6.1 Multiple Choice Questions
1) Regulation of enzyme activity occurs
- A) at the start of transcription.
- B) at the start of translation.
- C) posttranslationally.
- D) at any point on the enzymatic production pathway.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.15
2) When arginine is added to a culture already growing exponentially in a medium without arginine, what occurs?
- A) All cellular growth ceases.
- B) Growth continues, but the production of enzymes required for the synthesis of arginine ceases.
- C) Growth continues, but the production of enzymes required for the synthesis of arginine increases.
- D) The cell returns to the lag stage of growth to synthesize the proteins necessary for the metabolism of arginine.
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.2
3) Regulatory proteins
- A) are influenced by small molecules.
- B) bind to specific DNA sites.
- C) regulate transcription.
- D) regulate transcription, bind specific DNA sites, and can be influenced by small molecules.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
4) A protein region with a specific function and structure is called a
- A) conserved site.
- B) domain.
- C) locale.
- D) motif.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
5) Transcriptional regulators bind most frequently at the ________ site of DNA.
- A) major groove
- B) minor groove
- C) histone complex
- D) primary supercoil
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
6) Which type of regulator(s) specifically binds to operator regions of DNA?
- A) activators
- B) activators and inducers
- C) repressors
- D) repressors and corepressors
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
7) The lac operon is an example of ________ control in which the presence of an ________ is required for transcription to occur.
- A) negative / activator
- B) negative / inducer
- C) positive / activator
- D) positive / inducer
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
8) Enzyme induction occurs
- A) when the substrate is present.
- B) when the organism is environmentally stressed.
- C) continuously.
- D) when the substrate is depleted.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
9) Considering the catabolite repression mechanism, which observation would make you suspect it IS occurring?
- A) CRP bound to promoter sites
- B) elevated levels of transcripts for maltose and sucrose catabolism
- C) relatively low intracellular cyclic AMP levels
- D) RNA polymerase bound to biosynthetic promoter sequences
Answer: C
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.4
10) During a growth curve of Aliivibrio fischeri, when would you expect to see the strongest bioluminescence?
- A) lag phase
- B) early to middle log phase
- C) late log to early stationary phase
- D) middle to late stationary phase
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.8
11) What occurs when an inducer is added to an environment containing an organism with a metabolic pathway controlled by a repressor?
- A) The inducer combines with the repressor and activates the pathway.
- B) The inducer combines with the repressor and inactivates the pathway.
- C) The inducer combines with the substrate and blocks induction.
- D) The inducer does not combine with, but functions as a chaperone molecule for, the enzyme-substrate complex.
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.2
12) Which of the following do NOT bind to the promoter sequence during regulation?
- A) activators
- B) inducers
- C) repressors
- D) None of these bind directly to the promoter sequence.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
13) Cyclic AMP is synthesized from ATP by an enzyme called ________ which is involved in ________.
- A) adenylate cyclase / catabolite repression
- B) adenylate cyclase / transcriptional activation
- C) cAMP receptor protein (CRP) synthase / catabolite repression
- D) cAMP receptor protein (CRP) synthase / transcriptional activation
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.4
14) In negative control of transcription by the lac operon, how does the presence of an inducer affect transcription?
- A) The inducer binds to the operator.
- B) The inducer does not bind to the operator.
- C) The inducer causes the repressor to bind to the operator.
- D) The inducer prevents the repressor from binding to the operator.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
15) Mechanisms for controlling enzyme activity include all of the following EXCEPT
- A) degradation of the enzyme.
- B) feedback inhibition.
- C) covalent modification of the enzyme.
- D) addition of short sections of new amino acid sequence.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
16) The function of a kinase is
- A) methylation.
- B) response regulation.
- C) phosphorylation.
- D) glycosylation.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.6
17) The promoters of positively controlled operons require activator proteins because
- A) RNA polymerase easily recognizes the consensus sequence.
- B) they are required to inactivate the repressor proteins.
- C) the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter.
- D) they are needed to bind to the allosteric site of RNA polymerase.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.3
18) Transcriptional control in Archaea most closely resembles that in ________ and utilizes ________.
- A) bacteria; regulatory proteins
- B) eukaryotes; regulatory proteins
- C) bacteria; transcription factors
- D) eukaryotes; transcription factors
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.5
19) The most frequent way in which small regulatory RNA molecules exert their effects is by
- A) base pairing with other RNA molecules that have regions of complementary sequence.
- B) binding to a repressor and repressing enzyme transcription.
- C) acting as an inducer that then binds to an activator protein to allow transcription to proceed.
- D) signal transduction.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.11
20) ________ pathways typically rely on ________ proteins to inhibit mRNA synthesis.
- A) Anabolic; repressor
- B) Catabolic; repressor
- C) Anabolic; activator
- D) Catabolic; activator
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
21) ________ pathways typically utilize ________ proteins that stimulate binding of RNA polymerase to DNA.
- A) Anabolic; repressor
- B) Catabolic; repressor
- C) Anabolic; activator
- D) Catabolic; activator
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
22) In Bacteria, sensor kinases that respond to extracellular signals transfer this signal to the cytoplasmic machinery by typically phosphorylating the residues.
- A) histidine
- B) serine
- C) threonine
- D) tyrosine
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.6
23) Quorum sensing is a regulatory system that requires a certain cell density to work effectively. Each of the following activities utilizes quorum sensing EXCEPT
- A) Staph aureus
- B) transition of Candida albicans from budding yeast to elongated filaments.
- C) flagella synthesis in Proteus vulgaris.
- D) light emission by Aliivibrio fischeri.
Answer: C
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.8
24) Chemotaxis in bacteria occurs through the use of
- A) adaptation.
- B) quorum sensing.
- C) autoinduction.
- D) a modified two-component system.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.7
25) Which statement is TRUE of two separate regulators controlling one individual operon?
- A) The two regulators themselves must respond to different signals, which enables both to control the operon differently.
- B) One regulator will likely control the transcription of one section of the operon, whereas the other regulator will control the other component.
- C) One regulator will bind to the operator region whereas the other will bind to the promoter region so they can co-occur and co-regulate the operon.
- D) Two regulators trying to control the same operon will likely result in only one being maintained after several generations.
Answer: A
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.4
26) How would you design an sRNA to bind to a sequence?
- A) select six continuous nucleotides from the sequence
- B) take the complementary sequence of six continuous nucleotides
- C) select 200 continuous nucleotides from the sequence
- D) take the complementary sequence of 200 continuous nucleotides
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.11
27) When the nontemplate strand of a gene is transcribed into RNA, what is likely to result?
- A) A complementary sRNA will bind to it and form a functional ribozyme with secondary structure.
- B) It will complementarily bind to the gene sequence, form a hairpin loop, and transcriptionally repress the gene.
- C) The complementary mRNA transcribed from the template strand will bind to it and halt its translation.
- D) A global regulator will identify this as a stress, respond by inducing ribonuclease production, and it will be degraded.
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.11
28) Based on our understanding of the early stages of life, ________ is/are thought to be one of the earliest forms of metabolic regulation that evolved.
- A) attenuation
- B) feedback inhibition
- C) riboswitches
- D) transcription factors
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.12
29) Attenuation is a type of regulation that can control
- A) allosteric enzyme activity.
- B) transcriptional activity exclusively.
- C) translational activity exclusively.
- D) both transcriptional and translational activity.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.13
30) Which of the following is a characteristic of an isoenzyme?
- A) More than one enzyme is regulated by the same mechanism.
- B) The same reaction can be catalyzed by multiple enzyme variants.
- C) Multiple binding sites on the same enzyme enable multiple regulation mechanisms.
- D) More than one gene makes the same enzyme.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.14
31) Which regulatory mechanism does NOT depend on a conformational change in protein/enzyme structure to change activity?
- A) attenuation
- B) catabolite repression
- C) feedback inhibition
- D) negative control
Answer: A
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.13
32) An organism grown in nutrient rich broth to high turbidity always produces a blue pigment. When a large inoculum is transferred to a nutrient rich agar plate, it also appears blue. A researcher noticed that this never happens when very small colonies are grown on low nutrient agar plates, however. What is the most plausible conclusion?
- A) Large populations enabled the differentiation of a subpopulation of cells that created the blue pigment.
- B) Only high nutrient conditions provide enough energy for cells to produce this secondary metabolite that appears blue.
- C) Production of the blue pigment production is linked to quorum sensing.
- D) The strong gradient from very high to low nutrient bioavailability induces production of the blue metabolite.
Answer: C
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.8
33) A bacterium that either partially or fully catabolizes an acyl-homoserine lactone will likely disrupt
- A) attenuation.
- B) chemotaxis.
- C) endospore formation.
- D) quorum sensing.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.8
34) Phosphorylation of ________ regulates which direction a flagellum rotates, thus controlling whether an organism runs or tumbles.
- A) CheAW
- B) CheB
- C) CheY
- D) CheZ
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.7
35) Which organism would likely harbor the most two-component regulatory systems?
- A) an archaeon living in an extreme environment
- B) a bacterium occupying a heterogeneous niche with high nutrient mixing
- C) an organism capable of quorum sensing
- D) a parasitic bacterium living inside another organism
Answer: B
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.6
36) How is the activity of a riboswitch controlled?
- A) by other riboswitches
- B) metabolite binding can change its structure
- C) sigma factor binding alters its structure
- D) small RNA complementary binding disrupts its function
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.12
37) A mutation in the gene encoding the lactose repressor (lacI) that prevents lactose from binding to the LacI protein would result in
- A) constant expression of the lac operon in the absence of lactose.
- B) constant repression of the lac operon in the absence of lactose.
- C) constant expression of the lac operon in the presence of lactose.
- D) constant repression of the lac operon in the presence of lactose.
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.4
38) Why are two-component regulatory systems particularly useful for controlling gene expression in response to environmental signals?
- A) Two proteins controlling a gene means there are two chances to activate a gene.
- B) Two proteins delay the response time so the cell can be sure the change is permanent.
- C) One of the two proteins can be exposed to the external environment to receive a signal.
- D) Phosphorylation is a permanent change so genes are always turned on after signal.
Answer: D
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.6
39) In order for the helix-turn-helix motif to bind to DNA, the ________ must fit into the major groove of the DNA.
- A) homeotic switches
- B) recognition helix
- C) operator
- D) zinc fingers
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
40) Infection by the foodborne pathogen E coli O157:H7 involves all of the following host and pathogen signaling molecules EXCEPT
- A) AI-3.
- B) MCP.
- C) epinephrine.
- D) norepinephrine.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.8
41) Global regulatory systems include all of the following EXCEPT
- A) amino acid synthesis.
- B) transformation.
- C) heat shock response.
- D) anaerobic respiration.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.10
42) When more than one operon is under the control of a single regulatory protein, the operons are collectively called a(n)
- A) regulon.
- B) operator.
- C) autoinducer.
- D) riboswitch.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.3
43) Most riboswitches control
- A) transcription.
- B) translation.
- C) ribosome synthesis.
- D) tRNA synthesis.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.12
44) PII proteins are a family of signal-transducing proteins that regulate ________ during nitrogen metabolism.
- A) transcription factors.
- B) enzymes.
- C) membrane transport proteins.
- D) PII proteins regulate each of the above during nitrogen metabolism.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.15
45) All of the following are functions of heat shock proteins in bacteria EXCEPT
- A) prevention of inappropriate protein subunit aggregation.
- B) degradation of denatured proteins.
- C) responding to exposure to high levels of ethanol.
- D) stimulation of binary fission.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.10
6.2 True/False Questions
1) Proteins required at approximately the same level throughout a cellʹs growth cycle are often not subject to regulatory mechanisms and are constitutively synthesized.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
2) Short regions at the beginning and end of gene sequences are not translated into proteins.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
3) Small molecules usually act directly (rather than indirectly) in regulating transcription.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
4) A common structure for proteins that bind DNA is helix-turn-helix.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
5) The residues in a DNA-binding protein that interact with DNA usually correspond to each other according to the amino acids encoded by the DNA. For example, a DNA sequence containing AGC-AGA-CAG which encodes for Ser-Arg-Gln would likely have a DNA-binding protein with the sequence Ser-Arg-Gln.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.1
6) Some proteins that bind to DNA block transcription, whereas other proteins activate transcription.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
7) Enzyme repression typically affects catabolic pathways.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.2
8) Depending on the type of regulatory mechanism, activators and repressors can bind to operator regions which control transcription.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.2
9) A regulon is composed of at least two operons regulated by the same protein.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.3
10) The preferential use of glucose over other available carbon substrates for growth is mechanistically explained by catabolite repression.
Answer: TRUE
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.4
11) Proteins and RNA molecules that are needed in the cell at about the same level under all growth conditions, require constitutive expression.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.1
12) A two-component regulatory system usually involves both the sensor and response proteins being subject to phosphorylation.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.6
13) Cyclic AMP (cAMP) is involved in the global regulation of catabolic pathways in Escherichia coli, including the lac operon. A high activity of the enzyme adenylate cyclase which makes cAMP is suggestive of catabolite repression.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.4
14) Once cytoplasmic sensor proteins involved in chemotaxis regulation are phosphorylated, a cascade of other phosphate-transferring mechanisms provides ATP to rotate flagella.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.7
15) In most cases, the first product of a particular biosynthetic pathway represses the enzymes of the pathway.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.14
16) In catabolic repression, cells always use the least abundant carbon source first.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.4
17) The small size of prokaryotic cells limits their ability to respond to spatial gradients of a chemical.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.7
18) Adaptation is based on the level of phosphorylation of MCPs.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.7
19) Quorum sensing relies upon a large cell population which then turns on transcription.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.8
20) During chemotaxis, attractants increase the rate of autophosphorylation, whereas repellants decrease this rate.
Answer: FALSE
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.7
21) Heat shock proteins are produced in high number to minimize damaging effects during high heat conditions, in addition to other stress signals unrelated to heat such as ultraviolet radiation.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.10
22) Regulation of transcription in Archaea is more similar to that which occurs in eukaryotes than in bacteria.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.5
23) The stringent response is a mechanism used by bacteria that provides increased ability to survive in environments lacking nutrients.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.9
24) When environmental levels of amino acids become limited, synthesis of mRNA, rRNA, and tRNA increases immediately.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.9
25) In many environments, the limiting nutrient for bacterial growth is phosphorus.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.10
26) Multiple mechanisms for small RNA (sRNA)-mediated translational regulation exist, but all are unified by inactivating protein synthesis of the target mRNA.
Answer: FALSE
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.11
6.3 Essay Questions
1) Explain a common mechanism used by cells to ensure proper binding of DNA-binding proteins to DNA.
Answer: DNA-binding proteins are proteins composed of DNA-binding domains and that have a specific or general affinity for either single or double stranded DNA. One of the most common domains is the helix-turn-helix structure which consists of two segments of polypeptide chain that have an α-helix secondary structure connected by a short sequence forming the “turn.” The first helix is the recognition helix which interacts specifically with the DNA. The second helix is the stabilizing helix which stabilizes the first helix by interacting with it through hydrophobic interactions. The turn linking the two helices consists of three amino acids with glycine typically the first. Sequences are recognized by noncovalent interactions between the recognition helix of the protein and specific chemical in the sequence of base pairs on the DNA.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.1
2) Using the tryptophan operon as a model, describe the formation of a stem-loop structure and its function.
Answer: A pause in mRNA transcription occurs when tryptophan is in short supply and therefore requires the additional recruitment of a charged tryptophan tRNA for its synthesis. This lag or pause time provides the already synthesized mRNA component to bind to a different part of the transcript than would occur if there were no pause time. This different hairpin loop formed does not interact with the DNA sequence and therefore permits RNA polymerase to continue transcribing the entire mRNA, unlike what would occur if a lag time were absent.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.13
3) Explain the phenomenon known as diauxic growth. Hypothesize why this type of regulation might be important for the survival of a microbe in a mixed population.
Answer: Consider Escherichia coli as an example: when Escherichia coli is exposed simultaneously to both glucose and lactose, it will preferentially use glucose before lactose. Diauxic growth therefore explains how cells control gene expression, often as operons and regulons, to achieve the fastest growth rate. If a cell were to not have this mechanism, it might be outcompeted in the environment by other faster growing and replicating cells. Lacking a global catabolite regulation system might be advantageous for cells occurring in a mixed population where competition for a single substrate could provide poor growth yields.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.4
4) Describe the basic two-component regulatory system, and explain the function of each component.
Answer: The first component of the system involves a specific sensor kinase protein located in the cytoplasmic membrane which receives an external environmental signal. The kinase autophosphorylates itself once signaled and creates a phosphate ion for the second component. A response regulator protein within the cellʹs cytoplasm then binds the phosphate ion and ultimately binds DNA. The regulation in response to this DNA binding can either be positive or negative, depending on the system.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.6
5) Explain why is it unsurprising that two-component regulatory systems are all but absent in bacteria that live at the expense of a host.
Answer: Answers will vary, but a possible explanation is that an organism that has adapted to living inside of a host has such a narrow metabolic capacity that responses to nutrient limitation are less necessary than a bacterium growing in a dynamic community.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.6
6) Explain why one cell of a pathogenic bacterium by itself typically does not secrete a toxin, despite the species as a whole being characterized as toxin producing. To help in your explanation, predict what would happen to an individual pathogen that secreted a toxin inside a human.
Answer: One pathogenic bacterium is unlikely to be able to produce a sufficient number of toxin molecules to cause a disease. Once the bacterium grows to a sufficient quantity that it can cause disease through the production of toxins, coordination of toxin production by the population is performed by quorum sensing. The production of a toxin by a single cell would likely alert the immune system a foreign bacterium is present and would be killed by a healthy immune system.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.8
7) How is transcription in Archaea controlled?
Answer: Archaea have mechanisms to control transcription, which are most similar to those found in Bacteria rather than other mechanisms such as those in Eukarya. They contain both activators and repressors, which can regulate RNA polymerase activity in either a positive or negative way.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.5
8) Contrast prokaryotic and eukaryotic regulation of gene expression.
Answer: In Bacteria and Archaea, a common mechanism for regulation of RNA polymerase activity is to use DNA-binding proteins that either block RNA polymerase activity (repressor proteins) or stimulate RNA polymerase (activator proteins). An alternative mechanism found commonly in eukaryotes is to coordinate numerous DNA-binding proteins known as transcription factors to interact with RNA polymerase.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.5
9) Explain the function of cAMP in catabolite repression.
Answer: Adenylate cyclase synthesizes the regulatory nucleotide cAMP, and when the substrate glucose is present this process is inhibited (or repressed). This means glucose is a preferred substrate for catabolism in cells that contain this mechanism, such as E. coli. Once glucose has been used up, cAMP is then produced, which forms a cAMP—CAP—DNA complex, resulting in transcription of the genes that the CAP corresponds to, and catabolism of an additional substrate can be performed.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.4
10) A cocktail of 30 compounds was identified to harbor a corepressor, and subsequent separation and purification was performed to test individual compounds. Describe an experiment that would enable you to identify the specific compound that acts as a corepressor for a catabolic pathway. Be certain to explain what will be measured and how you could conclude which compound is a corepressor.
Answer: Experimental designs will vary, depending on whether transcripts, proteins, or enzyme function is quantified. Measurement of each would enable measuring activity of the repressor. In each case, a compound added to the system which shows significantly lower (basal) activity would indicate it is a corepressor. Compounds that do not act as a corepressor would result in no change in activity, meaning transcripts, proteins, and enzyme function would still be observed.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.2
11) In considering the function of heat shock proteins, why is it not surprising that these proteins are both highly conserved and very ancient? Provide your reasoning.
Answer: Answers will vary, but one idea is for an organism to survive in the environment, it must be able to survive unfavorable conditions. Recognizing that heat shock proteins generally facilitate the refolding or removal of misfolded proteins, whether it be from heat or other conditions, it makes sense these are so critical for a cell to survive that they would have needed to be present even in ancient organisms. Additionally, whether a protein is misfolded in a eukaryote or a prokaryote, it seems likely that a heat shock protein would be needed to refold it. These proteins may be similar, which explain the high conservation throughout all life forms.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.10
12) Provide evidence that supports the hypothesis that riboswitches are remnants of the RNA world when catalytic RNAs were the only self-replicating life forms.
Answer: Answers will vary, but one focus could be on their presence in bacteria that are considered to have evolved well before Eukarya, yet riboswitches are generally lacking in eukaryotes. If no other life forms were around and RNA was the only replicating form, the function of riboswitches could explain a (primitive) mechanism to control its own synthesis.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.12
13) Compare and contrast chemical, spatial, and temporal gradients as found in bacterial regulatory systems.
Answer: Bacteria cannot sense spatial gradients of chemicals due to their small size, so they are capable only of determining a chemical gradient over time. That is, they can sense the change in concentration of a chemical over time rather than the absolute concentration of the chemical stimulus. These temporal changes of a chemical are sensed with a two-component system. In the example of chemotaxis, an attractant or repellantʹs presence results in a change in activity of pre-existing flagella.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 6.7
14) Two common levels of regulation of protein activity are transcriptional control and posttranslational control. Discuss the differences between the two in terms of speed of response and energy efficiency.
Answer: Transcriptional control will be the slowest (the gene needs to be transcribed, the mRNA translated, and the protein correctly folded before protein activity) but is the most energy efficient because proteins are only made when they are needed. Posttranslational control is the fastest (the protein is already synthesized; it just needs a modification) but is less energy efficient because a protein might be made and degraded before it ever gets activated.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.14
15) Many of the virulence factors produced by the opportunistic pathogen Pseudomonas aeruginosa are the result of quorum sensing. What role do autoinducers play in this mechanism of bacterial pathogenicity?
Answer: Through the use of autoinducers, bacteria can regulate their behavior according to population density. The phenomenon of quorum sensing relies on the principle that when a single bacterium releases autoinducers (AIs) into the environment, their concentration is too low to be detected. However, when sufficient bacteria are present, autoinducer concentrations reach a threshold level that allows the bacteria to sense a critical cell mass and, in response, to activate or repress target genes including those of various virulence factors.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.8
16) Describe the difference between constitutive and inducible genes. Based on your knowledge of bacterial cells, give an example of each.
Answer: Constitutive genes are those which are transcribed at a relatively constant level regardless of the cell’s environmental conditions. These genes are sometimes referred to as ‘housekeeping genes’ and include genes for heat shock proteins such as dnaK in Escherichia coli. The expression of inducible genes is highly regulated and these genes must be able to be rapidly and specifically activated in response to environmental stimuli. Examples of inducible genes would include those coding for activator and repressor proteins.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.1
17) What are sigma factors? Explain the roles of the two different sigma factors known as RpoH and RpoS.
Answer: A sigma factor (σ factor) is a protein needed only for initiation of RNA synthesis. The specific sigma factor used to initiate transcription of a given gene will vary, depending on the gene and on the environmental signals needed to initiate transcription of that gene. RpoH is the heat shock sigma factor and is turned on when bacteria are exposed to heat. RpoS is highly expressed during the transition from the exponential to stationary phase of the bacterial growth curve.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 6.10
18) What would be the result of a mutation that results in the inactivation of the Tar MCP in Escherichia coli?
Answer: The mechanism of chemotaxis depends upon a signal cascade of multiple proteins. These sensory proteins allow the cell to monitor the concentration of various substances over time and are called methyl-accepting chemotaxis proteins (MCPs). The Tar MCP produced by Escherichia coli senses aspartate and maltose and the repellants cobalt and nickel. Binding of an attractant or repellant triggers interactions with the cytoplasmic proteins that affect flagellar rotation. A mutation in the gene for Tar MCP would result in inability to move toward aspartate and maltose or away from cobalt and nickel.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.7
19) Explain how adaptation of the chemotaxis system is achieved once an organism has responded to a stimulus.
Answer: Once an organism has successfully responded to a stimulus, it must stop responding and reset the sensory system until a further signal is received. During adaptation of the chemotaxis system, a feedback loop that relies on the response regulator CheB resets the system. This is accomplished by methylation and demethylation of the MCPs by CheR and phosphorylated CheB respectively.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 6.7
Brock Biology of Microorganisms, 15e (Madigan et al.)
Chapter 7 Molecular Biology of Microbial Growth
7.1 Multiple Choice Questions
1) Fluorescence microscopy allows visualization of all of the following EXCEPT
- A) cytoskeletal proteins.
- B) nuclear proteins.
- C) interactions between cytoplasmic proteins.
- D) All of these answer choices can be visualized.
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.1
2) Once replication of the chromosome has begun, DnaA is inactivated
- A) as a result of being blocked from oriC
- B) by increased expression of FtsZ.
- C) as a result of increased binding to ATP.
- D) by increased binding of DnaA-ATP to oriC.
Answer: A
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.2
3) ________ prohibits re-initiation of chromosome replication.
- A) DnaA-ATP
- B) SeqA
- C) oriC
- D) FtsZ
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
4) Prior to DNA replication, both strands of the chromosome are methylated on the ________ residue of the sequence-GATC-.
- A) A
- B) G
- C) C
- D) T
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
5) Hemimethylated DNA is found on
- A) the newly synthesized strand of DNA immediately after replication.
- B) the parental strand of DNA immediately after replication.
- C) both strands of dsDNA immediately after replication.
- D) neither strand of DNA immediately after replication.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
6) The typical time required for chromosome replication in Escherichia coli is
- A) 10 minutes.
- B) 20 minutes.
- C) 30 minutes.
- D) 40 minutes.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
7) The typical generation time of Escherichia coli is approximately ________ under optimal environmental conditions.
- A) 10 minutes
- B) 20 minutes
- C) 30 minutes
- D) 40 minutes
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
8) The Par system is necessary for
- A) formation of a replication fork in Escherichia coli.
- B) distribution of genetic material in replicating Caulobacter.
- C) septum formation.
- D) elongation of bacillus prior to cell division.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
9) Which of the following proteins is most active in divisome complexes?
- A) FtsZ
- B) crescentin
- C) MinCD
- D) MreB
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.3
10) A bacterium that lacks the mreB gene will have a ________ shape.
- A) bacillus
- B) coccoid
- C) short bacillus
- D) vibrio
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.4
11) What are the primary regulator units that control endospore formation?
- A) allosteric proteins
- B) antisense RNAs
- C) riboswitches
- D) sigma factors
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
12) All of the following are TRUE statements concerning binary fission of microbial cells EXCEPT
- A) the chromosome of the cell is replicated.
- B) elongation of the cell occurs and the chromosomes are pushed apart.
- C) a septum is formed across the midline of the cell.
- D) daughter cells produced can be of different sizes.
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.6
13) There are four basic stages to biofilm formation. Which of the following is the correct order of these stages?
- A) attachment, colonization, development, dispersal
- B) colonization, attachment, development, dispersal
- C) development, colonization, attachment, dispersal
- D) attachment, development, colonization, dispersal
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
14) Pseudomonas aeruginosa is a(n) ________ pathogen.
- A) opportunistic
- B) idiopathic
- C) pessimistic
- D) iatrogenic
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
15) Which of the following activities is a result of cyclic di-guanosine monophosphate synthesis?
- A) transition from planktonic to sessile growth during biofilm formation
- B) reduction in activity of flagellar motor
- C) biosynthesis of extracellular matrix during biofilm formation
- D) All of these answer choices activities are a result of cyclic di-guanosine monophosphate synthesis.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
16) The major symptoms of the genetic disorder cystic fibrosis are caused due to biofilm formation by
- A) Serratia marcesans.
- B) Pseudomonas aeruginosa.
- C) Vibrio cholera.
- D) Bacillus subtilis.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
17) All of the following are targets of currently used antibiotics EXCEPT
- A) DNA synthesis.
- B) biofilm formation.
- C) transcription.
- D) translation.
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.10
18) When a Bacillus anthracis population suddenly must form spores to survive a harsh nutrient poor environment, how do the cells obtain energy?
- A) Cells in a growth phase that have not used up all of their energy will be the only ones to make endospores, which is why relatively few endospores are often made from a large population.
- B) Intracellular energy reserves are quickly made available to produce endospores.
- C) Slow responding cells are cannibalized by others that already began spore formation.
- D) Global regulation is initiated to minimize energy waste in biosynthetic pathways and catabolic pathways are increased to consume remaining usable substrates to fuel spore formation.
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.6
19) Bacteria from the genus Caulobacter are used to model cellular differentiation in eukaryotes. The abundance of CtrA, DnaA, and GcrA separately control activity of other genes necessary for differentiation in Caulobacter. Thus, these three proteins can be classified as
- A) activating sensors.
- B) heterologous regulators.
- C) differentiating regulons.
- D) transcriptional regulators.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.7
20) A bacterium that either partially or fully catabolizes an acyl-homoserine lactone will likely disrupt
- A) attenuation.
- B) chemotaxis.
- C) endospore formation.
- D) quorum sensing.
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.9
21) Quorum sensing generally follows the mechanism of which type of regulation?
- A) feedback inhibition
- B) negative transcriptional regulation
- C) positive transcriptional regulation
- D) a two-component regulation system
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.9
22) Penicillins work by
- A) activating autolytic enzymes.
- B) blocking the transpeptidation reaction that forms peptidoglycan cross-links.
- C) stimulating autolysins to form holes in the plasma membrane.
- D) All of these answer choices are true.
Answer: D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
23) Differential gene expression occurs in each of the following bacterium EXCEPT
- A)
- B)
- C)
- D)
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
24) σF functions as a(n) ________ during endospore formation.
- A) transcription factor
- B) anti-sigma factor
- C) protease
- D) RNase
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
25) When Spo0A is highly phosphorylated
- A) heterocyst formation is initiated.
- B) stalk formation is initiated.
- C) endospore formation is initiated.
- D) photosynthesis is initiated.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
26) What is the function of heterocysts in cyanobacteria such as Anabaena?
- A) nitrogen fixation
- B) sexual reproduction
- C) asexual reproduction
- D) photosynthesis
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
27) The major shape-determining factor in Bacteria is the protein
- A) FtsZ.
- B) crescentin.
- C) MreB.
- D) MinD.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.3
28) In the process of binary fission in Bacteria, which action occurs first?
- A) formation of the divisome
- B) DNA replication
- C) cell elongation
- D) cytokinesis
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.3
29) FtsZ is related to ________, an important protein involved in cell division.
- A) actin
- B) myosin
- C) tubulin
- D) collagen
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.3
30) FtsA is related to ________, an important cytoskeletal protein in eukaryotes.
- A) actin
- B) myosin
- C) tubulin
- D) collagen
Answer: A
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.3
31) Endospore formation is stimulated when
- A) bacterial growth ceases due to limitation of an essential nutrient.
- B) the bacterium is undergoing binary fission.
- C) bacteria are dividing exponentially.
- D) nutrient levels rise.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
32) Proteins MinC, MinD, and MinE interact to
- A) help guide FtsZ to the cell midpoint.
- B) initiate peptidoglycan synthesis.
- C) assist chromosome segregation.
- D) determine cell shape.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.4
33) The curved-rod shape typical of Caulobacter is due to the protein(s)
- A) crescentin.
- B) MreB.
- C) crescentin and MreB.
- D) neither crescentin nor MreB.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.4
34) Which of the following traits is NOT characteristic of the Caulobacter life cycle?
- A) Cell division only occurs in the swarmer stage.
- B) Cell division only occurs in the stalked cell stage.
- C) Chromosome replication only occurs in the swarmer stage.
- D) Chromosome replication and cell division only occurs in the stalked cell stage.
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.7
35) Biofilm formation in Pseudomonas aeruginosa is triggered by ________ cell densities and repressed by ________ cell densities.
- A) high; high
- B) low; low
- C) high; low
- D) low; high
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.9
36) Biofilm formation in Vibrio cholera is triggered by ________ cell densities and repressed by ________ cell densities.
- A) high; high
- B) low; low
- C) high; low
- D) low; high
Answer: D
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.7
37) What is the function of bactoprenol?
- A) It is a hydrophobic alcohol that transports peptidoglycan precursors across the cytoplasmic membrane.
- B) It is responsible for forming the peptide cross-links between muramic acid residues in adjacent glycan chains.
- C) It triggers the recruitment of FtsZ and the initiation of the divisome.
- D) It supplies the energy necessary for transpeptidation to occur.
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.5
38) Shortage of key nutrients such as ________ results in transcription and secretion of a toxic protein that leads to cannibalization of neighboring cells of the same species.
- A) magnesium
- B) phosphate
- C) calcium
- D) iron
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
39) Heterocysts are specialized cells that undergo
- A) oxidative phosphorylation.
- B) photophosphorylation.
- C) nitrogen fixation.
- D) lysine decarboxylation.
Answer: C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.8
40) Which of the following is NOT a function of DnaA in Caulobacter?
- A) initiation of DNA replication
- B) initiation of swarming
- C) transcriptional regulation
- D) All of these answer choices are functions of DnaA.
Answer: B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.7
41) Why is the presence of a cell wall significant from a clinical standpoint?
- A) All types of cells have a cell wall, and it makes identification of the causative agent of disease difficult.
- B) The cell wall protects microorganisms from destruction by the immune system.
- C) Animal cells do not have cell walls, so antibiotics that target cell walls can selectively destroy invading microorganisms.
- D) Only gram-negative Bacteria have cell walls.
Answer: C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
42) Which of the following molecules functions as an intracellular signaling molecule during quorum sensing?
- A) acylated homoserine lactones
- B) hydrophilic lipids
- C) quinones
- D) proteorhodopsins
Answer: A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
43) Dormancy is a result of all of the following EXCEPT
- A) phenotypic heterogeneity.
- B) ineffective efflux pumps.
- C) TA modules.
- D) the stringent response.
Answer: B
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.11
7.2 True/False Questions
1) The purpose of hemimethylated DNA is to block chromosome replication.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
2) Plasmids are replicated in Escherichia coli using the Par system.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.2
3) PopZ plays a critical role in localizing proteins to the cell poles.
Answer: TRUE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.2
4) The Par system is analogous to spindle fibers in eukaryotic cells.
Answer: TRUE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.2
5) In the Caulobacter life cycle, the role of swarmer cells is strictly for reproduction.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.7
6) Biofilm formation is more likely to occur when Vibrio cholerae is found in low densities such as sea water.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
7) Multiple sigma factors are essential to induce the biosynthesis of endospores, consequently a complex regulatory mechanism such as this has a higher chance of mutation leading to incorrect functioning compared to a simple repression mechanism.
Answer: TRUE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.6
8) Generation of new bacillus-shaped cells starts with one cell elongating and terminates when it splits into two separate bacilli.
Answer: TRUE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.4
9) The activity of MinC and MinD direct whether a bacterial cell will be coccoid or bacillus shaped.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.4
10) In fast-growing Escherichia coli cells, multiple replication forks of genomic DNA allows binary fission to occur before the genome has been fully duplicated.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
11) Multiple sigma factors are essential to induce the formation of endospores.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
12) During elongation of a cell during binary fission, small gaps caused by transglycosylases are formed before cell membrane precursors can be inserted.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.3
13) Antibiotic resistance can develop in a bacterium as a result of spontaneous mutation.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
14) Natural antibiotics are produced by all bacterial genera.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
15) All members of the genus Pseudomonas respond similarly to c-di-GMP.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.9
16) The spherical shape of Staphylococcus aureus is a result of MreB function.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.4
17) Both crescentin and MreB play a role in formation of the vibrio-shaped bacterium Caulobacter.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.4
18) FtsZ is analogous to the protein tubulin in eukaryotic cells.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.4
19) In both cocci as well as rod-shaped cells, new peptidoglycan synthesis occurs in both directions from the central location of the FtsZ ring.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
20) Oxygenic phototrophs require the uptake of oxygen for the reactions of photosynthesis.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.8
21) Mobile resistance genes encode enzymes that inactivate an antibiotic by altering its structure.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
22) Unlike penicillin, methicillin targets an alternative penicillin binding protein called MecA in nonresistant Staphylococcus aureus.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.10
23) MRSA strains synthesize MecA only in the presence of methicillin or other β-lactam antibiotics.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
24) Antibiotic-sensitive bacteria sometimes produce cells that are transiently resistant to multiple antibiotics.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
25) When Mycobacterium tuberculosis is in the dormant state it is still susceptible to antibiotics, but the rate of death is slower.
Answer: FALSE
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.11
26) Persistence is a heritable trait.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.11
27) The toxin component of TA modules helps ensure survival during stressful growth conditions.
Answer: TRUE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.11
28) Free HipA toxin inhibits transcription.
Answer: FALSE
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.11
7.3 Essay Questions
1) Describe the functions of the FtsZ protein and the Z-ring in bacterial cells.
Answer: FtsZ is an essential cell division protein that forms a contractile ring structure (Z ring) at the future cell division site. The regulation of the ring assembly controls the timing and the location of cell division. One of the functions of the FtsZ ring is to recruit other cell division proteins to the septum to produce a new cell wall between the dividing cells.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.3
2) What is the purpose of the MreB protein in bacteria?
Answer: The bacterial actin homologue MreB is required for the maintenance of a rod-shaped cell and has been shown to form spirals that traverse along the longitudinal axis of Bacillus subtilis and Escherichia coli cells.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.4
3) What role does the Par system play in bacterial cell division?
Answer: The Par (partitioning) system can be found in many bacteria including Caulobacter which undergoes budding. Similar to mitotic spindle fibers which separate replicated chromosomes in eukaryotic cells, the Par system distributes chromosomes and plasmids equally to progeny cells during growth. This system is composed of the ParAATPase, the ParB chromosome-binding protein, and the PopZ complex, as well as a centromere-like parS sequence located near oriC.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.2
4) A researcher unexpectedly identified mutant bacteria deficient in the FtsZ protein. Their growth appeared filamentous, and displayed incomplete cell division. Explain the role of the FtsZ protein and how a deficiency would account for this altered growth.
Answer: FtsZ is an essential cell division protein that forms a contractile ring structure (Z ring) at the future cell division site. The regulation of the ring assembly controls the timing and the location of cell division. One of the functions of the FtsZ ring is to recruit other cell division proteins to the septum to produce a new cell wall between the dividing cells. Consequently, a mutation in this protein would result in incomplete cell division and a potentially filamentous appearance.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.3
5) How might a genetically modified strain of E.coli with a nonfunctional mreB gene compare to E. coli possessing a functional mreB gene?
Answer: MreB forms a simple cytoskeleton in Bacteria by forming patchlike filaments around the inside of the cell just below the cytoplasmic membrane. The MreB cytoskeleton is thought to define cell shape by recruiting other proteins that function in cell wall growth to group in a specific pattern. Inactivation of the gene encoding MreB in rod-shaped bacteria causes the cells to become coccus-shaped.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.3
6) Explain how FtsZ-type proteins being found in mitochondria and chloroplasts strengthens or weakens the endosymbiosis theory.
Answer: The protein FtsZ, encoded by the gene ftsZ, is specifically involved in prokaryotic cell division. FtsZ-like proteins identified in mitochondria and chloroplasts suggest the FtsZ-encoding gene was first present in prokaryotes and that these two organelles were subsequently developed and maintained in Eukarya. Ultimately this corroborates the ideas of the endosymbiotic theory and seemingly links chloroplast and mitochondrial origins to prokaryotes.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.3
7) Explain the role of CtrA in the formation of swarmer cells in Caulobacter.
Answer: CtrA is activated by phosphorylation in emerging swarmer cells in response to external signals. Once phosphorylated, CtrA-P activates genes that encode synthesis of the flagellum and other functions specific to swarmer cells. Conversely, CtrA-P represses the synthesis of the protein CcrA and so inhibits the initiation of DNA replication in swarmer cells by binding to and blocking the origin of replication.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.7
8) Explain how green fluorescent protein (GFP) could be utilized to determine the role of FtsZ in binary fission in Escherichia coli.
Answer: For visualizing molecular events, reporter genes that encode fluorescent products such as green fluorescent protein (GFP) are routinely used. In this case, the gene for the production of GFP would be incorporated into the genome of Escherichia coli in the region of the DNA that codes for the FtsZ gene so that both genes are controlled by the same regulatory sequence; that is, the gene’s regulatory sequence now controls the production of GFP, in addition to FtsZ. In cells where the gene is expressed, and the tagged FtsZ proteins are produced, GFP is produced at the same time. Thus, only those cells in which the tagged gene is expressed, or the target proteins are produced, will fluoresce when observed under fluorescence microscopy.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.3
9) Explain the impact spontaneous mutation of the gene minE in an Escherichia coli bacterium would have on future cell divisions.
Answer: The MinE protein is one of three proteins of the Min system encoded by the minB operon required to generate pole to pole oscillations prior to bacterial cell division as a means of specifying the midzone of the cell. If a spontaneous mutation were to occur resulting in inactivation of this protein, it would affect cell division in the next generation since the FtsZ ring would not form and cell division would not occur.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.3
10) Compare the process of new peptidoglycan formation in cocci versus rod-shaped cells.
Answer: In cocci, new cell wall material grows out in opposite directions from the FtsZ ring. In contrast, new cell wall grows at several locations along the length of a rod-shaped cell.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
11) The chemical composition of the peptidoglycan carrier bactoprenol is highly hydrophobic. Explain why this is necessary for bactroprenol to carry on its function.
Answer: Bactoprenol is a hydrophobic C55 alcohol that bonds to peptidoglycan precursors. Bactoprenol transports peptidoglycan precursors across the cytoplasmic membrane by rendering them sufficiently hydrophobic to pass through the fatty acid tails of the membrane interior. Once in the periplasm, bactoprenol interacts with enzymes called transglycosylases that insert cell wall precursors into the growing point of the cell wall and catalyze glycosidic bond formation.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
12) How does the antibiotic penicillin lead to the death of pathogens?
Answer: Several penicillin binding proteins have been identified in bacteria, including FtsI. When penicillin is bound to penicillin-binding proteins, the proteins lose their catalytic activity. In the absence of transpeptidation, the continued activity of autolysins so weakens the cell wall that the cell eventually bursts.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
13) Describe the steps of binary fission.
Answer: 1) Replication of the circular prokaryotic chromosome begins at the origin of replication and continues in both directions at once. 2) The cell begins to elongate. FtsZ proteins migrate toward the midpoint of the cell. 3) The duplicated chromosomes separate and continue to move away from each other toward opposite ends of the cell. FtsZ proteins form a ring around the periphery of the midpoint between the chromosomes. 4) The FtsZ ring directs the formation of a septum that divides the cell. Plasma membrane and cell wall materials accumulate. 5) After the septum is complete, the cell pinches in two, forming two daughter cells. FtsZ is dispersed throughout the cytoplasm of the new cells.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.3
14) Explain why the antibiotic penicillin is effective at destroying bacteria but does not have a similar effect in eukaryotic cells.
Answer: Animal cells do not possess a cell wall and therefore lack peptidoglycan. Consequently the drug can be administered in high doses and is typically nontoxic. Most pathogenic bacteria do have cell walls composed of peptidoglycan and are thus potential targets of the drug.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.5
15) The major trigger of endospore formation in Bacillus is nutrient deprivation. Without the appropriate nutrients, how can endospore formation occur?
Answer: Sporulating cells cannibalize cells of their own species to obtain the appropriate nutrients. Those Bacilli in which Spo0A has become activated secrete a toxic protein that lyses nearby Bacillus cells whose Spo0A has not yet become activated. The lytic protein is produced in addition to another protein that functions to delay sporulation of neighboring cells. Cells committed to sporulation also make an antitoxin protein to protect themselves against the effects of their own toxic protein. Once lysed, the sacrificed cells are used as a source of nutrients.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.6
16) Nitrogen fixation utilizes the enzyme nitrogenase, which is extremely sensitive to oxygen. How do cyanobacteria carry on both oxygenic photosynthesis and nitrogen fixation at the same time?
Answer: Some filamentous cyanobacteria such as Anabaena and Nostoc form specialized cells called heterocysts that are exclusively dedicated to nitrogen fixation. These cells lack photosystem II, the complex that produces O2 during oxygenic photosynthesis.
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section: 7.8
17) Your microbiology professor is interested in isolating a particular endospore-forming bacterium from its natural environment and asks you to devise a protocol to do this. Explain how you could go about separating the endospores from the other microbes in the population.
Answer: The most direct method for isolating the endospores would be to expose the mixed population to high heat for a brief period to induce sporulation. This would also serve the purpose of killing any non-endospore forming bacteria.
Bloom’s Taxonomy: 5-6: Evaluating/Creating
Chapter Section: 7.6
18) Various species of bacteria naturally produce antibiotics. What resistance mechanisms are utilized by bacteria to survive antibiotics produced by other microbes or from any antibiotics they may produce themselves?
Answer: Resistance mechanisms are genetically encoded and fall into four classes: (1) modification of the drug target; (2) enzymatic inactivation; (3) removal from the cell via efflux pumps, and (4) metabolic bypasses.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
19) Biofilm formation is known to increase the resistance of member bacteria to antibiotics. How does this happen?
Answer: There are several causes of increased antibiotic resistance in biofilms. These include decreased permeability of antibiotics due to the exo-polysaccharide matrix produced by the biofilm members. Efflux pumps also play a major role. In E. coli, genes encoding the AcrAB-To1C efflux pump are upregulated when cells enter biofilm growth. Pseudomonas aeruginosa encodes several multidrug efflux pumps that are more active when cells grow in an attached state.
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section: 7.10
