Genetics Quiz: Mutation Effects On Proteins

What this quiz covers

This quiz focuses on Mutation Effects On Proteins, 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. The functional consequence cannot be determined with certainty and could range from negligible to significant. (correct answer)
2. The mutation will be silent because it occurs on the protein's surface and not in the hydrophobic core.
3. The protein will be completely non-functional because a nonpolar residue was replaced by a polar one.
4. The protein's stability will definitively increase due to favorable interactions of threonine with water.

Explanation: When evaluating the functional impact of amino acid substitutions, you must consider multiple factors: the chemical properties of the amino acids involved, the location within the protein structure, and the complexity of protein function itself. The alanine-to-threonine substitution represents a significant chemical change—from a small, hydrophobic residue to a larger, polar one with a hydroxyl group. While the mutation occurs on the protein surface (away from functional sites), this doesn't automatically determine its impact. Surface residues can affect protein stability, folding kinetics, protein-protein interactions, and even distant conformational changes through allosteric effects. The polar threonine might form new hydrogen bonds with water or other residues, potentially stabilizing or destabilizing the protein, or it could disrupt existing surface interactions. Answer A correctly acknowledges this uncertainty—without experimental data, the functional consequence could range from completely negligible to significantly harmful, depending on the specific protein context and cellular environment. Answer B incorrectly assumes surface mutations are always silent. Surface residues frequently influence protein function through stability and interaction effects. Answer C represents an extreme overstatement—a single amino acid change, especially outside functional sites, rarely renders proteins completely non-functional. Answer D makes an unwarranted prediction about stability; while threonine can form favorable water interactions, it might also disrupt existing stabilizing interactions or create unfavorable steric clashes. Remember: predicting mutation effects requires experimental validation. Be wary of answer choices that make definitive claims about protein function without considering the inherent complexity and context-dependence of protein structure-function relationships.

Question 2

1. A nonsense mutation leading to a truncated protein.
2. A missense mutation changing glycine to alanine.
3. A silent mutation resulting in no change to the amino acid sequence. (correct answer)
4. A frameshift mutation altering the subsequent amino acid sequence.

Explanation: The DNA coding strand has the same sequence as the mRNA, with T instead of U. The original codon, corresponding to 5'-GGT-3', is GGU in mRNA. The mutated codon, corresponding to 5'-GGC-3', is GGC in mRNA. Due to the degeneracy of the genetic code (the 'wobble' in the third position), both GGU and GGC code for the amino acid glycine. Therefore, this substitution has no effect on the amino acid sequence and is a silent mutation. The distractors describe other types of mutations that do not apply here.

Question 3

1. A frameshift mutation leading to a completely non-functional protein.
2. A minor alteration in protein structure with a high probability of retained function. (correct answer)
3. A nonsense mutation causing premature truncation of the protein before position 150.
4. Complete loss of function due to the deletion of a single critical amino acid.

Explanation: A deletion of 3 base pairs removes exactly one codon. This is an 'in-frame' deletion, meaning the reading frame for all subsequent codons remains unchanged. The resulting protein will be missing one amino acid (phenylalanine) but will be otherwise identical to the wild-type protein. Since the deletion is in a non-critical, flexible loop, this small change is unlikely to disrupt the overall structure or function of the protein. Distractor A is the most common error, assuming any deletion causes a frameshift. Distractor C confuses a deletion with a nonsense mutation. Distractor D is unlikely given the location of the deletion specified in the stem.

Question 4

1. A full-length protein with a single amino acid substitution at position 298.
2. A protein truncated at or shortly after position 298, with an altered C-terminal sequence. (correct answer)
3. A completely non-functional protein due to a frameshift affecting the entire amino acid sequence.
4. No protein is produced, as the mutation triggers nonsense-mediated decay of the mRNA.

Explanation: A single nucleotide deletion causes a frameshift. This alters the reading frame from position 298 onward. The new reading frame will likely encounter a premature stop codon, leading to a truncated protein. The few amino acids translated after the deletion but before the new stop codon will also be incorrect. Because the mutation is so close to the natural C-terminus, the protein is only altered at its very end. Distractor A confuses a deletion with a substitution. Distractor C overstates the effect; a frameshift at the end of a gene is much less damaging than one at the beginning. Distractor D is incorrect because nonsense-mediated decay is typically not triggered by premature stop codons located in the final exon or very close to the natural stop codon.

Question 5

1. A stable, truncated polypeptide of 74 amino acids will be produced in large quantities.
2. The ribosome will read through the premature termination codon, producing a full-length protein.
3. The spliceosome will skip exon 2, resulting in an internally deleted, but potentially functional, protein.
4. The mutant mRNA transcript will be targeted for degradation, resulting in little or no protein production. (correct answer)

Explanation: Premature termination codons (PTCs) located upstream of the final exon-exon junction are typically recognized by the nonsense-mediated decay (NMD) surveillance pathway in mammalian cells. This pathway targets the faulty mRNA for rapid degradation to prevent the production of potentially harmful truncated proteins. Since the PTC at codon 75 in exon 2 is far from the final exon, NMD is the most likely outcome. Distractor A ignores the effect of NMD. Distractor B describes ribosomal read-through, which is a very rare event. Distractor C describes an alternative splicing outcome that is possible but less direct and certain than NMD activation.

Question 6

1. The protein is truncated because the deletion removes the tryptophan codon.
2. A frameshift replaces the C-terminal aspartic acid with a new sequence and extends the protein. (correct answer)
3. A missense mutation changes tryptophan, but the protein terminates at the original position.
4. The protein is unchanged because the mutation occurs immediately before the stop codon.

Explanation: The original sequence codes for Trp (UGG) - Asp (GAC) - Stop (UAG). Deleting the first G of the UGG codon results in the new sequence 5'-...UG GAC UAG...-3'. The reading frame shifts. The ribosome now reads the codons as UGG (Trp), followed by ACU (Threonine), followed by AG... (whatever follows in the 3' UTR). The original stop codon UAG is no longer in frame. This frameshift mutation not only changes the C-terminal amino acid from Asp to Thr but also causes the ribosome to read past the original termination site, extending the protein until a new stop codon is encountered in the shifted frame. Distractor A misunderstands how deletions affect the reading frame. Distractor C confuses a deletion with a substitution. Distractor D is incorrect.

Question 7

1. A frameshift mutation leading to premature termination. (correct answer)
2. A deletion mutation resulting in a slightly shorter, stable protein.
3. A simple nonsense mutation.
4. A missense mutation from codon 10 through codon 14.

Explanation: When you encounter questions about nucleotide deletions and their effects on protein synthesis, focus on how the deletion impacts the reading frame and what happens downstream. A single-nucleotide deletion in codon 10 shifts the reading frame for all subsequent codons. This frameshift causes the ribosome to read the mRNA sequence incorrectly from that point forward, creating entirely different amino acid sequences than intended. The fact that a stop codon appears at position 15 (where it shouldn't normally exist) confirms that this frameshift created a premature termination signal, cutting protein synthesis short. Answer A correctly identifies this as a frameshift mutation leading to premature termination - exactly what we observe here. Answer B is wrong because frameshift mutations don't produce "stable proteins." The completely altered amino acid sequence from codon 10 onward, plus early termination, would likely create a nonfunctional protein. Answer C incorrectly calls this a "simple nonsense mutation." Nonsense mutations involve a single base change that directly creates a stop codon, not a deletion that shifts the entire reading frame. Answer D mischaracterizes this as a missense mutation. While codons 10-14 do produce different amino acids due to the frameshift, this description ignores the crucial early termination and fails to recognize the frameshift mechanism. Remember: single-nucleotide insertions or deletions (except in multiples of three) always cause frameshifts. Look for downstream effects like premature stop codons to determine the full impact of the mutation.

Question 8

1. The insertion creates a premature stop codon, leading to termination. (correct answer)
2. The protein will have an extra amino acid inserted after tyrosine.
3. The amino acid sequence will be altered from the point of insertion onward, but translation will continue.
4. A missense mutation occurs, changing the fourth amino acid from glutamic acid to something else.

Explanation: The original sequence is read as AUG (Met), GCU (Ala), UAC (Tyr), GAG (Glu). The insertion of a U after UAC changes the sequence to 5'-AUG GCU UAC U GAG...-3'. The ribosome reads in triplets, so the new reading frame after UAC is UGA G... The codon UGA is a stop codon. Therefore, the insertion immediately creates a premature stop codon, and translation will terminate after the tyrosine is added. The resulting product would be a tripeptide (Met-Ala-Tyr). Distractor C fails to recognize the new stop codon. Distractors B and D misinterpret the effect of a frameshift-inducing insertion.

Question 9

1. Both mutations will have equally severe effects on kinase function.
2. Mutation Y will be more damaging as it alters the protein's surface interactions.
3. Mutation X will have a mild effect because serine and cysteine have similar structures.
4. Mutation X is likely to cause a severe loss of function, while Mutation Y is likely to be benign. (correct answer)

Explanation: Mutation X disrupts a disulfide bond, which is often crucial for maintaining the correct three-dimensional structure of a protein. Replacing cysteine with serine removes the sulfhydryl group necessary for this bond, likely leading to misfolding and a severe loss of function. Mutation Y is a conservative substitution (valine to isoleucine, both nonpolar) in a non-critical region (a surface loop). This type of change is very unlikely to have a significant impact on the protein's function. Therefore, Mutation X is predicted to be much more severe than Mutation Y. Distractor C is incorrect because despite some structural similarity, the functional difference of the side group (SH vs OH) is profound in this context.

Question 10

1. The reading frame will shift, producing a protein with an entirely different C-terminal sequence.
2. The protein will be missing two adjacent amino acids but the downstream reading frame will be correct. (correct answer)
3. The protein will be severely truncated due to a premature stop codon created by the deletion.
4. The protein will be completely normal, as the deletion of two amino acids is functionally insignificant.

Explanation: A deletion of 6 base pairs is a deletion of a multiple of 3. This means that exactly two codons are removed. This is an 'in-frame' deletion. The reading frame for the codons downstream of the deletion is not affected. Therefore, the resulting protein will be shorter by two amino acids, but the sequence of all other amino acids will be correct. Distractor A is incorrect because there is no frameshift. Distractor C is a possible secondary consequence if the deletion brings two parts of the protein together in a way that destabilizes it, but it is not the direct result of the deletion itself, which is in-frame. Distractor D makes an assumption about function that cannot be known for certain; the deletion could be catastrophic or benign depending on the location.

Question 11

1. There will be no significant effect, as it is only a single amino acid change.
2. The protein's stability will increase due to new favorable ionic bonds.
3. The protein will likely misfold or become unstable due to the unfavorable substitution. (correct answer)
4. The protein's function will be enhanced due to the increased polarity in the core.

Explanation: The hydrophobic core is a critical structural feature of globular proteins, stabilized by the exclusion of water and favorable interactions between nonpolar side chains. Leucine is a hydrophobic amino acid, well-suited for this environment. Glutamic acid is hydrophilic and negatively charged. Placing a charged, water-loving residue into the oily, nonpolar core is energetically very unfavorable. This is a non-conservative substitution in a critical location, and it will likely disrupt the normal folding pattern and decrease the overall stability of the protein. Distractor A underestimates the importance of chemical properties. Distractors B and D propose positive outcomes that are biochemically implausible in this context.

Question 12

1. A deletion of a single nucleotide at codon 55. (correct answer)
2. A silent substitution at codon 49.
3. A deletion of three nucleotides at codon 60.
4. A second single nucleotide insertion at codon 51.

Explanation: The original mutation is a +1 frameshift. To restore the reading frame, a compensating mutation that brings the frame back to normal is needed. A single nucleotide deletion (-1) downstream of the original insertion would cancel out the +1 shift, restoring the correct reading frame for the rest of the gene. This would result in a short segment of incorrect amino acids between codons 50 and 55, but the bulk of the protein would be correct, leading to a near-functional state. Distractor B would have no effect on the frameshift. Distractor C would remove an entire amino acid but would not correct the reading frame. Distractor D would create a +2 frameshift, which would not restore the original reading frame.

Question 13

1. A frameshift mutation will occur, leading to a completely altered C-terminal sequence.
2. A silent mutation will occur because the reading frame is maintained.
3. Two glutamine residues will be added to the protein's amino acid sequence. (correct answer)
4. A nonsense mutation will be created by the repeats, resulting in a truncated protein.

Explanation: The insertion consists of two CAG repeats, totaling 6 nucleotides. Since 6 is a multiple of 3, the insertion is 'in-frame' and does not cause a frameshift (eliminating A). The mRNA codon CAG codes for the amino acid glutamine. Adding two CAG repeats to the coding sequence will result in the addition of two glutamine residues to the polypeptide chain at that position. This is not a silent mutation, as the protein's primary structure is altered (eliminating B). While such an expansion can affect function, it does not directly create a stop codon (eliminating D). This mechanism is characteristic of trinucleotide repeat disorders.

Question 14

1. Mutation B will be more severe because it affects the C-terminal end, which is crucial for protein folding.
2. Mutation A is almost certain to produce a non-functional protein, while Mutation B is likely to have a minimal effect on function. (correct answer)
3. Both mutations will likely lead to a complete loss of function because any change to the primary sequence is deleterious.
4. Mutation A will result in a slightly shorter but functional protein, while Mutation B will cause the protein to misfold completely.

Explanation: Mutation A is a nonsense mutation very early in the coding sequence (codon 25 of 500). This will lead to a severely truncated protein that is almost certainly non-functional. Mutation B is a missense mutation that is 'conservative'—it replaces valine with isoleucine, which are both small, nonpolar amino acids with similar properties. Such a substitution, especially late in the protein sequence, is very likely to have little or no effect on the protein's structure and function. Therefore, Mutation A is far more severe than Mutation B. The other options contain incorrect reasoning about mutation severity.

Question 15

1. The protein will be truncated due to premature termination.
2. The protein will have a single amino acid substitution at its C-terminus.
3. The protein will be extended until the next in-frame stop codon is reached. (correct answer)
4. Translation will fail to terminate, resulting in a protein of random, excessive length.

Explanation: The original sequence has a UAA codon, which is a stop codon. The mutation changes this to UCA, which codes for the amino acid Serine. Therefore, translation will no longer terminate at this position. Instead, a serine will be added, and the ribosome will continue translating the 3' untranslated region (UTR) of the mRNA until it encounters the next in-frame stop codon. This results in an extended protein. Distractor A describes the effect of a nonsense mutation, not the mutation of a stop codon. Distractor B incorrectly assumes UAA coded for an amino acid. Distractor D is incorrect because termination will occur at a defined downstream stop codon, not randomly.

Question 16

1. The mutation has no effect on the protein as it is located within a non-coding intron.
2. Transcription of the gene is prematurely terminated at the site of the mutation.
3. Exon 3 is skipped during splicing, leading to a protein missing the amino acids coded by that exon.
4. Intron 2 is retained in the mature mRNA, likely causing a frameshift and creating a premature stop codon. (correct answer)

Explanation: The 5' splice site (typically a GT sequence in the DNA, corresponding to GU in the pre-mRNA) is essential for the spliceosome to recognize and remove an intron. Mutating this site will most likely cause the splicing machinery to fail to recognize the intron-exon boundary. As a result, intron 2 will be retained in the mature mRNA. Since introns do not typically maintain the reading frame and often contain stop codons, their inclusion in the mRNA usually leads to a frameshift and premature termination of translation, producing a severely altered and non-functional protein. Distractor A is a common misconception; while introns are non-coding, their boundary sequences are critical. Distractor B confuses splicing with transcription. Distractor C describes exon skipping, which can happen with splice site mutations, but intron retention is a very common and direct consequence of a non-functional 5' site.

Question 17

1. A frameshift mutation leading to a completely non-functional protein.
2. A minor alteration in protein structure with a high probability of retained function. (correct answer)
3. A nonsense mutation causing premature truncation of the protein before position 150.
4. Complete loss of function due to the deletion of a single critical amino acid.

Explanation: A deletion of 3 base pairs removes exactly one codon. This is an 'in-frame' deletion, meaning the reading frame for all subsequent codons remains unchanged. The resulting protein will be missing one amino acid (phenylalanine) but will be otherwise identical to the wild-type protein. Since the deletion is in a non-critical, flexible loop, this small change is unlikely to disrupt the overall structure or function of the protein. Distractor A is the most common error, assuming any deletion causes a frameshift. Distractor C confuses a deletion with a nonsense mutation. Distractor D is unlikely given the location of the deletion specified in the stem.

Question 18

1. A nonsense mutation leading to a truncated protein.
2. A missense mutation changing glycine to alanine.
3. A silent mutation resulting in no change to the amino acid sequence. (correct answer)
4. A frameshift mutation altering the subsequent amino acid sequence.

Explanation: The DNA coding strand has the same sequence as the mRNA, with T instead of U. The original codon, corresponding to 5'-GGT-3', is GGU in mRNA. The mutated codon, corresponding to 5'-GGC-3', is GGC in mRNA. Due to the degeneracy of the genetic code (the 'wobble' in the third position), both GGU and GGC code for the amino acid glycine. Therefore, this substitution has no effect on the amino acid sequence and is a silent mutation. The distractors describe other types of mutations that do not apply here.

Question 19

1. The insertion creates a premature stop codon, leading to termination. (correct answer)
2. The protein will have an extra amino acid inserted after tyrosine.
3. The amino acid sequence will be altered from the point of insertion onward, but translation will continue.
4. A missense mutation occurs, changing the fourth amino acid from glutamic acid to something else.

Explanation: The original sequence is read as AUG (Met), GCU (Ala), UAC (Tyr), GAG (Glu). The insertion of a U after UAC changes the sequence to 5'-AUG GCU UAC U GAG...-3'. The ribosome reads in triplets, so the new reading frame after UAC is UGA G... The codon UGA is a stop codon. Therefore, the insertion immediately creates a premature stop codon, and translation will terminate after the tyrosine is added. The resulting product would be a tripeptide (Met-Ala-Tyr). Distractor C fails to recognize the new stop codon. Distractors B and D misinterpret the effect of a frameshift-inducing insertion.

Question 20

1. A frameshift mutation will occur, leading to a completely altered C-terminal sequence.
2. A silent mutation will occur because the reading frame is maintained.
3. Two glutamine residues will be added to the protein's amino acid sequence. (correct answer)
4. A nonsense mutation will be created by the repeats, resulting in a truncated protein.

Explanation: The insertion consists of two CAG repeats, totaling 6 nucleotides. Since 6 is a multiple of 3, the insertion is 'in-frame' and does not cause a frameshift (eliminating A). The mRNA codon CAG codes for the amino acid glutamine. Adding two CAG repeats to the coding sequence will result in the addition of two glutamine residues to the polypeptide chain at that position. This is not a silent mutation, as the protein's primary structure is altered (eliminating B). While such an expansion can affect function, it does not directly create a stop codon (eliminating D). This mechanism is characteristic of trinucleotide repeat disorders.