Gene Regulation in Prokaryotes Chapter 16 Test Bank Docx

Genetics, 6e (Hartwell)

Chapter 16 Gene Regulation in Prokaryotes

1) A crucial step in the regulation of most bacterial genes occurs

A) at transcription initiation.

B) during RNA splicing.

C) during nuclear export of mRNA.

D) when nonsense suppressors translate mRNAs.

2) What would be the phenotype of a null mutation in the gene encoding Lac repressor?

A) constitutive expression of the lac operon

B) inducible expression of the lac operon

C) permanently repressed expression of the lac operon

D) cannot predict what would happen to expression of the lac operon

3) Transcription and translation can be coupled in bacteria but not eukaryotes because

A) no nuclear membrane exists in prokaryotes.

B) no nuclear membrane exists in eukaryotes.

C) the bacterial DNA is in a single chromosome.

D) eukaryotic chromosomes are found in nucleoids.

4) How do negative regulators such as the Lac repressor prevent RNA polymerase from initiating transcription?

A) by blocking the ribosome binding site

B) by forming a loop in the operator that restricts the passage of the polymerase

C) by physically blocking the DNA binding site of RNA polymerase

D) by binding to the polymerase to inhibit its catalytic activity

5) How does a positive regulator affect transcription by RNA polymerase in prokaryotes?

A) by allowing passage of the polymerase through an operator

B) by interacting with RNA polymerase to increase the frequency of transcription initiation

C) by causing the helix to unwind in the operator allowing easier initiation

D) by making the transcription start site more exposed to the polymerase

6) Proteins whose conformations change when they are bound to an effector molecule are called

A) enzymes.

B) allosteric proteins.

C) regulatory proteins.

D) activator proteins.

E) inhibitory proteins.

7) Catabolic pathways that break down complex substances into more usable units are usually regulated in response to the

A) end products of the pathway.

B) levels of the molecule that is to be broken down.

C) other metabolites that are limiting.

D) levels of enzymes in the pathway.

8) Anabolic pathways involved in the synthesis of essential molecules are usually regulated in response to

A) the end product of the pathway.

B) a substrate of the pathway.

C) other metabolites that are limiting.

D) the levels of enzymes in the pathway.

9) Which statement accurately describes global gene regulation in bacteria?

A) Sigma factors are not involved in gene regulation, only in attachment of the polymerase to the promoter.

B) Alternative sigma factors that recognize different promoters are a mechanism of global gene regulation.

C) E. coli cells devote more energy to the production of ribosomes during stress so that global gene regulation can occur.

D) All promoters are recognized by all sigma factors.

10) How is glucose involved in catabolite repression of the lac operon?

A) Glucose causes cAMP levels to increase, which leads to increased CRP binding and the lac operon is activated even when lactose is present.

B) Glucose causes cAMP levels to decrease, which leads to decreased CRP binding and the lac operon is repressed even when lactose is present.

C) Glucose is also a substrate for β-galactosidase and thus competes with lactose for this enzyme.

D) Glucose activates transcription of the lacI gene to increase the amount of Lac repressor in the cell.

11) The function of a helix-turn-helix motif in a transcription factor is to

A) recruit RNA polymerase.

B) form dimers with other transcription factors.

C) bind a specific sequence in the major groove of DNA.

D) unwind the double helix.

12) How does tryptophan, the end product of the trp operon, function in the regulation of the operon?

A) Trp binds to and inhibits the repressor, thus allowing transcription of the operon.

B) Trp binds to and activates the repressor, which then binds to DNA to allow transcription of the operon.

C) Trp binds directly to DNA and inhibits transcription of the operon.

D) Trp binds to and changes the conformation of the repressor, which can then bind DNA and block transcription of the operon.

13) What term describes regulation of the trp operon that can occur even in trpR− mutants, thus suggesting that the mechanism is repressor-independent?

A) modulation

B) derepression

C) attenuation

D) amplification

14) In E. coli the heat shock response, which alters the proteins produced by the cell at different temperatures, is mediated by

A) degradation of chaperone proteins at high temperatures.

B) denaturing of DNA in the promoters in the genes of heat-sensitive proteins.

C) synthesis of alternative sigma factors at high temperatures to activate transcription of heat-shock genes.

D) increasing the promoter affinity of already existing polymerase sigma factors at high temperatures.

15) As a general principle of gene regulation through operons, regulatory genes encode

A) trans-acting proteins that interact with cis-acting DNA elements.

B) cis-acting proteins that interact with cis-acting DNA elements.

C) cis-acting proteins that interact with trans-acting DNA elements.

D) trans-acting proteins that interact with trans-acting DNA elements.

16) In the trp operon, attenuation occurs through translation of two Trp codons in the leader sequence. What would happen if these two codons were mutated to stop codons?

A) Region 1 will bind to region 3 in the absence of Trp.

B) The antiterminator loop will form in the presence or absence of Trp.

C) Region 3 will bind to region 4 in and transcription will proceed in the presence or absence of Trp.

D) Region 3 will bind to region 1 in the presence of Trp.

17) Initiation of transcription by RNA polymerase involves the binding of which of the following subunits to the core enzyme?

A) delta

B) sigma

C) gamma

D) alpha

E) zeta

18) The transition from transcriptional initiation to elongation involves

A) binding of sigma factor.

B) release of sigma factor.

C) release of RNA polymerase from DNA.

D) binding of RNA polymerase to DNA.

E) binding of rho factor.

19) Which molecules are examples of effector molecules?

A) Allolactose and Trp

B) Lac repressor and Trp repressor

C) The promoter and operator

D) lacY and lacZ

E) Lactose and glucose

20) A single DNA unit that enables the simultaneous regulation of more than one coding region in response to environmental changes is called a(n)

A) promoter.

B) operator.

C) regulator.

D) inducer.

E) operon.

21) Alterations to DNA sites such as promoters and operators can act

A) only in cis.

B) only in trans.

C) either in trans or in cis.

D) These terms apply only to posttranscriptional regulation.

22) The scientists who proposed the operon theory are

A) Monod and Jacob.

B) Watson and Crick.

C) Hardy and Weinberg.

D) Darwin and Mendel.

E) Hershey and Chase.

23) Which of the following is an example of a reporter gene?

A) The linkers that are attached to RNA molecules to perform RNA-Seq

B) The lacZ coding sequence under the control of a prokaryotic promoter

C) The DNase I enzyme used to digest DNA for a footprint assay

D) A human protein expressed under the control of a prokaryotic promoter

24) The sigma factor that mediates a global heat shock response in E. coli is

A) sigma 70.

B) sigma 32.

C) sigma 34.

D) sigma 72.

E) sigma 36.

25) In the regulation of the trp operon, tryptophan acts as a(n)

A) repressor.

B) attenuator.

C) activator.

D) effector.

E) operator.

26) Fifteen different strains of bacteria are auxotrophic for maltose and their mutations do not complement each other. The mutations in these bacteria map to the same DNA region using transformation of random fragments of genomic DNA. Only two different enzymes participate in the maltose biosynthetic pathway, and the genes for both are within a single operon. What does this data suggest?

A) There are fifteen enzymes in maltose metabolism.

B) All the strains have mutations that prevent transcription of the operon.

C) Fifteen genes encode maltose biosynthesis enzymes, but only two genes are crucial.

D) Maltose is the preferred sugar source for the mutant bacteria.

27) Why is it advantageous for transcription factors to be multimeric and for their binding sites to be clustered?

A) The strength of the DNA-protein interactions is increased.

B) The DNA sequence specificity of each DNA-protein interaction is increased.

C) Each DNA-protein complex is better able to bind to RNA polymerase.

D) Competition for binding between different transcription factors for the same regulatory region is decreased.

28) Small RNAs can regulate gene expression by

A) binding target RNAs and inhibiting translation.

B) binding rho and preventing termination.

C) binding target promoter regions and preventing transcription.

D) binding target operator regions and preventing transcription.

29) What is the relationship between the promoters for antisense RNA and the promoters of their target genes?

A) The promoters are oriented so that they transcribe opposite strands of the same segment of DNA.

B) The promoters are found in different chromosomal regions but are controlled by the same sigma factors.

C) The sense and antisense RNAs use the same promoter, both RNAs are produced from a single transcript that is processed.

D) The sense and antisense RNAs are transcribed using the same promoter that is bidirectional.

30) What is an advantage of using the lac regulatory region to control the expression of exogenous proteins in bacterial cells?

A) Expression can be controlled by the addition of an inducer to the media.

B) This regulatory region functions in both bacterial and mammalian cells.

C) By using this regulatory region, eukaryotic genomic DNA can be expressed in bacterial cells.

D) Host regulatory mechanisms are not effective when this regulatory region is cloned into a plasmid.

31) Why is the analysis of transcriptomes useful for genetics research?

A) It can reveal many aspects of gene regulation.

B) It is easy to do without computer programs.

C) It can help discover nontranscribed DNA regions.

D) It directly identifies transcription factor binding sites.

32) How was E. coli was used to aid in the cloning of the bioluminescence genes from Vibrio fischeri?

A) Gene fragments from V. fischeri were transformed into E. coli and bioluminescent E. coli were identified.

B) Gene fragments from V. fischeri were cloned into plasmids, transformed into E. coli, and then retransformed back into V. fischeri.

C) Gene fragments from E. coli were used to disrupt the bioluminescence genes of V. fischeri by homologous recombination.

D) Transposons from E. coli were used to mutagenize the bioluminescence genes of V. fischeri.

33)  V. fischeri senses the density of V. fischeri cells in the environment by

A) secretion of a molecule that can then be internalized and bound to a receptor in the cytoplasm.

B) cell surface receptors that interact with other cell surface receptors on nearby cells.

C) a cell surface receptor that binds to a molecule secreted by cells in the environment.

D) a cytoplasmic receptor that binds to the LuxC protein.

34) What is one possible advantage to targeting quorum-sensing mechanisms for antibiotic development?

A) There may be no selective advantage to individual cells that are resistant to quorum-sensing inhibitors.

B) There is a selective advantage to those cells lacking a quorum-sensing system.

C) Quorum-sensing mechanisms are found in all pathogenic bacteria.

D) Quorum-sensing mechanisms are found only in human pathogens.

35) A mutation in a ribosome binding site in the lac operon would likely result in

A) reduction in the amount of LacZ, LacY, or LacA protein.

B) reduction in the amount of LacZ, LacY, and LacA proteins.

C) complete loss of transcription of the lac operon.

D) a lac mRNA that is shorter than wild-type.

36) Cells of which bacterial genotype would be able to utilize lactose as an energy source?

A) IS o+Z+Y+

B) I+o+Z+Y−/F′ (ISo+Z−Y+)

C) I+ocZ+Y−/F′ (I+o+Z−Y+)

D) I+o+Z−Y+

37) A fragment of DNA with a radioactive label is incubated with DNase in the presence and absence of a protein that acts as a positive regulator of transcription. The samples are then run on a gel and visualized with autoradiography. Which statement describes a possible result and conclusion from such an experiment?

A) The banding pattern on the gel is the same with and without the transcriptional regulator, therefore the protein binds the DNA fragment.

B) Some bands on the gel are absent in the sample with the transcriptional regulator, therefore the protein does not bind the DNA fragment.

C) Some bands on the gel are absent in the sample with the transcriptional regulator, therefore the protein binds the DNA fragment.

D) Some bands on the gel are absent in the sample without the transcriptional regulator, therefore the protein does not bind the DNA fragment.

38) An enzyme functions in a biosynthetic pathway in bacteria. You predict that the product of the reaction may regulate the levels of the protein by

A) binding an aptamer in the RNA leader sequence that alters the conformation to block the ribosome binding site.

B) binding an aptamer in the RNA leader sequence that alters the conformation to expose the ribosome binding site.

C) binding the promoter region to prevent transcription.

D) binding an aptamer in the RNA leader sequence that causes an antiterminator to form.

39) You have identified a 50 nucleotide deletion that is not within an open reading frame in an otherwise wild-type prokaryotic genome. You identify five mRNAs that are not translated in these mutant bacteria; the five mRNAs are transcribed from five different genes located far from the deletion. Select the most likely explanation for the effect of this deletion.

A) The deletion is within an antisense RNA that regulates translation of all five genes.

B) The deletion prevents expression of a small RNA that regulates translation of all five genes.

C) The deletion removes an aptamer in an RNA leader sequence.

D) The deletion removes an effector molecule that regulates transcription.

40) When RNA-Seq is performed before and after heat shock of bacterial cells, computer analysis would be expected to show

A) very few changes in the cDNAs sequenced because the response to heat shock occurs primarily at the protein level.

B) after heat shock, an increase in the average length of sequence reads of cDNAs that encode molecular chaperone proteins.

C) after heat shock, an increase in the number of sequence reads of cDNAs that encode molecular chaperone proteins.

D) after heat shock, a decrease in the number of sequence reads of cDNAs that encode molecular chaperone proteins.