populationGenetics

# populationGenetics - The Genetic Analysis of Populations...

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1 The Genetic Analysis of Populations and How They Evolve

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2 Outline of Chapter 21 The Hardy-Weinberg law A model for understanding allele, genotype, and phenotype frequencies for single gene traits in a genetically stable population Calculations beyond Hardy-Weinberg Measuring how selection and mutations change allele frequencies over time How to use tools of quantitative genetics to analyze inheritance of multifactorial traits
3 The Hardy-Weinberg law clarifies the relations between genotype and allele frequency within a generation and from one generation to the next. Five assumptions Infinitely large population Individuals mate at random. No new mutations appear in gene pool. No migration into or out of population No genotype-dependent differences in ability to survive and reproduce If all assumptions hold, population is in Hardy-Weinberg equilibrium.

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4 All natural populations violate one or more assumptions of Hardy-Weinberg law. However, equations derived based on assumptions are remarkably robust. Hardy-Weinberg law can be used as a null model.
5 Albanism example Population of 100,000 people 100 aa albinos = 200 a alleles 1,800 Aa carriers = 1,800 A alleles and 1,800 a alleles 98,100 AA individuals = 196,200 A alleles A allele frequency is 198,000/200,000 = 0.99 p = 0.99 a allele frequency is 2,000/200,000 = 0.01 • q = 0.01 This is also the allele frequencies of the gametes. Hardy-Weinberg equation for population p 2 + 2pq + q 2 = (0.99) 2 + 2(0.99 x 0.01) + (0.01) 2 = 0.9801 + 0.0198 + 0.0001 = 1 Next Generation of 100,000 people 100,000 X 0.9801 AA individuals 100,000 X 0.0198 Aa individuals 100,000 X 0.0001 aa individuals

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6 Frequency of p and q allele in next generation p + q = 1 Thus the frequency of the p allele in the next generation is • p 2 + ½[2p(1-p)] = p 2 + p(1-p) = p 2 + p – p 2 and the frequency of the q allele in the next generation is • q 2 + ½[2q (1-q)] = q 2 + q (1-q) = q 2 + q – q 2 For albinism where p = 0.99 and q = 0.01 The frequency of A allele in second generation is 0.98 + 0.99 – 0.98 = 0.99 The frequency of the a allele is 0.0001 + 0.01 – 0.0001 = 0.01 Genotype frequencies changed, but allele frequencies stay the same for both dominant and recessive alleles.
7 Calculating the frequency of heterozygous carriers when you only know the frequency of diseased individuals for a recessive trait PKU – phenylketonuria Autosomal recessive mutation 1 in 3600 Caucasians in USA have PKU q 2 = 1/3600 q = √q 2 = √1/3600 = 0.0167 p = 1-q = 0.9833 2pq = 2 X 0.0167 X 0.9833 = 0.0328 Frequency of carriers is thus about 3.3%

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8 Measuring how mutation and selection
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populationGenetics - The Genetic Analysis of Populations...

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