- A mutation in a bacterial gene creates a variant of Release Factor 1 (RF1) that now recognizes the tryptophan codon UGG in addition to its normal stop codons (UAA and UAG). What is the predicted outcome for a protein whose mRNA contains several internal UGG codons?
Answer: Translation will terminate prematurely at the first UGG codon encountered within the reading frame.
Normally, the UGG codon is recognized by a tRNA carrying tryptophan. Release factors recognize stop codons in the A site and trigger hydrolysis of the polypeptide from the P-site tRNA, terminating translation. If RF1 now recognizes UGG, it will compete with the Trp-tRNA for binding to the A site. When RF1 binds, it will trigger termination. This will result in the synthesis of a truncated polypeptide, with termination occurring at the site of the first internal UGG codon.
- The high fidelity of translation is maintained in part by a 'proofreading' step that occurs after initial codon recognition but before peptide bond formation. This mechanism primarily relies on which of the following?
Answer: The dissociation of an incorrectly paired aminoacyl-tRNA from the A site before it can be fully accommodated.
Translational proofreading involves kinetic selection. After EF-Tu hydrolyzes GTP, the aminoacyl-tRNA must rotate into the peptidyl transferase center ('accommodation'). A correct codon-anticodon pair forms a stable interaction that facilitates this rotation. An incorrect pair is less stable, significantly increasing the likelihood that the entire aminoacyl-tRNA will dissociate from the A site before the peptide bond can be formed. This step occurs after initial binding but before catalysis, acting as a final check.
- During the synthesis of a long polypeptide, a specific tRNA molecule that carries the 10th amino acid of the chain will pass through the ribosomal A, P, and E sites. Which sequence represents the correct order of sites occupied by this particular tRNA?
Answer: A site → P site → E site → Exit
When you encounter questions about tRNA movement through ribosomes, focus on the sequential nature of translation and how each tRNA must follow the same pathway regardless of which amino acid it carries.
During protein synthesis, every tRNA molecule follows an identical three-step journey through the ribosome. The tRNA carrying the 10th amino acid first enters the A site (aminoacyl site), where it arrives carrying its specific amino acid and pairs with the corresponding codon on mRNA. Next, the ribosome shifts during translocation, moving this tRNA to the P site (peptidyl site), where the growing polypeptide chain becomes attached to this tRNA's amino acid. Finally, another translocation event moves the now-empty tRNA to the E site (exit site) before it leaves the ribosome entirely.
Choice A incorrectly starts with the P site, suggesting the tRNA skips the initial A site entry. Choice B omits the A site completely and incorrectly places the P site before the E site, missing the fundamental entry point. Choice C places the E site before the P site, which reverses the actual order of translocation events and would be physically impossible given ribosome mechanics.
Choice D correctly shows the mandatory sequence: A site → P site → E site → Exit. This represents the universal pathway every tRNA follows during translation.
Remember that all tRNAs follow this same A→P→E pathway regardless of their position in the growing protein chain. The ribosome machinery operates like an assembly line with fixed stations that each tRNA must visit in order.
- A typical prokaryotic protein-coding sequence, from start codon to stop codon, is cloned into a standard eukaryotic expression vector that supplies a 5' cap and a poly-A tail. Despite correct transcription, no protein is synthesized. Which statement best explains this failure of translation?
Answer: The eukaryotic ribosome scans from the 5' cap but initiates inefficiently at the AUG codon due to the absence of a surrounding Kozak sequence.
Eukaryotic ribosomes typically initiate translation by binding to the 5' cap and scanning downstream to find the first AUG. The efficiency of initiation at this AUG is highly dependent on the surrounding nucleotide context, known as the Kozak sequence. Prokaryotic genes do not have a Kozak sequence because they use a different initiation mechanism (the Shine-Dalgarno sequence). Therefore, even if the eukaryotic ribosome finds the prokaryotic AUG, it will likely be a very inefficient start site, leading to little or no protein production.
- A researcher creates a modified ribosome in which the rRNA components of the large subunit are left intact, but all of its protein components are replaced with inert structural analogs that maintain the subunit's shape but lack native function. How will this modification most likely affect translation?
Answer: Peptide bond formation will occur, but the subsequent translocation of tRNAs and mRNA through the ribosome will be impaired.
The catalytic activity for peptide bond formation (peptidyl transferase activity) is a function of the ribosomal RNA (rRNA) in the large subunit, making the ribosome a ribozyme. Therefore, this catalytic function would remain intact. However, ribosomal proteins are essential for the large-scale conformational changes required for translocation and for interactions with protein-based elongation factors (like EF-G/eEF2). Replacing these proteins with inert analogs would severely impair the mechanical process of translocation.
- Programmed -1 ribosomal frameshifting often requires a 'slippery sequence' on the mRNA and a downstream RNA secondary structure. What is the primary function of this downstream secondary structure (e.g., a pseudoknot) in promoting the frameshift?
Answer: It causes the translating ribosome to pause, increasing the likelihood of tRNA slippage and realignment on the slippery sequence.
The downstream secondary structure acts as a physical barrier to the advancing ribosome. The ribosome must pause and use its helicase activity to unwind the structure. This pause increases the time that the A- and P-site tRNAs spend on the upstream slippery sequence. The extended dwell time provides a greater opportunity for the tRNAs to break their original codon-anticodon pairing and 'slip' backward by one nucleotide to an alternative, stable pairing in the -1 frame. The pause is therefore a critical kinetic component of the frameshifting mechanism.
- Excluding the cost of tRNA charging (amino acid activation), how many high-energy phosphate bonds (from GTP) are consumed to synthesize one complete 150-amino-acid prokaryotic polypeptide?
Answer: 301
The total number of GTP molecules hydrolyzed is calculated by summing the usage at each stage:
1. Initiation: 1 GTP is hydrolyzed by IF2 during the assembly of the 70S initiation complex.
2. Elongation: For a 150-amino-acid protein, there are 149 cycles of elongation. Each cycle consumes two GTPs: one by EF-Tu to deliver the aminoacyl-tRNA, and one by EF-G for translocation. This totals 149 × 2 = 298 GTP.
3. Termination: 1 GTP is hydrolyzed by RF3 to promote the dissociation of the release factors from the ribosome.
4. Ribosome Recycling: After the polypeptide is released, 1 more GTP is hydrolyzed by EF-G in conjunction with Ribosome Recycling Factor (RRF) to dissociate the ribosomal subunits from the mRNA.
Total = 1 (initiation) + 298 (elongation) + 1 (termination) + 1 (recycling) = 301 GTP.
- Puromycin is an antibiotic that structurally mimics the 3' end of an aminoacyl-tRNA. It enters the A site and is linked to the growing polypeptide chain. Given its mechanism, what is the direct consequence of puromycin action?
Answer: It causes the premature dissociation of the polypeptide chain from the ribosome.
Puromycin enters the A site and participates in peptide bond formation, resulting in the polypeptide being transferred to it from the P-site tRNA. However, the resulting peptidyl-puromycin molecule is not held firmly in the A site (as it's not a full tRNA) and cannot be translocated to the P site. Consequently, it dissociates from the ribosome, leading to the premature termination of translation and release of the truncated, puromycin-capped polypeptide.
- Regulated stop codon readthrough is a translational control mechanism used to produce C-terminally extended proteins. Which of the following is a well-established molecular feature that promotes this event at a specific UGA codon?
Answer: The presence of a specific mRNA secondary structure, such as a pseudoknot, immediately downstream of the stop codon.
Programmed stop codon readthrough often depends on cis-acting elements in the mRNA. A common mechanism involves a specific RNA structure, like a pseudoknot, located 3' to the stop codon. This structure causes the translating ribosome to pause. The pause alters the conformation of the ribosome at the A site and increases the time available for a near-cognate tRNA to bind to the stop codon, outcompeting the release factors and leading to the insertion of an amino acid instead of termination.
- During elongation, the ribosome's decoding center ensures fidelity but also accommodates non-Watson-Crick 'wobble' pairing. This balance between fidelity and flexibility is most critical at which point in the cycle?
Answer: When an aminoacyl-tRNA first enters the A site and is scrutinized before peptide bond formation.
The decoding center is located in the small ribosomal subunit and is responsible for monitoring the codon-anticodon interaction in the A site. This is the critical step where the ribosome must distinguish between correct and incorrect tRNAs. The geometry of the first two base pairs is strictly monitored, while the third (wobble) position is allowed more flexibility. This check occurs upon initial entry of the tRNA into the A site and before the irreversible step of peptide bond formation.
- Electron microscopy of a polysome reveals that ribosomes are spaced very closely near the 3' end of the mRNA, while they are more sparsely distributed near the 5' end. Which of the following is the most plausible explanation for this observation?
Answer: The rate of ribosomal translocation slows significantly as ribosomes approach the 3' end of the transcript.
The density of ribosomes on an mRNA is analogous to traffic on a highway. A high density (a 'traffic jam') indicates that the rate of movement has slowed down in that region. If ribosomes translocate more slowly near the 3' end—perhaps due to mRNA secondary structure, stretches of rare codons, or other obstacles—then ribosomes moving along behind them will catch up, resulting in closer spacing. A slow initiation rate (A) would lead to sparse spacing throughout. Inefficient termination (C) would cause a cluster only at the very end. 5' degradation (D) would remove ribosomes, not cause them to cluster at the other end.
- A cell line expresses a dominant-negative mutant of the eukaryotic initiation factor eIF4E. This mutant can bind to the 5' mRNA cap but cannot interact with eIF4G. What is the most precise description of the impact on cellular protein synthesis?
Answer: Cap-dependent translation will be severely inhibited, while translation initiated at Internal Ribosome Entry Sites (IRES) may be unaffected.
eIF4E's function is to bind the 5' cap and, through its interaction with eIF4G, recruit the rest of the initiation machinery (including the 40S subunit). The dominant-negative mutant sequesters the 5' caps of mRNAs, preventing the functional eIF4F complex from forming and blocking the initiation of cap-dependent translation. However, some viral and cellular mRNAs contain an Internal Ribosome Entry Site (IRES), which is an RNA structure that recruits the ribosome directly to the mRNA, bypassing the need for the 5' cap and eIF4E. Therefore, IRES-mediated, or cap-independent, translation would be largely unaffected by this specific mutant.
- Prokaryotic ribosomes stalled on a truncated mRNA lacking a stop codon are rescued by tmRNA. What are the two primary functions of the tmRNA in this ribosome rescue system?
Answer: It provides an open reading frame to add a degradation tag to the polypeptide and a stop codon to terminate translation.
When you encounter questions about prokaryotic translation rescue mechanisms, focus on understanding what happens when ribosomes get stuck and how the cell resolves this potentially harmful situation.
The tmRNA (transfer-messenger RNA) system is a elegant two-part solution to ribosome stalling. When a ribosome encounters truncated mRNA without a stop codon, it becomes stuck because there's no signal to terminate translation. The tmRNA molecule has a unique dual structure: it contains both tRNA-like and mRNA-like regions that work together to solve this problem.
Answer D correctly identifies tmRNA's two functions. First, its mRNA-like region provides an open reading frame that codes for a peptide tag (often called an SsrA tag) that gets added to the incomplete protein. This tag marks the protein for degradation by cellular proteases. Second, this same mRNA region contains a stop codon that allows the ribosome to properly terminate translation and dissociate from the mRNA.
Answer A is wrong because tmRNA doesn't directly recruit enzymes—it provides the degradation signal and termination sequence. Answer B incorrectly suggests tmRNA has hydrolase activity; it works through the ribosome's existing peptidyl transferase activity. Answer C is incorrect because tmRNA doesn't repair the original mRNA—instead, it provides an alternative template to complete translation with a degradation tag.
Remember that tmRNA is essentially a "molecular band-aid" that simultaneously tags defective proteins for destruction and allows stuck ribosomes to complete translation normally. Focus on its dual tRNA/mRNA nature when studying ribosome rescue systems.
- In an in vitro prokaryotic translation system, a non-hydrolyzable analog of GTP (GTP-γ-S) replaces GTP. The system contains all other necessary components. After the initiator fMet-tRNA has bound and the second aminoacyl-tRNA has successfully entered the A site, what will be the immediate consequence of the presence of GTP-γ-S?
Answer: The ribosome will stall with a dipeptidyl-tRNA in the A site and an uncharged tRNA in the P site.
The elongation cycle requires GTP hydrolysis for two key steps: 1) the delivery of an aminoacyl-tRNA to the A site by EF-Tu, and 2) the translocation of the ribosome along the mRNA by EF-G. The question states that the second aminoacyl-tRNA has already entered the A site, and a peptide bond can form (this is catalyzed by rRNA and does not require GTP). The next step is translocation, which is mediated by EF-G and requires GTP hydrolysis. With the non-hydrolyzable GTP-γ-S, EF-G cannot function, so translocation is blocked. This leaves the ribosome stalled with the newly formed dipeptidyl-tRNA in the A site and the now-uncharged initiator tRNA in the P site.
- A mutation in an aminoacyl-tRNA synthetase eliminates its editing domain, causing it to mischarge a fraction of tRNA-Ile molecules with Valine (Val). What is the direct consequence during translation when this mischarged Val-tRNA-Ile encounters an Isoleucine codon?
Answer: The ribosome will incorporate Valine into the polypeptide at a position specified for Isoleucine.
When you encounter questions about aminoacyl-tRNA synthetase errors, focus on understanding what happens at each step of translation and where quality control occurs.
Aminoacyl-tRNA synthetases normally have editing domains that catch mistakes when the wrong amino acid gets attached to a tRNA. Without this editing function, some tRNA-Ile molecules will carry Valine instead of Isoleucine. The key insight is that once translation begins, the ribosome primarily recognizes the tRNA's anticodon, not the amino acid it carries.
When the mischarged Val-tRNA-Ile encounters an Isoleucine codon, the anticodon perfectly matches the codon, so the ribosome accepts it into the A site. The ribosome then incorporates whatever amino acid the tRNA is carrying—in this case, Valine—into the growing protein chain. This creates a point mutation at the protein level: Valine appears where Isoleucine should be.
Answer A correctly describes this process. Answer B is wrong because the ribosome's proofreading mechanisms check codon-anticodon pairing, not whether the correct amino acid is attached to the tRNA. Answer C is incorrect because the mischarged tRNA will be accommodated normally since its anticodon matches the codon. Answer D is wrong because the peptidyl transferase center can form peptide bonds with any amino acid—it doesn't verify that the "correct" amino acid is present.
Remember: translation fidelity depends heavily on aminoacyl-tRNA synthetases getting it right initially. Once a charged tRNA enters the ribosome, codon-anticodon matching is the primary quality control mechanism, not amino acid verification.
- An in vitro experiment uses a hybrid ribosome assembled from the 30S small subunit of an E. coli ribosome and the 60S large subunit of a human ribosome. This chimera is placed in a translation system with all necessary factors from both organisms. Which step of protein synthesis is LEAST likely to be completed successfully?
Answer: Coordinated translocation of the hybrid ribosome along the mRNA.
Translocation is a complex mechanical process that requires precise, coordinated movement between the small and large ribosomal subunits. The interface between these subunits is critical for this conformational change. Additionally, translocation is driven by elongation factors (EF-G in bacteria, eEF2 in eukaryotes) that must bind to specific sites on the ribosome. A hybrid ribosome composed of subunits from different domains of life would likely have an incompatible interface, preventing the coordinated movement required for translocation. The respective elongation factors would also likely fail to function correctly with the hybrid.
- The Shine-Dalgarno sequence of a prokaryotic gene is 5'-AGGAGGU-3', which interacts with the 3' end of the 16S rRNA (3'-AUUCCUCCA-5'). Which of the following mutations in the Shine-Dalgarno sequence would most likely increase the rate of translation initiation?
Answer: 5'-AGGAGGA-3'
When you encounter questions about the Shine-Dalgarno sequence, focus on the principle of complementary base pairing strength. The Shine-Dalgarno sequence helps position the ribosome correctly on mRNA by binding to the 16S rRNA through Watson-Crick base pairing. Stronger binding between these sequences leads to more efficient translation initiation.
To determine which mutation increases translation, you need to analyze the base pairing between each mutated Shine-Dalgarno sequence and the 16S rRNA (3'-AUUCCUCCA-5'). Count the number of complementary base pairs formed:
Original sequence 5'-AGGAGGU-3' forms 6 base pairs with the rRNA. Option A (5'-AGGAGGA-3') creates 7 base pairs by changing the terminal U to A, which pairs with the terminal A of the rRNA. This additional base pair strengthens the interaction, increasing translation initiation efficiency.
Option B (5'-AGUAGGU-3') only forms 5 base pairs due to the G→U substitution that disrupts pairing with the rRNA's C. Option C (5'-ACGAGGU-3') also forms 5 base pairs because the G→C change eliminates pairing with the rRNA's U. Option D (5'-AGGAGGC-3') forms 6 base pairs, the same as the original sequence, since C cannot pair with the rRNA's terminal A.
Remember this pattern: in Shine-Dalgarno questions, more complementary base pairs generally mean stronger ribosome binding and increased translation efficiency. Always map out the base pairing systematically rather than guessing based on sequence similarity alone.
- A rare type of suppressor tRNA is found to recognize a four-base codon and correct a +1 frameshift mutation. For this suppression mechanism to successfully restore the proper reading frame for the rest of the mRNA, which corresponding action must be taken by the ribosome?
Answer: The ribosome must translocate by four nucleotides, instead of three, after the suppressor tRNA has been used.
A +1 frameshift means an extra base has been inserted. To correct this, the ribosome needs to 'skip' a base to get back into the original frame. A tRNA that reads four bases instead of three will cause the ribosome to advance its reading window by four nucleotides during translocation. This effectively moves the ribosome past the inserted base and restores the downstream reading frame. The other events are either prerequisites (like A site accommodation) or incorrect; the critical event that defines frameshift suppression is the altered translocation distance.
- In prokaryotic translation, a specific base-pairing interaction is essential for positioning the small ribosomal subunit at the correct start site. A mutation that disrupts this interaction would most directly interfere with which process?
Answer: The binding of the 30S subunit to the Shine-Dalgarno sequence on the mRNA.
The crucial interaction for initiating prokaryotic translation is the base pairing between the Shine-Dalgarno (SD) sequence on the mRNA and a complementary anti-SD sequence at the 3' end of the 16S rRNA within the 30S (small) ribosomal subunit. This interaction anchors the 30S subunit onto the mRNA, positioning the start codon in what will become the P site. A mutation disrupting this pairing would directly prevent this initial binding and alignment step.
