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35. Enzyme A on 11; B on 15; C on 18; D on 3; E on 7. Enzyme A is present in clones X and Y, and chromosome 11 is common to these two clones. Enzyme B is present only in X, and 15 is the only chromosome unique to X. Similar logic allows the assignment of the other genes. Critical Thinking Question: 1. Three-point crosses capture (allow us to see) double crossovers that have taken place in the two regions de ned by the three loci. However, any double crossovers that occur within one region or any crossovers involving more than two events will not be indicated correctly by random-strand analysis. 7 Linkage and Mapping in Prokaryotes and Bacterial Viruses 1. The prokaryotic chromosome is a double-stranded DNA circle that is small compared with most eukaryotic chromosomes. Viral chromosomes can be DNA or RNA. Viruses are obligate intracellular parasites. Whether they are alive depends on the de nition of the term alive. 3. A colony is a visible mass of cells derived usually from a single progenitor. A plaque is the equivalent growth of phages on a bacterial lawn, producing a cleared area lacking intact bacteria. 5. The bacterium could have survived and produced a colony if it was on a -free area, it became lysogenic (and thus resistant to further phage attack), or it was genetically resistant to phage . 7. 1, his arg ; 2, leu ; 3, lys ; 4, his met or his met ; 5, arg . 9. Where phages cannot grow: E. coli tonr, phage h . Where phage can grow: E. coli tons, phage h or h, or E. coli tonr, phage h.
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1)% 2.5 map units; b to cen3 1 1)% 7.5 map units.
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27. 12.5 map units.The only genotype that grows on minimal medium is arg ade . If the two genes were unlinked, 1/4 of the progeny should have this genotype; this is not seen. The genes must be linked; wildtype results from recombination between these two genes. The reciprocal class, arg ade , which has not been selected for, should be equally frequent, so: map units (2 25)/400 100 12.5. pab 23 6 pk 7 ad
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App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
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Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
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11. Far. If the selective locus is near, it passes into the F cell very early during conjugation. Consequently, there is a great reduction in the recovery of loci distal to the selective marker because both the Hfr and F members of a conjugation event can be killed by the selective agent (e.g., an antibiotic such as streptomycin). 13. See gures 7.8, 7.9, 7.15, 7.17, 7.18, and 7.26. 15. 9 min 1 min 8 min 7 min Origin az ton lac galB 17. For example, use an Hfr that is wild-type but str with the F factor integrated at minute 20. Use an F strain that is pyrD , purB , man , uvrC , his , and strr. Interrupt mating at one-minute intervals; plate cells on complete medium with streptomycin to kill Hfr cells and grow recombinant and nonrecombinant F cells. The next day, after colonies have grown up, replica-plate onto selective media. The following data would be generated:
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c. lys val his lys val his Both lys and his must be present to allow growth. d. lys val his lys val his We see no lys val his cells. e. lys and val are close together; they are cotransformed 75% of the time. Order could be lys val his or val lys his. f. val lys his. If the order is val lys his, val lys his should be rare, since this genotype results from a double exchange; and indeed, this class is the least frequent. 27. Mix the phages together with bacteria with increasing quantities of the two phages. Knowing the numbers of each in a particular case, it is possible to predict the proportion of cells doubly infected (product of probabilities).The recovery of recombinants should increase with that probability. In other words, recombination should occur only in doubly infected cells. 29. a 0.6 b 0.2 d 1.2 c
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