Basic concepts of population genetics - Bioversity International

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... with a codominant gene having multiple alleles Genotypes Genotype frequencies (expected) Number of individuals Genotype frequencies (observed) A 1 A 1 p 1 2 n 11 P 11 = n 11 /n A 1 A 2 2p 1 p 2 n 12 P 12 = n 12 /n p 1 = P 11 + ½ i 1 P 1j p 3 = P 33 + ½ j 3 P 3j A 1 A 3 2p 1 p 3 n 13 P 13 = n 13 /n p n = P nn + ½ j n P nj Copyright: IPGRI and Cornell University, 2003 Population genetics 26 26 A 2 A 2 p 2 2 n 22 P 22 = n 22 /n A 2 A 3 2p 2 p 3 n 23 P 23 = n 23 /n p 2 = P 22 + ½ j 2 P 2j p 4 = P 44 + ½ j 4 P 4j ... ... ... ... A n A n p n 2 n nn P nn = n nn /n Total (continued on next slide) This is the situation encountered when markers such as microsatellites are used. We have a locus A with n alleles A 1 , A 2 , A 3 , ..., A n and allele frequencies p 1 , p 2 , p 3 , ..., p n , respectively, such that A 1 = A 2 = A 3 = ... = A n 1 n 1
... with a codominant gene having multiple alleles (continued) Locus A Gel Ind.1 Ind.2 Ind.3 Ind.4 Ind.5 Ind.6 A1 1 1 1 0 0 0 A2 0 1 0 1 1 0 A3 0 0 1 0 1 1 M Genotypes A 1 A 1 Individuals 1 2 3 4 5 6 A 1A 2 A 1 A 3 Scoring bands Copyright: IPGRI and Cornell University, 2003 Population genetics 27 27 A 2A 2 Locus A A 2 A 3 A 3A 3 M 1 2 3 4 5 6 (1,0,0) (1,1,0) (1,0,1) (0,1,0) (0,1,1) (0,0,1) Again, with a codominant marker, the genotypes of the three genotypic classes can be observed. In the drawing above, top centre, we see a gel image with the banding pattern of a codominant marker with three alleles (A 1 , A 2 and A 3 ) in a diploid sample. We score each band (each row) independently, and transform them to a score of 1 if present or a score of 0 if not. We can do it by band (bottom left of slide) or by genotype (bottom right corner). In the table below, we can see the calculations of the expected and observed genotype frequencies, as well as the allele frequencies (p 1 , p 2 and p 3 ). (M = size marker.) Genotypes Genotype frequency (exp.) Number of individuals Genotype frequency (obs.) A 1 A 1 p 1 2 n 11 = 4 P 11 = n 11 /n = 0.17 A 1 A 2 2p 1 p 2 n 12 = 6 P 12 = n 12 /n = 0.25 A 1 A 3 2p 1 p 3 n 13 = 0 P 13 = n 13 /n = 0 A 2 A 2 p 2 2 n 22 = 10 P 22 = n 22 /n = 0.42 A 2 A 3 2p 2 p 3 n 23 = 2 P 23 = n 23 /n = 0.08 A 3 A 3 p 3 2 n 33 = 2 P 33 = n 33 /n = 0.08 p 1 = P 11 + ½P 12 + ½P 13 = P 11 + ½ j 1 P 1j = 0.17 + ½(0.25 + 0.00) = 0.30 p 2 = P 22 + ½P 21 + ½P 23 = P 22 + ½ j 2 P 2j = 0.42 + ½(0.25 + 0.08) = 0.59 p 3 = P 33 + ½P 31 + ½P 32 = P 33 + ½ j 3 P 3j = 0.08 + ½(0.00 + 0.08) = 0.12 Total 1 n = 24 1
Page 1 and 2: Genetic diversity analysis with mol
Page 3 and 4: Definitions: Population genetics T
Page 5 and 6: Phenotypic variation Pink Pale crea
Page 7 and 8: Genotype is … The description of
Page 9 and 10: What is polymorphism? ‘Presence
Page 11 and 12: Population structure Three levels o
Page 13 and 14: Allele frequency Allele frequency
Page 15 and 16: Genotype frequency This is the fre
Page 17 and 18: Demonstrating the H-W principle Gen
Page 19 and 20: The chi-square test This hypothesi
Page 21 and 22: Calculating allele frequencies: Exa
Page 23 and 24: Locus A B D … with a codominant m
Page 25: Locus A Locus B Locus D … with a
Page 29 and 30: Random mating Mating that takes pl
Page 31 and 32: Inbreeding coefficient It compares
Page 33 and 34: Forces shaping genetic diversity M
Page 35 and 36: Migration It is the movement of in
Page 37 and 38: Selection It is the inherited abili
Page 39 and 40: Effective population size Ne is th
Page 41 and 42: Consequences … (continued) A bot
Page 43 and 44: Appendix 1 Copyright: IPGRI and Cor
Page 45 and 46: By now you should know … The bas
Page 47:
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