1Population GeneticsMacrophageCCR5 CCR5-Δ322What accounts for thisvariation? Random? Pastepidemics (plague, smallpox)?What will happen to this variationin the future? Will Δ32 alleleincrease in frequency?These are the questions that “populationgenetics” is designed to addressHardy-Weinberg Principle1. Allele frequencies remain constantfrom generation to generation unlesssome outside force is acting to changethem2. When an allele is rare, there are manymore heterozygotes thanhomozygotes (if p is small, then ____is very small)3Assumptions of H-W1) Mating is random across the entire population.2) All genotypes have equal viability and fertility (noselection).3) Migration into the population can be ignored.4) Mutation does not occur, or is so rare it can beignored.5) Population is large enough that the allelefrequencies do not change from generation togeneration due to chance (random genetic drift).6) Allele frequencies are the same in females andmales.Usefulness of H-WIf you know the allele frequencies, you canpredict the genotype frequencies:Q: In S. France, the frequency of the Δ32 alleleis 10% (i.e., q=0.10). What proportion ofindividuals will be homozygous for the allele?What proportion will be heterozygous?aaAaAAq22pqp2Usefulness of H-WIf you know the frequency of one of thehomozygous genotypes, you can estimateallele frequencies, and predict the frequenciesof the other genotypes.Q: Among individuals of European descent,1/1700 newborns have cystic fibrosis (arecessive genetic disorder). What proportionof this population are heterozygous carriers?Hint: q2 = 1/1700 = 0.00059A:Multiple Alleles: ABO bloodtypesp = freq of A alleleq = freq of B alleler = freq of O alleleExpansion of [p + q + r]2 =p2 + q2 + r2+ 2pq + 2pr + 2qr4Assumptions of H-W1) Mating is random across the entire population.2) All genotypes have equal viability and fertility (noselection).3) Migration into the population can be ignored.4) Mutation does not occur, or is so rare it can beignored.5) Population is large enough that the allelefrequencies do not change from generation togeneration due to chance (random genetic drift).6) Allele frequencies are the same in females andmales.What happens when any ofthese assumptions areviolated?SelectionMutationNon-random mating______________________________If any of theseprocesses areoccurring, will tend toget ____________from H-W expectedproportionsHow can we detect deviationfrom H-W expectations?Do observed genotype frequencies matchHW expectations?Do observed genotype frequencies matchHW expectations?MMMNNNp22pqq20.2940.4960.209Ge not ypes Exp ect ed O bserv ed MM 294.3 298 MN 496.4 489 NN 209.3 213 p=.5425q=.45755Test for H-W Genotype FrequenciesGe not ypes Exp ect ed O bserv ed MM 294.3 298 MN 496.4 489 NN 209.3 213 Importance of H-WH-W is an important tool for populationgenetics.If assumptions are met, we can use it toestimate allele and genotypefrequencies that would otherwise bedifficult to measure.If assumptions are not met (can be testedstatistically), then we know that someoutside force is perturbing allele orgenotype frequencies.Change in allele frequenciesover generationsEvolution is defined as a changein allele (or genotype) frequenciesover generations, and evolutionwill be caused by violation of anyof the assumptions of H-W.Forces that cause deviationfrom H-W (evolution)1. Selection2. Mutation3. Genetic Drift4. Nonrandom Mating5. Gene Flow (Migration)6Genotype A has a constitutive mutation for enzyme production inthe lactose operon.B is the normal inducible lactose operon.A and B grown together in environment with limited lactose.p = 0.5, q = 0.520/95 a 25/95 A;25/95 a25/95 AGametecontribution2080% = 0.850100% = 125100% = 1Survival toreproductionaa25Aa50AA25GenotypesNumber:New allele frequencies?New genotype frequencies (assume randommating):p = = ____q = = ____ AA Aa aa0.28 0.50 0.227Consistent differences in survival orreproduction between genotypes =genotypic-specific differences infitnessWhen fitness values are expressedon a scale such that highestfitness=1, then the values arecalled relative fitnessTo conveniently calculate change inallele frequency due to selection,need concept of average fitnessChange in allele frequencyGenotype AA Aa aaGenotype Frequency p22pq q2Relative Fitness WAA WAa WaaW=average fitness= (p2WAA)+ (2pqWAa)+ (q2WAa)Freq of A after one gen. of selection:p' = p2 WAA/W + pqWAa/WFreq of a after one gen. of selection: (1-p’) or: q'= q2 Waa/W + pqWAa/WCCR5 Example; p(+)=0.9; q(Δ32)=0.1WΔ32/Δ32=1.0W+/Δ32=0.99W+/+=0.99Relative FitnessΔ32/Δ32q2=0.01+/Δ322pq=0.18+/+ p2=0.81Genotypefrequency:Average fitness W = 0. 81*0.99 + 0.18*0.99 + 0.01*1 = 0.9901q'=q2WΔ32/Δ32/W + pqW+/Δ32 /W=0.01009 +0.089991=0.100091p’= 1-q’ = 0.89999q20.010022pq0.18016p20.80998Next generationgenotype freq.qq28Selection will increase thefrequency of Δ32 alleleSelection is relatively weakThe favored allele is recessiveand the favored genotype is very rareThe change in allele frequency (response toselection) will be relatively slowResponse to selection can befast!Selection isstrongFavoredallele ispartiallydominantBoth allelesarecommonSelection is not always“Directional”Heterozygote advantageFrequency dependenceSelection varying in space or time9HeterozygoteadvantageFitnessA a a aA As=0.86t=0.12Selection coefficientsFitness (in symbols)Relative Fitness1-s11-t0.141.00.88HbS/HbSHbA/HbSHbA/HbARelative fitness of hemoglobin genotypes in YorubansEquilibrium frequencies:peq = s/(s+t) = 0.86/(0.12+0.86) = 0.88qeq = t/(s+t) = 0.12/(0.12+0.86) = 0.12Predict the genotype frequencies (at birth):HW proportions 0.774 0.211 0.0144Variable selection: genotypes have differentfitness effects in different environmentsFitnessFrequency-dependentselection10Other Examples of Freq-dep.
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