Genetics of Prokaryotes (2B) - Bio & Biochem

1. In a study of conjugation, an F+ E. coli donor carrying a plasmid with kanamycin resistance ($Kan^R$) and an intact tra operon was mixed with an F− recipient ($Kan^S$). After 30 min, cultures were plated on kanamycin. A parallel mixture included DNase I (100 U/mL) throughout the mating period. Colony counts (CFU/mL) on kanamycin were similar with and without DNase I. Which statement is most consistent with these observations?
   Answer: Transfer most likely occurred through direct cell-to-cell contact and plasmid DNA transfer that is DNase-resistant. The skill being tested is prokaryotic genetics. Conjugation is a mechanism of horizontal gene transfer in bacteria where DNA is passed directly from a donor to a recipient cell through a conjugative pilus, making it resistant to extracellular DNases. In this experiment, the F+ donor transfers a $Kan^R$ plasmid to the F- recipient, and the similar colony counts with or without DNase I indicate that the transfer does not rely on free extracellular DNA. Choice C aligns with this process because it describes direct cell-to-cell contact and DNase-resistant plasmid transfer, which explains why DNase does not affect the outcome. A common misconception is addressed in choice D, which suggests spontaneous mutation, but this fails because the setup involves mixing donor and recipient, implying gene transfer rather than independent mutation. To check similar questions, determine if the transfer is sensitive to DNase; conjugation is not, while transformation is. Additionally, confirm if cell contact is required, as conjugation depends on it unlike transduction or transformation.
2. A clinical isolate contains a small plasmid encoding a beta-lactamase. When the plasmid is cured (lost) by growth at elevated temperature, the strain becomes ampicillin sensitive but retains resistance to tetracycline encoded on the chromosome. This scenario most directly illustrates which principle about plasmid function in prokaryotes?
   Answer: Plasmid-borne traits can be conditionally maintained and lost without altering chromosomal markers. The skill being tested is prokaryotic genetics. Plasmids are extrachromosomal DNA molecules that can replicate independently and carry genes like antibiotic resistance, which can be lost without affecting the chromosome. In this scenario, curing the plasmid eliminates ampicillin resistance but leaves chromosomal tetracycline resistance intact, demonstrating plasmid autonomy. Choice A aligns because it highlights how plasmid-borne traits can be conditionally maintained and lost separately from chromosomal markers. Choice D is a distractor based on the misconception that plasmids encode essential host functions like RNA polymerase, but bacteria use chromosomal polymerases for all transcription. In similar questions, check if the trait is extrachromosomal by seeing if it can be cured independently. Confirm autonomy by noting unchanged chromosomal phenotypes post-curing.
3. In an experiment focused on conjugation, a donor strain is engineered to be F− but still carries a non-conjugative plasmid encoding resistance. When mixed with an F− recipient, no transfer is observed. Which statement is most consistent with the outcome?
   Answer: Non-conjugative plasmids generally require transfer functions (e.g., tra) provided in cis or trans to be mobilized. The skill being tested is prokaryotic genetics. Non-conjugative plasmids lack tra genes and cannot transfer independently but can be mobilized if transfer functions are provided. The failure of transfer between two F- strains shows the need for conjugative machinery. Choice A aligns because it notes that non-conjugative plasmids require tra functions for mobilization. Choice B is a distractor based on the misconception that competence enables conjugation, but they are distinct processes. In similar questions, check for tra presence to assess conjugative potential. Verify donor/recipient status to confirm mobilization needs.
4. A researcher studies conjugation between two E. coli strains. Donor cells carry an F plasmid encoding tetracycline resistance (tetR) and a functional pilus. Recipient cells lack the plasmid and are chloramphenicol-resistant (camR). After co-incubation, cells are plated on tetracycline + chloramphenicol; colonies are recovered. Which outcome is most consistent with conjugative plasmid transfer in this setup?
   Answer: Recipient cells acquired tetR by receiving a copy of the F plasmid from donors through a pilus-mediated contact. This question tests understanding of prokaryotic genetics, specifically bacterial conjugation. Conjugation is a process where DNA (typically plasmids) is transferred from donor to recipient cells through direct cell-to-cell contact via a pilus encoded by the F plasmid. The passage describes donor cells with F plasmid (tetR) and recipient cells without F plasmid (camR), and colonies grow on both antibiotics after co-incubation. Answer B correctly states that recipients acquired tetR by receiving the F plasmid through pilus-mediated contact, explaining why colonies are both tetR and camR. Answer A incorrectly reverses the direction—donors don't take up DNA during conjugation, they donate it. To identify conjugation, look for: F plasmid, pilus formation, direct cell contact, and unidirectional transfer from donor to recipient.
5. A plasmid carrying a toxin-antitoxin (TA) module is introduced into a bacterial population. After several generations without antibiotic selection, most cells still retain the plasmid. Which explanation is most consistent with plasmid maintenance under these conditions?
   Answer: TA systems can create post-segregational killing, enriching for plasmid-containing cells. The skill being tested is prokaryotic genetics. Toxin-antitoxin (TA) systems on plasmids promote maintenance by killing daughter cells that lose the plasmid, as the stable toxin persists while the labile antitoxin degrades. In this population, plasmid retention without selection suggests TA-mediated post-segregational killing enriches for plasmid carriers. Choice A aligns because it describes how TA systems ensure inheritance by eliminating plasmid-free cells. Choice B is incorrect based on the misconception that plasmids replicate faster than chromosomes, but replication rates vary and do not prevent loss without mechanisms like TA. In similar questions, look for absence of selection and high retention as indicators of addiction modules. Verify if the system targets plasmid-free cells post-division.
6. A bacterial strain becomes resistant after incubation with extracellular DNA, but only when CaCl$_2$ treatment and a brief heat shock are applied. No donor cells are present. Which statement is most consistent with the experimental requirement, given the focus on transformation?
   Answer: The treatment likely increases membrane permeability and DNA uptake efficiency, consistent with transformation. The skill being tested is prokaryotic genetics. Transformation in non-naturally competent strains often requires treatments like CaCl2 and heat shock to enhance membrane permeability and DNA uptake. The requirement for these conditions with extracellular DNA and no donors points to induced transformation. Choice D is correct as it explains how the treatment facilitates uptake in transformation. Choice B fails on the misconception that CaCl2 induces pili for conjugation, but no donors are present. For similar questions, identify competence induction methods to confirm transformation. Rule out other processes by noting absence of donors or phages.
7. An Hfr strain (chromosomal integration of the F factor) is mated briefly with an F− recipient. After 5 minutes, mating is disrupted mechanically. Among recipients, a chromosomal marker located near the integrated F origin is frequently acquired, but recipients rarely become fully F+. Which outcome is most likely based on conjugation principles?
   Answer: Recipients frequently acquire early chromosomal genes but remain F− because transfer is interrupted before the full F sequence is delivered This question tests understanding of prokaryotic genetics, specifically Hfr conjugation and interrupted mating. Hfr strains have the F factor integrated into the chromosome, allowing transfer of chromosomal DNA during conjugation in a linear, time-ordered fashion. The passage describes brief mating (5 minutes) between Hfr and F- strains, with early markers frequently transferred but recipients rarely becoming F+. The correct answer (B) explains that during interrupted mating, chromosomal genes near the origin transfer first, but the F factor sequences transfer last; brief mating allows early gene transfer but not the complete F factor, so recipients remain F-. Answer A incorrectly states the F factor transfers first, which contradicts the established order where chromosomal DNA adjacent to the origin transfers before the F sequences. To understand Hfr transfer, remember: genes transfer in linear order from the origin, F factor sequences transfer last, and interrupted mating yields partial transfer with recipients typically remaining F-.
8. A plasmid (pA) encodes tetracycline resistance (TetR) and has an origin of transfer (oriT) but lacks genes for pilus formation. When pA is introduced into a strain that also contains a conjugative helper plasmid (pHelper), TetR transfers to recipients during mixed culture. When pHelper is absent, no transfer is detected. This setup is intended to probe plasmid function in horizontal gene transfer. Which outcome is most likely when pHelper is present?
   Answer: pA can be mobilized because pHelper supplies the transfer machinery that acts on pA's oriT This question tests understanding of plasmid mobilization in prokaryotic genetics, where non-self-transmissible plasmids can be transferred with help from conjugative plasmids. Mobilizable plasmids contain an origin of transfer (oriT) but lack genes for pilus formation and mating apparatus; however, they can be transferred when conjugative helper plasmids provide these functions in trans. The passage describes pA containing oriT but lacking transfer genes, which can only transfer when pHelper is present, consistent with answer D - pHelper supplies the conjugation machinery that recognizes and acts on pA's oriT. Answer B incorrectly invokes transformation (which doesn't require helper plasmids or oriT), while C wrongly suggests transduction (phage-mediated transfer not mentioned). To identify mobilization scenarios, look for: plasmids with oriT but no transfer genes, requirement for helper plasmids, and conjugation-dependent transfer.
9. A lab tests natural transformation in a competent bacterial strain. Purified DNA carrying a kanamycin-resistance allele (kanR) is added to cultures under two conditions: DNase I present throughout incubation vs no DNase. After recovery, cells are plated on kanamycin. Colonies appear only when DNase is absent. Which statement is most consistent with the observed genetic change?
   Answer: kanR likely entered cells as free DNA and required extracellular persistence long enough for uptake and recombination. This question tests understanding of prokaryotic genetics, specifically the mechanism of natural transformation. Transformation involves the uptake of free DNA from the environment by competent bacterial cells, followed by incorporation into the genome through recombination. The passage describes an experiment where DNase I (which degrades extracellular DNA) prevents kanamycin-resistant colonies from appearing, indicating that the kanR gene must exist as free DNA outside the cells for successful transformation. Answer A correctly identifies that kanR entered as free DNA requiring extracellular persistence for uptake and recombination. Answer B incorrectly suggests conjugation, but DNase doesn't affect pilus formation—it only degrades DNA. To identify transformation questions, look for: competent cells, free/naked DNA, and sensitivity to DNase treatment.
10. Two bacterial cultures are separated by a 0.2-$\mu$m filter that allows passage of small particles and DNA but prevents cell-to-cell contact. After incubation, antibiotic resistance appears in the previously sensitive culture. Addition of DNase I to the medium abolishes transfer. Which statement is most consistent with the bacterial behavior described?
    Answer: Resistance spread by transformation, because extracellular DNA crossing the filter was required and DNase I eliminated it This question tests understanding of prokaryotic genetics, specifically distinguishing between horizontal gene transfer mechanisms. The 0.2-μm filter allows passage of DNA and small particles but prevents direct cell contact, while DNase I specifically degrades extracellular DNA. The passage shows antibiotic resistance transfer through the filter, abolished by DNase I addition. The correct answer (B) identifies transformation as the mechanism because only naked DNA can pass through the filter, and DNase I specifically prevents transfer by degrading this extracellular DNA. Answer A incorrectly suggests conjugation, but pili cannot extend through 0.2-μm filters to establish the cell-to-cell contact required for conjugation. To distinguish transformation from other mechanisms, look for: transfer through filters preventing cell contact, sensitivity to DNase I, and the ability of naked DNA to cross barriers that cells cannot.
11. To test transformation, a competent Streptococcus culture was incubated with linear DNA containing an erythromycin resistance cassette (ErmR) flanked by sequences homologous to a nonessential chromosomal locus. Parallel reactions included (i) no added DNA, (ii) added DNA, and (iii) added DNA plus excess unrelated salmon sperm DNA. ErmR colonies were abundant in (ii) but reduced in (iii). Which statement is most consistent with the observed reduction in (iii)?
    Answer: Unrelated DNA competitively inhibits uptake machinery, reducing acquisition of the homologous DNA fragment This question tests understanding of bacterial transformation and DNA uptake competition in prokaryotic genetics. Transformation involves competent bacteria taking up extracellular DNA through dedicated uptake machinery, which has limited capacity and can be saturated. The passage shows that excess unrelated salmon sperm DNA reduces ErmR transformants, consistent with answer D - the unrelated DNA competes for the same uptake machinery, reducing the frequency of cells acquiring the resistance cassette. Answer B incorrectly invokes transduction (phage-mediated transfer not present here), while C wrongly suggests conjugation (which requires cell contact, not naked DNA). To identify transformation competition, look for: competent bacteria, naked DNA addition, reduction in transformation with excess unrelated DNA, and the absence of phages or conjugative elements.
12. A donor strain carrying a conjugative plasmid with a functional relaxase gene is compared to an otherwise identical donor with a relaxase loss-of-function mutation. Both are mixed separately with the same plasmid-free recipient under conditions permitting cell contact. Only the functional-relaxase donor yields recipients that acquire the plasmid marker. Which statement is most consistent with the data and conjugation mechanism?
    Answer: Relaxase is required to nick plasmid DNA at oriT to initiate transfer of a single DNA strand into the recipient This question tests understanding of prokaryotic genetics, specifically the molecular mechanism of bacterial conjugation. Relaxase is an essential enzyme that initiates conjugative transfer by creating a site-specific nick at the origin of transfer (oriT) on the plasmid. The passage compares functional versus mutant relaxase donors, showing only functional relaxase enables plasmid transfer to recipients. The correct answer (A) accurately describes relaxase function: it nicks the plasmid at oriT to initiate rolling-circle replication and transfer of a single DNA strand through the conjugative pilus to the recipient. Answer B incorrectly assigns relaxase a role in transformation by degrading DNA, but relaxase specifically acts on plasmid DNA at oriT, not on extracellular DNA. To understand conjugation requirements, remember: relaxase nicks at oriT to start transfer, single-stranded DNA transfers through the pilus, and both donor and recipient synthesize complementary strands to complete the process.
13. A researcher observes that adding DNase I during a DNA acquisition experiment eliminates appearance of new antibiotic-resistant colonies. The setup involved bacteria incubated with purified DNA in the absence of any donor cells. The result is most consistent with which genetic process?
    Answer: Transformation, because extracellular DNA must persist long enough to be taken up. The skill being tested is prokaryotic genetics. Transformation is the uptake of free extracellular DNA, which is sensitive to DNase degradation, whereas conjugation and transduction are not. In this setup with purified DNA and no donor cells, DNase elimination of resistant colonies points to DNA uptake being blocked. Choice B is correct as it explains that transformation requires persistent extracellular DNA vulnerable to DNase. Choice A fails due to the misconception that DNase degrades the pilus in conjugation, but pili are protein structures not affected by DNase. For similar questions, assess DNase sensitivity to differentiate transformation from protected transfers. Confirm absence of donor cells or phages to rule out conjugation or transduction.
14. A hospital isolate contains a 90-kb plasmid (pR) encoding a beta-lactamase and a toxin–antitoxin maintenance system. After 50 generations of growth in antibiotic-free medium, plasmid retention was measured. Strain A (pR with intact toxin–antitoxin) retained pR in 94% of cells, whereas Strain B (pR with the antitoxin gene disrupted) retained pR in 12% of cells. Based on these results, which outcome is most likely in Strain B during antibiotic-free growth?
    Answer: Decreased plasmid retention because plasmid-free segregants are not eliminated by post-segregational killing This question tests understanding of plasmid maintenance systems in prokaryotic genetics, specifically toxin-antitoxin modules. Toxin-antitoxin systems ensure plasmid retention through post-segregational killing: the antitoxin (unstable) degrades faster than the toxin (stable), so plasmid-free segregants die when they lose the antitoxin gene. The passage shows that disrupting the antitoxin gene dramatically reduces plasmid retention (94% to 12%), consistent with answer B - without functional post-segregational killing, plasmid-free cells survive and accumulate during antibiotic-free growth. Answer A incorrectly reverses the mechanism (antitoxin loss would increase toxin activity, not prevent it), while C wrongly limits toxin-antitoxin function to conjugation events only. To recognize plasmid maintenance questions, look for: differential retention rates, toxin-antitoxin systems, antibiotic-free growth conditions, and the survival of plasmid-free segregants when the system is disrupted.
15. A clinical isolate carries a 6-kb plasmid encoding a beta-lactamase. When the plasmid is cured, the isolate becomes ampicillin-sensitive but remains viable. Sequencing shows the plasmid also contains an origin of transfer (oriT) but no chromosomal integration sequences. Which statement is most consistent with the plasmid's role in the population under intermittent antibiotic exposure?
    Answer: The plasmid is maintained because it can horizontally spread by conjugation and provides a selectable advantage only when ampicillin is present This question tests understanding of prokaryotic genetics, specifically plasmid maintenance and horizontal gene transfer. Plasmids are autonomous genetic elements that replicate independently of the chromosome and can provide selective advantages like antibiotic resistance. The passage describes a plasmid encoding beta-lactamase (ampicillin resistance) with an origin of transfer (oriT) but no integration sequences, and the host remains viable when the plasmid is lost. The correct answer (A) explains that the plasmid persists because it can spread horizontally via conjugation (enabled by oriT) and provides selective advantage only when ampicillin is present. Answer D incorrectly suggests oriT ensures equal segregation during cell division, but oriT functions in conjugative transfer, not plasmid partitioning. To analyze plasmid maintenance, consider: selective pressure (antibiotic presence), horizontal transfer capability (oriT for conjugation), and the conditional nature of the advantage (only beneficial under antibiotic exposure).
16. In a conjugation assay, donor cells carry a conjugative plasmid with an antibiotic resistance gene and a pilus gene. A mutant donor lacking the pilus gene is mixed with recipients; no resistant recipients are recovered. The key process is conjugation. Which statement is most consistent with the principle tested?
    Answer: Loss of pilus function is consistent with impaired cell-to-cell contact required for plasmid transfer to recipients. This question tests understanding of prokaryotic genetics, specifically the conjugation mechanism and pilus requirement. The pilus is the physical bridge enabling cell-to-cell contact necessary for DNA transfer during conjugation—without it, donors cannot establish the connection needed for plasmid transfer. The passage shows pilus gene deletion prevents resistance transfer, demonstrating its essential role. Answer D correctly identifies that loss of pilus function impairs the cell-to-cell contact required for plasmid transfer. Answer B incorrectly associates pili with transformation, but transformation uses competence machinery, not pili, for DNA uptake. To recognize conjugation requirements, look for: pilus necessity for cell contact, transfer failure without functional pili, and distinction from transformation which doesn't require pili.
17. A clinical isolate carries a small plasmid with a beta-lactamase gene (bla). When the plasmid is cured (removed) by growth under nonselective conditions, the isolate becomes sensitive to ampicillin while other phenotypes remain unchanged. Which statement is most consistent with plasmid function in this observation?
    Answer: Ampicillin sensitivity after curing is consistent with bla being plasmid-encoded and lost when the plasmid is eliminated. This question tests understanding of prokaryotic genetics, specifically plasmid function and maintenance. Plasmids are extrachromosomal genetic elements that replicate independently and can be lost from cells during division, especially without selective pressure. The passage shows that removing (curing) the plasmid causes loss of ampicillin resistance while other traits remain, indicating the bla gene is plasmid-encoded. Answer B correctly identifies that ampicillin sensitivity after curing confirms bla is on the plasmid and lost when eliminated. Answer A incorrectly suggests chromosomal location, which wouldn't be affected by plasmid curing. To analyze plasmid-related questions, check if: traits are lost with plasmid removal, other phenotypes remain unchanged, and the explanation fits extrachromosomal inheritance patterns.
18. A researcher notes that a resistance gene spreads rapidly through a bacterial population only when the gene is located on a conjugative plasmid, but spreads slowly when located on the chromosome. Which statement is most consistent with the observed difference in spread, focusing on one process?
    Answer: Conjugative plasmids can transfer horizontally between cells, accelerating dissemination relative to chromosomal inheritance. The skill being tested is prokaryotic genetics. Conjugative plasmids enable horizontal transfer between cells, spreading genes faster than vertical chromosomal inheritance. Rapid spread on plasmids vs. slow on chromosomes highlights conjugation's efficiency. Choice A is correct as it contrasts horizontal conjugation with vertical replication. Choice D fails due to the misconception that plasmids are only vertical, ignoring their horizontal mobility. For similar questions, compare spread rates between locations to infer transfer mode. Assess conjugation potential to explain dissemination differences.
19. A plasmid carrying a toxin–antitoxin module is introduced into a bacterial population. In antibiotic-free media, plasmid-bearing cells persist while plasmid-cured segregants decline. No additional growth advantage is measured for plasmid-bearing cells. Which statement is most consistent with the observed maintenance of the plasmid?
    Answer: The plasmid persists because post-segregational killing reduces survival of plasmid-free cells, increasing the relative frequency of plasmid-bearing cells This question tests understanding of prokaryotic genetics, specifically plasmid maintenance through toxin-antitoxin systems. Toxin-antitoxin modules encode a stable toxin and unstable antitoxin; when plasmids are lost during cell division, the antitoxin degrades faster than the toxin, killing plasmid-free cells. The passage describes plasmid persistence without growth advantage, with plasmid-cured cells declining in antibiotic-free media. The correct answer (C) explains post-segregational killing: cells losing the plasmid during division die from toxin activity when antitoxin production ceases, increasing the relative frequency of plasmid-bearing cells. Answer B incorrectly connects toxin-antitoxin systems to pilus formation and conjugation, which are unrelated functions. To recognize toxin-antitoxin maintenance, look for: plasmid persistence without selective pressure, decline of plasmid-free segregants, and maintenance through elimination of plasmid-free cells rather than growth advantage.
20. In a study of transformation, a naturally competent bacterial strain (Strain X) was incubated with purified DNA from Strain Y carrying a point mutation in gyrA that confers ciprofloxacin resistance (CipR). Cultures were treated with DNase I either during the incubation or after 30 minutes, then plated on ciprofloxacin. CipR colonies were recovered only when DNase I was added after 30 minutes. Which statement is most consistent with the bacterial behavior described?
    Answer: Uptake of extracellular DNA occurred before DNase I addition at 30 minutes, allowing chromosomal recombination to yield CipR cells This question tests understanding of bacterial transformation, a form of horizontal gene transfer in prokaryotic genetics. Transformation involves the uptake of naked DNA from the environment by naturally competent bacteria, which can then integrate this DNA into their chromosome through homologous recombination. The passage describes Strain X taking up purified DNA containing a gyrA mutation from Strain Y, with DNase I (which degrades extracellular DNA) added at different times. The correct answer C explains that DNA uptake occurred before the 30-minute DNase I addition, allowing time for the DNA to enter cells and undergo chromosomal recombination to produce CipR colonies. Answer A incorrectly invokes transduction (phage-mediated transfer), which wasn't described in the experiment, while B incorrectly suggests de novo mutations rather than DNA transfer. To identify transformation questions, look for: competent bacteria, purified/naked DNA, DNase sensitivity, and the absence of phages or cell-to-cell contact.

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