AP Biology Quiz: Mutations

What this quiz covers

This quiz focuses on Mutations, 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. Only codon 45 changes, producing a single amino acid substitution while downstream codons remain unchanged.
2. Translation stops immediately at codon 45 because single-base deletions always create a stop codon.
3. A frameshift alters the reading frame, changing many downstream amino acids and likely disrupting the active site. (correct answer)
4. The deletion is removed during RNA splicing because codons are recognized as introns by the spliceosome.
5. The mutant mRNA is not transcribed because RNA polymerase cannot read DNA that contains a deletion.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The single nucleotide deletion within codon 45 shifts the reading frame of the mRNA, causing all subsequent codons to be grouped differently and encoding a new sequence of amino acids. This frameshift mutation disrupts the enzyme's active site, which relies on precise downstream amino acids, likely rendering the protein non-functional. Molecularly, deletions not divisible by three alter the triplet codon reading frame starting from the mutation point, leading to widespread changes unless a compensatory insertion restores it. A tempting distractor is choice A, which claims only codon 45 changes with no downstream effects, stemming from the misconception that deletions affect only the immediate codon without shifting the frame. When evaluating insertion or deletion mutations, count the number of bases affected and assess if it disrupts the codon triplet grouping for downstream sequences.

Question 2

1. The enzyme's activity may decrease because a charged active-site residue is replaced with an uncharged residue. (correct answer)
2. The enzyme's activity must increase because all missense mutations strengthen substrate binding.
3. No change occurs because lysine and asparagine have identical side-chain charge and size.
4. Translation terminates at AAU because it is recognized as a stop codon by release factors.
5. A frameshift occurs at AAU, changing all downstream amino acids and preventing folding.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, specifically missense mutations in enzyme active sites. The mutation alters the codon from AAA (lysine, positively charged) to AAU (asparagine, polar but uncharged), replacing a charged residue critical for substrate interaction with an uncharged one, which may weaken binding or catalysis. The protein is produced at normal levels with this single substitution, but the loss of charge could impair the enzyme's activity depending on the active site's requirements. No frameshift or termination occurs, as it's a substitution within the coding sequence. A tempting distractor is choice C, which suggests no change due to identical side-chain properties, but this misconceives the key difference in charge between lysine and asparagine. To evaluate mutations in functional sites, compare the physicochemical properties of the amino acids and predict impacts on molecular interactions like catalysis.

Question 3

1. The protein sequence is unchanged because the mutation is synonymous for glutamic acid. (correct answer)
2. The protein is truncated because the mutation creates a stop codon in the mRNA.
3. A frameshift changes all downstream amino acids because one base was substituted.
4. The protein gains one extra amino acid because GAG signals ribosome pausing and insertion.
5. Transcription fails because RNA polymerase cannot bind when a codon changes.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, emphasizing silent or synonymous mutations. The point mutation changes the DNA codon from GAA to GAG, both of which transcribe to mRNA codons (GAA and GAG) that encode glutamic acid due to degeneracy in the genetic code. Translation proceeds normally, incorporating the same amino acid at that position, resulting in an unchanged protein sequence. No frameshift or truncation occurs because the mutation is a substitution that does not alter the reading frame or create a stop codon. A tempting distractor is choice B, which suggests the protein is truncated due to a new stop codon, but this misconceives that GAG codes for glutamic acid, not a stop like UAG. To assess point mutations, compare the original and mutated codons in the genetic code table to determine if the amino acid changes or remains the same.

Question 4

1. The polypeptide sequence is unchanged because the mutation is synonymous for arginine. (correct answer)
2. The polypeptide is shortened because CGU functions as a stop codon during translation.
3. A frameshift changes downstream amino acids because a substitution shifts the reading frame.
4. Transcription stops at the mutated codon because RNA polymerase cannot pass CGU.
5. The polypeptide gains an extra arginine because CGU is translated twice by the ribosome.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, highlighting synonymous substitutions. The point mutation changes the DNA codon from CGA to CGT, resulting in mRNA codons CGA and CGU, both of which encode arginine due to the redundant nature of the genetic code. Translation incorporates the same amino acid, leaving the polypeptide sequence unchanged without affecting length or frame. No transcriptional or translational halt occurs, as the mutation does not create a stop codon or disrupt reading. A tempting distractor is choice B, which claims the polypeptide is shortened because CGU is a stop codon, but this misconceives that CGU codes for arginine, unlike stop codons such as UGA. For suspected silent mutations, verify both codons in the genetic code to confirm if they specify the same amino acid.

Question 5

1. Lower transcription initiation is likely because altered promoter sequence can reduce RNA polymerase recruitment. (correct answer)
2. A missense mutation occurs, changing one amino acid in the protein's N-terminus.
3. Translation stops early because promoter mutations introduce stop codons into mRNA.
4. The mRNA reading frame shifts because promoter mutations change how ribosomes group codons.
5. Protein function increases because promoter mutations generally strengthen gene expression to meet cellular needs.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation in the TATA box from TATAAA to TATGAA impairs the promoter's ability to recruit RNA polymerase and transcription factors, likely reducing the initiation of transcription. This leads to lower mRNA production and consequently less protein synthesis, affecting cellular functions that require high levels of this protein. Molecularly, the TATA box is a core promoter element that positions the transcription machinery; alterations disrupt its consensus sequence and binding affinity. A tempting distractor is choice B, claiming a missense mutation in the protein's N-terminus, due to the misconception that promoter changes directly alter coding sequences rather than expression levels. When examining regulatory region mutations, evaluate their impact on transcription efficiency rather than direct changes to the protein sequence.

Question 6

1. Mitochondrial import may be reduced because replacing a positive residue with a negative one can disrupt targeting. (correct answer)
2. A premature stop codon forms, preventing synthesis of the targeting sequence and the rest of the protein.
3. No effect occurs because all point mutations in targeting sequences are removed by post-translational splicing.
4. A frameshift begins at the mutated codon because single-nucleotide substitutions shift the reading frame.
5. The mutation increases mitochondrial import because negatively charged residues bind more strongly to mitochondrial DNA.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The point mutation substitutes lysine (positively charged) with glutamate (negatively charged) in the mitochondrial targeting sequence, potentially disrupting import by altering the charge profile needed for recognition by import receptors. Although the full-length protein is translated, impaired targeting could reduce mitochondrial localization and function. Molecularly, mitochondrial signals rely on amphipathic helices with positive charges; charge-reversing mutations hinder translocation across membranes. A tempting distractor is choice B, suggesting a premature stop codon, arising from the misconception that amino acid codon changes like AAA to GAA create stops rather than missense substitutions. To evaluate mutations in targeting sequences, consider how amino acid properties affect signal recognition and predict localization outcomes.

Question 7

1. Less protein is produced because reduced transcription yields fewer mRNA molecules for translation. (correct answer)
2. A truncated protein is produced because promoter mutations create premature stop codons in mRNA.
3. The amino acid sequence changes because promoter mutations alter the reading frame of translation.
4. More protein is produced because weakening RNA polymerase binding increases translation efficiency.
5. No change occurs because promoter sequences affect only protein folding after translation.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, focusing on regulatory regions like promoters. The mutation alters the TATA box from TATAAA to TATGAA, weakening the binding of transcription factors and RNA polymerase II, which reduces the rate of transcription initiation. Consequently, fewer mRNA molecules are produced, leading to decreased translation and lower protein levels in the cell. The coding region remains unchanged, so any protein produced has the normal sequence, but the overall amount is reduced. A tempting distractor is choice D, which suggests more protein due to increased translation efficiency, but this misconceives that weakened promoter binding decreases, not increases, transcription. When mutations occur in promoters, assess their impact on transcription efficiency and downstream effects on mRNA and protein abundance.

Question 8

1. The intron is more likely to be retained or mis-spliced, altering the mRNA sequence. (correct answer)
2. Only a single amino acid changes because splice sites affect one codon at a time.
3. The mutation creates a new promoter, increasing transcription of the gene.
4. Translation will proceed normally because introns are always removed correctly.
5. The mutation will change the anticodon of a tRNA, altering translation fidelity.

Explanation: This question examines how mutations in splice sites affect mRNA processing in eukaryotes. The 3' splice acceptor site typically ends with AG, which is recognized by the spliceosome machinery for precise intron removal. When this AG changes to AA, the spliceosome cannot recognize the proper splice site, leading to intron retention or use of a cryptic splice site elsewhere in the sequence. This results in an altered mature mRNA that may include intron sequences or skip exon sequences, dramatically changing the protein product. Students often assume introns are always removed correctly (choice D), not understanding that specific sequences guide the splicing machinery. When analyzing splice site mutations, remember that even single nucleotide changes in conserved splice sequences can disrupt normal mRNA processing.

Question 9

1. Protein sequence changes because mutations in the 3' UTR alter the codons translated by ribosomes.
2. mRNA may be more stable, increasing the amount of protein produced without changing its amino acid sequence. (correct answer)
3. A frameshift occurs near the stop codon because 3' UTR sequences define the reading frame.
4. Transcription stops early because RNA polymerase terminates at AUUUA motifs in the DNA template.
5. The mutation prevents translation because ribosomes bind only to mRNAs containing AUUUA motifs.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation changes the 3' UTR motif from AUUUA, which promotes mRNA degradation, to AGGGA, potentially increasing mRNA stability by removing this destabilizing element. With unchanged transcription and translation signals, this leads to higher steady-state mRNA levels and more protein production without altering the amino acid sequence. Molecularly, 3' UTR motifs regulate post-transcriptional processes like decay; mutations can extend mRNA half-life, amplifying gene expression. A tempting distractor is choice A, claiming protein sequence changes, due to the misconception that UTR regions are translated into amino acids rather than serving regulatory roles. When analyzing UTR mutations, focus on their effects on mRNA stability, localization, or translation efficiency rather than direct coding changes.

Question 10

1. A silent mutation occurs because UGA can be read as arginine by most tRNAs in eukaryotic cells.
2. A premature stop codon produces a shortened protein lacking the C-terminal localization signal. (correct answer)
3. A missense mutation substitutes arginine with lysine, preserving protein length and localization.
4. A frameshift begins at codon 200 because single-base substitutions shift the reading frame.
5. Transcription stops at codon 200 because RNA polymerase recognizes stop codons in DNA.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation alters codon 200 from CGA (arginine) to UGA, a stop codon that terminates translation prematurely, producing a truncated protein. This shortened polypeptide lacks the C-terminal nuclear localization signal encoded after codon 240, preventing proper nuclear import. Molecularly, nonsense mutations introduce early stops, recruiting release factors and halting elongation before the full sequence is translated. A tempting distractor is choice C, proposing a missense substitution of arginine with lysine, stemming from the misconception that UGA codes for an amino acid instead of a termination signal. For suspected nonsense mutations, verify the mutated codon against the genetic code and assess the position relative to key functional domains.

Question 11

1. Translation initiates at the original site because ribosomes convert ACG to AUG during scanning.
2. Translation is likely reduced or begins at the downstream AUG, producing a shorter protein missing the signal peptide. (correct answer)
3. A frameshift occurs at codon 1, changing all amino acids and eliminating the stop codon.
4. Splicing removes the mutated start codon because start codons are recognized as introns.
5. The mutation increases transcription because start codons are binding sites for RNA polymerase.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation changes the start codon from AUG to ACG, which is not recognized as an initiation signal by eukaryotic ribosomes, potentially causing translation to skip to the next AUG at codon 25. This results in a shorter protein lacking the N-terminal signal peptide, impairing ER targeting and protein localization. Molecularly, initiation requires a specific AUG context; mutations abolish scanning recognition, leading to alternative start site usage if available downstream. A tempting distractor is choice A, suggesting ribosomes convert ACG to AUG, based on the misconception that translation machinery edits codons rather than strictly reading them. To assess start codon mutations, check for downstream alternative initiation sites and predict effects on protein length and functional domains.

Question 12

1. The reading frame shifts, changing many downstream amino acids and often introducing an early stop codon. (correct answer)
2. Only one amino acid changes because a single nucleotide deletion always causes a missense mutation.
3. The protein is unchanged because deletions are removed during RNA splicing in bacteria.
4. Transcription stops immediately because RNA polymerase cannot read a template with a deletion.
5. The mutation increases translation rate by shortening the mRNA without changing the amino acid sequence.

Explanation: This question requires analyzing how deletion mutations affect reading frames. When the second nucleotide is deleted from ATG AAA GGC..., the sequence becomes AGA AAG GCC..., shifting how the ribosome reads every subsequent codon by one position. This frameshift mutation changes all downstream amino acids and often introduces premature stop codons when the new reading frame encounters UAA, UAG, or UGA. Students choosing option B mistakenly think deletions only affect one amino acid, confusing them with substitution mutations. Remember that insertions or deletions of any number of nucleotides not divisible by three will cause frameshifts, altering the entire downstream sequence.

Question 13

1. A premature stop codon forms, producing a truncated polypeptide with fewer amino acids.
2. The amino acid sequence is unchanged because both codons specify phenylalanine.
3. One amino acid substitution occurs because the codon now specifies leucine instead of phenylalanine. (correct answer)
4. A frameshift occurs, altering all downstream codons and likely changing many amino acids.
5. The mRNA will not be produced because a single base substitution blocks transcription initiation.

Explanation: This question tests your ability to analyze the effects of point mutations on protein sequences. The mutation changes the coding strand from TTT to TTA, which means the mRNA will change from UUU to UUA. Looking at the genetic code, UUU codes for phenylalanine while UUA codes for leucine, resulting in a single amino acid substitution at that position. Students often incorrectly choose option B, thinking that similar-looking codons must code for the same amino acid, but the genetic code shows that even single nucleotide changes can alter the amino acid. To solve mutation problems, always transcribe the coding strand to mRNA, then use the genetic code table to determine the amino acid change.

Question 14

1. A longer protein is produced because stop codons cause ribosomes to add additional amino acids.
2. A truncated polypeptide is produced because translation terminates at the new stop codon. (correct answer)
3. No mRNA is made from the mutant allele because a stop codon prevents transcription.
4. The amino acid sequence is unchanged because TAG and CAG both code for glutamine.
5. A frameshift occurs because a single-base substitution shifts the reading frame.

Explanation: This question tests recognition of nonsense mutations and their effects on translation. The mutation changes the codon CAG (glutamine) to TAG, which transcribes to UAG in mRNA - one of the three stop codons. When the ribosome encounters this premature stop codon during translation, it releases the nascent polypeptide chain, producing a truncated protein missing all amino acids that would normally follow. Students choosing option D confuse DNA sequences with mRNA codons and fail to recognize TAG as corresponding to a stop codon in mRNA. To identify nonsense mutations, check if the mutation creates UAA, UAG, or UGA in the mRNA sequence.

Question 15

1. The protein will include one additional amino acid, with downstream amino acids remaining the same. (correct answer)
2. A frameshift will occur, changing all downstream amino acids and eliminating the stop codon.
3. The protein will be unchanged because insertions are removed during translation proofreading.
4. Transcription will stop at the insertion site because RNA polymerase cannot pass repeats.
5. The mutation will always increase enzyme activity because extra codons add functional domains.

Explanation: This question analyzes the effects of in-frame insertions on protein structure. When three nucleotides (one codon) are inserted into a coding sequence, the reading frame remains intact because the insertion is divisible by three. This results in one additional amino acid being incorporated at the insertion site, while all downstream amino acids remain unchanged since the reading frame is preserved. Students selecting option B confuse this with frameshift mutations, which only occur when insertions or deletions are not divisible by three. Remember that insertions or deletions of exactly three nucleotides (or multiples of three) maintain the reading frame while adding or removing amino acids.

Question 16

1. Translation initiation is likely reduced or abolished because the canonical start codon is missing. (correct answer)
2. A frameshift occurs immediately because ACG is interpreted as an insertion during translation.
3. The protein is unchanged because start codons only affect transcription, not translation.
4. The protein becomes longer because loss of ATG prevents termination at the normal stop codon.
5. The mRNA will be spliced differently because start codons determine intron removal.

Explanation: This question examines how mutations affecting the start codon impact translation initiation. The mutation changes ATG (which produces AUG in mRNA, the universal start codon) to ACG (producing ACG in mRNA). Since ribosomes typically require AUG to begin translation, losing this start codon will severely reduce or prevent translation initiation, as the ribosome may not recognize where to begin or may scan to a downstream AUG. Students selecting option C incorrectly think start codons affect transcription rather than translation, not understanding that transcription begins at promoters while translation begins at start codons. When analyzing start codon mutations, remember that AUG is essential for ribosome binding and translation initiation.

Question 17

1. Exon 3 is more efficiently included because the splice acceptor site becomes a stronger start codon signal.
2. mRNA translation proceeds normally because splice sites do not affect mature mRNA sequence.
3. The spliceosome may fail to recognize the acceptor site, causing exon 3 skipping or intron retention in the mature mRNA. (correct answer)
4. A frameshift occurs only within exon 3 because splice-site mutations change codon grouping inside exons.
5. RNA polymerase stops at the mutated splice site, producing a shorter pre-mRNA that cannot be capped.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation alters the splice acceptor site from CAG to CAA, disrupting the consensus sequence needed for the spliceosome to recognize and cleave the intron-exon boundary accurately. This can lead to exon 3 skipping or intron retention in the mature mRNA, omitting a critical catalytic residue and impairing protein function. Molecularly, splice sites contain specific motifs like the 3' AG dinucleotide, and mutations here prevent proper base-pairing with snRNAs, causing aberrant splicing outcomes. A tempting distractor is choice B, which states translation proceeds normally unaffected by splice sites, based on the misconception that splicing errors do not alter the final mRNA sequence used in translation. To analyze splicing mutations, identify changes in consensus splice site sequences and predict potential disruptions to exon inclusion or reading frame integrity.

Question 18

1. A premature stop codon forms at codon 10, preventing translation of most of the enzyme.
2. The amino acid sequence remains unchanged because both codons specify glycine, producing the same protein. (correct answer)
3. A frameshift begins at codon 10, altering the reading frame and changing all downstream amino acids.
4. The mutation prevents transcription because RNA polymerase cannot bind to genes with synonymous codons.
5. The enzyme becomes longer because silent mutations cause ribosomes to read through the normal stop codon.

Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation changes codon 10 from GGC to GGT, both of which encode glycine due to the redundancy of the genetic code, resulting in no alteration to the amino acid sequence. Consequently, the 300-amino acid enzyme is produced unchanged, maintaining its normal structure and function. Molecularly, this is a synonymous or silent mutation where the nucleotide change does not affect the translated protein because multiple codons specify the same amino acid. A tempting distractor is choice A, suggesting a premature stop codon at codon 10, arising from the misconception that any codon change involving G to T creates a stop signal like TGA. For point mutations in coding regions, compare the original and mutated codons using the genetic code to classify them as silent, missense, or nonsense.

Question 19

1. A longer polypeptide is produced because stop codons are read as tryptophan in eukaryotes.
2. A truncated polypeptide is produced because translation terminates at the new stop codon. (correct answer)
3. The polypeptide is unchanged because UGA and UGG encode the same amino acid.
4. A frameshift occurs at the mutation site, altering all downstream amino acids.
5. Transcription stops at the mutated codon, preventing mRNA formation past that point.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, particularly nonsense mutations in eukaryotic genes. The point mutation alters the codon from UGG (tryptophan) to UGA, a stop codon that is recognized by release factors, causing premature termination of translation. This results in a truncated polypeptide because the ribosome dissociates at the new stop codon, preventing synthesis of the full protein sequence downstream. No frameshift or transcriptional halt occurs, as the mutation is a single base change within the coding region. A tempting distractor is choice D, which claims a frameshift alters all downstream amino acids, but this misconceives that only insertions or deletions, not substitutions, cause frameshifts. For mutations creating stop codons, identify the codon change using the genetic code and predict the impact on polypeptide length and function.

Question 20

1. Translation initiates at the downstream AUG, producing a protein missing the first 20 amino acids. (correct answer)
2. A frameshift occurs at the mutation site, producing a truncated protein due to an early stop codon.
3. The mutation is silent because AUA and AUG both encode methionine at the start position.
4. Translation still initiates at the mutated codon, producing a protein with a single amino acid substitution.
5. mRNA transcription stops at the mutation site, preventing any protein from being produced.

Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, specifically point mutations in eukaryotic start codons. The mutation changes the mRNA start codon from AUG to AUA, which codes for isoleucine and is not recognized as a translation initiation signal by eukaryotic ribosomes. During the scanning process, the ribosome bypasses the mutated AUA and initiates translation at the next downstream AUG, located 60 nucleotides away, resulting in a protein that lacks the first 20 amino acids since 60 divided by 3 equals 20 codons. This truncation occurs because the downstream AUG is in the same reading frame, maintaining the sequence integrity after initiation but omitting the initial segment. A tempting distractor is choice C, which claims the mutation is silent because AUA and AUG both encode methionine, but this misconceives that AUA codes for isoleucine, not methionine, and ignores the special role of AUG in initiation. To evaluate similar mutations, always consult the genetic code table and consider how changes affect initiation, elongation, or termination stages of translation.