- A researcher wants to generate a library of gene knockouts in yeast. Her strategy is to induce frameshift mutations, which will lead to premature stop codons and truncated proteins. Which of the following substances would be the most appropriate choice to add to the yeast culture medium?
Answer: Ethidium bromide
The goal is to induce frameshift mutations. Ethidium bromide is a classic intercalating agent, a class of mutagens known to cause insertions and deletions (indels) by distorting the DNA helix and causing DNA polymerase to slip during replication. These indels result in frameshifts. The other options are not suitable: 5-Bromouracil (B) is a base analog that causes base substitutions (transitions). Hydroxylamine (C) and nitrous acid (D) are base-modifying agents that also cause base substitutions.
- The chemical structure shown is O⁶-methylguanine, a lesion formed by some alkylating mutagens. Due to the presence of the methyl group, this modified base alters its hydrogen-bonding properties. During replication, this change in structure most directly leads to which event? Refer to the chemical structure below.
Answer: Preferential base pairing with thymine instead of cytosine.
In normal guanine, the O⁶ position is a hydrogen bond acceptor that pairs with a donor on cytosine. The addition of a methyl group to this position in O⁶-methylguanine blocks this acceptor site. The resulting structure now mimics the hydrogen-bonding pattern of adenine, allowing it to form two stable hydrogen bonds with thymine. This mispairing during replication is the primary mutagenic mechanism, leading to a G:C to A:T transition. While replication can be slowed, it is not completely blocked (C), and cross-linking (B) and depurination (D) are different types of damage.
- Aflatoxin B1, a mycotoxin, is metabolized into a reactive epoxide that forms a bulky adduct with guanine bases in DNA. This adduct destabilizes the glycosidic bond, often leading to the formation of an apurinic (AP) site. When an unrepaired AP site is encountered during replication, DNA polymerases frequently insert an adenine opposite the gap. This entire process results in which characteristic mutation?
Answer: A G:C to T:A transversion
The process starts with a G:C pair. The guanine is modified and then lost, creating an apurinic (AP) site on that strand, opposite the original C. During replication, the strand with the C templates a normal G, preserving the sequence in one lineage. The strand with the AP site is used as a template, and a polymerase inserts an adenine (A) opposite the gap. This creates an A on the new strand. In the next round of replication, this A will template a thymine (T). The final result is a T:A pair where the original G:C pair was located. This G to T change (or C to A on the other strand) is a transversion.
- An intercalating agent such as acridine orange does not chemically alter any nucleotide. The mutagenic potential of such a compound relies on an error-prone response from which cellular process or enzyme?
Answer: DNA polymerase during replication
Intercalating agents distort the DNA helix by inserting between base pairs. This physical distortion, not a chemical change, is the source of mutation. During DNA replication, the DNA polymerase is the enzyme that reads the template and synthesizes the new strand. The distortion caused by the intercalating agent can cause the polymerase to 'slip,' leading it to either insert an extra base or skip a base on the template. This polymerase error is what fixes the frameshift mutation into the DNA sequence. Errors in transcription (B) or translation (C) would affect the protein product from that one transcript but would not be heritable mutations. The repair machinery (D) is meant to fix damage, not cause it, though errors in repair can also lead to mutation.
- Reactive oxygen species can damage DNA by converting guanine to 8-oxoguanine (8-oxoG). During replication, DNA polymerase often mispairs 8-oxoG with adenine. This leads to what specific type of point mutation?
Answer: A G:C to T:A transversion
The original base pair is G:C. Oxidative damage converts G to 8-oxoG. In the first round of replication, the strand containing 8-oxoG serves as a template. DNA polymerase inserts an adenine (A) opposite it, creating an 8-oxoG:A pair. In the second round of replication, this A on the new strand templates the incorporation of a thymine (T). The resulting base pair is T:A. Thus, the original G:C pair has been converted to a T:A pair. A change from a purine:pyrimidine pair (G:C) to a pyrimidine:purine pair (T:A) is a transversion.
- The base analog 5-bromouracil (5-BU) is mutagenic because it can undergo a tautomeric shift. It is typically incorporated into DNA opposite adenine, as it mimics thymine. However, its rare enol form can pair with guanine. This property allows 5-BU to induce which specific type of mutation?
Answer: A:T → G:C transitions
The primary mutagenic pathway for 5-BU involves its incorporation and a subsequent tautomeric shift. First, 5-BU (in its common keto form) is incorporated opposite adenine, replacing a thymine. The DNA now contains an A:5-BU pair. If, in a subsequent round of replication, the 5-BU shifts to its rare enol form, it will pair with guanine. This results in one daughter duplex having a G:5-BU pair. In the next round of replication, this G will pair with a C. The net result is the original A:T pair has been converted to a G:C pair, which is a transition. It can also cause the reverse G:C -> A:T transition, but the described mechanism leads to A:T -> G:C.
- A bacterial culture is treated with ethylmethanesulfonate (EMS), an alkylating agent. Subsequent sequencing of the bacterial genome reveals a significantly higher frequency of A:T base pairs at loci that were previously G:C pairs compared to untreated controls. Which mechanism best explains this observation?
Answer: EMS adds an ethyl group to guanine, causing it to preferentially pair with thymine during replication, which ultimately results in a G:C to A:T transition.
Ethylmethanesulfonate (EMS) is an alkylating agent that adds an ethyl group to guanine, forming O⁶-ethylguanine. This modified base has altered pairing properties and preferentially forms hydrogen bonds with thymine instead of cytosine. During the next round of replication, this thymine will then template the incorporation of an adenine, completing the G:C to A:T transition over two replication cycles. The other options describe incorrect mechanisms: intercalation causes frameshifts (A), deamination is the mechanism of agents like nitrous acid (C), and EMS is not a base analog (D).
- Nitrous acid is a deaminating agent that converts cytosine to uracil. If a single cytosine residue in the coding strand of a gene is deaminated, what is the resulting mutation after the DNA has undergone two rounds of replication?
Answer: A G:C to A:T transition
The original base pair is G:C. Nitrous acid deaminates cytosine (C) to uracil (U). The DNA now contains a G:U mismatch. During the first round of replication, the strand with G will template a normal C, preserving the wild-type sequence in one daughter molecule. The strand with U will template the incorporation of adenine (A), because U pairs with A. This results in a daughter molecule with an A:U pair. In the second round of replication, the strand with A will template the incorporation of thymine (T). The final result is an A:T base pair where the original G:C pair was. This entire process constitutes a G:C to A:T transition.
- An auxotrophic bacterial strain contains a UAG (amber) stop codon that arose from a UGG (tryptophan) codon. To select for revertants, a researcher needs to use a mutagen that can convert the UAG codon back to UGG. Which of the following mutagenic mechanisms would be required for this reversion?
Answer: An A:T → G:C transition on the template strand.
The mutation is from UGG (Trp) to UAG (Stop). This corresponds to a DNA change on the template strand from 3'-ACC-5' to 3'-ATC-5'. The change is C→T on the template strand, which is a G:C → A:T transition in the DNA duplex. To revert this mutation (UAG → UGG), the template strand must change back from 3'-ATC-5' to 3'-ACC-5'. This requires a T→C change on the template strand, which corresponds to an A:T → G:C transition in the DNA duplex. Frameshift mutations (D) would not restore the original codon. Transversions (C) would result in a different amino acid or stop codon.
- An E. coli strain that is deficient in nucleotide excision repair (NER) is exposed to a short burst of UV-C radiation and then grown in the dark. Which of the following represents the most likely mechanism by which mutations will be fixed in the population?
Answer: Replication forks will stall at dimer sites, triggering the SOS response and recruitment of error-prone translesion synthesis (TLS) polymerases to replicate across the damage.
UV radiation causes pyrimidine dimers. In the absence of NER, these bulky lesions are not removed. Since the bacteria are grown in the dark, photoreactivation by photolyase (A) cannot occur. When the replication fork encounters a dimer, it stalls. This stall activates the SOS response, which includes the expression of specialized, low-fidelity translesion synthesis (TLS) polymerases (e.g., Pol V). These polymerases can replicate across the dimer but frequently insert incorrect bases opposite the lesion, leading to point mutations. BER (B) repairs single base damage, not bulky dimers. MMR (D) repairs replication mismatches, not DNA damage like dimers.
- Spontaneous mutations can occur during DNA replication due to tautomeric shifts of the bases. If an adenine on the template DNA strand temporarily shifts to its rare imino tautomer just as the DNA polymerase is passing, what base will be incorporated into the newly synthesized strand, and what type of mutation will this eventually cause?
Answer: Cytosine will be incorporated, resulting in an A:T to G:C transition.
In its common amino form, adenine (A) pairs with thymine (T). However, in its rare imino tautomeric form, adenine's hydrogen bonding pattern changes, and it preferentially pairs with cytosine (C). If this happens on the template strand during replication, a C will be incorporated into the new strand. In the next round of replication, this C will serve as a template for a guanine (G). The net result is that the original A:T base pair becomes a G:C base pair, which is a transition mutation.
- Individuals with the genetic disorder Xeroderma Pigmentosum (XP) have a greatly increased risk of developing skin cancer upon exposure to sunlight. This is due to a defect in the nucleotide excision repair (NER) system. The high rate of mutation in the skin cells of XP patients is a direct consequence of the cell's inability to repair which specific type of mutagen-induced lesion?
Answer: Cyclobutane pyrimidine dimers caused by UV radiation.
Sunlight contains UV radiation, which causes the formation of covalent bonds between adjacent pyrimidine bases (thymine or cytosine) in a DNA strand, creating lesions such as cyclobutane pyrimidine dimers. These are bulky lesions that distort the DNA helix. The nucleotide excision repair (NER) pathway is the primary mechanism for repairing such bulky adducts. In XP patients, a defective NER system means these dimers persist, leading to errors (mutations) during DNA replication and consequently a high risk of skin cancer. Mismatch repair handles replication errors (A), other pathways handle double-strand breaks (B), and base excision repair handles apurinic sites (C).
- Two mutagens are used to treat parallel bacterial cultures. Mutagen X treatment results in a high frequency of proteins that are truncated and non-functional. Mutagen Y treatment results primarily in full-length proteins that have a single amino acid substitution and reduced, but not absent, enzymatic activity. Which statement is the most plausible identification of these mutagens?
Answer: Mutagen X is an intercalating agent; Mutagen Y is an alkylating agent.
Truncated proteins are typically the result of premature stop codons. Frameshift mutations (caused by insertions or deletions of 1 or 2 bases) frequently create premature stop codons. Intercalating agents are the classic cause of frameshifts. Therefore, Mutagen X is likely an intercalating agent. Full-length proteins with single amino acid changes are the result of missense mutations, which are caused by base-pair substitutions. Alkylating agents, base analogs, and UV light all cause base-pair substitutions. Therefore, Mutagen Y is a mutagen that causes substitutions. Option B correctly identifies Mutagen X as an intercalating agent and Mutagen Y as an alkylating agent (which causes substitutions).
- Proflavin, an acridine dye, is a potent mutagen for bacteriophages. Its mutagenic effect is not due to covalent modification of DNA bases. Instead, the primary molecular event during DNA replication that leads to proflavin-induced mutation is:
Answer: the stabilization of a looped-out nucleotide by the intercalated proflavin, leading to DNA polymerase slippage and an indel.
Proflavin is an intercalating agent. It inserts itself between the stacked base pairs of the DNA double helix, causing a distortion. This distortion can stabilize transiently looped-out single nucleotides on either the template or the newly synthesized strand during replication. This causes the DNA polymerase to 'slip,' either inserting an extra base or skipping a template base, resulting in a frameshift mutation (an insertion or deletion, i.e., an indel). Proflavin is not incorporated into the DNA (A), nor does it form cross-links (C) or deaminate bases (D).
- The mutagenic potential of intercalating agents like ethidium bromide is highly dependent on DNA sequence context. In which of the following template DNA sequences would such an agent be most likely to induce a +1 frameshift (insertion) mutation during replication?
Answer: 5'-TTTTTTTTTTTT-3'
Intercalating agents cause frameshift mutations by promoting DNA polymerase 'slippage'. This phenomenon is most pronounced in regions with monotonous runs of a single base (homopolymeric tracts). In a sequence like 5'-TTTTTTTTTTTT-3', the template and the newly synthesized strand can easily misalign by one or more bases while still maintaining stable hydrogen bonding. This misalignment can stabilize a looped-out base, leading to an insertion or deletion. The other sequences, while repetitive to some degree, offer a more rigid template that is less prone to this type of slippage.
- A histidine-requiring (His⁻) mutant of Salmonella is used in an Ames test. The mutation is known to be a G:C → A:T transition. A test compound is added to the culture, and a high number of His⁺ revertant colonies are observed. This result indicates that the test compound is mutagenic and most likely causes which specific type of DNA alteration?
Answer: A:T → G:C transitions
The Ames test measures the reversion of a mutation back to the wild-type state. The original mutation that caused the His⁻ phenotype was a G:C to A:T transition. Therefore, to revert to His⁺, the reverse mutation must occur: an A:T to G:C transition. Since the test compound caused a high number of revertants, it must be capable of inducing A:T to G:C transitions. A compound causing the original mutation (A) would not cause reversion. Transversions (C) or deletions (D) would not restore the original G:C pair.
