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Answers Lecture 15. 1. A 2. 3’-poly A tail, 5’-cap, splicing 3. D 4. D (E is the same as D – a mistake in typing) 5. A 6. if you are given 3’-CAT-5’ as the template strand of DNA, then the mRNA will be 5’GUA-3’. The mRNA will be 5’-CAU-3’ if it is the coding strand of DNA that you are given. 7. A 8. B 9. A 10. B 11. C 12. D 13. B 14. A 15. C 16. E 17. D 18. E 19. D 20. C 21. A 22. E 23. B 24. ribose vs. deoxyribose sugar, single stranded vs. double stranded, U vs T 25. recruits general transcription factors to the basal promoter, which brings in RNA polymerase 26. RNA polymerase, 5’ to 3’ 27. see text 28. E, P, A 29. stop codon in mRNA 30. nonsense mutation – introduces a stop codon; missense – changes an amino acid; silent – no change Lecture 16 1. D 2. D 3. E 4. C 5. A 6. general transcription factors (includes TAT binding protein – TBP, which binds to the TATA box and recruits the rest of the GTFs) and sequence specific transcription factors 7. introns 8. liver because it has the sequence-specific transcription factors that bind to the upstream portion of the promoter – this recruits the general transcription factors that bind to the basal promoter and recruit RNA polymerase; RNA polymerase synthesizes the mRNA - the first general transcription factor to bind is TBP, which recruits the remaining factors; this complex recruits RNA polymerase – then transcription can begin. 9. mRNA is degraded (microRNAs) or protein is degraded (ubiquitin and the proteasome) 10. see your lecture notes 11. reverse transcriptase Lecture 18. 1. D 2. A 3. D 4. C 5. C 6. no b-galactosidase is being made – thus DNA has been inserted into the lacZ gene; if the lacZ gene is intact (no insertion into it) the colony will be blue 7. A 8. B 9. D 10. reverse transcriptase 11. C 12. child 2 13. 3 14. see notes outlining Southern blotting 15. A- all B – only LB or LB-amp C – only LB D – reverse transcriptase reaction with mRNA 16. see notes for DNA sequencing