Slide 1POPULATION GENETICSProbability works for individuals but how about populations?DefinitionsWhat will the proportions of these colors be in the next generation?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 have we done so far?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 occurPerhaps a table would be helpful...Slide 10Probability of each genotype in the offspringSlide 12Slide 13Slide 14Slide 15Slide 16What are the allele frequencies?POPULATION GENETICS●Population Genetics is the study of genetics at the population level●Mendelian Population is a group of sexually reproducing organisms with a close degree of genetic relationship●Gene Pool is a mixture of the genetic units (Genes or Gametes) produced by a Mendelian population from which the next generation arises. Alleles occur in this pool●Evolution: Through events such as natural selection, migration, or mutation, the gene pool changes as new alleles enter or existing alleles exit the pool. These changes are the basis for evolutionProbability 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 in the 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 in the 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 in the 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 in the 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