chpt. 23 the evolution of populations-- population genetics

Download Chpt. 23 The Evolution of Populations-- Population Genetics

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Slide 2 Chpt. 23 The Evolution of Populations-- Population Genetics Slide 3 Individuals are selected Populations evolve Slide 4 Populations evolve Slide 5 Populations Populations = unit of evolution Natural selection Natural selection = mechanism of evolution Gradualism Gradualism = accumulation of small changes in gene poolover LONG periods of time Slide 6 Hardy-Weinberg Theorem Slide 7 examines the gene structure of a NON-evolving population Slide 8 Hardy-Weinberg Theorem Hardy-Weinberg Theorem examines the gene structure of a NON-evolving population. Obviously, this is not common, however, gives a base-line / model NULL HYPOTHESIS for determining if and why populations evolve Slide 9 Hardy-Weinberg Theorem Hardy-Weinberg Theorem even though alleles are shuffled and recombined during meiosis and random fertilization. This has no effect on the overall gene pool percentages. Slide 10 Not SWIMMING pool. A a A a A a A a a a A A A a GENE POOL!!! Slide 11 Hardy-Weinberg Theorem Gene pool frequencies (percentages) will remain unchanged if no mechanism that can cause evolution to occur acts on a population. Slide 12 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Mutations are not occurring Slide 13 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Natural selection is not occurring Slide 14 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: Population is LARGE Slide 15 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE breeds Slide 16 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE randomly mates Slide 17 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: EVERYONE produces the same number of offspring Slide 18 Hardy-Weinberg Theorem Gene pool frequencies will remain unchanged if: NOONE migrates in or out of the population everyone stays Slide 19 Hardy-Weinberg Theorem Slide 20 5 Agents of evolutionary change: MutationGene Flow (migration) Genetic Drift (same # of offspring) Selection Non-random mating Slide 21 Hardy-Weinberg Theorem remember: H.W. explains: the frequency of alleles remains constant in a population unless acted upon by agents OTHER THAN sexual recombination. Slide 22 Hardy-Weinberg Principle p2 + 2pq + q2 = 1 Mathematical statement about the relative frequency of alleles (genotypes) in a population. p +q = 1 Slide 23 Hardy-Weinberg Theorem Frequency of alleles remains constant in a population, unless acted upon by agents OTHER THAN sexual recombination. Inheritance does not cause changes in allele frequency. Slide 24 Hardy-Weinberg Principle p + q = 1 p = frequency of dominant allele Slide 25 Hardy-Weinberg Principle p + q = 1 q = frequency of recessive allele Slide 26 Hardy-Weinberg Principle in most cases, we only know the phenotypic frequencies Mathematical statement about the relative frequency of alleles (genotypes) in a population. Slide 27 Hardy-Weinberg Principle q 2 = # of aa individuals q 2 = frequency of homozygote recessive individuals Slide 28 Hardy-Weinberg Principle p 2 = # of AA individuals p 2 = frequency of homozygote dominant individuals Slide 29 Hardy-Weinberg Principle p = # of (AA) + 2 (# Aa) p = frequency of dominant allele Slide 30 Hardy-Weinberg Principle q = # of (aa) + 2 (# Aa) q = frequency of recessive allele Slide 31 Hardy-Weinberg Principle 2pq = # of Aa individuals 2pq = frequency of heterozygote individuals Slide 32 Hardy-Weinberg Principle by comparing genotypic frequencies from one generation to the next, you can learn whether or not evolution has occurred Slide 33 Hardy-Weinberg Principle if genotypic frequencies have changed from your original count evolution has occurred! Slide 34 Suppose there are 1,000 individuals in a population Genotype NumberGenotypic Frequency AA 4900.49 Aa 4200.42 aa 90 0.09 total 10001.00 Slide 35 Suppose there are 1,000 individuals in a population Genotypic Frequency 0.49 0.42 0.09 total1.00 Genotypic frequency = the proportion of a particular genotype found in a population AA Aa aa Slide 36 Suppose there are 1,000 individuals in a population Phenotype NumberPhenotypic Frequency dominant 9100.91 recessive 900.09 total 10001.00 Slide 37 Suppose there are 1,000 individuals in a population Phenotypic Frequency 0.91 0.09 total1.00 Phenotypic frequency = the proportion of a particular phenotype found in a population Slide 38 Suppose there are 1,000 individuals in a population Allele NumberAllele Frequency A 14000.7 a 4200.3 total 20001.00 Slide 39 480 Allele frequency Pssst(There are 1,000 copies of the flower color gene in this population of 500 total flowers) q = q = frequency of recessive allele However, we do not know how many as there are just by looking at phenotype Slide 40 480 Genotypic frequency q 2 = 20/500 q2 = frequency of recessive genotype Slide 41 480 Genotypic frequency q 2 =.04 q2 = frequency of recessive genotype Slide 42 480 Allele frequency q = .04 q = frequency of recessive allele Slide 43 480 Allele frequency q =.2 q = frequency of recessive allele Slide 44 480 Allele frequency q =.2 q = frequency of recessive allele p = frequency of dominant allele p + q = 1 p +.2 = 1 p = 1 -.2 p =.8 Slide 45 480 Some of the pink flowers will be AA and some will be Aa p2 + 2pq + q2 = 1 p =.8 q =.2 Slide 46 480 Some of the pink flowers will be AA and some will be Aa.64 +.32 +.04 = 1 Slide 47 480 How many of the pink flowers will be AA and how many will be Aa.64 X 500 individuals 320 individuals are AA Slide 48 480 How many of the pink flowers will be AA and how many will be Aa.32 X 500 individuals 160 individuals are Aa Slide 49 480 320 are AA 320 are AA 160 are Aa 160 are Aa 480 total Slide 50 Genetic structure of next generation.8 x.8 =.64 Slide 51 Genetic structure of next generation.2 x.2 =.04 Slide 52 Genetic structure of next generation.2 x.8 =.16.8 x.2 =.16.32 aA Aa Slide 53 Hardy-Weinberg Principle Under ideal conditions, the relative allele frequencies are the same in the offspring generation as in the parent generation. p 2 + 2pq + q 2 = 1 p 2 + 2pq + q 2 = 1 Slide 54 IF IF: frequencies in a population deviate from Hardy-Weinberg (these are set numbers) THEN THEN: we know the population is evolving. Slide 55 Oh Hardy, why did you quit your job at ABERCRO MBIE? Weinburg, I keep telling you, I got sick of changing genes!!! Slide 56 change over time is a result of changes in a populations frequency of genotypes / genetic Slide 57 Slide 58 Slide 59 Lets Hardy!! What percentage of the human population are carriers for the allele for PKU? 1 out of 10,000 babies are born with this recessive disease Most of the time you will begin by determining the frequency of the homozygous recessive genotype Find q2 (frequency of homo recessive) Slide 60 Lets Hardy!! What percentage of the human population are carriers for the allele for PKU? 1 out of 10,000 babies are born with this recessive disease Hint: q2 = 1 / 10,000 (frequency of aa) Find q Find p Find pq Slide 61 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed). Slide 62 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed). Slide 63 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed). Slide 64 AP Problems Using Hardy-Weinberg Solve for q 2 (% of total) Solve for q (equation) Solve for p (1- q) H-W is always on the national AP Bio exam (but no calculators are allowed) Slide 65 AP Problems Using Hardy-Weinberg population: 100 cats 84 black, 16 white How many of each genotype? population: 100 cats 84 black, 16 white How many of each genotype? q 2 (bb): 16/100 =.16 0.4 q (b): .16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 q 2 (bb): 16/100 =.16 0.4 q (b): .16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Slide 66 AP Problems Using Hardy-Weinberg bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Assuming H-W equilibrium Sampled data bbBbBB p 2 =.20 2pq=.64 q 2 =.16 How do you explain the data? Null hypothesis

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