Gene Regulation (Operons) - Microbiology

1. 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.
   Answer: The cell will not be able to produce any functional β-galactosidase. 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-.
2. 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?
   Answer: Transcription will occur only at a low, basal level. 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.
3. 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?
   Answer: The his operon will be transcribed constitutively, leading to wasteful overproduction of histidine. 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.
4. A mutant strain, unable to utilize a sugar called sorbitol, is isolated. When a plasmid containing the wild-type sor operon and its associated regulatory gene is introduced, the resulting merozygote strain can now metabolize sorbitol. This experiment strongly suggests that the original mutation was in which type of element?
   Answer: A gene encoding a trans-acting, diffusible regulatory protein. The ability of the wild-type genes on the plasmid to restore function to the cell is called complementation. Complementation works when the plasmid provides a functional, diffusible product (usually a protein) that was missing due to the original mutation. This defines a trans-acting factor. If the original mutation were in a cis-acting element like a promoter or operator on the chromosome, a functional copy on a plasmid could not fix it, because cis elements only control the DNA to which they are physically attached. While a structural gene product is also trans-acting, the most general and accurate conclusion is that the mutation affected a gene for a diffusible protein, which includes regulatory proteins.
5. Which statement accurately describes a key mechanistic difference between the negative regulation of the lac operon and the trp operon in E. coli?
   Answer: The lac operon repressor is synthesized in an active form, while the trp operon repressor is synthesized in an inactive form. The LacI repressor protein is synthesized in a form that is active by default and can bind to the operator. The inducer (allolactose) is required to inactivate it. In contrast, the TrpR repressor protein (an aporepressor) is synthesized in an inactive form that cannot bind to the operator. It requires the corepressor (tryptophan) to bind to it and induce a conformational change that activates it, allowing it to bind the operator. This is a fundamental difference between inducible and repressible systems.
6. 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?
   Answer: A glucose-induced decrease in cAMP levels, leading to the inactivation of CAP. 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.
7. 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?
   Answer: A low, basal level of transcription. 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.
8. 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?
   Answer: Expression will decrease, as fewer ribosomes will translate the leader peptide, favoring the terminator hairpin. 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.
9. A specific mutation in the lacI gene produces a super-repressor protein ($*lacI^S$*) that binds to the operator but is unable to bind allolactose. What is the resulting phenotype for the lac operon in this mutant strain?
   Answer: The operon will be non-inducible and thus permanently repressed. The LacI repressor protein normally prevents transcription by binding to the operator. This repression is relieved when the inducer, allolactose, binds to the repressor and causes it to detach from the operator. A super-repressor ($*lacI^S$*) can bind to the operator but not to the inducer. Therefore, the super-repressor will remain bound to the operator even in the presence of lactose, preventing transcription permanently. This makes the operon non-inducible.
10. In the context of the trp operon's attenuation mechanism, which event is the direct cause for the formation of the 2:3 anti-terminator hairpin structure?
    Answer: The stalling of the ribosome on the tandem tryptophan codons in region 1. When you encounter questions about the trp operon's attenuation mechanism, focus on the coupling between transcription and translation that creates different RNA secondary structures based on ribosome behavior. The 2:3 anti-terminator hairpin forms specifically when the ribosome stalls at the tandem tryptophan codons in region 1 of the leader sequence. Here's why: When tryptophan levels are low, tryptophanyl-tRNA is scarce, causing the ribosome to pause at these codons. This stalling leaves region 1 covered by the ribosome, preventing it from base-pairing with region 2. Consequently, region 2 becomes available to pair with region 3, forming the anti-terminator structure that allows transcription to continue through the structural genes. Let's examine why the other options are incorrect. Option B describes classical repression by TrpR-tryptophan binding to the operator, which is a separate regulatory mechanism that doesn't directly cause hairpin formation in attenuation. Option C (rapid translation) would actually prevent anti-terminator formation—when tryptophan is abundant, rapid translation allows region 1 to pair with region 2, freeing region 3 to pair with region 4 and form a terminator. Option D refers to a downstream outcome rather than the direct cause of hairpin formation. The correct answer is A because ribosome stalling is the immediate trigger that determines which regions are available for base-pairing. Study tip: Remember that attenuation works through ribosome-RNA coupling—the ribosome's position determines which hairpin structure forms. Slow ribosome = anti-terminator = transcription continues.
11. The ara operon in E. coli is controlled by the AraC protein. In the absence of arabinose, AraC acts as a repressor by binding to two sites and looping the DNA. In the presence of arabinose, AraC changes conformation and acts as an activator. A mutant is isolated with a mutation in the araC gene that prevents the protein from binding to DNA at all. How will this affect ara operon expression?
    Answer: The operon will be non-inducible and will have no expression. When analyzing operon regulation problems, focus on the dual role of regulatory proteins and what happens when they're completely non-functional. The ara operon requires AraC protein for both repression (without arabinose) and activation (with arabinose) - it's essential for any transcriptional control. The mutant AraC protein cannot bind DNA at all, which eliminates both its repressor and activator functions. Without AraC binding, the operon cannot respond to arabinose presence or absence. Most importantly, AraC is required as an activator for transcription to occur even when arabinose is present. Since the mutant protein cannot bind DNA to activate transcription, no expression will occur regardless of arabinose levels. This makes the operon non-inducible with no expression, confirming answer D. Answer A is wrong because permanent repression would require a functional repressor that's always bound - but this mutant can't bind at all. Answer B incorrectly assumes that losing the repressor function alone would cause high expression, but ignores that AraC's activator function is also lost. Answer C is incorrect because the operon won't be inducible at all since AraC cannot activate transcription; catabolite repression is irrelevant when there's no basal expression to begin with. For operon questions, always consider whether regulatory proteins have single or dual functions. When a protein is completely non-functional, trace through both its positive and negative regulatory roles to predict the phenotype. Don't assume losing repression automatically means high expression if activation is also required.
12. 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?
    Answer: XylR is a repressor, and the operon is under negative inducible control. 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.
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?
    Answer: The operon will be inducible by lactose. 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.
14. 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?
    Answer: No expression of β-galactosidase under any condition. 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.
15. 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?
    Answer: Expression will be constitutive. 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 $LacI^S$) 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.
16. 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?
    Answer: Transcription will be prematurely terminated, even when tryptophan is scarce. 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.
17. 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?
    Answer: Attenuation is eliminated, and transcription proceeds past the leader sequence regardless of tryptophan levels. 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.
18. 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?
    Answer: The mutant will have a low level of transcription, while the wild-type will have virtually none. 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.
19. 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?
    Answer: The LacI repressor is unbound from the operator, and the CAP-cAMP complex is bound to the promoter, resulting in maximal transcription. 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.

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