MCAT Biological and Biochemical Foundations of Living Systems Quiz: 2b Prokaryotic Genetics

What this quiz covers

This quiz focuses on 2b Prokaryotic Genetics, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Biological and Biochemical Foundations of Living Systems.

How to use this quiz

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.

Question 1

1. The resistance determinant was transferred by bacteriophage-mediated transduction, which would be insensitive to extracellular DNase I
2. Ciprofloxacin exposure induced de novo gyrA mutations in Strain X, so DNase I timing would not affect CipR recovery
3. Uptake of extracellular DNA occurred before DNase I addition at 30 minutes, allowing chromosomal recombination to yield CipR cells (correct answer)
4. Resistance requires direct cell-to-cell contact via a conjugative pilus, so DNase I would prevent transfer only when added late

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.

Question 2

1. Fluorescence indicates conjugation because plasmid DNA cannot enter cells unless a pilus is present in the recipient.
2. Fluorescence requires phage-mediated delivery of plasmid DNA, so a phage inhibitor should increase fluorescence by preventing lysis.
3. Fluorescence is most consistent with uptake of free plasmid DNA; blocking phage activity would not prevent this process. (correct answer)
4. 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.

Question 3

1. rifR colonies arose from bacteriophage-mediated DNA delivery, which is inhibited by CaCl2 treatment
2. rifR colonies arose because cells incorporated naked DNA during transformation, and DNase I prevented uptake when present during exposure (correct answer)
3. rifR colonies arose via conjugation, requiring direct donor–recipient contact that was disrupted by DNase I
4. rifR colonies arose because rifampin induced de novo mutations at the rif locus during selection

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).

Question 4

1. In the absence of selection, plasmid carriage can impose a fitness cost, favoring loss of the plasmid over time (correct answer)
2. Plasmid-bearing cells should outcompete plasmid-free cells because plasmids universally increase metabolic efficiency
3. The increased plasmid-free frequency indicates the resistance gene was transferred to the chromosome by generalized transduction
4. The slower growth proves the plasmid must be integrated into the chromosome, since extrachromosomal DNA cannot affect growth

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.

Question 5

1. Relaxase is required to nick plasmid DNA at oriT to initiate transfer of a single DNA strand into the recipient (correct answer)
2. Relaxase is required to degrade extracellular plasmid DNA so that competent recipients can transform
3. Relaxase is required in the recipient to integrate the plasmid into the chromosome before transfer can occur
4. Relaxase is required to package plasmid DNA into bacteriophage capsids during generalized transduction

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.

Question 6

1. The increased mRNA results from enhanced transcription initiation, leading to higher translation due to coupled transcription-translation (correct answer)
2. The increased ribosome occupancy indicates reduced transcription initiation and compensatory translation upregulation
3. The unchanged mRNA half-life implies the operon is regulated exclusively at the post-translational level
4. The promoter insertion most likely increased translation by adding a Shine-Dalgarno sequence upstream of each gene

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.

Question 7

1. Species 2 likely lacks competence under the tested conditions, preventing uptake of extracellular plasmid DNA (correct answer)
2. Species 2 must have received the plasmid but expressed KanR in the wrong direction relative to replication
3. Species 1 likely acquired KanR via specialized transduction, which would not require competence factors
4. 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.

Question 8

1. The treatment causes targeted chromosomal recombination without requiring exogenous DNA.
2. CaCl$_2$ and heat shock are required to induce pilus synthesis for conjugation.
3. The treatment activates prophage excision, enabling specialized transduction.
4. The treatment likely increases membrane permeability and DNA uptake efficiency, consistent with transformation. (correct answer)

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.

Question 9

1. Transfer likely reflects spontaneous mutation in Strain 2 triggered by proximity to Strain 1.
2. Transfer likely occurs via uptake of free DNA, which should pass through the filter.
3. Transfer likely occurs via bacteriophage particles, which cannot cross 0.22 µm filters.
4. Transfer likely requires cell-to-cell contact consistent with conjugation. (correct answer)

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.

Question 10

1. Erm^R colonies likely arose from transcriptional upregulation of a pre-existing chromosomal erm gene.
2. Heat-killed cells should transform more efficiently because membranes become permeable.
3. Erm^R colonies must result from conjugation because plasmids cannot enter cells by uptake.
4. Transformation requires active cellular processes for DNA uptake; dead cells cannot acquire plasmid DNA. (correct answer)

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.

Question 11

1. Conjugation depends on donor-encoded transfer machinery, including pilus formation and DNA processing. (correct answer)
2. Conjugation is mediated by bacteriophages, so donor tra genes are unnecessary.
3. Conjugation requires recipient competence genes, so donor mutations should not matter.
4. Conjugation transfers only chromosomal DNA, not plasmids, so tra genes are irrelevant.

Explanation: The skill being tested is prokaryotic genetics. Conjugation relies on donor-encoded tra genes that facilitate pilus formation, DNA processing, and transfer to the recipient. The tra knockout in the donor abolishes plasmid transfer, highlighting the donor's role in providing transfer machinery. Choice A aligns because it correctly states that conjugation depends on donor tra functions for pilus and DNA handling. Choice B is a distractor based on the misconception that conjugation is phage-mediated, but it is a direct cell-to-cell process independent of viruses. In similar questions, identify if mutations affect donor or recipient to pinpoint machinery origin. Check for pilus involvement to confirm conjugation over other transfers.

Question 12

1. TA systems can create post-segregational killing, enriching for plasmid-containing cells. (correct answer)
2. Plasmids are always replicated faster than chromosomes, so they cannot be lost.
3. DNase in the environment prevents plasmid loss by protecting extracellular DNA.
4. Only bacteriophages, not plasmids, can encode stable inheritance systems.

Explanation: 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.

Question 13

1. The increased mRNA is most consistent with generalized transduction, which packages mRNA and injects it into cells during starvation.
2. Carbon starvation increases mRNA by converting the plasmid into chromosomal DNA, which is transcribed at a fixed higher rate.
3. The increased mRNA indicates that starvation causes reverse transcription of protein into mRNA, bypassing transcriptional control.
4. A starvation-responsive regulator likely alters transcription from the plasmid promoter, changing mRNA abundance without changing DNA sequence. (correct answer)

Explanation: This question tests understanding of prokaryotic genetics, specifically transcriptional regulation of plasmid-borne genes. Prokaryotic gene expression often responds to environmental conditions through regulatory proteins that modulate transcription from promoters. The passage shows increased mRNA during carbon starvation without DNA sequence changes, indicating transcriptional regulation. Answer D correctly identifies that a starvation-responsive regulator alters transcription from the plasmid promoter, changing mRNA levels while DNA remains constant. Answer C incorrectly proposes reverse transcription from protein to mRNA, which violates the central dogma. To recognize transcriptional regulation, look for: condition-dependent mRNA changes, unchanged DNA sequences, and responses consistent with known regulatory mechanisms like catabolite repression.

Question 14

1. Differential plasmid retention is consistent with plasmids having different stability/partitioning, especially without selection. (correct answer)
2. P2 was lost because chromosomal genes always replace plasmid genes when antibiotics are absent, by directed recombination.
3. P2 was lost because transformation selectively degrades plasmids with kanR sequences in the extracellular medium.
4. P1 was retained because conjugation forces recipients to keep the first plasmid acquired and expel all later plasmids.

Explanation: This question tests understanding of prokaryotic genetics, specifically plasmid stability and maintenance. Different plasmids have varying stability based on their replication control, partitioning systems, and metabolic burden—some are lost more readily without selection. The passage describes differential retention where P1 persists but P2 is lost without antibiotic selection. Answer A correctly explains this as different stability/partitioning properties between plasmids, especially without selective pressure maintaining them. Answer B incorrectly suggests directed recombination replacing plasmid genes, which isn't a standard mechanism. To analyze plasmid stability, consider: different plasmids have different maintenance systems, selection pressure affects retention, and plasmid loss is stochastic without active partitioning.

Question 15

1. Stable chromosomal integration by homologous recombination during transformation, which does not require RecA
2. Generalized transduction, because transient phenotypes indicate phage-mediated genomic integration
3. Transient maintenance of non-replicating donor DNA acquired by transformation without RecA-dependent integration (correct answer)
4. Conjugative transfer of chromosomal DNA requiring oriT, producing unstable merodiploids

Explanation: This question tests understanding of prokaryotic genetics, specifically transformation and the role of RecA in DNA integration. RecA is essential for homologous recombination, which allows stable integration of foreign DNA into the chromosome. The passage describes a RecA-deficient strain showing transient expression of a wild-type allele from purified DNA, with the phenotype lost without selection. The correct answer (C) explains that without RecA, the transformed DNA cannot integrate via homologous recombination, so it exists transiently as non-replicating DNA that dilutes out during cell division. Answer A incorrectly suggests stable integration can occur without RecA, which contradicts the fundamental requirement for RecA in homologous recombination. To identify transient transformation, look for: RecA deficiency preventing integration, temporary phenotype expression, and loss of phenotype without selective pressure maintaining the non-integrated DNA.

Question 16

1. Conjugative plasmid transfer from F+ donor to F− recipient requiring a functional pilus (correct answer)
2. Transformation of recipients by uptake of free plasmid DNA released from donors, which requires pilus assembly proteins
3. Generalized transduction of AmpR by phage particles produced during co-culture, which depends on pilus genes for adsorption
4. Spontaneous acquisition of CamR by the donor strain, followed by selection on dual antibiotics

Explanation: This question tests understanding of bacterial conjugation, a key mechanism of horizontal gene transfer in prokaryotic genetics. Conjugation requires direct cell-to-cell contact mediated by a pilus structure encoded by F (fertility) factor genes, allowing transfer of plasmid DNA from F+ donors to F− recipients. The passage shows that AmpR (from the donor's plasmid) and CamR (from the recipient's chromosome) are both present in colonies selected on dual antibiotics, indicating the F− recipient acquired the AmpR plasmid through conjugation. The correct answer A accurately describes this process, while the loss of transfer when pilus genes are absent confirms conjugation's requirement for functional pili. Answer B incorrectly suggests transformation (which doesn't require pili), and C incorrectly invokes transduction (phage-mediated transfer not mentioned in the experiment). When identifying conjugation, look for: F+/F− strains, requirement for cell contact, pilus dependence, and plasmid transfer without DNase sensitivity.

Question 17

1. Resistance spread by conjugation, which requires a pilus that can extend through the filter pores
2. Resistance spread by transformation, because extracellular DNA crossing the filter was required and DNase I eliminated it (correct answer)
3. Resistance spread by specialized transduction, because DNase I degrades bacteriophage capsids
4. Resistance spread by spontaneous mutation induced by DNase I, which is blocked when DNase I is present

Explanation: 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.

Question 18

1. Increased plasmid retention because loss of antitoxin prevents toxin activity and reduces selection against plasmid-free cells
2. Decreased plasmid retention because plasmid-free segregants are not eliminated by post-segregational killing (correct answer)
3. No change in plasmid retention because toxin–antitoxin systems only function during conjugation events
4. Higher beta-lactamase expression because disruption of antitoxin increases plasmid copy number in all cells

Explanation: 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.

Question 19

1. pA becomes integrated into the recipient chromosome at high frequency because oriT functions as a recombination hotspot
2. pA transfer occurs by transformation because oriT sequences are recognized by competence proteins only in the presence of pHelper
3. pA transfer occurs by transduction because pHelper increases phage production and packages oriT-containing DNA selectively
4. pA can be mobilized because pHelper supplies the transfer machinery that acts on pA's oriT (correct answer)

Explanation: 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.

Question 20

1. Conjugation, because plasmid DNA must be transferred through a pilus to enable chromosomal insertion
2. Transformation followed by homologous recombination, resulting in chromosomal integration and loss of the original plasmid (correct answer)
3. Specialized transduction, because short homologous sequences are characteristic of prophage attachment sites
4. Spontaneous mutation, because selection on kanamycin induces targeted insertion events at homologous loci

Explanation: This question tests understanding of prokaryotic genetics, specifically chromosomal integration following transformation. Homologous recombination allows foreign DNA with sequence similarity to integrate into the chromosome at corresponding sites. The passage describes kanamycin resistance found integrated in the chromosome with homologous flanking sequences, while the original plasmid is absent. The correct answer (B) explains transformation followed by homologous recombination: the plasmid DNA was taken up, and the kanR gene integrated at a chromosomal site with sequence homology, while the non-homologous plasmid backbone was lost. Answer C incorrectly invokes specialized transduction, but the presence of homologous sequences in both donor and recipient indicates recombination, not phage-mediated insertion at specific attachment sites. To identify integration via homologous recombination, look for: presence of homologous sequences flanking the insertion, loss of vector sequences, and stable chromosomal integration of the selected marker.