Genetics Quiz: Mutation Types

What this quiz covers

This quiz focuses on Mutation Types, 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. 24 amino acids (correct answer)
2. 25 amino acids
3. 385 amino acids
4. 9 amino acids

Explanation: The first 9 codons are translated correctly, producing 9 amino acids. The 2-bp deletion at the 10th codon causes a frameshift. Translation continues in the new, incorrect frame for 15 codons before a stop codon is reached. Therefore, the total length of the resulting polypeptide will be the sum of the correctly translated amino acids and the frameshifted amino acids: 9 (correct) + 15 (frameshifted) = 24 amino acids.

Question 2

1. A silent point mutation that created a codon that is rarely used by the organism's translational machinery. (correct answer)
2. A nonsense mutation that was suppressed by a compensatory mutation in a tRNA gene.
3. An in-frame deletion of a single codon corresponding to a non-essential amino acid.
4. A missense mutation that was corrected by post-translational protein editing mechanisms.

Explanation: The mutation is silent at the protein sequence level but has a phenotypic consequence (reduced protein production). This can be explained by codon bias. The new codon, while coding for the same amino acid, may correspond to a rare tRNA in the cell. This would slow down the rate of translation, leading to lower protein levels. A suppressed nonsense mutation (B) would likely have some amino acid change or still be inefficient. An in-frame deletion (C) would alter the primary sequence. Post-translational editing (D) does not correct missense mutations at the primary sequence level.

Question 3

1. A transversion resulting in a non-conservative missense mutation. (correct answer)
2. A transition resulting in a non-conservative missense mutation.
3. A transversion resulting in a conservative missense mutation.
4. A transition resulting in a conservative missense mutation.

Explanation: When analyzing point mutations, you need to classify them on two dimensions: the type of nucleotide change and the functional impact on the protein. Looking at the codon change from GAA (glutamic acid) to GUA (valine), the single nucleotide substitution is A→U. This represents a purine (adenine) changing to a pyrimidine (uracil), which defines a transversion. If it were purine-to-purine or pyrimidine-to-pyrimidine, it would be a transition. The amino acid change is equally important. Glutamic acid is a negatively charged, hydrophilic amino acid, while valine is nonpolar and hydrophobic. This dramatic difference in chemical properties makes it a non-conservative substitution that significantly alters protein function—exactly what happens in sickle cell anemia, where this change causes hemoglobin to polymerize and distort red blood cells. Answer A correctly identifies both aspects: transversion + non-conservative missense mutation. Answer B incorrectly calls the A→U change a transition. Answer C correctly identifies the transversion but wrongly suggests the glutamic acid-to-valine substitution is conservative—the opposite charge and polarity properties make this clearly non-conservative. Answer D gets both classifications wrong, calling it a transition and conservative. Study tip: For mutation classification questions, always check both the nucleotide change (purine↔pyrimidine = transversion; within purine or pyrimidine families = transition) and amino acid properties (similar properties = conservative; different properties = non-conservative). Sickle cell anemia is a classic example of how a single non-conservative substitution can have devastating functional consequences.

Question 4

1. A conservative missense transition.
2. A non-conservative missense transversion. (correct answer)
3. A conservative missense transversion.
4. A non-conservative missense transition.

Explanation: First, classify the DNA change. The coding strand changes from T (a pyrimidine) to G (a purine). This is a transversion. Second, classify the amino acid change. Cysteine to Tryptophan is a change in amino acid, so it is a missense mutation. Although both are nonpolar, the significant difference in the size of their side chains (Tryptophan is much larger) makes this a non-conservative substitution, as it is likely to affect protein structure. Therefore, the most precise classification is a non-conservative missense transversion.

Question 5

1. It will be identical to the wild-type protein.
2. It will have nine incorrect amino acids and may have an altered length. (correct answer)
3. It will be completely non-functional due to a persistent frameshift.
4. It will have a single incorrect amino acid corresponding to the site of insertion.

Explanation: The insertion at codon 6 shifts the reading frame. The deletion at codon 15 restores the original reading frame from codon 15 onwards. However, the codons between the two mutations (codons 6 through 14, a total of 9 codons) will be translated in the incorrect frame, leading to nine incorrect amino acids. The overall length of the protein will likely be the same, unless one of the mutated codons becomes a stop codon.

Question 6

1. It replaces a hydrophobic amino acid with a hydrophilic one on the protein's surface.
2. It substitutes a proline residue into the middle of an alpha-helix.
3. It changes a cysteine residue involved in a critical disulfide bond.
4. It substitutes a valine for an isoleucine in a non-essential, flexible loop. (correct answer)

Explanation: This question asks to identify the most conservative missense mutation. Valine and isoleucine are both small, nonpolar, hydrophobic amino acids with very similar structures. A substitution of one for the other (a conservative substitution), especially in a non-essential region, is least likely to disrupt the protein's structure and function. In contrast, changing polarity (A), disrupting a secondary structure like an alpha-helix with proline (B), or breaking a required disulfide bond (C) would all be considered non-conservative changes with a high probability of causing a loss of function.

Question 7

1. A point mutation that converts the codon adjacent to the insertion into a stop codon.
2. An insertion of one additional nucleotide near the original insertion site. (correct answer)
3. A deletion of two nucleotides anywhere else in the gene.
4. A deletion of three nucleotides downstream of the insertion site.

Explanation: The initial insertion of two nucleotides causes a +2 frameshift. To restore the reading frame, a compensatory mutation is needed that brings the total number of inserted/deleted bases to a multiple of three. An insertion of one additional nucleotide (+1) would result in a total insertion of 2 + 1 = 3 nucleotides. This would restore the reading frame downstream from the second insertion, creating an in-frame insertion of one codon. While the intervening amino acids would be wrong, this has the potential to restore function. A stop codon (A) would not restore function. A deletion of two bases (C) would still leave the frame shifted. A deletion of three bases (D) would remove one amino acid but would not correct the original +2 frameshift.

Question 8

1. The 3' end of the mRNA is less stable and is rapidly degraded.
2. Fewer amino acids will be altered before a potential termination codon is reached. (correct answer)
3. Mutations at the 3' end are more easily repaired by the DNA repair machinery.
4. The ribosome is more likely to terminate translation prematurely at the 3' end.

Explanation: A frameshift mutation alters the reading frame from the point of the mutation onwards. If the mutation occurs near the 5' end (the beginning of the coding sequence), a large portion of the protein will be translated with an incorrect amino acid sequence. If it occurs near the 3' end (the end of the coding sequence), most of the protein will have been translated correctly, and only a small C-terminal portion will be altered. This makes the mutation more likely to result in a partially or fully functional protein.

Question 9

1. A transition resulting in a missense mutation.
2. A transversion resulting in a nonsense mutation.
3. A transition resulting in a nonsense mutation. (correct answer)
4. A transversion resulting in a frameshift mutation.

Explanation: The DNA change is from G to A. Both guanine (G) and adenine (A) are purines, so this is a transition. The mutation changes a codon for an amino acid (Tryptophan) into a stop codon, which is classified as a nonsense mutation. Therefore, the event is a transition resulting in a nonsense mutation.

Question 10

1. No change, because the mutation is in a non-coding intron.
2. A single amino acid substitution, as the ribosome reads through the mutated base.
3. Inclusion of the third intron's sequence in the mature mRNA, likely causing a frameshift. (correct answer)
4. Exon skipping, where the exon preceding the third intron is removed from the mature mRNA.

Explanation: The branch point is a critical cis-acting sequence required for the chemical reactions of splicing. A mutation at this site can disrupt the splicing process for that intron. The most direct consequence is failure to splice out the intron (intron retention). This leads to the inclusion of the intron's sequence in the final mRNA. Since intron sequences are not typically in a protein-coding frame and often contain stop codons, this usually results in a frameshift and/or a premature stop codon, leading to a non-functional protein. While exon skipping (D) can also result from splicing mutations, intron retention is a very common outcome of a disabled branch point.

Question 11

1. Intragenic suppression. (correct answer)
2. Intergenic suppression.
3. True reversion.
4. A silent mutation.

Explanation: When you encounter questions about mutations and reversions, focus on whether the suppression occurs within the same gene (intragenic) or involves a different gene (intergenic), and whether the original mutation is actually removed. In this scenario, you have a frameshift mutation caused by a G insertion that disrupts the reading frame and creates a nonfunctional protein. The revertant strain regains function not by removing the original insertion, but by acquiring a C deletion three codons downstream. This second mutation restores the correct reading frame, allowing production of a functional protein despite both mutations being present. This exemplifies intragenic suppression (A) because both the original mutation and the suppressing mutation occur within the same gene, and the suppressor mutation compensates for the harmful effects of the first mutation without removing it. (B) Intergenic suppression is incorrect because this would require the suppressing mutation to be in a different gene, such as a tRNA gene that could read through stop codons. Here, both mutations are in the same coding sequence. (C) True reversion is wrong because the original G insertion remains present. True reversion would require the exact reversal of the original mutation—deletion of the inserted G nucleotide. (D) A silent mutation is incorrect because both mutations significantly affect the protein sequence and function, just in compensatory ways. Study tip: Remember that intragenic suppression involves two mutations in the same gene that cancel each other's effects, while the original mutation stays put—it's genetic teamwork, not erasure.

Question 12

1. Missense mutation, and the second is a frameshift mutation.
2. Nonsense mutation, and the second is a suppressor mutation. (correct answer)
3. Frameshift mutation, and the second is a reversion mutation.
4. Silent mutation, and the second is an enhancer mutation.

Explanation: The first mutation caused a truncated protein, which is characteristic of a nonsense (stop) codon. The second mutation in a tRNA gene restores function. This is a classic example of intergenic suppression. The mutant tRNA has an altered anticodon that can now recognize the premature stop codon and insert an amino acid (in this case, glutamine), allowing translation to continue to produce a full-length protein. A reversion would be a change back to the original sequence in the same gene.

Question 13

1. A single-nucleotide deletion in the first exon.
2. A 3-bp in-frame deletion in a functionally critical domain.
3. A point mutation that converts a UAA stop codon to a UAC tyrosine codon.
4. A point mutation changing the third base of a codon that results in the same amino acid. (correct answer)

Explanation: A point mutation that does not change the amino acid sequence is a silent mutation (D). Such mutations usually have no phenotypic effect, although rare exceptions exist (e.g., affecting splicing or codon bias). A frameshift deletion (A) is typically catastrophic. An in-frame deletion in a critical domain (B) would likely destroy function by removing a key amino acid. A read-through mutation of a stop codon (C) would add extra amino acids to the C-terminus, which often destabilizes the protein or alters its function. Thus, the silent mutation is the most likely to be benign.

Question 14

1. A series of independent point mutations.
2. A frameshift mutation due to base insertion.
3. An in-frame insertion mutation. (correct answer)
4. A nonsense mutation leading to a truncated protein.

Explanation: The mutation is an increase in the number of CAG repeats. Since CAG is a trinucleotide (3 base pairs), the insertion is a multiple of three. Insertions of nucleotides in multiples of three do not alter the reading frame; they are called in-frame insertions. This specific mutation leads to the addition of multiple glutamine residues (polyglutamine tract) in the huntingtin protein. It is not a frameshift because the reading frame is maintained. It is not a series of point mutations, but rather a large-scale insertion event, often due to polymerase slippage.

Question 15

1. It will cause a frameshift, altering the entire amino acid sequence.
2. It will cause a single amino acid substitution at the N-terminus.
3. It will result in a silent mutation with no change to the protein.
4. It might alter the protein's expression level but will not change its amino acid sequence. (correct answer)

Explanation: The mutation occurs in the 5' UTR, which is a non-coding region upstream of the start codon. Therefore, it cannot cause a frameshift, missense, or nonsense mutation in the protein itself because it's not part of the translated sequence. However, the 5' UTR can contain regulatory elements that affect mRNA stability or translation initiation (e.g., ribosome binding sites). A mutation in this region could therefore alter how much protein is made, but the protein's amino acid sequence will be unchanged.

Question 16

1. A G to T transversion.
2. An A to T transversion.
3. A G to A transition. (correct answer)
4. A C to T transition.

Explanation: First, determine the DNA sequences. The mRNA codon changes from 5'-UCA-3' to 5'-UAA-3'. The corresponding DNA coding strand changes from 5'-TCA-3' to 5'-TAA-3'. The DNA template strand is complementary to the coding strand (and mRNA), so the original template strand sequence was 3'-AGT-5'. The new template strand sequence is 3'-ATT-5'. Therefore, the mutation on the template strand was a change from G to A. Both G and A are purines, so this is a transition.

Question 17

1. A mutation is a form of DNA damage that has not yet been repaired.
2. DNA damage, such as a thymine dimer, becomes a mutation only after it causes a heritable change in the DNA sequence during replication. (correct answer)
3. All point mutations arise from the direct chemical conversion of one base to another, which constitutes DNA damage.
4. A frameshift mutation is considered DNA damage, whereas a point mutation is a true mutation.

Explanation: This question tests the distinction between DNA damage (a physical or chemical alteration of the DNA structure, like a thymine dimer or an abasic site) and a mutation (a permanent, heritable change in the base sequence). DNA damage can often be repaired perfectly. However, if the replication machinery encounters the damage before it is repaired, it may insert an incorrect base opposite the lesion. Once this incorrect base is established and the DNA is replicated again, the change becomes permanent and heritable, at which point it is classified as a mutation. Therefore, damage is the precursor, and mutation is the heritable outcome.

Question 18

1. The 3' end of the mRNA is less stable and is rapidly degraded.
2. Fewer amino acids will be altered before a potential termination codon is reached. (correct answer)
3. Mutations at the 3' end are more easily repaired by the DNA repair machinery.
4. The ribosome is more likely to terminate translation prematurely at the 3' end.

Explanation: A frameshift mutation alters the reading frame from the point of the mutation onwards. If the mutation occurs near the 5' end (the beginning of the coding sequence), a large portion of the protein will be translated with an incorrect amino acid sequence. If it occurs near the 3' end (the end of the coding sequence), most of the protein will have been translated correctly, and only a small C-terminal portion will be altered. This makes the mutation more likely to result in a partially or fully functional protein.

Question 19

1. It will be identical to the wild-type protein.
2. It will have nine incorrect amino acids and may have an altered length. (correct answer)
3. It will be completely non-functional due to a persistent frameshift.
4. It will have a single incorrect amino acid corresponding to the site of insertion.

Explanation: The insertion at codon 6 shifts the reading frame. The deletion at codon 15 restores the original reading frame from codon 15 onwards. However, the codons between the two mutations (codons 6 through 14, a total of 9 codons) will be translated in the incorrect frame, leading to nine incorrect amino acids. The overall length of the protein will likely be the same, unless one of the mutated codons becomes a stop codon.

Question 20

1. A conservative missense transition.
2. A non-conservative missense transversion. (correct answer)
3. A conservative missense transversion.
4. A non-conservative missense transition.

Explanation: First, classify the DNA change. The coding strand changes from T (a pyrimidine) to G (a purine). This is a transversion. Second, classify the amino acid change. Cysteine to Tryptophan is a change in amino acid, so it is a missense mutation. Although both are nonpolar, the significant difference in the size of their side chains (Tryptophan is much larger) makes this a non-conservative substitution, as it is likely to affect protein structure. Therefore, the most precise classification is a non-conservative missense transversion.