Page 30 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition
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1.7.2  Examples of selection indices using different sources of information

         Data available on correlated traits

         Example 1.6
         Assume the following parameters were obtained for average daily gain (ADG) from
         birth to 400 days and lean per cent (LP) at the same age in a group of beef calves:

                                      Heritability     Standard deviation
                    ADG (g/day)          0.43               80.0
                    LP (%)               0.30                7.2
         The genetic and phenotypic correlations (r  and r ) between ADG and LP are 0.30
                                              g     p
         and −0.10, respectively. Construct an index to improve growth rate in the beef calves.
         Assuming ADG as trait 1 and LP as trait 2, then from the given parameters:
                    2
             p 11  = 80 =  6400
                 7 2 =
             p 22  = .  2  51 84
                          .
             p 12  = rp ( p )( p ) =- .  (  )( (  .
                                 0 1 6400 51 84) =-57 6.
                           22
                      11
             g  = h 2  p  ) = 0 43.  (6400 ) = 2752
              11    ( 11
                                    )
                                         .
             g  = h 2  (p  ) = 030.  (5184. ) = 15 552
              22      22
             g  = rg  g  )(g  ) = 62 064
                                  .
              12     ( 11  22
         The index equations to be solved are:
            ⎡ b ⎤ ⎡ p 11  p ⎤ − 1 ⎡ g ⎤
              1
                        12
                               11
                 =
            ⎢  ⎥ ⎢        ⎥  ⎢  ⎥
              2 b ⎣  ⎦ ⎣ p 21 p 22⎦  ⎣ g 21⎦
         Inserting appropriate values gives:
            é  1 b ù é 6400.00 - 57.60ù - 1 é 2752.000ù
                 =
            ê  ú ê                ú  ê        ú
              2 b ë  û ë  - 57.60  51.84 û  ë  62.0664 û
         The solutions are b  = 0.445 and b  = 1.692.
                         1             2
            The index therefore is:
                                ) + 1.692(LP − m )
            I = 0.445(ADG − m ADG             LP
                   and m  are herd averages for ADG and LP. Using Eqn 1.21:
         where m ADG    LP
            r = ( é 0 445.  (2752 ) 1 692.+  (62 064.  )) 2752/  ù = 0 695.
                                                   û
                 ë
         Using single records on individual and relatives

         Example 1.7
         Suppose the ADG for a bull calf (y ) is 900 g/day and the ADG for his sire (y ) and dam
                                     1                                  2
         (y ) are 800 g/day and 450 g/day, respectively. Assuming all observations were obtained
          3
          14                                                              Chapter 1
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