Regulation of Gene Transcription
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AP Biology › Regulation of Gene Transcription
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 becomes constitutively expressed because the activator cannot bind
Gene R protein increases because translation no longer requires an activator
Gene R transcription is unchanged because activators act only after transcription begins
Gene R transcription increases because low oxygen directly stabilizes mRNA
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.
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?
Increase proteasome activity to accelerate TF turnover in the nucleus
Add more ribosomal subunits to increase translation of existing mRNA
Inhibit DNA polymerase to keep chromatin in an unreplicated state
Mutate the gene Z poly(A) signal to lengthen the mRNA tail
Introduce a constitutively active kinase that phosphorylates TF without hormone signaling
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.
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?
An increase in the concentration of amino acids used in gene E protein synthesis
A mutation in the transcription factor that prevents dimerization after ligand binding
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.
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 prevents DNA replication, which is required for transcription initiation
miR-1 directly blocks sigma factor binding to the gene G promoter
miR-1 changes the amino acid sequence of gene G, lowering transcription rate
miR-1 increases enhancer activity by recruiting mediator to the nucleus
miR-1 acts post-transcriptionally by targeting mRNA, not promoter DNA
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.
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
A mutation that increases the stability of gene X mRNA in the cytoplasm
An increase in the number of ribosomes that initiate translation on gene X mRNA
A synonymous mutation in the coding region that changes one codon but not the amino acid
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.
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?
Mutate the promoter to increase RNA polymerase binding affinity
Increase ribosome concentration to raise translation initiation rates
Mutate the repressor’s sugar-binding site so S cannot bind it
Delete the operator sequence so the repressor cannot bind DNA
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.
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?
Protein levels of A–C rise only after DNA replication increases operon copy number
mRNA for genes A–C increases because the small molecule activates RNA polymerase
Genes A–C become permanently expressed due to operator sequence disruption
Transcription of genes A–C remains low because RNA polymerase is blocked
Translation of genes A–C increases because ribosomes bypass the operator
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.
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?
Increasing the rate of DNA replication at the gene H locus
Replacing the Trp codons in the leader with codons for a different amino acid
Deleting the ribosome-binding site of the downstream gene to slow translation
Increasing the number of peroxisomes to improve fatty acid breakdown
Adding a poly(A) tail to the gene H mRNA to enhance nuclear export
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.
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 changes the amino acid sequence of gene A without altering its promoter
A mutation that increases the number of mitochondria per cell
A mutation that increases the half-life of gene A protein in the cytosol
A mutation that increases the efficiency of translation initiation for gene A mRNA
A mutation that increases the affinity of the TF for the inhibitor protein
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.
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?
Recruiting a transcriptional activator that binds a distal enhancer and loops to the promoter
Increasing the rate of gene Z protein degradation to reduce feedback inhibition
Inhibiting RNA-dependent RNA polymerase to reduce siRNA production
Deleting introns from gene Z so splicing is no longer required
Increasing ribosome binding to gene Z mRNA by altering the 5' UTR sequence
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.