#### Transcript PPT

Midterm Exam: Weds. 15 March what’s covered on the test? • Lecture material through 14 March • Text reading assignments 1 2 Western Pyrénées National Park, France 3 Effective population size N = population size = total number of individuals Ne = effective population size = ideal population size that would have a rate of decrease in H equal to that of the actual population (N) number of individuals contributing gametes to the next generation 4 Effective population size cntd. If Ne/N 1, then rate of loss of H is minimum. The larger the Ne/N, the lower the rate of loss of H. Predictable loss of heterozygosity (H) in each generation for non-ideal populations 1 Rate of loss of H defined: 2Ne per generation 5 32b Examples of effective population size Taxon Drosophila Humans a snail species plants golden lion tamarins Ne .48 to .71 N .69 to .95 N .75 N lower .32 N (94 of 290) 6 32A-2 Assumptions of an ideal population • • • • • Infinitely large population random mating no mutation no selection no migration 7 31a -1 5 causes of microevolution (I.e., changes in allele frequencies) 1) genetic drift - stochastic variation in inheritance Expected F2: 9 - 3 - 3 - 1 Random deviation Observed F2: 9 - 3 - 2.8 - 1.2 2) Assortative (nonrandom) mating 3) Mutation 4) Natural selection 5) Migration (gene flow) 8 31a-2 Fixation of alleles Fn Parental generation for many populations A = .5 a = .5 p = q = .5 fixed A = 1.0 a=0 Genetic drift lost p = 1.0 A=0 a = 1.0 lost fixed q = 1.0 time 9 34A-1 The Effect on Nonrandom Mating on Ne: What is a formula for calculating the effect of unequal numbers of males and females (non-random breeding)on Ne? Ne = 4 MF M+F M = # of breeding males F = # of breeding females Population A M = 50 F = 50 Population B M = 10 F = 90 N = 100 Ne = 4 x 50 x 50 50 + 50 = 100 = 4 x 10 x 90 10 + 90 = 36 10 10f The effect of non-random mating on H Given 2 cases, with N = 150 and * Ht=1 = 1 - 1 2 Ne Ne = 100 (population A) Ne = 36 (population B) Ht = the proportion of heterozygosity remaining in the next (t=1) generation Population A: Ht = 1 - 1 2 x 100 Population B: Ht = 1 - 1 2 x 36 = 1 - .005 = .995 = 1 - .014 = .986 %H remaining after t=1 generations 11 36A-1 Generalized equation: * Ht = H0 1 - 1 2Ne t t = # of generations later H0 = original heterozygosity What is H after 5 more generations? Population A: H5 = H0 (.995) 5 = .995 (.995)5 = .970 Population B: H5 = H0 (.986) 5 = .995 (.986)5 = .919 12 36A-2 Formulae for calculating H and Ne 1 = proportion of H0 lost at each generation 2Ne 1 - 1 = proportion of H0 remaining after the first generation 2 Ne assuming H = 100% at the start. Ht = H0 1 - 1 2Ne Ne = 4 MF M+F t = the absolute amount of H0 remaining after t generations 1) unequal sex ratios or 2) nonrandom breeding decrease Ne 13 37A Mutation Nondisjunctive point mutations over short term: not important in changing allele frequencies f (A1) = 0.5 mutation rate A1 --> A2 = 1 mutation 105 generations over 2000 generations, f (A1) = 0.49 If f (A2) increases rapidly, selection must be involved Long-term, over evolutionary time mutation is critical - providing raw material for natural selection Mutation rate is independent of H, P, Ne but mutation can increase H and increase P 14 36A1