Genetics of Prokaryotes (2B)
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MCAT Biological and Biochemical Foundations of Living Systems › Genetics of Prokaryotes (2B)
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?
Uptake of extracellular DNA occurred before DNase I addition at 30 minutes, allowing chromosomal recombination to yield CipR cells
Resistance requires direct cell-to-cell contact via a conjugative pilus, so DNase I would prevent transfer only when added late
Ciprofloxacin exposure induced de novo gyrA mutations in Strain X, so DNase I timing would not affect CipR recovery
The resistance determinant was transferred by bacteriophage-mediated transduction, which would be insensitive to extracellular DNase I
Explanation
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.
A researcher adds purified plasmid DNA encoding a fluorescent protein to a bacterial culture under conditions known to induce competence. Fluorescent cells appear after incubation. When the same experiment is repeated with a bacteriophage inhibitor but without DNase, fluorescence still appears. The genetic process of interest is transformation. Which statement is most consistent with these results?
Fluorescence is most consistent with uptake of free plasmid DNA; blocking phage activity would not prevent this process.
Fluorescence requires phage-mediated delivery of plasmid DNA, so a phage inhibitor should increase fluorescence by preventing lysis.
Fluorescence indicates conjugation because plasmid DNA cannot enter cells unless a pilus is present in the recipient.
Fluorescence indicates chromosomal mutation because competence conditions increase replication errors that create fluorescent alleles.
Explanation
This question tests understanding of prokaryotic genetics, specifically distinguishing transformation from other horizontal gene transfer mechanisms. Transformation involves uptake of free DNA by competent cells and doesn't require phage activity—only DNA availability and cellular competence matter. The passage shows fluorescence appears under competence conditions even with phage inhibitor present, ruling out transduction. Answer C correctly identifies that fluorescence results from uptake of free plasmid DNA, which phage inhibitors wouldn't prevent. Answer B incorrectly suggests phage requirement for plasmid delivery, contradicting the experimental result. To identify transformation, look for: competence induction, free DNA uptake, independence from phage activity, and sensitivity to DNase but not phage inhibitors.
A lab incubated a naturally competent bacterial strain with purified linear DNA carrying a rifampin-resistance allele (rifR). After incubation, cells were treated with DNase I, then plated on rifampin. Colonies appeared only when cells had been exposed to CaCl2 and a brief heat shock; colonies were absent when DNase I was added before incubation with DNA. Which statement is most consistent with the genetic process responsible for rifR acquisition?
rifR colonies arose via conjugation, requiring direct donor–recipient contact that was disrupted by DNase I
rifR colonies arose because rifampin induced de novo mutations at the rif locus during selection
rifR colonies arose from bacteriophage-mediated DNA delivery, which is inhibited by CaCl2 treatment
rifR colonies arose because cells incorporated naked DNA during transformation, and DNase I prevented uptake when present during exposure
Explanation
This question tests understanding of prokaryotic genetics, specifically the mechanism of transformation. Transformation is the uptake of naked DNA from the environment by competent bacterial cells, which can be naturally competent or made competent through chemical treatment like CaCl2 and heat shock. The passage describes conditions where rifampin-resistant colonies only appeared when cells were treated with CaCl2 and heat shock, and no colonies formed when DNase I was added before DNA exposure. The correct answer (B) accurately describes transformation because DNase I degrades extracellular DNA, preventing its uptake when added before incubation. Answer A incorrectly invokes bacteriophage-mediated transduction, but CaCl2 actually enhances transformation, not inhibits phage activity. To identify transformation in similar questions, look for: uptake of naked DNA, sensitivity to DNase I treatment, and the requirement for competence (natural or induced).
A research group tracks plasmid function by measuring growth rates of isogenic bacteria with or without a large resistance plasmid (pLarge) in antibiotic-free medium. The plasmid-bearing strain grows 15% slower. After prolonged culture without antibiotics, plasmid-free cells increase in frequency. Which statement is most consistent with these findings?
In the absence of selection, plasmid carriage can impose a fitness cost, favoring loss of the plasmid over time
The increased plasmid-free frequency indicates the resistance gene was transferred to the chromosome by generalized transduction
The slower growth proves the plasmid must be integrated into the chromosome, since extrachromosomal DNA cannot affect growth
Plasmid-bearing cells should outcompete plasmid-free cells because plasmids universally increase metabolic efficiency
Explanation
This question tests understanding of plasmid fitness costs in prokaryotic genetics, a key factor affecting plasmid stability in bacterial populations. Large plasmids often impose metabolic burdens through replication costs, expression of plasmid genes, and interference with host processes, reducing the growth rate of plasmid-bearing cells compared to plasmid-free cells. The passage shows plasmid-bearing cells grow 15% slower and decrease in frequency without antibiotic selection, consistent with answer A - the fitness cost favors plasmid loss when the selective advantage (antibiotic resistance) is removed. Answer B incorrectly claims plasmids universally increase efficiency (contradicted by the slower growth), while C wrongly invokes chromosomal transfer without evidence. When analyzing plasmid stability, look for: growth rate differences, frequency changes over time, presence/absence of selection pressure, and the balance between plasmid benefits and costs.
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?
Relaxase is required to degrade extracellular plasmid DNA so that competent recipients can transform
Relaxase is required to nick plasmid DNA at oriT to initiate transfer of a single DNA strand into the recipient
Relaxase is required to package plasmid DNA into bacteriophage capsids during generalized transduction
Relaxase is required in the recipient to integrate the plasmid into the chromosome before transfer can occur
Explanation
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.
A researcher introduces a plasmid carrying a strong constitutive promoter upstream of a chromosomal operon via homologous recombination, leading to a 6-fold increase in the operon’s mRNA level by qPCR. Ribosome profiling shows increased ribosome occupancy across the operon without a change in mRNA half-life. This experiment is described as a prokaryotic gene expression change. Which statement is most consistent with the data?
The unchanged mRNA half-life implies the operon is regulated exclusively at the post-translational level
The increased ribosome occupancy indicates reduced transcription initiation and compensatory translation upregulation
The promoter insertion most likely increased translation by adding a Shine-Dalgarno sequence upstream of each gene
The increased mRNA results from enhanced transcription initiation, leading to higher translation due to coupled transcription-translation
Explanation
This question tests understanding of prokaryotic gene expression regulation, specifically the relationship between transcription and translation. In prokaryotes, transcription and translation are coupled - ribosomes bind mRNA as it's being synthesized, so increased transcription typically leads to increased translation. The passage describes inserting a strong promoter that increases mRNA levels 6-fold with corresponding increases in ribosome occupancy, consistent with answer A's explanation of enhanced transcription initiation driving higher translation. Answer B incorrectly suggests reduced transcription (contradicting the 6-fold mRNA increase), while D wrongly attributes the effect to Shine-Dalgarno sequences (which would be part of the original genes, not the promoter). When analyzing prokaryotic gene expression changes, look for: coordinated changes in mRNA and ribosome occupancy, the coupling of transcription-translation, and promoter effects on transcription initiation rather than translation efficiency.
In an experiment focused on transformation, a researcher adds a circular plasmid encoding kanamycin resistance (KanR) to two bacterial species under identical conditions. Species 1 yields KanR colonies; Species 2 yields none. Sequencing confirms the plasmid is intact in Species 1 transformants. No bacteriophage are detected in either culture. Which statement is most consistent with these observations?
Species 1 likely acquired KanR via specialized transduction, which would not require competence factors
Species 2 must have received the plasmid but expressed KanR in the wrong direction relative to replication
Species 2 likely lacks competence under the tested conditions, preventing uptake of extracellular plasmid DNA
Species 1 likely donated the plasmid to Species 2 through an F pilus, but selection prevented detection of transconjugants
Explanation
This question tests understanding of species-specific competence in bacterial transformation within prokaryotic genetics. Natural competence for DNA uptake varies dramatically between bacterial species and requires specific environmental conditions and genetic machinery; many bacteria are not naturally competent or require specific inducing conditions. The passage shows Species 1 successfully transforms with a plasmid while Species 2 does not under identical conditions, consistent with answer A - Species 2 likely lacks competence machinery or appropriate conditions for DNA uptake. Answer C incorrectly invokes transduction (explicitly ruled out by absence of phage), while D wrongly suggests conjugation from Species 1 to 2 (impossible as they were separately exposed to naked plasmid DNA). When evaluating transformation experiments, look for: species-specific competence differences, requirement for DNA uptake machinery, and the distinction between naturally competent and non-competent bacteria.
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?
CaCl$_2$ and heat shock are required to induce pilus synthesis for conjugation.
The treatment activates prophage excision, enabling specialized transduction.
The treatment causes targeted chromosomal recombination without requiring exogenous DNA.
The treatment likely increases membrane permeability and DNA uptake efficiency, consistent with transformation.
Explanation
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.
Two bacterial strains are mixed: Strain 1 is F+ and carries a plasmid encoding gentamicin resistance $(Gen^R$). Strain 2 is F− and $Gen^S$. After co-incubation, $Gen^R$ appears in Strain 2 only when cells are allowed direct contact; separating strains with a 0.22 µm filter prevents transfer. Which statement is most consistent with the observed genetic change?
Transfer likely reflects spontaneous mutation in Strain 2 triggered by proximity to Strain 1.
Transfer likely requires cell-to-cell contact consistent with conjugation.
Transfer likely occurs via uptake of free DNA, which should pass through the filter.
Transfer likely occurs via bacteriophage particles, which cannot cross 0.22 µm filters.
Explanation
The skill being tested is prokaryotic genetics. Conjugation requires direct cell-to-cell contact for DNA transfer via a pilus, which is blocked by physical barriers like filters. The transfer of $Gen^R$ only with direct contact and prevention by a 0.22 µm filter indicates contact-dependent mechanism. Choice D is correct as it links the need for contact to conjugation. Choice B fails on the misconception that free DNA can pass filters for transformation, but the filter size blocks bacteria while allowing DNA, yet transfer requires contact here. For similar questions, test physical separation to confirm contact necessity. Evaluate filter pore size to distinguish cellular from molecular transfers.
To test transformation, Bacillus subtilis was incubated with a circular plasmid encoding erythromycin resistance $(Erm^R$). In one condition, cells were heat-killed before plasmid addition; in another, live cells were used. Only the live-cell condition produced $Erm^R$ colonies. Which statement is most consistent with the data?
Heat-killed cells should transform more efficiently because membranes become permeable.
$Erm^R$ colonies must result from conjugation because plasmids cannot enter cells by uptake.
Transformation requires active cellular processes for DNA uptake; dead cells cannot acquire plasmid DNA.
$Erm^R$ colonies likely arose from transcriptional upregulation of a pre-existing chromosomal erm gene.
Explanation
The skill being tested is prokaryotic genetics. Transformation requires live, competent cells to actively uptake and integrate extracellular DNA, as dead cells lack the necessary metabolic activity and membrane integrity. In this Bacillus subtilis experiment, only live cells produce $Erm^R$ colonies, indicating active processes are essential. Choice D is correct because it emphasizes that dead cells cannot perform the energy-dependent uptake required for transformation. Choice B fails on the misconception that heat-killing enhances permeability for transformation, but actually, viability is crucial for DNA internalization. For similar questions, test cell viability to confirm active uptake mechanisms. Distinguish from passive processes by noting requirements for competence and energy.