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Tamarin: Principles of Genetics, Seventh Edition
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Back Matter
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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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17. Three genes. Gene A: mutants 1, 4, 8; gene B: mutants 2, 5; gene C: mutants 6, 7. Mutant 3 probably contains a deletion that spans genes A and C. Begin by nding mutants that do not complement. These should have mutations in the same gene. Mutants 2 and 5 are in the same gene. Initially, we suspect that mutants 1, 3, 4, and 8 are in the same gene, which is different from the gene that contains 6 and 7. If mutant 3 is in gene A, it should complement 6 and 7, and it does not. One explanation is that 3 is a deletion spanning genes A and C. Alternatively, mutant 3 could be in gene A but be a polar mutation. Either possibility implies that the order is B A C. 19. 3 6 Begin with deletions that yield mostly s.These must be large deletions that cover most of the other deletions. Mutations 1 and 5 are such mutations. Since they give no wild-type, they must overlap: 1 5 Now look at mutant 2. It gives wild-type recombinants with 1 but not 5. Therefore it must overlap the region deleted in 5. Mutant 3, by similar logic, must cover part of deletion 1. We can draw these results as follows. Broken lines indicate we do not know yet how long the deletion is: 1 3 -------------------------------------------------2 -------------------------------------------------5 1 4 5 2
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beginning portion of the gene is missing. The following genotypes will give no complementation: 1 2 ____/////////____ No functional product of gene 1 1 2 ____////////____ No functional product of gene 2
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We could also get a lack of complementation when mutants are in two genes if we have a bacterial operon in which one of the genes contains a polar mutation creating a transcription stop signal. Such a mutation eliminates all distal functions. Thus, if the operon is A B C D,
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and we construct the partial diploid A (polar) A B C D , B C D
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we will get no complementation because the top DNA is effectively A B C D . 27. The auxotroph probably contains a deletion. If a few bases are missing, nothing is available to cause transitions or transversions. It is highly unlikely that the correct number of missing bases could be spontaneously and correctly inserted. 29. Excision repair endonucleases can recognize dimerizations, mismatched bases, and apurinic-apyrimidinic sites. 31. See gure 12.38 for a diagram of recombination. Branch migration is shown in gure 12.39. Critical Thinking Question: 1. The gene is a linear entity that speci es the linear order of amino acids in a protein in a colinear fashion. Although scientists in the 1960s were convinced of colinearity, there were other alternatives possible. For example, DNA could be a branching structure. Or, transcription of DNA could take place such that the beginning, middle, and end of the gene were not in order. Thus colinearity supported our understanding of the shape and functioning of DNA. 13 Genomics, Biotechnology, and Recombinant DNA
Now look at 4. It gives (no wild-type recombinants) with both 1 and 5 and therefore must be in the region that 1 and 5 overlap. If 4 extended to and overlapped either 2 or 3, we expect to see with them. Since this prediction is not met, 4 must be a small deletion spanning at least part of the overlap of 1 and 5. Since mutant 6 gives no wild-type with 1, 4, or 5, it must be within the common region deleted in all three strains. 21. We know that the anticodon pairs with the codon, and we expect nonsense suppressors to contain an altered anticodon. The fact that the nonsense codon can be read by a transfer RNA with a normal anticodon but altered dihydrouridine loop suggests that the way in which this loop interacts with the ribosome causes the anticodon sequence to be misread. 23. x x: AT GC. y y: GC AT. This problem requires logic and a knowledge of how mutagens work. For x, the key is the response to HA, which only causes GC AT transitions. Mutant x is reverted by HA, therefore, x must be GC, and the normal x was AT. AP-induced mutations can also be reverted by AP. Since y is not reverted by HA, y must be AT. Therefore, y must be GC. 25. A deletion that spans regions of more than one gene. Consider the following two genes: 1 2
1. Type II endonucleases are valuable because they cut DNA at speci c points and many leave overlapping or sticky ends. 3. In DNA with a random sequence, a four-cutter will nd sites approximately once in 44 bases ( 1/256 0.0039). A six-cutter will 0.0002). nd sites approximately once in 46 bases ( 1/4096 An eight-cutter will nd sites approximately once in 48 bases ( 1/65,536 0.000015). 5. DNA can be joined by having compatible ends to begin with or by blunt-end ligation (linkers combine these methods). The appropriateness of a method depends on what DNA is to be cloned and how that DNA can be obtained. Having DNA with sticky ends created by the same restriction enzyme would be easiest but sometimes is not available. Adding linkers by blunt-end ligation with a particular restriction site is usually the best compromise. 7. A plasmid is a self-replicating circle of DNA found in many cells. Foreign DNA inserted into a vector forms an expression vector if that foreign DNA produces a protein product. Cosmids are plasmids that contain cos sites and are useful for cloning large segments of DNA (up to 50 kb). YACs, yeast arti cial chromosomes, have the loci to replicate in yeast (centromere, replication origin, and telomeres). They can be used to clone very large pieces of DNA, upwards of one million bases.
By de nition, mutations in gene 1 will complement mutations in gene 2 but not other mutations in gene 1. If we have a deletion, x, that covers part of both 1 and 2 (slashes), 1 2 ______///////////////________ x presumably gene 1 will be nonfunctional because it is missing the last part of the protein. Gene 2 will be nonfunctional because the
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