Mutations
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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?
The mutant mRNA is not transcribed because RNA polymerase cannot read DNA that contains a deletion.
A frameshift alters the reading frame, changing many downstream amino acids and likely disrupting the active site.
The deletion is removed during RNA splicing because codons are recognized as introns by the spliceosome.
Only codon 45 changes, producing a single amino acid substitution while downstream codons remain unchanged.
Translation stops immediately at codon 45 because single-base deletions always create a stop codon.
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.
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?
The enzyme’s activity may decrease because a charged active-site residue is replaced with an uncharged residue.
No change occurs because lysine and asparagine have identical side-chain charge and size.
The enzyme’s activity must increase because all missense mutations strengthen substrate binding.
A frameshift occurs at AAU, changing all downstream amino acids and preventing folding.
Translation terminates at AAU because it is recognized as a stop codon by release factors.
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.
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?
Transcription fails because RNA polymerase cannot bind when a codon changes.
The protein is truncated because the mutation creates a stop codon in the mRNA.
The protein sequence is unchanged because the mutation is synonymous for glutamic acid.
The protein gains one extra amino acid because GAG signals ribosome pausing and insertion.
A frameshift changes all downstream amino acids because one base was substituted.
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.
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?
Transcription stops at the mutated codon because RNA polymerase cannot pass CGU.
The polypeptide is shortened because CGU functions as a stop codon during translation.
A frameshift changes downstream amino acids because a substitution shifts the reading frame.
The polypeptide gains an extra arginine because CGU is translated twice by the ribosome.
The polypeptide sequence is unchanged because the mutation is synonymous for arginine.
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.
A gene’s promoter contains a TATA box sequence on the coding strand: 5'-TATAAA-3' located 30 bases upstream of the transcription start site. In a mutant, the TATA box is changed to 5'-TATGAA-3'. The coding region of the gene is unchanged. The protein product is normally synthesized in large amounts when the gene is transcribed.
Which outcome is most likely from this mutation?
Lower transcription initiation is likely because altered promoter sequence can reduce RNA polymerase recruitment.
Protein function increases because promoter mutations generally strengthen gene expression to meet cellular needs.
A missense mutation occurs, changing one amino acid in the protein’s N-terminus.
The mRNA reading frame shifts because promoter mutations change how ribosomes group codons.
Translation stops early because promoter mutations introduce stop codons into mRNA.
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.
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?
A premature stop codon forms, preventing synthesis of the targeting sequence and the rest of the protein.
A frameshift begins at the mutated codon because single-nucleotide substitutions shift the reading frame.
Mitochondrial import may be reduced because replacing a positive residue with a negative one can disrupt targeting.
No effect occurs because all point mutations in targeting sequences are removed by post-translational splicing.
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.
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?
More protein is produced because weakening RNA polymerase binding increases translation efficiency.
The amino acid sequence changes because promoter mutations alter the reading frame of translation.
Less protein is produced because reduced transcription yields fewer mRNA molecules for translation.
A truncated protein is produced because promoter mutations create premature stop codons in mRNA.
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.
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?
The mutation creates a new promoter, increasing transcription of the gene.
Only a single amino acid changes because splice sites affect one codon at a time.
The intron is more likely to be retained or mis-spliced, altering the mRNA sequence.
The mutation will change the anticodon of a tRNA, altering translation fidelity.
Translation will proceed normally because introns are always removed correctly.
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.
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?
A frameshift occurs near the stop codon because 3' UTR sequences define the reading frame.
Protein sequence changes because mutations in the 3' UTR alter the codons translated by ribosomes.
mRNA may be more stable, increasing the amount of protein produced without changing its amino acid sequence.
Transcription stops early because RNA polymerase terminates at AUUUA motifs in the DNA template.
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.
A gene encodes a nuclear protein. In the wild type, codon 200 in the coding strand is 5'-CGA-3' (mRNA 5'-CGA-3'), encoding arginine. A mutant changes codon 200 to 5'-TGA-3' (mRNA 5'-UGA-3'). The nuclear localization signal is located near the C-terminus, encoded after codon 240. No other sequence changes occur.
Which outcome is most likely from this mutation?
A premature stop codon produces a shortened protein lacking the C-terminal localization signal.
Transcription stops at codon 200 because RNA polymerase recognizes stop codons in DNA.
A frameshift begins at codon 200 because single-base substitutions shift the reading frame.
A missense mutation substitutes arginine with lysine, preserving protein length and localization.
A silent mutation occurs because UGA can be read as arginine by most tRNAs in eukaryotic cells.
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.