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Unformatted text preview: there is a sample of 20 people that reveals the following genotypes: (RR) (Rr) (RR) (Rr) (rr) (Rr) (RR) (RR) (Rr) (RR) (Rr) (rr) (Rr) (Rr) (RR) (RR) (Rr) (RR) (rr) (Rr) a. What percentage of the people are right handed? Left handed? 17/20 are right handed so 85% 3/20 are left handed so 15% b. Find p and q and interpret each in the context of the problem. p = 25/40 = .625 62.5% of the alleles in the sample are dominant R. q = 15/40 = .375 37.5% of the alleles in the sample are recessive r. Note: p and q are NOT telling us the percent of people that are right and left handed. They refer the allele frequency Now suppose that we took another sample of 10 people. This time we only know their phenotypes. (Right) (Left) (Right) (Right) (Right) (Right) (Right) (Right) (Left) (Right) c. What percentage of the people are right handed? Left handed? Right handed = 8/10 = 80% Left handed = 2/10 = 20% c. Can you find p and q exactly? Why? No. We can’t see the genotypes of each person. For each right handed person, we do not know if they are homozygous (RR) or heterozygous (Rr). d. Estimate p and q and interpret each in the context of the problem. q 2 = .20 so q = .447 We estimate that 44.7% of the alleles in the sample are recessive r. p + q = 1 so p = .553 We estimate that 55.3% of the alleles in the sample are dominant R. e. Estimate how many of the right handed people are homozygous and how many are heterozygous. Homozygous = RR = p 2 so (.553) 2 = .305 .305 X 10 = 3.05 so we would guess 3 are homozygous. Heterozygous = Rr = 2pq = 2(.553)(.447) = .494 X 10 = 4.94 so we would guess 5 are heterozygous. 3 homozygous + 5 heterozygous = 8 right handed people. Formulas: p 2 + 2pq + q 2 = 1 p = frequency of the dominant allele in a population p + q = 1 q = frequency of the recessive allele in a population Example problem: In 1990 the East Kentwood High School student body was made up of 90% right handed students. Being right handed (R) is the dominant trait over being left handed (r). a. What is p and q for the population of 1990 East Kentwood High School students. Interpret each. 90% right handed means 10% are left handed. q 2 = .10 so q = .316. p + q = 1 so p = .684. p = .684 68.4% of all the alleles at EKHS in 1990 are R. q = .316 31.6% of all the alleles at EKHS in 1990 are r. b. Find the percent of the student body in 1990 that are homozygous right handed, heterozygous right handed, and left handed. p 2 = (.684) 2 = .468 so 46.8% of students are homozygous right handed (RR) 2pq = 2(.316)(.684) = .432 so 43.2% of students are heterozygous right handed (Rr) q 2 = (.316) 2 = .10 so 10% of students are left handed (rr) Fast forward to today at East Kentwood. Mr. V. took a random sample of 100 East Kentwood students today and found that 18 of them were left handed....
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 Fall '11
 Boyle
 Biology, Population Ecology, Lefthandedness, Zygosity

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