5 Populations & Gene Pools a population is a localized group of interbreeding individualsthe gene pool is collection of alleles in the populationRemember the difference between alleles & genes!an allele frequency is how common the allele is in the population
6 Populations Evolve, Not Individuals Evolution is change in allele frequencies in a population from one generation to the next.What conditions would be necessary for a population to NOT evolve?no mutation (no genetic change)no migration (no gene flow in or out)random mating (no sexual selection)very large population size (no genetic drift)no natural selection (everyone is equally fit)
7 Hardy-Weinberg Equilibrium A hypothetical, non-evolving population where allele frequencies remain constantServes as a null hypothesis:Since natural populations are rarely in H-W equilibrium, the model is used to determine if evolutionary forces are acting on a population.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. HardymathematicianW. Weinbergphysician
8 Hardy-Weinberg Theorem Let’s start by counting alleles:Assume 2 alleles = B, bfrequency of dominant allele (B) = pfrequency of recessive allele (b) = qThese frequencies must add up to 1 (100%), so:p + q = 1BBBbbb
9 Hardy-Weinberg Theorem Now let’s count individuals:frequency of homozygous dominant: p x p = p2frequency of homozygous recessive: q x q = q2frequency of heterozygotes: (p x q) + (q x p) = 2pqThese frequencies must also add up to 1 (100%), so:p2 + 2pq + q2 = 1BBBbbb
11 Hardy-Weinberg Example In a population of 100 cats, there are84 black & 16 white.How many of each genotype?q2 (bb): 16/100 = .16q (b): √.16 = 0.4p (B): = 0.6p2=.362pq=.48q2=.16BBBbbbMust assume population is in H-W equilibrium!What are the genotype frequencies?
12 Hardy-Weinberg Example 2pq=.48q2=.16Assuming H-W equilibriumBBBbbbNull hypothesisp2=.20p2=.742pq=.642pq=.10q2=.16q2=.16Sampled data 1:Hybrids are in some way weaker.Immigration in from an external population that is predomiantly homozygous BNon-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.bbBbBBSampled dataHow do you explain the data?How do you explain the data?
13 Application of H-W Principle: Sickle Cell Anemia Caused by a mutation in gene coding for hemoglobin, an oxygen-carrying blood proteinnormal allele = Hbrecessive allele = HsLow oxygen levels causes RBC to sicklebreakdown of RBCclogging small blood vesselsdamage to organsoften lethal
14 Sickle Cell Frequency High frequency of heterozygotes 1 in 5 in Central Africans = HbHsunusual for allele with severe detrimental effects in homozygotes1 in 100 = HsHsusually die before reproductive ageSickle 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 CcWhy is the Hs allele maintained at such high levels in African populations?Suggests some selective advantage of being heterozygous…
15 MalariaSingle-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells123
16 Heterozygote Advantage In tropical Africa, where malaria is common:homozygous dominant (HbHb ): reduced reproduction due to malariahomozygous recessive (HsHs): reduced reproduction due to sickle cell diseaseheterozygotes (HbHs): resistant to malaria – sickle cell in benign conditionFrequency of sickle cell allele & distribution of malaria