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Name: In an analysis of five rII mutants, complementation testing yielded th
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Description: In an analysis of five rII mutants, complementation testing yielded the following results: Mutants tested1,21,31,41,5Which would be a possible arrangement of mutations on the two rll genes (A and B)?(1, 2, 3) and (4, 5)(1, 4, 5) and (2, 3)+Results (c

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Mutants tested 1,2 1,3 1,4 1,5 Which would be a possible arrangement of mutations on the two rll genes (A and B)? (1, 2, 3) and (4, 5) (1, 4, 5) and (2, 3) Results (complementation). + +

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter15: From Dna To Protein

Section: Chapter Questions

Problem 15TYK

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! Required information Hfr (thrt, azi, ton", lact, galt, str) and F¯ (thr¯, azi³, ton³, lac, ga, str) cells were mated in an interrupted mating experiment. The results of replica plating the Str Thr* exconjugants to test for the presence of the other four genes are shown in the graph. Percentage of cells with Hfr genetic markers among Str' Thr* recombinants 100 80 60 40 20 0 0 Multiple Choice I 30 azi' lac+ gal+ I 10 20 40 Minutes prior to interruption of conjugation ton' gene order could not be determined. I 50 the fraction of cells with each marker would be 100%. 90% 85% If a sample had been taken at 70 minutes after mating instead of at 1-minute intervals, then 40% 20% gene order could be determined because the final fraction of exconjugants with a marker decreases as distance from the origin of transfer increases. gene order could be determined because the final fraction of exconjugants with a marker increases with distance from the origin of transfer.
Complementation tests of recessive mutants a through f produced the data shown in the table below. "+" means wild-type phenotype; "-" means mutant phenotype. PSCO a b с 0 0 0 0 0 a e a f b + C + d e + + + + + - d e f A new mutant, g, is isolated and it fails to complement c. Which other mutant(s) would g also fail to complement? (Select all correct answers.) ++ f ++ + + -
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Practice Question 2 The beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 3'...TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...5' 5'..AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCOCTACGTGATAG...3' promoter A) Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3'
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Question Choose forward and reverse primers to amplify gene X 5'ATTAGTCATGCCAACTTGCACTGATATGA...geneX...CCAGGCGCTAACCTAGATCGCTAGTATCG3' Forward primer: A. 5'-AGTCACGTTCAACCGTACTG-3' B. 5'-GTCATGCCAACTTGCACTGA-3' C. 5'-TCAGTGCAAGTTGGCATGAC-3' D. 5'-CAGTACGGTTGAACGTGACT-3' E. other? Reverse primer: A. 5'-TGATCGCTAGATCCAATCGC-3' B. 5'-CGCTAACCTAGATCGCTAGT-3' C. 5'-ACTAGCGATCTAGGTTAGCG-3' D. 5'-GCGATTGGATCTAGCGATCA-3' E. other? * Sequences above are 20mers with 50%GC.
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15. Several mutants are isolated, all of which require com- pound G for growth. The compounds (A to E) in the bio- synthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. olord bos og elu Compound tested А В B CDEG Mutanf 1 a. What is the order of compounds A to E in the pathway? b. At which point in the pathway is each mutant blocked? c. Would a heterokaryon composed of double mutants 1,3 and 2,4 grow on a minimal medium? Would 1,3 and 3,4? Would 1,2 and 2,4 and 1,4? + + + + 3 45