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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Probability works for individuals but how about populations? •You are a plant breeder and were given a field with 1000 plants–450 red, 300 pink, and 250 white•Assuming these plants mate randomly, what will the proportions of these colors be in the next generation?Definitions•Frequency–The number (count) of an item within a population–Example: 450 red snapdragons•Relative Frequency–The proportion (fraction) of an item within a population–Example: 450 / 1000 = 0.45 = 45% red snapdragonsWhat will the proportions of these colors be next generation?•How do we solve this problem?–Determine the relative frequency of each genotype and allele–relative frequency of RR = x = #RR / #individuals (N)•x = 450/1000 = 0.45–relative frequency of RW = y = #RW / #individuals (N)•y = 300/1000 = 0.30–relative frequency of WW = z = #WW / #individuals (N)•z = 250/1000 = 0.25–Note: x + y + z = 1•Calculating relative allele frequency–Frequency of allele R = p–p(R) = total # of R alleles from each genotype divided by total # of alleles (2N)•p(R) = [(2 × # RR) + (# RW)] / (2N)•p(R) = [(2 × 450) + (300)] / (2 × 1000) = 0.6–Frequency of allele W = q–q(W) = total # of W alleles from each genotype divided by sample size (N)•q(W) = [(2 × # WW) + (# RW)] / (2N)•q(W) = [(2 × 250) + (300)] / (2 × 1000) = 0.4–Note: p + q = 1What will the proportions of these colors be next generation?What have we done so far?•Calculated the relative frequency of each genotype in the population (x, y, & z)•Calculated the relative frequency of each allele in the population (p & q)•What next?•Now examine all possible mating types. How many are there?–3 types of male (RR, RW, & WW)  3 types of female (RR, RW, & WW) = 9 possible crosses•Calculate the probability each type of cross will occurWhat will the proportions of these colors be next generation?Perhaps a table would be helpful...RR (0.45) RW (0.30) WW (0.25)RR (0.45)RW (0.30)WW (0.25)0.20250.13500.11250.13500.09000.07500.11250.07500.0625Males Females •What is the probability that heterozygotes will mate?•frequency RW males  frequency RW females •0.30  0.30 = 0.09 (this is the middle cell of the table)•Therefore, mating among heterozygotes is expected to occur 9% of the timeNote: All the cells add up to 1!•We’ve calculated the probability of each mating type•What next?•We need to determine what type of offspring will come from each mating type…We already know how to do this…What will the proportions of these colors be next generation?Probability of each genotype in the offspring•To predict genotype frequencies in the offspring we use:–frequency of each mating type•RW  RW = 0.3  0.3 = 0.09–frequency of offspring resulting from each mating type•25% RR, 50% RW, 25% WWProbability of each genotypein the offspringPARENTSMATING (type) FREQUENCY RR RW WWRR x RRRR x RWRR x WWRW x RWRW x WWWW x WWRESULTING GENOTYPE OF OFFSPRINGRR (0.45) RW (0.30) WW (0.25)RR (0.45) 0.20250.13500.1125RW (0.30) 0.13500.09000.0750WW (0.25) 0.11250.07500.0625Males FemalesProbability of each genotypein the offspringPARENTSMATING (type) FREQUENCY RR RW WWRR x RR 0.2025 0.2025RR x RWRR x WWRW x RWRW x WWWW x WWGENOTYPE FREQUENCY OF RESULTING OFFSPRINGRR (0.45) RW (0.30) WW (0.25)RR (0.45) 0.20250.13500.1125RW (0.30) 0.13500.09000.0750WW (0.25) 0.11250.07500.0625Males Females How do we combine these? (AND or OR) is the question:1. Probability: RW male & RR female AND RR male & RW female2. Probability: RW male & RR female OR RR male & RW femaleOR = add the probabilitiesProbability of each genotypein the offspringPARENTSMATING (type) FREQUENCY RR RW WWRR x RR 0.2025 0.2025RR x RW.135 + .135 = .27.5.27 = .135 .5.27=.135RR x WWRW x RWRW x WWWW x WW0.225 0.2250.09 .25.09=.0225.5.09=.045.25.09=.02250.15 .5.15=.075 .5.15=.0750.0625 0.06251 0.36 0.48 0.16 GENOTYPE FREQUENCY OF RESULTING OFFSPRING•If we consider all possible matings, the genotypic frequencies of the offspring will be:–x(RR) = 0.36–y(RW) = 0.48–z(WW) = 0.16What are the allele frequencies?•p(R) = (rel freq RR) + 0.5 × (rel freq RW)p(R) = 0.36 + (0.5 × 0.48) = 0.6•q(r) = (rel freq WW) + 0.5 × (rel freq RW)q(W) = 0.16 + (0.5 × 0.48) = 0.4•NOTE: THESE ARE THE SAME AS WE SAW IN THE PARENTS–They are in equilibrium•What will the relative genotypic frequencies be in the next generation?x(RR) = 0.36, y(RW) = 0.48, z(WW) = 0.16–Genotypic frequencies achieve equilibrium after one generation of random mating•Try this yourself at home to checkHardy-Weinberg Equilibrium•If there are no perturbations in the system, allele and genotype frequencies will remain constant through time•Mendel’s laws can result in stability of allele frequenciesHardy-Weinberg Equilibirum•Segregation of alleles and random mating means that probability of different genotypes (x, y, z) is solely determined by allele frequencies (p, q)•At equilibrium we expect genotypic frequencies of:(p + q)[from females] × (p + q)[from males] = 1 = p2 + 2pq + q2 = 1q × q = q2q × pqp × qp × p = p2femalespqmalespGametesFrom:RRWWRWRWRRWWHardy-Weinberg Equilibrium•At equilibrium we expect genotypic frequencies of:(p + q)[from females] × (p + q)[from males] = 1 = p2 + 2pq + q2 = 1p(R) = 0.6 q(W) = 0.4Expected genotypes in next generation:x(RR) = p2 = 0.6 × 0.6 = 0.36y(RW) = 2pq = 2 × 0.6 × 0.4 = 0.48z(WW) = q2 = 0.4 × 0.4 = 0.16These are the exact same values we got through the step-by-step analysis.Does HWE hold for more than two alleles?•YES•Example: Three alleles A, B, & C with frequencies p, q, & r (respectively)•p + q + r = 1•(p + q + r) × (p + q + r) = 1p2 + pq + pr + qp + q2 + qr + rp + rq + r2= 1p2 + 2pq + 2pr + q2 + 2qr + r2 = 1 AA AB AC BB BC CC•Infinite population size•No mutation•No selection•Closed population (no gene flow)•Random matingAssumptionsHow can we tell if a population is in equilibrium?•Prediction:–If the population is in equilibrium, observed and expected values should be the same•How do we test


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ASU MAT 294 - Introduction3

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