with an A2 allele in half of the progeny to make equal frequencies of A1A1 and A1A2 progeny. The mean value of these progeny, and hence the average effect of an A1 allele, would be 0.5(10.5) + 0.5(0.0) = 5.25 kg. The breeding value of an A1A1 genotype is twice the average effect of the A1 allele because it has two A1 alleles. Therefore, we double the average effect of an A1 allele to determine that the breeding value of the A1A1 genotype is 2(5.25) = 10.5 kg. Figure 10.3 also shows two examples of the breeding value for the IGF1 locus in dogs in relation to the genotypic scale of measurement and the population mean.

As another example, focus on cases (a) and (c) of Table 10.5 and consider why the heterozygotes both have breeding values of zero. When heterozygotes breed, they contribute equal proportions of A1 and A2 alleles to their progeny. With a dominance value of zero and equal allele frequencies, it makes intuitive sense in case (a) that the heterozygotes have a breed ing value of zero. Each of the A1 and A2 alleles that a heterozygote contributes to its progeny is paired with A1 and A2 alleles in equal frequency from the population. That would form an equal number of A1A1 and A2A2 genotypes with a mean value of zero whereas all heterozygous progeny would also have a mean value of zero since the heterozygote genotypic value is zero. In example (c), the progeny genotypes and frequencies are identical to example (a) since the allele frequencies are the same. However, there is now dominance such that the heterozygote has a genotypic value of 5.25 kg. Still the breeding value of a heterozygote is zero. An equal frequency of A1A1 and A2A2 genotypes in the progeny gives a mean value of zero. The 50% of progeny that are heterozygotes have a value of 5.25 because of dominance. Nonetheless, 0.5(5.25) = 2.625 exactly equals the population mean of 2.625 so the heterozygous progeny also do not alter the population mean.

Genotypic value +a = 10.5

Midpoint = 0


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