2008problemset7key

2008problemset7key - BICD100W06 ProblemSet#7KEY

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
BICD100 W’06 Problem Set #7 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 f(c) = 1 – p = 0.25 = q 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
b. Is the population in Hardy-Weinberg equilibrium? Indicate your hypothesis and the values of chi- square and p. Hypothesis: population in H-W equilibrium. If so, expect f(cc) = q
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/14/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

Page1 / 5

2008problemset7key - BICD100W06 ProblemSet#7KEY

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online