PS 2 answers, probs 9+10+11

PS 2 answers, probs 9+10+11 - p 2 (calculated) = 0.30 p 2...

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Problem 9. Assuming the population is at Hardy-Weinberg equilibrium, how many heterozygous tongue rollers are there in the population? Let p = allele frequency for rolling Let q = allele frequency for non-rolling q 2 = 2600/10,000 = 0.26 q = 0.51 p = 1 - q = 0.49 2pq = 2(0.49)0.51 = 0.4998 Number of heterozygous rollers = 0.4998(10,000) = 4998 (5000 would be close enough) Problem 10. 1. q = square root of 1/10,000 = 0.0100 2. 2pq = 0.0198 = 1.98% of U.S. population Problem 11. Is the baby population at Hardy-Weinberg equilibrium? There are two ways to do this problem. You can calculate expected allele frequencies from genotype frequencies and compare them to actual allele frequencies, as below: q 2 = 20,000/100,000 = 0.2 q = 0.45 p = 1 - q = 0.55
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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.

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