Measuring Evolution of Populations 2007-2008

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

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

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)

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 1908. 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

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

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

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

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): 1 - 0.4 = 0.6 p2=.36 2pq=.48 q2=.16 BB Bb bb Must assume population is in H-W equilibrium! What are the genotype frequencies?

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?

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

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…

Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

Heterozygote Advantage In tropical Africa, where malaria is common: homozygous dominant (normal) die or reduced reproduction from malaria: HbHb homozygous recessive die or reduced reproduction from sickle cell anemia: HsHs heterozygote carriers are relatively free of both: HbHs survive & reproduce 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

Any Questions?? 2005-2006

Hardy-Weinberg Lab Data Mutation Gene Flow Genetic Drift Selection Non-random mating 2006-2007

Hardy Weinberg Lab: Equilibrium Original population Case #1 F5 18 individuals 36 alleles p (A): 0.5 q (a): 0.5 total alleles = 36 p (A): (4+4+7)/36 = .42 q (a): (7+7+7)/36 = .58 AA 4 Aa 7 aa 7 AA .25 Aa .50 aa .25 AA .22 Aa .39 aa .39 How do you explain these data?

Hardy Weinberg Lab: Selection Original population Case #2 F5 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (9+9+6)/30 = .80 q (a): (0+0+6)/30 = .20 AA 9 Aa 6 aa AA .25 Aa .50 aa .25 AA .60 Aa .40 aa How do you explain these data?

Hardy Weinberg Lab: Heterozygote Advantage Original population Case #3 F5 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (4+4+11)/30 = .63 q (a): (0+0+11)/30 = .37 AA 4 Aa 11 aa AA .25 Aa .50 aa .25 AA .27 Aa .73 aa How do you explain these data?

Hardy Weinberg Lab: Heterozygote Advantage Original population Case #3 F10 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (6+6+9)/30 = .70 q (a): (0+0+9)/30 = .30 AA 6 Aa 9 aa AA .25 Aa .50 aa .25 AA .4 Aa .6 aa How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5-1 6 individuals 12 alleles p (A): 0.5 q (a): 0.5 total alleles = 12 p (A): (4+4+2)/12 = .83 q (a): (0+0+2)/12 = .17 AA 4 Aa 2 aa AA .25 Aa .50 aa .25 AA .67 Aa .33 aa How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5-2 5 individuals 10 alleles p (A): 0.5 q (a): 0.5 total alleles = 10 p (A): (0+0+4)/10 = .4 q (a): (1+1+4)/10 = .6 AA Aa 4 aa 1 AA .25 Aa .50 aa .25 AA Aa .8 aa .2 How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5-3 5 individuals 10 alleles p (A): 0.5 q (a): 0.5 total alleles = 10 p (A): (2+2+2)/10 = .6 q (a): (1+1+2)/10 = .4 AA 2 Aa 2 aa 1 AA .25 Aa .50 aa .25 AA .4 Aa .4 aa .2 How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5 5 individuals 10 alleles p (A): 0.5 q (a): 0.5 AA Aa aa p q 1 .67 .33 0 .83 .17 2 0 .8 .2 .4 .6 3 4 .4 .2 .6 .4 AA .25 Aa .50 aa .25 How do you explain these data?

Any Questions?? 2007-2008