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ASU BIO 340 Exam 3 Questions with
Answers
1. What will be the effect on transcription when an inhibitor is in play?
A. Allow transcription B. Prevent transcription: B Explanation: An inhibitor prevents activator from binding to DNA; activators are used in positive
regulation (increasing transcription) - thus, the inhibitor prevents the activator from positive regulating (increasing) transcription.
2. What will be the effect on transcription when an inducer is in play?
A. Allow transcription B. Prevent transcription: A Explanation: When an inducer is in play, it will allow transcription.
3. For a gene under positive control, which of the following would prevent transcription?
A. Repressor B. Corepressor C. Inducer D. Effector E. Inhibitor: E Explanation: Positive regulation includes activators (attected by ettectors and inhibitors); inhibitors prevent activator from binding to DNA, thus preventing transcription.
4. For a gene under negative control, which of the following would you expect
to find bound to the repressor if there is active transcription? A. Repressor B. Corepressor C. Inducer D. Effector E. Inhibitor: C Explanation: In negative regulation, there are repressors (attected by correpressors and inducers). If there were a corepressor, the repressor would successfully bind to the operator and prevent transcription. Inducers, however, prevent the binding of the repressor - thus causing a prevention of it doing its job, and allowing transcription.
5. Which molecule serves as the inducer for the lac operon?
A. cAMP B. Glucose C. Lactose D. Allolactose: D Explanation: Allolactose is the molecule that binds to repressor protein, making it unable to bind to the operator.
6. Which of the following conditions will result in the greatest levels of transcrip- tion of the lac operon?
A. Lactose absent, glucose absent B. Lactose absent, glucose present C. Lactose present, glucose present D. Lactose
2 / present, glucose absen: D Explanation: From the perspective of the cell, it is most eflcient to repress expression of the genes allowing for metabolism of lactose when it's absent, to allow it when present, but only to prioritize it when glucose (the preferred food source) is absent. In C, allolactose will also be present, allowing transcription, but positive regulation (i.e. increase in transcription) only happens as in D, when glucose is absent.
7. Which of the following mutations will result in the lowest level of transcription
of the lac operon when lactose is present?
A. An operator mutation that prevents binding of the lac repressor B. A mu-
tation in the lac repressor that prevents it from binding to the operator C. A mutation in CAP that prevents it from binding to the CAP binding site D. A mutation in the lac repressor that prevents it from binding allolactose: D Explanation: A - if repressor can't bind, the gene will be permanently turned on, there will be transcription. B - exactly the same ettect as A. C - "tempting but wrong"; CAP can't bind, will cause lower but not total prevention of transcription. D - correct answer; if the inducer is totally inettective, then the repressor will permanently bind to operator and totally prevent transcription.
8. The regulatory elements in the lac operon act to:
A. Repress transcription when there is no lactose present B. Repress transcrip- tion when there is lactose present C. Activate transcription when there is no glucose present D. Activate transcription when there is glucose present E. A and C: E Explanation: The cell wants to both repress transcription when there is no lactose present and activate it when glucose is absent (C "slightly less good" answer than A, but still true). Similar to Clicker Question 6
9. A bacterium needs to make a certain lipid that it cannot obtain from its environment. That same lipid is
involved in regulating the expression of the enzymes that make it. Which of the following is the most likely regulatory function of the lipid? A. Effector B. Activator C. Inducer D. Co-repressor E. Operator: D Explanation: If you have that lipid, you would want it to not make any more. So, if you have the lipid in the cell, you would want to turn ott the transcription, so it would likely be involved in negative regulation. Ettectors and activators are both
4 / histones? A. Acetylation of histone tails B. Deacetylation of histone tails: A. Acetylation of histone tails
14. Which of the following regulatory sequences may be located tens of thou- sands of nucleotides
away from the genes they regulate? A. Core promoters B. Enhancers C. Proximal elements D. TATA boxes: B. Enhancers
15. Which of the following enzymes can slide a histone octamer to a different position on a eukaryotic
chromosome? A. Chromatin modification complex B. Histone acetyltransferase C. Topoiso- merase D. Chromatin remodeling complex: D. Chromatin remodeling complex
16. Which regulation of gene expression mechanism is not found in eukaryotes?
A. Different strength promoters B. Activator binding to DNA C. Repressor bind- ing to operator D. Alternative splicing: C Explanation: Repressors bind to silencers in eukaryotes (operators specific to prokaryotes)
17. Which regulation of gene expression mechanism is not found in prokary-
otes? A. Different strength promoters B. Activator binding to DNA C. Repressor binding to operator D. Alternative splicing: D Explanation: RNA polymerase binding in prokaryotes
- will have promoters of dittering strengths, may have very strong or slightly weaker aflnity to sigma factor. Activator binding to DNA in both pro and eukaryotes; if to enhancer, that is specific to eukaryotes. Prokaryotes don't do splicing at all (thus, alternative splicing is not possible in prokaryotes.
18. What type of mutation is seen here? Wild type: 5'-TAC AAA ATA CAG CGG-3'
Mutation: 5'-TAC AAG ATA CAG CGG-3' A. Missense B. Nonsense C. Silent D. Frameshift: C Explanation: AAA would normally encode a Leu. AAG would also encode a Leu
19. : What type of mutation is seen here? Wild type: 5'-TAC AAA ATA CAG CGG-3' Mutation: 5'-TAC
AAG ATA CAG CGG-3' A. Transversion B. Deletion C. Transition
5 / D. Insertion: C Explanation: A is a purine and G is a purine, so this is a transition
20. Which type of mutation is most likely to produce a loss-of-function allele? A. The deletion of 6
nucleotides B. A point mutation that replaces an amino acid with one of similar chemical properties C. The insertion of an exon after the stop codon D. A transition in the third position of a codon E. An insertion of 4 nucleotides: E. an insertion of 4 nucleotides
21. DNA polymerase III adds nucleotides
. A. To the 3' end of the RNA primer B. To the 5' end of the RNA primer C. In the place of the primer RNA after it is removed D. To both ends of the RNA primer E. To internal sites in the DNA template: A. to the 3' end of the RNA primer
22. If the two oligonucleotides below are allowed to anneal and DNA polymerase and all substrates (
dNTPs, etc.) are added to the mix, what will be the first three nucleotides incorporated into the DNA molecule? A. 5' TAT 3' B. 5' TTG 3' C. 5' UUG 3' D. 5' ACT 3' 5' ATCCTGGAACACTGTACCATCGGTACCAATCACAGGTCCTTACAGT 3' 5' GGACCTGTGA 3': B Explanation: The primer will bind to the complementary sequence 5' TCACAGGTCC 3' on the DNA molecule. To the left of that sequence (where the 3' end of the primer is), the template reads AAC. Therefore, the complementary nucleotides TTG will be added.
23. You wish to amplify the following double-stranded DNA sequence by PCR.
Which primer pair would you choose? A. 5'-ATTGCCGC-3' and 5'-GAATTTCG-3' B. 5'-ATTGCCGC-3' and 5'-CTTAAAGC-3' C. 5'-TAACGGCG-3' and 5'-CTTAAAGC-3' D. 5'-TAACGGCG-3' and 5'-GAATTTCG-3' E. 5'-ATTGCCGC-3' and 5'-GCTTTAAT-3': A Explanation: If you want a primer that is complementary to the bottom strand, you can pick the beginning of the top
7 / D. EcoRI, SpeI, TaqI, Sau3A1 E. SphI and KpnI: E Explanation: When the text says 3' overhang, the overhang ends with an exposed 3' end. This means that the longer sequence is on the 5' strand.
30. Given the linear DNA molecule shown below, how many fragments would you expect if it was cut
with BamHI and BcII? A. 1 B. 2 C. 3 D. 4 E. 5: C Explanation: BamHI and BclI will cut the double-strand only once each, having three pieces of double-stranded DNA as a product.
31. Can the sticky ends produced by BamHI and BcII be ligated together?
A. Yes B. No: A Explanation: The overhangs left by each enzyme are complementary to each other
32. Can the resulting ligated sequences be cut by BamHI A. Yes B. No: B Explanation: The resulting ligated sequence no
longer has the recognition sequence for BamHI. So the enzyme can no longer cut it
33. Molecular cloning involves cutting a gene from its normal location, inserting it into a circular piece of
DNA, and then transferring the circle of DNA into a bacterial or yeast cell. Which of the following is used to cut the gene from its normal location? A. Restriction enzyme B. Plasmid C. Bacteriophage D. Vector E. Ligase: A. restriction enzyme
34. Molecular cloning involves cutting a gene from its normal location, inserting it into a circular piece of
DNA, and then transferring the circle of DNA into a bacterial or yeast cell. Which of the following describes the circular DNA from a bacterial cell? A. Restriction enzyme B. Plasmid C. Bacteriophage D. Vector E. Ligase: B A vector is a plasmid that is specifically used to carry a piece of DNA. Both B and D are correct for the purposes of this class
35. What extra elements are required in expression vectors that are not found in typical cloning vectors? A.
Sites for restriction nuclease cleavage B. Selectable markers, such as antibiotic resistance C. Replication origins D. Promoters E. Operators: D Promoters are required to express a gene, but not for cloning vectors.
36. In using Agrobacterium tumefaciens to transfer genes into plants, what is transferred from the
bacterium to the plant? A. Ti plasmid B. T-DNA C. Opines
8 / D. Only the recombining sequence E. The entire bacterium: B The whole plasmid doesn't go through the plant cell, so it is not A. The bacterium will make proteins from the Ti plasmid that allows for the T-DNA to be transferred to the plant cells
