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| Uitvouwen | ||
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The heritability for milk production is 0.3. The daughters produce on average 10,000 kg, and the population average is 9.500 kg. Step 1: the phenotype superiority = 10,000 – 9,500 = 500 kg. Step 2: the regression coefficient (see formula for offspring information in table 1) b= Step 3: the EBV for milk production of this bull is 1.78 * 500 = 890 kg Note: the maximum regression coefficient of a single parent (usually sire) on offspring is 2 because the sire passes half its genome on to the offspring. Turning that around, and assuming that the sire is mated to average dams, if you have information on the superiority of the offspring than that of the sire is that of the offspring times 2. |
Thus:
| Paneel | ||
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The maximum regression coefficient when using offspring information is 2, and not 1 |
| Uitvouwen | ||
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The heritability for slaughter weight is 0.4, the population average is 875 g/d, and that of the 20 full sibs is 900 g/d. The common environmental effect for full sibs (c2) = 0.45 Step 1: the phenotypic superiority = 900 – 875 = 25 g/d Step 2: the regression coefficient = (½* 20 * 0.4)/(1+(20-1)*( ½* 0.4 + 0.45)) = 4/13.35 = 0.30 Step 3: the EBV for average daily gain from 25 to 100 kg for this pig is 25 * 0.3 = 7.5 g/d. Note: the regression coefficient is lower than the heritability. Reason is that full sibs perform more alike because they have shared a common environment. Therefore, a smaller proportion of the phenotypic superiority can be assigned to shared genetics than without shared common environment. This is taken into account through the c2 when determining the regression coefficient for estimating the breeding value. |
Thus:
| Paneel | ||
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Thus: the presence of a common environmental effect has a reducing effect on the estimated breeding value |