Class03-ClassSlides-Spring 2011-03222011

Class03-ClassSlides-Spring 2011-03222011 - Genetic Analysis...

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1 Genetic Analysis II (385) – Population Genetics Lecture 03 Date: 03/22/11 Instructor: Professor Derek Gordon Office: Life Sciences Building, Room 128 Phone: 445-3386 Phone: 445 3386 E-mail: gordon@biology.rutgers.edu Office Hours: Tues, Thurs: 12:00pm – 1:15pm, and by appt. Assessment question Is it possible for a locus with more than two alleles to be in Hardy Weinberg Equilibrium? Is it possible for a gamete containing more than one locus to achieve Hardy Weinberg Equilibrium? 3/22/2011 PopGen 3 - 2011 2
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2 Assessment question Is it possible for a locus with more than two alleles to be in Hardy Weinberg Equilibrium? YES! Is it possible for a gamete containing more than one locus to achieve Hardy Weinberg Equilibrium? YES! 3/22/2011 PopGen 3 - 2011 3 HARDY-WEINBERG EQUILIBRIUM WITH 3 OR MORE ALLELES AT A LOCUS ALLELES allele array 2 p + q = 1 3 p + q + r = 1 4 p+q+r+s=1 3/22/2011 PopGen 3 - 2011 4 p + q + r + s = 1 5 p + q + r + s + t = 1
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3 HARDY-WEINBERG EQUILIBRIUM WITH 3 OR MORE ALLELES AT A LOCUS. In every case, the expected genotype array at HWE is obtained by squaring the allele array. For example, for obtained by squaring the allele array. For example, for three alleles: (p + q +r) 2 = p 2 (A 1 A 1 ) + 2pq(A 1 A 2 ) + 2pr(A 1 A 3 ) + q 2 (A 2 A 2 ) + 2qr(A 2 A 3 ) + r 2 (A 3 A 3 ) The general rule for calculating the allele array (allele 3/22/2011 PopGen 3 - 2011 5 The general rule for calculating the allele array (allele frequencies) from the observed genotype array is: HARDY-WEINBERG EQUILIBRIUM WITH 3 OR MORE ALLELES AT A LOCUS. p x = f(A x A x ) + 1/2 f(A x A y ) where p x is the freq. of allele A x and f(A x A x ) and f(A x A y ) are the homozygote and and and are the homozygote and heterozygote genotypes containing the allele A x . * 3/22/2011 PopGen 3 - 2011 6
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4 HARDY-WEINBERG EQUILIBRIUM WITH 3 OR MORE ALLELES AT A LOCUS. A 1 A 1 A 1 A 2 A 2 A 2 A 1 A 3 A 2 A 3 A 3 A 3 .10 .20 .30 .30 .08 .02 = 1.0 What are the frequencies of A 1 , A 2 and A 3 ? 3/22/2011 PopGen 3 - 2011 7 I'll do the first one for you. f( A 1 ) = p = f( A 1 A 1 ) + 1/2(f( A 1 A 2 )) + 1/2(f( A 1 A 3 )) = .10 + 1/2(.20) + 1/2(.30) = .35 3/22/2011 PopGen 3 - 2011 8
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5 Your turn f(A 2 ) = q = = f(A 2 A 2 ) + 1/2(f(A 1 A 2 )) + 1/2(f(A 2 A 3 ), f(A ) 1/2(f(A )) 1/2(f(A = = f (A 3 ) = r = = f(A 3 A 3 ) + 1/2(f(A 1 A 3 )) + 1/2(f(A 2 A 3 ), = 3/22/2011 PopGen 3 - 2011 9 = Check: p + q + r = HARDY-WEINBERG EQUILIBRIUM WITH TWO LOCI For two or more loci, even under conditions of panmixia , the approach to Hardy-Weinberg equilibrium is gradual. the approach to Hardy Weinberg equilibrium is gradual. Each locus considered separately will be in HWE; however, if the two loci are considered jointly they are in HWE disequilibrium . *Essential definition 3/22/2011 PopGen 3 - 2011 10
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6 HARDY-WEINBERG EQUILIBRIUM WITH TWO LOCI Consider two loci; A and B, each with two alleles A, a and B, b B, b The allele arrays are : p(A) + q(a) = 1. 3/22/2011 PopGen 3 - 2011 11 and r(B) + s(b) = 1. HARDY-WEINBERG EQUILIBRIUM WITH TWO LOCI We now need to develop the gametic array. The gametic array is the set of frequencies of all possible gametes.
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This note was uploaded on 04/04/2011 for the course GENETICS 385 taught by Professor Brennemanandgordon during the Spring '11 term at Rutgers.

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Class03-ClassSlides-Spring 2011-03222011 - Genetic Analysis...

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