Population Genetics

P q = 1 a rare recessive disorder has a frequency of

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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% Hardy-Weinberg 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 Chi-square Which are in HW Equilibrium? A) 55AA, 10Aa, 35aa B) 36AA, 48Aa, 16aa C) 100AA D) 81AA, 18Aa, 1aa E) 25AA, 50Aa, 25aa Estimating X-linked allele frequencies from population data q= 2C+B+CY/2N f +N m With variance of q(Vq)=q(1-q)/(2Nf+Nm) C=#recessive females B=#heterozygous females CY=#recessive males N f =#females observed N m =#males observed EB Speiss, Genes in Populations 2-locus 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 2-locus 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 2-locus 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 X-linked recessive traits • More common among males (q) than among females (q 2 )...
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p q = 1 A rare recessive disorder has a frequency of...

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