Mutations - AP Biology

1. A eukaryotic gene's coding DNA (template strand) includes the triplet 3′-TAC-5′, which is transcribed into mRNA codon 5′-AUG-3′ (start). A point mutation changes the template triplet to 3′-TAT-5′, producing mRNA 5′-AUA-3′ at the same position. Translation initiates at the first AUG encountered by the ribosome during scanning. No other mutations occur, and the next AUG in the mRNA is 60 nucleotides downstream in the same reading frame. Which outcome is most likely for the protein produced from the mutated mRNA?
   Answer: Translation initiates at the downstream AUG, producing a protein missing the first 20 amino acids. 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.
2. A mitochondrial protein is encoded in the nucleus and imported into mitochondria using an N-terminal targeting sequence rich in positively charged amino acids. A point mutation changes one codon in this targeting sequence from 5'-AAA-3' (mRNA 5'-AAA-3', lysine) to 5'-GAA-3' (mRNA 5'-GAA-3', glutamate). The rest of the protein-coding sequence is unchanged, and translation produces a full-length polypeptide. Which outcome is most likely from this mutation?
   Answer: Mitochondrial import may be reduced because replacing a positive residue with a negative one can disrupt targeting. 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.
3. A bacterial gene encodes a 300–amino acid enzyme. In the wild type, codon 10 in the coding strand is 5'-GGC-3' (mRNA 5'-GGC-3'), which encodes glycine. A mutant has codon 10 changed to 5'-GGT-3' (mRNA 5'-GGU-3'). No other DNA changes are present, and transcription and translation initiation occur normally. Which outcome is most likely from this mutation?
   Answer: The amino acid sequence remains unchanged because both codons specify glycine, producing the same protein. 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.
4. A gene encodes a membrane channel protein. A missense mutation changes one codon in the coding region so that the mRNA codon changes from 5′-GCU-3′ (alanine) to 5′-GAA-3′ (glutamic acid). The substituted amino acid lies within a transmembrane helix composed mostly of nonpolar residues. No other changes occur to transcription or translation. Which outcome is most likely at the protein level?
   Answer: The protein may misfold or insert into the membrane less effectively due to a polar residue in the helix. This question assesses the skill of analyzing mutations and their effects on protein synthesis, particularly missense mutations in functional protein domains. The missense mutation changes the codon from GCU (alanine, nonpolar) to GAA (glutamic acid, polar and charged), introducing a charged residue into a transmembrane helix that typically requires nonpolar amino acids for proper membrane insertion and stability. This substitution can disrupt the hydrophobic environment of the helix, leading to misfolding or inefficient membrane integration of the channel protein. Transcription and translation otherwise proceed normally, producing a full-length protein with this single change. A tempting distractor is choice C, which claims the mutation is silent due to identical side-chain properties, but this misconceives the distinct chemical properties of alanine (nonpolar) versus glutamic acid (charged). To analyze missense mutations, evaluate the chemical differences between the original and new amino acids and consider the structural context, such as membrane domains.
5. A eukaryotic gene normally produces a protein with an N-terminal signal peptide that targets it to the endoplasmic reticulum (ER). A point mutation changes the start codon in the coding strand from 5'-ATG-3' (mRNA 5'-AUG-3') to 5'-ACG-3' (mRNA 5'-ACG-3'). All downstream codons remain unchanged, including a second AUG at codon 25. Transcription produces normal amounts of mRNA. Which outcome is most likely from this mutation?
   Answer: Translation is likely reduced or begins at the downstream AUG, producing a shorter protein missing the signal peptide. 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.
6. A nuclear gene in a vertebrate cell is transcribed in the nucleus, and its mRNA is translated by cytosolic ribosomes using the standard genetic code. A point mutation in the coding region, upstream of the normal stop codon, changes one codon in the mRNA from UGG to UGA. All other codons remain unchanged and in frame. Which outcome is most likely for the polypeptide translated from this mutant mRNA?
   Answer: The polypeptide will be truncated because translation terminates at the new stop codon. This question examines nonsense mutations. In the standard genetic code used by cytosolic ribosomes, UGG codes for tryptophan, while UGA is one of the three stop codons (UAA, UAG, UGA). Because this mutation creates UGA upstream of the normal stop codon, the ribosome reaches the new stop codon early, a release factor binds, and the polypeptide is released. The product is a truncated polypeptide lacking all amino acids that would normally follow, so choice B is correct. Choice A is incorrect because stop codons terminate elongation rather than cause it to continue. Choice C describes a missense mutation, but UGA does not code for an amino acid in the standard code, so the change is a nonsense mutation, not an amino acid replacement. Choice D is incorrect because a substitution replaces one base without changing the number of nucleotides, so the reading frame is preserved. Choice E is incorrect because stop codons act during translation, not transcription, so the mutant mRNA is still produced. When analyzing a point mutation, determine whether the new codon specifies the same amino acid (silent), a different amino acid (missense), or a stop (nonsense).
7. A coding DNA strand includes the sequence 5′-CGA-3′, which is transcribed into mRNA 5′-CGA-3′ and translated as arginine. A point mutation changes the coding DNA codon to 5′-CGT-3′, producing mRNA 5′-CGU-3′. Both CGA and CGU specify arginine in the standard genetic code. Which outcome is most likely for the resulting polypeptide?
   Answer: The polypeptide sequence is unchanged because the mutation is synonymous for arginine. 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.
8. A gene's coding region contains the sequence 5'-…ATC GAA TTT…-3' on the coding strand, corresponding to mRNA 5'-…AUC GAA UUU…-3' and amino acids Ile–Glu–Phe. In a mutant, a single base substitution changes the middle codon from 5'-GAA-3' to 5'-GTA-3', producing mRNA 5'-GUA-3'. The protein's length remains the same, and translation reaches the normal stop codon. Which outcome is most likely from this mutation?
   Answer: A missense mutation substitutes valine for glutamate at that position, potentially altering protein folding or activity. This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The single base substitution changes the codon from GAA (glutamate) to GUA (valine), resulting in a missense mutation that replaces a charged amino acid with a nonpolar one at that position. This alteration could affect protein folding, stability, or activity without changing the overall length, as translation proceeds to the normal stop codon. Molecularly, missense mutations swap one amino acid for another, potentially disrupting interactions if the residues have differing properties. A tempting distractor is choice D, stating no change because both code for glutamate, based on the misconception of ignoring the actual codon specificity in the genetic code. For point mutations, decode both original and mutated codons to classify the type and predict functional impacts based on amino acid differences.
9. A cytosolic enzyme is encoded by a gene whose coding sequence includes codons 45–47: 5'-GAA CCT TTT-3', producing mRNA 5'-GAA CCU UUU-3' and amino acids Glu–Pro–Phe. In a mutant, one nucleotide is deleted from the DNA within codon 45, changing the sequence to 5'-GAC CTT TT…-3' from that point onward. No other changes occur in the gene. The enzyme's active site depends on amino acids encoded downstream of codon 45. Which outcome is most likely from this mutation?
   Answer: A frameshift alters the reading frame, changing many downstream amino acids and likely disrupting the active site. 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.
10. A gene produces an mRNA with a 3' untranslated region (3' UTR) that contains the sequence 5'-AUUUA-3', a motif that promotes rapid mRNA degradation. A mutation changes this motif to 5'-AGGGA-3' in the 3' UTR, while the coding region remains unchanged. Transcription rate of the gene is unchanged, and translation initiation signals in the mRNA are intact. Which outcome is most likely from this mutation?
    Answer: mRNA may be more stable, increasing the amount of protein produced without changing its amino acid sequence. 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.
11. A gene's promoter contains a consensus TATA box sequence that helps recruit transcription factors and RNA polymerase II. A mutation changes the TATA box from TATAAA to TATGAA, without altering the coding region. In cells with this mutated promoter, the mRNA produced from the gene is measured and found to be substantially lower than in cells with the normal promoter. Which outcome is most likely for the protein encoded by this gene in mutated cells?
    Answer: Less protein is produced because reduced transcription yields fewer mRNA molecules for translation. 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.
12. A eukaryotic gene contains an intron with the sequence ...CAG|GTAAGT... where "|" marks the exon–intron boundary and GTAAGT is part of the 5' splice site consensus. A point mutation changes the intron sequence to ...CAG|GTCAGT.... The coding region in the exons is otherwise unchanged. The pre-mRNA is transcribed normally. Which outcome is most likely at the cellular level?
    Answer: The mature mRNA will likely retain intronic sequence due to reduced splice site recognition. This question tests understanding of how mutations affect RNA splicing in eukaryotes. The mutation changes the 5' splice site consensus from GT to GC, which disrupts the spliceosome's ability to recognize this intron boundary. When splice sites are mutated, the spliceosome may fail to remove the intron, resulting in mature mRNA that retains intronic sequences, which can introduce stop codons or alter the reading frame. Students selecting option C incorrectly believe introns always protect against protein changes, not realizing that improper splicing can include intron sequences in the final mRNA. To analyze splicing mutations, check if the mutation affects the conserved GT-AG splice site sequences.
13. A gene encodes an enzyme whose active site includes a lysine residue. A mutation changes the mRNA codon for that residue from 5′-AAA-3′ (lysine) to 5′-AAU-3′ (asparagine). The rest of the mRNA sequence is unchanged, and the protein is produced at normal levels. Lysine is positively charged at cellular pH, while asparagine is polar but uncharged. Which outcome is most likely for the enzyme's function?
    Answer: The enzyme's activity may decrease because a charged active-site residue is replaced with an uncharged residue. 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.
14. A eukaryotic pre-mRNA intron begins with the conserved 5′ splice site sequence GU in the RNA. A mutation in the DNA changes the transcribed intron start from GU to AU in the pre-mRNA, while all exons remain unchanged. The spliceosome requires the GU sequence to recognize and remove the intron efficiently. Which outcome is most likely for the mature mRNA and resulting protein?
    Answer: The intron is retained in the mRNA, potentially introducing a premature stop codon during translation. This question assesses the skill of analyzing mutations and their effects on protein synthesis, specifically in RNA splicing of eukaryotic pre-mRNA. The mutation alters the 5' splice site from GU to AU, which is not recognized by the spliceosome, leading to inefficient intron removal and retention of the intron in the mature mRNA. This retained intron can introduce premature stop codons or disrupt the reading frame during translation, often resulting in a nonfunctional or truncated protein. The exons remain unchanged, but the splicing defect affects the overall mRNA sequence processed for translation. A tempting distractor is choice B, which claims only a single amino acid substitution occurs with normal splicing, but this misconceives that the mutation is in the intron boundary, impacting splicing rather than directly coding for an amino acid. When evaluating mutations in non-coding regions, consider their impact on processes like splicing and how that alters the final mRNA and protein.
15. A gene's coding region includes a repeated sequence of three nucleotides. During DNA replication, one additional triplet is inserted into the coding sequence (e.g., one extra codon is added), while the rest of the sequence remains unchanged and in-frame. Transcription and translation occur normally. Which outcome is most likely for the resulting protein compared with the original?
    Answer: The protein will include one additional amino acid, with downstream amino acids remaining the same. 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.
16. A diploid cell has two alleles of a gene encoding a membrane channel. One allele acquires a single-base substitution in the coding region so that one mRNA codon becomes UAG instead of CAG. The other allele remains unchanged. The mutant allele is transcribed, and its mRNA is translated using the standard genetic code, with translation beginning normally. Which outcome is most likely for the mutant allele's protein product?
    Answer: A truncated polypeptide is produced because translation terminates at the new stop codon. This question tests recognition of nonsense mutations and their effects on translation. In the standard genetic code, the mRNA codon CAG specifies glutamine, while UAG is one of the three stop codons (UAA, UAG, UGA). When the ribosome reaches this premature UAG codon, a release factor binds and the polypeptide is released, so the product is a truncated polypeptide lacking the amino acids that would have followed. Choice B is correct. Choice A is incorrect because stop codons end elongation rather than extend it. Choice C is incorrect because stop codons act during translation; they do not prevent transcription, which is why the mutant mRNA is made. Choice D is incorrect because UAG does not code for any amino acid in the standard code. Choice E is incorrect because a substitution replaces one base without changing the number of nucleotides, so the reading frame is preserved. To identify nonsense mutations, check whether the substitution creates UAA, UAG, or UGA in the mRNA.
17. A coding sequence contains the segment 5'-...GAA CTC GGT...-3', translated in-frame. A point mutation changes the middle codon from CTC to CTA, producing 5'-...GAA CTA GGT...-3'. The gene's expression level and mRNA processing are unchanged. Using the standard genetic code, which outcome is most likely for the translated protein?
    Answer: The amino acid sequence is unchanged because CTC and CTA both specify leucine. This question tests understanding of silent mutations and the degeneracy of the genetic code. The mutation changes the codon from CTC to CTA on the coding strand, which means the mRNA changes from CUC to CUA. Consulting the genetic code table, both CUC and CUA code for leucine, making this a silent mutation with no change to the amino acid sequence. Students choosing option A incorrectly assume all codon changes must alter amino acids, not recognizing that the genetic code's degeneracy means multiple codons can specify the same amino acid. To identify silent mutations, always check if both the original and mutated codons encode the same amino acid using the genetic code table.
18. A gene contains the coding DNA sequence 5′-CAA GCT GAA-3′, producing mRNA 5′-CAA GCU GAA-3′ and the amino acids Gln–Ala–Glu. A point mutation changes the third codon in the coding DNA from 5′-GAA-3′ to 5′-GAG-3′. Both GAA and GAG transcribe to mRNA codons that specify glutamic acid. Which outcome is most likely for the protein encoded by the mutated gene?
    Answer: The protein sequence is unchanged because the mutation is synonymous for glutamic acid. 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.
19. A gene encodes a cytosolic protein with an N-terminal signal peptide absent. A mutation changes the start codon on the coding strand from ATG to ACG, while the rest of the coding region remains unchanged and in-frame. The promoter is intact, and transcription initiates normally. Ribosomes in this cell typically begin translation at AUG codons. Which outcome is most likely for protein production from this mutant allele?
    Answer: Translation initiation is likely reduced or abolished because the canonical start codon is missing. 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.
20. A eukaryotic gene has an intron with the 3′ splice acceptor site ending in the DNA sequence 5′-...AG-3′ (coding strand). A mutation changes this acceptor site to 5′-...AA-3′, while the coding exons remain unchanged. The pre-mRNA is transcribed normally. Which outcome is most likely for the mature mRNA?
    Answer: The intron is more likely to be retained or mis-spliced, altering the mRNA sequence. 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.

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