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Biol3306_Exam2Formulae - ∆ p = p’-p • p is the...

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Finite population size N e = (4N m N f )/(N m + N f ), where N m and N f are the number of males and females, respectively N e = t , where N 1 , N 2 , ... N t are the population sizes over t generations 1/N 1 + 1/N 2 + ... + 1/N t H’ = H(1-1/(2N e )) The mean time to fixation for new beneficial alleles is: t = 4 ln(2N e )/s The mean time to fixation for new neutral alleles is: t = 4N e An allele is “effectively neutral” if s 1/(2N e ) f’(A 1 A 1 ) = p 2 (1-F) + pF; f’(A 1 A 2 ) = 2pq(1-F); f’(A 2 A 2 ) = q 2 (1-F) + qF; H’ = H(1-F) Definitions Consider a locus with two alleles, A 1 and A 2 , with frequencies p and q, respectively w 11 , w 12 , and w 22 are the fitnesses of A 1 A 1 , A 1 A 2 , and A 2 A 2 , respectively N e is the effective population size μ is the rate of mutation from A 1 to A 2 p’ is the value of p in the next generation;
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Unformatted text preview: ∆ p = p’-p • p is the equilibrium value of p • H is the population heterozygosity • ± is the coefFcient of inbreeding ˆ Infnite population size • w = p 2 w 11 + 2pqw 12 + q 2 w 22 • p’ = (p 2 w 11 + pqw 12 )/w _ Biol 3306 Evolutionary Biology Formulae ±or Exam 2 Two loci • D = g AB g ab- g Ab g aB , where D is the coefFcient of linkage disequilibrium and g X is the frequency of haploytpe X _ _ _ Heritability and Response to Selection • R = h 2 S where R is the response to selection, h 2 is the heritability and S is the selection differential • h 2 = R / S where h 2 is the heritability, R is the response to selection and S is the selection differential...
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