- The lac operon in E. coli consists of the genes lacZ, lacY, and lacA, transcribed as a single polycistronic mRNA. A researcher introduces a nonsense mutation early in the lacZ coding sequence. In the presence of an inducer, which of the following effects on protein expression is most likely to be observed due to polar effects?
Answer: A truncated LacZ protein will be produced, and the synthesis of LacY and LacA proteins will be significantly reduced.
A nonsense mutation introduces a premature stop codon, leading to the synthesis of a truncated protein (in this case, LacZ). This premature termination of translation can also have a polar effect on downstream genes in an operon. Ribosomes dissociate from the mRNA, which can expose sites for Rho-dependent transcriptional termination or simply lead to mRNA degradation, preventing efficient translation of the subsequent genes (lacY and lacA). Therefore, the most likely outcome is a truncated LacZ and significantly reduced levels of LacY and LacA. Choice A is incorrect because a nonsense mutation causes truncation, not a full-length protein. Choice B is too absolute; some residual transcription/translation of downstream genes may occur. Choice D ignores the well-documented phenomenon of polarity in bacterial operons.
- A scientist wants to quantitatively compare the in vivo activity of two promoters, P_strong_ and P_weak_. They construct two plasmids of identical copy number, one with P_strong_ fused to a promoterless green fluorescent protein (GFP) gene and the other with P_weak_ fused to the same GFP gene. After transforming these into identical bacterial host strains and growing them to mid-log phase under inducing conditions, what is the most appropriate measurement to determine the relative strength of the two promoters?
Answer: Measure the bulk fluorescence of each culture and normalize this value by the culture's optical density.
When comparing promoter strength in vivo, you need a quantitative method that directly reflects transcriptional activity while accounting for experimental variables like cell density. This is fundamentally about measuring gene expression output in living cells.
Option D provides the most practical and appropriate approach. Bulk fluorescence measurement captures the total GFP protein produced by each culture, which directly reflects the cumulative transcriptional activity of each promoter. Normalizing by optical density (OD) is crucial because it accounts for differences in cell number between cultures, giving you fluorescence per cell. This standardization allows for accurate comparison of promoter strength regardless of minor variations in culture density.
Option A, while technically sound, is unnecessarily complex and expensive for a straightforward promoter comparison. Single-molecule FISH measures mRNA directly but requires sophisticated equipment and doesn't provide the practical readout needed for routine promoter characterization.
Option B is fundamentally flawed because growth rate reflects metabolic burden, not promoter strength. Many factors beyond GFP expression affect growth, making this an unreliable proxy for transcriptional activity.
Option C examines promoter sequence rather than actual function. While consensus sequences can predict promoter strength, this approach ignores the complex cellular environment, regulatory factors, and chromatin context that influence real promoter performance in vivo.
Remember: when evaluating gene expression experimentally, always normalize your measurements to account for cell number differences, and choose methods that measure the actual biological output rather than indirect proxies.
- A researcher performs a comprehensive screen for mutants unable to grow on citrate as a sole carbon source, using transposon mutagenesis on a rich medium followed by replica plating. After screening tens of thousands of individual mutants, they find that all identified citrate-negative mutants have insertions located in one of two genes, citA or citB. Which of the following is the most sound and cautious interpretation of this result?
Answer: Other genes required for citrate utilization may also be essential for viability on the rich medium used for the initial mutagenesis.
When you encounter questions about genetic screens and mutagenesis, think critically about what the experimental design can and cannot reveal. This question tests your understanding of the limitations inherent in transposon mutagenesis screens.
The key insight is that this screen can only identify genes that are dispensable for growth on rich medium but essential for citrate utilization. If a gene is required for both citrate metabolism AND survival on the rich medium used for mutagenesis, insertional mutants in that gene would die during the initial growth phase and never make it to the replica plating step. Therefore, answer D correctly identifies this experimental blind spot—other genes required for citrate utilization may also be essential for viability on rich medium and would be missed entirely.
Answer A is too definitive. The screen cannot detect all genes involved in citrate utilization due to the experimental limitation described above. Answer B misinterprets the results as indicating transposon bias, when the pattern more likely reflects the experimental design's constraints. Answer C makes an unfounded assumption about gene organization. Two genes can be the only viable targets for disruption without being physically linked—their chromosomal arrangement isn't relevant to this observation.
The broader principle here is that genetic screens have inherent biases based on their selection conditions. Always consider what classes of mutants might be eliminated before they can be detected. On microbiology exams, look for questions that test whether you can distinguish between "these are the only genes involved" versus "these are the only genes we can detect with this particular method."
- A nonsense mutation in the trpA gene creates a premature UAG stop codon, resulting in a non-functional, truncated protein. A second, suppressor mutation is found in a tRNA gene, which allows the synthesis of a full-length, active TrpA protein. This suppressor tRNA is found to insert a glutamine residue at the UAG codon. What is the most likely change that occurred in the suppressor tRNA gene?
Answer: A mutation in the anticodon loop, changing it from 3'-GUC-5' to 3'-AUC-5'.
This describes an informational suppressor, specifically a nonsense suppressor tRNA. The premature stop codon in the mRNA is UAG (read 5' to 3'). To read this codon, a tRNA needs a complementary anticodon, which is 3'-AUC-5'. A normal glutamine tRNA reads either CAA or CAG codons; the tRNA reading CAG would have the anticodon 3'-GUC-5'. A mutation that changes this anticodon to 3'-AUC-5' would allow it to recognize the UAG stop codon and insert its amino acid (glutamine), permitting translation to continue. Choice A describes a misacylated tRNA but doesn't explain reading the stop codon. Overexpression (B) would not change codon recognition. While wobble base modifications (D) are important, a direct change in the anticodon sequence (C) is the most common mechanism for this type of suppression.
- A bacterial strain has a point mutation in fabI, an essential gene. The mutation results in an enzyme with reduced catalytic efficiency, leading to a slow-growth phenotype. Introducing a low-copy plasmid with the wild-type *fabI*⁺ gene restores the normal growth rate. Interestingly, introducing a high-copy plasmid carrying the mutant fabI allele into the original slow-growing mutant also restores the normal growth rate. What do these results most strongly suggest about the mutation?
Answer: The mutation significantly lowers the enzyme's specific activity (k_cat) but does not abolish it.
The key observation is that overproducing the mutant protein restores the wild-type phenotype. This is a classic gene dosage effect. If the mutation simply makes the enzyme less efficient (i.e., lower specific activity or k_cat), but doesn't completely destroy its function, then the cell can compensate by synthesizing a much larger quantity of this 'slow' enzyme. The increased enzyme concentration ([E]) raises the overall reaction rate (V_max = k_cat * [E]), restoring the necessary metabolic flux. A dominant negative mutation (A) would likely worsen the phenotype upon overexpression. While protein instability (B) could be overcome by overproduction, reduced specific activity (C) is a more precise biochemical explanation for a 'partially active' enzyme. A transcriptional mutation (D) is ruled out because the experiment involves expressing the mutant coding sequence from a plasmid promoter.
- A bacterial mutant exhibits constitutive high-level expression of the ara operon, even in the absence of the inducer arabinose. Genetic mapping shows the mutation is linked to the operon itself, but sequencing of the regulatory protein gene, araC, reveals a wild-type sequence. Which of the following experiments would be the most direct and effective next step to identify the mutation?
Answer: Sequence the *cis*-acting regulatory DNA sites of the ara operon in the mutant, including the operator (araO) and initiator (araI) sites.
The phenotype is constitutive expression, and the *trans*-acting factor (araC) is wild-type. This strongly implies the mutation is in a *cis*-acting DNA element where the AraC protein binds to regulate transcription. For the ara operon, these are the operator and initiator sites. A mutation in one of these sites could prevent AraC from binding properly to repress the operon in the absence of arabinose, leading to the observed constitutive expression. Therefore, the most direct way to find the mutation is to sequence these specific regulatory DNA regions (C). A complementation test (A) is not ideal for a *cis*-acting mutation. DNA affinity chromatography (B) is a method to find DNA binding proteins, not to identify a mutation in a known binding site. Measuring internal arabinose (D) is a less direct hypothesis.
- A histidine auxotroph of E. coli (His⁻) is plated on a minimal medium lacking histidine, and a small number of His⁺ colonies grow. To distinguish whether these revertants arose from a true back mutation within the his gene or from an unlinked, extragenic suppressor mutation, which of the following experimental approaches would be most informative?
Answer: Perform P1 transduction using phage grown on a revertant strain to infect the original His⁻ auxotroph, then select for an unrelated marker and screen for the His⁺ phenotype.
The most informative method is genetic back-crossing, for which P1 transduction is the standard tool in E. coli. If the suppressor mutation is unlinked to the original his mutation, P1 phage will often transfer the original *his*⁻ allele without transferring the suppressor allele. This would result in the regeneration of His⁻ auxotrophs among the transductants, indicating the presence of an extragenic suppressor. If it were a true revertant or a tightly linked intragenic suppressor, the His⁺ phenotype would almost always be transferred along with any nearby markers. Choice A is incomplete because it cannot rule out a linked intragenic suppressor and provides no information if a change is found. Choice C measures fitness but does not distinguish the genetic location of the mutation. Choice D describes a test for dominance, which is not the primary way to distinguish between these reversion mechanisms.
- A lethal mutation in the tryptophanyl-tRNA synthetase gene (trpS) causes the enzyme to mistakenly charge serine onto tRNA^(Trp). A second, suppressor mutation that restores viability is isolated. Sequencing reveals this suppressor mutation is located in the gene encoding tRNA^(Trp). What is the most plausible molecular mechanism for this suppression?
Answer: The mutation in tRNA^(Trp) alters a region recognized by the synthetase, restoring the correct specific interaction with the mutant TrpS protein.
The initial problem is a loss of specificity between the mutant TrpS enzyme and its cognate tRNA^(Trp). The suppressor mutation is in the tRNA itself and restores viability. This points to a restored interaction. A mutation in the tRNA (likely in the acceptor stem or other identity elements) could create a new conformation that the mutant synthetase can now recognize and charge correctly with tryptophan. This is a classic example of informational suppression via restored molecular interaction. Choice A is incorrect; changing the anticodon would cause widespread mistranslation at serine codons, which would also be lethal. Choice C is not a plausible mechanism; making more of the 'wrong' substrate doesn't fix the enzyme's error. Choice D would not solve the problem of incorporating tryptophan into proteins.
- A temperature-sensitive mutant of Bacillus subtilis grows normally at 30°C but fails to divide at 42°C. In a temperature-shift experiment, a culture growing exponentially at 30°C is moved to 42°C. The optical density (a measure of total cell mass) continues to increase at a near-normal rate for approximately one generation, but the count of viable cells (colony-forming units) plateaus immediately. This specific discrepancy between mass increase and cell number increase is most characteristic of a mutation affecting which process?
Answer: Initiation of chromosome replication.
The key observation is that cells continue to grow in size (increasing mass/OD) but cannot divide to produce more viable cells. This phenotype is classic for a defect in the initiation of DNA replication or in cell division (cytokinesis). Cells that have already initiated replication at 30°C can complete that round and grow larger, but they cannot start a new round at 42°C and therefore cannot divide. A defect in peptidoglycan synthesis (B) would likely cause cell lysis at the higher temperature, leading to a decrease in OD. A defect in ribosome formation (C) or ATP synthesis (D) would halt protein synthesis and overall metabolism, causing both mass and viable cell count to plateau almost immediately.
- To create a precise, unmarked deletion of the arcA gene in Salmonella, a microbiologist uses a suicide vector containing DNA regions flanking arcA and the sacB gene. After homologous recombination integrates the entire plasmid into the chromosome, the intermediate strain contains both a wild-type and a deleted arcA allele arranged in tandem. What is the specific purpose of the subsequent step, which involves plating this intermediate strain on a nutrient agar medium containing 5% sucrose?
Answer: To provide a strong positive selection for the rare cells that have undergone a second homologous recombination event, excising the plasmid backbone.
The sacB gene, derived from Bacillus subtilis, encodes the enzyme levansucrase, which converts sucrose into a toxic polymer in Gram-negative bacteria like Salmonella. Therefore, medium containing sucrose is lethal to any cell that contains the sacB gene. The intermediate strain has the entire plasmid, including sacB, integrated into its chromosome. A second recombination event can excise the plasmid backbone, which also removes the sacB gene. This excision event makes the cell resistant to sucrose. Thus, plating on sucrose is a powerful counter-selection method to specifically isolate cells that have undergone the desired second crossover, leading to either the wild-type or the desired deletion allele, while killing all cells that retain the integrated plasmid.
- A tryptophan auxotroph (*trpA*⁻) was isolated after EMS mutagenesis. A spontaneous revertant (Rev) is isolated that grows without tryptophan. To distinguish a true back mutation from an unlinked extragenic suppressor, a P1 transduction is performed. Phage grown on the Rev donor (which is also Kanᴿ due to a linked cysH::kan marker) are used to infect the original *trpA*⁻ auxotroph. Transductants are selected on plates with kanamycin. What phenotypic pattern among the Kanᴿ transductants would provide the strongest evidence for an unlinked, extragenic suppressor?
Answer: A significant fraction of Kanᴿ transductants (e.g., >20%) are found to be Trp⁻, while the remainder are Trp⁺.
When analyzing genetic reversions, you need to distinguish between true back mutations (which restore the original gene function at the same locus) and suppressor mutations (which compensate for the defect but occur at different genetic locations). P1 transduction helps make this distinction by testing whether the reversion and original mutation are physically linked on the chromosome.
In this experiment, if the reversion were a true back mutation at the trpA locus, it would be tightly linked to the cysH::kan marker since they're on the same chromosomal region. This means nearly all KanR transductants would also be Trp+ because the "fixed" trpA gene and the kanamycin resistance would transfer together as a unit.
However, if an unlinked extragenic suppressor is responsible for the Trp+ phenotype, the suppressor gene sits elsewhere on the chromosome, far from the cysH::kan marker. During transduction, you can inherit the KanR marker without necessarily getting the distant suppressor gene. This creates a mixed population: some KanR transductants receive both the marker and suppressor (Trp+), while others get only the marker but not the suppressor (Trp−).
Answer D correctly identifies this mixed phenotype as evidence for an unlinked suppressor. Answer A describes tight linkage characteristic of a back mutation. Answer B incorrectly suggests reduced transduction frequency. Answer C wrongly implies the suppressor never transfers—it should transfer sometimes, just not always with the KanR marker.
Remember: when distinguishing reversion types, look for co-inheritance patterns. Tight linkage suggests back mutation; independent assortment suggests unlinked suppression. - An E. coli auxotroph is unable to synthesize tryptophan. It is tested for growth on minimal medium (MM) supplemented with intermediates of the tryptophan biosynthetic pathway. The observed growth results are: MM alone (no growth); MM + chorismate (no growth); MM + anthranilate (growth); MM + indole (growth). The initial steps of the pathway are: Chorismate → Anthranilate → PRPP → ... → Indole. A mutation in the gene encoding which function would be consistent with this pattern?
Answer: A component of the anthranilate synthase complex, which converts chorismate to anthranilate.
In analyzing a metabolic pathway using nutritional supplementation, a block in the pathway is identified as being immediately before the first intermediate that restores growth. The mutant cannot grow when supplied with chorismate but can grow when supplied with anthranilate. This means the enzymatic step to convert chorismate to anthranilate is defective, but all subsequent steps are functional. The enzyme complex responsible for this conversion is anthranilate synthase, which is encoded by the trpE and trpG genes. A mutation in trpA (A) would prevent growth on indole. A mutation between anthranilate and indole (C) would not be consistent with growth on anthranilate. A simple loss-of-function mutation in the trpR repressor (D) would lead to constitutive expression of the pathway, not auxotrophy.
- A scientist wants to investigate the role of a specific active site residue, Asp-102, in the catalytic function of a protease. Which experimental strategy provides the most direct and unambiguous evidence for the role of this specific residue?
Answer: Use site-directed mutagenesis to change the codon for Asp-102 to Ala-102, then express and purify both the wild-type and mutant enzymes to compare their specific activities.
The question asks for the most direct way to test the function of a specific amino acid residue. Site-directed mutagenesis (B) is the ideal technique for this, as it allows the researcher to make a precise, intentional change (e.g., Asp to Ala) and then directly measure the effect of that single change on the protein's function (e.g., catalytic activity). Random mutagenesis (A) and screening for spontaneous mutants (D) are powerful for discovering important residues but are not direct tests of a pre-existing hypothesis about a specific one. A gene knockout (C) demonstrates the importance of the entire gene/protein, but provides no information about the roles of individual amino acids within it.
- A researcher hypothesizes that the protein RegA is a transcriptional activator for the lux operon. To test this, a reporter plasmid is constructed with the lux promoter fused to a lacZ gene. This plasmid is introduced into an E. coli strain that has a deletion of the regA gene (Δ*regA*). The activity of the reporter is then measured after introducing a second plasmid carrying either a wild-type regA gene or an empty vector control. Which outcome would most strongly support the hypothesis?
Answer: High β-galactosidase activity when the regA plasmid is present, and low activity when the empty vector is present.
The experiment is designed to test if RegA activates the lux promoter. The host strain lacks its own RegA. The reporter system (lux promoter driving lacZ) will only be active if an activator is present. If RegA is indeed the activator, adding it back on a plasmid should turn on the promoter, leading to lacZ expression and high β-galactosidase activity. The empty vector control, which does not provide RegA, should result in low activity. This outcome, described in choice A, directly demonstrates that RegA is necessary for activation. Choice B would suggest RegA is a repressor. Choice C would suggest the promoter is constitutively active and not regulated by RegA. Choice D would suggest RegA is not the activator or that some other required condition is missing.
- The protein DnaA forms a multimeric complex at the origin of replication to initiate DNA synthesis. A researcher discovers a mutant allele of dnaA (∗dnaA∗M) that produces a stable protein unable to bind ATP, a step essential for its activity. When this allele is introduced on a low-copy plasmid into a wild-type (*dnaA*⁺) cell, replication is severely inhibited, despite the presence of the normal DnaA protein from the chromosome. This type of mutational effect is best described as:
Answer: dominant negative.
A dominant negative (or antimorphic) mutation produces a mutant protein that not only is non-functional itself but also interferes with the function of the wild-type protein in the same cell. In this case, the mutant DnaA-M protein likely incorporates into the DnaA multimeric complex but, due to its inability to bind ATP, renders the entire complex inactive. This 'poisons' the function of the wild-type DnaA, leading to a dominant phenotype. It is not haploinsufficient (A), as the problem is not an insufficient amount of wild-type protein. It is not recessive (B), as the mutant phenotype is expressed in the presence of a wild-type allele. It is not a gain-of-function (D), as the protein has lost a function (ATP binding), and its effect is inhibitory.
- A gene encodes a 300-amino-acid protein essential for motility. Mutant 1 has a single nucleotide insertion at the codon for amino acid position 10. Mutant 2 has a single G-to-A nucleotide substitution at the same codon, resulting in a conservative change from glycine (GGA) to alanine (GCA). Which statement most accurately predicts the functional consequences for the resulting proteins?
Answer: Mutant 1 will likely produce a severely truncated, non-functional protein, while Mutant 2's protein may retain significant or full function.
When analyzing mutations, you need to consider both the type of mutation and its location to predict functional consequences. The key distinction here is between frameshift mutations (insertions/deletions) and point mutations (substitutions).
Mutant 1 has a single nucleotide insertion at position 10. This frameshift mutation shifts the reading frame for all subsequent codons, likely creating multiple stop codons downstream. The result is a severely truncated protein that terminates prematurely, losing most of its 300 amino acids and essential motility function.
Mutant 2 involves a conservative amino acid substitution from glycine to alanine. Both are small, nonpolar amino acids with similar chemical properties. Since this occurs at position 10 (early in the sequence but likely not in the active site of a 300-amino acid protein), and the change is chemically conservative, the protein will retain its full length and likely maintain significant or complete function.
Option A is incorrect because Mutant 1 won't produce a full-length protein due to the frameshift. Option B wrongly assumes that any early mutation causes complete function loss—this ignores the critical difference between frameshift and conservative substitution mutations. Option C reverses the severity, incorrectly suggesting the substitution is more damaging than the frameshift and wrongly assuming position 10 is the active site.
For mutation questions, always identify the mutation type first: frameshifts typically cause severe truncation and loss of function, while conservative substitutions, especially outside active sites, often preserve function. Location matters, but mutation type usually dominates the functional outcome.
- A screen for lactose-non-fermenting (Lac⁻) mutants of E. coli yields two distinct mutants. Mutant A reverts to a Lac⁺ phenotype at a low but detectable frequency when treated with the base analog 2-aminopurine. In contrast, Mutant B has never been observed to revert to Lac⁺, despite repeated attempts with powerful chemical mutagens and mutator strains. What is the most probable type of mutation in Mutant B?
Answer: A large deletion that removes part or all of the lacZ gene.
The key piece of information is the reversion frequency. Point mutations, such as nonsense (A) and missense (B) mutations, change a single base or a few bases and can be reverted by a second mutation that either restores the original sequence (true reversion) or compensates for the defect (suppression). Insertions of transposable elements (D) can also be revertible if the element excises. However, a large deletion (C) physically removes a segment of DNA. Since the genetic information is completely gone, there is no template for a simple reversion event to restore the original sequence, making such mutations functionally non-revertible. The inability of Mutant B to revert strongly suggests it carries a large deletion.
- A wild-type bacterial strain (WT) and an isogenic mutant (Mut) with a missense mutation are co-cultured in a chemostat. The initial ratio of Mut to WT is 1:1. After 20 generations of growth, the ratio of Mut to WT is found to be 1:16. The relative fitness of the mutant (w) compared to the wild-type is related to the selection coefficient (s) by the formula w=1−s. Based on these data, what is the approximate selection coefficient (s) against the mutant?
Answer: 0.14
The change in the ratio of two competing strains over time is a function of their relative fitness. The final ratio (Rf) is related to the initial ratio (Ri) and the relative fitness (w) over 't' generations by the formula Rf = Ri * wt. Here, Ri = 1, Rf = 1/16, and t = 20. So, 1/16 = 1 * w20. To solve for w, we take the 20th root of both sides: w = (1/16)1/20. We can solve this using logarithms: ln(w) = (1/20) ln(1/16) ≈ (1/20) (-2.77) ≈ -0.1385. For small selection coefficients, ln(w) = ln(1 - s) ≈ -s. Therefore, s ≈ 0.1385. The closest answer choice is 0.14.
