2. AP Biology Evolution of Populations Doonesbury - Sunday February 8, 2004

3. AP Biology Populations evolve  Natural selection acts on individuals  differential survival  “survival of the fittest”  differential reproductive success  who bears more offspring  Populations evolve  genetic makeup of population changes over time  favorable traits (greater fitness) become more common Presence of lactate dehydrogenase Mummichog

4. AP Biology Changes in populations Bent Grass on toxic mine site Pocket Mice in desert lava flows Pesticide molecule Insect cell membrane Target site Resistant target site Insecticide resistance Target site Decreased number of target sites

5. AP Biology Individuals DON’T evolve!!!

6. AP Biology Individuals DON’T evolve… Individuals survive or don’t survive… Populations evolve Individuals are selected

7. AP Biology Fitness  Survival & Reproductive success  individuals with one phenotype leave more surviving offspring Body size & egg laying in water striders

8. AP Biology Natural selection  Natural selection adapts a population to its environment  a changing environment  climate change  food source availability  new predators or diseases  combinations of alleles that provide “fitness” increase in the population

9. AP Biology Variation impacts natural selection  Natural selection requires a source of variation within the population  there have to be differences  some individuals must be more fit than others

10. AP Biology 5 Agents of evolutionary change MutationGene Flow Genetic DriftSelection Non-random mating

11. AP Biology 1. Mutation & Variation  Mutation creates variation  new mutations are constantly appearing  Mutation changes DNA sequence  changes amino acid sequence?  changes protein?  changes structure?  changes function?  changes in protein may change phenotype & therefore change fitness

12. AP Biology 2. Gene Flow  Movement of individuals & alleles in & out of populations  seed & pollen distribution by wind & insect  migration of animals  sub-populations may have different allele frequencies  causes genetic mixing across regions  reduce differences between populations

13. AP Biology Human evolution today  Gene flow in human populations is increasing today  transferring alleles between populations Are we moving towards a blended world?

14. AP Biology 3. Non-random mating  Sexual selection

15. AP Biology Sex & Variation  Sex spreads variation  one ancestor can have many descendants  sex causes recombination  offspring have new combinations of traits = new phenotypes  Sexual reproduction recombines alleles into new arrangements in every offspring 14

16. AP Biology Warbler finch Tree finches Ground finches 4. Genetic drift  Effect of chance events  founder effect  small group splinters off & starts a new colony  bottleneck  some factor (disaster) reduces population to small number & then population recovers & expands again

17. AP Biology Founder effect  When a new population is started by only a few individuals  some rare alleles may be at high frequency; others may be missing  skew the gene pool of new population  human populations that started from small group of colonists  example: colonization of New World

18. AP Biology Distribution of blood types  Distribution of the O type blood allele in native populations of the world reflects original settlement

19. AP Biology Distribution of blood types  Distribution of the B type blood allele in native populations of the world reflects original migration

20. AP Biology Out of Africa Likely migration paths of humans out of Africa Many patterns of human traits reflect this migration 50,000ya 10-20,000ya

21. AP Biology Bottleneck effect  When large population is drastically reduced by a disaster  famine, natural disaster, loss of habitat…  loss of variation by chance event  alleles lost from gene pool  not due to fitness  narrows the gene pool

22. AP Biology Cheetahs  All cheetahs share a small number of alleles  less than 1% diversity  as if all cheetahs are identical twins  2 bottlenecks  10,000 years ago  Ice Age  last 100 years  poaching & loss of habitat

23. AP Biology Conservation issues  Bottlenecking is an important concept in conservation biology of endangered species  loss of alleles from gene pool  reduces variation  reduces adaptability Breeding programs must consciously outcross Peregrine Falcon Golden Lion Tamarin

24. AP Biology 5. Natural selection  Differential survival & reproduction due to changing environmental conditions  climate change  food source availability  predators, parasites, diseases  toxins  combinations of alleles that provide “fitness” increase in the population  adaptive evolutionary change

25. AP Biology 5 Agents of evolutionary change MutationGene Flow Genetic DriftSelection Non-random mating

26. AP Biology Measuring Evolution of Populations 25

27. AP Biology Populations & gene pools  Concepts  a population is a localized group of interbreeding individuals  gene pool is collection of alleles in the population  remember difference between alleles & genes!  allele frequency is how common is that allele in the population  how many A vs. a in whole population 26

28. AP Biology Evolution of populations  Evolution = change in allele frequencies in a population  hypothetical: what would it be like if allele frequencies didn’t change?  non-evolving population 1. very large population size (no genetic drift) 2. no migration (movement in or out) 3. no mutation (no genetic change) 4. random mating (no sexual selection) 5. no natural selection (no selection) 27

29. AP Biology Hardy-Weinberg equilibrium  Hypothetical, non-evolving population  preserves allele frequencies  Serves as a model  natural populations rarely in H-W equilibrium  useful model to measure if forces are acting on a population  measuring evolutionary change W. Weinberg physician G.H. Hardy mathematician 28

30. AP Biology Hardy-Weinberg theorem  Alleles  assume 2 alleles = B, b  frequency of dominant allele (B) = p  frequency of recessive allele (b) = q  frequencies must add to 100%, so: p + q = 1 bbBbBB 29

31. AP Biology Hardy-Weinberg theorem  Individuals  frequency of homozygous dominant: p x p = p 2  frequency of homozygous recessive: q x q = q 2  frequency of heterozygotes: (p x q) + (q x p) = 2pq  frequencies of all individuals must add to 100%, so: p 2 + 2pq + q 2 = 1 bbBbBB 30

32. AP Biology Using Hardy-Weinberg equation q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each genotype? population: 100 cats 84 black, 16 white How many of each genotype? bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Must assume population is in H-W equilibrium! 31

33. AP Biology Using Hardy-Weinberg equation bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Assuming H-W equilibrium Sampled data bbBbBB p 2 =.74 2pq=.10 q 2 =.16 How do you explain the data? p 2 =.20 2pq=.64 q 2 =.16 How do you explain the data? Null hypothesis 32

34. AP Biology How do allele frequencies change? Think of all the factors that would keep a population out of H-W equilibrium! 33

35. AP Biology Real world application of H-W  Frequency of allele in human population  Example:  What % of human population carries allele for PKU (phenylketonuria)  Should you screen prospective parents?  ~ 1 in 10,000 babies born in the US is born with PKU  results in mental retardation, if untreated  disease is caused by a recessive allele  PKU = homozygous recessive (aa) 34

36. AP Biology H-W & PKU disease  frequency of homozygous recessive individuals q 2 (aa) = 1 in 10,000 = 0.0001 0.01  frequency of recessive allele (q): q = √0.0001 = 0.01  frequency of dominant allele (p): p (A) = 1 – 0.01 = 0.99  frequency of carriers, heterozygotes: 2pq = 2 x (0.99 x 0.01) = 0.0198 = ~2%  ~2% of the US population carries the PKU allele 300,000,000 x.02 = 6,000,000 people 35

37. AP Biology Hardy-Weinberg Lab data 36

38. AP Biology Hardy Weinberg Lab: No Selection total alleles = 48.6 p (A): (6+6+18)/48 =.6.4 q (a): 18/48 =.4 total alleles = 48.6 p (A): (6+6+18)/48 =.6.4 q (a): 18/48 =.4 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population AA6 Aa18 aa0 How do you explain these data? Case #1 37

39. AP Biology Hardy Weinberg Lab: Selection total alleles = 48.9 p (A): (19+19+5)/48 =.9.1 q (a): 5/48 =.1 total alleles = 48.9 p (A): (19+19+5)/48 =.9.1 q (a): 5/48 =.1 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population AA19 Aa5 aa0 How do you explain these data? Case #2 38

40. AP Biology Hardy Weinberg Lab: total alleles = 48.7 p (A): (9+9+15)/48 =.7.3 q (a): 15/48 =.3 total alleles = 48.7 p (A): (9+9+15)/48 =.7.3 q (a): 15/48 =.3 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 24 individuals 48 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population AA9 Aa15 aa0 How do you explain these data? Case #3 Heterozygote Advantage 39

41. AP Biology Hardy Weinberg Lab: Genetic Drift 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population How do you explain these data? AA310 Aa563 aa015 pq Case #4 40

42. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 16.7 p (A): (3+3+5)/16 =.7.3 q (a): 5/16 =.3 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population How do you explain these data? AA310 Aa554 aa015 pq Case #4.7.3 41

43. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 16.7 p (A): (3+3+5)/16 =.7.3 q (a): 5/16 =.3 total alleles = 16.7 p (A): (3+3+5)/16 =.7.3 q (a): 5/16 =.3 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population How do you explain these data? total alleles = 14.5 p (A): (1+1+5)/14 =.5.5 q (a): (5+1+1)/14 =.5 total alleles = 14.5 p (A): (1+1+5)/14 =.5.5 q (a): (5+1+1)/14 =.5 AA310 Aa554 aa015 p.7.5 q.3.5 Case #4 42

44. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 16.7 p (A): (3+3+5)/16 =.7.3 q (a): 5/16 =.3 total alleles = 16.7 p (A): (3+3+5)/16 =.7.3 q (a): 5/16 =.3 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 8 individuals 16 alleles 0.5 A: 0.5 0.5 a: 0.5 Original population How do you explain these data? total alleles = 14.5 p (A): (1+1+5)/14 =.5.5 q (a): (5+1+1)/14 =.5 total alleles = 14.5 p (A): (1+1+5)/14 =.5.5 q (a): (5+1+1)/14 =.5 total alleles = 18.2 p (A): 4/18 =.2.8 q (a): (4+5+5)/18 =.8 total alleles = 18.2 p (A): 4/18 =.2.8 q (a): (4+5+5)/18 =.8 AA310 Aa554 aa015 p.7.5.2 q.3.5.8 Case #4 43

45. AP Biology Essential Questions  How do populations change over time?  What factors can cause changes in populations over time?  How did modern understandings of genetics impact evolutionary thought? 44