chapter 7 Flashcards
Which factor best explains why RNA can assume a wide variety of three-dimensional shapes in the cell?
A. RNA molecules are typically larger than DNA molecules.
B. RNA uses the base uracil instead of thymine.
C. RNA is single-stranded and can fold back to form intramolecular base pairs.
D. RNA is always chemically modified immediately after transcription.
Answer: C
Explanation: RNA’s single-stranded nature permits regions within the same molecule to base-pair with complementary stretches, creating loops and other structural motifs that allow for diverse three-dimensional conformations.
What primarily accounts for the chemical difference between DNA and RNA that causes RNA to be more reactive and thus often less stable in cells?
A. The use of ribose in RNA, which has an extra 2′-OH group.
B. The replacement of adenine with hypoxanthine in RNA.
C. The presence of phosphodiester bonds only in DNA.
D. The formation of hairpin loops in RNA.
Answer: A
Explanation: RNA contains ribose, whose 2′-OH group renders the molecule more chemically reactive (and more prone to hydrolysis) than DNA, which contains deoxyribose lacking that hydroxyl group.
Which factor best explains why RNA polymerase can initiate RNA synthesis without the assistance of a primer?
A. RNA polymerase has a proofreading subunit with advanced exonuclease function.
B. The 2′-OH group in ribose enables direct base-pairing without an RNA primer.
C. RNA polymerase can bind promoter sequences directly and position the first ribonucleotide for catalysis.
D. RNA polymerase dimerizes with single-stranded DNA-binding proteins to start synthesis.
Answer: C
Explanation: RNA polymerase recognizes and binds to specific promoter sequences in the DNA, positioning the first ribonucleotide correctly. It does not require a pre-existing 3′-OH end (i.e., a primer) to begin adding nucleotides.
Which of the following best distinguishes RNA transcription from DNA replication in eukaryotes?
A. RNA transcription involves both parental DNA strands simultaneously.
B. RNA transcription requires a short RNA primer to begin.
C. Transcription occurs in a 3′ to 5′ direction, whereas replication occurs 5′ to 3′.
D. The newly formed RNA strand disassociates rapidly from the DNA template, allowing multiple polymerases to work in quick succession.
Answer: D
Explanation: During transcription, the nascent RNA strand separates from its DNA template, allowing multiple RNA polymerases to follow closely behind one another on the same gene. This contrasts with replication, where the newly synthesized DNA remains base-paired to its complementary strand.
Which statement best distinguishes eukaryotic mRNA from bacterial mRNA with respect to gene organization?
A. Eukaryotic mRNA lacks 5′ and 3′ untranslated regions, whereas bacterial mRNA does not.
B. Eukaryotic mRNA typically contains multiple protein-coding regions per transcript, whereas bacterial mRNA generally encodes a single protein.
C. Eukaryotic mRNA typically encodes one protein per transcript (monocistronic), whereas bacterial mRNA often encodes several proteins (polycistronic).
D. Eukaryotic mRNA and bacterial mRNA are identical in their organization and protein-coding capacity.
Answer: C
Explanation: Bacterial mRNAs can be polycistronic—encoding multiple proteins within a single transcript. In contrast, eukaryotic mRNAs typically code for only one protein per mRNA.
Which of the following best describes gene expression when the final product is a noncoding RNA?
A. Transcription but no translation, because RNA itself is the functional product.
B. Translation only, because noncoding RNA bypasses transcription.
C. No transcription or translation is required.
D. Transcription initiates on the antisense strand, but translation occurs on the sense strand.
Answer: A
Explanation: If a gene’s final product is an RNA (e.g., rRNA, tRNA, miRNA), then gene expression involves only transcription; no translation is needed because the RNA itself performs the relevant function.
Which characteristic of a bacterial promoter ensures that transcription proceeds in a single, defined direction for a given gene?
A. The presence of the 2′-OH group on ribose.
B. The polarity (5′→3′ sequence orientation) of the promoter upstream of the transcription start site.
C. Its ability to recruit helicase to unwind the entire chromosome.
D. Its requirement for a DNA primer to initiate RNA synthesis
Answer: B
Explanation: Each promoter has distinct, asymmetric sequences. Because RNA polymerase must add nucleotides in the 5′→3′ direction, these sequences dictate which DNA strand is used as the template and thus the direction of transcription.
Which best describes the role of the σ (sigma) factor in bacterial transcription initiation?
A. It provides energy for RNA strand elongation by cleaving phosphate bonds.
B. It specifically recognizes terminator sequences, causing RNA polymerase to detach.
C. It binds and unwinds the DNA replication origin, initiating DNA synthesis.
D. It detects specific base-pair features on the promoter, allowing polymerase to bind tightly and open the double helix.
Answer: D
Explanation: The σ factor in bacteria scans the DNA for a promoter sequence by recognizing particular nucleotide features on the outside of the double helix. This enables stable binding of RNA polymerase and the subsequent local unwinding of DNA to start transcription.
Why is the TATA box critical to initiating eukaryotic transcription?
A. It encodes the start codon for translation.
B. It serves as a binding site for TFIID, causing local DNA distortion and guiding assembly of general transcription factors.
C. It codes for a specialized sigma factor that unwinds DNA at the promoter.
D. It stabilizes the RNA–DNA hybrid during elongation.
Answer: B
Explanation: The TATA box is recognized by TFIID, which distorts the DNA and helps recruit the rest of the initiation complex, making it essential for orienting and assembling the transcription machinery.
What is the direct role of TFIIH’s protein kinase activity in transcription initiation?
A. It dissociates the RNA–DNA hybrid by cleaving phosphodiester bonds in the nascent RNA strand.
B. It phosphorylates ribonucleotides so they can be linked into the nascent RNA.
C. It phosphorylates the C-terminal tail of RNA polymerase II, allowing the polymerase to escape the initiation complex and begin elongation.
D. It catalyzes the formation of peptide bonds at the ribosome.
Answer: C
Explanation: TFIIH’s kinase subunit modifies the “tail” of RNA polymerase II by phosphorylation. This event frees the polymerase from most of the general transcription factors and initiates RNA-chain elongation.
Which of the following most directly explains why eukaryotic mRNAs need 5′ capping and 3′ polyadenylation?
A. They protect the protein’s amino acid sequence from mutations.
B. They promote intron removal by recruiting ribosomes.
C. They increase mRNA stability and assist in its export from the nucleus, as well as signaling that the transcript is complete.
D. They allow bacteria to initiate translation simultaneously as the RNA is transcribed.
Answer: C
Explanation: The 5′ cap and poly-A tail both help stabilize the mRNA, promote nuclear export, and ensure that the message is recognized as complete before translation begins.
What is the main difference between bacterial and eukaryotic mRNA processing that explains why eukaryotic transcripts need to be transported out of the nucleus?
A. Bacterial mRNAs lack UTP and therefore remain in the nucleus.
B. Eukaryotic mRNAs undergo capping and polyadenylation in the nucleus, whereas bacterial mRNAs do not and are synthesized directly in the cytosol.
C. Bacterial RNA polymerase cannot form phosphodiester bonds.
D. Eukaryotic ribosomes exclusively function in the nucleus to initiate translation on unprocessed RNAs.
Answer: B
Explanation: In bacteria, there is no nucleus, so transcription and translation occur in the same compartment without the need for mRNA export or capping and polyadenylation. In eukaryotes, these modifications happen in the nucleus, and only after processing can mRNAs be exported to the cytosol for translation.
What feature of snRNPs underlies their ability to recognize intron–exon boundaries during splicing?
A. Their protein components bind nonspecifically to all nucleotides.
B. Their RNA components base-pair with specific consensus sequences at the splice sites.
C. They possess helicase activity that unwinds the entire length of the transcript.
D. They degrade the pre-mRNA outside of the intron–exon boundaries.
Answer: B
Explanation: The RNA in the snRNPs is complementary to specific splice-site sequences in the pre-mRNA. This precise base-pairing allows the snRNPs to locate and catalyze the removal of introns.
Which statement best explains how alternative splicing increases the coding potential of the eukaryotic genome?
A. It generates multiple mRNAs by rearranging the exon order in the final transcript.
B. It combines exons from different genes to produce mRNAs that span multiple chromosomes.
C. It allows some exons to be skipped, producing different mature mRNAs from the same primary transcript.
D. It attaches a poly-A tail at both the 5′ and 3′ ends of the same mRNA.
Answer: C
Explanation: Alternative splicing permits certain exons to be selectively omitted or included in the final mRNA—without changing their original order—thereby allowing multiple protein isoforms to be produced from one gene
Which of the following best explains why eukaryotic RNA-processing enzymes cluster with RNA polymerase II in “factories”?
A. It ensures that transcription cannot begin until all RNAs have been fully processed.
B. It spatially concentrates the enzymes and substrates, allowing newly transcribed RNA to be capped, spliced, or polyadenylated rapidly and efficiently.
C. It permits the DNA to be moved from the nucleus to the cytosol before splicing occurs.
D. It separates ribosomal RNA synthesis from mRNA synthesis entirely.
Answer: B
Explanation: By grouping transcription and processing factors together, the cell ensures that each nascent RNA is quickly and accurately modified, increasing the efficiency and coordination of gene expression.
How do these nuclear “factories” exemplify a general principle of cellular organization?
A. They illustrate how large membrane-bound compartments strictly segregate all metabolic reactions.
B. They demonstrate that ribosomes must always cluster around DNA polymerases.
C. They show that specialized tasks—like DNA replication, repair, or RNA processing—frequently occur in condensates or aggregates of proteins and nucleic acids.
D. They prove that proteins always fold improperly unless they are in a “factory.”
Answer: C
Explanation: The principle is that cells organize certain biochemical processes into specialized aggregates or condensates, thereby bringing together all the necessary molecular machinery for efficient function.
Which factor primarily ensures that only properly processed eukaryotic mRNAs exit the nucleus for translation?
A. The presence of DNA polymerase near the nuclear pore.
B. Nuclear pore complexes selectively recognize the set of bound proteins (cap-binding proteins, poly-A–binding proteins, and exon junction complexes) on a mature mRNA.
C. The nucleolus degrades all mature RNAs in the nucleus.
D. The absence of the poly-A tail signals export readiness.
Answer: B
Explanation: Properly processed mRNAs have distinct protein markers at the 5′ cap, spliced exon junctions, and 3′ poly-A tail. Nuclear pore complexes detect these features, allowing only mature mRNAs to pass into the cytosol.
Which statement best explains why short-lived mRNAs are advantageous for certain cellular functions?
A. They produce polypeptides that cannot be properly folded.
B. They minimize the need for intron removal.
C. They allow protein levels to rapidly adjust in response to changing conditions.
D. They ensure that mRNAs outcompete other RNAs in the nucleus.
Answer: C
Explanation: Short-lived mRNAs enable the cell to quickly change the amount of protein produced in response to environmental or developmental signals, because these mRNAs are degraded rapidly and do not linger to continually make protein.
Which statement best explains why the genetic code is described as “redundant”?
A. There are more amino acids than codons, so some codons get skipped.
B. Multiple different codons can specify the same amino acid.
C. The same codon can specify multiple amino acids.
D. The protein synthesis machinery uses both DNA and RNA as templates simultaneously.
Answer: B
Explanation: The genetic code is redundant because there are 64 possible codons but only 20 amino acids, so several codons can encode the same amino acid.
Which feature of tRNA best explains how it can recognize multiple codons for the same amino acid?
A. The 5′ phosphate end of tRNA forms a covalent bond with the mRNA.
B. The anticodon loop of some tRNAs can base-pair flexibly (wobble) at the third codon position.
C. The 3D “L-shape” of tRNA blocks most binding sites on the ribosome.
D. tRNAs undergo immediate degradation after attaching the amino acid.
Answer: B
Explanation: Many tRNAs can “wobble” at the third codon position, thereby matching more than one codon that codes for the same amino acid.
Why is the tRNA’s 3′ end essential for accurate protein synthesis?
A. It joins with the ribosomal RNA to catalyze peptide bond formation directly.
B. It contains the anticodon loop that pairs with the mRNA codon.
C. It is the site where the corresponding amino acid is covalently attached, ensuring the correct amino acid is delivered for each codon.
D. It anchors the tRNA to the nuclear envelope before export.
Answer: C
Explanation: The 3′ end of the tRNA binds the amino acid specific to that tRNA’s anticodon, thus guaranteeing that each codon in the mRNA is matched to the correct amino acid.
Which statement best describes how aminoacyl-tRNA synthetases contribute to the fidelity of protein synthesis?
A. They check for wobble base-pairing at the ribosome to prevent misreading of codons.
B. They hydrolyze GTP to move the ribosome along the mRNA.
C. They specifically match each tRNA with its correct amino acid by recognizing unique nucleotides in the anticodon loop and 3′ acceptor stem.
D. They degrade improperly folded proteins.
Answer: C
Explanation: Aminoacyl-tRNA synthetases identify both the correct amino acid and specific structural features on the tRNA (particularly in the anticodon loop and acceptor stem), ensuring the fidelity of the translation process.
Why is ATP hydrolysis necessary in the reaction catalyzed by aminoacyl-tRNA synthetases?
A. It unfolds the tRNA so that the amino acid can bind freely.
B. It provides the energy to form a high-energy bond between the tRNA and amino acid, which later drives peptide bond formation.
C. It signals the ribosome to begin translation before the synthetase reaction is complete.
D. It powers the import of amino acids from the nucleus to the cytoplasm.
Answer: B
Explanation: The energy from ATP hydrolysis is used to create a covalent, high-energy linkage between the amino acid and the tRNA. This stored energy helps catalyze the subsequent peptide bond formation in the growing polypeptide chain.
Which describes the main function of the large ribosomal subunit during protein synthesis?
A. It recognizes the start codon and recruits initiation factors.
B. It catalyzes the formation of peptide bonds between adjacent amino acids.
C. It matches tRNAs to codons via anticodon–codon base-pairing.
D. It cleaves the mRNA once the protein is fully synthesized.
Answer: B
Explanation: The large subunit contains the peptidyl transferase center, which catalyzes the formation of peptide bonds.
