Page 152 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition
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Then the solution for additive genetic effect for animals 6 using Eqn 3.8 is:
             ˆ
            u  = w ((uˆ  + uˆ )/2) + w (YD )
             6    1  1   5      2    6
               = w ((−0.3300 + −0.2449)/2) + w (−0.9844) = −0.8367
                  1                        2
         with w = 2(0.672)/6.344, w  = 5/6.344 and 6.344 = the sum of the numerators of
               1                  2
         w  and w .
          1      2
            For animal 8 with ten TD records, the solution for additive genetic effect is:
            u  = w ((uˆ  + uˆ )/2) + w (YD )
             ˆ
             8    1  1   7      2    8
               = w ((−0.3300 + 1.1477)/2) + w (0.3746) = 0.3786
                  1                       2
         with w = 2(0.672)/11.344, w  = 10/11.344 and 11.344 = the sum of the numerators
               1                  2
         of w  and w . The weights on YDs were 0.7882 and 0.8815 for animals 6 and 8,
             1      2
         respectively. This illustrates the fact that as the number of TD increases, more emphasis
         is placed on performance records of the animal. Considering animal 4 with ten TD
         records and a progeny, her breeding value can be calculated as:
             ˆ
            u  = w ((uˆ  + uˆ )/2) + w (YD ) + w (uˆ  − 0.5uˆ )
             4    1  1   2      2    4    3  7     3
               = w ((−0.3300 + −0.1604)/2) + w (−0.0226)
                  1                        2
                 + w (2(1.1477) − 0.4934) = 0.0043
                    3
         where w  = 2(0.672)/11.68, w  = 10/11.68 and w  = 0.5(0.672)/11.68 and 11.68 is
                1                  2                3
         the sum of the numerators of w , w  and w . There was a slight reduction to the
                                     1   2      3
         weight given to parent average from 0.1185 (animal 8) to 0.1151 (animal 4) due to
         the additional information from progeny.
            The solution for pe of an animal can be calculated as in Section 4.2.2, using Eqn 4.4.
         Here, the correction of the TD records is for the estimates for HTD effects and fixed
         regressions and animal effect. Thus for cow 6, pˆe  can be calculated as:
                                                   6
                  ⎛  ⎡ 10.4⎤  ⎡ 3.1893⎤  ⎡ 10.0835⎤  − ⎡  0.8367⎤⎞
                  ⎜  ⎢ 12.3 ⎥  ⎢ 3.33099 ⎥  ⎢ 12.2001 ⎥  ⎢ − 0.8367 ⎥⎟
                  ⎜  ⎢   ⎥  ⎢      ⎥  ⎢        ⎥  ⎢       ⎥⎟
                            ⎢
                     ⎢
            ˆ pe  =  ⎜ t′ 13.2⎥ − 3.3897⎥ − ⎢ 12..6254⎥ − − ⎢  0.83667⎥⎟  5.4380
              6
                  ⎜  ⎢   ⎥  ⎢      ⎥  ⎢        ⎥  ⎢ −     ⎥⎟
                  ⎜  ⎢ 11.6 ⎥  ⎢ 0.6751 ⎥  ⎢ 12.2077 ⎥  ⎢  0.8367 ⎥⎟
                  ⎜ ⎝  ⎢ ⎣  8.4 ⎥ ⎦  ⎢ ⎣ 0.0000 ⎥ ⎦  ⎢ ⎣ 11.5679 ⎥ ⎦  ⎢ ⎣ − 0.8367 ⎠
                                                          ⎥⎟
                                                          ⎦
                  −
                                  −
               = − 9.1650/5.4380= 1.6853
         where t is a column vector of order 5 (number of TD records for the animal), with
         all elements equal to one. However, in contrast to pe estimates in Example 4.1, these
         pe estimates represent permanent environmental factors affecting TD records within
         lactation.


         9.3  Random Regression Model

         In Section 9.2, the advantage of including fixed regressions on days in milk in the
         model was to account for the shape of the lactation curve for different groups of
         cows. However, the breeding values estimated represented genetic differences
         between animals at the height of the curves. Although different residual variances


          136                                                             Chapter 9
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