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Health

Why overdominance ain't all that

Gene ExpressionBy Razib KhanApril 19, 2007 2:25 AM

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A few days ago I posted about how overdominance, the fitness advantage of a heterozygote (an Aa genotype instead of an AA or aa genotype), can maintain polymorphism (genetic variation) within a population at a locus. Roughly, the equilibrium ratio between the two alleles is determined by their respective fitnesses in the homozygote state. For example, if AA & aa are of equal fitness and of lower fitness than the heterozygote then the final equilibrium frequencies will be 0.5. Heterozygote advantage is intuitively comprehensible to many people, after all we've all heard of "hybrid vigor." During the genesis of theoretical evolutionary genetics some thinkers promoted the idea that balancing selection of various kinds, including heterozygote advantage, would result in a non-trivial proportion of extant genetic variation within a population. Coalescing around Sewall Wright and Theodosius Dobzhansky this was the "Balance School," which stood in contrast with the "Classical School" which took its lead from R.A. Fisher, who emphasized directional selection constraining polymorphism and either fixing or eliminating mutant alleles within a population. In 1966 Richard Lewontin and J.L. Hubby published A molecular approach to the study of genic heterozygosity in natural populations. II. Amount of variation and degree of heterozygosity in natural populations of Drosophila pseudoobscura (full paper at link),

which showed that the extant of genetic variation as inferred from allozyme diversity was far greater than either the Balance or Classical School

has predicted. At first Lewontin & Hubby offered the idea that selection for heterozyogosity might be maintaining the variation that they observed (Lewontin was a student of Dobzhansky), but soon enough they rejected this on theoretical grounds. The logic is simple, so I will first go through the formal motions. Consider a population whose fitness can modeled as such: mean fitness = 1 - (selection against p1)*(p1^2) - (selection against p2)*(p2^2) Roughly, this model shows the decrement in fitness from the heterozygote maximum which is generated by resegregation of homozogyotes each generation within a population at equilibrium.^1 The fitness of homozygotes is weighted by their frequency within the population, and this is removed from the ideal maximum, normalized to 1. But this is only at one locus. The reality is that organisms exhibit a complex genetic architecture and innumerable loci. So let's assume that fitness is multiplicative, that is, the sum total fitness of an organism is the product of fitness across individual loci. Ergo: mean fitness across all loci = product of mean fitnesses across all loci Let's assume that the fitness of the two homozygotes across a number of loci is the same so that the proportion of alleles 1 & 2 at each diallelic locus is 0.5. Our relation simplifies to^2: mean fitness across all loci = (1 - 0.5*(homozygote fitness))^(number of loci) Now, by definition we know that homozygote fitness is going to be less than 1 because we're modeling overdominance. Let's assume that the homozygotes are all of fitness 0.85 (selection against = 0.15), and assume that there are 100 such loci. We obtain a mean fitness of: 0.0004 = (1 - 0.5*0.150)^100 So the you see the problem, if we assume widespread heterozygote advantage very few individuals within a population would resegregate across independent loci so as to attain the maximum fitness. In fact, if you assume that some of the homozygotes are lethal you are trapped by the same problem that you encounter in inbreeding: it becomes difficult for a zygote to run the gauntlet of the myriad possibilities of a fatal genetic combination and be viable. Then why variation? Neutral Theory and its heirs! 1 - First, homozygote matings produce only homozygote matings. Second, matings of heterozygotes with homozygotes and other heterozygotes also produce homozygotes. 2 - The formalism would be mean(w) = 1 - s1p1^2 - s2p2^2, where s1 = s2 and p1^2 = p2^2. This means you really have 1 - sp^2 - sp^2, where p = 0.5, so algebraically it is equivalent to 1 - 0.5(s).

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