AP Biology Quiz: Transcription And Rna Processing

What this quiz covers

This quiz focuses on Transcription And Rna Processing, 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.

Question 1

1. RNA polymerase removed introns during elongation by excising intron sequences from the DNA template strand.
2. Ribozymes in the cytosol degraded intron regions of the RNA after the transcript exited the nucleus.
3. The spliceosome excised introns from the pre-mRNA and ligated adjacent exons to form a continuous RNA. (correct answer)
4. DNA polymerase replaced intron sequences with exon sequences, producing a shorter RNA during transcription.
5. RNA polymerase initiated transcription at exon 2 in later rounds, generating transcripts lacking intron sequences.

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.

Question 2

1. Different RNA polymerases transcribe different exons, so cell type 2 fails to transcribe exon 3.
2. Alternative splicing in cell type 2 excludes exon 3 during processing of the same primary transcript. (correct answer)
3. Cell type 2 edits the DNA to remove exon 3 after transcription, yielding a shorter RNA template.
4. Cell type 2 adds a longer 5′ cap that replaces exon 3 sequence during RNA modification.
5. Cell type 2 degrades exon 3 using ribosomal enzymes, then ligates the remaining RNA in the cytosol.

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.

Question 3

1. The mutated 3′ splice site prevents removal of intron 1, so intron 1 is retained while later introns are spliced. (correct answer)
2. The mutation causes RNA polymerase to skip exon 2 during transcription, fusing exon 1 directly to exon 3.
3. The mutation converts intron 1 into an exon, so exon 1 is deleted during processing.
4. The mutation blocks 5′ capping, so the RNA cannot be spliced and all introns are retained.
5. The mutation changes the template strand, so RNA polymerase inserts intron sequences into the DNA.

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.

Question 4

1. Splicing and other processing can act on pre-existing pre-mRNA molecules even when new transcription stops. (correct answer)
2. RNA polymerase I compensates by transcribing the same gene, producing pre-mRNA for processing.
3. DNA repair enzymes synthesize RNA directly from the mature mRNA template when transcription is blocked.
4. Cytoplasmic enzymes add introns back to mRNA, making it appear shorter as it cycles through the nucleus.
5. 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.

Question 5

1. The deleted upstream sequence is a promoter required for assembly of transcription factors and RNA polymerase II. (correct answer)
2. The deleted upstream sequence is an intron required for spliceosome binding before transcription can begin.
3. The deleted upstream sequence encodes the poly(A) tail, which must be transcribed to initiate RNA synthesis.
4. The deleted upstream sequence is a terminator that prevents RNA polymerase from stalling at the start site.
5. The deleted upstream sequence is an exon that must be removed by splicing to activate RNA polymerase II.

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.

Question 6

1. Changing the branch point disrupts lariat formation, reducing spliceosome-catalyzed excision of intron 3. (correct answer)
2. Changing the branch point prevents RNA polymerase II from binding the promoter, lowering transcription initiation.
3. Changing the branch point blocks addition of the 5′ cap, which is required to define intron boundaries.
4. Changing the branch point causes poly(A) polymerase to add the tail within intron 3, deleting exon 4.
5. Changing the branch point converts intron 3 into a template for DNA replication, increasing intron copy number.

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.

Question 7

1. Reduced TBP binding lowers the frequency of transcription initiation, decreasing the number of nascent RNAs made. (correct answer)
2. Reduced TBP binding prevents spliceosome assembly, causing rapid degradation of all primary transcripts.
3. Reduced TBP binding blocks poly(A) polymerase recruitment, so RNA polymerase II cannot elongate.
4. Reduced TBP binding converts RNA polymerase II into RNA polymerase III, altering transcript abundance.
5. Reduced TBP binding increases transcription termination at the start codon, shortening all primary transcripts.

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.

Question 8

1. Cytosolic RNAs are synthesized by RNA polymerase I, which adds a different 5′ modification.
2. Cytosolic RNAs lack 5′ caps because capping occurs only after export from the nucleus.
3. Cytosolic RNAs often have 5′ ends protected by bound cap-binding proteins that block enzyme access. (correct answer)
4. Nuclear RNAs are double-stranded, so the enzyme can remove caps only in the nucleus.
5. 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.

Question 9

1. Mutation of AAUAAA reduces CPSF binding, preventing cleavage and subsequent polyadenylation at that site. (correct answer)
2. Mutation of AAUAAA causes the spliceosome to remove the entire 3′ end as an intron.
3. Mutation of AAUAAA converts RNA polymerase II into RNA polymerase I, extending the transcript.
4. Mutation of AAUAAA increases capping efficiency, which blocks poly(A) tail addition.
5. 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.

Question 10

1. The TATA mutation reduces binding of transcription factors needed to recruit RNA polymerase II. (correct answer)
2. The TATA mutation prevents intron removal by blocking spliceosome assembly on the pre-mRNA.
3. The TATA mutation prevents addition of the poly(A) tail, stopping transcription initiation.
4. The TATA mutation changes the RNA codons, causing early termination of RNA processing.
5. 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.

Question 11

1. RNA polymerase skipped exon 2 by reading through intron 1 without incorporating complementary ribonucleotides.
2. Alternative splicing produced different exon combinations by selecting distinct splice sites during intron removal. (correct answer)
3. DNA methylation removed exon 2 from the gene in Population 2, preventing its transcription into RNA.
4. Polyadenylation cleaved exon 2 from Population 2 transcripts, leaving exon 1 directly adjacent to exon 3.
5. tRNA editing replaced exon 2 with exon 3 sequences, generating an mRNA lacking exon 2 information.

Explanation: This question tests understanding of transcription and RNA processing, specifically alternative splicing in eukaryotic gene expression. The correct answer is B because alternative splicing allows the same pre-mRNA to be processed differently by selecting different splice sites, resulting in mature mRNAs with different exon combinations—in this case, one population includes all three exons while the other skips exon 2. This mechanism generates protein diversity from a single gene without changing the DNA sequence, as both populations come from identical DNA. The absence of introns in both populations confirms that splicing has occurred, just with different outcomes. Answer C is incorrect because DNA methylation affects transcription regulation but doesn't physically remove exons from the gene—this represents a misconception that epigenetic modifications alter DNA sequence. The key strategy is to recognize that different mRNA products from the same gene indicate alternative splicing rather than DNA changes.

Question 12

1. Mutation of the 5′ splice site reduced spliceosome recognition at intron 1, preventing its excision. (correct answer)
2. Mutation of the 5′ splice site increased RNA polymerase speed, causing intron 1 to be transcribed twice.
3. Mutation of the 5′ splice site blocked poly(A) addition, which is required to remove intron 1 only.
4. Mutation of the 5′ splice site converted intron 1 into an exon, so it was transcribed from DNA again.
5. Mutation of the 5′ splice site prevented export to the cytosol, where intron 1 is normally removed.

Explanation: This question tests understanding of transcription and RNA processing, specifically how splice site mutations affect intron removal. The correct answer is A because the spliceosome requires specific consensus sequences at both the 5' splice site (beginning of intron) and 3' splice site (end of intron) to recognize and remove introns—mutation of the 5' splice site at the exon 1-intron 1 boundary prevents the spliceosome from recognizing intron 1 as an intron. Without proper recognition, intron 1 cannot be excised even though intron 2, with intact splice sites, is removed normally. This selective retention demonstrates that each intron is independently recognized and processed based on its own splice signals. Answer E is incorrect because it suggests introns are removed in the cytosol, but splicing occurs exclusively in the nucleus—this represents a misconception about the cellular location of RNA processing. The strategy is to connect specific splice site mutations with retention of the corresponding intron.

Question 13

1. A poly(A) polymerase added adenines after cleavage at a polyadenylation signal, producing uniform 3′ ends. (correct answer)
2. RNA polymerase used adenine-rich DNA as a template, so the transcript's 3′ end became polyadenylated.
3. The spliceosome inserted adenines at the 3′ end while removing introns, fixing termination positions.
4. A DNA ligase added an adenine-only exon at the gene's 3′ end, shortening transcription variability.
5. A ribosome stalled at the stop codon and protected the 3′ end from degradation, equalizing lengths.

Explanation: This question tests understanding of transcription and RNA processing, specifically 3' end formation and polyadenylation in eukaryotes. The correct answer is A because poly(A) polymerase adds approximately 200 adenine residues to the 3' end of pre-mRNA after cleavage at the polyadenylation signal, creating a uniform termination point that is not encoded in the DNA. This processing step occurs in the nucleus and explains why transcripts in the second condition have consistent 3' ends with poly(A) tails, unlike the variable termination in the first condition. The poly(A) tail is added post-transcriptionally, which is why the adenines are not found in the DNA sequence. Answer B is incorrect because it suggests the poly(A) tail comes from the DNA template, representing a misconception that all RNA sequences must be complementary to DNA—polyadenylation is a post-transcriptional modification. The strategy is to recognize that sequences present in RNA but absent from DNA indicate post-transcriptional modifications.

Question 14

1. Reduced sigma factor binding decreased transcription initiation frequency, lowering production of the full-length RNA. (correct answer)
2. Reduced sigma factor binding prevented spliceosome assembly, causing removal of the intergenic region from RNA.
3. The promoter mutation increased polyadenylation at the 3′ end, shortening transcripts before gene Y.
4. The promoter mutation blocked ribosome binding, causing rapid degradation of the RNA spanning X and Y.
5. The promoter mutation caused RNA polymerase to use the coding strand as template, reducing detectable RNA.

Explanation: 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.

Question 15

1. Eukaryotic genes often contain introns that are transcribed into pre-mRNA and later removed by splicing. (correct answer)
2. Bacterial RNA polymerase removes internal RNA segments during elongation, so no extra sequences appear.
3. Eukaryotic RNA polymerase copies both DNA strands, creating extra internal sequences that are later deleted.
4. Bacteria add a 5' cap that blocks transcription of internal regions, preventing extra sequences in primary RNA.
5. Eukaryotes use DNA polymerase to synthesize pre-mRNA, which introduces extra sequences not present in DNA.

Explanation: This question assesses understanding of transcription and RNA processing in eukaryotes. The extra internal sequences in eukaryotic primary transcripts are introns, which are transcribed from the DNA template but removed by splicing to form mature mRNA, unlike bacterial genes that typically lack introns. In bacteria, the primary transcript matches the final mRNA because there are no introns to remove, highlighting a key difference in gene structure between prokaryotes and eukaryotes. Both use DNA as the template, so the difference arises from the presence of introns in eukaryotic genes. A tempting distractor is choice C, which incorrectly states that eukaryotic RNA polymerase copies both DNA strands to create extra sequences, stemming from the misconception that transcription involves both strands simultaneously. To compare transcription across organisms, remember that eukaryotic RNA processing includes intron removal, which is absent in most bacterial genes.

Question 16

1. The altered 3′ splice site reduced spliceosome recognition, preventing excision of that intron from pre-mRNA. (correct answer)
2. The altered 3′ splice site caused RNA polymerase to terminate early, producing a longer transcript with extra intron.
3. The altered 3′ splice site increased DNA replication errors, inserting intron sequence multiple times into the gene.
4. The altered 3′ splice site redirected the transcript to the nucleolus, where introns are retained by default.
5. The altered 3′ splice site blocked addition of the 5′ cap, which is required for transcription elongation.

Explanation: This question tests understanding of transcription and RNA processing, specifically the importance of splice site consensus sequences. The correct answer is A because the AG dinucleotide at the 3' splice site is essential for spliceosome recognition and intron removal—changing AG to AA prevents the spliceosome from recognizing this as a valid 3' splice site. Without proper recognition of both the 5' splice site (GU) and 3' splice site (AG), the spliceosome cannot excise this intron, leading to its retention in the RNA. Other introns with intact splice sites are removed normally, demonstrating the sequence-specific nature of splicing. Answer C is incorrect because it suggests mutations in RNA affect DNA sequence through replication errors, representing a misconception about the direction of genetic information flow. The key strategy is to connect splice site consensus sequence mutations with failure of intron removal.

Question 17

1. The spliceosome fails to recognize the mutated splice site, so the intron is retained in the processed RNA. (correct answer)
2. RNA polymerase transcribes past the normal termination site because the splice site mutation alters DNA topology.
3. The mutation converts the intron into an exon, so RNA polymerase copies additional DNA downstream.
4. The mutation blocks addition of the 5' cap, preventing exon ligation and increasing transcript length.
5. The mutation causes the intron to be removed twice, duplicating adjacent exon sequence in the cytoplasmic RNA.

Explanation: 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.

Question 18

1. Without the polyadenylation signal, cleavage is inefficient, so RNA polymerase often continues transcription downstream. (correct answer)
2. Without the polyadenylation signal, the 5′ cap cannot be added, so transcription initiation shifts downstream.
3. Without the polyadenylation signal, introns cannot be recognized, so splicing adds extra downstream exons.
4. Without the polyadenylation signal, the DNA template strand is degraded, causing polymerase to copy beyond the gene.
5. Without the polyadenylation signal, RNA polymerase switches to using RNA as a template, extending the transcript.

Explanation: This question tests understanding of transcription and RNA processing, specifically the coupling of 3' end formation with transcription termination. The polyadenylation signal serves dual functions: it directs cleavage/polyadenylation of the RNA and triggers transcription termination by RNA polymerase II. Without this signal, the cleavage and polyadenylation machinery cannot recognize where to process the 3' end, so poly(A) tails are rarely added. More importantly, RNA polymerase II lacks the signal to terminate transcription and continues elongating far past the normal gene endpoint, producing aberrantly long transcripts. Choice B incorrectly links the poly(A) signal to 5' capping, but capping occurs co-transcriptionally at the beginning of transcription and is independent of 3' end signals. When analyzing 3' end formation, remember that the polyadenylation signal coordinates both RNA processing (cleavage and poly(A) addition) and transcription termination.

Question 19

1. Reduced snRNA base-pairing lowers spliceosome recognition of the 5′ splice site, decreasing intron 4 excision. (correct answer)
2. Reduced snRNA base-pairing increases RNA polymerase binding to intron 4, extending transcription through the intron.
3. Reduced snRNA base-pairing prevents addition of the poly(A) tail, which normally removes introns from the RNA.
4. Reduced snRNA base-pairing causes exon 4 to be removed as an intron, eliminating exon 4 from all transcripts.
5. Reduced snRNA base-pairing converts the intron into DNA, so intron 4 is inserted back into the genome.

Explanation: This question tests understanding of transcription and RNA processing, specifically the base-pairing requirement for spliceosome assembly. The U1 snRNA within the U1 snRNP must base-pair with the 5' splice site consensus sequence for spliceosome assembly to begin. When mutations reduce this complementarity at intron 4's 5' splice site, U1 snRNP binding is weakened, decreasing the efficiency of spliceosome assembly specifically at intron 4. This leads to increased retention of intron 4 in the mature RNA, while other introns with normal 5' splice sites are recognized and removed efficiently. Choice D incorrectly suggests that poor snRNA binding would cause exon removal, but reduced spliceosome function leads to intron retention, not exon skipping. To predict splicing outcomes, remember that snRNA-splice site complementarity determines splicing efficiency—reduced base-pairing leads to intron retention, not exon loss.

Question 20

1. RNA polymerase removes introns during transcription by skipping intron regions on the DNA template.
2. Spliceosomes remove introns from the primary transcript and ligate exons to form mature mRNA. (correct answer)
3. Ribozymes in the cytoplasm cut out exons and retain introns to create a shorter RNA molecule.
4. DNA polymerase excises intron sequences from the gene after transcription to match the cytoplasmic RNA.
5. Helicase separates RNA strands in the nucleus, causing intron sequences to degrade before export.

Explanation: 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.