## Info

P

q

ma2 = Pd - qa

as shown in Math box 10.1. The average effect of the

based on the difference between the mean values of genotypes produced by A2 and the population

Math box 10.1 The average effect of the A1 allele

Start with the difference between the mean value of all genotypes produced by the A1 allele in Table 10.2 and the population mean in equation 10.4:

Then expand a(p - q) to give a1 = pa + qd - pa + qa - 2pqd (10.16)

Then factor out q:

Inside the parentheses, two terms contain d in common that can be factored to give a1 = q(a + d(1 - 2p)) (10.19)

The final step is to notice that p + q = 1 so that (p + q) can be substituted:

After adding p to -2p, the simplified equation for the average effect of the A1 allele is a1 = q(a + d(q - p)) (10.21)

Working through several examples based on the IGF1 locus in dogs will help illustrate the average effect and what it measures. Table 10.3 gives four examples of the average effect for the A1 allele for the four combinations of two allele frequencies and two levels of dominance. In all of the examples, bear in mind that the A1-containing genotypes are A1A1 and A1A2 and that all values are relative to a midpoint value of 19.5 kg.

In example (a) in Table 10.3, the A1 and A2 allele frequencies are equal. Since there is no dominance, the heterozygotes have a mean value equal to the midpoint. The two homozygotes are equally frequent and their average values are equal but have opposite signs. This results in a population mean value of zero. Given an A1 allele, when sampling a second allele from this population to make genotypes it is equally likely to obtain either allele. Thus, 50% of the A1 containing genotypes are A1A1 with a value of +a = 10.5 and 50% are A1A2 with a value of d = 0. In total, the mean value of the A1-containing genotypes is 0.5(a) or 5.25 kg. This is the same as the average effect since the mean value of all three genotypes is zero. At these allele frequencies, the population mean is exactly at the midpoint and the A1-containing genotypes serve to increase the average value of the population by 5.25 kg.

The situation is different in example (b) in Table 10.3 since the frequency of A1 is high and the frequency of A2 is low. Given an A1 allele, when a second allele is drawn at random from this population to make genotypes, it is much more likely to be another A1 allele rather than an A2 allele. Thus, the mean of the A1-containing genotypes is nearly the same as a (9.45 kg), the genotypic value of A1A1. However, the average effect is small (1.05 kg) since the mean value of all three genotypes is also large (8.4 kg). At these allele frequencies, the population mean is near its upper limit due to the low frequency of the A2 allele and the resulting low frequency of A2-containing genotypes that would reduce the average value of the population.

Comparing cases (a) and (c) in Table 10.3 is informative since allele frequencies are identical but the degree of dominance is different. With dominance in example (c), 50% of the A1-containing genotypes are A1A1 with a value of +a = 10.5, and 50% are A1A2 with a value of d = 5.25. The mean value of the A1-containing genotypes is now larger at 0.5(a) + 0.5(d) or 7.875 kg. However, because of dominance, the mean value of the total population is also greater at 2.625 kg. Thus, the difference between the mean value of the A1-containing genotypes and the mean value of the entire population remains at 5.25 kg, exactly as it was in example A. Comparing examples

A1 M^ = pa + qd = (0.5)(10.5) + (0.5)(0.0) = 5.25 a1 = MA - M = 5.25 - 0.0 = 5.25 a, = q(<a + d(q - p)) = 0.5(10.5 + 0.0(0.5 - 0.5)) = 5.25 a = a + d(q - p) = 10.5 + 0.0(0.5 - 0.5) = 10.5 a1 = qa = (0.5)(10.5) = 5.25

A1 M^ = pa + qd = (0.9)(10.5) + (0.1)(0.0) = 9.45 a1 = MA - M = 9.45 - 8.4 = 1.05 a1 = q(a + d(q - p)) = 0.1(10.5 + 0.0(0.1 - 0.9)) = 1.05 a = a + d(q - p) = 10.5 + 0.0(0.1 - 0.9) = 10.5 a1 = qa = (0.1)(10.5) = 1.05

A1 M^ = pa + qd = (0.5)(10.5) + (0.5)(5.25) = 7.875 a1 = MA - M = 7.875 - 2.625 = 5.25 a1 = q(a + d(q - p)) = 0.5(10.5 + 5.25(0.5 - 0.5)) = 5.25 a = a + d(q - p) = 10.5 + 5.25(0.5 - 0.5) = 10.5 a1 = qa = (0.5)(10.5) = 5.25

A1 M^ = pa + qd = (0.9)(10.5) + (0.1)(5.25) = 9.975 a1 = MA - M = 9.975 - 9.345 = 0.630 a1 = qfa + d(q - p)) = 0.1(10.5 + 5.25(0.1 - 0.9)) = 0.630 a = a + d(q - p) = 10.5 + 5.25(0.1 - 0.9) = 6.3 a1 = qa = (0.1)(6.3) = 0.63

Table 10.3 Examples of the average effect for the IGF1 locus in dogs. All cases assume that a = 10.5 kg as shown in the genotypic scale in Fig. 10.1. For each set of allele frequencies and dominance, the table shows the population mean (M), the mean value of all genotypes that contain an A1 allele (MA ), the average effect of an allelic replacement (a), and the average effect of an A1 allele (a1). Values are all in kilograms and relative to the midpoint value of 19.5 kg.

(b) and (d) in Table 10.3 illustrates that the average effect changes with a shift in the dominance value. Both the population mean and the mean of the A1-containing genotypes change with a change in dominance at those allele frequencies.

A critical lesson to take from these examples is the contextual nature of the average effect. Average effects depend on allele and genotype frequencies and are therefore specific to the population and time point when they are measured. This is sometimes a difficult point to grasp when considering the genetic basis of phenotypic variation. Even though the genotypic values and the degree of dominance may remain constant among populations, the average effect of an allele shifts depending on the allele and genotype frequencies. This is a part of the major distinction between identifying genes and alleles that impact phenotype viewed from the perspective of Mendelian genetics (e.g. an allele causes a certain phenotype) and understanding how such alleles shape pheno-typic distributions within and between populations (e.g. an allele has a large average effect).

Another way to understand the average effect when there are two alleles at a locus is to suppose that it is possible to randomly sample genotypes from a population and then be able to substitute one of the alleles in a genotype for another. When there are two alleles, it would be possible to replace one A2 allele initially present in the genotype with one A1 allele or vice versa. The average effect can be thought of in terms of the change in the mean value of the population when one allele replaces another in all the sampled genotypes. To measure the average effect in this manner requires determining the change in value that comes about by allelic replacement as well as the frequency with which the change in value occurs. Summing these frequency-weighted value changes will give the mean change in value that is the average effect.

Imagine we elect to replace an A2 allele with an A1 allele in genotypes sampled at random from the population that contain at least one A2 allele (see Fig. 10.2). Let p be the frequency of the A1 allele and q be the frequency of the A2 allele. Changing an A1A2 genotype

A2 -Original allele

Replacement allele

0 0