Problem Set 6 Key

Problem Set 6 Key - BICD100 FA2011 Problem Set#6 KEY 1 In humans the drug isoniazid is used to treat tuberculosis The drug is inactivated in the

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BICD100 FA2011 Problem Set #6 KEY 1) . In humans, the drug isoniazid is used to treat tuberculosis. The drug is inactivated in the liver by the enzyme acetyl transferase. Two alleles of the gene for this enzyme exist in human populations. Enzymes encoded by the a s and a r alleles inactivate the drug slowly and rapidly, respectively. Homozygous a s a s individuals respond well to isoniazid treatment, a r a s heterozygotes less so, and a r a r homozygotes respond poorly. Reponse rates to isoniazid treatment for tuberculosis in different racial groups in a 1964 study is as follows: RESPONSE TO ISONIAZID population good intermediate poor total # individuals Japanese 20 81 108 209 Caucasians 61 37 7 105 Use the chi squared test to determine whether each of these populations is in Hardy-Weinberg equilibrium with respect to acetyl transferase alleles. Let a r a r (poor responder) be AA, let a s a s (good responder) be aa, and a r a s (intermediate) be Aa. q = f(aa) + 1/2f(Aa) and p = 1 – q Hypothesis: population is in H-W equilibrium -> expect f(aa) = q 2 , f(Aa) = 2pq, f(AA) = p 2 For Japanese, q = 0.29 so q 2 = 0.08 X 209 -> expect ~17 good responders p = 0.71 so p 2 = 0.50 X 209 -> expect ~105 poor responders 2pq = 2(0.29)(0.71) X 209 -> expect ~86 intermediate responders χ 2 = (O – E) 2 /E = (20 – 17) 2 /17 + (81 – 86) 2 /86 + (108 – 105) 2 /105 = 0.906 At one degree of freedom, 0.5 > p > 0.1 so accept hypothesis that population is in H-W equilibrium. For Caucasian population, follow exactly the same process -> χ 2 = 0.287 so 0.9 > p > 0.5 so accept hypothesis for this population too. 2). Warfarin is a rat poison that acts by hindering blood coagulation. Heavy use of the poison has resulted in the evolution of resistance in most rat populations due to a mutant allele R (call the normal, sensitive allele S). S mutates to R at a rate of 1 x 10 -6 . In the absence of warfarin, SS and RS rats both have a fitness of 1, while the fitness of RR homozygotes is 0.46. In the presence of warfarin, the relative fitnesses of SS, RS and RR genotypes are 0.68, 1.00 and 0.37, respectively. Among rats populating the NY subway system, the frequency of the R allele is 0.05. a. Assuming random mating among these rats, if warfarin use were totally discontinued, what would be the expected frequency of rats of each genotype at equilibrium due to mutation/selection balance? In absence of warfarin, R is recessive, and W for RR genotype is 0.46 so s = 0.54. q equil = √μ /s = (1 x 10 -6 )/0.54 = 0.0014 so p equil = 0.9986 f(SS) = p 2 = 0.9972 f(RS) = 2pq = 0.0028 f(RR) = q 2 = 1.9 x 10 -6 b. Assuming random mating among these rats, if a massive rat eradication program is lauched 1
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This note was uploaded on 01/16/2012 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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Problem Set 6 Key - BICD100 FA2011 Problem Set#6 KEY 1 In humans the drug isoniazid is used to treat tuberculosis The drug is inactivated in the

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