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