Microbiology Quiz: Plasmids And Mobile Elements
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Plasmids And Mobile ElementsQuestion 1 of 20

An F' plasmid, carrying a functional trpA⁺ allele, is formed by the improper excision of an F factor from an Hfr chromosome. This F' plasmid is conjugated into a recipient E. coli strain that has a non-functional trpA⁻ allele on its chromosome. The resulting cell is grown on a minimal medium lacking tryptophan. The cell grows successfully. This phenomenon is best described as:

Gene conversion, where the F' plasmid uses its trpA⁺ allele as a template to permanently repair the chromosomal trpA⁻ allele.
Suppression, where a mutation in a different gene on the plasmid circumvents the need for the TrpA protein.
Reversion, where the chromosomal trpA⁻ allele spontaneously mutates back to the functional trpA⁺ form.
Complementation, where the plasmid-borne trpA⁺ allele provides the function missing from the chromosomal allele.
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Microbiology Quiz

Microbiology Quiz: Plasmids And Mobile Elements

Practice Plasmids And Mobile Elements in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Plasmids And Mobile Elements, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An F' plasmid, carrying a functional trpA⁺ allele, is formed by the improper excision of an F factor from an Hfr chromosome. This F' plasmid is conjugated into a recipient E. coli strain that has a non-functional trpA⁻ allele on its chromosome. The resulting cell is grown on a minimal medium lacking tryptophan. The cell grows successfully. This phenomenon is best described as:

  1. Gene conversion, where the F' plasmid uses its trpA⁺ allele as a template to permanently repair the chromosomal trpA⁻ allele.
  2. Suppression, where a mutation in a different gene on the plasmid circumvents the need for the TrpA protein.
  3. Reversion, where the chromosomal trpA⁻ allele spontaneously mutates back to the functional trpA⁺ form.
  4. Complementation, where the plasmid-borne trpA⁺ allele provides the function missing from the chromosomal allele. (correct answer)
Explanation: When you encounter questions about bacterial conjugation and gene function, focus on understanding what happens when functional genes are introduced into cells with defective alleles. In this scenario, an F' plasmid carries a functional trpA⁺ gene into a recipient cell whose chromosomal trpA⁻ gene is non-functional. The TrpA protein is essential for tryptophan biosynthesis, so cells normally need it to grow on minimal medium without tryptophan. Here, the cell grows successfully because the plasmid's trpA⁺ gene produces functional TrpA protein, compensating for the defective chromosomal copy. This is complementation (D) – when one functional gene copy rescues the phenotype of a defective allele by providing the missing gene product. Both alleles remain unchanged; the plasmid simply supplies what the chromosome cannot. Option A (gene conversion) is incorrect because the chromosomal trpA⁻ allele isn't repaired or changed – it remains defective while the plasmid provides function. Option B (suppression) would involve a mutation in a different gene that bypasses the need for TrpA entirely, but here we're directly providing TrpA function. Option C (reversion) would mean the chromosomal trpA⁻ mutated back to trpA⁺, but complementation occurs without any chromosomal changes. Remember this key distinction: complementation provides function through an additional functional copy, while other mechanisms either repair the original gene or work around the defect entirely. Look for scenarios where functional genes are introduced alongside defective ones – that's your cue for complementation.

Question 2

Upon insertion of a transposon into a new site on a bacterial chromosome, a specific DNA sequence from the target site is found to be duplicated, appearing as short direct repeats flanking the newly inserted element. For example, the sequence 5'-ATTCG-[Transposon]-ATTCG-3' is observed, where the target site was originally just 5'-ATTCG-3'. What is the direct cause of this target site duplication?

  1. The transposase uses the target sequence as a primer to synthesize a copy of the transposon during replicative transposition.
  2. The transposon itself carries the duplicated sequence at its ends, which are then integrated into the chromosome.
  3. Host cell DNA repair machinery recognizes the insertion as damage and duplicates a flanking sequence as part of the repair process.
  4. The transposase makes staggered nicks in the two strands of the target DNA, and DNA polymerase fills in the resulting single-stranded gaps. (correct answer)
Explanation: When you encounter questions about transposon insertion and target site duplication, focus on the molecular mechanism of how transposases cut DNA and how cells repair the resulting gaps. The target site duplication occurs because of the specific way transposases cut target DNA. The enzyme makes staggered cuts on opposite strands at slightly offset positions, creating single-stranded overhangs rather than blunt cuts. When the transposon inserts into this gap, the host cell's DNA polymerase fills in the single-stranded regions on both sides, effectively duplicating the original target sequence. This is why you see identical sequences flanking the transposon - they're both copies of the original target site. Answer A is incorrect because the target sequence isn't used as a primer for transposon synthesis; transposition involves moving existing DNA, not synthesizing new copies of the transposon itself during simple transposition. Answer B misunderstands the source of duplication - the repeated sequences come from the target site, not from sequences carried by the transposon. Answer C incorrectly suggests this is purely a repair response to "damage," when actually the duplication is an inevitable consequence of the staggered cutting mechanism, not an error-correction process. Remember that target site duplications are a hallmark of transposon insertion. When you see direct repeats flanking mobile elements in exam questions, think about the staggered cutting mechanism - it's the key to understanding how these duplications arise during the normal insertion process.

Question 3

A researcher wishes to cure a strain of E. coli of its R plasmid. It is known that the replication of this plasmid is slightly more sensitive to the inhibition of DNA gyrase than is the replication of the host chromosome. Which strategy would most likely be effective for selectively eliminating the plasmid without killing the host cells?

  1. Transforming the strain with a second plasmid from the same incompatibility group.
  2. Growing the cells in a medium containing a carefully titrated, sub-lethal concentration of novobiocin. (correct answer)
  3. Exposing the culture to a brief pulse of high-intensity ultraviolet radiation to damage the plasmid.
  4. Culturing the cells for many generations in the absence of the antibiotic for which the plasmid confers resistance.
Explanation: When you encounter questions about plasmid curing, focus on exploiting differences between plasmid and chromosomal DNA replication without killing the host cell. The key insight here is that this R plasmid's replication is more sensitive to DNA gyrase inhibition than the host chromosome. DNA gyrase is essential for relieving supercoiling tension during DNA replication. Novobiocin specifically inhibits DNA gyrase, so at carefully controlled sub-lethal concentrations, it will preferentially block plasmid replication while allowing chromosomal replication to continue. This selective pressure will cure the cells of their plasmids over several generations while keeping the host cells viable. Let's examine why the other options won't work effectively. Option A involves incompatibility groups - while introducing a plasmid from the same incompatibility group can displace the original plasmid, this doesn't cure the cell of plasmids entirely; it just replaces one plasmid with another. Option C suggests UV radiation, but UV damage is non-specific and would harm both plasmid and chromosomal DNA equally, likely killing the host cells. Option D describes simply removing antibiotic selection pressure, but this won't actively eliminate the plasmid - cells will continue carrying it without selective disadvantage. The correct answer is B because it exploits the specific vulnerability mentioned in the question stem. Remember: successful plasmid curing requires exploiting a differential sensitivity between plasmid and chromosome. Always look for strategies that selectively target the plasmid's unique characteristics while preserving host cell viability.

Question 4

Broad-host-range plasmids, such as those from the IncP group, can replicate in a diverse array of bacterial species, whereas narrow-host-range plasmids like ColE1 are much more restricted. Which factor is the most critical molecular determinant of a plasmid's host range?

  1. The ability of the plasmid's conjugation system to form a mating bridge with different bacterial species.
  2. The type of antibiotic resistance gene carried, which determines which hosts can be selected for.
  3. The specificity of the interaction between the plasmid's replication initiation protein (Rep) and the host's DNA replication machinery. (correct answer)
  4. The plasmid's ability to evade the restriction-modification systems present in various potential hosts.
Explanation: When you encounter questions about plasmid host range, focus on the fundamental molecular mechanisms that control where plasmids can successfully replicate and maintain themselves. The correct answer is C because plasmid replication depends entirely on the compatibility between the plasmid's replication initiation protein (Rep) and the host cell's DNA replication machinery. Broad-host-range plasmids like IncP have Rep proteins that can interact with the replication machinery of many different bacterial species, while narrow-host-range plasmids like ColE1 have Rep proteins with much more restrictive compatibility. This Rep-host interaction is the primary bottleneck that determines whether a plasmid can establish itself in a new host. Let's examine why the other options are secondary factors: A is incorrect because conjugation ability affects plasmid transfer between cells, not whether the plasmid can replicate once it arrives in a new host. A plasmid could conjugate successfully but still fail to replicate. B is wrong because antibiotic resistance genes are selection tools for laboratory detection—they don't determine the fundamental ability of a plasmid to replicate in different hosts. D is incorrect because while restriction-modification systems can degrade foreign DNA, many plasmids have evolved ways around these defenses, and this isn't the primary determinant of host range. Remember: host range questions usually come down to replication compatibility. The Rep protein acts like a molecular key that must fit the host's replication "lock"—without this match, nothing else matters for plasmid establishment.

Question 5

The copy number of the pBR322 plasmid is negatively regulated by an antisense RNA molecule called RNA I. RNA I binds to the 5' end of a primer transcript, RNA II, preventing it from forming a structure required for DNA polymerase I to initiate replication. A mutation in the gene for RNA I reduces its binding affinity for RNA II. What is the most likely consequence of this mutation?

  1. The plasmid will switch from theta replication to rolling circle replication to compensate for the defect.
  2. The plasmid copy number will decrease because RNA I can no longer effectively regulate replication.
  3. The plasmid will be lost from the host cell immediately because replication cannot be initiated at all.
  4. The plasmid copy number will increase significantly because the inhibition of replication initiation is relieved. (correct answer)
Explanation: When you encounter questions about plasmid copy number regulation, focus on understanding how regulatory molecules affect the balance between promoting and inhibiting replication. In pBR322, RNA I acts as a negative regulator by binding to RNA II and preventing the formation of structures needed for DNA polymerase I to initiate replication. This creates a feedback loop: as plasmid copy number increases, more RNA I is produced, which increasingly inhibits further replication. When the mutation reduces RNA I's binding affinity for RNA II, this inhibitory mechanism becomes less effective. The correct answer is D because weakened RNA I binding means less interference with RNA II function. RNA II can more readily form the structures required for replication initiation, leading to increased plasmid copy number. The negative feedback system is compromised, allowing more rounds of replication to occur. Option A is incorrect because the replication mechanism itself isn't damaged—only the regulatory system is affected. The plasmid will continue using theta replication. Option B represents backward thinking; if the inhibitor (RNA I) becomes less effective, replication increases rather than decreases. Option C is too extreme—the mutation affects regulation, not the fundamental ability to replicate. Some RNA I binding likely still occurs, and even without it, replication could proceed. Remember that in regulatory systems, when negative regulators are weakened or removed, the process they normally inhibit will increase in activity. Always consider whether a regulatory molecule promotes or inhibits the process in question.

Question 6

A clinical isolate of Pseudomonas aeruginosa harbors a large, 150 kb R plasmid conferring resistance to five different classes of antibiotics. When this isolate is serially subcultured for 100 generations in a rich, antibiotic-free medium, sequence analysis reveals that a majority of the population has lost the entire R plasmid. What is the most probable evolutionary reason for this observation?

  1. Replicating and expressing the genes on the large plasmid imposes a significant fitness cost, allowing plasmid-free cells to outcompete plasmid-bearing cells. (correct answer)
  2. The rich medium contains acridine orange or a similar intercalating agent that specifically inhibits the replication of R plasmids.
  3. The plasmid is a temperature-sensitive replicon that is unstable at the standard incubation temperature of 37°C.
  4. The plasmid spontaneously integrates into the host chromosome, a process that is favored in the absence of antibiotic selection.
Explanation: Carrying large plasmids with many genes incurs a metabolic or 'fitness' cost on the host bacterium. The cell must expend energy and resources to replicate the plasmid DNA and express its genes. In the absence of selective pressure (i.e., no antibiotics), cells that spontaneously lose the plasmid can replicate faster and will eventually dominate the population.

Question 7

A researcher attempts to introduce two different plasmids into a single E. coli cell. Plasmid A carries an ampicillin resistance gene and Plasmid B carries a tetracycline resistance gene. Both plasmids utilize the ColE1 origin of replication and its associated RNA-based copy number control system. After successful transformation, the cell is grown for many generations in a medium containing both ampicillin and tetracycline. What is the most likely composition of plasmids in the resulting cell population?

  1. Most cells will contain only Plasmid A or only Plasmid B, leading to a heterogeneous population with respect to resistance. (correct answer)
  2. Most cells will stably maintain both plasmids due to the strong selective pressure from both antibiotics.
  3. The two plasmids will undergo homologous recombination to form a single, larger plasmid carrying both resistance genes.
  4. Both plasmids will be forced out of the cells because their replication control systems interfere with each other, leading to cell death.
Explanation: Plasmids that share the same replication control mechanism belong to the same incompatibility (Inc) group and cannot be stably maintained together in a bacterial lineage. The ColE1 origin is one such system. Random partitioning during cell division will lead to daughter cells that inherit only one type of plasmid or the other. Even with selection, the population will become a mixture of cells with either Plasmid A or Plasmid B, as a cell losing one plasmid will be killed, but a cell that never received both at the start or loses one early on will propagate as a single-plasmid lineage.

Question 8

Sequence analysis of a bacterial chromosome reveals a kanamycin resistance gene (kanR) flanked by two identical copies of an insertion sequence, IS50. The two IS50 elements are oriented as inverted repeats relative to each other. The IS50 element itself encodes a transposase. What is the most likely mechanism by which the kanR gene can be mobilized to a new DNA location?

  1. The transposase recognizes the outermost inverted repeats of the two IS50 elements, mobilizing the entire segment as a composite transposon. (correct answer)
  2. Homologous recombination between the two identical IS50 elements excises the kanR gene as a circular DNA molecule, which is then lost.
  3. The kanR gene is transcribed and then reverse-transcribed into DNA for insertion at a new locus, typical of a retrotransposon.
  4. The transposase from one IS50 element mobilizes only that single element, leaving the kanR gene behind on the chromosome.
Explanation: This structure describes a composite transposon. The transposase, encoded by the IS elements, can recognize the inverted repeats at the very ends of the entire structure (the 'outside' ends). This allows the entire segment of DNA between these ends, including the kanR gene, to be moved as a single unit to a new location.

Question 9

While bacterial chromosomes and plasmids are both replicons found in bacteria, a fundamental distinction is made between them. Which statement provides the most accurate and universally applicable distinction?

  1. A chromosome carries genes essential for viability under all growth conditions, whereas a plasmid carries non-essential, accessory genes. (correct answer)
  2. A chromosome is always a single, large circular molecule, whereas plasmids are always small, high-copy-number molecules.
  3. Chromosomal replication is bidirectional, while plasmid replication universally proceeds via a rolling circle mechanism.
  4. Chromosomes are vertically inherited, whereas plasmids are exclusively transferred horizontally between cells.
Explanation: The defining characteristic of a chromosome is that it contains the 'housekeeping' genes necessary for the basic life functions of the cell (e.g., core metabolism, replication, transcription, translation). A cell cannot survive without its chromosome. Plasmids, by contrast, carry 'accessory' genes that may provide a selective advantage in specific environments (e.g., antibiotic resistance, virulence) but are not required for survival under normal laboratory conditions.

Question 10

A microbiologist studying the replication of a novel staphylococcal plasmid observes replication intermediates using electron microscopy. The predominant structures seen are circular DNA molecules with a linear, single-stranded 'tail' of various lengths extending from the circle. These observations are most consistent with which mode of plasmid replication?

  1. Reverse transcription replication
  2. Theta (θ) replication
  3. D-loop replication
  4. Rolling circle replication (correct answer)
Explanation: When you encounter questions about DNA replication mechanisms, focus on the distinctive structural features each process creates. The key clue here is the observation of circular DNA molecules with linear, single-stranded "tails" of varying lengths. Rolling circle replication (D) perfectly explains these observations. In this mechanism, one strand of the circular plasmid is nicked, creating a 3'-OH end that serves as a primer. DNA polymerase extends this strand while displacing the 5' end, which forms the characteristic single-stranded tail. As replication proceeds, the tail grows longer, creating exactly what the microbiologist observed—circular molecules with linear tails of various lengths depending on how far replication has progressed. Let's examine why the other options don't fit: Reverse transcription replication (A) involves RNA-dependent DNA synthesis and wouldn't produce these specific structures. Theta (θ) replication (B) creates a structure resembling the Greek letter theta, with two replication forks moving bidirectionally around the circle—you'd see a "bubble" or figure-eight shape, not linear tails. D-loop replication (C) forms a displacement loop structure where newly synthesized DNA displaces one strand, but this creates a localized loop within the circle, not the extended linear tails described. Remember this pattern: when you see "circular DNA with single-stranded tails" in electron microscopy descriptions, think rolling circle replication immediately. The length variation of the tails is the diagnostic feature—it reflects replication caught at different stages of completion.

Question 11

The tra operon of the F plasmid, which encodes the machinery for conjugation, is subject to transcriptional repression in established F⁺ cells. The FinOP system ensures that the TraJ activator is inhibited, keeping the tra operon silent most of the time. What is the primary selective advantage for the F plasmid and its host to tightly regulate and repress conjugation?

  1. Preventing the F plasmid from integrating into the chromosome, which would be lethal to the host cell.
  2. Conserving significant cellular energy that would be spent synthesizing the large pilus structure and transfer proteins. (correct answer)
  3. Ensuring the plasmid maintains a low copy number, which reduces the metabolic burden on the host.
  4. Blocking the entry of other, incompatible plasmids that might use the same conjugation machinery.
Explanation: When you encounter questions about bacterial conjugation and plasmid regulation, focus on the evolutionary costs and benefits of these energy-intensive processes. The F plasmid's tight regulation of its tra operon reflects a fundamental trade-off between reproduction and survival. Conjugation is extraordinarily expensive for bacterial cells. Building the conjugative pilus requires synthesizing numerous large proteins, including pilin subunits, transfer proteins, and DNA processing enzymes. The pilus itself is a massive structure that can extend several cell lengths, and the entire conjugation process diverts substantial cellular resources from growth and maintenance. The FinOP regulatory system evolved because uncontrolled conjugation would drain the cell's energy reserves, making both the plasmid and host less competitive. Option A is incorrect because F plasmid integration (forming Hfr cells) isn't lethal—it's actually a normal part of the F plasmid lifecycle. Option C misses the mark since tra operon regulation doesn't control plasmid copy number; that's managed by separate replication control systems. Option D describes plasmid incompatibility, but this involves competition for replication machinery, not conjugation regulation. The correct answer is B because the primary selective pressure is energy conservation. Cells that wastefully produce conjugation machinery when it's not needed would be outcompeted by more efficient variants. Remember that in microbiology, regulation questions often center on resource allocation. Bacteria are extremely efficient organisms, so any energetically expensive process like conjugation, sporulation, or antibiotic production is typically tightly controlled to maximize cellular fitness.

Question 12

An Hfr strain with genotype pro⁺ leu⁺ lac⁺ is mated with an F⁻ strain with genotype pro⁻ leu⁻ lac⁻. The genes are transferred in the order pro, leu, lac. The mating is interrupted after a time sufficient for the transfer of pro and leu but not lac. Analysis of the recipient cells shows that many have become pro⁺ leu⁺, but none have become F⁺. Which statement best describes the genetic state of the pro⁺ leu⁺ recipient cells?

  1. They are F⁻ cells that have integrated the pro⁺ and leu⁺ alleles into their chromosome through homologous recombination. (correct answer)
  2. They are F' cells carrying a new plasmid with the pro⁺ and leu⁺ genes derived from the donor chromosome.
  3. They are F⁺ cells because the origin of transfer (oriT) of the F factor is always transferred first.
  4. They are transiently pro⁺ leu⁺ due to the presence of a linear DNA fragment that will not be replicated.
Explanation: In an Hfr mating, the F factor is integrated into the chromosome, and the chromosome is transferred linearly from the origin of transfer. The majority of the F factor genes are transferred last. Interrupted mating allows only the initial part of the chromosome to be transferred. The recipient cell remains F⁻ because it does not receive the full F factor. The transferred linear DNA can be incorporated into the recipient's chromosome via homologous recombination, leading to a stable change in genotype.

Question 13

A low-copy-number bacterial plasmid contains a partitioning (par) locus consisting of a centromere-like DNA site (parS) and genes for two proteins, ParA and ParB. ParB binds to parS, and ParA is an ATPase that positions the ParB-parS complex. A mutation completely deletes the parS site but leaves the parA and parB genes intact and expressed. What is the most likely consequence for this mutant plasmid?

  1. The ParA and ParB proteins will now cause the plasmid to integrate into the host chromosome.
  2. The copy number of the plasmid will increase to ensure at least one copy enters each daughter cell.
  3. The plasmid will be rapidly and randomly lost during cell division due to failed segregation. (correct answer)
  4. The ParB protein will bind nonspecifically to chromosomal DNA, leading to filamentation and cell death.
Explanation: When you encounter questions about bacterial plasmid maintenance, focus on the essential role of partitioning systems in ensuring equal distribution during cell division. The par locus functions like a miniature chromosome segregation system, where ParB acts as an adaptor protein that must bind to the specific parS DNA sequence to form a nucleoprotein complex that ParA can then position correctly. Without the parS site, ParB has no specific binding target on the plasmid. This breaks the entire partitioning mechanism because ParA cannot position what it cannot recognize. Low-copy-number plasmids are particularly vulnerable to segregation defects because there are fewer copies to randomly distribute between daughter cells. The result is rapid, random loss during cell division as plasmids fail to segregate properly and may end up concentrated in just one daughter cell while the other receives none. Choice A is incorrect because integration requires specific recombination mechanisms, not partitioning proteins. Choice B misunderstands plasmid biology—copy number is controlled by replication machinery, not partitioning systems, and cells don't compensate for segregation defects by increasing replication. Choice D assumes ParB will bind nonspecifically to chromosomal DNA, but ParB typically has high specificity for parS sequences and wouldn't cause the dramatic effects described. Remember that partitioning systems are about proper distribution, not replication or integration. When you see par locus mutations, immediately think about segregation failure and the resulting plasmid instability, especially in low-copy-number systems where random distribution is insufficient.

Question 14

The tra operon of the F plasmid, which encodes the machinery for conjugation, is subject to transcriptional repression in established F⁺ cells. The FinOP system ensures that the TraJ activator is inhibited, keeping the tra operon silent most of the time. What is the primary selective advantage for the F plasmid and its host to tightly regulate and repress conjugation?

  1. Preventing the F plasmid from integrating into the chromosome, which would be lethal to the host cell.
  2. Conserving significant cellular energy that would be spent synthesizing the large pilus structure and transfer proteins. (correct answer)
  3. Ensuring the plasmid maintains a low copy number, which reduces the metabolic burden on the host.
  4. Blocking the entry of other, incompatible plasmids that might use the same conjugation machinery.
Explanation: When you encounter questions about bacterial conjugation and plasmid regulation, focus on the evolutionary costs and benefits of these energy-intensive processes. The F plasmid's tight regulation of its tra operon reflects a fundamental trade-off between reproduction and survival. Conjugation is extraordinarily expensive for bacterial cells. Building the conjugative pilus requires synthesizing numerous large proteins, including pilin subunits, transfer proteins, and DNA processing enzymes. The pilus itself is a massive structure that can extend several cell lengths, and the entire conjugation process diverts substantial cellular resources from growth and maintenance. The FinOP regulatory system evolved because uncontrolled conjugation would drain the cell's energy reserves, making both the plasmid and host less competitive. Option A is incorrect because F plasmid integration (forming Hfr cells) isn't lethal—it's actually a normal part of the F plasmid lifecycle. Option C misses the mark since tra operon regulation doesn't control plasmid copy number; that's managed by separate replication control systems. Option D describes plasmid incompatibility, but this involves competition for replication machinery, not conjugation regulation. The correct answer is B because the primary selective pressure is energy conservation. Cells that wastefully produce conjugation machinery when it's not needed would be outcompeted by more efficient variants. Remember that in microbiology, regulation questions often center on resource allocation. Bacteria are extremely efficient organisms, so any energetically expensive process like conjugation, sporulation, or antibiotic production is typically tightly controlled to maximize cellular fitness.

Question 15

A Class 1 integron from an E. coli isolate contains an array of three different gene cassettes: aadA (streptomycin resistance), dfrA1 (trimethoprim resistance), and catB (chloramphenicol resistance). The expression of the protein products from all three of these cassettes is primarily dependent on which single feature of the integron?

  1. A common promoter, Pc, located within the 5'-conserved segment (5'-CS) upstream of the cassette array. (correct answer)
  2. Each gene cassette containing its own individual promoter that is activated upon insertion into the integron.
  3. The integrase enzyme (IntI1), which also functions as a transcriptional activator for the inserted genes.
  4. The presence of a strong ribosomal binding site (RBS) within the attC recombination site of each cassette.
Explanation: A key feature of integrons is their ability to capture and express promoterless gene cassettes. Expression is driven by a promoter (most commonly the Pc promoter) located in the 5'-conserved segment, which is part of the integron itself, not the individual cassettes. This promoter transcribes a single long mRNA that includes the genes from the inserted cassettes, which are then translated.

Question 16

A researcher attempts to introduce two different plasmids into a single E. coli cell. Plasmid A carries an ampicillin resistance gene and Plasmid B carries a tetracycline resistance gene. Both plasmids utilize the ColE1 origin of replication and its associated RNA-based copy number control system. After successful transformation, the cell is grown for many generations in a medium containing both ampicillin and tetracycline. What is the most likely composition of plasmids in the resulting cell population?

  1. Most cells will contain only Plasmid A or only Plasmid B, leading to a heterogeneous population with respect to resistance. (correct answer)
  2. Most cells will stably maintain both plasmids due to the strong selective pressure from both antibiotics.
  3. The two plasmids will undergo homologous recombination to form a single, larger plasmid carrying both resistance genes.
  4. Both plasmids will be forced out of the cells because their replication control systems interfere with each other, leading to cell death.
Explanation: Plasmids that share the same replication control mechanism belong to the same incompatibility (Inc) group and cannot be stably maintained together in a bacterial lineage. The ColE1 origin is one such system. Random partitioning during cell division will lead to daughter cells that inherit only one type of plasmid or the other. Even with selection, the population will become a mixture of cells with either Plasmid A or Plasmid B, as a cell losing one plasmid will be killed, but a cell that never received both at the start or loses one early on will propagate as a single-plasmid lineage.

Question 17

The copy number of the pBR322 plasmid is negatively regulated by an antisense RNA molecule called RNA I. RNA I binds to the 5' end of a primer transcript, RNA II, preventing it from forming a structure required for DNA polymerase I to initiate replication. A mutation in the gene for RNA I reduces its binding affinity for RNA II. What is the most likely consequence of this mutation?

  1. The plasmid will switch from theta replication to rolling circle replication to compensate for the defect.
  2. The plasmid copy number will decrease because RNA I can no longer effectively regulate replication.
  3. The plasmid will be lost from the host cell immediately because replication cannot be initiated at all.
  4. The plasmid copy number will increase significantly because the inhibition of replication initiation is relieved. (correct answer)
Explanation: When you encounter questions about plasmid copy number regulation, focus on understanding how regulatory molecules affect the balance between promoting and inhibiting replication. In pBR322, RNA I acts as a negative regulator by binding to RNA II and preventing the formation of structures needed for DNA polymerase I to initiate replication. This creates a feedback loop: as plasmid copy number increases, more RNA I is produced, which increasingly inhibits further replication. When the mutation reduces RNA I's binding affinity for RNA II, this inhibitory mechanism becomes less effective. The correct answer is D because weakened RNA I binding means less interference with RNA II function. RNA II can more readily form the structures required for replication initiation, leading to increased plasmid copy number. The negative feedback system is compromised, allowing more rounds of replication to occur. Option A is incorrect because the replication mechanism itself isn't damaged—only the regulatory system is affected. The plasmid will continue using theta replication. Option B represents backward thinking; if the inhibitor (RNA I) becomes less effective, replication increases rather than decreases. Option C is too extreme—the mutation affects regulation, not the fundamental ability to replicate. Some RNA I binding likely still occurs, and even without it, replication could proceed. Remember that in regulatory systems, when negative regulators are weakened or removed, the process they normally inhibit will increase in activity. Always consider whether a regulatory molecule promotes or inhibits the process in question.

Question 18

Upon insertion of a transposon into a new site on a bacterial chromosome, a specific DNA sequence from the target site is found to be duplicated, appearing as short direct repeats flanking the newly inserted element. For example, the sequence 5'-ATTCG-[Transposon]-ATTCG-3' is observed, where the target site was originally just 5'-ATTCG-3'. What is the direct cause of this target site duplication?

  1. The transposase uses the target sequence as a primer to synthesize a copy of the transposon during replicative transposition.
  2. The transposon itself carries the duplicated sequence at its ends, which are then integrated into the chromosome.
  3. Host cell DNA repair machinery recognizes the insertion as damage and duplicates a flanking sequence as part of the repair process.
  4. The transposase makes staggered nicks in the two strands of the target DNA, and DNA polymerase fills in the resulting single-stranded gaps. (correct answer)
Explanation: When you encounter questions about transposon insertion and target site duplication, focus on the molecular mechanism of how transposases cut DNA and how cells repair the resulting gaps. The target site duplication occurs because of the specific way transposases cut target DNA. The enzyme makes staggered cuts on opposite strands at slightly offset positions, creating single-stranded overhangs rather than blunt cuts. When the transposon inserts into this gap, the host cell's DNA polymerase fills in the single-stranded regions on both sides, effectively duplicating the original target sequence. This is why you see identical sequences flanking the transposon - they're both copies of the original target site. Answer A is incorrect because the target sequence isn't used as a primer for transposon synthesis; transposition involves moving existing DNA, not synthesizing new copies of the transposon itself during simple transposition. Answer B misunderstands the source of duplication - the repeated sequences come from the target site, not from sequences carried by the transposon. Answer C incorrectly suggests this is purely a repair response to "damage," when actually the duplication is an inevitable consequence of the staggered cutting mechanism, not an error-correction process. Remember that target site duplications are a hallmark of transposon insertion. When you see direct repeats flanking mobile elements in exam questions, think about the staggered cutting mechanism - it's the key to understanding how these duplications arise during the normal insertion process.

Question 19

Broad-host-range plasmids, such as those from the IncP group, can replicate in a diverse array of bacterial species, whereas narrow-host-range plasmids like ColE1 are much more restricted. Which factor is the most critical molecular determinant of a plasmid's host range?

  1. The ability of the plasmid's conjugation system to form a mating bridge with different bacterial species.
  2. The type of antibiotic resistance gene carried, which determines which hosts can be selected for.
  3. The specificity of the interaction between the plasmid's replication initiation protein (Rep) and the host's DNA replication machinery. (correct answer)
  4. The plasmid's ability to evade the restriction-modification systems present in various potential hosts.
Explanation: When you encounter questions about plasmid host range, focus on the fundamental molecular mechanisms that control where plasmids can successfully replicate and maintain themselves. The correct answer is C because plasmid replication depends entirely on the compatibility between the plasmid's replication initiation protein (Rep) and the host cell's DNA replication machinery. Broad-host-range plasmids like IncP have Rep proteins that can interact with the replication machinery of many different bacterial species, while narrow-host-range plasmids like ColE1 have Rep proteins with much more restrictive compatibility. This Rep-host interaction is the primary bottleneck that determines whether a plasmid can establish itself in a new host. Let's examine why the other options are secondary factors: A is incorrect because conjugation ability affects plasmid transfer between cells, not whether the plasmid can replicate once it arrives in a new host. A plasmid could conjugate successfully but still fail to replicate. B is wrong because antibiotic resistance genes are selection tools for laboratory detection—they don't determine the fundamental ability of a plasmid to replicate in different hosts. D is incorrect because while restriction-modification systems can degrade foreign DNA, many plasmids have evolved ways around these defenses, and this isn't the primary determinant of host range. Remember: host range questions usually come down to replication compatibility. The Rep protein acts like a molecular key that must fit the host's replication "lock"—without this match, nothing else matters for plasmid establishment.

Question 20

The transposon Tn3 is a non-composite transposon. It possesses short inverted repeats at its ends and encodes its own transposase (tnpA) and resolvase (tnpR) between them. Its mechanism of transposition is fundamentally different from that of composite transposons flanked by IS elements. A key intermediate structure formed during Tn3 transposition is:

  1. An RNA-DNA hybrid molecule formed during reverse transcription of the transposon message.
  2. A free, circularized transposon element that is excised from the donor molecule before insertion.
  3. A cointegrate, where the donor and target replicons are temporarily fused into a single larger molecule. (correct answer)
  4. A hairpin structure formed by the excision of the transposon via the 'cut-and-paste' mechanism.
Explanation: When you encounter questions about transposon mechanisms, focus on understanding the fundamental differences between composite and non-composite transposons. Tn3 represents a classic non-composite transposon with a unique replicative transposition mechanism that's quite different from the "cut-and-paste" approach of many composite transposons. Tn3 transposition works through a replicative mechanism where the transposon duplicates itself during the process. The key intermediate is a cointegrate structure - a fusion molecule where both the donor DNA (containing the original transposon) and the target DNA become temporarily joined into one larger replicon. This cointegrate contains two copies of Tn3 at the junctions between the donor and target sequences. The resolvase enzyme (tnpR) then acts on specific resolution sites to separate this cointegrate back into two distinct molecules, each now containing a copy of the transposon. Option A incorrectly describes retrotransposition, which involves RNA intermediates and reverse transcriptase - not relevant to Tn3. Option B describes excision-based transposition where the element becomes free before insertion, but Tn3 never exists as a free circular intermediate. Option D references hairpin structures from cut-and-paste mechanisms typical of composite transposons like Tn10, not the replicative mechanism of Tn3. Remember this distinction: composite transposons typically use cut-and-paste mechanisms, while non-composite transposons like Tn3 use replicative mechanisms involving cointegrate intermediates. The presence of resolvase is often your clue that cointegrate resolution is involved.