Genetics Quiz: Translation Stages

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It recruits a protease to degrade the nascent peptide and a nuclease to degrade the faulty mRNA.
2. It hydrolyzes the bond linking the polypeptide to the tRNA and catalyzes the dissociation of the ribosomal subunits.
3. It repairs the broken mRNA by ligating a new 3' end and re-initiates translation from the point of stalling.
4. It provides an open reading frame to add a degradation tag to the polypeptide and a stop codon to terminate translation. (correct answer)

Explanation: 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.

Question 2

1. A specialized editing site on EF-Tu that hydrolyzes the amino acid from a mismatched tRNA.
2. The dissociation of an incorrectly paired aminoacyl-tRNA from the A site before it can be fully accommodated. (correct answer)
3. The peptidyl transferase center, which can sense a mismatch and refuse to catalyze peptide bond formation.
4. The ribosome reversing translocation by one codon to allow for a second attempt at tRNA binding.

Explanation: 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.

Question 3

1. A specialized editing site on EF-Tu that hydrolyzes the amino acid from a mismatched tRNA.
2. The dissociation of an incorrectly paired aminoacyl-tRNA from the A site before it can be fully accommodated. (correct answer)
3. The peptidyl transferase center, which can sense a mismatch and refuse to catalyze peptide bond formation.
4. The ribosome reversing translocation by one codon to allow for a second attempt at tRNA binding.

Explanation: 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.

Question 4

1. Eukaryotic ribosomes require a Shine-Dalgarno sequence, which is absent from the prokaryotic coding sequence.
2. The prokaryotic start codon AUG is not recognized by the eukaryotic initiator Met-tRNA.
3. The eukaryotic ribosome scans from the 5' cap but initiates inefficiently at the AUG codon due to the absence of a surrounding Kozak sequence. (correct answer)
4. The lack of introns in the prokaryotic sequence prevents proper processing and export of the mRNA from the eukaryotic nucleus.

Explanation: 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.

Question 5

1. Binding of the 30S subunit to a prokaryotic mRNA's Shine-Dalgarno sequence.
2. Formation of the first peptide bond between fMet and the second amino acid.
3. Coordinated translocation of the hybrid ribosome along the mRNA. (correct answer)
4. Recognition of an mRNA codon in the A site by its cognate tRNA.

Explanation: 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.

Question 6

1. When an aminoacyl-tRNA first enters the A site and is scrutinized before peptide bond formation. (correct answer)
2. When the peptidyl-tRNA is held in the P site, maintaining the reading frame before translocation.
3. As the uncharged tRNA moves through the E site just prior to exiting the ribosome.
4. At the moment of peptide bond catalysis by the peptidyl transferase center in the large subunit.

Explanation: 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.

Question 7

1. The rate of translation initiation at the 5' end is slow, causing a backup of ribosomes at the 3' end.
2. The rate of ribosomal translocation slows significantly as ribosomes approach the 3' end of the transcript. (correct answer)
3. The termination of translation at the stop codon is inefficient, preventing ribosomes from dissociating quickly.
4. The mRNA is being actively degraded from the 5' end, forcing the ribosomes to cluster at the 3' end.

Explanation: 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.

Question 8

1. It acts as a binding site for a specific protein factor that actively pushes the ribosome into the new reading frame.
2. It base-pairs with the rRNA in the E site, which destabilizes the P-site tRNA and encourages realignment.
3. It melts upon contact with the ribosome, releasing a burst of energy that powers the unconventional translocation event.
4. It causes the translating ribosome to pause, increasing the likelihood of tRNA slippage and realignment on the slippery sequence. (correct answer)

Explanation: 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.

Question 9

1. The ribosome will pause at UGG codons, slowing translation but ultimately producing a full-length protein.
2. Translation will terminate prematurely at the first UGG codon encountered within the reading frame. (correct answer)
3. The full-length protein will be produced, but with a different amino acid substituted for tryptophan at UGG positions.
4. The ribosome will fail to terminate at the gene's authentic UAA or UAG stop codon, producing an extended protein.

Explanation: 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.

Question 10

1. P site → A site → E site → Exit
2. P site → E site → Exit
3. A site → E site → P site → Exit
4. A site → P site → E site → Exit (correct answer)

Explanation: 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.

Question 11

1. All translation will cease because the 40S ribosomal subunit cannot be recruited to any mRNA molecules.
2. The 60S large subunit will be unable to join the initiation complex, stalling translation after start codon recognition.
3. Cap-dependent translation will be severely inhibited, while translation initiated at Internal Ribosome Entry Sites (IRES) may be unaffected. (correct answer)
4. The ribosome will be recruited to mRNA but will be unable to scan for the start codon, leading to widespread initiation failure.

Explanation: 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.

Question 12

1. 5'-AGGAGGA-3' (correct answer)
2. 5'-AGUAGGU-3'
3. 5'-ACGAGGU-3'
4. 5'-AGGAGGC-3'

Explanation: 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.

Question 13

1. The binding of the 30S subunit to the Shine-Dalgarno sequence on the mRNA. (correct answer)
2. The catalytic formation of the first peptide bond by the 50S subunit.
3. The recruitment of the initiator fMet-tRNA to the ribosomal P site.
4. The hydrolysis of GTP by initiation factor 2 (IF2).

Explanation: 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.

Question 14

1. The formation of a dipeptide will be blocked because the peptidyl transferase center requires GTP hydrolysis.
2. The ribosome will stall with a dipeptidyl-tRNA in the A site and an uncharged tRNA in the P site. (correct answer)
3. The third incoming aminoacyl-tRNA, complexed with EF-Tu and GTP-γ-S, will bind irreversibly to the A site.
4. The ribosomal subunits will fail to form a stable initiation complex, preventing any elongation from occurring.

Explanation: 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.

Question 15

1. 151
2. 299
3. 300
4. 301 (correct answer)

Explanation: 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.

Question 16

1. The formation of a dipeptide will be blocked because the peptidyl transferase center requires GTP hydrolysis.
2. The ribosome will stall with a dipeptidyl-tRNA in the A site and an uncharged tRNA in the P site. (correct answer)
3. The third incoming aminoacyl-tRNA, complexed with EF-Tu and GTP-γ-S, will bind irreversibly to the A site.
4. The ribosomal subunits will fail to form a stable initiation complex, preventing any elongation from occurring.

Explanation: 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.

Question 17

1. The ribosome will pause at UGG codons, slowing translation but ultimately producing a full-length protein.
2. Translation will terminate prematurely at the first UGG codon encountered within the reading frame. (correct answer)
3. The full-length protein will be produced, but with a different amino acid substituted for tryptophan at UGG positions.
4. The ribosome will fail to terminate at the gene's authentic UAA or UAG stop codon, producing an extended protein.

Explanation: 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.

Question 18

1. The A site must expand to accommodate the physically larger anticodon loop of the suppressor tRNA.
2. The peptidyl transferase center must operate at a slower rate to allow for the four-base pairing.
3. The ribosome must translocate by four nucleotides, instead of three, after the suppressor tRNA has been used. (correct answer)
4. The E site must hold the suppressor tRNA for a longer duration to ensure the new frame is established.

Explanation: 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.

Question 19

1. The binding of the 30S subunit to the Shine-Dalgarno sequence on the mRNA. (correct answer)
2. The catalytic formation of the first peptide bond by the 50S subunit.
3. The recruitment of the initiator fMet-tRNA to the ribosomal P site.
4. The hydrolysis of GTP by initiation factor 2 (IF2).

Explanation: 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.

Question 20

1. It competitively inhibits the binding of EF-G, stalling the ribosome prior to translocation.
2. It causes the premature dissociation of the polypeptide chain from the ribosome. (correct answer)
3. It cross-links the A and P sites, permanently inactivating the ribosome.
4. It prevents the hydrolysis of GTP by EF-Tu, blocking entry of all subsequent tRNAs.

Explanation: 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.