
6. Information on own performance, EBV of the sire (acc=0.9) and 25 half sibs
7. Information on own performance, 25 half sibs and 50 progeny
Use the symbols VA for additive genetic variance and VP for phenotypic variance.
Note that for a single trait prediction you can also substitute these by V P = 1 and VA = h2
You can plug the formulas from above in Excel, and for these cases calculate
the index weights and the accuracy, assuming a certain value for h 2..
For some guidance on matrix calculation in Excel, see the end of this exercise sheet.
You can check your answers with STSELIND.XLS
The code looks like
vrep = rep + (1 - rep) / nrep
vdam = rep + (1 - rep) / ndam
vsir = rep + (1 - rep) / nsir
vfs = tfs + (1 - tfs) / nfs
fsgrsize = nfs + 1
fsOD = (nfsgr - 1) * (fsgrsize * (fsgrsize - 1))
hsOD = (nhs * (nhs - 1)) - fsOD
vhs = (nhs + fsOD * tfs + hsOD * ths) / (nhs * nhs)
vprog = ths + (1 - ths) / nprog
P(1, 1) = vrep
P(1, 2) = 0.5 * h2
P(1, 3) = 0.5 * h2
P(1, 4) = tfs
P(1, 5) = 0.25 * h2
P(1, 6) = 0.5 * h2
P(2, 2) = vdam
P(2, 3) = 0
P(2, 4) = 0.5 * h2
P(2, 5) = 0
P(2, 6) = 0.25 * h2
P(3, 3) = vsir
Exercise Day1
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P(3, 4) = 0.5 * h2
P(3, 5) = 0.5 * h2
P(3, 6) = 0.25 * h2
P(4, 4) = vfs
P(4, 5) = 0.25 * h2
P(4, 6) = 0.25 * h2
P(5, 5) = vhs
P(5, 6) = 0.125 * h2
P(6, 6) = vprog
For i = 2 To 6
For j = 1 To i
P(i, j) = P(j, i)
RHS(i, j) = RHS(j, i)
RFS(i, j) = RFS(j, i)
Next
Next
G(1, 1) = h2
G(2, 1) = 0.5 * h2
G(3, 1) = 0.5 * h2
G(4, 1) = 0.5 * h2
G(5, 1) = 0.25 * h2
G(6, 1) = 0.5 * h2
Exercise Day1
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Exercise 2
Correlations between relatives’ EBV
Consider the following cases, and for each case, calculate the correlation between EBVs on full
sibs and half sibs.
You can use STSELIND, and use the P-matrix and the index weights to work out this problem.
1. Information known on EBV of the sire (acc=0.9) and dam (acc=0.5)
2. One own performance record
3. Information known on own performance and an EBV of the sire (acc=0.9) and dam (acc=0.5)
4. Information on own performance, EBV of the sire (acc=0.9), dam (acc=0.5) and 50 progeny
See the new stselind:
' R matrices for correlation among siblings
RFS = P
RHS = P
RFS(1, 1) = tfs
RFS(6, 6) = h2 / 8
RHS(1, 1) = ths
RHS(1, 2) = 0
RHS(1, 4) = ths
RHS(2, 2) = 0
RHS(2, 6) = 0
RHS(4, 4) = ths
RHS(6, 6) = h2 / 16
Exercise Day1
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Exercise 3
Pseudo BLUP
In real life, parents have not just their own records, but they have an estimated breeding value
with a certain accuracy, using BLUP. This accuracy is based on ancestor information, their own
siblings and perhaps their offspring. The STSELIND program does not take these into account,
other than that the progeny of the parents are the sibs of the animals. Also, BLUP corrects for the
records of the mates of sires, when their progeny are evaluated.
The amount of ancestral information can be derived from a given population structure.
The program BLUPEBV.XLS does a full Pseudo-BLUP prediction of EBV accuracy, given genetic parameters,
and a certain population structure (Half-sib and full-sib family size). In the following, use BLUPEBV to
calculate EBV accuracy, and compare that with the more simple STSELIND approach:
h2=0.25, FS family size =2, HS family =40
and
h2=0.10 FS family size =4, HS family =80
Explain the negative weight on EBV of mates
Vary h2 and look at weights on parental EBV
Exercise Day1
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Exercises Day 1
Part 2: Selection response
Exercise 1
Consider the following breeding program that is being initiated in a large breeding herd:
 Selection is based on EBV derived from own performance only, for a repeated measures trait
that is evaluated on both males and females at 6 months of age and at 18 months of age.
 Trait parameters are:
h2=.30
repeatability = 0.6
Phenotypic st. dev. = 10 Each
year 500 males and 500 females are born and available for selection at 6 months of age.
 Males stay in the herd for 1 year and are used for breeding for one breeding season (year).
Each year, the top 50 6-month-old sires are selected based on EBV derived from own
performance at 6 months of age. They produce their progeny when they are 1 year of age
 Females stay in the herd for 2 years but only 75% of all 1-year old females survive to 18
months of age; 25% of all 1-year-old females are culled from the herd for reasons unrelated to
the trait under selection. So each year 500 6-month olds and 375 18-month-olds are available.
 Each year the top 125 out of the 500 6-month-old females and the top 75 out of the 375 18month-old females are selected for breeding. 6- and 18-month-olds are selected based on EBV
from own performance at 6 months and on the average performance at 6- and 18-months,
respectively. They produce progeny when they are 1 and 2 years of age, respectively.
In year 0, the herd is unselected and consists of 500 6-month-old males, 500 6-month-old
females, and 375 18- month-old females (NOTE: the herd is unselected, thus all age groups have
the same genetic mean in year 0)
Our aim is to predict the genetic progress that will be made in this herd over the next 2 years.
a) Predict the average genetic level of progeny that are born in year 1 (i.e. based on selection of
the 6-month-old males and the 6-month-old and the 18-month-old females that are present in
year zero). Hint: predict the average genetic level of the selected 6-month-old sires, the
selected 6-month-old dams, and the selected 18-month-old dams, and based on those, predict
the genetic level of the progeny.
b) Predict the average genetic level of progeny that are produced in year 2 (i.e. based on
selection of sires that are 6 months old in year 1 and females that are 6 and 18 months old in
year 1). Hint: use the same procedure as under a) but now take into account that the males
(females) that are now 6-months old, are the progeny whose mean you calculated under a).
c) Predict the asymptotic rate of progress per year for the breeding program described above
and compare it to the progress that is made from year to year based on the calculations made
in a) and b). Discuss your results.
In year 1, the breeder now decides to optimize the proportions selected from 6- and 18-monthold females, for a total of 200 dams selected, rather than using 125 6-month-olds and 75 18month-olds, in order to maximize the genetic level of progeny in year 2.
d) Using TRUNCSEL.XLS, determine the optimal proportions selected from the two age groups of
females that are available in year 1. To do this, specify the mean and standard deviations of
the EBV and the proportion of available selection candidates (females) that are 6- and 18months old, and enter this in the spreadsheet.
e) Using the proportions determined in d), compute the genetic level of progeny born in year 3
and compare this with results obtained in c).
Exercise Day1
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Exercise 2
The response per year can be given by the formula of Rendel and Robertson: Ryr = S/L.
For simplicity assume equal selection intensities in males and females).
We can maximize selection response by truncation section across age classes, (assuming the
selection criteria are comparable across age classes). Truncation selection across age classes
maximizes the mean of the selected parents.
Show algebraically that maximizing the mean of selected parents results in maximizing the
response per year (hence, optimizes selection across age classes)
Exercise 3
Using the Excel Spreadsheet ‘Genetic_gain.xls’, evaluate the impact of the percentage of cows
inseminated by young bulls and progeny group size on genetic gain. Find the optimal combination
of these two variables in order to maximize genetic gain.
Exercise Day1
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Exercises Day 1
Part 3: Change of Variance
Exercise 1
Consider a trait with heritability equal to 0.25 and phenotypic standard deviation equal to 20. The
trait is expressed in females only. Assume the infinitesimal genetic model. In a population of
infinite size, each generation the best 10% males and the best 50% females are selected and
mated at random to produce the next generation. Males are selected based on the average
performance of 5 half-sisters. Females are selected on an optimal index of own performance and
the average performance of 4 half-sisters. There are no common environmental effects among
sibs. All half sisters are themselves half sibs of each other.
a.
Starting from an unselected population, derive accuracies of selection and response to
selection in the initial generation (=generation 1).
b.
Taking into account the Bulmer effect, derive the genetic variance among progeny that
result from selection in that first generation.
c.
Taking into account the effect of selection on genetic variance and accuracy of selection
due to the Bulmer effect, predict the accuracies of selection (i.e. set up the appropriate
indexes based on the new genetic variances), responses to selection, and genetic variance
of progeny that result from selection in the second generation. Discuss the results.
d.
Repeat b. and c. but now assume that selection is on Animal Model BLUP EBV. The
accuracies of EBV used for selection are 0.25 for males and 0.529 for females. Note: these
accuracies are the accuracies that are obtained from the inverse of the Mixed Model
Equations without accounting for the effect of selection on genetic variance (i.e. they are
accuracies based on an unselected population).
e.
Derive the asymptotic genetic variance, accuracy, heritability and response to selection for
the breeding program under d, using Animal Model BLUP EBV.
f.
Compare results from a. through e. to those you get from using the program SelAction.
Exercise 2: Pseudo BLUP EBV with the Bulmer effect
a. Use BLUPEBV.XLS and compare accuracy and index weights with and without Bulmer? You can
also evaluate this by changing the % selected and see how this changes index weights.
b. Evaluate reduction in accuracy due to Bulmer with different heritabilities
c. How does the variance of parental EBV change with the proportion selected?
d. Calculate the accuracy of a parental average EBV with and without selection (Bulmer
correction) for different proportions selected. Look also at the variance of parental EBV.
PA = ½ EBVsire + ½ EBVdam
Var(PA) = ¼Var(EBVsire) + ¼Var(EBVdam)
Accuracy = sqrt(Var(PA)/VA)
Exercise Day1
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Matrix calculations using Excel
You can do some basic matrix calculations with MS Excel.
First put in the values of your matrices
To multiply two matrices:
- select an area of the size of the resulting matrix
- type: =MMULT(
- select the area of the first matrix
- type a comma (,)
- select area of the second matrix
- type a close bracket )
- press: Ctrl_Shift_Enter
To add or subtract a matrix (vector):
- select an area of the size of the resulting matrix
- type: = (
- select the area of the first matrix
- type a + or - select area of the second matrix
- type a close bracket )
- press: Ctrl_Shift_Enter
To invert a matrix:
- select an area of the size of the resulting matrix
- type: =MINVERSE(
- select the area of the first matrix
- type a close bracket )
- press: Ctrl_Shift_Enter
To transpose a matrix (vector):
- select an area of the size of the resulting matrix
- type: =TRANSPOSE(
- select the area of the first matrix
- type a close bracket )
- press: Ctrl_Shift_Enter
A more specialized matrix calculation program is MATLAB. It contains many more matrix functions
and mathematical function than excel. MATLAB allows you to make and run programs, draw
graphs, and run simulation). A MATLAB student version is very well suitable for animal breeding
problems and quite easy to use.
Exercise Day1
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