Genetics Quiz: Operon Logic

What this quiz covers

This quiz focuses on Operon Logic, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

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.

Question 1

1. The expression of lacZ will become uninducible, and it will not be produced.
2. The expression of lacZ will become constitutive. (correct answer)
3. The regulation of lacZ will be normal, as the operator is still on the same DNA molecule.
4. lacZ will be expressed, but transcription will terminate at the new operator site.

Explanation: 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.

Question 2

1. The operon will be constitutively expressed at high levels, regardless of tryptophan concentration.
2. Transcription will terminate prematurely at the attenuator, leading to low expression levels. (correct answer)
3. The operon will exhibit normal regulation, with high expression in the absence of tryptophan.
4. The operon will no longer be subject to repression by the TrpR protein, but attenuation will function normally.

Explanation: 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.

Question 3

1. The operon will be permanently repressed, regardless of tryptophan levels. (correct answer)
2. The operon will be constitutively expressed, but still subject to attenuation.
3. The operon will show normal regulation, as attenuation is the primary control.
4. The operon will be uninducible by low tryptophan levels.

Explanation: 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.

Question 4

1. Tryptophan will be synthesized at a high rate.
2. Tryptophan will be synthesized at a basal rate.
3. Tryptophan synthesis will be regulated by tryptophan levels via attenuation.
4. Tryptophan synthesis will be repressed. (correct answer)

Explanation: 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.

Question 5

1. Introducing an F' plasmid carrying I^+ P^+ O^+ Z^-.
2. Introducing an F' plasmid carrying I^- P^+ O^+ Z^+.
3. Introducing an F' plasmid carrying I^+ P^+ O^C Z^+.
4. Introducing an F' plasmid carrying I^+ P^+ O^+ Z^+. (correct answer)

Explanation: 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.

Question 6

1. The operon would be transcribed constitutively.
2. The operon would be inducible, but only in the complete absence of glucose.
3. The operon would be transcribed at a basal level regardless of lactose presence.
4. The operon would be uninducible, with no transcription under any condition. (correct answer)

Explanation: 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.

Question 7

1. A mutation in lacI that prevents repressor binding to the operator.
2. A mutation in the operator (O^C) that prevents repressor binding.
3. A mutation in the CAP binding site that allows RNA polymerase to bind with high affinity without CAP. (correct answer)
4. A combination of an I^- mutation and a constitutively active adenylate cyclase.

Explanation: 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.

Question 8

1. No transcription, as glucose is a direct inhibitor of the lac promoter.
2. Basal transcription, as the presence of glucose prevents high-level expression.
3. High-level transcription, as high cAMP allows CAP to activate the promoter. (correct answer)
4. Constitutive transcription, as the regulatory roles of both glucose and lactose are bypassed.

Explanation: 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.

Question 9

1. Production will be constitutive, regardless of lactose or glucose levels.
2. Production will be inducible by lactose but repressed by glucose.
3. No functional enzyme will be produced under any condition. (correct answer)
4. Basal levels of the enzyme will be produced, but it cannot be induced to high levels.

Explanation: 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.

Question 10

1. Constitutive production of TrpE.
2. No production of TrpE. (correct answer)
3. Inducible production of TrpE (i.e., produced only when tryptophan is low).
4. Basal level production of TrpE.

Explanation: 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.

Question 11

1. Tryptophan synthesis will be constitutive and high, as termination is always prevented.
2. The operon will be fully repressed in high tryptophan, but transcription will be higher than wild-type in low tryptophan. (correct answer)
3. The operon will be uninducible, resulting in no tryptophan synthesis, as the anti-terminator loop cannot form.
4. Regulation will be identical to wild-type, as the TrpR repressor is the primary regulator.

Explanation: 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.

Question 12

1. The operon will show increased expression compared to wild-type.
2. The operon will show decreased expression compared to wild-type. (correct answer)
3. The operon's regulation by attenuation will be unaffected, but repression will be lost.
4. The operon will be constitutively expressed at high levels.

Explanation: 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.

Question 13

1. A negatively controlled, inducible operon where metanose acts as an inducer. (correct answer)
2. A negatively controlled, repressible operon where metanose acts as a co-repressor.
3. A positively controlled, inducible operon where RegM is an activator.
4. A positively controlled, repressible operon where RegM is an activator.

Explanation: 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.

Question 14

1. Attenuation allows the cell to bypass repression entirely when tryptophan levels are extremely low.
2. Attenuation provides a mechanism for responding to a graded range of tryptophan concentrations, which simple repression does not. (correct answer)
3. Attenuation serves as a backup system that functions only if the TrpR repressor protein is mutated or absent.
4. Attenuation links the rate of tryptophan synthesis directly to the availability of glucose for energy.

Explanation: 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'.

Question 15

1. Constitutive expression, as the O^C allele allows transcription from the second chromosome.
2. Uninducible expression, as the wild-type repressor will bind to the constitutive operator.
3. Normal inducible and repressible expression, as the wild-type alleles are dominant. (correct answer)
4. No expression, as the O^C allele is cis-linked to a non-functional Z^- gene.

Explanation: 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.

Question 16

1. A negatively controlled, inducible operon where metanose acts as an inducer. (correct answer)
2. A negatively controlled, repressible operon where metanose acts as a co-repressor.
3. A positively controlled, inducible operon where RegM is an activator.
4. A positively controlled, repressible operon where RegM is an activator.

Explanation: 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.

Question 17

1. Tryptophan synthesis will be constitutive and high, as termination is always prevented.
2. The operon will be fully repressed in high tryptophan, but transcription will be higher than wild-type in low tryptophan. (correct answer)
3. The operon will be uninducible, resulting in no tryptophan synthesis, as the anti-terminator loop cannot form.
4. Regulation will be identical to wild-type, as the TrpR repressor is the primary regulator.

Explanation: 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.

Question 18

1. The operon will show increased expression compared to wild-type.
2. The operon will show decreased expression compared to wild-type. (correct answer)
3. The operon's regulation by attenuation will be unaffected, but repression will be lost.
4. The operon will be constitutively expressed at high levels.

Explanation: 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.

Question 19

1. Transcription will be at a high, activated level because glucose is low.
2. Transcription will occur at a low, basal level. (correct answer)
3. No transcription will occur because CAP is required for RNA polymerase to bind.
4. Transcription will be constitutive, as positive control is eliminated.

Explanation: 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.

Question 20

1. Attenuation allows the cell to bypass repression entirely when tryptophan levels are extremely low.
2. Attenuation provides a mechanism for responding to a graded range of tryptophan concentrations, which simple repression does not. (correct answer)
3. Attenuation serves as a backup system that functions only if the TrpR repressor protein is mutated or absent.
4. Attenuation links the rate of tryptophan synthesis directly to the availability of glucose for energy.

Explanation: 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'.