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Unformatted text preview: p+q = 1 A rare recessive disorder has a frequency of 1/10,000. Which of the following most closely approximates the frequency of heterozygotes in this population? a) 28% b) 10% c) 5% d) 2% e) 1% p+q = 1 A rare recessive disorder has a frequency of 1/10,000. Which of the following most closely approximates the frequency of heterozygotes in this population? q 2 = 0.0001, q=0.01, 2pq=2q=0.02 a) 28% b) 10% c) 5% d) 2% e) 1% HardyWeinberg Equilibrium • allows calculation of predicted genotype frequencies in a population • allows probability calculations using Product Rule • population might not be in HWE if there are major deviations from assumptions Hardy  Weinberg Equilibrium freq(A 1 ) = p 1 freq(A 2 ) = p 2 A 1 A 1 A 1 A 2 A 2 A 2 p 1 2 p 2 2 2p 1 p 2 A 1 A 2 A 1 A 2 A 1 A 2 A 1 A 2 A 2 A 2 A 1 A 1 p 1 2 p 1 p 2 p 1 p 2 p 2 2 (p 1 + p 2 ) 2 = p 1 2 + 2p 1 p 2 + p 2 2 Tests to Determine if Allele Frequency Data Meet H/W Expectations Exact test Likelihood ratio test Heterozygote/homozygote ratio Chisquare Which are in HW Equilibrium? A) 55AA, 10Aa, 35aa B) 36AA, 48Aa, 16aa C) 100AA D) 81AA, 18Aa, 1aa E) 25AA, 50Aa, 25aa Estimating Xlinked allele frequencies from population data q= 2C+B+CY/2N f +N m With variance of q(Vq)=q(1q)/(2Nf+Nm) C=#recessive females B=#heterozygous females CY=#recessive males N f =#females observed N m =#males observed EB Speiss, Genes in Populations 2locus genotype observed frequencies BB Bb bb Sum AA 72 284 284 640 p=.8,q=.2 Aa 36 142 142 320 u=.33,v=.67 aa 4 18 18 40 Sum 112 444 444 1000 2locus genotypes expected gametic output AB= 72+1/2(284+36)+1/4(142) = 267.5 4/15 Ab= 284+1/2(284+142)+1/4(142)=532.5 8/15 aB= 4+1/2(36+18)+1/4(142) = 66.5 1/15 ab= 18+1/2(18+142)+1/4(142) =133.5 2/15 g= 4/15 8/15 0.0711 0.2844 0.2844 1/15 2/15 zygote matrix (Z) = 0.0356 0.1422 0.1422 0.0044 0.0177 0.0177 Result: expected values closely approximate observed values, therefore 2 loci are in linkage equilibrium Result: can use product rule to compute overall 2locus combined genotype frequency Approach to equilibrium • If allelic frequencies are not equal in males and females, for autosomal genes, the average of two parent frequencies is established after random mating among first generation progeny More complicated for X linked genes, as • 1) progeny males’ X chromosome allele frequencies are determined by those of their mothers • 2) average of two parents is NOT simply half the sum of the parent’s , but is average weighted by dosage of X chromosomes (2X+1X/3) Approach to equilibrium More complicated for X linked genes, as • 1) progeny males’ X chromosome allele frequencies are determined by those of their mothers • 2) average of two parents is NOT simply half the sum of the parent’s , but is average weighted by dosage of X chromosomes (2X+1X/3) • Progeny males’ f(A) : r’ A = p A • Progeny females’ f(A) : p’ A =p A +r A /2 Xlinked recessive traits • More common among males (q) than among females (q 2 )...
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 Spring '09
 Genetics, Population Genetics, population allele frequencies

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