FA2008 - BILD 3 - Lecture 05 (Population Genetics - 02)

FA2008 - BILD 3 - Lecture 05 (Population Genetics - 02) -...

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10/16/08 1 Lecture 5 The Genetics of Populations Reading Ch 23 Is a population in Hardy- Weinberg? 1) genotype frequencies (text p. 475, solution A-19) # frequencies AA – 16 16/120 = 0.13 Aa – 92 92/120 = 0.77 aa – 12 12/120 = 0.10 Total – 120 1.00 1) genotype frequencies (text p. 475, solution A-19) # frequencies AA – 16 16/120 = 0.13 Aa – 92 92/120 = 0.77 aa – 12 12/120 = 0.10 Total – 120 1.00 2) calculate allele frequencies Frequency A (p) = freq (AA) + ½ frequency (Aa) = (0.13) + ½ (0.77) = 0.52 = p Frequency of a (q) = freq (aa) + ½ frequency (Aa) = (0.10) + ½ (0.77) = 0.48 = q 1) genotype frequencies (text p. 475, solution A-19) # frequencies AA – 16 16/120 = 0.13 Aa – 92 92/120 = 0.77 aa – 12 12/120 = 0.10 Total – 120 1.00 2) calculate allele frequencies Frequency A (p) = freq (AA) + ½ frequency (Aa) = (0.13) + ½ (0.77) = 0.52 = p Frequency of a (q) = freq (aa) + ½ frequency (Aa) = (0.10) + ½ (0.77) = 0.48 = q 3) expected (H-W) vs. observed AA (p 2 ) Aa (2pq) aa (q 2 ) exp’d freq. (0.52) 2 = 0.27 2(0.52)(0.48) = 0.50 (0.48) 2 = 0.23 exp’d # (0.27)(120) = 32 (0.50)(120) = 60 (0.23)(120) = 28 observed # 16 92 12 Simulation of drawing an allele from a gene pool True frequency of allele is 0.6. When the number of samples (zygotes) is small,
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This note was uploaded on 03/31/2009 for the course BILD 632282 taught by Professor Henter during the Spring '08 term at UCSD.

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FA2008 - BILD 3 - Lecture 05 (Population Genetics - 02) -...

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