Info
Using the A1A1 genotype breeding value of 2qa where a = a + d(q  p), the difference between the genotypic value relative to the population mean and the breeding value is
A1A1 Dominance deviation = 2q(a  dp)  2q(a + d(q  p)) (10.41)
This equation can be expanded to
A1A1 Dominance deviation = 2qa  2qdp  2qa  2q2d + 2qdp (10.42)
and then canceling terms gives
A1A1 Dominance deviation = 2q2d (10.43)
Table 10.6 gives the expressions for all three genotypic values relative to the population mean as well as the dominance deviations derived using this same reasoning.
The dominance deviations for the four IGF1 examples are shown in Table 10.7. Examination of Table 10.7 shows that the genotypic values of the heterozygotes (measured from the population mean) and breeding values in the table are always zero when there is no dominance. In both cases (a) and (b), the A1A2 genotypic value is at the midpoint (d = 0), giving dominance deviations of zero regardless of the allele frequencies. Another way to describe this is to say that when there is no dominance, a genotype has a breeding value that is identical to its genotypic value (measured relative to the population mean) since genotypic values are determined by the addition of average effects alone. Without dominance, genotypic values in progeny as measured by the breeding value can be predicted perfectly from the combination of average effects.
In cases (c) and (d) the A1A2 genotypic values are not at the midpoint (d ^ 0), resulting in dominance deviations that are not zero. Let's examine the dominance deviation for the A1A1 genotype in case (c) of Table 10.7 as illustrated in Fig. 10.3a. The genotypic value of the A1A1 genotype when measured relative to the population mean is 10.5  2.625 = 7.875. The breeding value of the A1A1 genotype is 10.5 based on the average effects of its two alleles. With this breeding value, the mean of the progeny from an A1A1 genotype would be outside the largest phenotypic value, which is physically impossible. The breeding value has in essence assumed that the population mean is zero, as it would be at p = q = 0.5 without dominance. But there is dominance, so the population mean is greater than it would be without dominance. The A1A1breeding value needs to be adjusted downward so that it does not exceed the largest phenotypic value. This adjustment is the dominance deviation. In this particular case, the population mean is 2.625 rather than zero because of dominance shown by the A1A2 genotype (the A1A2 genotype has a value of 5.25 and makes up 50% of the population at p = q = 0.5). Therefore, the dominance deviation is 2.625. Note that in general the magnitude of dominance deviations changes with population allele frequencies just as the magnitude of genotypic values measured from the population mean and breeding values do.
The dominance deviation for the A1A1 genotype in case (d) of Table 10.7 is also diagrammed in Fig. 10.3.
Table 10.7 Genotypic values, breeding values, and dominance deviations for the three IGF 1 locus genotypes  
in dogs. Genotypic values, breeding value and dominance deviation values are all given relative to the 

Post a comment