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ANSC*6370 - QG and Animal Breeding
Fall 2003 - Assignment 12 - Answers

The data below are exactly as given in the notes, plus one additional cow (9 with parents 7 and 2), and shown below.

Cow	Visit	Age	Age Group	Obs.
1	1	22	1	224
2	1	30	2	244
3	1	28	1	224
1	2	34	2	236
2	2	42	3	247
3	2	40	3	242
4	2	20	1	220
1	3	47	3	239
2	3	55	4	241
4	3	33	2	234
2	4	66	4	244
4	4	44	3	228
9	4	25	1	217

Let the model be

$$\begin{eqnarray*}y_{ijkm:t} & = & V_{i} + AG_{j} \\ & & + (a_{k0}z_{0} + a_{k1... ...0}z_{0} + p_{k1} z_{1} + p_{k2} z_{2}) \\ & & + e_{ijkm:t} \\ \end{eqnarray*}$$

Thus, the model uses age groups (4 of them) in place of the fixed regressions to account for the shape of the phenotypic relationship between the observations and age.

Use the same parameters as given in the notes for ${\bf G}$and ${\bf P}$. The residual variance, however, will be different for each age group, as follows: 0.8 for age group 1, 0.92 for age group 2, 1.05 for age group 3, and 1.15 for age group 4. Thus, ${\bf R}$ is diagonal, but the diagonals differ depending on the age group to which the cow's observation belongs.

Construct the MME and solve. Calculate EBVs for age 24 months and rank the animals. Compare the age group solutions to the fixed regressions in the notes.

The first thing to do is to compute the Legendre polynomials for the new animal. Standardize age 25 months into a number between -1 and +1, as

w₂₅ = -1 + 2 ((25 - 18)/(68-18)) = -.72

then form the vector ${\bf w}$ as

$${\bf w} = \left( \begin{array}{l} 1.00 \\ -0.72 \\ 0.5184 \end{array} \right).$$

The Legendre polynomials are calculated as

$${\bf z} = \left( \begin{array}{rrr} 0.7071 & 0 & 0 \\ 0 & 1.... ...in{array}{l} 0.7071 \\ -0.8818 \\ 0.4389 \end{array} \right).$$

The inverse of the relationship matrix has to be increased by one row and column for animal 9. This should be easy to do by now.

The ${\bf X}$ matrix for the Age Group effects of the model is

$${\bf Xb} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & ... ...{l} AG_{1} \\ AG_{2} \\ AG_{3} \\ AG_{4} \end{array} \right).$$

The ${\bf W}$ matrix for the random Visit effects is the same as in the notes except one row is added for animal 9.

With nine animals, the ${\bf Z}$ matrix has dimensions 13 rows by 27 columns (3 coefficients for each animal). The ${\bf R}$matrix is diagonal with diagonal elements equal to .8, .92, .8, .92, 1.05, 1.05, .8, 1.05, 1.15, .92, 1.15, 1.05, and .8, respectively.

The solutions are as follows:

i	$\hat{V}_{i}$	j	$\hat{AG}_{j}$
1	-0.3282	1	223.9322
2	0.3492	2	237.7465
3	0.5371	3	238.5614
4	-0.5582	4	233.9345

Animal	$\hat{a}_{0}$	$\hat{a}_{1}$	$\hat{a}_{2}$	$\hat{p}_{0}$	$\hat{p}_{1}$	$\hat{p}_{2}$
1	-1.8504	-0.4685	0.0074	0.6975	-0.0328	0.0010
2	3.6973	0.9567	-0.0067	7.0483	0.1862	0.0025
3	0.4092	0.1323	-0.0015	2.0097	0.2583	-0.0016
4	-5.0511	-1.5219	0.0181	-4.9692	-0.5676	0.0030
9	-2.7117	0.8091	-0.0051	-4.9111	0.2064	-0.0038
5	0.1887	-0.0874	0.0014	0	0	0
6	3.0420	0.5601	-0.0045	0	0	0
7	-1.7315	0.2330	0.0002	0	0	0
8	-1.9056	-0.5558	0.0061	0	0	0

To compute an EBV for 24 months of age, multiply the solutions above by the Legendre polynomials for age 24 months. Thus for animal 1, the EBV at 24 months would be

EBV₂₄ = 0.7071(-1.8504)-0.8328(-0.4685)+0.3061(0.0074) = -0.9160.

The EBV at 48 months for animal 1 would be

EBV₄₈ = 0.7071(-1.8504)+0.2449(-0.4685)-0.6957(0.0074) = -1.4283.

The EBV can be expressed for any month between 18 and 68.

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