1. Chapter 3: Deviations from the Hardy- Weinberg equilibrium Systematic deviations Selection, migration and mutation Random genetic drift Small effective population size

2. Deviations from the Hardy-Weinberg law Systematic deviations: Migration Selection Mutation

3. Deviations from the Hardy-Weinberg law Selection

4. Deviations from the Hardy- Weinberg law Migration

5. Deviations from the Hardy-Weinberg law Mutation Small population size, random changes

6. Mutation: The selection coefficient has the symbol s The mutation frequency has the symbol  Selection mutations equilibrium occurs when: q 2  s =  for the recessive genes pq  s = p  s =  for the dominant genes Deviations from the Hardy-Weinberg law

7. Genetic load Selection can cause the death of some individuals or make them unable to reproduce This cost is called a genetic load Belgian Blue cattle

8. Selection against the recessive Genotype EE Ee ee Total Frequency p 2 2pq q 2 1,00 Fitness 1 1 1-s Proportion p 2 2pq q 2 (1-s) 1-sq 2 after selection Fitness is constant (1-s). That is the opposite of selection, s Genetic load = sq 2

9. Selection against the recessive: Example Example: q =0,25 and s=1: Genotype EE Ee ee Total Frekvens 0.5625 0.375 0.0625 1.00 Fitness 1 1 0 Proportion 0.5625 0.375 0 0.9375 after selection q’ = (2pq/2 + q 2 (1-s))/(1-sq 2 ) = (0.375/2 + 0)/ 0.9375 = 0.20

10. Selection against the recessive: Several generations Formula for the calculating of gene frequency in the following generation q’ = (2pq/2 + q 2 (1-s))/(1-sq 2 ) q

11. Selection against the recessive: Formula for s=1 Expansion to n generations for s=1: q n = q 0 /(1+n  q 0 ) n can be isolated n = 1/q n - 1/q 0 Example: Gene frequency changes from 0.01 to 0.005 n = 1/0.005 - 1/0.01 = 200 - 100 = 100 generations

12. Bedlington-terrier, example

13. Genotype EE Ee ee Total Frekvens p 2 2pq q 2 1,00 Fitness 1-s 1 1 1-s 2 Proportion p 2 (1-s 1 ) 2pq q 2 (1-s 2 ) 1-p 2 s 1 - q 2 s 2 after selection Genetic load = p 2 s 1 + q 2 s 2 Selection for heterozygotes

14. Selection for heterozygotes: Equilibrium frequency After selection the gene frequency is calculated by use of the gene counting method: q' = (q 2 (1-s 2 ) + pq)/(1-p 2 s 1 - q 2 s 2 ) And equilibrium occurs at:  q = pq(ps 1 - qs 2 )/(1-p 2 s 1 - q 2 s 2 ) = 0 for: ps 1 - qs 2 = 0 q= s 1 / (s 1 + s 2 ) which is the equilibrium frequency ^

15. Selection for heterozygotes: Fitness-graph Relative fitness by over dominance

16. In the population 5% is born with sickle cell anaemia. q 2 = 0.05  q = 0.22 s 2 = 1 Equilibrium occurs at: p= s 2 / (s 1 + s 2 ) = 1 - q Which solved gives: s 1 = (s 2 /(1 - q)) - s 2 = 0.285 ^ Sickle cell anaemia in malaria areas Selection for heterozygotes: Example

17. Selection against heterozygotes The gene couting method gives: q' = (q 2 (1-s 2 ) + pq)/(1-p 2 s 1 - q 2 s 2 ) Equilibrium occurs at:  q = pq(ps 1 - qs 2 )/(1-p 2 s 1 - q 2 s 2 ) = 0 for q = s 1 / (s 1 + s 2 ) The equilibrium is unstable ^

18. Small populations

19. Binominal variance on p or q: s 2 = (p  q)/(2  N), 2  N is equal to the number of genes, drawn from the population to form the new generation Small populations: Variance on the gene frequency

20. Small populations, continued The standard deviation of the gene frequency Number of genes

21. Effective population size (N e ) The number of sires and dams for the new generation has significance for N e N e the harmonic mean of the two sexes 4/N e = 1/N males + 1/N females

22. 10 males and 10 females 4/N e = 1/10 + 1/10 which gives N e = 20 1 male and 10 females 4/N e = 1/1 + 1/10 which gives N e = 3.7 100 males and 100000 females 4/N e = 1/100 + 0 which gives N e = 400 Effective population size (N e ): Examples

23. Increase in the degree of inbreeding (  F) The increase in inbreeding per generation is dependent on the effective population size (N e )  F = 1/(2  N e ) In a population with N e = 20, the increase in each generation is delta F = 2.5 %. The inbreeding coefficient F is defined in the next chapter.