Microbiology Quiz: Measuring Mutation Effects

20 questions · exam conditions

0:00

Measuring Mutation EffectsQuestion 1 of 20

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?

The genes citA and citB are the only genes in the genome that are essential for citrate utilization.

The transposon used for mutagenesis exhibits an extremely high degree of insertional specificity for the citA and citB genes.

The genes citA and citB must be adjacent in a single operon, making them a large contiguous target for insertion.

Other genes required for citrate utilization may also be essential for viability on the rich medium used for the initial mutagenesis.

← Back to quizzes

Microbiology Quiz

Microbiology Quiz: Measuring Mutation Effects

Practice Measuring Mutation Effects in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Measuring Mutation Effects, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

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.

All questions

Question 1

The genes citA and citB are the only genes in the genome that are essential for citrate utilization.
The transposon used for mutagenesis exhibits an extremely high degree of insertional specificity for the citA and citB genes.
The genes citA and citB must be adjacent in a single operon, making them a large contiguous target for insertion.
Other genes required for citrate utilization may also be essential for viability on the rich medium used for the initial mutagenesis. (correct answer)

Explanation: 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."

Question 2

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?

Both mutations will produce full-length proteins, but Mutant 1 will be non-functional while Mutant 2 will be fully functional.
Both mutations will likely result in a complete loss of protein function because they occur so early in the coding sequence.
Mutant 2 will result in a complete loss of function due to the change in the active site, while Mutant 1 will have a minimal effect on motility.
Mutant 1 will likely produce a severely truncated, non-functional protein, while Mutant 2's protein may retain significant or full function. (correct answer)

Explanation: 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.

Question 3

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 (ΔregA). 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?

High β-galactosidase activity when the regA plasmid is present, and low activity when the empty vector is present. (correct answer)
Low β-galactosidase activity when the regA plasmid is present, and high activity when the empty vector is present.
Consistently high β-galactosidase activity in the presence of either the regA plasmid or the empty vector.
Consistently low β-galactosidase activity in the presence of either the regA plasmid or the empty vector.

Explanation: 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.

Question 4

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?

To select for cells that have spontaneously cured themselves of the entire integrated plasmid without a second recombination.
To provide a strong positive selection for the rare cells that have undergone a second homologous recombination event, excising the plasmid backbone. (correct answer)
To induce high-level expression from the sacB gene, which is required to activate the second recombination event.
To enrich for cells that have acquired a suppressor mutation that bypasses the toxic effect of the sacB gene product.

Explanation: 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.

Question 5

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?

The mutation is dominant negative, interfering with the chromosomal copy.
The mutation primarily causes the enzyme to become unstable and rapidly degraded.
The mutation significantly lowers the enzyme's specific activity (k_cat) but does not abolish it. (correct answer)
The mutation is in a regulatory sequence, leading to reduced transcription of the fabI gene.

Explanation: 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.

Question 6

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?

Sequence the original his gene in the revertant colonies; the absence of any change from the original mutant sequence would confirm an extragenic suppressor.
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. (correct answer)
Measure the growth rate of the revertant strain in minimal medium supplemented with varying concentrations of histidine to detect subtle differences in enzyme efficiency.
Transform the revertant strain with a plasmid carrying the wild-type his gene and observe if this causes a dominant negative phenotype, indicating a suppressor interaction.

Explanation: 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.

Question 7

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:

haploinsufficient.
recessive lethal.
dominant negative. (correct answer)
gain-of-function.

Explanation: 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.

Question 8

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?

Nearly 100% of the Kanᴿ transductants are Trp⁺, showing tight linkage of the reversion to the original locus.
The frequency of Kanᴿ transductants is dramatically reduced, suggesting the reversion event is incompatible with the recipient.
All Kanᴿ transductants are Trp⁻, indicating the suppressor mutation was never transferred by the phage.
A significant fraction of Kanᴿ transductants (e.g., >20%) are found to be Trp⁻, while the remainder are Trp⁺. (correct answer)

Explanation: 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 Kan^R 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 Kan^R marker without necessarily getting the distant suppressor gene. This creates a mixed population: some Kan^R 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 Kan^R marker. Remember: when distinguishing reversion types, look for co-inheritance patterns. Tight linkage suggests back mutation; independent assortment suggests unlinked suppression.

Question 9

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?

The mutation in tRNA^(Trp) changes its anticodon to one that base-pairs with a serine codon.
The mutation in tRNA^(Trp) alters a region recognized by the synthetase, restoring the correct specific interaction with the mutant TrpS protein. (correct answer)
The mutation in tRNA^(Trp) increases its cellular concentration, which competitively inhibits the mischarging reaction.
The mutation in tRNA^(Trp) allows it to be correctly charged by the seryl-tRNA synthetase as a compensatory mechanism.

Explanation: 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.

Question 10

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?

A nonsense mutation early in the lacZ gene.
A missense mutation in a critical active site residue of β-galactosidase.
A large deletion that removes part or all of the lacZ gene. (correct answer)
An insertion of an IS element (transposon) into the lac promoter.

Explanation: 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.

Question 11

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?

Treat a bacterial culture with a chemical mutagen like EMS, screen for mutants with reduced protease activity, and then sequence the protease gene in those mutants.
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. (correct answer)
Generate a complete gene knockout of the protease using homologous recombination and observe the resulting global changes in the bacterial proteome.
Perform a genetic screen to isolate spontaneous mutants that show enhanced protease activity and determine the location of the mutations by whole-genome sequencing.

Explanation: 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.

Question 12

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?

A mutation in the acceptor stem that allows it to be charged with tryptophan instead of glutamine.
A mutation in its promoter that causes massive overexpression of the glutamine tRNA.
A mutation in the anticodon loop, changing it from 3'-GUC-5' to 3'-AUC-5'. (correct answer)
A mutation that prevents modification of the wobble base, allowing it to read UAG.

Explanation: 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.

Question 13

A mutation in the acceptor stem that allows it to be charged with tryptophan instead of glutamine.
A mutation in its promoter that causes massive overexpression of the glutamine tRNA.
A mutation in the anticodon loop, changing it from 3'-GUC-5' to 3'-AUC-5'. (correct answer)
A mutation that prevents modification of the wobble base, allowing it to read UAG.

Question 14

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?

Initiation of chromosome replication. (correct answer)
Peptidoglycan synthesis and cross-linking.
Formation of the 50S ribosomal subunit.
Synthesis of ATP via the electron transport chain.

Explanation: 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.

Question 15

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?

0.14 (correct answer)
0.25
0.50
0.80

Explanation: The change in the ratio of two competing strains over time is a function of their relative fitness. The final ratio (R_f) is related to the initial ratio (R_i) and the relative fitness (w) over 't' generations by the formula R_f = R_i * w^t. Here, R_i = 1, R_f = 1/16, and t = 20. So, 1/16 = 1 * w^20. 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.

Question 16

To select for cells that have spontaneously cured themselves of the entire integrated plasmid without a second recombination.
To provide a strong positive selection for the rare cells that have undergone a second homologous recombination event, excising the plasmid backbone. (correct answer)
To induce high-level expression from the sacB gene, which is required to activate the second recombination event.
To enrich for cells that have acquired a suppressor mutation that bypasses the toxic effect of the sacB gene product.

Question 17

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?

The enzyme that converts indole to tryptophan.
A component of the anthranilate synthase complex, which converts chorismate to anthranilate. (correct answer)
An enzyme required for an intermediate step between anthranilate and indole.
The TrpR repressor protein that regulates the expression of the entire trp operon.

Explanation: 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.

Question 18

The mutation is dominant negative, interfering with the chromosomal copy.
The mutation primarily causes the enzyme to become unstable and rapidly degraded.
The mutation significantly lowers the enzyme's specific activity (k_cat) but does not abolish it. (correct answer)
The mutation is in a regulatory sequence, leading to reduced transcription of the fabI gene.

Question 19

Both mutations will produce full-length proteins, but Mutant 1 will be non-functional while Mutant 2 will be fully functional.
Both mutations will likely result in a complete loss of protein function because they occur so early in the coding sequence.
Mutant 2 will result in a complete loss of function due to the change in the active site, while Mutant 1 will have a minimal effect on motility.
Mutant 1 will likely produce a severely truncated, non-functional protein, while Mutant 2's protein may retain significant or full function. (correct answer)

Question 20

Nearly 100% of the Kanᴿ transductants are Trp⁺, showing tight linkage of the reversion to the original locus.
The frequency of Kanᴿ transductants is dramatically reduced, suggesting the reversion event is incompatible with the recipient.
All Kanᴿ transductants are Trp⁻, indicating the suppressor mutation was never transferred by the phage.
A significant fraction of Kanᴿ transductants (e.g., >20%) are found to be Trp⁻, while the remainder are Trp⁺. (correct answer)