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10.2 Mean genotypic value in a population
• Deriving the population mean phenotypic value.
• The population mean phenotypic value under random mating.
• The population mean phenotypic value with nonrandom mating.
The mean phenotypic value of a population (symbolized M) is dictated by the genotypic values and the genotype frequencies. Table 10.1 shows the frequencies, genotypic values, and frequencyweighted genotypic values for the three genotypes under the assumption of random mating. The average phenotype of the entire population is just the sum of the phenotypic values of all individuals divided by the number of individuals. If the average environmental deviation for all genotypes is zero, then the mean phenotype of the population is the sum the three frequencyweighted genotypic values:
This can be simplified by factoring an a out of the first and third terms:
Then notice that (p2  q2) = (p + q)(p  q) and also that p + q = 1 for a diallelic locus. Making this substitution leads to
This population mean genotypic value is identical to the population mean phenotypic value if the mean environmental deviation is zero. It is important to note that M is measured as a deviation from the midpoint on the scale of genotypic values. Therefore,
M must be added to the midpoint value to obtain the absolute population genotypic mean value.
This expression for the mean genotypic value in a population leads to two important biological conclusions. First, it shows that the mean of a quantitative trait depends on allele frequencies in the population because these dictate genotype frequencies. Second, the division of the expression for the mean phenotype into two terms is informative. The a(p  q) term shows that changes in allele frequencies shift the mean up or down by some fraction of a depending on which of the two homozygotes is more frequent. At the extremes of allele frequency, when p = 1 then the mean phenotype is equal to +a and if q = 1 then the mean phenotype is a. The 2pqd term shows that the frequency of the heterozygote alone determines the impact of dominance on the mean phenotype. When the phenotypic value of the heterozygote is exactly at the midpoint between the homozygotes (d = 0), the heterozygote genotype has no impact on the population mean. Completely additive gene action exists when d = 0.
The role of allele frequency on the population mean can be seen in an example. Imagine a population of dogs where mating is random and the IGF1+ and IGF1~ alleles are both segregating. For IGF1 in dogs, a = 10.5 kg, d = 5.25 kg, and the midpoint is 19.5 kg (see Problem box 10.1). If the two IGF1 alleles are at equal frequencies then p = q = 0.5. The mean phenotypic value would be
M = 10.5(0.5  0.5) + 2(0.5)(0.5)(5.25) = 2.625 kg (10.5)
as a deviation from the midpoint. In absolute terms, the mean phenotype in the population would be 2.625 + 19.5 = 22.125 kg. Since both homozygotes are equally frequent their effect on the average cancels out. That leaves 50% of the population as heterozygotes, shifting the mean above the midpoint by V2d. If the two IGF1 alleles are at unequal frequencies, say p = 0.9 and q = 0.1, the mean phenotypic value would be
as a deviation from the midpoint and 28.845 kg in absolute terms. At these allele frequencies the AXAX homozygote composes 81% of the population while the A2A2 homozygote is only 1% of the population so the balance is strongly in favor of +a. The remaining 18% of the population is made up of heterozygotes, which also shift the average toward +a since d is positive.
The role of gene action on the population mean phenotype can also be seen in this example. Suppose that the IGF1 alleles are p = 0.9 and q = 0.1 but that d = 0. The mean phenotypic value is then will impact the mean phenotype if there is dominance because the frequency of heterozygotes will change by the amount f2 pq.
Extending the IGF1 example further, imagine there is consanguineous mating so that f = 0.2 when allele frequencies are p = 0.9 and q = 0.1, and d = 5.25. This would lead to f2pq = 0.036 or a 3.6% deficit of heterozygotes and a 1.8% excess of both homozygotes compared to random mating. The mean phenotype is then
M = (0.81 + 0.018)(10.5) + (0.18  0.036)(5.25) + (0.01 + 0.018)(10.5) (10.9)
which equals 9.156 kg as a deviation from the midpoint, or 28.656 kg. This final example shows that changes in genotype frequencies caused by nonrandom mating, even when allele frequencies remain constant, can impact the population mean phenotype.
as a deviation from the midpoint and 2 7.9 kg in absolute terms. This population mean phenotype is lower than the result in equation 10.6 because the genotypic value of the heterozygotes is zero when d = 0. Therefore, the 18% of the population made up of heterozygotes does not contribute to any deviation from the midpoint in addition to that caused by the frequencies of the two homozygotes.
So far, random mating has been assumed when predicting the mean phenotype. However, consanguineous mating is also common in populations and will change the mean phenotype in predictable ways. Using the results of equation 2.20, the expected genotype frequencies can include the impact of nonrandom mating. The mean phenotype with the possibility of nonrandom mating is given by
10.3 Average effect of an allele
• Deriving the average phenotypic effect of an allele in a population.
• The average effect as a deviation from the population mean.
• The average effect as substitution of one allele in a genotype for another.
To describe the inheritance of quantitative phenotypes across generations, it is necessary to think in terms of alleles. Although genotypes determine genotypic values, alleles and not genotypes are inherited by individuals. Thus, a new measure of value is needed that can be used to link the genotypic values of one generation with the genotypic values of their progeny in the next generation. The concept of average effect is used to assign a value to an allele and predict how it impacts the mean genotypic value of the population. As we will see, the average effect depends on the genotypic values a and d as well as the allele and genotype frequencies in the population.
where f is the inbreeding coefficient. Changes in autozygosity (f ^ 0) alter all genotype frequencies but will only impact the population mean phenotype when there is some degree of dominance (d ^ 0). The impact of the inbreeding coefficient is identical on each of the homozygotes, increasing or decreasing them both by the same amount fpq with no impact on the mean. In contrast, changes in the autozygosity
Average effect of an allele The mean phenotypic deviation from the population mean of that group of individuals which received a particular allele from one parent and the other allele from a parent drawn at random from the population.
One way to visualize the basis of the average effect is to think of a slot machine used in gambling (also called a poker or fruit machine). Mechanical slot machines normally have three wheels that are each labeled with symbols having different frequencies on the wheels. When the wheels are spun, they will each stop and display a random symbol. An imaginary averageeffect slot machine has just two wheels that represent the two alleles in a diploid genotype. One of the wheels is broken and does not spin, so that a single symbol is always displayed. For the average effect of the Ax allele for example, Ax would always appear on the broken wheel. The other wheel has symbols for all of the alleles in the population in proportion to their frequency in the population. Spinning this slot machine many times with one allele on the broken wheel and recording the value of the genotype for each spin would give expected frequencies of all of the genotypes in the population that contain an Ax allele under random mating. The average of all of the genotypic values from many spins would give the average value of all the genotypes that contain a given allele. This average value of genotypes containing one allele may very well be different than the average value of the population. This difference in averages is the average effect of the allele that is present on the broken wheel.
The derivation of the mean value of all genotypes that contain either an Ax (MA ) or an A2 (MA ) allele is shown in Table 10.2. This logic could be used for a locus with any number of alleles. If mating is random, we expect an A1 allele to be paired in a genotype with another A1 allele p percent of the time and with an A2 allele q percent of the time. The average value of all genotypes that contain an A1 allele is the frequencyweighted sum of the values of each genotype that has at least one A1 allele. In symbols, this mean value is for the A1 allele. This quantity can be thought of as the mean value of a large number of individuals that all inherit the same allele from the same parent but that inherit the other allele in their genotype from a parent drawn at random from the population. Notice that HardyWeinberg expected genotype frequencies are a key assumption in Table 10.2, since expected genotype frequencies would be different if any of the HardyWeinberg assumptions did not hold.
The average effect is a deviation that measures the difference between the value of all genotypes that contain a given allele and the population mean. Therefore, the population mean must be subtracted from the mean values of genotypes produced by a given allele that are shown in Table 10.2. The average effect is then a = Ma  M
using the a to represent the average effect and x to indicate any allele. The average effect is determined by a regression line (see Fig. 10.4, below) that minimizes the deviations (rather than the squared deviations) of individual values from the line. In many cases, when genotypes are in HardyWeinberg frequencies, the simple relationship of equation 10.11 suffices to describe the average effect. In more complicated situations such as when HardyWeinberg is not met (f ^ 0), the line that minimizes the deviations is determined by a leastsquares fit.
Substituting the expression for the population mean from equation 10.4 and the mean value of all genotypes containing the A1 allele from Table 10.2, the average effect of the A1 allele is a1 = pa + qd  (a(p  q) + 2pqd) (10.12) which simplifies to
Allele 

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