- Based on the information in the passage, the
met operon is most likely: Answer: A negatively controlled, inducible operon where metanose acts as an inducer.
Observation 1 shows that without a functional RegM protein, the operon is on. This means RegM's normal function is to turn the operon off, making it a repressor (negative control). Observation 2 shows that metanose turns the operon on. This means metanose prevents the repressor (RegM) from working. A substance that inactivates a repressor is an inducer. This logic matches a negatively controlled, inducible system like the lac operon.
- A merodiploid E. coli strain has the genotype F'
trpR+ O^C trpE- / trpR- O+ trpE+. What is the phenotype of this strain regarding the production of the TrpE enzyme when grown in a medium rich in tryptophan? Answer: No production of TrpE.
The trpR+ allele is dominant and trans-acting, producing a functional repressor. In high tryptophan, this repressor is activated. The O^C operator is cis-acting and linked to a non-functional trpE- gene, so no enzyme is made from the plasmid. The active repressor (produced from the plasmid) binds to the wild-type operator (O+) on the chromosome, repressing transcription of the functional trpE+ gene. Thus, no functional enzyme is made from either copy.
- Consider an E. coli strain with a wild-type
trpR gene. A mutation is introduced that deletes region 4 of the trpL leader sequence. How will this strain regulate tryptophan biosynthesis compared to a wild-type strain? Answer: The operon will be fully repressed in high tryptophan, but transcription will be higher than wild-type in low tryptophan.
The trp operon is regulated by both repression (via TrpR) and attenuation (via trpL). Deleting region 4 of the leader sequence prevents the formation of the 3-4 terminator hairpin, abolishing the attenuation mechanism. However, the trpR repressor system is still intact. In high tryptophan, the TrpR-tryptophan complex will bind the operator and prevent transcription. In low tryptophan, the repressor is inactive, and transcription proceeds at a high rate because the attenuation 'brake' is disabled.
- A mutation in the
trpL gene of the trp operon replaces the two adjacent tryptophan codons with codons for alanine. How would this mutation most likely affect the regulation of the operon in a medium lacking tryptophan? Answer: Transcription will terminate prematurely at the attenuator, leading to low expression levels.
In the wild-type operon, a lack of tryptophan causes the ribosome to stall at the Trp codons in the leader sequence. This stalling allows the 2-3 anti-terminator loop to form, permitting transcription to continue. By replacing the Trp codons with Ala codons, the ribosome no longer stalls in low-tryptophan conditions because alanine is typically abundant. The ribosome proceeds quickly, allowing the 3-4 terminator loop to form, causing attenuation (premature termination) even when tryptophan is scarce.
- A point mutation in the
trpL leader sequence of E. coli prevents the formation of the 2-3 hairpin, but does not affect the ability of the 1-2 or 3-4 hairpins to form. What is the most likely phenotype of this mutant in a low-tryptophan environment? Answer: The operon will show decreased expression compared to wild-type.
The 2-3 hairpin is the anti-terminator. In low tryptophan, its formation prevents the 3-4 terminator hairpin from forming, allowing transcription. If the 2-3 hairpin cannot form, region 3 will be available to pair with region 4, forming the terminator hairpin even when the ribosome stalls at region 1. This causes premature termination (attenuation) even in low tryptophan, leading to decreased expression.
- If a mutation occurred in E. coli that caused ribosomes to translate mRNA significantly slower than normal, but with no change in accuracy, how would this affect the regulation of the trp operon in the presence of high tryptophan?
Answer: The operon would be expressed at a higher level because the slow ribosome would mimic low-tryptophan conditions.
When you encounter questions about the trp operon, focus on how attenuation relies on the precise timing between transcription and translation. The trp operon uses a sophisticated mechanism where ribosome speed on the leader sequence determines whether transcription continues.
In normal conditions with high tryptophan, charged tryptophanyl-tRNA is abundant, so ribosomes translate quickly through the tryptophan codons in the leader sequence. This fast translation allows the ribosome to cover region 1 before region 2 is fully transcribed, enabling regions 3 and 4 to form a terminator hairpin that stops transcription. However, if ribosomes translate significantly slower due to mutation, they would stall at the tryptophan codons even when tryptophan is abundant - mimicking the low-tryptophan scenario where ribosomes pause due to lack of charged tryptophanyl-tRNA.
This slow ribosome movement would prevent the 3-4 terminator hairpin from forming and instead allow the 2-3 anti-terminator structure to form, permitting transcription to continue. Therefore, answer D is correct - the slow ribosome creates conditions that mimic tryptophan starvation, leading to higher operon expression.
A is wrong because ribosome speed critically affects which hairpins can form. B incorrectly suggests attenuation would be completely lost rather than altered. C misunderstands the mechanism - the issue isn't reaching a stop codon but rather the timing of ribosome movement relative to RNA folding.
Remember: attenuation mechanisms depend on the kinetic coupling between transcription and translation, so any change in ribosome speed affects the regulatory outcome.
- A bacterial mutant has a constitutively active adenylate cyclase, resulting in constantly high intracellular cAMP levels. If this mutant is grown in a medium containing both high glucose and high lactose, what is the expected level of lac operon transcription?
Answer: High-level transcription, as high cAMP allows CAP to activate the promoter.
High-level transcription of the lac operon requires two conditions: lactose must be present to inactivate the LacI repressor, and cAMP levels must be high to activate CAP. In this mutant, cAMP levels are always high due to the active adenylate cyclase, so CAP is always active. Since lactose is also present, the LacI repressor is inactivated. With the repressor removed and CAP activated, RNA polymerase will bind efficiently, leading to high-level transcription, bypassing the usual effect of glucose.
- In the trp operon, a mutation in the
trpR gene results in a repressor that cannot bind tryptophan but can still bind to the operator. What is the regulatory consequence of this mutation? Answer: The operon will be permanently repressed, regardless of tryptophan levels.
The wild-type TrpR repressor is synthesized in a form that cannot bind the operator; it requires tryptophan as a co-repressor to become active. This mutant repressor binds the operator without tryptophan. Since it cannot bind tryptophan, its state is not influenced by tryptophan levels. It will continuously bind the operator and repress transcription, making the cell unable to synthesize tryptophan even when levels are low.
- If the gene for the LacI repressor (
lacI) were mutated such that the repressor could still bind to the operator but could no longer bind allolactose, what would be the effect on the transcription of the lac operon? Answer: The operon would be uninducible, with no transcription under any condition.
When analyzing lac operon regulation, focus on the dual role of the LacI repressor: it must bind DNA to block transcription AND respond to lactose by releasing from DNA when the sugar is present.
In normal lac operon function, the LacI repressor sits on the operator sequence, blocking RNA polymerase and preventing transcription. When lactose (or allolactose) is present, it binds to LacI, causing a conformational change that releases the repressor from the operator, allowing transcription to proceed. This makes the operon "inducible" by lactose.
However, this mutant LacI can still bind the operator but cannot bind allolactose. This means the repressor will permanently occupy the operator sequence, regardless of whether lactose is present or absent. Since RNA polymerase cannot access the promoter when the repressor blocks the operator, no transcription can occur under any conditions. The operon becomes completely uninducible, making D correct.
Option A is wrong because constitutive transcription would require the repressor to be unable to bind the operator. Option B incorrectly suggests the operon remains inducible—but without allolactose binding, the repressor cannot be removed regardless of glucose levels. Option C is incorrect because "basal level" transcription implies some transcription occurs, but the permanently bound repressor prevents any transcription.
Remember this pattern: when analyzing operon mutants, trace through both binding events separately. Ask yourself what happens to DNA binding AND what happens to inducer binding, then determine the combined effect on transcription.
- Which of the following describes the primary reason for the existence of attenuation as a second layer of regulation for the trp operon, in addition to repression?
Answer: Attenuation provides a mechanism for responding to a graded range of tryptophan concentrations, which simple repression does not.
The TrpR repression system is largely an on/off switch that responds to high levels of tryptophan. Attenuation, however, is a more sensitive mechanism. The degree to which the ribosome stalls on the leader sequence is proportional to the scarcity of charged tRNA-Trp. This allows the cell to fine-tune the rate of transcription across a wider, more graded range of intermediate tryptophan concentrations, rather than being simply 'on' or 'off'.
- You have two mutant E. coli strains that cannot express β-galactosidase. One strain has a promoter mutation (
P-), and the other has a super-repressor mutation (I^S). Which of the following merodiploids would allow you to distinguish between the two original strains by testing for lactose metabolism? Answer: Introducing an F' plasmid carrying I^+ P^+ O^+ Z^+.
The I^S mutation is dominant and trans-acting, while P- is recessive and cis-acting. Introducing a wild-type operon (I^+ P^+ O^+ Z^+) will complement the P- mutation (the plasmid's operon will be functional), resulting in a Lac+ phenotype. However, it will not complement the I^S mutation because the super-repressor protein produced from the chromosome will repress the operon on the plasmid as well, resulting in a Lac- phenotype. This allows for differentiation.
- The table shows β-galactosidase activity in several E. coli strains under different growth conditions. Which strain most likely has a mutation rendering the Catabolite Activator Protein (CAP) non-functional?
Answer: Strain Z
A strain with a non-functional CAP (cap-) cannot activate transcription to high levels, even when conditions are optimal (low glucose, high lactose). The repressor is still functional, so the operon is off without lactose. With lactose, the repressor is removed, allowing for basal transcription. Thus, the phenotype is inducibility to a low/basal level, but never to a high level. Strain Z shows no expression (-) without lactose and only basal expression (+) with lactose, regardless of glucose, matching this phenotype perfectly. Wild-type (W) shows high expression (+++) when lactose is present and glucose is absent. Strain X is constitutive. Strain Y is uninducible.
- A mutation in the
crp gene results in a non-functional Catabolite Activator Protein (CAP). How will this mutation affect the transcription of the lac operon in an environment containing low glucose and high lactose? Answer: Transcription will occur at a low, basal level.
High-level lac operon transcription requires active CAP to recruit RNA polymerase. In a cap- mutant, the CAP-cAMP complex cannot form, even if cAMP levels are high (due to low glucose). With lactose present, the repressor is inactivated, so RNA polymerase can bind the promoter inefficiently without CAP's help. This results in a low, basal level of transcription, not the high, activated level seen in wild-type cells under these conditions.
- Consider a lac operon merodiploid with the genotype
I^+ P^- O^+ Z^+ / I^- P^+ O^+ Z^-. What is the phenotype with respect to β-galactosidase production when grown in the presence of lactose and absence of glucose? Answer: No functional β-galactosidase will be produced.
This question requires analyzing each chromosome separately. The first chromosome has a non-functional promoter (P-), so no transcription of Z^+ can occur from it. The second chromosome has a functional promoter (P+) and could be transcribed, but it carries a non-functional β-galactosidase gene (Z^-). Therefore, no functional enzyme can be produced from either chromosome under any condition. The status of the I gene is irrelevant to the final outcome.
- What is the fundamental logical difference between the negative control of the lac operon and the negative control of the trp operon?
Answer: The lac repressor is synthesized in an active form, while the trp repressor is synthesized in an inactive form.
Both are negatively controlled by repressor proteins that bind to operators. The key difference is their default state and how they interact with their respective small molecules. The LacI repressor is synthesized in an active state that binds the operator by default; the inducer (allolactose) inactivates it. The TrpR repressor is synthesized as an inactive aporepressor; the co-repressor (tryptophan) must bind to it to activate it, allowing it to bind the operator. This reflects their function: turn on genes to break down lactose when it appears vs. turn off genes to synthesize tryptophan when it's already present.
- Through genetic engineering, the lac operator sequence is deleted from its normal position and re-inserted downstream of the
lacA gene. The rest of the operon is wild-type. How will this modification affect the expression of β-galactosidase (lacZ)? Answer: The expression of lacZ will become constitutive.
The operator's function is to block RNA polymerase movement from the promoter into the structural genes. By moving the operator downstream of all structural genes, the repressor can no longer prevent RNA polymerase from transcribing lacZ, lacY, and lacA. Transcription will initiate and proceed through the structural genes regardless of the repressor's state, resulting in constitutive expression.
- A bacterial strain has the genotype F'
I^+ O^C Z^- / I^S O^+ Z^+ for the lac operon. How will this strain regulate the production of functional β-galactosidase? Answer: No functional enzyme will be produced under any condition.
The I^S allele produces a trans-acting super-repressor that cannot be inactivated by lactose. It will bind to the wild-type operator (O^+) on the second chromosome, permanently repressing the Z^+ gene. The O^C allele is cis-acting and constitutive, but it is on a chromosome with a non-functional Z^- gene, so no functional enzyme can be produced from this locus. Therefore, no functional β-galactosidase is ever synthesized.
- A researcher creates a fusion where the lac operator and promoter are placed upstream of the trp structural genes (
trpE, D, C, B, A). The native trp promoter/operator is removed, and the strain has a wild-type lacI gene. How will the cell regulate tryptophan synthesis in a medium containing glucose but no lactose? Answer: Tryptophan synthesis will be repressed.
When analyzing gene regulation problems involving promoter/operator swaps, focus on which regulatory system now controls the structural genes. In this fusion, the trp structural genes are now under control of the lac regulatory elements, completely replacing their native tryptophan-responsive regulation.
The lac operon is negatively regulated by the LacI repressor protein. In the absence of lactose (the inducer), LacI binds tightly to the lac operator and blocks transcription. Since this medium contains glucose but no lactose, the LacI repressor will bind to the operator sequence and prevent RNA polymerase from transcribing the trp structural genes. This means tryptophan synthesis will be repressed regardless of the cell's actual need for tryptophan.
Choice A is incorrect because high-rate synthesis would only occur if lactose were present to inactivate the LacI repressor. Choice B is wrong because basal transcription occurs when there's leaky repression, but LacI provides tight repression in the absence of lactose. Choice C represents a common misconception—attenuation is a tryptophan-responsive mechanism that requires the native trp leader sequence and regulatory elements, which have been completely removed in this fusion construct.
Remember that in regulatory fusion experiments, the structural genes adopt the regulatory behavior of whatever promoter/operator system controls them. The genes "forget" their original regulation and respond only to their new regulatory environment.
- In a merodiploid strain
I^+ O^+ Z^+ / I^- O^C Z^-, what is the expression pattern of functional β-galactosidase? Answer: Normal inducible and repressible expression, as the wild-type alleles are dominant.
The first chromosome (I^+ O^+ Z^+) is a complete, wild-type operon. The second chromosome (I^- O^C Z^-) has a non-functional repressor (I^- is recessive to I^+), a constitutive operator (O^C), and a non-functional structural gene (Z^-). The I^+ gene on the first chromosome produces functional repressor that acts in trans, controlling the O^+ operator on its own chromosome. The second chromosome will be transcribed constitutively due to O^C, but since it has Z^-, it produces no functional enzyme. Therefore, the cell's phenotype is determined entirely by the first, wild-type chromosome, which exhibits normal inducible and repressible expression.
- Which of the following scenarios in the lac operon would result in constitutive synthesis of β-galactosidase that is NOT subject to catabolite repression (i.e., high expression even with glucose)?
Answer: A mutation in the CAP binding site that allows RNA polymerase to bind with high affinity without CAP.
To have constitutive expression, the repressor must be neutralized (e.g., I- or O^C). To be insensitive to catabolite repression, the need for CAP-cAMP activation must be bypassed. A mutation in the CAP binding site that makes the promoter inherently strong, allowing high-affinity binding of RNA polymerase without help from CAP, would achieve both. Transcription would be high whenever the operator is free. If this is combined with an I- or O^C mutation, expression would be high and constitutive regardless of lactose or glucose. Option C is the most direct way to achieve insensitivity to catabolite repression. Options A, B, and D all result in a system that is still repressed by glucose because the promoter still requires CAP-cAMP for high-level activation.
