Lecture notes Null hypothesis: Genetic basis for variation at multiple loci I. Linkage disequilibrium A. Overview of assumptions and predictions 1. This theory is an extension of the Hardy-Weinberg Theorem, applied to 2 loci
2. Involves a priori assumptions about the associations between loci (we therefore talk about haplotype or gamete frequencies)
3. If these associations are completely random (i.e., in "linkage equilibrium", "coupling" and "repulsion" gametes are equally common, and the coefficient (measure) of linkage Disequilibrium, D, is 0; where D = pabpAB - pAbpaB
4. Even if loci are not linked, they may be nonrandomly associated under certain conditions
5. Furthermore, if the loci are linked, the rate at which linkage equilibrium is approached is related to the recombination distance: Dn = D0(1-r)n B. Case 1: Loci assort independently (i.e., as Mendelian factors)- 1. Assume 2 alleles at 2 loci-4 possible gametes (haplotypes): AB, Ab, aB, ab
2. Also, assume a "Gamete pool": f(AB)=pAB ; f(Ab)=pAb ; etc.
3. Allele frequencies in the gamete pool: pa = pab + paB ;
pA = pAb + pAB = 1 - pa ;
pb = pab + pAb ;
pB = paB + pAB = 1 - pb ; 4. As in the Hardy-Weinberg Theorem, although we can get allele frequencies given gamete frequencies, we can not do the reverse, unless we assume independence (no linkage) between the two loci,and random mating (independent assortment); then: pab = papb ; pAb = pApb ; pAB = pApB ; paB = papB (i.e., frequency of the gamete with the combination Ab is predicted merely by the frequency of A and the frequency of b, since they associate indendently, or randomly) 5. If these equalities hold, then there is "Linkage Equilibrium" 6. If the genes do not segregate independently, then: pab = papb + D ; D = pab - papb where D is the departure from linkage equilibrium, or amount of Disequilibrium; i.e., D is the difference between the actual frequency of the ab gamete, pab, and its expected frequency, papb based on independent assortment 7. Given this, then we can derive the other gamete frequencies: pa = pab + paB ; and thus paB = pa - pab Because pab = papb + D , substitution gives: paB = pa - (papb + D ) = pa(1-pb) - D Since pB = 1 - pb , paB = papB - D Similarly, pab = papb + D ; paB = papb - D ; pAB = pApB + D ; pAb = pApb - D Then, pABpab = (pApB + D )(papb + D ) , and through some simple algebraic manipulations of this equation, it can be shown that: pABpab = pAbpaB + D , and thus, D = pABpab - pAbpaB That is, D is the difference between the frequency of "coupling" (AB and ab) and "repulsion" (Ab and aB) gametes. If these frequencies are equal, D = 0 and there is linkage equilibrium. 8. If a population has just been formed by mixing 2 homozygous populations (AABB x aabb), then the next (F1) generation is formed by AB and ab gametes only, and D = pABpab - pAbpaB = (1/2 x 1/2) - (0 x 0) = 1/4 = 0.25 That is, there is maximum "gametic excess" and linkage disequilibrium, even if the genes are not actually linked! 9. If the loci are linked, it will take longer to establish linkage equilibrium, depending on the frequency of recombination between the loci, as described next. (Return to top of page.) C. Case 2: Two loci are linked 1. To find out how much longer it will take to establish linkage equilibrium, we must incorporate recombination frequency (distance between linked loci) in the calculation 2. First, we have to estimate the frequency of a particular gamete in the next generation: Pab is the frequency of the ab gamete that will be produced from the following (parental) zygotes (each of which exists at a particular frequency, f): Pab = f(ab/ab) [each ab/ab zygote produces only ab gametes]
+ 0.5f(ab/aB) [each ab/aB zygote produces half ab gametes, and recombination
+ 0.5f(ab/Ab) yields the same pairs of gametes]
+ 0.5(1-r)f(ab/AB) [each ab/AB zygote produces the ab gamete at a frequency
that is only half the time that recombination does not occur]
+ 0.5(r f(Ab/aB)) [the ab gamete only arises from half the recombinations in this zygote] These parental zygotes were produced by the following gametic combinations, respectively: ab x ab ; ab x aB and aB x ab) ; ab x Ab and Ab x ab ; ab x AB and AB x ab ; Ab x aB and aB x Ab. The frequency of gamete ab is thus related to the respective frequencies of these zygotes as follows: Pab = pab2 + 0.5(2pabpaB) + 0.5(2pabpAb) + 0.5(1-r)(2pabpAB) + 0.5(r (2pAbpaB)) Using a substitution with D = pABpab - pAbpaB , Pab = pab - rD and similarly for the other gametes:
PaB = paB + rD
PAb = pAb + rD
PAB = pAB - rD 3. That is, the gamete frequencies change each successive generation, depending on r. 4. Because D is dependent on the gamete frequencies (and thus on r), it too will change each successive generation. 5. If we carry the algebra through just a bit more, we find that D depends on r as follows: Dn = D0 (1-r)n , where n is the number of generations following the original population (0). D. Conclusions about linkage disequilibrium 1. If the two loci are not linked (r=0.5), then D decreases by half each generation. 2. If the two loci are linked (r<0.5), then D decreases more slowly, depending on the value of r. (Return to top of page.) II. Continuous variation and quantitative inheritance A. Assumptions and predictions 1. Even if genes are "particulate", if each has a slight effect, and if these effects are additive in some way, then they can determine "continuous" variation 2. If a trait is additive, the magnitude of the phenotypic variance of the population is directly proportional to the frequency of heterozygotes in the population 3. If the phenotype is a simple sum of the contributions of all the loci, then the individual variances are also additive 4. Then also, phenotypic variance is the sum of the variances of both genetic and evironmental effects B. Comparison of qualitative vs. quantitative inheritance 1. Follow 3 generations (P0, F1 and F2) of two highly inbred (isogenic) parental (P0) strains with different phenotypes for a particular trait that is determined by either discrete or continuous inheritance: |
