Microbiology Quiz: Types Of Mutations

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Types Of MutationsQuestion 1 of 20

A temperature-sensitive (ts) mutant of a bacteriophage forms plaques on a bacterial lawn at 30°C (permissive temperature) but fails to form plaques at 42°C (restrictive temperature). The most probable molecular explanation for this phenotype is a:

nonsense mutation in a gene essential for DNA replication.

missense mutation leading to a protein that is conformationally unstable at 42°C.

frameshift mutation in a gene encoding a structural capsid protein.

deletion of the promoter for a gene required for host cell lysis.

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Microbiology Quiz

Microbiology Quiz: Types Of Mutations

Practice Types Of Mutations in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

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

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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.

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Question 1

nonsense mutation in a gene essential for DNA replication.
missense mutation leading to a protein that is conformationally unstable at 42°C. (correct answer)
frameshift mutation in a gene encoding a structural capsid protein.
deletion of the promoter for a gene required for host cell lysis.

Explanation: Temperature-sensitive phenotypes are typically caused by missense mutations. The resulting amino acid substitution creates a protein that can fold into its active conformation at the lower, permissive temperature but misfolds and becomes inactive at the higher, restrictive temperature. Nonsense mutations, frameshift mutations, and promoter deletions would likely result in a non-functional product at all temperatures.

Question 2

A researcher analyzes thousands of spontaneous point mutations within the coding regions of the E. coli genome. A strong statistical trend emerges: substitutions in the third position of a codon are far less likely to result in an amino acid change than substitutions in the first or second position. This is a direct consequence of the fact that:

the genetic code is degenerate, with many amino acids specified by multiple codons differing at this position. (correct answer)
the mismatch repair system is more efficient at correcting errors in the third codon position.
DNA polymerase is less likely to make an error when replicating the third base of a codon.
mutations in the third position are more likely to create a stop codon, and are thus removed by selection.

Explanation: When you encounter questions about mutation patterns in coding sequences, think about how the genetic code's structure affects the impact of different base changes. The key insight here lies in the degeneracy of the genetic code. Most amino acids are encoded by multiple codons that differ specifically in their third position - this is called "wobble base pairing." For example, leucine is coded by UUA, UUG, CUU, CUC, CUA, and CUG. Notice how the third position can vary widely while still specifying the same amino acid. This means a mutation in the third position often results in a "silent" or "synonymous" mutation that doesn't change the protein sequence. Answer A correctly identifies this phenomenon - the genetic code's degeneracy, particularly at the third codon position, explains why these mutations are less likely to cause amino acid changes. Answer B is incorrect because mismatch repair systems don't preferentially target specific codon positions based on their location within codons. These systems recognize structural distortions in DNA, not codon position significance. Answer C misrepresents DNA polymerase function. The enzyme doesn't have position-specific error rates based on codon structure - it operates at the DNA level without "knowing" about codon boundaries. Answer D contradicts the observed data. Stop codons are actually less likely to be created by third-position mutations, and the question specifically notes that amino acid changes are less frequent, not that mutations are being selected against. Remember: the genetic code's redundancy at the third position is evolution's buffer against harmful mutations - a critical concept linking molecular biology to evolutionary fitness.

Question 3

A single base substitution in an E. coli gene changes a tryptophan codon (UGG) to a stop codon (UGA) at position 50 of a 400-amino acid protein. Which of the following accurately describes the primary molecular consequence in this bacterial cell?

The mutant mRNA will be rapidly degraded by the nonsense-mediated decay (NMD) pathway, preventing translation.
A full-length, non-functional protein will be synthesized containing an incorrect amino acid at position 50.
A truncated, 49-amino acid polypeptide will be synthesized, which is likely non-functional. (correct answer)
Transcription will be prematurely terminated at the site of the mutation, resulting in a shorter mRNA transcript.

Explanation: A nonsense mutation creates a stop codon within the coding sequence. When the ribosome encounters this premature stop codon (UGA), it terminates translation. The result is the synthesis of a shortened, or truncated, polypeptide. In this case, the protein would be 49 amino acids long instead of 400. Nonsense-mediated decay (NMD) is a quality control mechanism primarily found in eukaryotes, not bacteria. The mutation affects translation, not transcription, so the mRNA length is not affected.

Question 4

A single guanine deletion occurs at position +7 in the coding sequence of a bacterial gene shown below. The start codon (ATG) is at positions +1 to +3.

5'-ATG GGC TTA GCA CGC TAA-3'

Which of the following describes the most probable consequence for the resulting polypeptide?

A truncated, nonfunctional protein due to the immediate creation of a nonsense codon.
A full-length protein with a single amino acid substitution at the site of the mutation.
A protein with a radically altered amino acid sequence downstream of the mutation's location. (correct answer)
A silent mutation with no change in the protein's primary structure due to codon redundancy.

Explanation: The original coding sequence is 5'-ATG GGC TTA GCA CGC TAA-3', which translates to Met-Gly-Leu-Ala-Arg-Stop. Deleting the guanine at position +7 (the first G of the GCA codon) results in the new sequence 5'-ATG GGC TAC ACG CTA A..-3'. The reading frame is now shifted. The new codons are ATG (Met), GGC (Gly), TAC (Tyr), ACG (Thr), CTA (Leu), etc. The amino acid sequence is completely different after the second amino acid (Glycine). This is a frameshift mutation. While a premature stop codon might eventually appear in the new frame, the guaranteed and primary consequence is the alteration of the downstream amino acid sequence.

Question 5

A microbiologist identifies two distinct point mutations in the coding sequence of a bacterial enzyme. Mutation 1 changes a GCA codon to a GCG codon. Mutation 2 changes an AAA codon to an AGA codon. (Relevant codons: GCA/GCG = Alanine; AAA = Lysine; AGA = Arginine). Which statement best classifies these two mutations?

Both are silent mutations because they result in functionally similar amino acids.
Mutation 1 is a silent mutation, while Mutation 2 is a conservative missense mutation. (correct answer)
Mutation 1 is a transversion and Mutation 2 is a transition, making Mutation 1 more impactful.
Both are neutral mutations because Lysine and Arginine have similar chemical properties.

Explanation: Mutation 1 (GCA → GCG) changes the codon but not the encoded amino acid (both code for Alanine); this is a silent mutation. Mutation 2 (AAA → AGA) changes the encoded amino acid from Lysine to Arginine; this is a missense mutation. Because Lysine and Arginine are both basic, positively charged amino acids, it is considered a conservative missense mutation. A neutral mutation refers to the effect on fitness, not the molecular change. A silent mutation is by definition silent at the protein level, while a conservative missense may or may not be neutral with respect to fitness.

Question 6

A researcher is studying a 1500 bp bacterial gene that codes for a 500-amino-acid polypeptide. Two different nonsense mutations are identified. Mutant 1 has a nonsense mutation in the 10th codon. Mutant 2 has a nonsense mutation in the 490th codon. Which statement most accurately predicts the phenotypes?

Both mutants will produce nonfunctional proteins and have identical null phenotypes because any truncation is deleterious.
Both mutations will be targeted by nonsense-mediated decay, resulting in no protein product and identical phenotypes.
Mutant 2 will have a more severe phenotype because the longer, aberrant protein is more likely to be toxic.
Mutant 1 will likely have a null phenotype, while Mutant 2 may retain partial or full protein function. (correct answer)

Explanation: When analyzing nonsense mutations, the key factor is where the premature stop codon occurs relative to functional protein domains. The location determines whether any useful protein function can be salvaged. Mutant 1 has a nonsense mutation at codon 10, producing only the first 9 amino acids before termination. This tiny fragment (less than 2% of the full protein) cannot possibly retain the protein's structure or function - it's essentially a null mutation. Mutant 2, however, has its nonsense mutation at codon 490, allowing production of 489 amino acids (about 98% of the full protein). This near-complete protein may retain partial or even full function if the missing C-terminal region isn't critical for the protein's active site or essential structural elements. Choice A is wrong because truncation severity depends entirely on location - losing 10 amino acids from the C-terminus is very different from losing 490 from the N-terminus. Choice B incorrectly assumes nonsense-mediated decay will eliminate both transcripts equally. While this quality control mechanism exists in eukaryotes, it's less prominent in bacteria, and even when active, its efficiency varies with mutation position. Choice C reverses the actual relationship - shorter truncated proteins are typically less problematic than longer ones missing critical regions. The correct answer is D because mutation location determines functional impact: early nonsense mutations create null phenotypes, while late ones may preserve function. Remember: with nonsense mutations, ask yourself "how much functional protein structure remains?" Early truncations are almost always null; late truncations may retain activity.

Question 7

A large chromosomal inversion occurs in a bacterium, flipping a 50-kb segment of DNA that contains several operons. The breakpoints of the inversion do not occur within any coding sequences or promoters. Which of the following is the most likely consequence?

A null phenotype for all genes within the segment due to their reversed orientation relative to the origin of replication.
Frameshift mutations in the genes located nearest to the two breakpoints of the inversion.
Minimal phenotypic change, but the altered gene order would be detectable by whole-genome sequencing. (correct answer)
Failure of PCR amplification for all genes within the inverted segment using their standard primers.

Explanation: If an inversion does not disrupt any genes or their regulatory elements at its breakpoints, it can be phenotypically silent. All the genes are still present and can be transcribed normally, regardless of their orientation on the chromosome. The primary change is to the gene order on the chromosome. This large-scale rearrangement would not be detected by simple PCR of an internal gene but would be evident from techniques that map gene order, such as whole-genome sequencing or restriction mapping.

Question 8

A Tn5 transposon, which contains a kanamycin resistance gene and strong transcriptional terminators, inserts into the middle of the trpB gene within the polycistronic E. coli trp operon (trpE-trpD-trpC-trpB-trpA). The bacterium is grown in a rich medium containing both tryptophan and kanamycin. What is the most likely phenotype related to the trp operon genes?

The cell will produce functional TrpE, TrpD, and TrpC, but not TrpB or TrpA. (correct answer)
The cell will produce a functional TrpB-Kanamycin resistance fusion protein.
The cell will be unable to grow due to the inactivation of the essential trpB gene.
The cell will express all trp genes normally, as tryptophan in the medium represses the operon.

Explanation: The insertion of the transposon into trpB directly disrupts the coding sequence, preventing the production of a functional TrpB protein. Because the operon is transcribed as a single polycistronic mRNA, the transcriptional terminators within the transposon will cause transcription to stop prematurely. This polar effect will prevent the transcription of any downstream genes, in this case, trpA. The upstream genes (trpE, trpD, trpC) will be transcribed normally up to the insertion point. The cell will grow because the medium supplies tryptophan.

Question 9

A bacterial enzyme functions as a homotetramer, composed of four identical polypeptide subunits. A specific missense mutation in the subunit gene results in a protein that can still assemble into the tetramer but renders any complex containing even one mutant subunit catalytically inactive. In a hypothetical diploid bacterium heterozygous for this mutation, what would be the expected level of enzyme activity, and what is this type of mutation called?

~50% activity; a hypomorphic mutation.
<10% activity; an antimorphic (dominant negative) mutation. (correct answer)
~100% activity; a recessive loss-of-function mutation.
100% activity; a hypermorphic mutation.

Explanation: This mutation is antimorphic, also known as dominant negative, because the mutant protein product interferes with the function of the wild-type product. In a heterozygote producing 50% wild-type and 50% mutant subunits, the subunits assemble randomly. The probability of forming a tetramer composed of only wild-type subunits is (0.5)^4 = 0.0625, or 6.25%. All other combinations will contain at least one mutant subunit and be inactive. Therefore, the total enzyme activity will be severely reduced to less than 10% of the wild-type level.

Question 10

A mutation occurs in the E. coli gene gltT, which encodes a tRNA for glutamate (tRNA-Glu). The mutation changes the anticodon from 5'-UUC-3' (which recognizes the GAA codon for glutamate) to 5'-UUA-3'. What is the most likely consequence of this mutation?

The cell will insert the amino acid glutamate at UAA stop codons, producing abnormally long proteins. (correct answer)
All glutamate codons (GAA) will now be misread as leucine codons, altering many proteins.
The mutation will be silent, as the cell has multiple redundant tRNA genes for glutamate.
The mutant tRNA will fail to be charged with glutamate, leading to a general halt in protein synthesis.

Explanation: When you encounter tRNA mutation questions, focus on the anticodon-codon pairing rules and remember that anticodons are read 3' to 5' while codons are read 5' to 3'. The mutant tRNA-Glu has anticodon 5'-UUA-3', which pairs with codon 5'-UAA-3' (the amber stop codon). Since the tRNA is still charged with glutamate by the same aminoacyl-tRNA synthetase (which recognizes the tRNA's structure, not its anticodon), this creates a "suppressor" tRNA that inserts glutamate at UAA stop codons. This allows translation to continue past normal termination signals, producing abnormally long proteins with additional amino acids beyond the intended stop point. Option A correctly describes this suppressor tRNA mechanism. Option B is wrong because the cell still has other functional tRNA-Glu molecules that continue reading GAA codons normally - this mutation affects only one tRNA species out of multiple. Option C misses the point entirely; while redundancy exists for normal glutamate incorporation, this mutant tRNA gains a new, problematic function rather than simply losing its old one. Option D is incorrect because aminoacyl-tRNA synthetases recognize tRNA structure through the acceptor stem and variable regions, not the anticodon, so the tRNA will still be charged with glutamate. For microbiology exams, remember that suppressor tRNA questions test whether you understand that: (1) tRNA identity for charging depends on structural elements beyond the anticodon, and (2) anticodon mutations can create new codon recognition patterns while maintaining the original amino acid specificity.

Question 11

To generate mutants for a genetic screen, a microbiologist treats a culture of Salmonella enterica with ethidium bromide. This chemical is a planar molecule that inserts itself between the stacked bases of a DNA helix. What is the primary type of mutation this treatment is expected to induce?

Transitions, by chemically modifying guanine to cause mispairing with thymine.
Transversions, by creating abasic sites that are filled in randomly during repair.
Frameshifts, by causing single-base insertions or deletions during DNA replication. (correct answer)
Thymine dimers, by using light energy to form covalent bonds between adjacent pyrimidines.

Explanation: Ethidium bromide is a classic intercalating agent. Its insertion between DNA bases distorts the helix. During replication, this distortion can cause DNA polymerase to either slip and insert an extra base or skip a base entirely. This leads to single-base insertions or deletions, which, in a coding region, result in frameshift mutations. Other mutagens cause transitions/transversions, and UV light causes thymine dimers.

Question 12

A strain of E. coli has a loss-of-function mutation in the mutS gene. The MutS protein is essential for initiating the mismatch repair (MMR) pathway by recognizing and binding to mispaired bases in newly synthesized DNA. This mutant strain would be expected to exhibit a higher rate of:

frameshift mutations specifically within short repetitive DNA sequences.
large chromosomal deletions and inversions after exposure to ionizing radiation.
thymine dimer formation following exposure to UV light.
spontaneous point mutations, including both transitions and transversions. (correct answer)

Explanation: The mismatch repair (MMR) system, initiated by MutS, corrects errors made by DNA polymerase during replication. These errors include base-base mismatches (which lead to substitutions) and small insertion/deletion loops. A defect in MMR leads to a 'mutator' phenotype, characterized by a dramatic increase in the rate of all types of spontaneous point mutations (transitions and transversions) as well as small frameshifts. The most general and accurate description of the consequence is an increase in spontaneous point mutations. MMR is not involved in repairing large rearrangements or UV-induced dimers.

Question 13

A point mutation in the hisD gene of Salmonella renders it auxotrophic for histidine (His-). To characterize the mutation, the strain is treated with two different mutagens: hydroxylamine (which specifically induces G:C to A:T transitions) and proflavin (an intercalating agent that causes frameshifts). The His- strain shows a high rate of reversion to His+ when treated with hydroxylamine, but no increase in reversion rate with proflavin. What was the most likely original mutation in the hisD gene?

A single base pair deletion.
A G:C to A:T transition.
An A:T to G:C transition. (correct answer)
An A:T to T:A transversion.

Explanation: The logic of reversion analysis is key here. The strain is revertible by hydroxylamine, which causes G:C → A:T transitions. This means the reversion mutation is a change from G:C to A:T. For this to restore the wild-type function, the original mutation must have been the opposite: an A:T → G:C transition. The lack of reversion with proflavin confirms the original mutation was a base substitution, not a frameshift. Choice B is the reversion event itself, not the original mutation, a common error in interpreting these experiments.

Question 14

A strain of E. coli has a nonsense mutation in the lacZ gene, resulting in a premature UAG stop codon and an inability to metabolize lactose. A subsequent spontaneous mutation in a different gene restores the Lac+ phenotype, even though the original lacZ mutation is still present. This second mutation most likely occurred in a gene encoding:

a ribosomal protein that increases the rate of translation termination.
a transfer RNA (tRNA) whose anticodon has been altered to recognize the UAG codon. (correct answer)
the sigma factor responsible for initiating transcription of the lac operon.
a DNA glycosylase that specifically recognizes and excises the original mutation from the DNA.

Explanation: This scenario describes intergenic suppression. The second mutation compensates for the first one without correcting it. A nonsense mutation creates a premature stop codon (UAG). A mutation in a tRNA gene can alter its anticodon so that it now recognizes the stop codon and inserts an amino acid, allowing translation to continue. This is a common mechanism for nonsense suppression. Increased termination (A) would worsen the problem. Altering transcription (C) cannot fix a problem in the mRNA sequence. DNA repair (D) is ruled out because the stem states the original mutation is still present.

Question 15

The coding sequence for a bacterial adhesin protein contains a region with five tandem repeats of the codon GCT (Alanine): ...GCT GCT GCT GCT GCT.... A researcher compares two mutants. Mutant A has a 3-base-pair deletion within this region. Mutant B has a 4-base-pair deletion within this region. What is the most likely outcome for the proteins produced by these mutants?

Both mutants will produce a severely truncated, nonfunctional protein due to a frameshift.
Mutant A will produce a slightly shorter but likely functional protein, while Mutant B will produce a nonfunctional protein with an altered C-terminus. (correct answer)
Mutant A will be phenotypically wild-type due to the repetitive sequence, while Mutant B will have a single amino acid substitution.
Both mutants will shift the reading frame, but the effect will be more severe in Mutant B due to the larger deletion.

Explanation: The critical factor is whether the number of deleted bases is a multiple of three. Mutant A has a 3-bp deletion, which removes exactly one codon. This is an in-frame deletion that will result in a protein missing one amino acid (Alanine) but with the correct reading frame downstream. This protein may still be functional. Mutant B has a 4-bp deletion, which is not a multiple of three. This will cause a frameshift mutation, altering the entire downstream reading frame and almost certainly leading to a truncated, nonfunctional protein.

Question 16

Both mutants will produce nonfunctional proteins and have identical null phenotypes because any truncation is deleterious.
Both mutations will be targeted by nonsense-mediated decay, resulting in no protein product and identical phenotypes.
Mutant 2 will have a more severe phenotype because the longer, aberrant protein is more likely to be toxic.
Mutant 1 will likely have a null phenotype, while Mutant 2 may retain partial or full protein function. (correct answer)

Question 17

During DNA replication in E. coli, an adenine base on the template strand undergoes a spontaneous, transient tautomeric shift to its imino form at the moment the replication fork passes. What is the most likely mutation to be incorporated into the newly synthesized strand and subsequently established in the genome after the next round of replication?

An A:T to G:C transition. (correct answer)
An A:T to T:A transversion.
An A:T to C:G transversion.
A single nucleotide deletion at the site of the tautomer.

Explanation: Normally, adenine (A) pairs with thymine (T). The rare imino tautomer of adenine (A*) preferentially base-pairs with cytosine (C). During the first round of replication, the A* on the template strand will cause a C to be incorporated into the new strand, creating an A*:C mismatch. In the second round of replication, this new strand with C will serve as a template, and a guanine (G) will be incorporated opposite it. The final result is that the original A:T base pair has been permanently changed to a G:C base pair, which is a transition mutation.

Question 18

A single base pair deletion.
A G:C to A:T transition.
An A:T to G:C transition. (correct answer)
An A:T to T:A transversion.

Question 19

frameshift mutations specifically within short repetitive DNA sequences.
large chromosomal deletions and inversions after exposure to ionizing radiation.
thymine dimer formation following exposure to UV light.
spontaneous point mutations, including both transitions and transversions. (correct answer)

Question 20

the genetic code is degenerate, with many amino acids specified by multiple codons differing at this position. (correct answer)
the mismatch repair system is more efficient at correcting errors in the third codon position.
DNA polymerase is less likely to make an error when replicating the third base of a codon.
mutations in the third position are more likely to create a stop codon, and are thus removed by selection.