Question 1

Retroviral life cycle — Retroviruses replicate via which type of nucleic acid intermediate inside infected cells?

Accepted Answer

Introduction: Retroviruses package RNA genomes but integrate into host chromosomes. This question targets the central dogma reversal step that defines retroviral replication. Given Data / Assumptions: Retroviruses (e.g., HIV) carry an RNA genome. They encode reverse transcriptase. Integration into host DNA is required for productive infection. Concept / Approach: Reverse transcriptase synthesizes complementary DNA (cDNA) from the viral RNA template, yielding a double-stranded DNA intermediate. This proviral DNA integrates into host chromosomal DNA via integrase. Transcription by host RNA polymerase II then produces viral mRNAs and genomic RNA for assembly. Step-by-Step Solution: Viral entry delivers RNA and reverse transcriptase.cDNA is synthesized → double-stranded DNA intermediate forms.Integrase inserts this DNA into the host genome (provirus).Therefore, replication proceeds through a DNA intermediate. Verification / Alternative check: Antiretroviral drugs targeting reverse transcriptase (e.g., NRTIs, NNRTIs) block DNA intermediate formation, validating its essential role in the life cycle. Why Other Options Are Wrong: RNA/mRNA alone: these are templates, not the integrated intermediate. rDNA: refers to ribosomal DNA repeats, unrelated to retroviral replication. Protein–RNA hybrid: not the key replicative intermediate. Common Pitfalls: Assuming RNA viruses always replicate only via RNA intermediates; retroviruses are the classic exception due to reverse transcription. Final Answer: DNA.

Question 2

RNA chemistry — Which RNA class characteristically contains unusual (modified) purines and pyrimidines as part of its normal structure?

Accepted Answer

Introduction: Many RNAs undergo post-transcriptional modification. This question asks which RNA species is especially rich in unusual bases—modified purines and pyrimidines that fine-tune structure and function. Given Data / Assumptions: Common RNA classes: mRNA, rRNA, tRNA, small RNAs. Modified bases include methylations, thiolations, and complex derivatives. Function of modifications is to stabilize structure and support decoding. Concept / Approach: Transfer RNAs (tRNAs) are heavily modified, often containing 10–20 modified nucleotides per molecule. Examples include pseudouridine, dihydrouridine, inosine, 1-methylguanosine, and queuosine. These modifications stabilize the L-shaped tertiary structure and modulate anticodon–codon pairing fidelity during translation. Step-by-Step Solution: Identify which RNA needs precise structural features at small size: tRNA must fold into a conserved 3D shape for aminoacylation and decoding.Recognize that extensive base modifications help maintain tRNA stability and correct wobble base pairing.Therefore, tRNA is the RNA class most characteristically enriched in unusual bases.Other RNAs have modifications (e.g., rRNA methylations and pseudouridines), but not as densely as tRNA. Verification / Alternative check: tRNA sequencing maps consistently reveal numerous modified nucleotides, particularly near the anticodon loop and D/T arms, underscoring their functional importance in translation accuracy. Why Other Options Are Wrong: mRNA: relatively few internal base modifications in canonical eukaryotic mRNA beyond the 5′ cap and occasional internal marks. rRNA: contains modifications but less densely than tRNA. nRNA/pre-mRNA: primarily processed by splicing and capping; not heavily modified internally. siRNA: short regulatory RNAs without extensive base modification patterns. Common Pitfalls: Conflating the mRNA 5′ cap modification with high internal modification density; tRNA is the standout for unusual bases throughout the molecule. Final Answer: tRNA.

Question 3

Central dogma — What is the primary role of messenger RNA (mRNA) in gene expression?

Accepted Answer

Introduction: Gene expression involves transcription of DNA into RNA and translation of RNA into protein. This question identifies the specific function of messenger RNA within that flow of information. Given Data / Assumptions: DNA encodes genes. RNA species include mRNA, rRNA, and tRNA with distinct roles. Protein synthesis occurs on ribosomes using mRNA templates. Concept / Approach: mRNA is synthesized by RNA polymerase as a complementary RNA copy of the coding region of a gene (using the template strand). After processing (5′ cap, splicing, poly(A) tail) in eukaryotes, the mature mRNA carries the codon sequence that directs amino acid order during translation. Step-by-Step Solution: Transcription: DNA → pre-mRNA → processed mRNA.Export and translation: mRNA binds ribosomes; tRNAs decode codons into amino acids.Therefore, mRNA functions as the direct genetic message specifying the protein sequence.Other RNAs provide structural or adaptor roles, not the primary message content. Verification / Alternative check: mRNA sequencing (RNA-seq) reads out expressed gene copies, and in vitro translation using purified mRNA demonstrates that its sequence alone encodes the protein product. Why Other Options Are Wrong: Carries amino acids: tRNA, not mRNA, carries amino acids as aminoacyl-tRNA. Ribosome component: rRNA and ribosomal proteins form the ribosome. Genetic material of some organisms: RNA genomes exist in some viruses, but that is not the role of cellular mRNA. Catalyzes peptide bond formation: peptidyl transferase activity resides in rRNA within the large subunit. Common Pitfalls: Confusing the roles of different RNA classes during translation; mRNA is the template, tRNA is the adaptor, and rRNA is the catalytic/structural core. Final Answer: It is a direct copy of a gene.

Question 4

Translation fundamentals — The anticodon sequence is a defining structural feature found on which molecule?

Accepted Answer

Introduction: Decoding of mRNA into protein depends on accurate base pairing between codons and anticodons. This question asks you to identify the macromolecule that carries the anticodon used in this process. Given Data / Assumptions: Codons are triplets on mRNA. Anticodons are complementary triplets that recognize codons. Translation occurs on ribosomes with help from adaptors. Concept / Approach: The anticodon resides on transfer RNA (tRNA). Each tRNA carries a specific amino acid at its 3′ end and has an anticodon loop that recognizes the corresponding codon on mRNA, ensuring incorporation of the correct amino acid into the growing polypeptide chain. Step-by-Step Solution: mRNA provides codons.tRNA provides anticodons and carries amino acids.rRNA forms the catalytic ribosomal core.Therefore, the anticodon is on tRNA. Verification / Alternative check: Crystal structures of tRNA bound to ribosomes clearly show the anticodon loop base pairing with mRNA codons in the decoding center. Why Other Options Are Wrong: mRNA: contains codons, not anticodons. Ribosome/rRNA: provide scaffolding and catalysis, not the anticodon. DNA polymerase: functions in DNA replication, unrelated to translation decoding. Common Pitfalls: Mixing up codon vs anticodon terminology; codon = mRNA, anticodon = tRNA. Final Answer: tRNA.

Question 5

mRNA capping — Which modified base is found at the 5′ end (cap) of a typical eukaryotic mRNA?

Accepted Answer

Introduction: Eukaryotic mRNAs are modified co-transcriptionally at the 5′ end with a specialized cap that protects the transcript and promotes translation. This question asks you to identify the specific modified base present in that cap structure. Given Data / Assumptions: The 5′ cap is linked via a unique 5′–5′ triphosphate bridge. Cap structure enhances stability, nuclear export, and ribosome recruitment. Cap-binding proteins recognize the methylated base. Concept / Approach: The canonical eukaryotic cap (cap 0) contains 7-methyl-guanosine (m7G) attached to the first transcribed nucleotide via a 5′–5′ triphosphate linkage. Additional 2′-O-methylations on the first one or two riboses (cap 1 and cap 2) may occur, but the defining modified base at the cap itself is m7G. Step-by-Step Solution: Identify hallmark: methylation at the N7 position of guanosine.Recognize linkage: m7G connected 5′–5′ to the mRNA body.Therefore, the correct modified base is 7-methyl-guanosine.Other methylations (e.g., 2′-O) are additional sugar methylations on internal nucleotides adjacent to the cap. Verification / Alternative check: Cap-binding complex eIF4E specifically recognizes the m7G cap, facilitating translation initiation and interactions with eIF4G and the poly(A)-binding protein loop. Why Other Options Are Wrong: 1-methyl-adenosine / 1-methyl-guanosine: not the defining cap base. 2′-O-methyl-guanosine: refers to ribose methylation, not the signature N7 base methylation. pAp: a metabolite, not the cap base. Common Pitfalls: Confusing sugar methylations near the cap with the essential N7 methylation of guanosine that defines the mRNA cap. Final Answer: 7-methyl-guanosine.

Question 6

RNA ligation — Which enzyme can ligate a nicked RNA molecule under standard biochemical conditions?

Accepted Answer

Introduction: Joining breaks in nucleic acids is central to cloning and RNA repair workflows. This question asks which enzyme is specifically used to ligate RNA ends or seal nicks in RNA substrates. Given Data / Assumptions: Enzymes considered are standard molecular biology tools. Substrate is RNA with a nick or suitable termini. Reaction requires appropriate cofactors (often ATP) and termini (e.g., 5′ phosphate, 3′ hydroxyl). Concept / Approach: T4 RNA ligase catalyzes the formation of a phosphodiester bond between RNA 5′ phosphate and 3′ hydroxyl termini. Variants (e.g., Rnl1, Rnl2) are widely used for adapter ligation in small RNA library prep. T4 DNA ligase primarily ligates DNA nicks or cohesive ends; DNA polymerase fills gaps but does not ligate nicks; RNase H cleaves RNA in RNA–DNA hybrids rather than joining it. Step-by-Step Solution: Identify substrate: RNA, not DNA.Match enzyme to substrate: T4 RNA ligase is designed for RNA ligation.Exclude non-ligases: DNA polymerase fills but does not seal; RNase H cleaves.Therefore, T4 RNA ligase is correct. Verification / Alternative check: Small RNA sequencing protocols universally employ T4 RNA ligases to attach 3′ and 5′ adapters to microRNAs and other small RNAs, confirming enzyme specificity and utility. Why Other Options Are Wrong: DNA polymerase: lacks ligase activity. T4 DNA ligase: efficient for DNA; limited utility on pure RNA ends under standard conditions. All of these: overinclusive; only T4 RNA ligase is appropriate. RNase H: endonuclease that degrades RNA in RNA–DNA hybrids. Common Pitfalls: Assuming DNA ligases will work equivalently on RNA. Substrate specificity and reaction conditions differ markedly between DNA and RNA ligases. Final Answer: T4 RNA ligase.

Question 7

Retroviral genomes — What is the genetic material of retroviruses such as HIV within virions?

Accepted Answer

Introduction: Viruses vary in their genomic composition. Retroviruses are a distinctive class that package one type of nucleic acid in virions yet replicate via a different intracellular intermediate. This question asks for the genome type present inside the viral particle of HIV and related retroviruses. Given Data / Assumptions: Focus is on the packaged genome inside virions, not the intracellular intermediate. Retroviruses carry reverse transcriptase. Integration into host DNA occurs after infection. Concept / Approach: Retroviruses package positive-sense single-stranded RNA (two copies per virion). Upon entry, reverse transcriptase synthesizes a double-stranded DNA copy that integrates as a provirus. Thus, the virion genome is RNA, even though a DNA intermediate is essential in the life cycle. Step-by-Step Solution: Identify virion contents: ssRNA genome plus reverse transcriptase and integrase.Clarify replication: RNA → DNA (by reverse transcriptase) → integration → transcription to produce more RNA.Therefore, the genetic material inside virions is RNA.Do not confuse with the DNA intermediate used post-entry. Verification / Alternative check: Biochemical purification of retroviral particles reveals genomic RNA and reverse transcriptase activity; antiretroviral drugs target reverse transcriptase to block conversion of RNA to DNA. Why Other Options Are Wrong: DNA: present only after reverse transcription inside the host cell. Protein: structural components exist, but not the genetic material. All of these / DNA–RNA hybrid: not accurate descriptions of the packaged genome. Common Pitfalls: Conflating the packaged genome (RNA) with the replication intermediate (DNA). Retroviruses are defined by RNA genomes plus reverse transcription. Final Answer: RNA.

Question 8

RNA chemistry — Why is RNA inherently less stable than DNA in alkaline (basic) solutions?

Accepted Answer

Introduction: RNA is notoriously more labile than DNA in alkaline solutions. This question probes the chemical reason behind that instability, a concept that underpins many laboratory techniques (e.g., RNase-free handling) and explains why DNA, not RNA, is nature’s long-term information store. Given Data / Assumptions: RNA contains ribose; DNA contains 2′-deoxyribose. RNA uses uracil; DNA uses thymine. Alkaline pH provides hydroxide ions that can act as bases/nucleophiles. Backbone integrity depends on phosphodiester bonds connecting 3′-OH and 5′-phosphate. Concept / Approach: The key structural difference is the 2′-hydroxyl on ribose. Under basic conditions, this 2′-OH can be deprotonated to form a 2′-O⁻ that intramolecularly attacks the adjacent phosphorus, forming a cyclic 2′,3′-phosphodiester intermediate and cleaving the backbone. DNA lacks the 2′-OH, so this pathway is blocked, making DNA far more stable in alkali. Step-by-Step Solution: 1) In base, OH⁻ abstracts a proton from the 2′-OH of ribose → 2′-O⁻ forms.2) The 2′-O⁻ attacks the adjacent phosphate → a 2′,3′-cyclic phosphate intermediate forms.3) Hydrolysis of this intermediate yields a mixture of 2′ or 3′-monophosphate fragments → backbone cleavage.4) DNA lacks a 2′-OH, preventing this intramolecular transesterification and therefore resisting alkali. Verification / Alternative check: Classic lab demonstrations show that RNA incubated at high pH rapidly smears on gels (fragmentation), while DNA remains largely intact. Mild alkaline hydrolysis is even used deliberately to degrade RNA contaminants during DNA prep. Why Other Options Are Wrong: a) Adenine is not the determinant; the sugar is.c) Uracil does not depurinate; depurination refers to purines, and base-catalyzed cleavage is sugar-driven.d) Single-strandedness may expose bases but does not cause the specific alkali lability.e) The phosphodiester bond is covalent in both DNA and RNA. Common Pitfalls: Confusing base-catalyzed cleavage with acid-catalyzed depurination; attributing instability to uracil rather than the ribose 2′-OH. Final Answer: The ribose 2′-OH enables base-catalyzed intramolecular attack on the phosphate, cleaving the RNA backbone.

Question 9

Translation — Which two structural features of a tRNA directly convert a triplet mRNA codon into its corresponding amino acid during protein synthesis?

Accepted Answer

Introduction: tRNA is the adaptor that “reads” mRNA codons and delivers the matching amino acid. This question asks which two parts of tRNA are directly responsible for decoding and amino acid attachment, the core of the adaptor hypothesis in translation. Given Data / Assumptions: tRNA has a cloverleaf secondary structure and L-shaped tertiary fold. Each tRNA ends with a conserved 3′-CCA sequence. Ribosomes align tRNA anticodons with mRNA codons. Concept / Approach: The anticodon loop base-pairs with the complementary mRNA codon to ensure proper decoding. The 3′ CCA acceptor end is the site where aminoacyl-tRNA synthetases esterify the cognate amino acid to the 3′-terminal ribose, creating aminoacyl-tRNA for peptide bond formation. Step-by-Step Solution: 1) Anticodon loop ↔ codon interaction mediates genetic code recognition.2) 3′ CCA end ↔ amino acid attachment by aminoacyl-tRNA synthetase.3) Together, these link the nucleotide code to a specific amino acid during translation. Verification / Alternative check: Mutations in anticodon change decoding; editing and recognition elements near the acceptor stem control charging specificity, confirming these features’ direct roles. Why Other Options Are Wrong: a,b,e) TψC, D, or variable loops help structure and recognition but do not both decode and carry the amino acid.d) The listed features do participate directly. Common Pitfalls: Assuming all loops have equal roles; confusing synthetase recognition sites with ribosomal decoding sites. Final Answer: Anticodon loop + 3′ CCA acceptor end.

Question 10

tRNA tertiary structure — Which statements about the 3D architecture of yeast tRNA (and most tRNAs) are accurate?

Accepted Answer

Introduction: tRNAs adopt a conserved tertiary architecture that enables precise positioning in the ribosome during translation. Yeast tRNA was among the first solved RNA 3D structures and serves as a model for tRNA folding principles. Given Data / Assumptions: tRNAs fold into an L-shaped 3D structure from a cloverleaf secondary structure. Stacking and noncanonical base pairs contribute substantially to stability. Tertiary interactions bridge distant loops and stems. Concept / Approach: Evaluate each statement against known structural features: base stacking, conserved L-shape across tRNAs, and many non–Watson–Crick interactions (e.g., G•U wobble, base triples) that maintain the compact fold and correct geometry for ribosomal binding. Step-by-Step Solution: 1) Base stacking between adjacent bases stabilizes helical stems and coaxial stacking.2) Comparative structures show most tRNAs share an L-shaped fold suited for A- and P-site fit.3) Noncanonical pairs and tertiary contacts (e.g., D loop–TψC loop interactions) are crucial for the final 3D shape. Verification / Alternative check: X-ray and cryo-EM data of multiple tRNAs and tRNA–ribosome complexes confirm all three features consistently. Why Other Options Are Wrong: e) tRNAs are not unfolded in vivo; they are highly structured for function. Common Pitfalls: Assuming only Watson–Crick pairing matters; overlooking stacking and tertiary contacts in RNA folding. Final Answer: All of the above.

Question 11

Transcription directionality — During RNA synthesis, what RNA sequence is produced from the DNA template strand 5′-T p A p G p C-3′ (written 5′→3′ on the template)?

Accepted Answer

Introduction: Transcription produces RNA complementary to the DNA template and antiparallel to it. Getting orientation correct is essential for predicting RNA sequences from DNA and vice versa. Given Data / Assumptions: Template strand is explicitly given 5′→3′ as T–A–G–C. RNA uses U instead of T. Polymerases synthesize RNA 5′→3′ while reading the template 3′→5′. Concept / Approach: Because the provided template is written 5′→3′, we must first reverse its orientation to align with the polymerase reading direction. Complement bases accordingly and then express the RNA in the standard 5′→3′ direction. Step-by-Step Solution: 1) Template (given) 5′→3′: T–A–G–C.2) Reverse to 3′→5′ for reading: C–G–A–T.3) Complement to RNA (5′→3′ product): G–C–U–A.4) Therefore, RNA sequence is 5′-G C U A-3′ (option b, noting U for uracil). Verification / Alternative check: Check against the non-template (coding) strand: it would be 5′-G C T A-3′ in DNA, so replacing T with U gives 5′-G C U A-3′ in RNA, matching the result. Why Other Options Are Wrong: a) Uses T instead of U and has wrong orientation.c) Contains T, not U, and mismatches base pairing.d) Incorrect base order; A U C G would pair with T A G C in the same orientation, which is not antiparallel.e) Not reported in the standard 5′→3′ product orientation requested. Common Pitfalls: Forgetting antiparallel orientation; writing T in RNA; failing to reverse the template before complementing. Final Answer: 5′-G p C p U p A-3′.

Question 12

RNA base pairing — In cellular RNA, uracil (U) base-pairs with which nucleotide during Watson–Crick pairing?

Accepted Answer

Introduction: Canonical base pairing underlies the accuracy of transcription and translation. In RNA, uracil replaces thymine and forms specific hydrogen bonds with its complementary base. This question tests recall of standard pairing rules in RNA. Given Data / Assumptions: RNA uses U where DNA uses T. Watson–Crick rules apply to standard RNA duplexes. Wobble pairing can occur at tRNA anticodons but is not the canonical rule. Concept / Approach: Apply Watson–Crick pairing: A pairs with U in RNA (analogous to A–T in DNA), and G pairs with C. Although inosine can wobble pair in tRNA, it is not the universal canonical partner for U in standard duplex RNA. Step-by-Step Solution: 1) Identify the RNA base in question: uracil (U).2) Recall canonical pairs: A–U and G–C.3) Therefore, U pairs with adenine in standard Watson–Crick pairing. Verification / Alternative check: Structures of RNA duplexes and mRNA–tRNA interactions confirm A–U hydrogen-bonding patterns analogous to A–T in DNA. Why Other Options Are Wrong: b,d) Cytosine pairs with guanine; guanine pairs with cytosine.c) Thymine occurs mainly in DNA, not RNA.e) Inosine participates in wobble at the tRNA anticodon position, not the canonical RNA duplex partner for U. Common Pitfalls: Overgeneralizing wobble rules; forgetting that U replaces T in RNA. Final Answer: Adenine.

Question 13

Eukaryotic mRNA processing — The mature mRNA from which source would be expected to carry a 3′ poly(A) tail?

Accepted Answer

Introduction: Polyadenylation is a hallmark of most eukaryotic mRNAs, enhancing stability, nuclear export, and translation. This question distinguishes eukaryotic mRNA processing from bacterial and phage transcripts. Given Data / Assumptions: Eukaryotic pre-mRNAs are typically 5′-capped, spliced, and 3′-polyadenylated. Bacterial and bacteriophage mRNAs usually lack long 3′ poly(A) tails that function like eukaryotic poly(A); short polyadenylation in bacteria often marks decay. Human insulin is a eukaryotic gene expressed in human cells. Concept / Approach: Identify the organismal origin of the mRNA. Eukaryotic mRNAs (e.g., human insulin mRNA) receive a poly(A) tail via poly(A) polymerase after cleavage at a consensus site; bacterial/phage transcripts generally do not have this eukaryotic-style modification. Step-by-Step Solution: 1) Determine source: human vs bacterial/phage.2) Apply processing rule: eukaryotes → poly(A); bacteria/phage → typically no such tail for stability/translation.3) Conclude human insulin mRNA has a poly(A) tail. Verification / Alternative check: mRNA isolation protocols (oligo-dT selection) enrich eukaryotic polyadenylated transcripts such as insulin mRNA, confirming the presence of a poly(A) tail. Why Other Options Are Wrong: a,b,d) Prokaryotic/phage mRNAs lack canonical eukaryotic polyadenylation; any bacterial poly(A) often promotes decay.e) Many archaeal transcripts are not polyadenylated like eukaryotic mRNAs. Common Pitfalls: Assuming all RNAs have poly(A); conflating bacterial polyadenylation (RNA turnover) with eukaryotic stabilization. Final Answer: Human insulin mRNA.

Question 14

tRNA size — What is the typical average length of a mature tRNA molecule (reported in nucleotides/bp)?

Accepted Answer

Introduction: tRNAs are among the smallest functional RNAs in cells, optimized for delivery of amino acids to ribosomes. Knowing their approximate size helps interpret gels, blots, and sequencing results in molecular biology. Given Data / Assumptions: Mature tRNAs are compact and heavily modified. Reported lengths cluster around 73–93 nucleotides depending on species and isoacceptor. Conventional “average” cited in texts is near 76–85 nt. Concept / Approach: Choose the option that best reflects the accepted average length. Values much larger (100–140) are typical of small mRNAs or snRNAs, not tRNAs; values far smaller (40) are too short to form the cloverleaf and L-shaped structures required for function. Step-by-Step Solution: 1) Recall typical range: ~73–93 nt.2) Identify the nearest average: ~80 nt.3) Select the option that states ≈ 80 nucleotides. Verification / Alternative check: tRNA gene annotations and tRNA-seq datasets routinely report lengths near 75–90 nt with post-transcriptional CCA addition at the 3′ end, consistent with an ~80-nt average. Why Other Options Are Wrong: b–d) Overestimate typical tRNA length; these sizes do not fit standard tRNA architecture.e) Too short to encode the necessary structural elements. Common Pitfalls: Confusing tRNA with snRNA or miRNA size classes; ignoring that “bp” here tracks nucleotide count for single-stranded RNA length. Final Answer: ≈ 80 nucleotides.

Question 15

RNA size classes — Which of the following RNA types is generally the smallest by length in cells?

Accepted Answer

Introduction: Cellular RNAs come in distinct size classes. Recognizing which are smallest helps in experimental design (e.g., gel percentage, library preparation) and in understanding RNA biology broadly. Given Data / Assumptions: tRNAs are ~73–93 nt. mRNAs vary widely, often hundreds to thousands of nucleotides. rRNAs are among the largest and most abundant RNAs (e.g., 5S ~120 nt, 16S/18S and 23S/28S are much larger). Concept / Approach: Compare typical lengths: although 5S rRNA is small (~120 nt), most rRNA species and typical mRNA molecules are larger than tRNA. Therefore, in general cellular context, tRNA is the smallest of the listed major classes. Step-by-Step Solution: 1) List characteristic sizes: tRNA ~80 nt; 5S rRNA ~120 nt; 16S/18S rRNA ~1500–1900 nt; mRNA varies but usually >500 nt.2) Identify the overall smallest class: tRNA.3) Choose tRNA accordingly. Verification / Alternative check: Denaturing PAGE of total RNA shows a distinct small tRNA band below 5S rRNA, confirming its smaller average size. Why Other Options Are Wrong: a) mRNAs are typically longer.c) rRNAs (except 5S) are much larger; even 5S exceeds most tRNAs.d) Sizes are not comparable across these classes.e) lncRNAs are usually long (>200 nt). Common Pitfalls: Focusing on 5S rRNA size and forgetting that the question asks in general among the listed classes. Final Answer: Transfer RNA (tRNA).

RNA Structure Questions