Transcription and RNA Processing
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AP Biology › Transcription and RNA Processing
In a eukaryotic cell, researchers isolate nuclear RNA from a gene containing three exons separated by two introns. A newly synthesized transcript is detected that hybridizes to probes complementary to exon 1, intron 1, exon 2, intron 2, and exon 3. After 10 minutes, the predominant nuclear RNA hybridizes only to exon probes, and sequencing shows exon 1 directly joined to exon 2 and exon 2 joined to exon 3. No changes are detected in the DNA sequence of the gene. Which explanation best accounts for the change in hybridization pattern over time?
DNA polymerase replaced intron sequences with exon sequences, producing a shorter RNA during transcription.
The spliceosome excised introns from the pre-mRNA and ligated adjacent exons to form a continuous RNA.
RNA polymerase removed introns during elongation by excising intron sequences from the DNA template strand.
RNA polymerase initiated transcription at exon 2 in later rounds, generating transcripts lacking intron sequences.
Ribozymes in the cytosol degraded intron regions of the RNA after the transcript exited the nucleus.
Explanation
This question tests understanding of transcription and RNA processing, specifically how introns are removed from pre-mRNA in eukaryotic cells. The correct answer is C because the spliceosome is the molecular machinery responsible for removing introns and joining exons together in the nucleus, which explains why the RNA initially contains both exons and introns but later contains only exons. The observation that exons are directly joined together (exon 1 to exon 2 to exon 3) is the hallmark of splicing, where introns are excised and adjacent exons are ligated. Answer A is incorrect because RNA polymerase transcribes the DNA template as written and cannot selectively remove sequences during elongation—this represents a misconception that transcription and splicing occur simultaneously. The key strategy is to recognize that changes in RNA sequence after transcription, without DNA changes, indicate post-transcriptional processing like splicing.
An investigator compares nuclear RNA from two eukaryotic cell types and finds that both produce the same primary transcript from a gene with four exons. In cell type 1, the predominant processed RNA contains exons 1-2-3-4. In cell type 2, the predominant processed RNA contains exons 1-2-4, with exon 3 absent; splice junctions match known splice-site sequences. The DNA sequence of the gene is identical in both cell types. Which explanation best accounts for the different processed RNAs?
Alternative splicing in cell type 2 excludes exon 3 during processing of the same primary transcript.
Cell type 2 edits the DNA to remove exon 3 after transcription, yielding a shorter RNA template.
Cell type 2 degrades exon 3 using ribosomal enzymes, then ligates the remaining RNA in the cytosol.
Different RNA polymerases transcribe different exons, so cell type 2 fails to transcribe exon 3.
Cell type 2 adds a longer 5′ cap that replaces exon 3 sequence during RNA modification.
Explanation
This question assesses understanding of transcription and RNA processing in eukaryotes, particularly how cell-specific factors influence mRNA isoforms. Both cell types produce the same primary transcript with four exons, but cell type 2 excludes exon 3 in processed RNA with matching splice junctions and identical DNA, suggesting a regulated processing difference. Alternative splicing allows exclusion of specific exons based on cellular context, yielding the 1-2-4 isoform in cell type 2. Choice B is correct because it accounts for exon skipping during splicing of the shared transcript. A tempting distractor is choice A, which wrongly attributes exclusion to different RNA polymerases, stemming from the misconception that transcription enzymes select exons. To address variability in RNA products, verify if primary transcripts are identical and consider alternative splicing as a mechanism for diversity.
A scientist introduces a point mutation in the 3′ splice site (AG) of intron 1 in a eukaryotic gene. The 5′ splice site of intron 1 and all splice sites of other introns remain unchanged. When nuclear RNA is analyzed, a new RNA species appears that includes exon 1, intron 1, and exon 2 in a single continuous segment, while downstream introns are still removed normally. Which explanation best accounts for the structure of this new RNA species?
The mutation blocks 5′ capping, so the RNA cannot be spliced and all introns are retained.
The mutated 3′ splice site prevents removal of intron 1, so intron 1 is retained while later introns are spliced.
The mutation changes the template strand, so RNA polymerase inserts intron sequences into the DNA.
The mutation converts intron 1 into an exon, so exon 1 is deleted during processing.
The mutation causes RNA polymerase to skip exon 2 during transcription, fusing exon 1 directly to exon 3.
Explanation
This question tests understanding of transcription and RNA processing, specifically how splice site mutations prevent intron removal. The 3' splice site (AG dinucleotide) is essential for the second step of splicing, where the free 3' OH of the upstream exon attacks the 3' splice site to join exons and release the intron lariat. When this site is mutated in intron 1, the spliceosome cannot complete removal of intron 1, even though it can recognize the 5' splice site and form the lariat intermediate. The result is retention of intron 1 between exons 1 and 2, while downstream introns with intact splice sites are removed normally. Choice B incorrectly suggests that RNA polymerase can skip exons during transcription, but polymerase transcribes the DNA template continuously and cannot selectively omit sequences. When analyzing splicing mutations, remember that both 5' and 3' splice sites must be functional for intron removal—mutation of either site causes intron retention.
A eukaryotic pre-mRNA is synthesized and then processed. When a drug that inhibits RNA polymerase II is added, no new pre-mRNA is produced. However, existing pre-mRNA molecules in the nucleus still become shorter over time and then appear in the cytoplasm with internal sequences removed. The drug does not inhibit enzymes in the nuclear extract responsible for RNA processing. Which explanation best accounts for why mature mRNA still appears after transcription is inhibited?
DNA repair enzymes synthesize RNA directly from the mature mRNA template when transcription is blocked.
Cytoplasmic enzymes add introns back to mRNA, making it appear shorter as it cycles through the nucleus.
Splicing and other processing can act on pre-existing pre-mRNA molecules even when new transcription stops.
RNA polymerase I compensates by transcribing the same gene, producing pre-mRNA for processing.
RNA polymerase II inhibition causes introns to be skipped during transcription, producing mature mRNA immediately.
Explanation
This question assesses understanding of transcription and RNA processing in eukaryotes. Mature mRNA continues to appear after RNA polymerase II inhibition because splicing and other processing steps act on pre-existing pre-mRNA molecules in the nucleus, independent of new transcription. The drug blocks new pre-mRNA synthesis but does not affect processing enzymes, allowing accumulated pre-mRNA to be shortened by intron removal and exported to the cytoplasm. This demonstrates that RNA processing occurs post-transcriptionally and can proceed on transcripts made before inhibition. A tempting distractor is choice E, which wrongly claims inhibition causes introns to be skipped during transcription, based on the misconception that processing is part of transcription itself. In experiments with transcription inhibitors, note that post-transcriptional steps like splicing can continue, explaining ongoing production of mature RNA.
A student examines transcription of a eukaryotic gene and finds that RNA synthesis begins only when a specific DNA sequence upstream of the coding region is present. When that upstream sequence is deleted, no RNA transcript is produced, even though the coding region and termination signals remain intact. The student uses purified general transcription factors and RNA polymerase II in a cell-free system and gets the same result. Which explanation best accounts for why deletion of the upstream sequence prevents transcription?
The deleted upstream sequence is a promoter required for assembly of transcription factors and RNA polymerase II.
The deleted upstream sequence is a terminator that prevents RNA polymerase from stalling at the start site.
The deleted upstream sequence is an exon that must be removed by splicing to activate RNA polymerase II.
The deleted upstream sequence is an intron required for spliceosome binding before transcription can begin.
The deleted upstream sequence encodes the poly(A) tail, which must be transcribed to initiate RNA synthesis.
Explanation
This question assesses understanding of transcription and RNA processing in eukaryotes. The deleted upstream sequence is the promoter, essential for assembling general transcription factors and RNA polymerase II to initiate transcription, explaining why its removal prevents RNA synthesis despite intact coding and termination regions. The cell-free system with purified components confirms the promoter's direct role in transcription initiation, independent of other cellular factors. Without the promoter, the transcription machinery cannot bind and start synthesizing RNA. A tempting distractor is choice B, which wrongly identifies the sequence as an intron needed for spliceosome binding before transcription, reflecting the misconception that splicing precedes transcription. When deletions affect transcription, identify core elements like promoters first, as they are required for initiating RNA synthesis in eukaryotes.
In eukaryotes, spliceosomes recognize conserved sequences at intron boundaries and a branch-point adenosine within the intron. A mutation changes the branch-point adenosine of intron 3 to cytosine, without altering the 5′ or 3′ splice sites. In mutant cells, intron 3 is frequently retained in the RNA, while other introns are removed normally. Which explanation best accounts for how the branch-point mutation alters RNA processing?
Changing the branch point causes poly(A) polymerase to add the tail within intron 3, deleting exon 4.
Changing the branch point disrupts lariat formation, reducing spliceosome-catalyzed excision of intron 3.
Changing the branch point converts intron 3 into a template for DNA replication, increasing intron copy number.
Changing the branch point blocks addition of the 5′ cap, which is required to define intron boundaries.
Changing the branch point prevents RNA polymerase II from binding the promoter, lowering transcription initiation.
Explanation
This question tests understanding of transcription and RNA processing, specifically the role of the branch point in splicing. The branch-point adenosine is crucial for the first step of splicing, where its 2' OH attacks the 5' splice site to form the lariat structure. When this adenosine is mutated to cytosine in intron 3, the spliceosome cannot efficiently form the lariat intermediate, preventing proper excision of intron 3 even though the splice sites remain intact. This results in frequent retention of intron 3 in the mature RNA, while other introns with normal branch points are spliced out correctly. Choice C incorrectly claims that the branch point mutation affects 5' capping, but capping occurs co-transcriptionally at the 5' end of the RNA and is independent of splicing signals within introns. To understand splicing mechanisms, remember that three conserved sequences are required: the 5' splice site, the branch point, and the 3' splice site—disruption of any one prevents that intron's removal.
During transcription initiation by RNA polymerase II, general transcription factors assemble at a promoter to form a preinitiation complex. A mutation in a promoter reduces binding of TATA-binding protein (TBP) but does not change the coding region or RNA processing signals. In mutant cells, the amount of primary transcript produced from this gene decreases, yet the primary transcripts that are produced show normal splicing and normal 5′ capping. Which explanation best accounts for the decreased RNA output in the mutant cells?
Reduced TBP binding increases transcription termination at the start codon, shortening all primary transcripts.
Reduced TBP binding blocks poly(A) polymerase recruitment, so RNA polymerase II cannot elongate.
Reduced TBP binding lowers the frequency of transcription initiation, decreasing the number of nascent RNAs made.
Reduced TBP binding prevents spliceosome assembly, causing rapid degradation of all primary transcripts.
Reduced TBP binding converts RNA polymerase II into RNA polymerase III, altering transcript abundance.
Explanation
This question tests understanding of transcription and RNA processing, specifically how transcription initiation affects RNA production. TBP (TATA-binding protein) is essential for recruiting RNA polymerase II to the promoter and forming the preinitiation complex. When TBP binding is reduced due to the promoter mutation, fewer preinitiation complexes form, resulting in less frequent transcription initiation events and therefore fewer primary transcripts produced overall. The transcripts that do get made undergo normal processing (capping and splicing) because these processes depend on different factors that recognize sequences within the RNA, not the promoter. Choice B incorrectly links TBP to spliceosome assembly, but TBP functions only in transcription initiation, not in post-transcriptional splicing. To understand transcriptional regulation, remember that promoter mutations affect the quantity of transcripts produced, while mutations in processing signals affect the quality or structure of those transcripts.
A researcher isolates nuclear RNA from eukaryotic cells and treats it with an enzyme that specifically removes the 5′ cap but does not affect the RNA backbone elsewhere. The researcher then repeats the treatment on cytosolic RNA from the same cells. Both samples contain transcripts produced by RNA polymerase II. The 5′ cap is normally added co-transcriptionally after the nascent RNA reaches about 20–30 nucleotides in length. Which explanation best accounts for why most cytosolic RNA molecules resist this decapping treatment compared with nuclear RNA molecules?
Cytosolic RNAs often have 5′ ends protected by bound cap-binding proteins that block enzyme access.
Nuclear RNAs are double-stranded, so the enzyme can remove caps only in the nucleus.
Cytosolic RNAs are synthesized by RNA polymerase I, which adds a different 5′ modification.
Cytosolic RNAs lack 5′ caps because capping occurs only after export from the nucleus.
The 5′ cap is added by the spliceosome, so only unspliced nuclear RNAs can be decapped.
Explanation
This question tests understanding of transcription and RNA processing, specifically the protection of mature mRNAs in the cytoplasm. The 5' cap is added co-transcriptionally in the nucleus to all RNA polymerase II transcripts, so both nuclear and cytoplasmic RNAs initially have caps. However, cytoplasmic mRNAs that are being translated have cap-binding proteins (like eIF4E) bound to their 5' caps, which would physically block access of the decapping enzyme to its substrate. Nuclear RNAs, especially those still undergoing processing, have less stable cap-binding protein associations, making their caps more accessible to the decapping enzyme. Choice B incorrectly claims that capping occurs after nuclear export, but capping actually happens co-transcriptionally when the RNA is only 20-30 nucleotides long. To analyze RNA modifications, consider not just when modifications are added but also what proteins interact with those modifications in different cellular compartments.
A eukaryotic pre-mRNA is transcribed and then processed at its 3′ end. In a wild-type cell, a cleavage and polyadenylation specificity factor (CPSF) recognizes the AAUAAA sequence in the pre-mRNA and promotes cleavage downstream, followed by addition of a poly(A) tail. In mutant cells, the AAUAAA sequence is changed to AACAAA, but transcription through the region continues. Compared with wild type, the predominant RNA from mutant cells is longer and lacks a poly(A) tail. Which explanation best accounts for these observations?
Mutation of AAUAAA reduces CPSF binding, preventing cleavage and subsequent polyadenylation at that site.
Mutation of AAUAAA increases capping efficiency, which blocks poly(A) tail addition.
Mutation of AAUAAA causes the spliceosome to remove the entire 3′ end as an intron.
Mutation of AAUAAA converts RNA polymerase II into RNA polymerase I, extending the transcript.
Mutation of AAUAAA changes the DNA template strand, so the RNA is synthesized in reverse orientation.
Explanation
This question tests understanding of transcription and RNA processing, specifically 3' end formation through cleavage and polyadenylation. The correct answer is A because the AAUAAA hexanucleotide is the binding site for CPSF (cleavage and polyadenylation specificity factor), and mutating it to AACAAA prevents CPSF recognition, blocking cleavage and poly(A) addition. Without proper 3' end processing, RNA polymerase II continues transcribing past the normal termination site, producing longer RNAs lacking poly(A) tails. Answer B incorrectly suggests the spliceosome would remove the 3' end, but spliceosomes recognize specific intron boundaries with GU-AG sequences, not polyadenylation signals—this reflects confusion between splicing and 3' end processing mechanisms. To predict 3' end processing outcomes, identify whether key sequence elements (AAUAAA, downstream elements) and protein factors (CPSF, CstF) can interact properly.
A researcher maps the transcription start site of a eukaryotic gene and finds that RNA synthesis begins at a specific nucleotide downstream of the promoter. A point mutation occurs in the TATA box of the promoter, but the coding region and splice sites remain unchanged. In vitro transcription assays show greatly reduced production of the primary transcript from this promoter. Which explanation best accounts for the reduced primary transcript production?
The TATA mutation prevents intron removal by blocking spliceosome assembly on the pre-mRNA.
The TATA mutation reduces binding of transcription factors needed to recruit RNA polymerase II.
The TATA mutation prevents addition of the poly(A) tail, stopping transcription initiation.
The TATA mutation changes the RNA codons, causing early termination of RNA processing.
The TATA mutation increases rRNA transcription, which competitively removes exons from mRNA.
Explanation
This question tests understanding of transcription and RNA processing, specifically transcription initiation. The TATA box is a core promoter element that binds transcription factors (like TFIID) needed to recruit RNA polymerase II to the transcription start site. A mutation in the TATA box reduces binding of these transcription factors, which decreases RNA polymerase II recruitment and thus reduces production of the primary transcript. The coding region and splice sites remain unchanged, so this is purely a transcription initiation defect, not a processing defect. Choice B incorrectly suggests TATA mutations affect splicing, but the TATA box functions in transcription initiation at the DNA level, not in RNA processing. When analyzing promoter mutations, focus on their effects on transcription factor binding and RNA polymerase recruitment.