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Measuring Evolution of Populations

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Presentation on theme: "Measuring Evolution of Populations"— Presentation transcript:

1 Measuring Evolution of Populations

2 5 Agents of evolutionary change
Mutation Gene Flow Non-random mating Genetic Drift Selection

3 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

4 Evolution of populations
Evolution = change in allele frequencies in a population hypothetical: what conditions would cause allele frequencies to not change? non-evolving population REMOVE all agents of evolutionary change very large population size (no genetic drift) no migration (no gene flow in or out) no mutation (no genetic change) random mating (no sexual selection) no natural selection (everyone is equally fit)

5 Hardy-Weinberg equilibrium
Hypothetical, non-evolving population preserves allele frequencies Serves as a model (null hypothesis) natural populations rarely in H-W equilibrium useful model to measure if forces are acting on a population measuring evolutionary change G.H. Hardy (the English mathematician) and W. Weinberg (the German physician) independently worked out the mathematical basis of population genetics in Their formula predicts the expected genotype frequencies using the allele frequencies in a diploid Mendelian population. They were concerned with questions like "what happens to the frequencies of alleles in a population over time?" and "would you expect to see alleles disappear or become more frequent over time?" G.H. Hardy mathematician W. Weinberg physician

6 Hardy-Weinberg theorem
Counting Alleles assume 2 alleles = B, b frequency of dominant allele (B) = p frequency of recessive allele (b) = q frequencies must add to 1 (100%), so: p + q = 1 BB Bb bb

7 Hardy-Weinberg theorem
Counting Individuals frequency of homozygous dominant: p x p = p2 frequency of homozygous recessive: q x q = q2 frequency of heterozygotes: (p x q) + (q x p) = 2pq frequencies of all individuals must add to 1 (100%), so: p2 + 2pq + q2 = 1 BB Bb bb

8 H-W formulas B b B BB Bb b Bb bb Alleles: p + q = 1 Individuals:
p2 + 2pq + q2 = 1 B b BB Bb bb BB Bb bb

9 Using Hardy-Weinberg equation
population: 100 cats 84 black, 16 white How many of each genotype? q2 (bb): 16/100 = .16 q (b): √.16 = 0.4 p (B): = 0.6 p2=.36 2pq=.48 q2=.16 BB Bb bb Must assume population is in H-W equilibrium! What are the genotype frequencies?

10 Using Hardy-Weinberg equation
p2=.36 2pq=.48 q2=.16 Assuming H-W equilibrium BB Bb bb Null hypothesis p2=.20 p2=.74 2pq=.10 2pq=.64 q2=.16 q2=.16 Sampled data 1: Hybrids are in some way weaker. Immigration in from an external population that is predomiantly homozygous B Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats. Sampled data 2: Heterozygote advantage. What’s preventing this population from being in equilibrium. bb Bb BB Sampled data How do you explain the data? How do you explain the data?

11 Application of H-W principle
Sickle cell anemia inherit a mutation in gene coding for hemoglobin oxygen-carrying blood protein recessive allele = HsHs normal allele = Hb low oxygen levels causes RBC to sickle breakdown of RBC clogging small blood vessels damage to organs often lethal

12 Sickle cell frequency High frequency of heterozygotes
1 in 5 in Central Africans = HbHs unusual for allele with severe detrimental effects in homozygotes 1 in 100 = HsHs usually die before reproductive age Sickle Cell: In tropical Africa, where malaria is common, the sickle-cell allele is both an advantage & disadvantage. Reduces infection by malaria parasite. Cystic fibrosis: Cystic fibrosis carriers are thought to be more resistant to cholera: 1:25, or 4% of Caucasians are carriers Cc Why is the Hs 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
In tropical Africa, where malaria is common: homozygous dominant (normal) die of malaria: HbHb homozygous recessive die of sickle cell anemia: HsHs heterozygote carriers are relatively free of both: HbHs survive more, more common in population Hypothesis: In malaria-infected cells, the O2 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 p2 + 2 pq + q2 = 1

16 Black (b) is recessive to white (B)
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 q2 = 4/16 or 0.25 q = = 0.5

17 Once you know q you can figure out p . . . p + q = 1 p + q = 1
Once you know q you can figure out p p + q = 1 p + q = 1 p = 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

You know pp from problem bb or q2 = 4/16 = 0.25 BB or p2 = (0.5)2 = 0.25 Bb = 2pq = 2 (0.5) (0.5) = 0.5 25% of population are bb 25% of population are BB 50% of population are Bb

19 Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. q2 = 0.4 q = = p = = aa = 0.4 = 40% Aa = 2 (0.632) (0.368) = =46.5% AA = (0.3676) (0.3676) = .135 = 13.5% Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006

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 q2 = 1/1700 q = q = 0.024 p = 1 – = 0.976 Frequency of C = 97.6% Frequency of c = 2.4% NOW FIND THE GENOTYPIC FREQUENCIES

22 CC or p2 = (0.976)2 = .953 Cc or 2pq = 2 (0.976) (0.024) = cc = 1/1700 = 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? (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? q2 = 9/10000 q = q = 0.03 p = 1 – 0.03= 0.97 Frequency of C = 97% Frequency of c = 3%

25 CC or q2 = (0.976)2 = .953 Cc or 2pq = 2 (0.976) (0.024) = cc = 1/1700 = CC = 95.3% of population Cc = 4.68% of population cc = .06% of population

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