Genetics Quiz: Mrna Processing

What this quiz covers

This quiz focuses on Mrna Processing, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

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. Use of an alternative start codon 150 bp upstream of the normal one.
2. A tandem duplication of the 150 bp intron in the genomic DNA.
3. Retention of the 150 bp intron between E1 and E2 due to a splicing defect. (correct answer)
4. Nonsense-mediated decay of the 250 bp transcript, leading to accumulation of a precursor.

Explanation: When you encounter RT-PCR questions, focus on what happens during mRNA processing and how PCR amplifies the final mRNA product, not the genomic DNA. In healthy tissue, the RT-PCR produces a 250 bp product from mRNA where the intron has been properly spliced out. The diseased tissue shows both this normal 250 bp band plus a new 400 bp band. Since the difference is exactly 150 bp (the intron size), this strongly suggests that some mRNA molecules retain the intron due to defective splicing. The 400 bp product represents: E1 + intron + E2, while the 250 bp product represents: E1 + E2 (intron removed). This makes C correct. Option A is wrong because alternative start codons don't create the observed pattern—RT-PCR primers target specific exon sequences, and using a different start codon wouldn't add 150 bp between the primer sites. Option B incorrectly focuses on genomic DNA changes, but RT-PCR analyzes mRNA, so genomic duplications wouldn't directly create this banding pattern in the cDNA product. Option D misunderstands nonsense-mediated decay, which degrades faulty transcripts rather than creating larger, stable products that would show up as distinct bands. Study tip: In RT-PCR questions, always trace what happens to the mRNA, not the genomic DNA. When you see two bands differing by exactly an intron's length, think splicing defects—it's a classic pattern where some transcripts are properly processed while others aren't.

Question 2

1. Covalent circularization of the mature mRNA, ligating the 3' end to the 5' end to prevent exonuclease access.
2. Formation of a stable stem-loop structure at the 3' end of the transcript, which is bound by a protective protein. (correct answer)
3. The spliceosome remains permanently attached to the 3' end of the final exon after splicing is complete.
4. A second, modified 7-methylguanosine cap is added to the 3' end of the transcript, mirroring the 5' end.

Explanation: When you encounter questions about mRNA processing and stability, focus on the essential functions each modification serves. The 5' cap prevents degradation by 5' exonucleases, while poly(A) tails typically protect against 3' exonucleases and aid translation. If an organism lacks polyadenylation but maintains stable mRNAs, it must have evolved alternative 3' end protection mechanisms. The most plausible alternative is answer B: formation of stable stem-loop structures bound by protective proteins. This mechanism mimics how some viruses and certain cellular transcripts (like histone mRNAs) achieve stability without poly(A) tails. The secondary structure creates a physical barrier preventing 3' to 5' exonuclease access, while protein binding provides additional protection and potentially regulatory functions. Answer A is implausible because covalent circularization would prevent ribosome release and recycling during translation termination. Answer C incorrectly assumes the spliceosome could function as a protective complex—spliceosomes disassemble after completing their catalytic function and wouldn't provide stable 3' protection. Answer D is biochemically unrealistic because the 7-methylguanosine cap requires specific 5' triphosphate chemistry that doesn't exist at processed 3' ends. For genetics exams, remember that when organisms lack standard molecular machinery, they often evolve functionally equivalent alternatives. Focus on the underlying purpose of each modification (protection from nucleases, translation enhancement) rather than memorizing only the canonical mechanisms. This approach helps you evaluate novel scenarios logically.

Question 3

1. The branch point adenosine within the first intron of the pre-mRNA.
2. The polyadenylation signal and the 3' untranslated region.
3. The entire spliceosome complex, as trans-splicing is self-catalyzed.
4. The 5' cap and the 5' untranslated region of the first exon. (correct answer)

Explanation: When you encounter questions about RNA processing in parasites like trypanosomes, focus on how their mechanisms differ from conventional eukaryotic systems and what functions are being replaced or modified. Trans-splicing in trypanosomes is a unique process where a pre-formed Spliced Leader (SL) RNA containing a 5' cap is joined to the 5' end of pre-mRNAs. Since this 39-nucleotide SL RNA already carries the 5' cap structure and includes untranslated sequences, it essentially replaces the need for the conventional 5' cap and 5' UTR that would normally be present on the first exon of a typical pre-mRNA. The SL RNA provides these essential 5' end modifications in trans (from a separate molecule) rather than in cis (on the same molecule). This is why answer D is correct. Let's examine why the other options are incorrect: A is wrong because the branch point adenosine is still required for the trans-splicing mechanism to work—the chemistry of splicing still needs this key nucleotide. B is incorrect because trans-splicing affects the 5' end processing, not the 3' end modifications like polyadenylation. C is false because trans-splicing still requires spliceosome components; it's not a self-catalyzed ribozyme reaction. Remember that trans-splicing questions often test whether you understand that this process specifically addresses 5' end modifications. The key insight is recognizing that when an external RNA provides the 5' cap and UTR, the original pre-mRNA no longer needs these features.

Question 4

1. Slower elongation allows the nascent RNA to form complex secondary structures that physically mask strong splice sites, forcing the spliceosome to use weaker ones.
2. Slower elongation reduces the frequency of transcriptional errors, which would otherwise be misinterpreted as aberrant splice sites by the spliceosome.
3. A slow polymerase has a less phosphorylated C-terminal domain (CTD), which has a higher affinity for the specific splicing factors that recognize weak exons.
4. A slow polymerase allows more time for splicing factors to be recruited to the weak splice sites on the nascent transcript before stronger downstream sites are synthesized. (correct answer)

Explanation: When you encounter questions about RNA processing and transcription dynamics, focus on the kinetic competition between different cellular processes occurring simultaneously. The key insight here is that transcription and splicing happen co-transcriptionally—while the RNA is still being made. This creates a race between the polymerase synthesizing new sequence and the spliceosome assembling on existing splice sites. When RNA Polymerase II moves slowly, it gives splicing factors more time to recognize and bind to weak splice sites before stronger, more competitive sites downstream are even transcribed. Think of it like giving someone extra time to notice a quiet conversation before a louder one starts in the same room. Option A is incorrect because RNA secondary structures don't selectively mask strong splice sites while exposing weak ones—if anything, structures would more likely interfere with weak site recognition. Option B misses the point entirely; transcriptional errors aren't the mechanism driving alternative splicing patterns, and the spliceosome doesn't confuse errors with splice sites. Option C contains a factual error—slower elongation doesn't necessarily correlate with less CTD phosphorylation, and CTD phosphorylation status affects which processing factors are recruited, not their affinity for weak versus strong sites. The correct answer is D because it captures the fundamental principle: kinetic coupling between transcription and splicing creates a timing-dependent competition for splice site recognition. Study tip: For RNA processing questions, always consider the temporal relationship between transcription and RNA processing—they're coupled processes, not sequential ones.

Question 5

1. The protein components of the snRNPs that recognize splice sites.
2. ATP-dependent helicases that are required to rearrange the spliceosome.
3. The catalytic core formed by the snRNAs, such as U2 and U6 snRNA. (correct answer)
4. The C-terminal domain of RNA Polymerase II that coordinates splicing with transcription.

Explanation: When you encounter questions about RNA splicing mechanisms, focus on identifying which components perform the actual catalytic chemistry versus those that provide regulatory or structural support. Group II self-splicing introns are remarkable because they can remove themselves from RNA transcripts without any protein machinery. The key insight is that these introns fold into complex three-dimensional structures that create their own catalytic center. This ribozyme activity directly replaces the catalytic function normally performed by the spliceosome's RNA components. In normal spliceosome-mediated splicing, the catalytic chemistry is carried out by the RNA portions of small nuclear ribonucleoproteins (snRNPs), particularly U2 and U6 snRNAs. These snRNAs form the active site that catalyzes both transesterification reactions needed to remove introns and form lariat intermediates. In Group II self-splicing introns, the intron's own folded RNA structure creates an equivalent catalytic center, making answer C correct. Answer A is incorrect because while snRNP proteins do help recognize splice sites in spliceosomes, Group II introns have their own sequence elements for splice site recognition built into their structure. Answer B is wrong because although ATP-dependent helicases facilitate spliceosome assembly and rearrangement, self-splicing introns don't require these energy-dependent conformational changes. Answer D is incorrect because the C-terminal domain of RNA Polymerase II coordinates splicing with transcription but isn't directly involved in the catalytic mechanism itself. Remember: in splicing questions, distinguish between components that perform catalysis (the actual chemistry) versus those that provide recognition, regulation, or coordination functions.

Question 6

1. The branch point sequence.
2. The Kozak consensus sequence.
3. The 5' splice site consensus sequence.
4. The polyadenylation signal sequence. (correct answer)

Explanation: The polyadenylation signal (typically AAUAAA) is a sequence transcribed from the DNA that is located in the final exon of a gene. After transcription, this signal resides in the pre-mRNA upstream of where the transcript will be cleaved. Following cleavage and polyadenylation, this signal sequence becomes part of the 3' untranslated region (3' UTR) of the mature mRNA. The branch point and 5' splice site are located within introns and are removed during splicing. The Kozak sequence surrounds the start codon and is part of the 5' UTR.

Question 7

1. The mutation altered an exonic splicing enhancer (ESE), causing the entire exon to be skipped.
2. The mutation inadvertently created a new 5' splice site (a cryptic splice site) within the exon. (correct answer)
3. The change in codon usage caused by the mutation led to premature termination of translation.
4. The silent mutation disrupted a binding site for a miRNA, leading to improper mRNA degradation.

Explanation: When you encounter a question about silent mutations causing unexpected protein changes, think about how mutations can affect RNA processing beyond just changing amino acids. The key insight is that exonic sequences don't just code for proteins—they also contain regulatory elements that guide splicing. The correct answer is B because a G-to-A mutation can create a new splice site within the exon. Splice sites follow specific sequence patterns, particularly the GT dinucleotide at 5' splice sites. If the mutation creates a sequence that resembles a splice site (called a cryptic splice site), the splicing machinery may recognize and use this new site instead of the normal one. This would cause premature cleavage of the exon, removing part of the coding sequence and resulting in a shorter, truncated protein when translated. Let's examine why the other options don't fit: A is incorrect because skipping the entire exon would likely cause a frameshift or remove essential protein domains, but the question specifies the mutation is in the center of the exon, suggesting partial rather than complete exon removal. C is wrong because silent mutations by definition don't change amino acids, so codon usage alone wouldn't cause premature termination. D doesn't work because miRNA binding typically occurs in 3' UTRs, not within exons, and wouldn't directly cause protein truncation. Remember: Silent mutations aren't always truly "silent"—they can disrupt splicing regulatory sequences. When you see unexpected effects from synonymous changes, consider impacts on RNA processing rather than just protein coding.

Question 8

1. The physical handoff of the pre-mRNA from RNA Polymerase II to the U1 snRNP.
2. The ATP-dependent displacement of U1 snRNP from the 5' splice site and its replacement by U6 snRNP. (correct answer)
3. The base pairing between the U2 snRNA and the branch point sequence within the intron.
4. The binding of the U2AF protein complex to the polypyrimidine tract and the 3' splice site.

Explanation: The spliceosome is highly dynamic. While U1 snRNP performs the initial recognition of the 5' splice site, this is not the final check. Before the first catalytic step, the spliceosome undergoes a major conformational change where U1 is displaced and U6 snRNA, a core component of the catalytic center, base-pairs with the 5' splice site instead. This second binding event by U6 acts as a proofreading step. If the 5' splice site sequence is not a good match for U6, the active conformation cannot be achieved, and the splicing reaction is aborted. This two-step verification greatly enhances splicing fidelity.

Question 9

1. The pre-mRNA would be cleaved at the poly(A) site but would lack any added adenosine tail. (correct answer)
2. The pre-mRNA would fail to be cleaved at the poly(A) site because PAP is required to recruit the endonuclease complex.
3. Transcription would fail to terminate, resulting in a long, unprocessed read-through transcript attached to the DNA.
4. The pre-mRNA would be fully spliced and capped but would be immediately degraded from the 3' end by nuclear exonucleases.

Explanation: 3' end formation is a two-step process. First, an endonuclease complex (containing CPSF and CstF) recognizes the polyadenylation signal and cleaves the pre-mRNA downstream. Second, poly(A) polymerase (PAP) binds to the newly created 3' end and adds the poly(A) tail. A specific inhibitor of PAP's catalytic activity would block the second step but not the first. Therefore, the transcript would be cleaved correctly but would not be polyadenylated. This cleaved but non-polyadenylated transcript would then be a substrate for degradation, but its immediate state post-cleavage is as described in A.

Question 10

1. Pre-mRNAs are synthesized at a normal rate but are inefficiently capped, spliced, and polyadenylated. (correct answer)
2. Transcription initiation is enhanced, but elongation is completely blocked immediately after promoter clearance.
3. Pre-mRNAs are produced and fully processed correctly but cannot be efficiently exported from the nucleus.
4. The synthesis of tRNAs and rRNAs is inhibited, while mRNA synthesis remains unaffected.

Explanation: The phosphorylated CTD of RNA Polymerase II serves as a binding platform that recruits the machinery for 5' capping, splicing, and 3' polyadenylation. This physically couples transcription with mRNA processing, ensuring these modifications occur efficiently as the transcript emerges. Truncating the CTD eliminates this platform, uncoupling the processes. As a result, capping, splicing, and polyadenylation still occur, but with greatly reduced efficiency and accuracy. Distractor D is incorrect because RNA Pol II transcribes mRNA, not tRNA/rRNA. Distractor C points to a downstream defect, but the primary problem is in processing. Distractor B is incorrect as the CTD is also important for elongation.

Question 11

1. The intron will be retained in the mature mRNA, as the first transesterification step of splicing is blocked. (correct answer)
2. The spliceosome will fail to recognize the 5' splice site, causing the upstream exon to be skipped.
3. Splicing will proceed normally, but the excised intron will form an unstable linear molecule instead of a lariat.
4. The 3' splice site of the preceding intron will be used, resulting in the deletion of the exon between the two introns.

Explanation: The 2'-OH group of the branch point adenosine is essential for initiating the first transesterification reaction, where it attacks the 5' splice site. Mutating this adenosine to a guanosine prevents this nucleophilic attack, thus blocking lariat formation and the entire splicing process for that intron. This failure to splice leads to the retention of the intron in the mature mRNA, which typically introduces a frameshift or a premature stop codon during translation.

Question 12

1. Pre-mRNAs are synthesized at a normal rate but are inefficiently capped, spliced, and polyadenylated. (correct answer)
2. Transcription initiation is enhanced, but elongation is completely blocked immediately after promoter clearance.
3. Pre-mRNAs are produced and fully processed correctly but cannot be efficiently exported from the nucleus.
4. The synthesis of tRNAs and rRNAs is inhibited, while mRNA synthesis remains unaffected.

Explanation: The phosphorylated CTD of RNA Polymerase II serves as a binding platform that recruits the machinery for 5' capping, splicing, and 3' polyadenylation. This physically couples transcription with mRNA processing, ensuring these modifications occur efficiently as the transcript emerges. Truncating the CTD eliminates this platform, uncoupling the processes. As a result, capping, splicing, and polyadenylation still occur, but with greatly reduced efficiency and accuracy. Distractor D is incorrect because RNA Pol II transcribes mRNA, not tRNA/rRNA. Distractor C points to a downstream defect, but the primary problem is in processing. Distractor B is incorrect as the CTD is also important for elongation.

Question 13

1. A 2'-5' phosphodiester bond between the guanine at the 5' end of the intron and the adenine at the branch point. (correct answer)
2. A 5'-5' triphosphate linkage between the guanine at the 5' end of the intron and the terminal guanine of the upstream exon.
3. A 3'-5' phosphodiester bond between the adenine at the 3' end of the intron and the branch point adenine.
4. A peptide bond between a lysine residue in a splicing factor and the 5' phosphate of the intron.

Explanation: The lariat structure is formed during the first transesterification reaction of splicing. The 2'-hydroxyl group of a specific adenosine residue within the intron (the branch point) performs a nucleophilic attack on the phosphodiester bond at the 5' splice site. This reaction cleaves the upstream exon from the intron and simultaneously forms a novel 2'-5' phosphodiester bond, linking the 5' end of the intron (typically a G) to the branch point adenosine, creating the characteristic loop of the lariat.

Question 14

1. Use of an alternative start codon 150 bp upstream of the normal one.
2. A tandem duplication of the 150 bp intron in the genomic DNA.
3. Retention of the 150 bp intron between E1 and E2 due to a splicing defect. (correct answer)
4. Nonsense-mediated decay of the 250 bp transcript, leading to accumulation of a precursor.

Explanation: When you encounter RT-PCR questions, focus on what happens during mRNA processing and how PCR amplifies the final mRNA product, not the genomic DNA. In healthy tissue, the RT-PCR produces a 250 bp product from mRNA where the intron has been properly spliced out. The diseased tissue shows both this normal 250 bp band plus a new 400 bp band. Since the difference is exactly 150 bp (the intron size), this strongly suggests that some mRNA molecules retain the intron due to defective splicing. The 400 bp product represents: E1 + intron + E2, while the 250 bp product represents: E1 + E2 (intron removed). This makes C correct. Option A is wrong because alternative start codons don't create the observed pattern—RT-PCR primers target specific exon sequences, and using a different start codon wouldn't add 150 bp between the primer sites. Option B incorrectly focuses on genomic DNA changes, but RT-PCR analyzes mRNA, so genomic duplications wouldn't directly create this banding pattern in the cDNA product. Option D misunderstands nonsense-mediated decay, which degrades faulty transcripts rather than creating larger, stable products that would show up as distinct bands. Study tip: In RT-PCR questions, always trace what happens to the mRNA, not the genomic DNA. When you see two bands differing by exactly an intron's length, think splicing defects—it's a classic pattern where some transcripts are properly processed while others aren't.

Question 15

1. The protein components of the snRNPs that recognize splice sites.
2. ATP-dependent helicases that are required to rearrange the spliceosome.
3. The catalytic core formed by the snRNAs, such as U2 and U6 snRNA. (correct answer)
4. The C-terminal domain of RNA Polymerase II that coordinates splicing with transcription.

Explanation: When you encounter questions about RNA splicing mechanisms, focus on identifying which components perform the actual catalytic chemistry versus those that provide regulatory or structural support. Group II self-splicing introns are remarkable because they can remove themselves from RNA transcripts without any protein machinery. The key insight is that these introns fold into complex three-dimensional structures that create their own catalytic center. This ribozyme activity directly replaces the catalytic function normally performed by the spliceosome's RNA components. In normal spliceosome-mediated splicing, the catalytic chemistry is carried out by the RNA portions of small nuclear ribonucleoproteins (snRNPs), particularly U2 and U6 snRNAs. These snRNAs form the active site that catalyzes both transesterification reactions needed to remove introns and form lariat intermediates. In Group II self-splicing introns, the intron's own folded RNA structure creates an equivalent catalytic center, making answer C correct. Answer A is incorrect because while snRNP proteins do help recognize splice sites in spliceosomes, Group II introns have their own sequence elements for splice site recognition built into their structure. Answer B is wrong because although ATP-dependent helicases facilitate spliceosome assembly and rearrangement, self-splicing introns don't require these energy-dependent conformational changes. Answer D is incorrect because the C-terminal domain of RNA Polymerase II coordinates splicing with transcription but isn't directly involved in the catalytic mechanism itself. Remember: in splicing questions, distinguish between components that perform catalysis (the actual chemistry) versus those that provide recognition, regulation, or coordination functions.

Question 16

1. The mutation altered an exonic splicing enhancer (ESE), causing the entire exon to be skipped.
2. The mutation inadvertently created a new 5' splice site (a cryptic splice site) within the exon. (correct answer)
3. The change in codon usage caused by the mutation led to premature termination of translation.
4. The silent mutation disrupted a binding site for a miRNA, leading to improper mRNA degradation.

Explanation: When you encounter a question about silent mutations causing unexpected protein changes, think about how mutations can affect RNA processing beyond just changing amino acids. The key insight is that exonic sequences don't just code for proteins—they also contain regulatory elements that guide splicing. The correct answer is B because a G-to-A mutation can create a new splice site within the exon. Splice sites follow specific sequence patterns, particularly the GT dinucleotide at 5' splice sites. If the mutation creates a sequence that resembles a splice site (called a cryptic splice site), the splicing machinery may recognize and use this new site instead of the normal one. This would cause premature cleavage of the exon, removing part of the coding sequence and resulting in a shorter, truncated protein when translated. Let's examine why the other options don't fit: A is incorrect because skipping the entire exon would likely cause a frameshift or remove essential protein domains, but the question specifies the mutation is in the center of the exon, suggesting partial rather than complete exon removal. C is wrong because silent mutations by definition don't change amino acids, so codon usage alone wouldn't cause premature termination. D doesn't work because miRNA binding typically occurs in 3' UTRs, not within exons, and wouldn't directly cause protein truncation. Remember: Silent mutations aren't always truly "silent"—they can disrupt splicing regulatory sequences. When you see unexpected effects from synonymous changes, consider impacts on RNA processing rather than just protein coding.

Question 17

1. PABP directly catalyzes the re-addition of adenine nucleotides to counteract the action of deadenylases.
2. PABP interacts with the cap-binding protein eIF4G, promoting a closed-loop structure that protects the 5' cap from removal. (correct answer)
3. PABP acts as a physical shield, sterically hindering the access of both endonucleases and exonucleases to the entire mRNA molecule.
4. PABP recruits a methyltransferase that modifies the poly(A) tail, making it unrecognizable to deadenylase enzymes.

Explanation: PABP plays a crucial role in enhancing translation and stabilizing mRNA through the 'closed-loop' model. PABP bound to the 3' poly(A) tail also binds to eIF4G, a component of the cap-binding complex at the 5' end. This protein bridge effectively circularizes the mRNA, which is thought to promote efficient re-initiation of translation. Critically, this structure also protects the 5' cap from decapping enzymes, which is often the first step in mRNA decay. When the poly(A) tail becomes too short to bind PABP, this protective loop is lost, and the mRNA becomes a target for degradation.

Question 18

1. Splicing will halt completely after Intron 1 is removed, yielding a partially processed E1-E2-I2-E3-I3-E4 transcript.
2. Intron 2 will be retained, but Intron 3 will be spliced out normally, creating an E1-E2-I2-E3-E4 mRNA.
3. The entire pre-mRNA transcript will be degraded by the nuclear exosome due to the splicing defect.
4. Exon 3 will be skipped, resulting in an mRNA where Exon 2 is ligated directly to Exon 4. (correct answer)

Explanation: When you encounter splicing mutations, understanding the "exon definition" model is crucial. This model explains how splicing machinery recognizes exons by identifying both the 3' splice site (AG) at the end of an intron and the 5' splice site (GT) at the beginning of the next intron as a coordinated unit. When the 3' splice site of intron 2 is destroyed, the splicing machinery cannot properly define where exon 3 begins. Since exon definition requires recognition of both splice sites flanking an exon, the inability to identify exon 3's boundaries leads to exon 3 being skipped entirely. The splicing machinery will instead connect exon 2 directly to exon 4, creating an E1-E2-E4 transcript. This phenomenon is called "exon skipping." Option A incorrectly suggests splicing halts completely, but the exon definition model shows that other exons can still be recognized independently. Option B proposes intron retention with normal downstream splicing, but this misunderstands how exon definition works—without proper 3' splice site recognition, exon 3 cannot be defined, making this scenario unlikely. Option C suggests complete transcript degradation, but while some quality control mechanisms exist, exon skipping typically occurs before degradation pathways are triggered. Remember this pattern: when splice sites are mutated, think "exon definition" first. A destroyed 3' splice site typically leads to skipping of the downstream exon, not retention of the upstream intron. This distinction is key for genetics exam success.

Question 19

1. The intron will be retained in the mature mRNA, as the first transesterification step of splicing is blocked. (correct answer)
2. The spliceosome will fail to recognize the 5' splice site, causing the upstream exon to be skipped.
3. Splicing will proceed normally, but the excised intron will form an unstable linear molecule instead of a lariat.
4. The 3' splice site of the preceding intron will be used, resulting in the deletion of the exon between the two introns.

Explanation: The 2'-OH group of the branch point adenosine is essential for initiating the first transesterification reaction, where it attacks the 5' splice site. Mutating this adenosine to a guanosine prevents this nucleophilic attack, thus blocking lariat formation and the entire splicing process for that intron. This failure to splice leads to the retention of the intron in the mature mRNA, which typically introduces a frameshift or a premature stop codon during translation.

Question 20

1. Addition of a 7-methylguanosine cap via a standard 5'-3' phosphodiester bond.
2. Enzymatic removal of the 5' triphosphate to leave a 5' monophosphate.
3. Addition of a guanosine to the 5' end via a 5'-5' triphosphate linkage, followed by methylation at the N7 position. (correct answer)
4. Increasing the length of the 3' poly(A) tail to over 500 nucleotides while leaving the 5' end unmodified.

Explanation: The standard eukaryotic mRNA 5' cap has two primary functions: protection from 5' exonucleases (stability) and recruitment of the ribosome via the cap-binding complex eIF4E (translation). The unmodified 5' triphosphate end fails at both. The correct modification is the enzymatic addition of the 7-methylguanosine cap, which critically involves a unique 5'-5' triphosphate linkage. Choice A specifies the wrong linkage. Choice B would only partially solve the stability issue and not the translation issue. Choice D would improve stability from the 3' end but would not fix the critical defect in translation initiation at the 5' end.