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Molecular Biology 1st Edition Cox Doudna ODonnell Test Bank

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Molecular Biology 1st Edition Cox Doudna ODonnell Test Bank

ISBN-13: 978-0716779988

ISBN-10: 0716779986

 

Description

Molecular Biology 1st Edition Cox Doudna ODonnell Test Bank

ISBN-13: 978-0716779988

ISBN-10: 0716779986

 

 

 

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Chapter 12 – Test Bank

 

 

Section 12.1

 

  1. Point mutations in the DNA sequence don’t cause:

 

  1. silent mutation.
  2. missense mutation.
  3. nonsense mutation.
  4. frameshift mutations.
  5. transversion mutations.

 

Ans: D

 

  1. The exchange of a purine-pyrimidine base pair for the other purine-pyrimidine pair is:

 

  1. a silent mutation.
  2. a transition mutation.
  3. a frameshift mutation.
  4. a nonsense mutation.
  5. a transversion mutation.

 

Ans: B

 

  1. Tumor suppressor genes:

 

  1. encode proteins that drive the cell cycle forward.
  2. encode proteins that suppress development.
  3. encode proteins that suppress cell division.
  4. encode proteins that suppress DNA repair.
  5. encode proteins that suppress metabolism.

 

Ans: C

 

  1. Indels:

 

  1. are mutations that arise due to large duplications of DNA.
  2. are caused by aberrant recombination or by template slippage during replication.
  3. are mutations in which one base pair is exchanged for another.
  4. are mutations that inactivate oncogenes, which cause cells to lose control of cell growth.
  5. are the result of replication errors.

 

Ans: B

 

  1. Insertions and deletions of three base pairs:

 

  1. change one codon and therefore, only one amino acid in the protein sequence.
  2. do not disrupt the reading frame so the protein ends up completely normal.
  3. do not disrupt the reading frame so the protein will have an almost normal sequence.
  4. cause frameshifts that will change the reading frame and therefore the sequence of the protein.
  5. usually result in the expression of a truncated protein.

 

Ans: C

 

  1. Polyglutamine (polyQ) diseases:

 

  1. result from template slippage during replication.
  2. result when CAG repeats increase above the disease threshold.
  3. is a triple expansion disease.
  4. include Huntington’s disease and fragile X syndrome.
  5. All of the above make correct statements.

 

Ans: E

 

  1. In which of the following large scale mutations would the cell lose genetic information?

 

  1. Translocation
  2. Inversion
  3. Deletion
  4. Duplication
  5. None of the above.

 

Ans: C

 

  1. Why is the formation of a BCR-ABL fusion gene a carcinogenic event?

 

  1. The tyrosine kinase activity of the ABL gene becomes inactive.
  2. The tyrosine kinase activity of the ABL gene is now controlled by a weak promoter resulting in less tyrosine kinase activity.
  3. The tyrosine kinase activity of the ABL gene becomes unregulated leading to uncontrollable cell division.
  4. The tyrosine kinase activity of the ABL gene is reduced due to truncation of the protein product.
  5. The tyrosine kinase activity of the BCR gene is increased in the fusion protein product.

 

Ans: C

 

  1. 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

 

  1. A transversion mutation leading to a nonsense codon
  2. A frameshift mutation leading to a missense amino acid substitution
  3. A transition mutation resulting in a silent amino acid substitution
  4. A transition mutation leading to a nonsense codon
  5. An insertion mutation resulting in a frameshift of the amino acid sequence

 

Ans: A

 

  1. Which of the following techniques is used to test for genetic diseases such as fragile X syndrome?

 

  1. RNA-Seq
  2. ChIP-Seq
  3. PCR
  4. Western blotting
  5. Immunoprecipitation

 

Ans: C

 

  1. A mutant protein has been designated as G189A. What does this designation reveal about the mutation?

 

  1. That a G nucleotide has been exchanged for an A nucleotide 189 bases from the 5’ end of the coding region.
  2. That a glycine in position 189 of the protein is now an alanine.
  3. That an A nucleotide has been exchanged for a G nucleotide 189 bases from the 5’ end of the coding region.
  4. That a glycine is now in position 189 of the protein instead of an alanine.
  5. That a nonsense mutation occurred at position 189 of the protein.

 

Ans: B

 

Short Answer

 

  1. 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.

 

Section 12.2

 

  1. Which of the following is a common type of hydrolysis damage in the cell?

 

  1. Deamination of thymidine.
  2. Breakage of the glycosidic linkage between the pyrimidines and the DNA backbone.
  3. Removal of cytosine from the DNA backbone to create an abasic site.
  4. Deamination of cytosine to uracil.
  5. All of these are common.

 

Ans: D

 

  1. Which of the following is a correct pairing of a base to the product of its deamination?

 

  1. Thymidine → uracil
  2. Cytosine → thymine
  3. Adenine → xanthine
  4. Guanine → hypoxanthine
  5. 5-methylcytosine → thymine

 

Ans: E

 

  1. Which bases are more susceptible to hydrolysis of the glycosidic linkage that connects them to the DNA backbone?

 

  1. Adenine and guanine
  2. Adenine and thymine
  3. Cytosine and guanine
  4. Cytosine and adenine
  5. Cytosine, thymine, and uracil

 

Ans: A

 

  1. Which of the following does not contribute significantly to DNA damage?

 

  1. Deamination due to hydrolysis
  2. Alkylation of bases
  3. Deamination due to superoxide radicals
  4. Oxidation by hydroxyl radicals
  5. Oxidation by hydrogen peroxide (H2O2)

 

Ans: C

 

  1. 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?

 

  1. The compound in the disk is so toxic that it inhibits almost all growth on the plate.
  2. The few colonies that grew are resistant to the compound.
  3. The few colonies that grew contain reversion mutations that allow them to grow on media without histidine.
  4. The compound in the disk is not mutagenic so there was no increase in the reversion rate.
  5. Answers C and D are both correct.

 

Ans: E

 

  1. An alkylating agent will cause which of the following types of DNA damage?

 

  1. insertions and deletions of 1-4 bases
  2. base substitutions
  3. double-stranded breaks in the DNA
  4. thymidine dimers
  5. frameshift mutations

 

  1. Which of the following is true of chemotherapeutic agents?

 

  1. They create broken chromosomes or stalled replication forks.
  2. The cytotoxic effect of these DNA damaging agents requires the cell to be actively dividing.
  3. These agents are toxic to cancer cells because they are actively dividing to form a tumor.
  4. The adverse side-effects of chemotherapeutics are strongest in cell types that divide rapidly (hair, blood, and the gastrointestinal tract).
  5. All of the above are true.

 

Ans: E

 

  1. Which of the following can cause double-stranded breaks in DNA?

 

  1. Ultraviolet radiation
  2. Cosmic rays
  3. Gamma rays
  4. Microwaves
  5. All of the above

 

Ans: E

 

 

 

 

 

 

  1. Which of the following are the result of external mutagens?

 

  1. Template slippage during replication
  2. Addition of bulky adducts to the DNA
  3. Depurination
  4. Recombination errors
  5. Replication errors

 

Ans: B

 

  1. Compounds that kill the cell are said to be:

 

  1. cytotoxic.
  2. genotoxic.
  3. mutagenic.
  4. carcinogenic.
  5. auxotrophic.

 

Ans: A

 

Short Answer

 

  1. Why is repairing a single-stranded break easier to repair than a double-stranded break?

 

Ans:  Single-stranded breaks can be repaired by 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.

 

Section 12.3

 

  1. Which of the following is not part of the mismatch repair (MMR) mechanism?

 

  1. MutS and MutL form a complex at the site of the mismatch and while scanning in both directions along the DNA form a loop.
  2. Mismatched bases are recognized by MutS.
  3. When the MutS/MutL complex finds a GATC site, MutH binds to the complex and cleaves the methylated strand.
  4. Helicase and exonuclease activities unwind and degrade the strand containing the mismatched base.
  5. Pol III fills the gap and ligase seals the DNA strand.

 

Ans: C

 

  1. What is the main difference between mismatch repair (MMR) in eukaryotes and prokaryotes?

 

  1. Eukaryotes lack a functional Mut H and Dam methylase.
  2. Eukaryotes have proteins homologous to MutS and MutH but not MutL.
  3. Eukaryotes only repair the lagging strand.
  4. Prokaryotic MMR can repair small loops of unpaired nucleotides, whereas eukaryotes cannot.
  5. Both use exonucleases to eliminate the mismatched strand but eukaryotes only require 3’-5’ exonucleases.

Ans: A

 

  1. Which of the following repair enzymes is targeted for degradation after it reacts one time?

 

  1. Photolyase
  2. AP endonulease
  3. MutH
  4. DNA glycosylase
  5. Methyltransferase

 

Ans: E

 

  1. Photolyases are present in all cells except:

 

  1. bacteria.
  2. archaea.
  3. humans.
  4. yeast.
  5. both C and D.

 

Ans: C

 

  1. Which repair mechanism requires light energy?

 

  1. Nucleotide excision repair (NER)
  2. Base excision repair (BER)
  3. Mismatch repair (MMR)
  4. Photoreactivation
  5. Transcription coupled repair

 

Ans: D

 

  1. Which of the following is true about base excision repair (BER) in bacteria?

 

  1. BER is used to repair abasic sites created by hydrolysis or by DNA glycosylases.
  2. BER requires the action of a unique exinuclease activity to cleave 5’ of the abasic site.
  3. At the nick created by the nuclease, Pol III can excise the damaged strand by nick translation.
  4. BER can be used to excise benzopyrene adducts.
  5. BER requires binding of the MutS protein to identify the abasic site.

 

Ans: A

 

  1. Which of the following is not true about base excision repair (BER) in eukaryotes?

 

  1. Since eukaryotes lack a polymerase with 5’-3’ exonuclease activity the abasic strand can’ t be degraded from a nick.
  2. Long-patch repair eliminates the strand containing the abasic site by displacement to create a “flap” that is removed by the flap endonuclease.
  3. Short-patch repair only removes the 5’deoxyribose phosphate and replaces it with a nucleotide.
  4. BER in eukaryotes also uses glycosylase to remove inappropriate or damaged bases.
  5. BER in eukaryotes cannot be used to repair abasic sites created by hydrolysis.

 

Ans: E

 

  1. Why does mismatch repair not lead to mutation upon repair of heteroduplex DNA?

 

  1. The incorrect base is recognized because it is methylated.
  2. The strand containing the misincorporated base is methylated.
  3. The strand containing the misincorporated base is not methylated.
  4. The strand containing the misincorporated base is phosphorylated.
  5. Actually, mismatch repair does lead to mutation about 50% of the time.

 

Ans: C

 

  1. Which of the following is not a step in nucleotide excision repair (NER)?

 

  1. The UvrA2UvrB complex creates a single-stranded bubble at the damage site to which UvrA subunits are tightly bound and the UvrB subunit dissociates.
  2. The UvrA2UvrB complex scanned the DNA for damage.
  3. Pol I fills the gap created by the removal of the damaged strand and the DNA is sealed by ligase.
  4. UvrC (exinuclease) nicks the damaged strand on either side of the damage.
  5. UvrD (helicase) unwinds and releases the damaged strand.

 

Ans: D

 

  1. Why is transcription-coupled repair (TCR) considered particularly efficient when compared to nucleotide excision repair (NER)?

 

  1. The exinuclease employed in TCR removes a smaller fragment containing the damage.
  2. TCR targets transcriptionally active regions for repair over damage in less active parts of the genome.
  3. It uses error-prone polymerases so it can move through damaged areas faster.
  4. It directly repairs the damage rather than removing and replacing the damaged strand.
  5. TCR requires less protein interactions.

 

Ans: B

 

  1. Thymidine dimers cannot be repaired in humans by which of the following?

 

  1. Nucleotide excision repair (NER)
  2. Transcription-coupled repair (TCR)
  3. Translesion synthesis (TLS) using Pol η
  4. Photoreactivation
  5. All of the above can be used to repair thymidine dimers in human cells.

 

Ans: D

 

  1. Which of the following is NOT true of translesion synthesis (TLS)?

 

  1. Uses polymerases with wider-than-normal active site architecture to accommodate abnormal nucleotides in the template.
  2. Lacks the 3’-5’ exonuclease activity usually required for proofreading.
  3. 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.
  4. TLS can be used to repair lesions such as pyrimidine dimers, bulky adducts, and double-stranded breaks.
  5. All of the above are true.

 

Ans: D

 

  1. What is the most critical difference between global nucleotide excision repair (NER) and transcription-coupled repair (TCR)?

 

  1. The initiation step; in NER the XPC protein recognizes the damage but in TCR it is the RNA polymerase that recognizes the damage.
  2. The excision step; NER uses UvrC (exinuclease) to nick the DNA but TCR uses the AP endonuclease.
  3. The removal step; NER uses UvrD (helicase) to unwind and release the damaged strand but in TCR the damaged strand is degraded by Pol I.
  4. The final step; NER uses Pol I and ligase to fill the gap but TCR uses Pol III and ligase.
  5. None of the above are true.

 

Ans: A

 

  1. 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 by removing the T than to remove the G and form an A-T pair?

 

  1. G-C pairs can form three hydrogen bonds between them and are therefore stronger.
  2. Oxidative damage to guanine by alkylation is the most common reason for G-T pairs so removal of the T will maintain the original G-C pair.
  3. T bases easily tautomerize into C basesRemember, removing the T will make sure the pair stays in its original form.
  4. Answers B and C are true.
  5. Answers A, B, and C are true.

 

Ans: B

 

Short Answer

 

  1. 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 by depurination.

 

How We Know

 

  1. What key element was created by 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.

 

  1. How did Renato Delbecco determine that the repair of UV damage could be increased by 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.

 

 

 

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