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POPULATION GENETICS ● Population Genetics is the study of genetics at the population level ● Mendelian Population is a group of sexually reproducing organisms.

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Presentation on theme: "POPULATION GENETICS ● Population Genetics is the study of genetics at the population level ● Mendelian Population is a group of sexually reproducing organisms."— Presentation transcript:

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2 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 evolution

3 Probability 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?

4 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 snapdragons

5 What 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

6 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 = 1 What will the proportions of these colors be in the next generation?

7 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?

8 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 occur What will the proportions of these colors be in the next generation?

9 Perhaps a table would be helpful... RR (0.45)RW (0.30)WW (0.25) RR (0.45) RW (0.30) WW (0.25) 0.2025 0.1350 0.1125 0.1350 0.0900 0.0750 0.1125 0.0750 0.0625 Males 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 time Note: All the cells add up to 1!

10 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?

11 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% WW

12 Probability of each genotype in the offspring PARENTS MATING (type)FREQUENCYRRRWWW RR x RR RR x RW RR x WW RW x RW RW x WW WW x WW RESULTING GENOTYPE OF OFFSPRING

13 RR (0.45)RW (0.30)WW (0.25) RR (0.45)0.2025 0.1350 0.1125 RW (0.30)0.1350 0.0900 0.0750 WW (0.25)0.1125 0.0750 0.0625 Males Females

14 Probability of each genotype in the offspring PARENTS MATING (type)FREQUENCYRRRWWW RR x RR0.2025 RR x RW RR x WW RW x RW RW x WW WW x WW GENOTYPE FREQUENCY OF RESULTING OFFSPRING

15 RR (0.45)RW (0.30)WW (0.25) RR (0.45)0.2025 0.1350 0.1125 RW (0.30)0.1350 0.0900 0.0750 WW (0.25)0.1125 0.0750 0.0625 Males Females How do we combine these? (AND or OR) is the question: 1.Probability: RW male & RR female AND RR male & RW female 2.Probability: RW male & RR female OR RR male & RW female OR = add the probabilities

16 Probability of each genotype in the offspring PARENTS MATING (type)FREQUENCYRRRWWW RR x RR0.2025 RR x RW.135 +.135 =.27.5 .27 =.135 RR x WW RW x RW RW x WW WW x WW 0.225 0.09.25 .09=.0225.5 .09=.045.25 .09=.0225 0.15.5 .15=.075 0.0625 10.360.480.16 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) = 0.16

17 What 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 check


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