BIOB 272 1st EditionLecture 16 Outline of Last Lecture Phylogenies and Tree of LifeI. Comparison of Genetic Linkage and Physical MapsII. Complete Genome MapsIII. Genome Annotationa. Compare to Known Genesb. Predict GenesIV. Phylogenies and Tree of LifeV. Reconstructing PhylogenyVI. Monophyletic Group vs. Paraphyletic GroupVII. Reading a Phylogenetic Treea. Types of Trees:i. Unrootedii. Rootedb. Extant Speciesc. Taxonomic Units VIII. Phylogenetic Reconstructiona. Choosing Characters:b. Homologousc. Convergent Evolution- Analogous Characterd. How to Choose the Correct Phlylogenyi. Principle of Parsimony: Occam’s Razore. Other Phylogenetic Methodsi. Distance-Based Methodsii. Model-Based MethodsOutline of Current Lecture Population GeneticsI. Causes of Changes in Genealogy1. Reversals (back mutations)2. Parallel SubstitutionsII. Phylogenetics in Practice- HomoplasyThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.III. Darwin’s Four Motivation Sources1. Evidence for descent with modification2. Geology and Uniformitarianism3. Artificial Selection4. Struggle for existence within poplutionsIV. Darwin’s Four Postulates for Evolution by Natural Selection1. Individuals vary2. At least some variation is heritable3. Some individuals leave more progeny than others4. Variation among individuals in survival, reproduction not random, but depend onheritable trait variationV. The Equilibrium Population- Hardy Weinberg Equilibrium ModelVI. Testing for Hardy-Weinberg ProportionsVII. Occurrence of Cystic Fibrosis: Autosomal Recessive (cc)Current LecturePopulation GeneticsI. Causes of Changes in Genealogy1. Reversals (back mutations): can remove synapomorphies from a genealogy- DNA sequence changes between the ancestral DNA sequence and the new DNA sequence2. Parallel Substitutions: Lead to convergent evolution in DNA sequencesII. Phylogenetics in Practice- Homoplasy= caused by both convergence and reversal- this similarity is not due to shared ancestry- All real datasets will have some homoplasic characters- conflicting evolutionary signals- To deal with homoplasy- carefully choose characters (DNA/proteins) and use LOTS of characters-employ complex algorithms to alternate tree topologies= LOTS of tree topologies are possibleIII. Darwin’s Four Motivation Sources1. Evidence for descent with modification- fossils, finches, orchids, etc.2. Geology and Uniformitarianism: “forces that acted in the past are similar to forces acting in the present”3. Artificial Selection: on domesticated plants and animals4. Struggle for existence within populations: human society is constantly struggling for limited resources- based on Malthus’ basic theory= hit a point of crisis- “An Essay on the Principle of Population, 1798”IV. Darwin’s Four Postulates for Evolution by Natural Selection1. Individuals vary2. At least some variation is heritable3. Some individuals leave more progeny than others4. Variation among individuals in survival, reproduction not random, but depend onheritable trait variation** Outcome: Genetic variants with higher survival and/or reproduction increase in frequency in population- Key Points:o In the absence of heritable variation, natural selection will NOT result in evolutionary changeo Natural selection acts through variation in the survival and reproduction of individuals, but evolutionary change occurs in populationsV. The Equilibrium Population- Hardy Weinberg Equilibrium Model- Assumptions1. Random Mating2. No Mutation3. Large (infinite) population size (no genetic drift)4. No differential survival or reproduction (no natural selection)5. Not Migration (no gene flow)- If these assumptions met:1. The population will NOT evolve- allele frequencies will remain constant from generation to generation= Hardy-Weinberg equilibrium2. Genotypes will be in binomial proportions=Hardy-Weinberg Proportions=p^2+ 2pq+ q^2Zygot genotypes will be in binomial proportions:Gametes Zygote (diploid) GenotypesA a AA Aa aap q p^2+ 2pq+ q^2** With this model you can:1. Calculate genotype frequencies form allele frequencies2. Calculate allele frequencies from genotype frequenciesVI. Testing for Hardy-Weinberg Proportions: to test whether a population is in Hardy-Weinbery proportions= it is in H-W equilibrium1. Determine allele frequencies form observed genotype numbers2. Calculate expected genotype numbers under HW model3. Compare observed numbers to expected numbers (chi-square test)Example:Genotype: MM MN NN TotalObserved: 47 46 7 100Step 1: Freq (M)= p= (2(MM) + (MN))/ 2(total) = (94+46)= 2(100) = 0.70Step 2: Expected genotype numbers:MM MN NNp^2n 2pqN q^2N(0.7)^2*100=49 2(0.7)(0.3)100=42 (0.3)^2*100= 9Genotype: MM MN NN TotalObserved: 47 46 7 100Expected: 49 42 9 100Step 3: Chi-Square Testx^2= (Observed- expected)^2 / expectedDegrees of freedom (df)= #genotypes- #alleles =1x^2= ((47-49)^2 / 49) + ((46-42)^2)/42) + ((7-9)^2)/ 9) = 0.910.91< 3.84 so we ACCEPT these as H-W proportions because H-W IS the null modelVII. Occurrence of Cystic Fibrosis: Autosomal Recessive (cc)- 1/3200 European Americans- 1/9200 Hispanics- 1/3500 African Americans- 1/31000 Asian AmericansHardy-Weinberg Equilibrium Example: Assume this class of 220 student comes from a population in 1/3200 European Americans people are afflicted (cc) with cystic fibrosis:Step 1:CC Cc ccp^2 2pq q^2(3199/3200) 1/3200q^2= 1/3200square root of q^2=q=square root (1/3200)= 0.018Step 2: p= 1-q= 0.982CC Cc ccp^2 2pq q^20.965 0.035 about 0=0.000312pq N= (0.035) (220)= 7.7 carriers in classUM= 420