Genetics Quiz: Mutations And Translation

What this quiz covers

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

How to use this quiz

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

Question 1

1. Both mutations are non-conservative and will likely cause a complete loss of protein function.
2. The Gly -> Asp change is conservative, while the Arg -> Lys change is non-conservative and more likely to be damaging.
3. Both mutations are conservative substitutions and are unlikely to have any significant effect on the protein's function.
4. The Gly -> Asp change is non-conservative and likely disruptive, while the Arg -> Lys change is conservative and may have a minor effect. (correct answer)

Explanation: When evaluating the functional impact of amino acid substitutions, you need to consider how dramatically the chemical properties change between the original and substituted amino acids. Conservative substitutions involve amino acids with similar properties, while non-conservative substitutions involve dramatically different properties. Let's analyze each mutation. The Gly → Asp change substitutes a small, nonpolar amino acid with an acidic, negatively charged one. This represents a major shift in both size and charge properties, making it highly non-conservative and likely to disrupt protein structure or function. The Arg → Lys change substitutes one positively charged, basic amino acid with another positively charged, basic amino acid. While these aren't identical, they share the crucial property of positive charge, making this a relatively conservative substitution that may have minimal functional impact. Looking at the wrong answers: Choice A incorrectly assumes both mutations are non-conservative and will cause complete function loss, ignoring that Arg → Lys maintains similar charge properties. Choice B reverses the analysis entirely, incorrectly calling the dramatic Gly → Asp change conservative. Choice C wrongly categorizes both as conservative substitutions, missing the significant property changes in the Gly → Asp mutation. The correct answer is D because it properly identifies the Gly → Asp change as non-conservative and potentially disruptive due to the dramatic shift from small/nonpolar to charged/acidic, while recognizing the Arg → Lys change as conservative since both amino acids share positive charge. Remember: focus on charge, polarity, and size when predicting substitution effects—dramatic property changes usually mean greater functional impact.

Question 2

1. Translation will initiate at the AUA codon, incorporating Isoleucine as the first amino acid.
2. Translation will be completely abolished, as AUA can never serve as an initiation codon.
3. The ribosome will stall at the AUA codon, leading to the degradation of the mRNA transcript.
4. The ribosome will likely scan past the AUA codon and initiate at the downstream AUG, producing an N-terminally truncated protein. (correct answer)

Explanation: When you encounter questions about translation initiation, focus on the scanning mechanism that eukaryotic ribosomes use to find start codons. The ribosome binds to the 5' cap and scans downstream until it finds the first AUG in a favorable context (like a strong Kozak sequence). The mutation from AUG to AUA disrupts this process because AUA cannot serve as a start codon for translation initiation. While AUA codes for isoleucine during elongation, it lacks the special recognition sequences needed to recruit the initiator tRNA ($Met-tRNA^i$) and initiation factors required to begin translation. The ribosome's scanning mechanism will simply skip over the AUA codon and continue searching downstream until it encounters the next AUG at position +15, where normal translation initiation can occur. This produces a protein that's missing its first 15 amino acids—an N-terminally truncated version. Answer A is incorrect because AUA cannot function as an initiation codon, even though it codes for isoleucine during normal translation. Answer B overstates the effect—translation isn't completely abolished since there's a downstream AUG available. Answer C describes a stalling mechanism that doesn't occur with this type of mutation; the ribosome simply continues scanning rather than getting stuck. Remember that eukaryotic translation follows the "first AUG rule"—ribosomes scan for the first available start codon. When the original start site is mutated to a non-start codon, the ribosome continues scanning to find the next functional AUG, often resulting in truncated proteins rather than complete translation failure.

Question 3

1. The substituted amino acid destabilizes the protein's tertiary structure, causing it to unfold at the higher, non-permissive temperature. (correct answer)
2. The mRNA transcript containing the mutation is stable at 30°C but is rapidly degraded by cellular nucleases at 37°C.
3. The mutation creates a premature stop codon that is only recognized by the ribosome at the higher temperature, leading to a truncated protein.
4. The mutation affects a splice site, leading to correct splicing at 30°C but aberrant splicing and a frameshift at 37°C.

Explanation: Temperature-sensitive mutations are often missense mutations that cause subtle defects in protein folding. The mutant protein can fold correctly enough to be functional at a lower (permissive) temperature. However, the increased thermal energy at a higher (non-permissive) temperature is sufficient to overcome the weaker interactions in the mutant protein's structure, causing it to misfold, denature, and lose activity. The location in the hydrophobic core makes it particularly sensitive to substitutions that disrupt packing and stability.

Question 4

1. The protein will be shorter than wild-type due to the skipping of Exon 1 during mRNA processing.
2. The protein will be identical to the wild-type protein, but its expression level will be significantly reduced.
3. The translated protein will be significantly longer than wild-type, containing an additional 95 amino acids.
4. The protein will likely be truncated due to a frameshift and a premature stop codon within the retained intron sequence. (correct answer)

Explanation: Disruption of the 5' splice site will most likely cause the splicing machinery to fail to recognize and excise Intron 1. This leads to intron retention. Because the length of Intron 1 (95 bp) is not a multiple of three, its inclusion in the mature mRNA will shift the reading frame. This frameshift will alter all downstream codons and almost certainly introduce a premature stop codon, leading to a truncated protein.

Question 5

1. There will be no consequence, as the amino acid sequence of the protein remains unchanged.
2. The mutation will cause premature termination of translation because the ribosome does not recognize the rare codon.
3. Translation may pause at the rare GGA codon due to low cellular concentration of the corresponding tRNA, potentially causing protein misfolding. (correct answer)
4. The GGA codon will be incorrectly read as a different amino acid, resulting in a missense mutation and loss of function.

Explanation: Codon usage bias is a known phenomenon where some synonymous codons are used more frequently than others. The cellular concentration of tRNAs often matches this bias. A switch to a rare codon can slow down the rate of translation at that specific point because the ribosome has to wait for the scarce corresponding tRNA. This pause can disrupt the normal rhythm of co-translational folding, potentially leading to a misfolded, non-functional protein, even though the primary amino acid sequence is identical to the wild-type.

Question 6

1. The entire intron will be retained in the mRNA, leading to a frameshift.
2. The exon following the mutated intron will be skipped during splicing.
3. The mature mRNA will contain an additional 12 nucleotides from the 3' end of the intron. (correct answer)
4. The mRNA will be identical to the wild-type, as the cryptic site perfectly compensates for the mutation.

Explanation: When a canonical splice site is inactivated, the splicing machinery may use a nearby suboptimal sequence, known as a cryptic splice site. In this case, the splicing machinery will use the cryptic site located 12 bp upstream of the original site. This means that the splice will occur earlier than normal, and the 12 nucleotides between the cryptic site and the original splice site will be incorrectly included in the mature mRNA, effectively becoming part of the downstream exon. This insertion will likely cause a frameshift in the resulting protein.

Question 7

1. A mutation in the gene for tryptophanyl-tRNA synthetase that alters its specificity.
2. A frameshift mutation downstream of the nonsense codon that restores the original reading frame.
3. A mutation that greatly increases the overall rate of translation for all cellular mRNAs.
4. A mutation in the anticodon of a tRNA for Tryptophan ($tRNA^Trp$), changing it from 5'-CCA-3' to 5'-UCA-3'. (correct answer)

Explanation: This describes a nonsense suppressor tRNA. The wild-type tRNA^Trp has an anticodon 5'-CCA-3' which recognizes the 5'-UGG-3' codon for Tryptophan. A mutation in the tRNA gene that changes the anticodon to 5'-UCA-3' allows this modified tRNA to recognize the 5'-UGA-3' stop codon. The ribosome then inserts Tryptophan at the position of the nonsense mutation, allowing translation to continue and produce a full-length, functional protein. This is a classic example of extragenic (or intergenic) suppression.

Question 8

1. A truncated protein A will be produced, but protein B will be produced at normal levels.
2. A truncated protein A will be produced, and the translation of protein B will be severely inhibited. (correct answer)
3. Both protein A and protein B will be produced at full length but will contain incorrect amino acid sequences.
4. The entire polycistronic mRNA will be degraded, resulting in no production of either protein.

Explanation: The insertion in geneA will cause a frameshift, leading to an incorrect amino acid sequence and, most likely, a premature stop codon. This results in a truncated, non-functional protein A. In translational coupling, the ribosome that finishes translating geneA is in the ideal position to immediately initiate translation of geneB. Because the ribosome terminates prematurely within geneA, it dissociates from the mRNA before reaching the start of geneB. This uncoupling severely reduces or abolishes the initiation of translation for protein B.

Question 9

1. A truncated protein of 103 amino acids, due to a frameshift introducing a premature stop codon.
2. A full-length protein with a completely altered amino acid sequence from position 104 onward.
3. A protein of 399 amino acids that is likely to retain a significant level of its original function. (correct answer)
4. A protein of 400 amino acids where the leucine at position 104 is replaced by a different amino acid.

Explanation: The deletion of three consecutive nucleotides constitutes an in-frame deletion. It removes a single codon, resulting in the removal of one amino acid from the polypeptide chain, but it does not alter the reading frame for the rest of the gene. Since the deleted amino acid is in a flexible loop rather than a critical active site or structural core, the resulting 399-amino-acid protein has a high probability of folding correctly and retaining most or all of its original function.

Question 10

1. The mutation, although synonymous, creates an exonic splicing silencer site, leading to incorrect splicing of 50% of the transcripts.
2. The CUA codon is a much rarer codon than CUG in humans, and the resulting translational pausing leads to premature termination or protein degradation. (correct answer)
3. The G-to-A change in the mRNA destabilizes the transcript, causing half of it to be degraded before translation can occur.
4. The mutation must be a polar mutation, affecting the transcription of downstream genes in an operon.

Explanation: This question highlights that even synonymous ('silent') mutations can have significant phenotypic effects. The levels of mRNA are normal, ruling out transcriptional or splicing defects (choices A and D, with D being irrelevant to eukaryotes). The discrepancy is between mRNA and protein levels, pointing to a translational issue. The most plausible explanation is codon usage bias. If CUA is a rare codon, the cell has a lower concentration of the corresponding tRNA. This causes ribosomes to pause at that site, which can lead to lower overall translation efficiency, ribosome drop-off, or co-translational protein misfolding and degradation, all of which would result in reduced protein levels.

Question 11

1. Within the gene's promoter region to increase the rate of transcription.
2. At the position corresponding to codon 280, near the C-terminus.
3. At the position corresponding to codon 54, a short distance downstream of the insertion. (correct answer)
4. Within the first intron, to alter the splicing of the pre-mRNA.

Explanation: The initial insertion causes a frameshift, altering the reading frame for all subsequent codons and likely leading to a premature stop. A single nucleotide deletion downstream of the insertion can act as a suppressor by restoring the original reading frame. For this to be effective, it must occur relatively close to the original mutation. A deletion at codon 54 would restore the correct reading frame for the majority of the protein (codons 55-300). The amino acids between 50 and 54 would be incorrect, but if this region is not critical, overall function can be restored. A deletion at codon 280 would leave the bulk of the protein (codons 50-280) out-of-frame.

Question 12

1. Patient A's mutant mRNA will likely be degraded via nonsense-mediated decay (NMD), while Patient B's mRNA is more likely to produce a stable, truncated protein. (correct answer)
2. Both patients will produce stable, truncated proteins of different lengths that are detectable on a Western blot.
3. Patient B's mutation will be more functionally detrimental, as the premature stop codon occurs later in the transcript.
4. Patient A's transcript will be stabilized due to the early stop codon, leading to an overproduction of a short, non-functional peptide.

Explanation: Nonsense-mediated decay (NMD) is a surveillance pathway that degrades mRNAs containing premature termination codons (PTCs). A PTC typically triggers NMD if it is located more than 50-55 nucleotides upstream of the final exon-exon junction. Patient A's mutation at position 350 is far upstream of the junction at 1,300, making its mRNA a target for NMD. Patient B's mutation at 1,500 is in the final exon, downstream of the last junction, so the NMD pathway is not triggered, and a stable, truncated protein is likely to be produced.

Question 13

1. Mutant 2 will be more disruptive to protein function because the complete removal of an amino acid is more severe than a single nucleotide change.
2. Both mutants will produce non-functional proteins of a similar, severely truncated length due to premature termination.
3. Mutant 1's protein will have a completely altered amino acid sequence downstream of position 11, while Mutant 2's protein will be missing one amino acid. (correct answer)
4. Both mutants will likely result in functional proteins, as the mutations occur early in the sequence and can be compensated for.

Explanation: Mutant 1, with a 1-bp deletion, will cause a frameshift. This alters the reading frame for all subsequent codons, leading to a completely different amino acid sequence from that point onward and likely a premature stop codon. Mutant 2, with a 3-bp deletion, is an in-frame deletion. It removes exactly one codon, resulting in a protein that is missing a single amino acid but maintains the correct reading frame for the rest of the sequence. The frameshift in Mutant 1 is almost always a more severe defect than an in-frame deletion of a single amino acid.

Question 14

1. A frameshift leads to immediate termination of translation due to the creation of a stop codon.
2. The protein will have an altered C-terminus and be extended in length by approximately 20 amino acids. (correct answer)
3. The protein will be truncated by one amino acid as the ribosome stalls at the mutated site.
4. The mutation will be silent as it occurs just before the natural stop codon.

Explanation: The deletion of a single nucleotide causes a frameshift. The original reading frame was UGC (Cys) | GAG (Glu) | UAA (Stop). The new frame is UGC (Cys) | AGU (Ser) | AA... The original UAA stop codon is no longer in frame. The ribosome will continue translating into the 3' UTR sequence in this new frame. Since the 3' UTR is 60 nucleotides long, this will add approximately 60/3 = 20 new, incorrect amino acids to the C-terminus before translation eventually terminates, resulting in a longer protein with an altered C-terminus.

Question 15

1. A silent mutation, because the substitution occurs at the third position of the tyrosine codon, which is typically a wobble position.
2. A nonsense mutation, resulting in the premature termination of translation after the tryptophan residue. (correct answer)
3. A missense mutation, changing the tyrosine residue to a different amino acid but allowing translation to continue.
4. A frameshift mutation, because the altered codon disrupts the ribosomal reading frame.

Explanation: The DNA coding strand has the same sequence as the mRNA, with T instead of U. The original DNA codon 5'-TAC-3' is transcribed into the mRNA codon 5'-UAC-3' (Tyrosine). The mutant DNA codon 5'-TAG-3' is transcribed into the mRNA codon 5'-UAG-3', which is a stop codon. Therefore, this mutation converts a codon for an amino acid into a stop codon, which is a nonsense mutation. Translation will terminate prematurely.

Question 16

1. The ribosome will fail to initiate at the mutated AUG codon and will scan downstream to the next available AUG, resulting in an N-terminally truncated protein.
2. The rate of translation initiation will be reduced because the mutation weakens the Kozak sequence, leading to lower overall protein yield. (correct answer)
3. The mutation will be silent because it occurs in the 5' untranslated region and does not change the start codon itself.
4. Translation will initiate normally, but the first amino acid incorporated will be isoleucine instead of methionine.

Explanation: The Kozak sequence (GCCRCCAUGG, where R is a purine) surrounds the start codon and influences the efficiency of translation initiation. The nucleotide at the -4 position is particularly important, with a purine (A or G) being optimal. The mutation changes C to A at the -4 position, which actually strengthens the Kozak sequence slightly. However, the overall context still represents a suboptimal Kozak sequence compared to the ideal consensus, so translation initiation efficiency may be somewhat reduced compared to an optimal sequence.

Question 17

1. The protein will be shorter than wild-type due to the skipping of Exon 1 during mRNA processing.
2. The protein will be identical to the wild-type protein, but its expression level will be significantly reduced.
3. The translated protein will be significantly longer than wild-type, containing an additional 95 amino acids.
4. The protein will likely be truncated due to a frameshift and a premature stop codon within the retained intron sequence. (correct answer)

Explanation: Disruption of the 5' splice site will most likely cause the splicing machinery to fail to recognize and excise Intron 1. This leads to intron retention. Because the length of Intron 1 (95 bp) is not a multiple of three, its inclusion in the mature mRNA will shift the reading frame. This frameshift will alter all downstream codons and almost certainly introduce a premature stop codon, leading to a truncated protein.

Question 18

1. Patient A's mutant mRNA will likely be degraded via nonsense-mediated decay (NMD), while Patient B's mRNA is more likely to produce a stable, truncated protein. (correct answer)
2. Both patients will produce stable, truncated proteins of different lengths that are detectable on a Western blot.
3. Patient B's mutation will be more functionally detrimental, as the premature stop codon occurs later in the transcript.
4. Patient A's transcript will be stabilized due to the early stop codon, leading to an overproduction of a short, non-functional peptide.

Explanation: Nonsense-mediated decay (NMD) is a surveillance pathway that degrades mRNAs containing premature termination codons (PTCs). A PTC typically triggers NMD if it is located more than 50-55 nucleotides upstream of the final exon-exon junction. Patient A's mutation at position 350 is far upstream of the junction at 1,300, making its mRNA a target for NMD. Patient B's mutation at 1,500 is in the final exon, downstream of the last junction, so the NMD pathway is not triggered, and a stable, truncated protein is likely to be produced.

Question 19

1. Mutant 2 will be more disruptive to protein function because the complete removal of an amino acid is more severe than a single nucleotide change.
2. Both mutants will produce non-functional proteins of a similar, severely truncated length due to premature termination.
3. Mutant 1's protein will have a completely altered amino acid sequence downstream of position 11, while Mutant 2's protein will be missing one amino acid. (correct answer)
4. Both mutants will likely result in functional proteins, as the mutations occur early in the sequence and can be compensated for.

Explanation: Mutant 1, with a 1-bp deletion, will cause a frameshift. This alters the reading frame for all subsequent codons, leading to a completely different amino acid sequence from that point onward and likely a premature stop codon. Mutant 2, with a 3-bp deletion, is an in-frame deletion. It removes exactly one codon, resulting in a protein that is missing a single amino acid but maintains the correct reading frame for the rest of the sequence. The frameshift in Mutant 1 is almost always a more severe defect than an in-frame deletion of a single amino acid.

Question 20

1. The ribosome will fail to initiate at the mutated AUG codon and will scan downstream to the next available AUG, resulting in an N-terminally truncated protein.
2. The rate of translation initiation will be reduced because the mutation weakens the Kozak sequence, leading to lower overall protein yield. (correct answer)
3. The mutation will be silent because it occurs in the 5' untranslated region and does not change the start codon itself.
4. Translation will initiate normally, but the first amino acid incorporated will be isoleucine instead of methionine.

Explanation: The Kozak sequence (GCCRCCAUGG, where R is a purine) surrounds the start codon and influences the efficiency of translation initiation. The nucleotide at the -4 position is particularly important, with a purine (A or G) being optimal. The mutation changes C to A at the -4 position, which actually strengthens the Kozak sequence slightly. However, the overall context still represents a suboptimal Kozak sequence compared to the ideal consensus, so translation initiation efficiency may be somewhat reduced compared to an optimal sequence.