AP Biology Quiz: Regulation Of Gene Transcription

20 questions · exam conditions

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Regulation Of Gene TranscriptionQuestion 1 of 20

In bacteria, gene R is transcribed from a promoter that requires an activator bound upstream to help recruit RNA polymerase. Under low oxygen, the activator binds DNA and transcription increases. A point mutation occurs in the activator's DNA-binding domain so it cannot bind the upstream site, but the promoter sequence is unchanged. Which result is most likely under low oxygen?

Gene R transcription decreases because RNA polymerase recruitment is reduced

Gene R transcription increases because low oxygen directly stabilizes mRNA

Gene R protein increases because translation no longer requires an activator

Gene R becomes constitutively expressed because the activator cannot bind

Gene R transcription is unchanged because activators act only after transcription begins

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AP Biology Quiz

AP Biology Quiz: Regulation Of Gene Transcription

Practice Regulation Of Gene Transcription in AP Biology 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 Regulation Of Gene Transcription, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

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.

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Question 1

Gene R transcription decreases because RNA polymerase recruitment is reduced (correct answer)
Gene R transcription increases because low oxygen directly stabilizes mRNA
Gene R protein increases because translation no longer requires an activator
Gene R becomes constitutively expressed because the activator cannot bind
Gene R transcription is unchanged because activators act only after transcription begins

Explanation: This question assesses transcriptional regulation in prokaryotes, focusing on activator-dependent promoter activation. Gene R transcription relies on the activator binding upstream to recruit RNA polymerase, which occurs under low oxygen to increase expression. The mutation prevents the activator from binding DNA, so RNA polymerase recruitment fails, resulting in decreased transcription even under low oxygen conditions. Consequently, the gene does not respond to the environmental cue as it cannot form the necessary activation complex. A tempting distractor is option D, suggesting constitutive expression because the activator cannot bind, but this confuses activation with repression, misconstruing that loss of an activator leads to derepression rather than reduced expression. In solving activator-related questions, distinguish between positive and negative regulation and evaluate how binding defects impact polymerase engagement.

Question 2

In a eukaryotic nucleus, transcription factor TF binds a promoter-proximal element of gene Z and increases transcription when TF is phosphorylated. A signaling kinase phosphorylates TF in response to hormone H. In cells treated with H, TF is present but gene Z mRNA does not increase. Sequencing shows TF is unchanged, but the kinase has a loss-of-function mutation. Which change would most likely increase transcription of gene Z in response to H?

Introduce a constitutively active kinase that phosphorylates TF without hormone signaling (correct answer)
Increase proteasome activity to accelerate TF turnover in the nucleus
Mutate the gene Z poly(A) signal to lengthen the mRNA tail
Add more ribosomal subunits to increase translation of existing mRNA
Inhibit DNA polymerase to keep chromatin in an unreplicated state

Explanation: This question examines transcriptional regulation in eukaryotes, specifically signal-dependent activation via phosphorylation. The transcription factor TF requires phosphorylation by a kinase activated by hormone H to stimulate gene Z transcription, but the kinase mutation prevents this, so mRNA does not increase with H. Introducing a constitutively active kinase phosphorylates TF independently of H signaling, ensuring TF activation and increasing gene Z transcription when H is present despite the original kinase defect. This bypasses the loss-of-function mutation by providing constant phosphorylation, restoring transcriptional output in hormone-treated cells. A tempting distractor is option D, adding ribosomal subunits, but this affects translation rather than transcription, arising from the misconception that post-transcriptional changes can fix a transcriptional defect. To tackle similar issues, pinpoint the broken step in the regulatory cascade and select fixes that directly compensate for it.

Question 3

In an animal cell, gene E is regulated by a transcription factor that binds the promoter only when it is dimerized. A ligand binds the transcription factor and promotes dimer formation. Without ligand, chromatin at the promoter is accessible but gene E transcription is low. After ligand addition, gene E primary transcript levels increase. Which change would most likely block ligand-induced transcription of gene E?

A mutation in the transcription factor that prevents dimerization after ligand binding (correct answer)
An increase in the concentration of amino acids used in gene E protein synthesis
A mutation that increases the catalytic rate of the proteasome
A mutation that increases splicing efficiency of gene E pre-mRNA
A mutation that increases the stability of gene E protein once translated

Explanation: This question assesses understanding of transcriptional regulation, specifically how ligand-induced dimerization enables transcription factor binding in eukaryotes. The ligand promotes TF dimerization, which is necessary for promoter binding and increased transcription, despite accessible chromatin. A mutation preventing dimerization after ligand binding would keep the TF monomeric and unable to bind the promoter, blocking the transcriptional increase. This disrupts the key activation step in the regulatory pathway. A tempting distractor is choice D, which improves splicing, but this is wrong because it affects pre-mRNA processing, reflecting the misconception that splicing efficiency controls promoter activity. To approach similar problems, trace how mutations affect the structural requirements for TF-DNA interactions.

Question 4

In a human cell line, transcription of gene G decreases when a specific microRNA (miR-1) is introduced. miR-1 has partial complementarity to a sequence in the 3' UTR of gene G mRNA and recruits proteins that reduce translation and can promote mRNA degradation. RNA polymerase II binding at the gene G promoter is unchanged after miR-1 introduction. Which statement best accounts for why gene G transcription is not directly reduced by miR-1?

miR-1 acts post-transcriptionally by targeting mRNA, not promoter DNA (correct answer)
miR-1 increases enhancer activity by recruiting mediator to the nucleus
miR-1 directly blocks sigma factor binding to the gene G promoter
miR-1 prevents DNA replication, which is required for transcription initiation
miR-1 changes the amino acid sequence of gene G, lowering transcription rate

Explanation: This question assesses understanding of transcriptional regulation, specifically distinguishing it from post-transcriptional mechanisms like microRNA action. miR-1 binds the 3' UTR of gene G mRNA, promoting degradation and reducing translation, which decreases mRNA levels without altering RNA polymerase II binding at the promoter. Thus, transcription initiation remains unchanged, as miR-1 does not interact with promoter DNA or transcriptional machinery. The observed decrease in transcription likely refers to reduced mRNA as a proxy, but it's not a direct effect on transcription. A tempting distractor is choice B, which suggests miR-1 increases enhancer activity, but this is wrong because miRNAs typically act post-transcriptionally, reflecting the misconception that they regulate enhancers. To approach similar problems, differentiate between effects on transcription initiation versus mRNA stability or translation.

Question 5

In E. coli, gene X is preceded by a promoter and an operator that overlaps the promoter. A repressor protein binds the operator and blocks RNA polymerase binding. When lactose is added, it is converted to allolactose, which binds the repressor and reduces its affinity for the operator. In cells grown without lactose, gene X mRNA levels are low; after lactose addition, gene X mRNA increases within minutes. Which change would most likely prevent the lactose-dependent increase in transcription of gene X?

A mutation that prevents allolactose from binding the repressor protein (correct answer)
A synonymous mutation in the coding region that changes one codon but not the amino acid
An increase in the number of ribosomes that initiate translation on gene X mRNA
A mutation that increases the stability of gene X mRNA in the cytoplasm
A deletion of a transcription terminator located downstream of the gene X coding sequence

Explanation: This question assesses understanding of transcriptional regulation, specifically how repressors and inducers control gene expression in prokaryotes like the lac operon. The repressor binds the operator to block RNA polymerase, preventing transcription in the absence of lactose, but allolactose binding to the repressor releases it, allowing transcription to increase. A mutation preventing allolactose from binding the repressor would keep the repressor bound to the operator even with lactose present, thus blocking the lactose-dependent increase in gene X transcription. This directly disrupts the induction mechanism at the transcriptional level by maintaining repression. A tempting distractor is choice C, which increases translation initiation, but this is wrong because it affects post-transcriptional processes, reflecting the misconception that translation rates directly control transcription. To approach similar problems, always identify whether the change targets the promoter, regulatory proteins, or DNA elements involved in transcription initiation.

Question 6

In cultured bacteria, gene X encodes a sugar transporter. A repressor protein binds a specific operator sequence overlapping the promoter and blocks RNA polymerase binding. When sugar S is added, S binds the repressor and reduces its affinity for the operator. After S addition, levels of gene X mRNA increase within minutes. Which change would most likely prevent the increase in transcription after adding sugar S?

Delete the operator sequence so the repressor cannot bind DNA
Mutate the promoter to increase RNA polymerase binding affinity
Mutate the repressor's sugar-binding site so S cannot bind it (correct answer)
Increase ribosome concentration to raise translation initiation rates
Duplicate gene X coding sequence without changing promoter or operator

Explanation: This question assesses understanding of transcriptional regulation in prokaryotes, focusing on repressor-mediated control of gene expression in response to environmental signals. In this system, the repressor binds the operator to block RNA polymerase, but sugar S binds the repressor, causing it to release the operator and allow transcription to proceed. Mutating the repressor's sugar-binding site prevents S from binding, so the repressor remains attached to the operator, maintaining the block on transcription even after S is added. Consequently, the expected increase in gene X mRNA levels does not occur because the regulatory switch fails to activate. A tempting distractor is option A, deleting the operator sequence, but this would lead to constitutive transcription by preventing repressor binding altogether, stemming from the misconception that removing the operator mimics repression rather than derepression. To solve similar problems, identify the role of each regulatory element and trace how a change disrupts the signal-response pathway.

Question 7

A bacterial operon contains genes A–C transcribed from one promoter. A regulatory protein binds an operator between the promoter and gene A, preventing transcription. When amino acid Q is scarce, a small molecule accumulates and binds the regulatory protein, causing it to release the operator. In Q-scarce conditions, a mutation is introduced that changes the operator DNA sequence so the regulatory protein binds it tightly even when the small molecule is present. Which outcome is most likely in Q-scarce conditions?

Transcription of genes A–C remains low because RNA polymerase is blocked (correct answer)
Translation of genes A–C increases because ribosomes bypass the operator
mRNA for genes A–C increases because the small molecule activates RNA polymerase
Genes A–C become permanently expressed due to operator sequence disruption
Protein levels of A–C rise only after DNA replication increases operon copy number

Explanation: This question tests comprehension of transcriptional regulation in bacterial operons, emphasizing repressor-operator interactions. Normally, the regulatory protein binds the operator to block transcription, but the small molecule releases it when Q is scarce, allowing RNA polymerase to transcribe genes A–C. The mutation causes the protein to bind tightly even with the small molecule present, so in Q-scarce conditions, the operator remains occupied, preventing RNA polymerase access and keeping transcription low. Thus, genes A–C mRNA levels do not increase as they would in wild type under the same conditions. A tempting distractor is option C, suggesting mRNA increases because the small molecule activates RNA polymerase, but this ignores that the mutation specifically disrupts repressor release, misconstruing the small molecule's role as direct polymerase activation rather than indirect derepression. When approaching operon problems, map out the regulatory logic and predict mutation effects by simulating the system's state under given conditions.

Question 8

A prokaryotic gene H is controlled by attenuation. The leader region of the mRNA contains codons for tryptophan and can form either a terminator hairpin that stops transcription or an antiterminator that allows transcription to continue. When tryptophan is abundant, ribosomes translate the leader quickly, favoring terminator formation and lowering downstream transcription. When tryptophan is scarce, ribosomes stall at the Trp codons, favoring antiterminator formation and increasing transcription. Which change would most likely decrease transcription of gene H during tryptophan scarcity?

Replacing the Trp codons in the leader with codons for a different amino acid (correct answer)
Increasing the rate of DNA replication at the gene H locus
Deleting the ribosome-binding site of the downstream gene to slow translation
Adding a poly(A) tail to the gene H mRNA to enhance nuclear export
Increasing the number of peroxisomes to improve fatty acid breakdown

Explanation: This question assesses understanding of transcriptional regulation, specifically attenuation mechanisms in prokaryotic genes like the trp operon. Ribosome stalling at Trp codons during scarcity favors antiterminator formation, allowing continued transcription, while quick translation during abundance forms the terminator. Replacing Trp codons with others prevents stalling even in scarcity, favoring terminator formation and decreasing transcription. This disrupts the attenuation logic by removing the sensing mechanism for tryptophan levels. A tempting distractor is choice C, which deletes a ribosome-binding site, but this is wrong because it affects downstream translation, reflecting the misconception that translation of structural genes controls attenuation. To approach similar problems, analyze how leader sequence features link nutrient sensing to transcription termination.

Question 9

In a plant cell, gene A transcription increases after exposure to a hormone. The gene's promoter contains a specific response element recognized by a hormone-activated transcription factor (TF). In the absence of hormone, the TF is bound by an inhibitor protein that masks the TF's DNA-binding domain. Hormone binding causes the inhibitor to dissociate, allowing the TF to bind the response element and recruit RNA polymerase II. Which change would most likely decrease hormone-induced transcription of gene A?

A mutation that increases the affinity of the TF for the inhibitor protein (correct answer)
A mutation that increases the efficiency of translation initiation for gene A mRNA
A mutation that increases the half-life of gene A protein in the cytosol
A mutation that changes the amino acid sequence of gene A without altering its promoter
A mutation that increases the number of mitochondria per cell

Explanation: This question assesses understanding of transcriptional regulation, specifically how hormone signaling modulates transcription factor activity through inhibitor dissociation. The hormone causes the inhibitor to release the TF, allowing it to bind the response element and recruit RNA polymerase II for increased transcription. A mutation increasing the TF's affinity for the inhibitor would make dissociation harder, keeping the TF masked and unable to activate transcription even with hormone present. This directly impairs the hormone-induced activation at the regulatory level. A tempting distractor is choice B, which boosts translation initiation, but this is wrong because it affects protein production post-transcriptionally, reflecting the misconception that increasing mRNA translation impacts transcription rates. To approach similar problems, evaluate how changes alter interactions between regulators, signals, and DNA-binding domains.

Question 10

A mammalian gene Z is normally transcribed at low levels. The promoter contains a CpG island. After treatment with a DNA methyltransferase activator, methylation at the CpG island increases, and gene Z mRNA decreases. Histone deacetylase (HDAC) activity is unchanged. Which change would most likely increase transcription of gene Z despite high CpG methylation at its promoter?

Increasing ribosome binding to gene Z mRNA by altering the 5' UTR sequence
Deleting introns from gene Z so splicing is no longer required
Recruiting a transcriptional activator that binds a distal enhancer and loops to the promoter (correct answer)
Inhibiting RNA-dependent RNA polymerase to reduce siRNA production
Increasing the rate of gene Z protein degradation to reduce feedback inhibition

Explanation: This question assesses understanding of transcriptional regulation, specifically how epigenetic modifications like DNA methylation repress gene expression and how enhancers can counteract them. High CpG methylation at the promoter typically condenses chromatin, reducing accessibility for transcription factors and RNA polymerase. Recruiting a transcriptional activator to a distal enhancer allows it to loop to the promoter, facilitating RNA polymerase recruitment and overriding methylation-induced repression. This enhances initiation despite the methylated promoter by providing an alternative activation pathway. A tempting distractor is choice A, which improves ribosome binding, but this is wrong because it affects translation efficiency, reflecting the misconception that translational changes can compensate for transcriptional repression. To approach similar problems, distinguish between interventions that act at chromatin, promoter, or enhancer levels to modulate transcription.

Question 11

A eukaryotic gene D shows reduced transcription after a signaling pathway activates a corepressor complex. The corepressor recruits histone deacetylase (HDAC) to the promoter region, and chromatin accessibility near the transcription start site decreases. RNA polymerase II protein levels remain constant. Which change would most likely counteract the decrease in transcription of gene D?

Inhibiting HDAC activity to maintain histone acetylation near the promoter (correct answer)
Increasing amino acid availability to speed translation elongation of gene D mRNA
Mutating the stop codon of gene D to extend the protein coding sequence
Decreasing the rate of protein ubiquitination in the cytosol
Increasing the number of nuclear pores to enhance mRNA export

Explanation: This question assesses understanding of transcriptional regulation, specifically how corepressors and HDACs repress eukaryotic genes through chromatin modifications. The corepressor recruits HDAC to deacetylate histones, condensing chromatin and reducing promoter accessibility for RNA polymerase II. Inhibiting HDAC would preserve histone acetylation, maintaining open chromatin and counteracting the repression to sustain transcription. This directly opposes the deacetylation mechanism at the epigenetic level. A tempting distractor is choice E, which increases nuclear pores, but this is wrong because it enhances mRNA export, reflecting the misconception that export efficiency regulates transcription initiation. To approach similar problems, consider how changes target chromatin-modifying enzymes or complexes involved in repression.

Question 12

In a eukaryotic cell line, gene Y transcription requires an activator that binds an enhancer 2 kb upstream. The activator recruits a histone acetyltransferase (HAT), increasing acetylation of nearby histone tails and correlating with higher gene Y mRNA. A drug that inhibits the HAT is added; the activator still binds DNA, but histone acetylation near the enhancer decreases. Which change would most likely restore high transcription of gene Y in the presence of the drug?

Overexpress a histone deacetylase to increase chromatin compaction
Increase tRNA abundance to accelerate elongation during translation
Mutate the gene Y start codon to improve ribosome recognition
Recruit a different coactivator with HAT activity to the enhancer-bound activator (correct answer)
Delete the introns from gene Y to reduce mRNA processing time

Explanation: This question evaluates knowledge of transcriptional regulation in eukaryotes, particularly how chromatin modifications influence gene activation. The activator normally recruits a histone acetyltransferase (HAT) to acetylate histones, loosening chromatin and promoting transcription, but the drug inhibits this HAT, reducing acetylation and transcription. Recruiting a different coactivator with HAT activity bypasses the inhibited HAT, restoring acetylation near the enhancer and allowing high gene Y transcription despite the drug. This works because the new coactivator provides the necessary enzymatic function to modify chromatin without relying on the original inhibited HAT. A tempting distractor is option A, overexpressing a histone deacetylase, but this would further compact chromatin and decrease transcription, based on the misconception that deacetylation aids activation rather than repression. For such questions, consider how epigenetic modifiers like HATs and HDACs alter chromatin accessibility and test interventions that compensate for specific defects.

Question 13

A mammalian gene has a CpG-rich promoter. When promoter cytosines are methylated, a methyl-binding protein recruits chromatin remodeling factors that increase nucleosome packing, and mRNA levels are low. In a treated cell population, promoter methylation increases while transcription decreases. Which change would most likely increase transcription without altering the gene's coding sequence?

Increase DNA methyltransferase activity to add additional methyl groups
Introduce a mutation that removes CpG sites from the promoter region (correct answer)
Add a translation inhibitor to reduce ribosome movement along mRNA
Increase availability of RNA nucleotides to speed RNA chain elongation
Duplicate the gene's exons to increase the number of possible proteins

Explanation: This question investigates transcriptional regulation via epigenetic mechanisms in mammals, particularly DNA methylation's role in gene silencing. Methylation of CpG sites in the promoter recruits proteins that compact chromatin, reducing transcription, and increased methylation correlates with decreased mRNA. Mutating to remove CpG sites prevents methylation, avoiding recruitment of repressive factors and allowing more open chromatin for higher transcription without changing the coding sequence. This eliminates the sites targeted for silencing, bypassing the methylation-dependent repression. A tempting distractor is option A, increasing DNA methyltransferase activity, but this would enhance methylation and further repress transcription, stemming from the misconception that more methylation activates rather than silences genes. To address epigenetics problems, identify modifiable elements like CpG sites and consider alterations that disrupt repressive marks.

Question 14

In bacteria, gene M is regulated by an inducible system. A repressor binds the operator and prevents transcription when inducer I is absent. When I is present, it binds the repressor and causes the repressor to release the operator, increasing transcription. A mutation changes the repressor so it binds the operator but cannot release it after binding I. Which condition will show the largest decrease in gene M transcription compared with wild type?

Cells grown with inducer I present, because repression cannot be relieved (correct answer)
Cells grown without inducer I, because repression is already absent
Cells grown with inducer I present, because RNA polymerase is degraded
Cells grown without inducer I, because translation initiation is blocked
Cells grown with inducer I present, because mRNA splicing is inhibited

Explanation: This question evaluates understanding of transcriptional regulation in inducible bacterial systems, comparing mutant and wild-type behaviors. In wild type, the repressor blocks transcription without inducer I, but I binds the repressor to release it, increasing transcription. The mutation prevents release even with I, so repression persists, causing a large decrease in transcription specifically when I is present compared to wild type. This condition highlights the mutant's failure to derepress, while without I, both are repressed similarly. A tempting distractor is option B, cells without I showing the decrease, but transcription is already low in both, misconstruing that the mutation affects the repressed state rather than the induced one. For inducible system questions, compare expression across conditions and quantify differences from wild type to identify the most impacted scenario.

Question 15

A eukaryotic gene is controlled by a transcriptional repressor that binds a silencer element and recruits a corepressor complex. The corepressor includes a protein that directly interacts with the mediator complex to reduce RNA polymerase II initiation. A mutation deletes the repressor's corepressor-interaction domain, but the repressor still binds the silencer. Which effect is most likely on transcription of the gene?

Transcription increases because silencer binding no longer recruits the inhibitory corepressor (correct answer)
Transcription decreases because the repressor binds DNA more strongly without the domain
Transcription is unchanged because corepressors act only during translation
mRNA is unchanged because mediator regulates splicing rather than initiation
Protein levels increase only if the mutation also changes the ribosome-binding site

Explanation: This question tests knowledge of transcriptional regulation in eukaryotes, emphasizing repressor-corepressor dynamics. The repressor binds the silencer and recruits a corepressor to inhibit RNA polymerase II initiation via mediator interaction, reducing transcription. Deleting the corepressor-interaction domain allows the repressor to bind but prevents corepressor recruitment, eliminating the inhibitory effect and increasing transcription. Thus, the gene is no longer effectively silenced despite repressor binding. A tempting distractor is option B, suggesting decreased transcription because the repressor binds more strongly, but this ignores that repression requires corepressor recruitment, based on the misconception that DNA binding alone suffices for inhibition. When analyzing repressor mutations, separate DNA-binding from effector functions and assess impacts on downstream regulatory steps.

Question 16

Delete the operator sequence so the repressor cannot bind DNA
Mutate the promoter to increase RNA polymerase binding affinity
Mutate the repressor's sugar-binding site so S cannot bind it (correct answer)
Increase ribosome concentration to raise translation initiation rates
Duplicate gene X coding sequence without changing promoter or operator

Question 17

In bacteria, a gene's promoter is followed immediately by a leader region that can form either a terminator hairpin (stopping transcription early) or an anti-terminator (allowing transcription to continue). When an amino acid is abundant, a ribosome rapidly translates the leader peptide, favoring terminator formation and reducing full-length mRNA. A mutation slows ribosome movement through the leader peptide without changing the promoter. Which outcome is most likely when the amino acid is abundant?

Full-length transcription increases because slower translation favors anti-terminator formation (correct answer)
Full-length transcription decreases because RNA polymerase cannot bind the promoter
Full-length transcription is unchanged because attenuation affects only translation
mRNA levels drop because the mutation increases DNA methylation at the promoter
Protein levels rise because the mutation increases tRNA charging efficiency directly

Explanation: This question examines transcriptional regulation via attenuation in bacteria, linking translation speed to RNA structure formation. Normally, abundant amino acid allows fast leader peptide translation, favoring terminator hairpin formation and early transcription stop, reducing full-length mRNA. The mutation slows ribosome movement, mimicking low amino acid conditions where slower translation promotes anti-terminator formation, allowing continued transcription and increased full-length mRNA even when amino acid is abundant. This shifts the balance toward the anti-terminator, bypassing the usual attenuation signal. A tempting distractor is option B, decreased transcription because RNA polymerase cannot bind, but this overlooks that attenuation acts post-initiation, misconstruing it as a promoter-level effect rather than elongation control. For attenuation problems, consider how ribosome dynamics influence RNA folding and apply that to predict structural outcomes under altered conditions.

Question 18

A gene in yeast is normally silenced when a repressor binds its promoter and recruits a histone deacetylase (HDAC), decreasing histone acetylation and reducing transcription. A mutant strain has the same repressor and promoter sequence, but gene mRNA is high even when the repressor is bound. Protein assays show the HDAC is absent. Which change would most likely return transcription to low levels?

Increase RNA polymerase concentration to saturate the promoter binding site
Express a functional HDAC that can be recruited by the repressor (correct answer)
Mutate the coding sequence to introduce an early stop codon
Increase amino acid availability to accelerate peptide elongation
Add a spliceosome inhibitor to prevent intron removal from pre-mRNA

Explanation: This question explores transcriptional regulation through chromatin modifications in eukaryotes, highlighting repressor-HDAC interactions. The repressor normally recruits HDAC to deacetylate histones, compacting chromatin and silencing the gene, but in the mutant, HDAC absence leads to persistent acetylation and high transcription. Expressing a functional HDAC allows the repressor to recruit it again, restoring deacetylation, chromatin compaction, and low transcription levels. This directly addresses the missing component, reinstating the repressive mechanism without altering other elements. A tempting distractor is option A, increasing RNA polymerase concentration, but this would not overcome chromatin-based silencing, based on the misconception that more polymerase can bypass epigenetic barriers. For these problems, focus on the specific regulatory mechanism and identify restorations that rebuild the original inhibitory pathway.

Question 19

In yeast, gene Y has an enhancer located 1 kb upstream. An activator binds the enhancer and recruits a mediator complex that helps RNA polymerase II assemble at the promoter. In nutrient-rich conditions, the activator is phosphorylated and localized in the nucleus; under starvation, the activator remains unphosphorylated in the cytosol. Starved cells show reduced gene Y pre-mRNA levels compared with nutrient-rich cells. Which change would most likely restore high transcription of gene Y during starvation?

A mutation that adds a nuclear localization signal to the activator protein (correct answer)
A mutation that removes the poly(A) signal from the gene Y transcript
A mutation that increases tRNA abundance for codons used in gene Y
A mutation that changes one amino acid in gene Y without affecting the promoter
An increase in proteasome activity that degrades RNA polymerase II

Explanation: This question assesses understanding of transcriptional regulation, specifically how activator localization and phosphorylation influence eukaryotic gene expression. The activator must be in the nucleus to bind the enhancer and recruit the mediator complex for RNA polymerase II assembly, but during starvation, it remains in the cytosol. Adding a nuclear localization signal would force the activator into the nucleus even without phosphorylation, restoring its ability to enhance transcription under starvation conditions. This bypasses the phosphorylation-dependent localization, directly enabling transcriptional activation. A tempting distractor is choice B, which removes the poly(A) signal, but this is wrong because it impairs mRNA stability and export, reflecting the misconception that post-transcriptional modifications control initiation rates. To approach similar problems, focus on whether the intervention restores the presence or activity of transcriptional regulators at the promoter.

Question 20

In bacteria, genes B and C are in an operon with a single promoter. A regulatory protein binds upstream of the promoter and increases RNA polymerase binding when glucose is absent. When glucose is present, intracellular cAMP decreases, and the regulatory protein does not bind DNA. With glucose present, mRNA from the operon is low even if the promoter sequence is unchanged. Which change would most likely increase transcription of the operon in the presence of glucose?

A mutation that prevents cAMP breakdown, keeping intracellular cAMP high (correct answer)
A mutation that removes the Shine-Dalgarno sequence upstream of gene B
A mutation that reduces the activity of ribosomal peptidyl transferase
A mutation that increases the frequency of homologous recombination in the chromosome
A mutation that shortens the operon mRNA poly(A) tail

Explanation: This question assesses understanding of transcriptional regulation, specifically how cAMP and regulatory proteins mediate catabolite repression in bacterial operons. Low cAMP in the presence of glucose prevents the regulatory protein from binding DNA, reducing RNA polymerase recruitment and operon transcription. A mutation preventing cAMP breakdown would maintain high cAMP levels, allowing the regulatory protein to bind and activate transcription even with glucose. This circumvents glucose repression by sustaining the activation signal. A tempting distractor is choice B, which removes the Shine-Dalgarno sequence, but this is wrong because it impairs translation initiation, reflecting the misconception that ribosomal binding sites control transcription. To approach similar problems, identify interventions that mimic or sustain the environmental signals required for promoter activation.