infinitésimo Posted May 16, 2007 Posted May 16, 2007 Hi everybody! Could anybody explain me the relation between the selection coefficient, the Malthusian fitness and the Darwinian fitness? I'm reviewing some papers and the authors uses one or another without explaining anything. How can I convert this parameters? Thanks!
lucaspa Posted May 25, 2007 Posted May 25, 2007 http://www.zoology.ubc.ca/~otto/Reprints/Day2001.pdf "For example, Darwinian fitness (W) is measured over one organismal generation, while Malthusian fitness (m) is a measure of fitness over a very small period of time (‘instantaneous’)." "For example, consider a single locus at which there are two alleles, A and a, under natural selection. A diploid population consists of three types of individuals: AA, Aa and aa. Let us denote the absolute fitness of each of these types by W(AA), W(Aa), and W(aa), respectively. Relative fitnesses can be obtained in any number of ways. For example, one can divide the absolute fitness of each type by the average absolute fitness of all types. Alternatively, one can choose a reference genotype and divide the absolute fitness of each genotype by the absolute fitness of the reference type (Table 1). Relative fitnesses are often written as 1 for AA, 11hs for Aa, and 11s for aa individuals, where s and h are known as selection and dominance coefficients, respectively. The selection coeffi- cient measures the strength of selection favouring the a allele." 1
infinitésimo Posted June 5, 2007 Author Posted June 5, 2007 Thanks lucaspa, that link is interesting, but it is mainly theoretical, luckily, I`ve found the answer (a more practical answer) I need in this other: http://myxo.css.msu.edu/ecoli/srvsrf.html Essentially, the Malthusian fitness has the form Ln[N(t)/N(0)]/t; and is likely the 'absolute' fitness of a genotype. The Darwinian (W) and the selection coefficient (s) could be derived from it, easily: W = mA/mB s = mA-mB where A and B refers to two competing genotypes.
lucaspa Posted June 6, 2007 Posted June 6, 2007 Thanks lucaspa, that link is interesting, but it is mainly theoretical, luckily, I`ve found the answer (a more practical answer) I need in this other: http://myxo.css.msu.edu/ecoli/srvsrf.html Essentially, the Malthusian fitness has the form Ln[N(t)/N(0)]/t; and is likely the 'absolute' fitness of a genotype. The Darwinian (W) and the selection coefficient (s) could be derived from it, easily: W = mA/mB s = mA-mB where A and B refers to two competing genotypes. I'm glad you found something better. Thanks for the new link. Although the equations for W and s don't look right. What's "t" in the Malthusian equation? Let me caution you about the use of the word "absolute". ALL fitness is relative to a specific environment. So to evolutionary biologists there is no "absolute" fitness; it's all relative. However, what this may be is the "theoretical" fitness of the genotype --derived solely from the math of the Malthusian equations. I'll have to look closer. I've never seen an evolutionary biology textbook use the Malthusian fitness. Instead, it's always w and s.
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