Microbiology Quiz: Gene Regulation Operons
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Gene Regulation OperonsQuestion 1 of 20

An E. coli mutant has a defective adenylate cyclase gene (cya), preventing cAMP production. This mutant is grown in a medium containing a high concentration of lactose and no glucose. What is the expected level of transcription of the lac operon?

A high, fully activated level of transcription.
No transcription, as CAP-cAMP binding is essential.
A low, basal level of transcription.
A constitutive level of transcription, independent of lactose.
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Microbiology Quiz

Microbiology Quiz: Gene Regulation Operons

Practice Gene Regulation Operons 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 Gene Regulation Operons, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

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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.

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Question 1

An E. coli mutant has a defective adenylate cyclase gene (cya), preventing cAMP production. This mutant is grown in a medium containing a high concentration of lactose and no glucose. What is the expected level of transcription of the lac operon?

  1. A high, fully activated level of transcription.
  2. No transcription, as CAP-cAMP binding is essential.
  3. A low, basal level of transcription. (correct answer)
  4. A constitutive level of transcription, independent of lactose.
Explanation: Transcription of the lac operon requires two conditions for high-level expression: 1) lactose must be present to inactivate the LacI repressor, and 2) glucose must be absent so that cAMP levels are high, allowing the CAP protein to bind and activate transcription. In this mutant, lactose is present, so the LacI repressor is removed from the operator. However, the cya mutation prevents cAMP synthesis. Without cAMP, the CAP protein cannot bind to the promoter to activate transcription. RNA polymerase can still bind the promoter weakly and initiate transcription at a very low, basal level.

Question 2

A microbiologist constructs a merozygote with the genotype F' lacI- P+ O+ Z+ / lacI+ P+ O+ Z-. Which of the following correctly predicts the expression of functional β-galactosidase by this strain?

  1. The operon will be expressed constitutively because of the lacI- allele.
  2. The operon will be non-functional as both DNA molecules have a mutation.
  3. The operon will be inducible by lactose. (correct answer)
  4. The operon will be permanently repressed due to conflicting signals.
Explanation: The lacI+ gene on the chromosome produces a functional repressor protein. This protein is a trans-acting factor, meaning it is diffusible and can act on any appropriate operator in the cell. Therefore, the functional repressor from the chromosome will bind to the operator on the F' plasmid, which is linked to a functional lacZ+ gene. This restores normal regulation to the F' plasmid's operon. Consequently, the operon will be repressed in the absence of lactose and induced in its presence.

Question 3

A mutation in the crp gene renders the catabolite activator protein (CAP) unable to bind to cAMP, although it is still produced. How would this mutation affect the transcription of the lac operon in an environment containing lactose but no glucose?

  1. Transcription will be at a maximal, fully induced level.
  2. There will be no transcription of the operon.
  3. Transcription will be constitutive regardless of the presence of lactose.
  4. Transcription will occur only at a low, basal level. (correct answer)
Explanation: In the presence of lactose and absence of glucose, two conditions are met in a wild-type cell: the LacI repressor is inactivated, and cAMP levels are high. High cAMP allows formation of the CAP-cAMP complex, which binds the promoter and strongly activates transcription. In this mutant, CAP cannot bind cAMP. Therefore, even though cAMP levels are high and the repressor is inactive, the active CAP-cAMP complex cannot form. Without this positive activation, RNA polymerase binds only weakly to the promoter, resulting in a low, basal level of transcription.

Question 4

A mutation in the trpL gene disrupts base pairing in region 3 of the leader sequence, preventing it from forming a hairpin with either region 2 or region 4. What is the regulatory consequence of this mutation for the trp operon?

  1. Transcription will always terminate prematurely due to the stable formation of a 1:2 hairpin.
  2. Attenuation is eliminated, and transcription proceeds past the leader sequence regardless of tryptophan levels. (correct answer)
  3. The operon becomes solely dependent on the TrpR repressor, with attenuation fixed in the 'off' position.
  4. The operon becomes hyper-sensitive to low tryptophan levels, leading to increased expression.
Explanation: The attenuation mechanism relies on the formation of one of two mutually exclusive hairpins: the 2:3 anti-terminator or the 3:4 terminator. The terminator hairpin is essential for terminating transcription in the presence of high tryptophan. If region 3 cannot base-pair, the 3:4 terminator hairpin can never form. Without this termination signal, RNA polymerase will continue transcription past the leader sequence regardless of ribosome position. This effectively eliminates attenuation as a regulatory mechanism.

Question 5

A partial diploid E. coli has the genotype F' lacI+ P- O+ Z+ / lacI+ P+ O+ Z-. The P- allele denotes a non-functional promoter. Predict this cell's ability to produce functional β-galactosidase when grown in the presence of lactose and absence of glucose.

  1. The cell will produce β-galactosidase constitutively.
  2. The cell will produce β-galactosidase at a high, inducible level.
  3. The cell will not be able to produce any functional β-galactosidase. (correct answer)
  4. The cell will produce β-galactosidase at a low, basal level.
Explanation: The promoter is a cis-acting element, meaning it controls the transcription of adjacent genes on the same DNA molecule. On the F' plasmid, the promoter (P-) is non-functional, so the linked lacZ+ gene cannot be transcribed. On the chromosome, the promoter (P+) is functional and the operon will be transcribed in the presence of lactose, but the structural gene (lacZ-) is mutated and cannot produce a functional β-galactosidase enzyme. Since neither DNA molecule can produce the functional enzyme, the cell will be phenotypically Lac-.

Question 6

Consider an E. coli strain with a null mutation in the trpR gene, rendering the TrpR repressor protein non-functional. How will the rate of transcription of the trp operon in this mutant compare to a wild-type strain when both are grown in a medium containing excess tryptophan?

  1. Both strains will exhibit maximal transcription of the operon.
  2. Both strains will exhibit virtually no transcription of the operon.
  3. The mutant will have a high level of transcription, while the wild-type will have none.
  4. The mutant will have a low level of transcription, while the wild-type will have virtually none. (correct answer)
Explanation: In a wild-type strain with excess tryptophan, two regulatory mechanisms are active: the TrpR repressor is activated by tryptophan and binds the operator, blocking most transcription; and any transcripts that are initiated are terminated by attenuation. This results in virtually no expression. In the trpR- mutant, the repressor is non-functional, so repression is lifted. However, the high tryptophan levels will still enable the attenuation mechanism to function, causing most transcripts to terminate prematurely. This results in a low, attenuated level of expression that is nonetheless significantly higher than the repressed and attenuated level in the wild-type.

Question 7

A mutation in the trpL gene of the trp operon changes the two adjacent tryptophan codons in the leader peptide sequence to codons for alanine. How will this mutation affect the regulation of the operon in response to low intracellular concentrations of tryptophan?

  1. Regulation will be normal, as the TrpR repressor is unaffected.
  2. The operon will be constitutively expressed at a high level.
  3. Transcription will be prematurely terminated, even when tryptophan is scarce. (correct answer)
  4. The operon will become inducible by tryptophan instead of repressible.
Explanation: The attenuation mechanism of the trp operon depends on the ribosome stalling at the two Trp codons in the leader sequence when tryptophan levels are low. This stalling allows an anti-terminator hairpin to form. By changing the Trp codons to Ala codons, the ribosome will no longer pause in response to low tryptophan levels (assuming alanine is not limiting). It will translate the leader sequence quickly, which favors the formation of the 3:4 terminator hairpin, leading to premature termination of transcription. Thus, the operon will be inappropriately turned off even when the cell needs to synthesize tryptophan.

Question 8

An E. coli mutant has a defective adenylate cyclase gene (cya), preventing cAMP production. This mutant is grown in a medium containing a high concentration of lactose and no glucose. What is the expected level of transcription of the lac operon?

  1. A high, fully activated level of transcription.
  2. No transcription, as CAP-cAMP binding is essential.
  3. A low, basal level of transcription. (correct answer)
  4. A constitutive level of transcription, independent of lactose.
Explanation: Transcription of the lac operon requires two conditions for high-level expression: 1) lactose must be present to inactivate the LacI repressor, and 2) glucose must be absent so that cAMP levels are high, allowing the CAP protein to bind and activate transcription. In this mutant, lactose is present, so the LacI repressor is removed from the operator. However, the cya mutation prevents cAMP synthesis. Without cAMP, the CAP protein cannot bind to the promoter to activate transcription. RNA polymerase can still bind the promoter weakly and initiate transcription at a very low, basal level.

Question 9

A mutation in the trpL gene of the trp operon changes the two adjacent tryptophan codons in the leader peptide sequence to codons for alanine. How will this mutation affect the regulation of the operon in response to low intracellular concentrations of tryptophan?

  1. Regulation will be normal, as the TrpR repressor is unaffected.
  2. The operon will be constitutively expressed at a high level.
  3. Transcription will be prematurely terminated, even when tryptophan is scarce. (correct answer)
  4. The operon will become inducible by tryptophan instead of repressible.
Explanation: The attenuation mechanism of the trp operon depends on the ribosome stalling at the two Trp codons in the leader sequence when tryptophan levels are low. This stalling allows an anti-terminator hairpin to form. By changing the Trp codons to Ala codons, the ribosome will no longer pause in response to low tryptophan levels (assuming alanine is not limiting). It will translate the leader sequence quickly, which favors the formation of the 3:4 terminator hairpin, leading to premature termination of transcription. Thus, the operon will be inappropriately turned off even when the cell needs to synthesize tryptophan.

Question 10

A newly discovered bacterial operon, the xyl operon, is involved in xylose catabolism. The operon's expression is controlled by a regulatory protein, XylR. A mutation that results in a non-functional XylR protein leads to constitutive expression of the xyl operon's structural genes, even in the absence of xylose. Based on this observation, what is the role of XylR and the nature of the operon's control?

  1. XylR is an activator, and the operon is under positive inducible control.
  2. XylR is a repressor, and the operon is under negative inducible control. (correct answer)
  3. XylR is a repressor, and the operon is under negative repressible control.
  4. XylR is an activator, and the operon is under positive repressible control.
Explanation: In the wild-type state, the operon is turned off in the absence of xylose. The mutation that removes functional XylR causes the operon to be turned on. This indicates that the normal function of XylR is to turn the operon off. A protein that binds to DNA and inhibits transcription is a repressor, so this is a form of negative control. Since the operon codes for catabolism of xylose, it is logical that the presence of xylose would induce it by inactivating the repressor. This describes a negative inducible system.

Question 11

Consider an E. coli strain with a null mutation in the trpR gene, rendering the TrpR repressor protein non-functional. How will the rate of transcription of the trp operon in this mutant compare to a wild-type strain when both are grown in a medium containing excess tryptophan?

  1. Both strains will exhibit maximal transcription of the operon.
  2. Both strains will exhibit virtually no transcription of the operon.
  3. The mutant will have a high level of transcription, while the wild-type will have none.
  4. The mutant will have a low level of transcription, while the wild-type will have virtually none. (correct answer)
Explanation: In a wild-type strain with excess tryptophan, two regulatory mechanisms are active: the TrpR repressor is activated by tryptophan and binds the operator, blocking most transcription; and any transcripts that are initiated are terminated by attenuation. This results in virtually no expression. In the trpR- mutant, the repressor is non-functional, so repression is lifted. However, the high tryptophan levels will still enable the attenuation mechanism to function, causing most transcripts to terminate prematurely. This results in a low, attenuated level of expression that is nonetheless significantly higher than the repressed and attenuated level in the wild-type.

Question 12

A partial diploid E. coli has the genotype F' lacI+ P- O+ Z+ / lacI+ P+ O+ Z-. The P- allele denotes a non-functional promoter. Predict this cell's ability to produce functional β-galactosidase when grown in the presence of lactose and absence of glucose.

  1. The cell will produce β-galactosidase constitutively.
  2. The cell will produce β-galactosidase at a high, inducible level.
  3. The cell will not be able to produce any functional β-galactosidase. (correct answer)
  4. The cell will produce β-galactosidase at a low, basal level.
Explanation: The promoter is a cis-acting element, meaning it controls the transcription of adjacent genes on the same DNA molecule. On the F' plasmid, the promoter (P-) is non-functional, so the linked lacZ+ gene cannot be transcribed. On the chromosome, the promoter (P+) is functional and the operon will be transcribed in the presence of lactose, but the structural gene (lacZ-) is mutated and cannot produce a functional β-galactosidase enzyme. Since neither DNA molecule can produce the functional enzyme, the cell will be phenotypically Lac-.

Question 13

A microbiologist constructs a merozygote with the genotype F' lacI- P+ O+ Z+ / lacI+ P+ O+ Z-. Which of the following correctly predicts the expression of functional β-galactosidase by this strain?

  1. The operon will be expressed constitutively because of the lacI- allele.
  2. The operon will be non-functional as both DNA molecules have a mutation.
  3. The operon will be inducible by lactose. (correct answer)
  4. The operon will be permanently repressed due to conflicting signals.
Explanation: The lacI+ gene on the chromosome produces a functional repressor protein. This protein is a trans-acting factor, meaning it is diffusible and can act on any appropriate operator in the cell. Therefore, the functional repressor from the chromosome will bind to the operator on the F' plasmid, which is linked to a functional lacZ+ gene. This restores normal regulation to the F' plasmid's operon. Consequently, the operon will be repressed in the absence of lactose and induced in its presence.

Question 14

A mutation in the trpL gene disrupts base pairing in region 3 of the leader sequence, preventing it from forming a hairpin with either region 2 or region 4. What is the regulatory consequence of this mutation for the trp operon?

  1. Transcription will always terminate prematurely due to the stable formation of a 1:2 hairpin.
  2. Attenuation is eliminated, and transcription proceeds past the leader sequence regardless of tryptophan levels. (correct answer)
  3. The operon becomes solely dependent on the TrpR repressor, with attenuation fixed in the 'off' position.
  4. The operon becomes hyper-sensitive to low tryptophan levels, leading to increased expression.
Explanation: The attenuation mechanism relies on the formation of one of two mutually exclusive hairpins: the 2:3 anti-terminator or the 3:4 terminator. The terminator hairpin is essential for terminating transcription in the presence of high tryptophan. If region 3 cannot base-pair, the 3:4 terminator hairpin can never form. Without this termination signal, RNA polymerase will continue transcription past the leader sequence regardless of ribosome position. This effectively eliminates attenuation as a regulatory mechanism.

Question 15

A mutation in the lac operon's operator sequence, lacO^C, allows for constitutive expression. A second mutation occurs in the promoter of the same operon, lacP-, which prevents RNA polymerase from binding. What will be the phenotype of a haploid cell carrying both the lacO^C and lacP- mutations?

  1. No expression of β-galactosidase under any condition. (correct answer)
  2. Inducible expression of β-galactosidase.
  3. Constitutive expression of β-galactosidase.
  4. Expression only in the absence of glucose.
Explanation: When analyzing operon mutations, you need to consider how each component affects gene expression and remember that both transcription initiation and regulation must function for proper gene expression. The lac operon requires a functional promoter for RNA polymerase binding and transcription initiation. The lacO^C mutation prevents the lac repressor from binding to the operator, which would normally allow constitutive expression when the promoter is functional. However, the lacP- mutation eliminates RNA polymerase binding to the promoter, completely blocking transcription initiation. Since transcription cannot begin without promoter function, no mRNA will be produced regardless of the regulatory state of the operator. Think of it this way: even if the "brake" (repressor) is removed by the lacO^C mutation, the "engine" (promoter) is broken due to the lacP- mutation, so the transcription machinery cannot start. Answer A is correct because the non-functional promoter prevents any β-galactosidase expression under all conditions. Answer B is wrong because inducible expression requires both a functional promoter and normal operator regulation, neither of which exists here. Answer C incorrectly assumes the lacO^C mutation can override the promoter defect—constitutive expression still requires transcription to occur. Answer D misses the point entirely, as glucose regulation (catabolite repression) is irrelevant when the promoter cannot initiate transcription. Remember: in operon analysis, promoter function is absolutely required for transcription. Regulatory mutations are meaningless if the basic transcription machinery cannot engage.

Question 16

The addition of glucose to a culture of E. coli growing on lactose leads to a rapid decrease in the transcription of the lac operon. This phenomenon, known as catabolite repression, is primarily mediated by which molecular event?

  1. A glucose-induced decrease in cAMP levels, leading to the inactivation of CAP. (correct answer)
  2. Glucose transport into the cell inhibiting the activity of lactose permease (LacY).
  3. Glucose binding directly to the LacI repressor, increasing its affinity for the operator.
  4. A glucose-mediated increase in the degradation rate of lacZ mRNA.
Explanation: When you encounter questions about the lac operon and catabolite repression, focus on the central role of cAMP and CAP (catabolite activator protein) in glucose regulation. The lac operon requires both the absence of glucose and the presence of lactose for maximum transcription. Catabolite repression occurs because glucose prevents cAMP synthesis through the phosphoenolpyruvate (PEP) phosphotransferase system. When glucose is abundant, cAMP levels drop dramatically. Since CAP requires cAMP binding to become active, low cAMP means inactive CAP. Without the CAP-cAMP complex binding to the CAP binding site upstream of the lac promoter, RNA polymerase has poor affinity for the promoter, drastically reducing transcription even when lactose is present and the LacI repressor is inactivated. This makes choice A correct. Choice B is wrong because glucose doesn't directly inhibit lactose permease activity—the regulation occurs at the transcriptional level, not through competitive inhibition of transport proteins. Choice C misrepresents the mechanism entirely; glucose never binds directly to LacI repressor—that's allolactose's role, and glucose doesn't enhance LacI binding. Choice D incorrectly suggests post-transcriptional regulation through mRNA degradation, but catabolite repression specifically operates by preventing transcription initiation. Remember this hierarchy: glucose regulation (via cAMP-CAP) overrides lactose regulation (via LacI repressor). Even with lactose present, glucose addition will shut down the lac operon through the cAMP-CAP mechanism. Focus on understanding this two-tiered control system for exam success.

Question 17

An E. coli culture growing in a low-tryptophan medium is treated with a low dose of kasugamycin, an antibiotic that inhibits translational initiation. What is the predicted effect of this treatment on the expression of the trp operon via the attenuation mechanism?

  1. Expression will increase, as slower ribosome movement favors the anti-terminator hairpin.
  2. Expression will decrease, as fewer ribosomes will translate the leader peptide, favoring the terminator hairpin. (correct answer)
  3. Expression will be unaffected, as attenuation is regulated by termination, not initiation.
  4. Expression will decrease, as the antibiotic's general inhibition of protein synthesis reduces cell viability.
Explanation: The attenuation mechanism depends on the ribosome's position on the leader transcript. The anti-terminator (2:3) hairpin forms when the ribosome stalls in region 1. The terminator (3:4) hairpin forms when the ribosome quickly passes region 1 and blocks region 2. If kasugamycin inhibits translational initiation, fewer ribosomes will load onto the leader transcript. With no ribosome covering region 1, the default mRNA folding will occur, which includes the formation of a 1:2 hairpin. This allows the 3:4 terminator hairpin to form, leading to premature termination and decreased expression of the operon.

Question 18

A partial diploid strain of E. coli has the genotype F' lacI^S O+ Z- / lacI+ O^C Z+. The lacI^S allele produces a super-repressor that cannot bind the inducer, while the lacO^C allele is a constitutive operator to which repressors cannot bind. How will β-galactosidase be expressed in this strain?

  1. Expression will be inducible by lactose.
  2. Expression will be constitutive. (correct answer)
  3. No expression will occur under any condition.
  4. Expression will occur only in the absence of glucose.
Explanation: The lacO^C mutation is cis-acting, meaning it only affects the expression of genes on the same DNA molecule. In this case, it is on the chromosome linked to a functional lacZ+ gene. Since repressors (both LacI+ and LacISLacI^S) cannot bind to O^C, the chromosomal lacZ+ gene will be transcribed constitutively. The F' plasmid has a lacZ- allele, so it cannot produce functional β-galactosidase regardless of its regulation. Therefore, the cell's phenotype is determined by the constitutively expressed chromosomal allele.

Question 19

A bacterium synthesizes the amino acid histidine using the his operon, which is regulated by a repressor protein, HisR, and histidine as a corepressor. A mutation occurs in the his operator sequence that prevents the HisR-histidine complex from binding. What is the most likely consequence of this mutation on bacterial growth in a histidine-rich medium?

  1. The his operon will be permanently repressed, leading to histidine starvation.
  2. The his operon will be transcribed constitutively, leading to wasteful overproduction of histidine. (correct answer)
  3. The his operon will become inducible by histidine, reversing its normal regulation.
  4. Regulation will be unaffected because the hisR gene is unchanged.
Explanation: This is a repressible operon. Normally, high levels of histidine (the corepressor) would cause the HisR repressor to bind to the operator and shut down transcription. The described operator mutation is analogous to lacO^C; it prevents the repressor from binding. Because the repressor cannot bind, the operon will be transcribed continuously, even when histidine is abundant in the medium. This leads to wasteful energy expenditure and overproduction of an amino acid that is already available.

Question 20

An E. coli cell is grown in a medium containing a high concentration of lactose and a very low concentration of glucose (essentially glucose-starved conditions). Which statement best describes the state of the lac operon's regulatory proteins and the resulting transcriptional activity?

  1. The LacI repressor is bound to the operator, and CAP is unbound from the promoter, resulting in no transcription.
  2. The LacI repressor is unbound from the operator, but CAP is also unbound from the promoter, resulting in basal transcription.
  3. The LacI repressor is unbound from the operator, and the CAP-cAMP complex is bound to the promoter, resulting in maximal transcription. (correct answer)
  4. The LacI repressor is bound to the operator, and the CAP-cAMP complex is bound to the promoter, resulting in no transcription.
Explanation: A high concentration of lactose means allolactose is present to bind and inactivate the LacI repressor, so it is unbound from the operator. Very low glucose concentration (glucose starvation) results in high intracellular cAMP levels. This cAMP binds to the catabolite activator protein (CAP), forming an active complex that binds to the promoter. The combination of an unoccupied operator (negative control relieved) and an active CAP-cAMP complex at the promoter (positive control activated) leads to maximal transcription.