AP Biology Quiz: Translation

What this quiz covers

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

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. The polypeptide will be shorter because termination occurs at codon 4 after release factor binding. (correct answer)
2. The polypeptide will be longer because the ribosome skips the stop codon and continues elongation.
3. No polypeptide will form because RNA polymerase cannot transcribe an mRNA containing UAA.
4. The same-length polypeptide will form because codon 4 changes only the DNA template strand.
5. A different amino acid will be inserted at codon 4 because UAA pairs with a complementary tRNA anticodon.

Explanation: 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.

Question 2

1. A polypeptide of four amino acids is released when the ribosome reaches UGA. (correct answer)
2. A polypeptide of five amino acids is released because UGA codes for the final amino acid.
3. No polypeptide is produced because AUG is transcribed only in the nucleus, not translated.
4. A polypeptide of three amino acids is produced because translation starts at GCU, not AUG.
5. A polypeptide of four amino acids is produced because the ribosome reads codons as overlapping pairs.

Explanation: 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.

Question 3

1. One amino acid substitution occurs at the position specified by the mutated codon. (correct answer)
2. All downstream amino acids change because any substitution causes a shift in the reading frame.
3. The polypeptide becomes longer because a substitution inserts an extra codon into the mRNA.
4. The polypeptide becomes shorter because substitutions are removed by the ribosome during elongation.
5. No polypeptide forms because a single codon change prevents ribosome binding at the start codon.

Explanation: This question assesses the skill of analyzing translation by examining the impact of a missense mutation on the polypeptide sequence. The single-base substitution changes one codon to another that codes for a different amino acid, but it does not alter the reading frame or the positions of the start and stop codons. The ribosome translates the mRNAs sequentially, incorporating the altered amino acid only at the position of the mutated codon, while all other amino acids remain the same. Thus, the mutant polypeptide differs by a single amino acid substitution at that specific position compared to the original. A tempting distractor is B, which states all downstream amino acids change due to a reading frame shift, but this is incorrect due to the misconception that nucleotide substitutions cause frameshifts like insertions or deletions do. When analyzing point mutations, use a codon table to check if the change affects only one codon and amino acid, without shifting the frame for subsequent codons.

Question 4

1. A longer polypeptide is produced because translation continues to the next in-frame stop codon. (correct answer)
2. A shorter polypeptide is produced because mutating a stop codon prevents initiation at AUG.
3. No polypeptide is produced because a sense codon cannot be read by the ribosome.
4. An identical polypeptide is produced because stop codons do not affect translation length.
5. A longer mRNA is produced because translation adds nucleotides after the mutated stop codon.

Explanation: 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.

Question 5

1. The mutant polypeptide is identical because ribosomes remove inserted nucleotides during translation.
2. The mutant polypeptide differs because the insertion shifts the reading frame for downstream codons. (correct answer)
3. The mutant polypeptide is longer because insertions always eliminate stop codons from mRNA.
4. The mutant polypeptide is unchanged because tRNA anticodons determine the mRNA sequence.
5. No mutant polypeptide forms because insertions prevent transcription of the mRNA in the nucleus.

Explanation: 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.

Question 6

1. Longer polypeptides accumulate because tRNAs remain bound and add amino acids without peptide bonds.
2. Short peptide fragments fail to elongate because peptide bond formation is required for chain growth. (correct answer)
3. More mRNA is produced because blocking translation increases RNA polymerase activity.
4. Normal-length polypeptides form because codon-anticodon pairing alone creates covalent links.
5. Translation proceeds normally because the small subunit catalyzes peptide bonds during elongation.

Explanation: This question assesses the skill of analyzing translation by predicting the consequences of inhibiting peptide bond formation during elongation. The drug blocks the large ribosomal subunit's peptidyl transferase activity, preventing the transfer of the growing polypeptide to the incoming aminoacyl-tRNA despite codon-anticodon pairing. Although initiation and tRNA binding can occur, the chain cannot grow beyond the initial amino acid, leading to stalled ribosomes and short peptide fragments. In treated cells, full-length polypeptides cannot form, resulting in a failure to produce functional proteins compared to untreated cells. A tempting distractor is D, which claims normal-length polypeptides form because codon-anticodon pairing creates covalent links, but this is wrong due to the misconception that base pairing alone forms peptide bonds without ribosomal catalysis. To evaluate similar inhibitor effects, identify which step of translation is blocked and trace how it halts the process, preventing downstream outcomes like chain elongation.

Question 7

1. The polypeptides will differ by one amino acid at position 5 because different tRNAs deliver different amino acids. (correct answer)
2. The polypeptides will be identical because the ribosome ignores codon differences unless a stop codon is present.
3. mRNA Y will not be translated because only one codon is permitted at each position in a reading frame.
4. Both polypeptides will be shorter because any codon change causes release factor binding and termination.
5. The polypeptides will differ in length because the altered codon changes the number of nucleotides in the mRNA.

Explanation: This question analyzes how codon differences affect polypeptide sequences during translation. When position 5 contains different codons in the two mRNAs, different tRNAs will bind at that position, each delivering its specific amino acid (M for mRNA X, N for mRNA Y), resulting in polypeptides that differ by exactly one amino acid at position 5. All other positions remain identical because the codon sequences are the same, recruiting the same tRNAs and amino acids. Students choosing B incorrectly think ribosomes ignore codon differences, but the genetic code strictly determines which tRNA binds each codon. To predict polypeptide differences, compare codon sequences position by position and use the genetic code to determine amino acid changes.

Question 8

1. Ribosomes will incorporate a different amino acid at that codon because codon-anticodon pairing changes.
2. Polypeptides will frequently stall at that codon because uncharged matching tRNA cannot donate an amino acid. (correct answer)
3. Polypeptides will be unchanged because amino acids bind directly to codons without requiring tRNAs.
4. Transcription will fail for those mRNAs because synthetases are required to synthesize RNA nucleotides.
5. Translation will terminate at that codon because the codon becomes a stop codon when tRNA is uncharged.

Explanation: 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.

Question 9

1. Ribosomes will transcribe a complementary RNA strand to unwind the hairpin and continue translation.
2. Elongation will slow or pause near codon 6 because ribosome translocation is hindered. (correct answer)
3. The amino acid sequence will be unchanged and synthesized at the same rate because hairpins affect only DNA replication.
4. Translation will start at codon 6 because hairpins convert upstream codons into introns that are spliced out.
5. Termination will occur at codon 6 because a hairpin is recognized as a stop codon by release factors.

Explanation: 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.

Question 10

1. Translation will initiate normally because any codon can serve as a start codon if tRNAs are abundant.
2. Translation initiation will be greatly reduced because the initiator tRNA does not recognize the altered codon. (correct answer)
3. Elongation will proceed but will skip the first amino acid because the ribosome begins at codon 2 automatically.
4. The mRNA will be converted into DNA before translation, restoring AUG at the start position.
5. Termination will occur at the first codon because AUC is recognized by release factors as a stop signal.

Explanation: This question tests understanding of translation initiation requirements. The initiator tRNA specifically recognizes the AUG start codon through complementary base pairing, and changing AUG to AUC prevents this recognition, greatly reducing or eliminating translation initiation at that site. Without proper initiation complex formation at the intended start site, the ribosome cannot begin elongation from the correct position. Students choosing A incorrectly think any codon can serve as a start codon, but prokaryotic and eukaryotic ribosomes specifically require AUG (occasionally GUG or UUG in bacteria) for efficient initiation. When analyzing translation efficiency, remember that initiation is a highly regulated step requiring specific sequences and factors.

Question 11

1. The polypeptide must change because any base substitution forces the ribosome to shift the reading frame.
2. The polypeptide may be unchanged if the substituted codon recruits a tRNA carrying the same amino acid. (correct answer)
3. Translation will stop because a third-base substitution is interpreted as a termination signal by the ribosome.
4. The mRNA will not bind the ribosome because substitutions prevent formation of phosphodiester bonds in RNA.
5. The amino acid sequence upstream of the mutation changes because codons are read from the 3′ end first.

Explanation: This question analyzes the redundancy (degeneracy) of the genetic code. Many amino acids are encoded by multiple codons that differ only in the third position (wobble position), so a substitution at this position may recruit a different tRNA that carries the same amino acid, resulting in no change to the polypeptide sequence. For example, both UCU and UCC code for serine, so this third-position change wouldn't alter the protein. Students choosing A incorrectly think any base change must alter the amino acid, not understanding genetic code redundancy. When analyzing point mutations, check the genetic code table to determine whether the original and mutated codons encode the same or different amino acids.

Question 12

1. mRNA X produces a shorter polypeptide than mRNA Y because termination occurs earlier on X. (correct answer)
2. mRNA X produces a longer polypeptide than mRNA Y because UAA recruits an extra tRNA.
3. Both mRNAs produce identical polypeptides because stop codons do not affect elongation.
4. Neither mRNA is translated because UAA prevents initiator tRNA binding at AUG.
5. mRNA Y is transcribed more slowly because stop codons reduce RNA polymerase speed.

Explanation: 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.

Question 13

1. Ribosomes read mRNA 5′ to 3′; antiparallel pairing occurs within each codon-anticodon interaction. (correct answer)
2. Ribosomes read mRNA 3′ to 5′, but only when a stop codon is present in the first position.
3. Codons are read on the DNA template strand, so mRNA directionality is irrelevant to translation.
4. Anticodons pair parallel to codons, so the reading direction is determined by tRNA orientation.
5. The amino acid sequence is determined by amino acid availability, not by codon order on mRNA.

Explanation: This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is A because the ribosome reads mRNA in the 5′ to 3′ direction, advancing one codon at a time, while each codon-anticodon pairing is antiparallel within that framework. This means the overall amino acid sequence is determined by the 5′ to 3′ codon order, not by reversing the reading direction. The student's error overlooks that antiparallel pairing occurs locally for each codon, but the ribosome's movement maintains the 5′ to 3′ progression. A tempting distractor is B, which falsely states ribosomes read 3′ to 5′ with stop codons, based on the misconception that directionality reverses under specific conditions. A transferable strategy is to visualize the ribosome's 5′ to 3′ movement along mRNA, integrating local antiparallel base pairing without altering the overall direction.

Question 14

1. Translation terminates normally because stop codons cannot base-pair with any anticodon.
2. Translation continues past UAG, adding an amino acid at that position before proceeding. (correct answer)
3. Transcription of the mRNA stops at UAG because stop codons terminate RNA polymerase.
4. The ribosome moves 3′ to 5′ to avoid the stop codon and maintain termination.
5. The first codon is skipped because a modified tRNA prevents initiator tRNA binding at AUG.

Explanation: 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.

Question 15

1. A different amino acid will be inserted at the original codon because the mRNA codon determines amino acid identity directly.
2. The same amino acid may be inserted at a new codon because the tRNA carries its original amino acid but reads a different codon. (correct answer)
3. Translation will stop at the new codon because altered anticodons are recognized as release factors.
4. All proteins will be unchanged because anticodons do not participate in translation once initiation is complete.
5. Transcription will change because altered tRNA anticodons cause RNA polymerase to substitute different nucleotides in mRNA.

Explanation: 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).

Question 16

1. A dipeptide forms: methionine linked to alanine, then translation terminates at UAA. (correct answer)
2. A tripeptide forms because UAA recruits a tRNA that inserts a third amino acid.
3. No peptide forms because the ribosome cannot bind mRNA unless UAA is absent.
4. Only alanine is produced because GCU is the first codon translated by the ribosome.
5. A longer peptide forms because the ribosome skips UAA and continues to the next codon.

Explanation: 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.

Question 17

1. 5′-UAC-3′, because codon and anticodon pair in the same 5′ to 3′ direction.
2. 3′-UAC-5′, because anticodons base-pair antiparallel to codons in the ribosome. (correct answer)
3. 5′-AUG-3′, because identical triplets pair to stabilize the ribosome during elongation.
4. 3′-CAU-5′, because anticodons must be identical to codons to recruit amino acids.
5. 5′-CAU-3′, because RNA base pairing requires C–A and U–G matches in translation.

Explanation: This question assesses the skill of analyzing translation by determining the correct orientation of a tRNA anticodon for base pairing with an mRNA codon. The mRNA codon 5′-AUG-3′ requires an anticodon that pairs antiparallel, meaning the anticodon is oriented 3′-UAC-5′ to allow complementary base pairing: A with U, U with A, and G with C. This antiparallel alignment ensures stable hydrogen bonding in the ribosome during codon-anticodon recognition. In contrast, writing the anticodon in the same 5′ to 3′ direction would not allow proper pairing, as the bases would not complement correctly. A tempting distractor is A, which proposes 5′-UAC-3′ because codon and anticodon pair in the same direction, but this is wrong due to the misconception that RNA strands pair parallel rather than antiparallel like DNA. When solving similar problems, always reverse the codon's orientation and apply complementary base rules to derive the anticodon sequence.

Question 18

1. The polypeptide is unchanged because deleting one nucleotide removes exactly one amino acid.
2. The polypeptide sequence changes after the deletion because the reading frame shifts downstream. (correct answer)
3. Translation stops immediately because any deletion converts AUG into a stop codon.
4. The polypeptide is identical because tRNAs correct the mRNA sequence during elongation.
5. A longer polypeptide forms because deletions always delay release factor binding at UAA.

Explanation: 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.

Question 19

1. A polypeptide identical to mRNA 1 forms because internal codons determine initiation.
2. No polypeptide forms because initiation is unlikely without a start codon to set the frame. (correct answer)
3. A shorter polypeptide forms because ribosomes always begin at the second codon present.
4. A longer polypeptide forms because lacking AUG prevents stop codon recognition.
5. The mRNA is converted into DNA first because AUG is required for reverse transcription.

Explanation: 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.

Question 20

1. Translation stops at CCU because any codon change after AUG creates a stop signal.
2. The amino acid sequence is unchanged because CCU and CCG recruit proline-charged tRNAs. (correct answer)
3. The mRNA is not transcribed because codon substitutions prevent RNA polymerase binding.
4. The ribosome inserts a different amino acid at CCU because only one codon can specify proline.
5. The polypeptide becomes longer because CCU prevents recognition of the UGA stop codon.

Explanation: 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.