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Lect 26 - 1 Population Genetics Bio Sci 97 Lecture 26 Dr...

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©2003 C. Brachmann 1 Population Genetics Bio Sci 97 Lecture 26 Dr. Carrie Brachmann [email protected] reading lect 27: H& J ch 15
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©2003 C. Brachmann 2 Outline Hardy Weinberg equation Assumptions for H-W H-W and multiple alleles and X- linkage Deviations from H-W: inbreeding Relatedness
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©2003 C. Brachmann 3 Population genetics is the application of Mendelian principles to gene frequencies in populations
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©2003 C. Brachmann 4 Some terms Population: a group of organisms of the same species living within a prescribed geographic location Sub-population: a population defined using smaller geographic locations or other criteria such as race or ethnicity Gene pool: all alleles that occur in a given population--the complete set of genetic information Genotype frequencies: proportion (or relative abundance) of individuals in a population that have a particular genotype Allele frequencies: proportion (or relative abundance) of a specific allele in a population
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©2003 C. Brachmann 5 Population genetics is the application of Mendelian principles to gene frequencies in populations
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©2003 C. Brachmann 6 A first look at allele frequencies A population of cattle segregating N and R alleles 756 white cattle NN: 1512 N alleles 3780 roan cattle NR: 3780 N alleles + 3780 R alleles 4169 red cattle RR: 8338 R alleles total 5292 N alleles + 12118 R alleles (17410) Frequency of N = 5292/ 17410 = .304 Frequency of R = 12118/ 17410 = .696
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©2003 C. Brachmann 7 Or, one could first calculate genotype frequencies White NN 756/ 8705 = .0869 Roan RN 3780/ 8705 = .4342 Red RR 4169/ 8705 = .4789 1.0000 then calculate the allele frequencies Frequency of N = .0869 + (0.5)(.4342) = .304 Frequency of R = .4789 + (0.5)(.4342) = .696 Frequency of N = (2 x 756) + 3780 = 5292/ 17410 = .304 Frequency of R = (2 x 4169) + 3780 = 12118/ 17410 = .696
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©2003 C. Brachmann 8 Hardy-Weinberg Principle If mating in a population occurs at random, the genotype frequencies and therefore the allele frequencies remain constant from generation to generation
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©2003 C. Brachmann 9 Hardy-Weinberg equation p = frequency of allele 1 (dominant allele) q = frequency of allele 2 (recessive allele) p + q = 1 Example - Gene pool of R and N alleles in cows Frequency of N = q = 5592/ 17410 = .304 Frequency of R = p = 12118/ 17410 = .696 p + q = .304 + .696 = 1.000
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©2003 C. Brachmann 10 Hardy-Weinberg equation can predict the frequency of different genotypes in the next generation New example from H& J: A gene with two alleles in the population: A and a . 1000 people: 795 AA, 190 Aa, 15 aa allele freq. of A = .89, allele freq. of a = .11 In the population, random mating (“like drawing female and male gametes at random from a large container”) is the random union of gametes. Gametes are either A or a and are male or female
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©2003 C. Brachmann 11 A population mates p 2 AA a Female gametes Male gametes ©1999 Lee Bardwell A A q p a p q pq Aa q 2 aa pq Aa (p+q) 2 = p 2 +2pq+q 2 = 1 Genotype freq. of aa genotype is q 2
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©2003 C. Brachmann 12 A population mates p 2 AA a Female gametes Male gametes ©1999 Lee Bardwell A A a p=.89 q=.11 pq Aa q 2 aa pq Aa p 2 + 2pq + q 2 = 1 .7921 + .1958 + .0121 = 1 .7921 .0979 .0121 .0979 p=.89 q=.11 AA Aa aa
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©2003 C.
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