Genetics Quiz: Mutagens And Mechanisms

What this quiz covers

This quiz focuses on Mutagens And Mechanisms, 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. A:T → G:C transitions (correct answer)
2. G:C → A:T transitions
3. A:T → T:A transversions
4. Single base-pair deletions

Explanation: 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.

Question 2

1. A G:C → A:T transition on the template strand.
2. An A:T → G:C transition on the template strand. (correct answer)
3. An A:T → C:G transversion on the template strand.
4. A single nucleotide insertion to restore the reading frame.

Explanation: 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.

Question 3

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

Explanation: 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.

Question 4

1. A G:C to A:T transition
2. A G:C to C:G transversion
3. A G:C to T:A transversion (correct answer)
4. A deletion of the G:C base pair

Explanation: 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.

Question 5

1. EMS intercalates between G:C base pairs, causing DNA polymerase to skip a base during replication, resulting in a deletion that is later filled by an A:T pair.
2. 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. (correct answer)
3. EMS deaminates cytosine, converting it to uracil, which pairs with adenine in the next round of replication, leading to a G:C to A:T transition.
4. EMS acts as a base analog for guanine and is incorporated opposite cytosine, but its structure leads to mispairing with adenine in subsequent replication cycles.

Explanation: 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).

Question 6

1. DNA polymerase during replication (correct answer)
2. RNA polymerase during transcription
3. Aminoacyl-tRNA synthetase during translation
4. The nucleotide excision repair machinery

Explanation: 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.

Question 7

1. Thymine will be incorporated, resulting in no mutation.
2. Cytosine will be incorporated, resulting in an A:T to G:C transition. (correct answer)
3. Guanine will be incorporated, resulting in an A:T to C:G transversion.
4. Adenine will be incorporated, resulting in a single-base insertion.

Explanation: 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.

Question 8

1. the incorporation of proflavin into the new DNA strand in place of a purine base, causing subsequent mispairing.
2. the stabilization of a looped-out nucleotide by the intercalated proflavin, leading to DNA polymerase slippage and an indel. (correct answer)
3. the formation of covalent cross-links between adjacent guanine bases, which stalls the replication fork and induces error-prone repair.
4. the direct conversion of cytosine to uracil through a chemical reaction catalyzed by the dye molecule.

Explanation: 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).

Question 9

1. EMS intercalates between G:C base pairs, causing DNA polymerase to skip a base during replication, resulting in a deletion that is later filled by an A:T pair.
2. 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. (correct answer)
3. EMS deaminates cytosine, converting it to uracil, which pairs with adenine in the next round of replication, leading to a G:C to A:T transition.
4. EMS acts as a base analog for guanine and is incorporated opposite cytosine, but its structure leads to mispairing with adenine in subsequent replication cycles.

Explanation: 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).

Question 10

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

Explanation: 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.

Question 11

1. A G:C → A:T transition on the template strand.
2. An A:T → G:C transition on the template strand. (correct answer)
3. An A:T → C:G transversion on the template strand.
4. A single nucleotide insertion to restore the reading frame.

Explanation: 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.

Question 12

1. Ethidium bromide (correct answer)
2. 5-Bromouracil
3. Hydroxylamine
4. Nitrous acid

Explanation: 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.

Question 13

1. A G:C to A:T transition
2. A G:C to C:G transversion
3. A G:C to T:A transversion (correct answer)
4. A deletion of the G:C base pair

Explanation: 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.

Question 14

1. Thymine will be incorporated, resulting in no mutation.
2. Cytosine will be incorporated, resulting in an A:T to G:C transition. (correct answer)
3. Guanine will be incorporated, resulting in an A:T to C:G transversion.
4. Adenine will be incorporated, resulting in a single-base insertion.

Explanation: 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.

Question 15

1. G:C → A:T transitions
2. A:T → G:C transitions (correct answer)
3. G:C → T:A transversions
4. Single base-pair deletions

Explanation: 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.

Question 16

1. Photolyase will directly reverse the majority of pyrimidine dimers using energy from ambient light, preventing most mutations.
2. The base excision repair (BER) pathway will recognize the distorted helix and excise the entire pyrimidine dimer, followed by accurate resynthesis.
3. 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. (correct answer)
4. The mismatch repair (MMR) system will identify the pyrimidine dimers as incorrect structures and attempt to repair them, often introducing deletions.

Explanation: 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.

Question 17

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

Explanation: 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.

Question 18

1. 5'-GATTACAGATTACA-3'
2. 5'-ATGCATGCATGC-3'
3. 5'-AGAGAGAGAGAG-3'
4. 5'-TTTTTTTTTTTT-3' (correct answer)

Explanation: 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.

Question 19

1. G:C → A:T transitions
2. A:T → G:C transitions (correct answer)
3. G:C → T:A transversions
4. Single base-pair deletions

Explanation: 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.

Question 20

1. Ethidium bromide (correct answer)
2. 5-Bromouracil
3. Hydroxylamine
4. Nitrous acid

Explanation: 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.