Genetic Toxicology Flashcards
Oncogenes:
a. maintain normal cellular growth and development.
b. exert their action in a genetically recessive fashion.
c. are often formed via translocation to a location with a more active promoter.
d. can be mutated to form proto-oncogenes.
e. include growth factors and GTPases, but not transcription factors.
Correct Answer: C
Explanation:
Oncogenes are mutated or dysregulated forms of proto-oncogenes, which are normal genes involved in cell growth and division. A common mechanism of oncogene activation is chromosomal translocation, where a proto-oncogene is relocated to a new genomic site that places it under the control of a stronger or more active promoter. This results in overexpression and contributes to uncontrolled cell proliferation — a hallmark of cancer.
Breakdown of Other Options:
a. Incorrect: This describes proto-oncogenes, not oncogenes.
b. Incorrect: Oncogenes exert a dominant effect — a single mutated copy can drive carcinogenesis.
d. Incorrect: It’s the other way around — proto-oncogenes mutate to become oncogenes.
e. Incorrect: Oncogenes can include transcription factors — e.g., MYC, not just growth factors or GTPases.
Which of the following is NOT one of the more common sources of DNA damage?
a. ionizing radiation.
b. UV light.
c. electrophilic chemicals.
d. DNA polymerase error.
e. x-rays.
Correct Answer: d. DNA polymerase error.
Explanation:
While replication errors can cause mutations, ionizing radiation, UV light, electrophilic chemicals, and X-rays are more prominent and direct sources of DNA damage.
Which of the following pairs of DNA repair mechanisms is most likely to introduce
mutations into the genetic composition of an organism?
a. nonhomologous end-joining (NHEJ) and base excision repair.
b. nonhomologous end-joining and homologous recombination.
c. homologous recombination and nucleotide excision repair.
d. nucleotide excision repair and base excision repair.
e. homologous recombination and mismatch repair.
Correct Answer: B
Explanation:
Nonhomologous end-joining (NHEJ) is error-prone because it ligates DNA ends without using a homologous template, often leading to insertions or deletions.
Homologous recombination is generally high-fidelity but can still introduce mutations when misalignment occurs or if recombination happens between non-identical sequences (e.g., in repeated elements or pseudogenes).
Together, this pairing has a higher risk of introducing mutations into the genome compared to combinations involving nucleotide excision repair, base excision repair, or mismatch repair, which are more precise and typically fix rather than create mutations.
Why not A?
While NHEJ is error-prone, base excision repair (BER) is highly accurate and not typically mutation-inducing. So this combo is less likely overall to introduce mutations than B.B(A)
Which of the following DNA mutations would NOT be considered a frameshift mutation?
a. insertion of 5 nucleotides.
b. insertion of 7 nucleotides.
c. deletion of 18 nucleotides.
d. deletion of 13 nucleotides.
e. deletion of 1 nucleotide.
Correct Answer: c. deletion of 18 nucleotides.
Explanation:
Frameshift mutations occur when the insertion or deletion is not a multiple of 3. 18 is divisible by 3, so the reading frame remains intact.
Which of the following base pair mutations is properly characterized as a transversion
mutation?
a. T → C.
b. A → G.
c. G → A.
d. T → U.
e. A → C
Correct Answer: e. A → C.
Explanation:
A transversion mutation is a purine to pyrimidine or vice versa. A (purine) → C (pyrimidine) fits this category.
All of the following statements regarding nondisjunction during meiosis are true EXCEPT:
a. Nondisjunction events can happen during meiosis I or meiosis II.
b. All gametes from nondisjunction events have an abnormal chromosome number.
c. Trisomy 21 (Down syndrome) is a common example of nondisjunction.
d. In a nondisjunction event in meiosis I, homologous chromosomes fail to separate.
e. The incorrect formation of spindle fibers is a common cause of nondisjunction during
meiosis.
Correct Answer: b. All gametes from nondisjunction events have an abnormal chromosome number.
Explanation:
Not all gametes are abnormal. If nondisjunction occurs in meiosis I or II, only some of the resulting gametes will be aneuploid.
Which of the following diseases does NOT have a recessive inheritance pattern?
a. phenylketonuria.
b. cystic fibrosis.
c. Tay–Sachs disease.
d. sickle cell anemia.
e. Huntington’s disease.
Correct Answer: e. Huntington’s disease.
Explanation:
Huntington’s disease is autosomal dominant. The others (e.g., PKU, cystic fibrosis, Tay-Sachs, sickle cell anemia) follow recessive inheritance.
What is the purpose of the Ames assay?
a. to determine the threshold of UV light bacteria can receive before having mutations in
their DNA.
b. to measure the frequency of aneuploidy in bacterial colonies treated with various
chemicals.
c. to determine the frequency of a reversion mutation that allows bacterial colonies to
grow in the absence of vital nutrients.
d. to measure the rate of induced recombination in mutagen-treated fungi.
e. to measure induction of phenotypic changes in Drosophila
Correct Answer: c. to determine the frequency of a reversion mutation that allows bacterial colonies to grow in the absence of vital nutrients.
Explanation:
The Ames assay uses bacteria with a mutation that prevents growth without histidine. Reversion (mutation repair) allows growth, indicating mutagenic potential of a chemical.
In mammalian cytogenic assays, chromosomal aberrations are measured after treatment of
the cells at which sensitive phase of the cell cycle?
a. interphase.
b. M phase.
c. S phase.
d. G1.
e. G2
Correct Answer: c. S phase.
Explanation:
DNA replication occurs during S phase. Damage during this time can lead to chromosomal aberrations.
Which of the following molecules is used to gauge the amount of a specific gene being
transcribed to mRNA?
a. protein.
b. mRNA.
c. DNA.
d. cDNA.
e. CGH.
Correct Answer:
d. cDNA
Explanation:
While mRNA is the molecule that directly reflects transcription activity, in practical lab applications (e.g., RT-PCR, microarrays), complementary DNA (cDNA) synthesized from mRNA is what is actually measured.
Here’s the breakdown:
mRNA (b) is indeed the direct product of transcription. However, it’s unstable and degrades quickly.
cDNA (d) is synthesized from mRNA using reverse transcriptase. It is stable and easily quantifiable, so it’s the molecule used in most gene expression assays.
DNA (c) reflects genomic content, not expression levels.
Protein (a) reflects translation, not transcription.
CGH (e) is used for detecting copy number variations in DNA, not transcriptional activity.
