2006problemset6key

# 2006problemset6key - BICD100 Genetics W'06 Problem Set#6...

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BICD100: Genetics W’06 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). Suppose that you are interested in determining whether or not random mating occurs within a wild population of pea plants where the difference between purple and white flower color is controlled by alleles of a single gene, C. Homozygous cc plants have white flowers whereas CC and Cc plants have purple flowers. In an analysis of this population, you find 32 plants with white flowers, and 168 plants with purple flowers. When the purple flower plants were selfed, 132 were true-breeding for flower color and the remaining 36 were not. This info tells you that f(aa) = 32/200 = 0.16, f(Aa) = 36/200 = 0.18, and f(AA) = 132/200 = 0.66 a. What is the frequency of the C allele? the c allele? f(C) = f(CC) + 1/2 f(Cc) = 0.66 + (1/2)(0.18) = 0.75 = p

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## This note was uploaded on 04/17/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

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2006problemset6key - BICD100 Genetics W'06 Problem Set#6...

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