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- 1. Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM KIM FOGLIA@explorebiology.com
- 2. 5 Agents of evolutionary change Mutation Gene Flow Genetic Drift Selection Non-random mating
- 3. Populations & gene pools <ul><li>Concepts </li></ul><ul><ul><li>a population is a localized group of interbreeding individuals </li></ul></ul><ul><ul><li>gene pool is collection of alleles in the population </li></ul></ul><ul><ul><ul><li>remember difference between alleles & genes ! </li></ul></ul></ul><ul><ul><li>allele frequency is how common is that allele in the population </li></ul></ul><ul><ul><ul><li>how many A vs. a in whole population </li></ul></ul></ul>
- 4. Evolution of populations <ul><li>Evolution = change in allele frequencies in a population </li></ul><ul><ul><li>hypothetical : what conditions would cause allele frequencies to not change? </li></ul></ul><ul><ul><li>non-evolving population </li></ul></ul><ul><ul><ul><li>REMOVE all agents of evolutionary change </li></ul></ul></ul><ul><ul><ul><li>very large population size (no genetic drift ) </li></ul></ul></ul><ul><ul><ul><li>no migration (no gene flow in or out) </li></ul></ul></ul><ul><ul><ul><li>no mutation (no genetic change) </li></ul></ul></ul><ul><ul><ul><li>random mating (no sexual selection) </li></ul></ul></ul><ul><ul><ul><li>no natural selection (everyone is equally fit) </li></ul></ul></ul>
- 5. Hardy-Weinberg equilibrium <ul><li>Hypothetical, non-evolving population </li></ul><ul><ul><li>preserves allele frequencies </li></ul></ul><ul><li>Serves as a model ( null hypothesis ) </li></ul><ul><ul><li>natural populations rarely in H-W equilibrium </li></ul></ul><ul><ul><li>useful model to measure if forces are acting on a population </li></ul></ul><ul><ul><ul><li>measuring evolutionary change </li></ul></ul></ul>W. Weinberg physician G.H. Hardy mathematician
- 6. Hardy-Weinberg theorem <ul><li>Counting Alleles </li></ul><ul><ul><li>assume 2 alleles = B , b </li></ul></ul><ul><ul><li>frequency of dominant allele ( B ) = p </li></ul></ul><ul><ul><li>frequency of recessive allele ( b ) = q </li></ul></ul><ul><ul><ul><li>frequencies must add to 1 (100 % ), so: </li></ul></ul></ul><ul><ul><li>p + q = 1 </li></ul></ul>bb Bb BB
- 7. Hardy-Weinberg theorem <ul><li>Counting Individuals </li></ul><ul><ul><li>frequency of homozygous dominant : p x p = p 2 </li></ul></ul><ul><ul><li>frequency of homozygous recessive : q x q = q 2 </li></ul></ul><ul><ul><li>frequency of heterozygotes : ( p x q ) + ( q x p ) = 2p q </li></ul></ul><ul><ul><ul><li>frequencies of all individuals must add to 1 (100 % ), so: </li></ul></ul></ul><ul><ul><li> p 2 + 2pq + q 2 = 1 </li></ul></ul>bb Bb BB
- 8. H-W formulas <ul><li>Alleles: </li></ul><ul><li>p + q = 1 </li></ul><ul><li>Individuals: </li></ul><ul><li>p 2 + 2p q + q 2 = 1 </li></ul>bb Bb BB B b B BB Bb b Bb bb BB B b Bb bb
- 9. Using Hardy-Weinberg equation What are the genotype frequencies? q 2 (bb): 16/100 = .16 q (b): √.16 = 0.4 p (B): 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each genotype? bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Must assume population is in H-W equilibrium!
- 10. Using Hardy-Weinberg equation bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Assuming H-W equilibrium Sampled data 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 bb Bb BB
- 11. Application of H-W principle <ul><li>Sickle cell anemia </li></ul><ul><ul><li>inherit a mutation in gene coding for hemoglobin </li></ul></ul><ul><ul><ul><li>oxygen-carrying blood protein </li></ul></ul></ul><ul><ul><ul><li>recessive allele = H s H s </li></ul></ul></ul><ul><ul><ul><ul><li>normal allele = H b </li></ul></ul></ul></ul><ul><ul><li>low oxygen levels causes RBC to sickle </li></ul></ul><ul><ul><ul><li>breakdown of RBC </li></ul></ul></ul><ul><ul><ul><li>clogging small blood vessels </li></ul></ul></ul><ul><ul><ul><li>damage to organs </li></ul></ul></ul><ul><ul><li>often lethal </li></ul></ul>
- 12. Sickle cell frequency <ul><li>High frequency of heterozygotes </li></ul><ul><ul><li>1 in 5 in Central Africans = H b H s </li></ul></ul><ul><ul><li>unusual for allele with severe detrimental effects in homozygotes </li></ul></ul><ul><ul><ul><li>1 in 100 = H s H s </li></ul></ul></ul><ul><ul><ul><li>usually die before reproductive age </li></ul></ul></ul>Why is the H s allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous…
- 13. Malaria Single-celled eukaryote parasite ( Plasmodium ) spends part of its life cycle in red blood cells 1 2 3
- 14. Heterozygote Advantage <ul><li>In tropical Africa, where malaria is common: </li></ul><ul><ul><li>homozygous dominant (normal) die of malaria: H b H b </li></ul></ul><ul><ul><li>homozygous recessive die of sickle cell anemia: H s H s </li></ul></ul><ul><ul><li>heterozygote carriers are relatively free of both: H b H s </li></ul></ul><ul><ul><ul><li>survive more, more common in population </li></ul></ul></ul>Hypothesis : In malaria-infected cells, the O 2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Frequency of sickle cell allele & distribution of malaria
- 15. HARDY-WEINBERG PRACTICE PROBLEMS p + q = 1 p 2 + 2 pq + q 2 = 1
- 16. <ul><li>Black (b) is recessive to white (B) </li></ul>Bb and BB pigs “look alike” so can’t tell their alleles by observing their phenotype. ALWAYS START WITH RECESSIVE alleles. p= dominant allele q = recessive allele 4/16 are black. So bb or q 2 = 4/16 or 0.25 q = 0.25 = 0.5
- 17. <ul><li>Once you know q </li></ul><ul><li>you can figure out p </li></ul><ul><li>. . . p + q = 1 </li></ul>p + q = 1 p + 0.5 = 1 p = 0.5 Now you know the allele frequencies. The frequency of the recessive (b) allele q = 0.5 The frequency of the dominant (B) allele p = 0.5 http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
- 18. WHAT ARE THE GENOTYPIC FREQUENCIES? <ul><li>You know pp from problem bb or q 2 = 4/16 = 0.25 </li></ul><ul><li>BB or p 2 = (0.5) 2 = 0.25 </li></ul><ul><li>Bb = 2pq = 2 (0.5) (0.5) = 0.5 </li></ul><ul><li>25% of population are bb </li></ul><ul><li>25% of population are BB </li></ul><ul><li>50% of population are Bb </li></ul>http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
- 19. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006 q 2 = 0.4 q = 0.4 = 0.6324 p = 1 - 0.6324 = 0.3676 aa = 0.4 = 40% Aa = 2 (0.632) (0.368) = 0.465 =46.5% AA = (0.3676) (0.3676) = .135 = 13.5%
- 20. PRACTICE HARDY WEINBERG 1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis. Calculate the allele frequencies for C and c in the population Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006
- 21. 1/1700 have cystic fibrosis q 2 = 1/1700 q = 0.00059 q = 0.024 p = 1 – 0.024 = 0.976 Frequency of C = 97.6% Frequency of c = 2.4% NOW FIND THE GENOTYPIC FREQUENCIES
- 22. CC or p 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population
- 23. Now you can answer questions about the population: How many people in this population are heterozygous? It has been found that a carrier is better able to survive diseases with severe diarrhea. What would happen to the frequency of the "c" if there was a epidemic of cholera or other type of diarrhea producing disease? 0.0468 (1700) = 79.5 ~ 80 people are Cc Cc more likely to survive than CC. c will increase in population
- 24. The gene for albinism is known to be a recessive allele. In Michigan, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming hardy-Weinberg equilibrium, what is the allele frequency for the dominant pigmentation allele in this population? q 2 = 9/10000 q = 0.0009 q = 0.03 p = 1 – 0.03= 0.97 Frequency of C = 97% Frequency of c = 3%
- 25. CC or q 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population

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