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Transcription and RNA Processing - AP Biology

  1. A researcher mutates the 5' splice site of an intron in a eukaryotic gene but leaves all exon sequences unchanged. Nuclear RNA from the mutant cell includes the intron sequence, and cytoplasmic RNA contains a longer transcript than in wild type. Sequencing shows the intron remains between the same two exons in the cytoplasmic RNA. The amount of transcription initiation measured at the promoter is unchanged. Which explanation best accounts for the longer cytoplasmic RNA in the mutant?
    Answer: The spliceosome fails to recognize the mutated splice site, so the intron is retained in the processed RNA. This question assesses understanding of transcription and RNA processing in eukaryotes. The mutation at the 5' splice site prevents the spliceosome from recognizing and removing the intron, leading to its retention in the processed RNA and a longer cytoplasmic transcript. Transcription initiation remains unchanged, and the intron stays between the same exons, indicating the issue is specifically with splicing rather than transcription or DNA alterations. This results in an abnormal mRNA that includes the intron sequence, increasing its length compared to wild type. A tempting distractor is choice B, which incorrectly claims the mutation causes transcription past the termination site due to altered DNA topology, stemming from the misconception that splice sites directly affect transcription elongation. For mutations affecting RNA length, evaluate impacts on splicing first, as it commonly alters transcript composition without changing transcription rates.
  2. In a eukaryotic nucleus, an enzyme that adds a 7-methylguanosine cap to the 5′ end of nascent RNA is inactivated. Transcription initiation and elongation still occur, and introns are removed from the primary transcript. When the 5′ ends of the processed RNAs are sequenced, they begin with the first transcribed nucleotide rather than a modified guanine linked by a 5′–5′ triphosphate bond. Which explanation best accounts for this sequencing result?
    Answer: The 5′ cap is added after transcription begins by a capping enzyme, so inactivation prevents the modification. This question assesses understanding of transcription and RNA processing in eukaryotes, emphasizing the role of the 5′ cap in mRNA maturation. Inactivating the capping enzyme leads to RNAs starting with the first transcribed nucleotide instead of a modified guanine, while transcription and splicing continue, indicating capping is an independent modification added early in transcription. The absence of the 5′–5′ triphosphate-linked cap in sequencing confirms failed enzymatic addition without halting RNA synthesis. Choice B is correct as it explains that the capping enzyme adds the structure post-initiation, accounting for the sequencing result. A tempting distractor is choice A, which erroneously suggests the cap is transcribed from a promoter, based on the misconception that caps are DNA-templated like RNA sequences. To solve such questions, remember that the 5′ cap is a non-templated modification added enzymatically, and check if other processes like transcription persist to isolate the affected step.
  3. In bacteria, a gene is transcribed into an RNA that is immediately usable as mRNA without intron removal. A student claims that bacterial mRNAs commonly undergo spliceosome-mediated intron excision in the cytoplasm. In an experiment, RNA from a bacterial culture is analyzed and shows continuous sequence matching the coding region with no missing internal segments. Which explanation best accounts for the absence of spliceosome-mediated splicing in the bacterial RNA sample?
    Answer: Bacteria generally lack spliceosomes, and most bacterial genes do not contain introns to remove. This question tests understanding of transcription and RNA processing differences between prokaryotes and eukaryotes. Bacterial RNA shows continuous sequence matching the coding region with no missing segments because bacteria generally lack spliceosomes and most bacterial genes do not contain introns to remove. Bacterial genes typically have continuous coding sequences, so their mRNA is immediately functional after transcription without requiring intron removal. This is a fundamental difference from eukaryotic genes, which commonly contain introns that must be removed by spliceosomes. Choice E incorrectly states bacterial RNA is synthesized in the nucleus, but bacteria lack a nucleus and perform transcription in the cytoplasm. When comparing prokaryotic and eukaryotic gene expression, remember that introns and spliceosome-mediated splicing are primarily eukaryotic features.
  4. A eukaryotic gene can produce two different mature mRNAs from the same primary transcript. In isoform 1, exon 2 is joined to exon 3. In isoform 2, exon 2 is absent and exon 1 is joined directly to exon 3. The DNA sequence of the gene is identical in both cell types, and transcription start and stop sites are unchanged. Which explanation best accounts for the production of the two different mature mRNAs?
    Answer: Alternative splicing selects different combinations of exons from the same pre-mRNA during RNA processing. This question assesses understanding of transcription and RNA processing in eukaryotes. The two mRNA isoforms arise from alternative splicing, where different exon combinations are selected from the same pre-mRNA, such as skipping exon 2 in one isoform to join exon 1 directly to exon 3. The DNA sequence and transcription sites are identical in both cell types, indicating the variation occurs during RNA processing rather than at the DNA or transcription level. This allows a single gene to produce multiple protein variants through regulated splicing patterns. A tempting distractor is choice B, which incorrectly suggests RNA polymerase switches template strands to change transcribed exons, based on the misconception that transcription direction alters exon inclusion. To explain mRNA variants from the same gene, consider alternative splicing as a key mechanism for generating diversity without genomic changes.
  5. A bacterial operon is transcribed from a single promoter into one continuous RNA that includes coding regions for genes X and Y. A mutation is introduced into the promoter that reduces binding of the sigma factor but does not alter the coding sequences. When cells are shifted to conditions that normally activate this operon, the amount of RNA spanning the X–Y intergenic region decreases relative to wild type. No introns are present in the operon. Which explanation best accounts for the reduced RNA level in the mutant?
    Answer: Reduced sigma factor binding decreased transcription initiation frequency, lowering production of the full-length RNA. This question tests understanding of transcription and RNA processing, specifically bacterial transcription initiation. The correct answer is A because sigma factor is essential for RNA polymerase binding to promoters in bacteria, and reduced sigma factor binding due to the promoter mutation decreases the frequency of transcription initiation, resulting in less RNA production from the operon. Since bacterial operons produce polycistronic mRNA containing multiple genes, reduced initiation affects the entire transcript including the X-Y intergenic region. The mutation specifically affects promoter recognition without changing the coding sequences or transcription elongation. Answer B is incorrect because bacteria lack spliceosomes and introns—this represents a misconception that confuses eukaryotic RNA processing with bacterial systems. The strategy is to recognize that promoter mutations affecting transcription factor binding reduce RNA production in bacteria.
  6. 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?
    Answer: Changing the branch point disrupts lariat formation, reducing spliceosome-catalyzed excision of intron 3. 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.
  7. 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?
    Answer: Alternative splicing in cell type 2 excludes exon 3 during processing of the same primary transcript. 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.
  8. A scientist engineers a eukaryotic gene so that the polyadenylation signal in the transcribed region is mutated. Transcription initiation still occurs, and a primary transcript is produced. Compared with wild type, the RNA isolated from the nucleus is longer at its 3' end and shows reduced addition of a poly(A) tail. Splicing patterns within the transcript are unchanged. Which explanation best accounts for the longer 3' end observed in the mutant nuclear RNA?
    Answer: Mutation of the polyadenylation signal reduces cleavage at the normal 3' processing site, extending the transcript. This question assesses understanding of transcription and RNA processing in eukaryotes. Mutating the polyadenylation signal prevents proper cleavage and poly(A) tail addition at the normal 3' site, causing RNA polymerase to continue transcribing downstream sequences and resulting in a longer 3' end in the nuclear RNA. Transcription initiation occurs normally, but the lack of the signal reduces 3' processing, leading to extended transcripts without affecting splicing patterns. The reduced poly(A) tail confirms the mutation's impact on end processing. A tempting distractor is choice B, which incorrectly links the mutation to preventing 5' capping and extra transcription, reflecting the misconception that 3' signals control 5' modifications. For mutations in processing signals, evaluate their specific roles in transcript termination and modification to explain changes in RNA length.
  9. A eukaryotic gene has two alternative 3′ splice sites within the same intron. In cell type 1, splicing uses the upstream 3′ splice site; in cell type 2, splicing uses the downstream 3′ splice site. The 5′ splice site is identical in both cell types, and transcription rates are equal. When mature RNAs are compared, cell type 2 produces an mRNA that is shorter by 18 nucleotides. Which explanation best accounts for the 18-nucleotide difference?
    Answer: Cell type 2 uses a downstream 3′ splice site, removing an additional 18 nucleotides from the transcript. This question tests understanding of transcription and RNA processing, specifically alternative splicing at different 3' splice sites. The correct answer is A because using the downstream 3' splice site in cell type 2 removes 18 additional nucleotides compared to using the upstream site in cell type 1. Since the 5' splice site is identical and only the 3' splice site choice differs, the 18-nucleotide difference represents the sequence between the two alternative 3' splice sites that is included in cell type 1 but removed in cell type 2. Answer C incorrectly suggests alternative splicing affects promoter selection, but splice-site choice occurs post-transcriptionally and cannot change which promoter was used—this reflects confusion between transcriptional and post-transcriptional regulation. When analyzing alternative splicing outcomes, map the splice sites used to predict which sequences will be included or excluded.
  10. A scientist designs an antisense oligonucleotide that binds to the branch point sequence within an intron of a eukaryotic pre-mRNA. After treatment, sequencing of nuclear RNA shows that this intron is frequently retained, while other introns in the same transcript are removed normally. The 5′ cap and poly(A) tail are still detected on the RNA molecules. Which explanation best accounts for the intron retention after treatment?
    Answer: Blocking the branch point prevented lariat formation required for spliceosome-mediated intron removal at that site. This question tests understanding of transcription and RNA processing, specifically the mechanism of spliceosome-mediated splicing. The correct answer is A because the branch point adenine within the intron is essential for the first step of splicing, where it attacks the 5' splice site to form a lariat intermediate—blocking this sequence with an antisense oligonucleotide prevents lariat formation and thus prevents intron removal. Without access to the branch point, the spliceosome cannot catalyze the transesterification reactions needed for splicing, causing retention of that specific intron. Other introns with accessible branch points are removed normally, and the presence of 5' cap and poly(A) tail indicates other processing steps proceed. Answer D is incorrect because exonucleases would degrade RNA from the ends inward and cannot selectively remove internal exons—this represents a misconception about RNA degradation mechanisms. The strategy is to recognize that blocking key splicing signals prevents intron removal.
  11. 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?
    Answer: The spliceosome excised introns from the pre-mRNA and ligated adjacent exons to form a continuous RNA. 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.
  12. 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?
    Answer: Splicing and other processing can act on pre-existing pre-mRNA molecules even when new transcription stops. 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.
  13. In a eukaryotic cell, a gene contains three exons separated by two introns. A researcher isolates nuclear RNA shortly after transcription and observes that the RNA includes sequences complementary to both introns and exons. Later, RNA isolated from the cytoplasm contains the exon sequences joined together with the intron sequences absent. No changes are detected in the DNA sequence of the gene. Which explanation best accounts for the difference between the nuclear RNA and cytoplasmic RNA?
    Answer: Spliceosomes remove introns from the primary transcript and ligate exons to form mature mRNA. This question assesses understanding of transcription and RNA processing in eukaryotes. The nuclear RNA includes both intron and exon sequences because it is the primary transcript directly produced by RNA polymerase, which copies the entire gene region including introns. The cytoplasmic RNA lacks introns and has exons joined together due to RNA splicing, where spliceosomes remove introns and ligate exons to form mature mRNA, explaining the difference without any changes to the DNA sequence. This process occurs post-transcriptionally in the nucleus before export to the cytoplasm. A tempting distractor is choice A, which incorrectly suggests that RNA polymerase skips introns during transcription, reflecting the misconception that introns are not transcribed at all. To approach similar questions, remember that eukaryotic genes contain introns that are transcribed but removed during RNA processing to produce functional mRNA.
  14. A researcher isolates nuclei from eukaryotic cells and measures RNA lengths from a single gene. The primary transcript averages 2,400 nucleotides and contains one intron of 900 nucleotides flanked by two exons. After RNA processing, the most abundant RNA from this gene averages 1,500 nucleotides and contains the same exon sequences as the primary transcript. No changes are detected in the DNA sequence of the gene. Which explanation best accounts for the 900-nucleotide decrease in RNA length after processing?
    Answer: The intron was excised during splicing, and exons were ligated together. This question tests understanding of transcription and RNA processing, particularly the mechanism of intron removal. The primary transcript is 2,400 nucleotides long and contains a 900-nucleotide intron, while the processed RNA is 1,500 nucleotides (exactly 900 nucleotides shorter). This 900-nucleotide decrease precisely matches the intron size, indicating that the intron was excised during splicing and the flanking exons were ligated together by the spliceosome. The DNA sequence remains unchanged, confirming this is an RNA processing event, not a DNA alteration. Choice A incorrectly suggests RNA polymerase can skip DNA regions during transcription, but RNA polymerase must transcribe the template continuously. To solve RNA length problems, calculate the difference between primary and mature RNA lengths and compare this to known intron sizes.
  15. 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?
    Answer: Cytosolic RNAs often have 5′ ends protected by bound cap-binding proteins that block enzyme access. 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.
  16. During transcription of a eukaryotic gene, RNA polymerase moves along the DNA template strand and synthesizes RNA. A student is given a short DNA template-strand sequence: 3'-T A C G G A T-5'. The student predicts the RNA sequence produced from this template in the 5' to 3' direction, before any RNA processing occurs. Which RNA sequence best matches the transcript segment synthesized from the template provided?
    Answer: 5'-A U G C C U A-3' This question assesses understanding of transcription and RNA processing in eukaryotes. The RNA sequence is synthesized complementary to the DNA template strand in the 5' to 3' direction, so for the template 3'-TACGGAT-5', the RNA is 5'-AUGCCUA-3', matching the pre-processing transcript segment. RNA polymerase reads the template from 3' to 5', adding nucleotides antiparallel with U replacing T. This direct transcription product occurs before any splicing or modifications. A tempting distractor is choice D, which is 5'-UAC GGA U-3', incorrect because it matches the coding strand sequence instead of the complementary transcript, reflecting the misconception of confusing template and coding strands. When predicting RNA from DNA, always transcribe complementary to the template strand in 5' to 3' orientation to avoid strand confusion.
  17. A eukaryotic gene contains an intron with the sequence 5′-GU…AG-3′ at its ends. A mutation changes the 3′ splice site from AG to AA, while leaving the rest of the pre-mRNA unchanged. Transcription and 5′ capping occur normally. After processing, most RNA molecules still contain the intron sequence. Which explanation best accounts for intron retention in the processed RNA?
    Answer: The spliceosome requires an AG dinucleotide at the 3′ splice site; changing it disrupts intron removal. This question tests understanding of transcription and RNA processing, specifically the requirement for conserved splice-site sequences. The correct answer is A because the spliceosome requires an AG dinucleotide at the 3' splice site for the second transesterification reaction that ligates exons together, and changing AG to AA prevents intron removal. The GU at the 5' splice site and AG at the 3' splice site are nearly invariant sequences essential for spliceosome recognition and catalysis. Answer E incorrectly suggests poly(A) polymerase would add adenines at the splice junction, but polyadenylation occurs only at the 3' end of the transcript after cleavage, not at internal splice sites—this reflects confusion between splicing and 3' end processing locations. To predict splicing outcomes, check whether both the 5' GU and 3' AG consensus sequences are intact.
  18. A eukaryotic gene is transcribed into a primary RNA transcript containing exons and introns. A point mutation changes the 5′ splice site of one intron from GU to AU in the RNA sequence produced. The amount of primary transcript synthesized is unchanged compared with wild type. When processed RNA is analyzed, a longer RNA species accumulates that retains the mutated intron sequence. Which explanation best accounts for the accumulation of the longer RNA species?
    Answer: The mutated splice site reduces recognition by spliceosome components, preventing efficient intron removal from the transcript. This question assesses understanding of transcription and RNA processing in eukaryotes, specifically how splice site mutations impact mRNA maturation. The GU to AU mutation at the 5′ splice site leads to accumulation of longer RNA retaining the intron, with unchanged primary transcript levels, pointing to defective intron removal rather than reduced synthesis. Splice sites are critical for spliceosome recognition, so altering them prevents efficient excision and causes intron retention. Choice A is correct because it describes how the mutation impairs spliceosome function, explaining the longer RNA species. A tempting distractor is choice E, which incorrectly involves ribosomes skipping introns, arising from the misconception that translation machinery handles splicing. When evaluating mutations, consider their location in splice sites and effects on processing steps like intron removal to predict outcomes.
  19. A eukaryotic cell line is treated with a chemical that inhibits poly(A) polymerase in the nucleus. Shortly afterward, newly synthesized RNA from several genes is still detected, and intron removal proceeds normally. However, the 3′ ends of these RNAs lack long stretches of adenosines that are present in untreated cells. No changes are observed in the DNA sequences of the genes. Which outcome best accounts for the observed difference in the RNA molecules?
    Answer: Poly(A) tails are added post-transcriptionally to the 3′ end of pre-mRNA by poly(A) polymerase. This question assesses understanding of transcription and RNA processing in eukaryotes, particularly the addition of poly(A) tails to mRNA. Inhibiting poly(A) polymerase results in RNAs lacking long adenosine stretches at the 3′ end, despite normal transcription and intron removal, showing that polyadenylation is a separate post-transcriptional step not encoded in the DNA. The unchanged DNA sequences confirm the absence of adenosines is due to failed enzymatic addition rather than a template change. Choice C is correct because poly(A) polymerase adds the tail after transcription, explaining the observed difference in treated cells. A tempting distractor is choice D, which mistakenly claims adenosine-rich DNA is transcribed into the tail, arising from the misconception that poly(A) sequences are directly templated like exons. When analyzing RNA modifications, distinguish between transcribed features and post-transcriptional additions like polyadenylation to identify the correct mechanism.
  20. In a eukaryotic nucleus, a gene contains three exons separated by two introns. After transcription, the primary RNA transcript is analyzed and found to include all five regions in the same 5′→3′ order as the DNA template. A second RNA sample from the same gene is shorter and lacks the sequences that correspond to the introns, while the exon sequences remain contiguous. No changes are detected in the DNA sequence. Which explanation best accounts for the difference between the two RNA samples?
    Answer: A spliceosome removed intron sequences from the primary transcript and ligated adjacent exons together. This question assesses understanding of transcription and RNA processing in eukaryotes, specifically how primary transcripts are modified to form mature mRNA. The primary RNA transcript includes both exons and introns in the same order as the DNA template, reflecting that transcription copies the entire gene region without skipping introns. The shorter RNA lacks intron sequences with contiguous exons, indicating post-transcriptional removal of non-coding regions while preserving coding sequences, and no DNA changes confirm this is an RNA-level process. Choice B is correct because a spliceosome excises introns from the primary transcript and joins exons, accounting for the observed difference between the samples. A tempting distractor is choice A, which wrongly suggests RNA polymerase skips introns during transcription, stemming from the misconception that introns are not transcribed at all. To approach similar questions, remember that eukaryotic transcription produces a pre-mRNA with introns, which are removed during RNA splicing to yield mature mRNA.

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