- A researcher compares translation of two mRNAs that differ only in the position of a stop codon. mRNA X: 5′-AUG AAA UAA GGG-3′. mRNA Y: 5′-AUG AAA GGG UAA-3′. In both cases, AUG recruits the initiator tRNA, and UAA is recognized by a release factor that triggers polypeptide release. The ribosome translates codons sequentially from 5′ to 3′ until it reaches a stop codon. Which outcome is most likely when comparing the two polypeptides?
Answer: mRNA X produces a shorter polypeptide than mRNA Y because termination occurs earlier on X.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is A because mRNA X has the UAA stop codon earlier, after AUG and AAA, leading to termination and a shorter polypeptide compared to mRNA Y, where UAA is after an additional GGG codon. Both initiate at AUG with the initiator tRNA, but the ribosome translates sequentially until reaching the stop, releasing the chain via a release factor. This position difference results in mRNA Y producing a longer chain with an extra amino acid. A tempting distractor is B, which falsely claims mRNA X is longer due to UAA recruiting a tRNA, based on the misconception that stop codons add amino acids like regular codons. A transferable strategy is to count the codons between start and stop when comparing mRNAs, as stop position directly determines polypeptide length.
- Translation requires that the ribosome maintain a reading frame, grouping nucleotides into codons from the start codon. A wild-type mRNA segment is 5′-AUG AAA CCG UUU GGA UAA-3′. In a mutant, a single nucleotide is inserted immediately after the start codon, producing 5′-AUG UAAA ACC GUU UGG AUAA-3′ when regrouped from the same start site. Assume the ribosome initiates at AUG in both cases and continues until it encounters the first in-frame stop codon. Which outcome is most likely for the mutant compared with the wild type?
Answer: The mutant polypeptide differs because the insertion shifts the reading frame for downstream codons.
This question assesses the skill of analyzing translation by evaluating the impact of a frameshift mutation on the reading frame and polypeptide sequence. The insertion of a single nucleotide after the start codon shifts the grouping of all downstream nucleotides into codons, changing the sequence from the wild-type AUG-AAA-CCG-UUU-GGA-UAA to the mutant AUG-UAA-AAC-CGU-UUG-GAU-AA in the provided regrouping. This frameshift alters the amino acids specified by the new codons and may change when the ribosome encounters an in-frame stop codon, resulting in a different polypeptide. Consequently, the mutant polypeptide differs from the wild-type due to the shifted reading frame affecting all codons after the insertion point. A tempting distractor is A, which states the polypeptide is identical because ribosomes remove inserted nucleotides, but this is incorrect due to the misconception that ribosomes edit mRNA during translation rather than reading it as is. To approach similar mutation problems, redraw the codon groupings starting from the start codon to visualize how insertions or deletions shift the frame and alter the sequence.
- A student compares two mRNAs translated in the same cytosol. mRNA 1 begins with AUG and has many codons before a stop codon. mRNA 2 lacks an AUG near its 5′ end but contains the same internal codons as mRNA 1. Translation requires an initiator tRNA to pair with a start codon to establish the reading frame, after which the ribosome matches tRNAs to codons sequentially. Which outcome is most likely for mRNA 2?
Answer: No polypeptide forms because initiation is unlikely without a start codon to set the frame.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because mRNA 2 lacks an AUG start codon near its 5′ end, making initiation unlikely as the initiator tRNA requires it to set the reading frame. Without proper initiation, the ribosome does not reliably begin translation, even if internal codons match those in mRNA 1. Translation on mRNA 1 proceeds normally from AUG to the stop codon, but mRNA 2's absence of a start prevents polypeptide synthesis. A tempting distractor is A, which falsely claims an identical polypeptide forms, arising from the misconception that internal codons can independently trigger initiation. A transferable strategy is to check for the presence and position of the start codon when comparing mRNAs, as it is crucial for establishing the translation reading frame.
- An mRNA is translated starting at the first AUG encountered by the ribosome, establishing the reading frame. A mutation changes the mRNA sequence upstream of the coding region, but the AUG start codon and all codons after it are unchanged. The ribosome binding and initiation still occur at the same AUG, and elongation proceeds codon by codon until the same stop codon. No RNA editing occurs. Which outcome is most likely for the polypeptide produced from the mutant mRNA compared with the original?
Answer: The polypeptide is identical because codons translated after the start codon are unchanged.
This question assesses the skill of analyzing translation by assessing how upstream mutations outside the coding region affect the polypeptide. The mutation is upstream of the start codon AUG, leaving the coding sequence from AUG to the stop codon unchanged. The ribosome initiates at the same AUG, establishing the same reading frame, and translates the identical codons into the same amino acid sequence. Therefore, the polypeptide produced from the mutant mRNA is identical to the original, as translation depends only on the sequence from the start codon onward. A tempting distractor is B, which suggests a frameshift because upstream mutations change all downstream codons, but this is incorrect due to the misconception that mutations before the start codon alter the reading frame within the coding region. For similar analyses, focus on whether the mutation affects the sequence between the start and stop codons, as upstream changes typically do not impact the translated product if initiation remains the same.
- During translation, the ribosome moves along mRNA in the 5′ to 3′ direction, decoding each codon in order. A researcher introduces an mRNA in which the start codon AUG is unchanged, but the stop codon is mutated to a sense codon that is recognized by a tRNA. The mRNA sequence downstream contains additional codons before a later stop codon. Assume initiation occurs normally and the ribosome remains bound to the mRNA. Which outcome is most likely for the translated product compared with the original mRNA?
Answer: A longer polypeptide is produced because translation continues to the next in-frame stop codon.
This question assesses the skill of analyzing translation by predicting the effect of mutating a stop codon to a sense codon on polypeptide length. Mutating the stop codon to a sense codon allows a tRNA to bind and add an amino acid at that position, preventing normal termination. The ribosome continues translating downstream codons until it reaches the next in-frame stop codon further along the mRNA. As a result, the translated product is a longer polypeptide incorporating additional amino acids beyond the original termination site. A tempting distractor is B, which claims a shorter polypeptide because the mutation prevents initiation at AUG, but this is incorrect due to the misconception that downstream mutations affect upstream initiation rather than only elongation and termination. To evaluate similar mutations, trace the translation path from the unchanged start codon and note how changes in stop signals extend or shorten the coding sequence.
- A ribosome has an A site (acceptor) and P site (peptidyl) that coordinate elongation. In a cell-free system, a drug prevents peptidyl transferase activity, so peptide bonds cannot form between the amino acid in the A site and the growing chain in the P site. Initiation and codon-anticodon pairing still occur, and tRNAs can still enter the A site. Which outcome is most likely during translation in the presence of the drug?
Answer: Elongation halts because amino acids cannot be covalently linked into a growing polypeptide chain.
This question analyzes how blocking peptide bond formation affects translation. Peptidyl transferase activity in the ribosome catalyzes the formation of peptide bonds between the amino acid in the A site and the growing polypeptide in the P site. When this activity is blocked by the drug, tRNAs can still enter and bind codons, but no peptide bonds form, preventing chain elongation beyond the initiator methionine. Students choosing A incorrectly think tRNAs form peptide bonds, but this catalysis occurs within the ribosome's peptidyl transferase center. When analyzing translation inhibitors, identify which specific step is blocked and trace the downstream consequences for polypeptide synthesis.
- An mRNA contains the sequence 5′-AUG GAA CUA UAG-3′, where UAG is a stop codon. A researcher introduces a modified tRNA whose anticodon base-pairs with UAG and is charged with an amino acid. In normal cells, release factors bind stop codons and promote polypeptide release. Assume the modified tRNA competes successfully with the release factor at UAG. Which outcome is most likely for translation of this mRNA?
Answer: Translation continues past UAG, adding an amino acid at that position before proceeding.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because the modified tRNA base-pairs with the UAG stop codon and adds its charged amino acid, competing with the release factor and allowing translation to continue past UAG. Normally, stop codons like UAG trigger release factors for termination, but the suppressor tRNA enables read-through by incorporating an amino acid instead. This results in elongation proceeding beyond the intended stop, potentially extending the polypeptide if more codons follow. A tempting distractor is A, which incorrectly asserts termination occurs normally, based on the misconception that stop codons cannot pair with any anticodon even if a modified tRNA is present. A transferable strategy is to consider how alterations like suppressor tRNAs can override normal stop codon function, evaluating competition with release factors.
- A student models translation using an mRNA with a start codon followed by several sense codons and then a stop codon. They note that tRNAs bind codons through complementary base pairing between the codon (mRNA) and anticodon (tRNA), and the ribosome catalyzes peptide bond formation as each amino acid is added. At the stop codon, no tRNA anticodon pairs; instead, a release factor binds and hydrolyzes the bond between the polypeptide and the tRNA in the ribosome. Which explanation best accounts for how translation terminates at a stop codon?
Answer: A release factor binds the stop codon and triggers polypeptide release from the ribosome.
This question assesses the skill of analyzing translation by explaining the mechanism of termination at a stop codon. When a stop codon like UAA, UAG, or UGA enters the ribosome's A site, no tRNA anticodon complements it, preventing the addition of an amino acid. Instead, a release factor protein binds to the stop codon in the A site, triggering the hydrolysis of the bond between the completed polypeptide and the tRNA in the P site. This action releases the polypeptide from the ribosome, disassembles the translation complex, and ends the process. A tempting distractor is B, which suggests a specialized tRNA binds the stop codon to add a final amino acid, but this is wrong due to the misconception that stop codons are decoded like sense codons rather than recognized by release factors. For understanding termination in similar scenarios, recall that stop codons signal release without amino acid incorporation, distinguishing them from sense codons that pair with tRNAs.
- A researcher provides a ribosome with an mRNA that begins 5′-AUG AAA GGC UAA-3′. In one treatment, a drug prevents the initiator tRNA from binding the start codon AUG, but it does not affect elongation once initiation occurs. In a second treatment, the drug does not affect initiation but prevents peptide bond formation between amino acids. In untreated cells, UAA is recognized by a release factor to terminate translation. Which outcome is most likely in the first treatment condition?
Answer: No polypeptide forms because the ribosome cannot initiate translation at AUG.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because the drug prevents the initiator tRNA from binding the AUG start codon, blocking the initiation step essential for ribosome assembly and reading frame establishment. Without initiation, elongation cannot occur, even though the mRNA has codons like AAA and GGC that could be translated if initiation happened. The UAA stop codon would normally terminate translation via a release factor, but no polypeptide is synthesized in this treatment. A tempting distractor is A, which incorrectly claims a full-length polypeptide forms without initiation, arising from the misconception that elongation can proceed independently of start codon recognition. A transferable strategy is to distinguish between initiation and elongation steps, verifying that disruptions to initiation halt the entire translation process.
- During translation, a ribosome begins at the start codon and reads mRNA in groups of three nucleotides (codons). A student provides an mRNA segment: 5′-AUG GCU UAC GAA UGA-3′. The tRNA carrying methionine binds the start codon, and each subsequent codon is matched by a tRNA with a complementary anticodon, adding one amino acid per codon. Translation ends when a stop codon enters the ribosome, and no tRNA binds that codon. Assume this segment is in-frame and is translated without skipping or editing. Which outcome is most likely for the polypeptide produced from this mRNA segment?
Answer: A polypeptide of four amino acids is released when the ribosome reaches UGA.
This question assesses the skill of analyzing translation by interpreting an mRNA sequence to determine the resulting polypeptide. The ribosome initiates at the start codon AUG, which codes for methionine, and then reads subsequent codons GCU (alanine), UAC (tyrosine), and GAA (glutamic acid), adding each corresponding amino acid to the growing chain. Translation proceeds until the stop codon UGA enters the ribosome, at which point no tRNA binds, and the polypeptide is released without adding an amino acid for the stop codon. Thus, the mRNA segment produces a polypeptide with four amino acids: methionine-alanine-tyrosine-glutamic acid. A tempting distractor is B, which claims a five-amino-acid polypeptide because UGA codes for a final amino acid, but this is incorrect due to the misconception that stop codons encode amino acids rather than signaling termination. To solve similar problems, always count the number of codons from the start codon up to but not including the stop codon, as stop codons do not add amino acids.
- A researcher engineers an mRNA with a strong secondary structure (a stable hairpin) that forms within the coding region. The hairpin begins just after codon 6 and physically impedes ribosome movement along the mRNA, but it does not change the codon sequence itself. Initiation occurs normally at the start codon, and tRNAs are available. Which outcome is most likely for translation of this mRNA?
Answer: Elongation will slow or pause near codon 6 because ribosome translocation is hindered.
This question tests understanding of how mRNA secondary structures affect ribosome movement. A stable hairpin structure in the mRNA creates a physical barrier that impedes the ribosome's translocation along the mRNA during elongation, causing translation to slow or pause when the ribosome encounters this structure around codon 6. The ribosome can eventually unfold the hairpin through its helicase activity, but this takes time and energy, reducing translation speed. Students choosing C incorrectly think RNA structures don't affect translation, confusing this with DNA replication, but ribosomes must physically move along mRNA and can be blocked by stable structures. To analyze translation efficiency, consider both the sequence information and physical accessibility of the mRNA.
- A mutation occurs in a tRNA gene such that the anticodon is altered, but the aminoacyl-tRNA synthetase still attaches the same amino acid to that tRNA as before. The altered anticodon now base-pairs with a different codon on mRNA. Ribosomes and mRNAs are otherwise normal. Which outcome is most likely in proteins translated in cells carrying this mutation?
Answer: The same amino acid may be inserted at a new codon because the tRNA carries its original amino acid but reads a different codon.
This question tests understanding of how tRNA mutations affect the genetic code. When a tRNA's anticodon mutates but the synthetase still charges it with the original amino acid, this tRNA will now deliver that same amino acid to a different codon position (wherever its new anticodon base-pairs). This creates a codon reassignment where the new codon now codes for the amino acid carried by the mutant tRNA instead of its normal amino acid. Students choosing A incorrectly think the mRNA codon directly determines amino acid identity, but amino acids are determined by which aminoacyl-tRNA binds the codon. To analyze tRNA mutations, consider both the anticodon change (affecting where it binds) and the synthetase specificity (affecting which amino acid it carries).
- During translation, a ribosome reads an mRNA segment 5′-AUG GCU UAA-3′. A tRNA with an anticodon complementary to GCU delivers alanine. The AUG codon is recognized by an initiator tRNA carrying methionine, and UAA is a stop codon that is recognized by a release factor rather than a tRNA. The ribosome moves along the mRNA one codon at a time and forms peptide bonds between adjacent amino acids as matching tRNAs bind. Which outcome is most likely for the synthesized polypeptide?
Answer: A dipeptide forms: methionine linked to alanine, then translation terminates at UAA.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is A because translation initiates at the AUG codon with methionine, followed by the GCU codon delivering alanine via its complementary tRNA, forming a dipeptide. The ribosome then encounters the UAA stop codon, which is recognized by a release factor rather than a tRNA, leading to termination without adding another amino acid. This results in the release of the Met-Ala dipeptide, as the ribosome moves codon by codon, forming peptide bonds between adjacent amino acids. A tempting distractor is B, which incorrectly assumes UAA recruits a tRNA to insert a third amino acid, stemming from the misconception that stop codons encode amino acids like regular codons. A transferable strategy is to map out the codon sequence and recall that stop codons signal termination via release factors, not tRNA binding.
- A ribosome translates an mRNA segment 5′-AUG CCC AAA GGG UAA-3′ in the shown reading frame. A mutation deletes the second nucleotide of the second codon, changing the sequence to 5′-AUG CCA AAG GGU AA-3′. Translation initiates at AUG and then continues by reading codons in the new frame until a stop codon is reached. No additional nucleotides are added, and the ribosome does not realign to the original frame. Which outcome is most likely for the mutant translation product?
Answer: The polypeptide sequence changes after the deletion because the reading frame shifts downstream.
This question assesses the skill of analyzing translation by determining the effect of a single nucleotide deletion on the reading frame and polypeptide. Deleting the second nucleotide of the second codon shifts the grouping of all subsequent nucleotides, changing the codons from the original AUG-CCC-AAA-GGG-UAA to the new frame AUG-CCA-AAG-GGU-AA, potentially continuing until a new stop codon. This frameshift alters the amino acid sequence after the deletion point and may change the polypeptide length depending on when a stop codon appears in the new frame. Consequently, the mutant polypeptide sequence changes downstream of the deletion compared to the wild type. A tempting distractor is A, which claims the polypeptide is unchanged because deleting one nucleotide removes exactly one amino acid, but this is wrong due to the misconception that deletions affect only the immediate codon without shifting the frame. To solve similar deletion problems, regroup the nucleotides into codons starting from the start codon and compare the new sequence to identify frameshift effects and potential early or late stops.
- A ribosome begins translating an mRNA that normally contains the codons 5′-AUG CCG GAA UGA-3′. In a mutant, the second codon changes from CCG to CCU. Both CCG and CCU are codons that can be matched by tRNAs carrying the same amino acid (proline), because different codons can specify the same amino acid. The start codon AUG still recruits the initiator tRNA, and UGA is still a stop codon recognized by a release factor. Which outcome is most likely for the polypeptide produced from the mutant mRNA?
Answer: The amino acid sequence is unchanged because CCU and CCG recruit proline-charged tRNAs.
This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because both CCG and CCU code for proline due to the degeneracy of the genetic code, where multiple codons can specify the same amino acid via tRNAs. The mutation from CCG to CCU thus does not change the amino acid inserted at that position, and translation proceeds from the AUG start to the UGA stop, producing the same polypeptide sequence. The start codon still initiates with methionine, and the stop codon terminates normally, unaffected by the substitution. A tempting distractor is D, which wrongly suggests that only one codon specifies proline, based on the misconception that the genetic code lacks redundancy. A transferable strategy is to consult the genetic code table for codon-amino acid assignments when evaluating mutations, focusing on whether the change alters the specified amino acid.
- A ribosome translates an mRNA by matching each codon with a tRNA anticodon, adding the amino acid carried by that tRNA to the growing polypeptide. A researcher introduces a mutation that deletes one nucleotide from the mRNA within the second codon of this sequence: 5′-AUG AAA GGC UCU-3′. Translation still initiates at AUG, and the ribosome continues reading triplets from that start. Assume no early stop codon is created in the region shown. Which outcome is most likely after the deletion?
Answer: The amino acid sequence from the deletion onward is altered because the reading frame shifts.
This question requires analyzing how frameshift mutations affect translation. When one nucleotide is deleted from the second codon, the reading frame shifts for all subsequent codons because ribosomes read mRNA in fixed triplets starting from the initiation site. After the deletion, what was originally read as AAA|GGC|UCU becomes AA?|G?C|?CU (where ? represents shifted positions), completely changing which three nucleotides form each codon downstream. Students choosing A incorrectly believe ribosomes can somehow realign to the original frame, but ribosomes lack this ability and mechanically read triplets from their starting point. When analyzing deletions or insertions, remember that unless the number is a multiple of 3, the reading frame shifts and all downstream amino acids change.
- During translation, a ribosome begins at a start codon and reads an mRNA 5′→3′ in sets of three nucleotides. A student compares two mRNAs that are identical except at codon 4: mRNA 1 has 5′-AUG GCU UAC GAA CCG-3′, and mRNA 2 has 5′-AUG GCU UAC UAA CCG-3′. In this system, UAA is a stop codon recognized by a release factor rather than a tRNA. All other codons shown are sense codons that recruit tRNAs carrying amino acids. Which outcome is most likely for translation of mRNA 2 compared with mRNA 1?
Answer: The polypeptide will be shorter because termination occurs at codon 4 after release factor binding.
This question tests your ability to analyze how stop codons affect translation outcomes. In mRNA 2, codon 4 changes from GAA (which codes for glutamate) to UAA, which is a stop codon that recruits release factors instead of tRNAs. When the ribosome encounters UAA at codon 4, the release factor binds and triggers termination, producing a polypeptide containing only the amino acids from codons 1-3 (Met-Ala-Tyr). Students choosing B incorrectly think ribosomes can skip stop codons, but release factors specifically recognize stop codons and cause termination. To solve translation problems, identify stop codons (UAA, UAG, UGA) and remember they always trigger termination when encountered.
- In an in vitro translation reaction, tRNAs are charged with amino acids by aminoacyl-tRNA synthetases before entering the ribosome. A mutation reduces the ability of one synthetase to attach its amino acid to its specific tRNA, but the tRNA can still base-pair normally with its matching codon in the ribosome. Ribosomes, mRNA, and all other charged tRNAs are abundant. Which outcome is most likely for polypeptides produced from mRNAs containing that codon?
Answer: Polypeptides will frequently stall at that codon because uncharged matching tRNA cannot donate an amino acid.
This question tests understanding of how uncharged tRNAs affect translation elongation. When a specific aminoacyl-tRNA synthetase cannot charge its tRNA with the appropriate amino acid, that tRNA enters the ribosome without an amino acid attached. While the uncharged tRNA can still base-pair with its codon in the A site, it cannot donate an amino acid for peptide bond formation, causing the ribosome to stall at that position. Students choosing A incorrectly think a different amino acid would be incorporated, but synthetases are highly specific and other synthetases won't charge this tRNA. To analyze translation efficiency problems, consider whether all components (charged tRNAs, release factors, elongation factors) are functional at each step.
- A ribosome translates an mRNA by matching codons with tRNA anticodons. A student writes an mRNA codon as 5′-AUG-3′ and proposes a tRNA anticodon written 5′-UAC-3′. In the ribosome, base pairing between codon and anticodon is antiparallel. The anticodon must align so that complementary bases pair correctly across the three positions. Which anticodon orientation is most likely to correctly pair with the codon 5′-AUG-3′ during translation?
Answer: 3′-UAC-5′, because anticodons base-pair antiparallel to codons in the ribosome.
Choice B is correct. Codon and anticodon pair antiparallel, so the anticodon's 3′ end lines up with the codon's 5′ end. Pairing 5′-AUG-3′ base by base (A with U, U with A, G with C) gives an anticodon of 3′-UAC-5′, which can also be written 5′-CAU-3′. Choice A is incorrect because codon and anticodon do not pair in the same 5′ to 3′ direction; 5′-UAC-3′ aligned antiparallel would place C opposite A at the first codon position. Choice C is incorrect because identical triplets are not complementary; A cannot pair with A or G with G. Choice D is incorrect because anticodons are complementary to codons, not identical, and 3′-CAU-5′ would place C opposite A at the first codon position. Choice E is incorrect because C pairs with G, not with A; 3′-CAC-5′ would leave a C–A mismatch at the first codon position. Strategy: write the complementary base for each codon position, then label the anticodon so its 3′ end faces the codon's 5′ end.
- A scientist introduces a premature stop codon into an mRNA at codon 3. In this system, stop codons are bound by release factors that trigger hydrolysis of the bond between the polypeptide and the tRNA in the P site, releasing the polypeptide. The ribosome then dissociates from the mRNA. Which explanation best accounts for why the polypeptide produced is truncated?
Answer: Release factors bind the stop codon and catalyze polypeptide release, ending elongation before downstream codons are read.
This question analyzes the mechanism of translation termination at stop codons. When the ribosome encounters a stop codon at position 3, release factors (not tRNAs) bind to the A site and catalyze hydrolysis of the bond between the polypeptide and the tRNA in the P site, releasing the truncated polypeptide and causing ribosome dissociation. This prevents translation of any codons downstream of the stop codon, producing a shortened protein. Students choosing B incorrectly confuse transcription with translation, thinking RNA polymerase is involved, but stop codons only affect the ribosome during translation, not mRNA synthesis. To understand truncated proteins, trace what happens when release factors bind stop codons versus when aminoacyl-tRNAs bind sense codons.
