1. 1 Midterm Exam: Weds. 15 March what’s covered on the test? Lecture material through 14 March Text reading assignments

2. 2 From Encarta

3. 3 Western Pyrénées National Park, France From Encarta

4. 4 Effective population size N = population size = total number of individuals N e = 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 32A-1

5. 5 Effective population size cntd. If N e /N 1, then rate of loss of H is minimum. The larger the N e/ 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: 2N e per generation 32b

6. 6 Examples of effective population size TaxonN e Drosophila.48 to.71 N Humans.69 to.95 N a snail species.75 N plantslower golden lion tamarins.32 N (94 of 290) 32A-2

7. 7 Assumptions of an ideal population Infinitely large population random mating no mutation no selection no migration 31a -1

8. 8 5 causes of microevolution (I.e., changes in allele frequencies) 1) genetic drift - stochastic variation in inheritance Expected F 2 : 9 - 3 - 3 - 1 Observed F 2 : 9 - 3 - 2.8 - 1.2 2) Assortative (nonrandom) mating 3) Mutation 4) Natural selection 5) Migration (gene flow) 31a-2 Random deviation

9. 9 Fixation of alleles Parental generation for many populations A =.5 a =.5 p = q =.5 fragmentation FnFn A = 1.0 a = 0 fixed lost fixed A = 0 a = 1.0 lost q = 1.0 p = 1.0 Genetic drift 34A-1 time

10. 10 What is a formula for calculating the effect of unequal numbers of males and females (non-random breeding)on N e ? N e = 4 MFM = # of breeding males M + F F = # of breeding females Population APopulation B M = 50M = 10 F = 50F = 90 N = 100 N e = 4 x 50 x 50 = 4 x 10 x 90 50 + 50 10 + 90 = 100 = 36 10f The Effect on Nonrandom Mating on Ne:

11. 11 The effect of non-random mating on H Given 2 cases, with N = 150 andN e = 100 (population A) N e = 36 (population B) H t=1 = 1 - 1 H t = the proportion of heterozygosity 2 N e remaining in the next (t=1) generation Population A: H t = 1 - 1 = 1 -.005 =.995 2 x 100 Population B: H t = 1 - 1 = 1 -.014 =.986 2 x 36 % H remaining after t=1 generations * 36A-1

12. 12 36A-2 Generalized equation: H t = H 0 1 - 1 t t = # of generations later 2N e H 0 = original heterozygosity What is H after 5 more generations? Population A: H 5 = H 0 (.995) 5 =.995 (.995) 5 =.970 Population B: H 5 = H 0 (.986) 5 =.995 (.986) 5 =.919 *

13. 13 Formulae for calculating H and N e 37A 1 = proportion of H 0 lost at each generation 2N e 1 - 1 = proportion of H 0 remaining after the first generation 2 N e assuming H = 100% at the start. H t = H 0 1 - 1 t = the absolute amount of H 0 remaining after 2N e t generations N e = 4 MF 1) unequal sex ratios or M + F2) nonrandom breeding decrease N e

14. 14 Mutation Nondisjunctive point mutations over short term: not important in changing allele frequencies f (A 1 ) = 0.5 mutation rate A 1 --> A 2 = 1 mutation 10 5 generations over 2000 generations, f (A 1 ) = 0.49 If f (A 2 ) 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, N e but mutation can increase H and increase P 36A1