Molecular Biology 1st Edition Cox Doudna ODonnell Test Bank
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Chapter 12 – Test Bank
- Point mutations in the DNA sequence don’t cause:
- silent mutation.
- missense mutation.
- nonsense mutation.
- frameshift mutations.
- transversion mutations.
- The exchange of a purine-pyrimidine base pair for the other purine-pyrimidine pair is:
- a silent mutation.
- a transition mutation.
- a frameshift mutation.
- a nonsense mutation.
- a transversion mutation.
- Tumor suppressor genes:
- encode proteins that drive the cell cycle forward.
- encode proteins that suppress development.
- encode proteins that suppress cell division.
- encode proteins that suppress DNA repair.
- encode proteins that suppress metabolism.
- are mutations that arise due to large duplications of DNA.
- are caused aberrant recombination or template slippage during replication.
- are mutations in which one base pair is exchanged for another.
- are mutations that inactivate oncogenes, which cause cells to lose control of cell growth.
- are the result of replication errors.
- Insertions and deletions of three base pairs:
- change one codon and therefore, only one amino acid in the protein sequence.
- do not disrupt the reading frame so the protein ends up completely normal.
- do not disrupt the reading frame so the protein will have an almost normal sequence.
- cause frameshifts that will change the reading frame and therefore the sequence of the protein.
- usually result in the expression of a truncated protein.
- Polyglutamine (polyQ) diseases:
- result from template slippage during replication.
- result when CAG repeats increase above the disease threshold.
- is a triple expansion disease.
- include Huntington’s disease and fragile X syndrome.
- All of the above make correct statements.
- In which of the following large scale mutations would the cell lose genetic information?
- None of the above.
- Why is the formation of a BCR-ABL fusion gene a carcinogenic event?
- The tyrosine kinase activity of the ABL gene becomes inactive.
- The tyrosine kinase activity of the ABL gene is now controlled a weak promoter resulting in less tyrosine kinase activity.
- The tyrosine kinase activity of the ABL gene becomes unregulated leading to uncontrollable cell division.
- The tyrosine kinase activity of the ABL gene is reduced due to truncation of the protein product.
- The tyrosine kinase activity of the BCR gene is increased in the fusion protein product.
- Which of the statements best describes the mutational event pictured below?
Wild-type ACC CAC UCU GGA UUU AAG GCA
thr his ser gly leu lys ala
Mutant ACC CAC UCU UGA UUU AAG GCA
thr his ser stop phe lys ala
- A transversion mutation leading to a nonsense codon
- A frameshift mutation leading to a missense amino acid substitution
- A transition mutation resulting in a silent amino acid substitution
- A transition mutation leading to a nonsense codon
- An insertion mutation resulting in a frameshift of the amino acid sequence
- Which of the following techniques is used to test for genetic diseases such as fragile X syndrome?
- Western blotting
- A mutant protein has been designated as G189A. What does this designation reveal about the mutation?
- That a G nucleotide has been exchanged for an A nucleotide 189 bases from the 5’ end of the coding region.
- That a glycine in position 189 of the protein is now an alanine.
- That an A nucleotide has been exchanged for a G nucleotide 189 bases from the 5’ end of the coding region.
- That a glycine is now in position 189 of the protein instead of an alanine.
- That a nonsense mutation occurred at position 189 of the protein.
- Why is mutation of DNA repair genes the most harmful to the cell?
Ans: Without DNA repair, mutations will accumulate in the cell rather than be repaired. Chances increase that a gene controlling the cell cycle will be damaged, which can lead to cancer.
- Which of the following is a common type of hydrolysis damage in the cell?
- Deamination of thymidine.
- Breakage of the glycosidic linkage between the pyrimidines and the DNA backbone.
- Removal of cytosine from the DNA backbone to create an abasic site.
- Deamination of cytosine to uracil.
- All of these are common.
- Which of the following is a correct pairing of a base to the product of its deamination?
- Thymidine → uracil
- Cytosine → thymine
- Adenine → xanthine
- Guanine → hypoxanthine
- 5-methylcytosine → thymine
- Which bases are more susceptible to hydrolysis of the glycosidic linkage that connects them to the DNA backbone?
- Adenine and guanine
- Adenine and thymine
- Cytosine and guanine
- Cytosine and adenine
- Cytosine, thymine, and uracil
- Which of the following does not contribute significantly to DNA damage?
- Deamination due to hydrolysis
- Alkylation of bases
- Deamination due to superoxide radicals
- Oxidation hydroxyl radicals
- Oxidation hydrogen peroxide (H2O2)
- The Ames test is used to identify potential carcinogens. A compound tested resulted in a few scattered colonies growing on the media, similar to the control plate with no disk. What does this result imply?
- The compound in the disk is so toxic that it inhibits almost all growth on the plate.
- The few colonies that grew are resistant to the compound.
- The few colonies that grew contain reversion mutations that allow them to grow on media without histidine.
- The compound in the disk is not mutagenic so there was no increase in the reversion rate.
- Answers C and D are both correct.
- An alkylating agent will cause which of the following types of DNA damage?
- insertions and deletions of 1-4 bases
- base substitutions
- double-stranded breaks in the DNA
- thymidine dimers
- frameshift mutations
- Which of the following is true of chemotherapeutic agents?
- They create broken chromosomes or stalled replication forks.
- The cytotoxic effect of these DNA damaging agents requires the cell to be actively dividing.
- These agents are toxic to cancer cells because they are actively dividing to form a tumor.
- The adverse side-effects of chemotherapeutics are strongest in cell types that divide rapidly (hair, blood, and the gastrointestinal tract).
- All of the above are true.
- Which of the following can cause double-stranded breaks in DNA?
- Ultraviolet radiation
- Cosmic rays
- Gamma rays
- All of the above
- Which of the following are the result of external mutagens?
- Template slippage during replication
- Addition of bulky adducts to the DNA
- Recombination errors
- Replication errors
- Compounds that kill the cell are said to be:
- Why is repairing a single-stranded break easier to repair than a double-stranded break?
Ans: Single-stranded breaks can be repaired DNA ligase. Double-stranded breaks require homologous recombination repair or nonhomologous end joining, which often result in the loss of bases and may cause mutations.
- Which of the following is not part of the mismatch repair (MMR) mechanism?
- MutS and MutL form a complex at the site of the mismatch and while scanning in both directions along the DNA form a loop.
- Mismatched bases are recognized MutS.
- When the MutS/MutL complex finds a GATC site, MutH binds to the complex and cleaves the methylated strand.
- Helicase and exonuclease activities unwind and degrade the strand containing the mismatched base.
- Pol III fills the gap and ligase seals the DNA strand.
- What is the main difference between mismatch repair (MMR) in eukaryotes and prokaryotes?
- Eukaryotes lack a functional Mut H and Dam methylase.
- Eukaryotes have proteins homologous to MutS and MutH but not MutL.
- Eukaryotes only repair the lagging strand.
- Prokaryotic MMR can repair small loops of unpaired nucleotides, whereas eukaryotes cannot.
- Both use exonucleases to eliminate the mismatched strand but eukaryotes only require 3’-5’ exonucleases.
- Which of the following repair enzymes is targeted for degradation after it reacts one time?
- AP endonulease
- DNA glycosylase
- Photolyases are present in all cells except:
- both C and D.
- Which repair mechanism requires light energy?
- Nucleotide excision repair (NER)
- Base excision repair (BER)
- Mismatch repair (MMR)
- Transcription coupled repair
- Which of the following is true about base excision repair (BER) in bacteria?
- BER is used to repair abasic sites created hydrolysis or DNA glycosylases.
- BER requires the action of a unique exinuclease activity to cleave 5’ of the abasic site.
- At the nick created the nuclease, Pol III can excise the damaged strand nick translation.
- BER can be used to excise benzopyrene adducts.
- BER requires binding of the MutS protein to identify the abasic site.
- Which of the following is not true about base excision repair (BER) in eukaryotes?
- Since eukaryotes lack a polymerase with 5’-3’ exonuclease activity the abasic strand can’ t be degraded from a nick.
- Long-patch repair eliminates the strand containing the abasic site displacement to create a “flap” that is removed the flap endonuclease.
- Short-patch repair only removes the 5’deoxyribose phosphate and replaces it with a nucleotide.
- BER in eukaryotes also uses glycosylase to remove inappropriate or damaged bases.
- BER in eukaryotes cannot be used to repair abasic sites created hydrolysis.
- Why does mismatch repair not lead to mutation upon repair of heteroduplex DNA?
- The incorrect base is recognized because it is methylated.
- The strand containing the misincorporated base is methylated.
- The strand containing the misincorporated base is not methylated.
- The strand containing the misincorporated base is phosphorylated.
- Actually, mismatch repair does lead to mutation about 50% of the time.
- Which of the following is not a step in nucleotide excision repair (NER)?
- The UvrA2UvrB complex creates a single-stranded bubble at the damage site to which UvrA subunits are tightly bound and the UvrB subunit dissociates.
- The UvrA2UvrB complex scanned the DNA for damage.
- Pol I fills the gap created the removal of the damaged strand and the DNA is sealed ligase.
- UvrC (exinuclease) nicks the damaged strand on either side of the damage.
- UvrD (helicase) unwinds and releases the damaged strand.
- Why is transcription-coupled repair (TCR) considered particularly efficient when compared to nucleotide excision repair (NER)?
- The exinuclease employed in TCR removes a smaller fragment containing the damage.
- TCR targets transcriptionally active regions for repair over damage in less active parts of the genome.
- It uses error-prone polymerases so it can move through damaged areas faster.
- It directly repairs the damage rather than removing and replacing the damaged strand.
- TCR requires less protein interactions.
- Thymidine dimers cannot be repaired in humans which of the following?
- Nucleotide excision repair (NER)
- Transcription-coupled repair (TCR)
- Translesion synthesis (TLS) using Pol η
- All of the above can be used to repair thymidine dimers in human cells.
- Which of the following is NOT true of translesion synthesis (TLS)?
- Uses polymerases with wider-than-normal active site architecture to accommodate abnormal nucleotides in the template.
- Lacks the 3’-5’ exonuclease activity usually required for proofreading.
- It is activated during replication when the polymerase stalls at a lesion. The TLS polymerase takes over long enough to extend the DNA over the lesion.
- TLS can be used to repair lesions such as pyrimidine dimers, bulky adducts, and double-stranded breaks.
- All of the above are true.
- What is the most critical difference between global nucleotide excision repair (NER) and transcription-coupled repair (TCR)?
- The initiation step; in NER the XPC protein recognizes the damage but in TCR it is the RNA polymerase that recognizes the damage.
- The excision step; NER uses UvrC (exinuclease) to nick the DNA but TCR uses the AP endonuclease.
- The removal step; NER uses UvrD (helicase) to unwind and release the damaged strand but in TCR the damaged strand is degraded Pol I.
- The final step; NER uses Pol I and ligase to fill the gap but TCR uses Pol III and ligase.
- None of the above are true.
- Many eukaryotes have a DNA glycosylase that specifically removes T residues from DNA when paired with G. Why is it better to repair the G-T pairing to G-C removing the T than to remove the G and form an A-T pair?
- G-C pairs can form three hydrogen bonds between them and are therefore stronger.
- Oxidative damage to guanine alkylation is the most common reason for G-T pairs so removal of the T will maintain the original G-C pair.
- T bases easily tautomerize into C bases, so removing the T will make sure the pair stays in its original form.
- Answers B and C are true.
- Answers A, B, and C are true.
- In mammalian cells, depurination occurs at a high rate. As many as 1 in 105 purines are lost from DNA in a 24-hour period. What repair mechanism is used to repair the DNA?
Ans: The cell uses base excision repair (BER) to repair the abasic sites created depurination.
How We Know
- What key element was created Modrich and coworkers that was used to demonstrate that MutS and MutL were required for the activation of MutH to cleave the nonmethylated strand?
Ans: They created a circular DNA containing a single mismatch and a GATC site nearby. This construct was also hemimethylated on one strand versus the other.
- How did Renato Delbecco determine that the repair of UV damage could be increased exposure to visible light?
Ans: T phage DNA was irradiated with UV light, which decreased its viability in E. coli. Subsequent exposure of the phage to visible light had no photoreversal effect; however, when the UV-irradiated phage were inside E. coli cells and then exposed to visible light repair in the form of increased phage viability was observed.