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1. Joint Equilibrium
After one generation of random mating, the alleles at any single locus will be in Hardy-Weinberg Equilibrium. However, when two or more loci are considered jointly, they may not be in joint equilibrium.
Other terms have been used in place of joint equilibrium, such as gametic phase equilibrium and linkage equilibrium. Linkage equilibrium does not have anything to do with genes that are linked, and is therefore, probably a bad choice of term. Try to use joint equilibrium.
Consider two loci each with two alleles. Let pA=0.2 be the frequency of the A1 allele at the first locus, and let pB=0.3 be the frequency of the B1 allele at the second locus. Then the possible genotypes and their expected frequencies, genotypic values would be as shown in the table below.
The genotypic values assume that there is no epistasis, i.e. no gene interactions between these two loci. However, epistasis could exist and each genotype could have a potentially different epistatic effect. Epistasis will be ignored for a little while in these notes.
The genetic mean, ignoring epistasis, is
The gametic frequencies from a population in joint equilibrium are (and hence the term gametic phase equilibrium)
In the single locus situation with a small population, drift carried the population towards the A1A1 genotype in some strains with frequency pA and towards the A2A2 genotype in other strains with frequency qA. With two loci, strains will move towards one of the four gamete types with the frequencies shown in the table above.
Joint disequilibrium can be generated easily by selection on the population. Suppose we select parents that have either genotype A1A1B1B1 with frequency p=0.3 or genotype A2A2B2B2 with frequency q=0.7. These parents are allowed to randomly mate, then the expected progeny genotypes and their frequencies are as follows:
|Parent 1||Parent 2||Offspring||Frequency|
With two loci, each with two alleles, there should be nine possible genotypes in the offspring generation if they were in joint equilibrium. However, there were only three genotypes in this offspring generation. Note that at the A-locus there is Hardy-Weinberg equilibrium, and at the B-locus there is Hardy-Weinberg equilibrium. However, when both loci are considered jointly, the progeny generation is not in joint equilibrium. The parent population was created by selection, by removing all but two genotypes. Thus, selection has caused the joint disequilibrium. Mixing populations of animals with different gene frequencies will also cause joint disequilbrium, and in small populations, chance can affect gametic frequencies and thereby cause joint disequilibrium. Joint disequilibrium can change genotypic variance.
2.1 Measuring Disequilibrium
To measure the amount of disequilibrium we need to look at the difference between expected gametic frequencies if the parent population was in joint equilibrium, versus the actual gametic frequencies, as in the table below.
|A1 B1||pApB||r=p||D= r-pApB|
|A1 B2||pAqB||s=0||-D = s - pAqB|
|A2 B1||qApB||t=0||-D = t - qApB|
|A2 B2||qAqB||u=q||D = u - qAqB|
pA=pB=p= 0.3, then
Now consider the progeny generation that resulted from the random mating of the selected parents. We need to determine the gametic frequencies of these offspring, as in the table below.
Note that D is equal to .105, which is one half the D of the parent generation.
2.2 Linked Loci
If we allow for physical linkage between two loci, and let crepresent the recombination rate, then we need to determine how
A1B1 gametes can be produced and their frequency.
From a nonrecombinant genotype
A1B1/- -, the
A1B1gamete is produced at a frequency of r(1-c). From a
A1 - /- B1, the
gamete is produced with frequency
pApBc. The total
A1B1 gametes is therefore,
3. Population Parameters
Below are the frequencies for genotypes at two loci and their genotypic values.
The genetic means for each loci are as for the single loci situation,
Likewise, the genotypic variances for each loci are
Several simplifications occur if dA and dB can be assumed
to be zero. Then
Below is a table of genotypic values of nine genotypes that are possible from two loci, each with two alleles.
The values of k1, k2, and k3 are all zero in the case when there are no interactions between the two loci. An additive by additive interaction could exist whenever A1 and B1 occur together. Thus, in the genotype A1A1B1B1 there could be four such interactions. If , then the genotypic value of that genotype would be different from the sum of the genotypic values at each loci.
Another possible interaction is an additive by dominance interaction. For example, whenever A1A2 genotype at the A-locus occurs with the B1 allele, then the genotypic value of the A1A2B1 - genotype would be altered from the usual sum of genotypic values at each locus. In this case, k2 would be different from zero.
The third and last possible interaction between two loci is the dominance by dominance interaction which occurs only when A1A2 is with B1B2, or the double heterozygote. Then k3 would be different from zero.
To determine means and variances under the existence of epistasis would require assuming one or more of these types of gene interactions to exist. This gets extremely messy and complex when extended to more than two loci, because more types of gene interactions are possible. Perhaps at a third loci, the presence of C1 might nullify or turn-off the effects of any genotypes at the A and B loci. Also, if loci have more than two alleles, then derivation of simple formulas becomes impossible for epistatic situations.
5. Many Loci
Assume a large, random mating population. Assume that
the number of loci affecting a particular trait approaches .
Assume that there is no epistasis (gene interactions) among loci.
The total genotypic value of an individual is then
The Central Limit Theorem can be applied to show that G asymptotically approaches a Normal distribution, and so do and , as long as the loci are statistically independent (disequilibrium value is 0 for all pairs of loci), and there is no epistasis. Animal breeders assume normality in nearly all cases, and the assumptions that are implied above. If n is small, then departures from normality can be large.
If the assumptions of the previous section are true, then
phenotypes can be described by a linear model as follows:
In matrix notation, phenotypes of all animals can be collectively
The assumptions that go along with this model are
This LaTeX document is available as postscript or asAdobe PDF.Larry Schaeffer