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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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13. EF-Tu brings a charged transfer RNA to the A site at the ribosome. EF-Ts is involved in recharging EF-Tu (see g. 11.14). The eukaryotic equivalents are eEF1 and eEF1 . 15. A signal peptide is a sequence of amino acids at the amino-terminal end of a protein that signals that the protein should enter a membrane (see g. 11.25). Although the concept is the same, the situation in eukaryotes is somewhat more complex because there are many different membrane-bound organelles, each having their own membrane-speci c requirements. Signal peptides are usually cleaved off the protein after the protein enters or passes through the membrane. 17. NH2-FGKICABHLNOEDJM-COOH 19. a. 5 -AUG AUU GAA UGC GAG CGG AGU-3 b. N-met-ile-glu-cys-glu-arg-ser First determine the sequence of the RNA complementary to the given DNA strand. Don t forget about polarity; as the strand is written, the 5 end of the RNA will be on the left. Blocking off successive groups of three bases allows the determination of the codons. Use the code to determine the amino acid sequence. 21. 12/27 phenylalanine (UUU, [2/3]3; UUC, [2/3]2[1/3]); 6/27 serine (UCU, [2/3]2[1/3]; UCC, [2/3][1/3]2); 6/27 leucine (CUU, [2/3]2[1/3]; CUC, [2/3][1/3]2); 3/27 proline (CCU, [2/3][1/3]2; CCC, [1/3]3 ). 23. The table could look the same (see table 11.4) except that the position would be left side rst position (5 end); top third position (3 end); right side second position. For example, the codons for valine (currently 5 -GUU-3 , 5 -GUC-3 , 5 -GUA-3 , and 5 -GUG3 ) would be 5 -GUU-3 , 5 -GCU-3 , 5 -GAU-3 , and 5 -GGU-3 . 25. We are mixing two RNA strands that are complementary; these strands will form a double-stranded RNA molecule. Since we observed the incorporation of no amino acids, the ribosome must not be able to read a double-stranded molecule. 27. If we write out (GUA)n as GUA GUA GUA GUA . . ., we see that we could use any of three different reading frames: GUA, UAG, or AGU. Since we see only two amino acids incorporated, either two of the possible codons code for the same amino acid, or one of the codons is a stop codon. If you look at the code, you will see that UAG is a stop codon. 29. The stop codon has probably mutated to give a codon for the amino acid leucine. The longer-than-normal protein suggests that the original stop was not read. Numerous possibilities exist. If the second letter of a stop codon were changed to a U, we would have UUA or UUG leucine codons. Alternatively, the insertion of a C before the U would yield CUA, CUG, or CUA as leucine codons. Similarly, an insertion of a U next to the rst G yields UUA or GUG as leucine codons. Since the next amino acid is phenylalanine, the next codon must be UUC or UUU. If a base were added, as above, the next codon would have to begin with A or G, and phenylalanine does not begin with A or G. Therefore, the most likely explanation is a change of the second letter from an A to a U. 31. Either CAU or CAC. Write down all possible codons for each amino acid. his CAU/C tyr UAU/C gln CAA/G pro CCX leu CUX
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Critical Thinking Question: 1. In general, transcriptional and translational signals are independent. We could look at this by asking the question, how does changing one of the signals affect the other process In other words, if we changed a translational signal such as a start or stop codon, would that affect the transcription of that gene In general, the answer is no. 12 DNA: Its Mutation, Repair, and Recombination
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1. For example, if the twenty individual cultures of table 12.1 had values of 15, 13, 15, 20, 17, 14, 21, 19, 16, 13, 27, 14, 15, 26, 12, 21, 14, 17, 12, 14, then the mutation theory would not have been supported because the variation between the individual and bulk cultures would not have been different. 3. Reading across each row, we gather more and more information. Row 1: 1, 6, and 7 are part of one complementation group. Row 2: 2 and 5 are part of one complementation group. Row 3: 3 and 4 are part of one complementation group. Row 4: no new information. Row 5: no new information. Row 6: reinforces that 6 and 7 are part of the same complementation group. Row 7: no new information. Thus, we conclude that there are three complementation groups present: 1, 6, and 7 are mutually noncomplementing, as are 2 and 5, and 3 and 4. The half-table missing is a mirror image because a cross of 1 and 3 is the same as a cross of 3 and 1 (reciprocity). The diagonal always contains negative elements because every mutant is a functional allele of itself. 5. All mutants should be crossed in pairwise combinations yielding heterozygote daughters. (Presumably, earlier crosses indicated that these are X-linked loci.) All F1 daughters will be wild-type: The mutations complement and therefore are not alleles. When eosin ies are crossed in a similar fashion, daughters will be wild-type except when the parents were eosin and white. In that case, daughters will be mutant, showing the lack of complementation and hence that the mutations are alleles. 7. Prokaryotic and phage genes generally do not have intervening sequences. Benzer and Yanofsky worked at a time when introns were unknown and it was assumed that the length of a gene was transcribed and then translated. If the genes had introns, Benzer and Yanofsky would have been unaware of them; introns would not affect colinearity or mapping. Introns would affect physical measures of the lengths of DNA, with which Benzer and Yanofsky were not involved. 9. The rex gene of phage represses growth of phage T4 rII mutants. 11. In replicating DNA, a transition mutation can occur by tautomerization of a base in the template strand (template transition) or entering the progeny strand (substrate transition). 13. 5-bromouracil (pyrimidine analogue) and 2-aminopurine (purine analogue) are incorporated into DNA as thymine and adenine, respectively. However, each undergoes tautomeric shifts more frequently than the normal base. Both cause transitions. Nitrous acid also promotes transitions by converting cytosine into uracil, which acts like thymine, and adenine into hypoxanthine, which acts like guanine. Pro avin induces insertions and deletions by intercalating and buckling DNA. Ethyl ethane sulfonate removes purine rings and thus promotes transitions and transversions. 15. See gure 12.25.
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For leu, note that UUA/G cannot result from a single change in the his codon. Therefore, leucine must be CUX. All of the other amino acids could result from single changes in either the rst or second base, and we are left with either codon being the one for his.
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