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Unformatted text preview: p 2 (calculated) = 0.30 p 2 (actual) = 20,000/100,000 = 0.2 2pq(calculated) = 2(0.45)0.55 = 0.50 2pq(actual) = 60,000/100,000 = 0.6 Answer: No Or you could use observed allele frequencies to calculate expected genotype frequencies and compare these to actual genotype frequencies, as below: q = 100,000/200,000 = 0.5; p = 0.5 expected genotype frequencies: q 2 = 0.5 2 = 0.25; expected homozygous sickle cell is 0.25x100,000, or 25,000 2pq = 2 x 0.5 x 0.5x = 0.5; expected heterozygotes: 0.5 x 100,000, or 50,000 p 2 = 0.5 2 = 0.25; expected homozygous normal is 0.25x100,000, or 25,000 compare actual to calculated genotypes: 25,000 vs 20,000 sickle cell sufferers; 50,000 vs 60,000 hets, 25,000 vs 20,000 normal Answer: No...
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This note was uploaded on 02/15/2012 for the course PMB 13 taught by Professor Freeling during the Spring '09 term at University of California, Berkeley.
 Spring '09
 Freeling
 Evolution

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