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UMass Amherst BIOLOGY 152 - Evolution III

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BIOLOGY 152 1st Edition Lecture 21Outline of Last LectureI. Genetic Variation A. Examples of Allele Diversity B. Sources of Genetic Variation C. Measuring Genetic Diversity Outline of Current Lectureb. Genetic Variation Continued c. Hardy Weinberg Equilibrium Current LectureEvolution III**Clicker Question**Population of butterflies have following genotype frequencies: 50% AA, 30% Aa, 20% aa. What is the frequency of the “a” allele?• answer = 0.35 • 20% —> 0.2 and add to that 1/2 of the heterozygous condition (whichwould be 30% or 0.3) • so 0.2 + 1/2(0.3) = 0.2 + 0.15 = 0.35 • think about a population of 100, so with 200 alleles Minnow Question: 1000 individuals were sampled and the following genotype frequencies were found: AA = .08, Aa = .28, aa = .64. What is theallele frequency of the “A” allele?• .08 +.1/2(.28) = .08 + 0.14 = 0.22**Clicker Question**Eastern gray squirrels occur in 3 morphs (gray, gray/black, brown). This is due to varied expression of a single gene — the allelic variation is actually in the promoter (little expression = gray, lots of expression = black) this hasto do with melanin. Black allele is rare but dominant. Will back squirrels eventually disappear or become more common?• no reason for B to be true: if there was selective pressure, a recessive These 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.gene could become more common. Natural selection determines this, notdominance or recessiveness of the gene • answer would be C, not enough information, because we are not told how many babies the black squirrels have so we know nothing about them from a natural selection standpoint Hardy-Weinberg EquilibriumAssumptions:• random mating • no mutation • no gene flow — don’t have new individuals coming into population • no genetic drift (infinite population size) • no natural selection If there is no selective pressure on a gene, then the alleles will stay at equilibrium if you have all these assumptions• if evolution is not happening, then your alleles will stay at equilibriumand it has nothing to do with dominance or recessiveness • if there is some sort of change in the gene pool, then they cannot be atequilibrium Hardy-Weinberg Equilibrium is a test to see if selection for an allele is occurring. For now we’ll look at two alleles but it does apply to more alleles. Lets use A and aRandom mating:Blue gametes = A green gametes= a• use what looks like a punnet square, but in which you are comparingfrequency instead of just the alleles • if you did the allele frequencies and found that everything is equal to 1,then they are in equilibrium and no evolution is happening • if not equal to 1, then evolution is happening and the allelesare not at equilibrium Example: HW in MN blood group of humans. Is there selection happening in human populations for these blood groups?MM : 0.835 MN: 0.156 NN: 0.009• First calculate allele frequency (already know the genotypes)• M = 0.835 + 1/2(0.156) = 0.913 • N = 0.009 + 1/2(0.156) = 0.087 • square 0.913 and 0.087 (or square p and q) and then find what 2pq is • then you compare the numbers you get and see if they’re close to thesame as the ones in the table • so we can assume that this population is in Hardy-Weinberg equilibriumand there is no evolution or natural selection So what is the allele frequency for the Ainu population for M?• 0.43 • N = 1-0.43 = 0.57 What is the expected value for MM if the allele is in Hardy Weinberg Equilibrium• all you need to do is square p which gives you the frequencyEven if there was strong selection on these alleles at one point, one generation of random mating and no evolutionary mechanisms acting on these alleles will result in genotype frequencies that conform to H-W. True or False?• True. If you have random mating and no selection, in a large population your allele frequencies will stay the same. The actual value doesn’t matter as long as they fit the equation. If there was selection in a population for MN, then you would get allele frequencies that were close to those numbers • if a generation is pushed out of equilibrium values for a generation, itwill return after only one generation to its equilibrium values Cystic FibroisisIf 9 in 10,000 newborns have the disease, what is the expected frequencies of the dominant allele if the genotypes conform to the Hardy Weinberg model9 in 10,000 = 0.0009 = q^2take the square root of that so that is your q allele frequency of a so A would be 1-0.03 =


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UMass Amherst BIOLOGY 152 - Evolution III

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