BIO 242 Microbiology Exam 3
Course: BIO 242: General Microbiology
Exam: Exam 3 (Modules: Microbial Genetics, Control of Microbial Growth, Antimicrobial
Therapy, Pathogenesis, and Immunology)
Total Questions: 75 Multiple Choice Questions
Part 1: Microbial Genetics, DNA Replication, and Gene Expression
(Questions 1–15)
1. Which of the following enzymes is primarily responsible for unwinding the DNA double
helix during bacterial replication by breaking hydrogen bonds between base pairs?
A) Topoisomerase (DNA Gyrase)
B) Helicase
C) DNA Polymerase III
D) Primase
Answer: B
Rationale: Helicase breaks the hydrogen bonds between complementary nitrogenous bases to
unwind and separate the two strands of the DNA double helix at the replication fork. DNA Gyrase
(topoisomerase) relieves supercoiling tension ahead of the fork, DNA Polymerase III synthesizes the
new strand, and Primase synthesizes RNA primers.
2. During DNA replication in Escherichia coli, Okazaki fragments are joined together into
a continuous strand by which enzyme?
A) DNA Polymerase I
B) RNA Polymerase
C) DNA Ligase
D) Topoisomerase IV
Answer: C
Rationale: DNA ligase catalyzes the formation of phosphodiester bonds between adjacent Okazaki
fragments on the lagging strand after DNA Polymerase I has removed the RNA primers and replaced
them with DNA nucleotides.
3. What is the role of the promoter region in bacterial transcription?
A) It marks the site where DNA replication begins.
, B) It codes for the ribosomal binding site on mRNA.
C) It serves as the recognition and binding site for RNA polymerase via the sigma factor.
D) It signals the termination of mRNA synthesis.
Answer: C
Rationale: The promoter is a specific DNA sequence located upstream of a gene where RNA
polymerase (facilitated by the sigma factor in prokaryotes) binds to initiate transcription. It is not
translated into protein and does not serve as a replication origin.
4. A point mutation that results in the substitution of one amino acid for another in the
resulting polypeptide chain is classified as a:
A) Silent mutation
B) Nonsense mutation
C) Missense mutation
D) Frameshift mutation
Answer: C
Rationale: A missense mutation is a nucleotide substitution that alters a codon so that it codes for a
different amino acid. A silent mutation alters the codon without changing the amino acid, while a
nonsense mutation creates a premature stop codon.
5. In the lac operon of E. coli, what occurs when lactose is present and glucose is
absent?
A) The repressor binds to the operator, blocking transcription.
B) Allolactose binds the repressor, inactivating it, while elevated cAMP allows CAP to bind the
promoter, leading to high transcription.
C) CAP is inactivated by high levels of cAMP, preventing RNA polymerase binding.
D) Transcription is completely shut down due to catabolite repression.
Answer: B
Rationale: When lactose is present, allolactose inactivates the repressor protein so it releases the
operator. Low glucose elevates intracellular cAMP, forming a cAMP-CAP complex that binds the
promoter and recruits RNA polymerase for maximal transcription.
6. The transfer of naked DNA from a donor cell to a competent recipient cell in
environment is known as:
A) Conjugation
B) Generalized Transduction
C) Specialized Transduction
D) Transformation
Answer: D
, Rationale: Transformation involves the uptake of free, naked extracellular DNA by a physiologically
competent bacterial cell. Conjugation requires cell-to-cell contact via a pilus, and transduction
requires a bacteriophage vector.
7. Which mechanism of horizontal gene transfer specifically requires a bacteriophage
(bacterial virus) to transport bacterial genes from one host to another?
A) Transduction
B) Transformation
C) Conjugation
D) Transposition
Answer: A
Rationale: Transduction is the transfer of bacterial DNA from a donor to a recipient cell mediated by
a bacteriophage during either the lytic cycle (generalized) or lysogenic cycle (specialized).
8. An Hfr (High Frequency of Recombination) bacterial cell is characterized by:
A) Carrying an autonomous F plasmid separate from the chromosome.
B) Having an F plasmid integrated directly into its bacterial chromosome.
C) Lacking the genes necessary to produce a sex pilus.
D) Containing multiple antibiotic-resistant R plasmids.
Answer: B
Rationale: An Hfr cell has integrated its F (fertility) plasmid into the circular bacterial chromosome via
homologous recombination, allowing high-frequency transfer of chromosomal genes during
conjugation.
9. Which Ames test outcome indicates that a chemical compound is a potential mutagen
and carcinogen?
A) No bacterial colonies grow on histidine-deficient media.
B) A significantly higher number of revertant Salmonella colonies grow on histidine-deficient
media compared to the control.
C) The bacterial colonies require ampicillin to survive on nutrient agar.
D) The test strain loses its cell wall and converts into an L-form.
Answer: B
Rationale: The Ames test utilizes a histidine-auxotrophic strain of Salmonella enterica. If a compound
induces a high rate of back-mutations (reversion to His+), allowing colonies to grow without added
histidine, the chemical is mutagenic.
10. Thymine dimers are covalent linkages between adjacent pyrimidine bases caused
Course: BIO 242: General Microbiology
Exam: Exam 3 (Modules: Microbial Genetics, Control of Microbial Growth, Antimicrobial
Therapy, Pathogenesis, and Immunology)
Total Questions: 75 Multiple Choice Questions
Part 1: Microbial Genetics, DNA Replication, and Gene Expression
(Questions 1–15)
1. Which of the following enzymes is primarily responsible for unwinding the DNA double
helix during bacterial replication by breaking hydrogen bonds between base pairs?
A) Topoisomerase (DNA Gyrase)
B) Helicase
C) DNA Polymerase III
D) Primase
Answer: B
Rationale: Helicase breaks the hydrogen bonds between complementary nitrogenous bases to
unwind and separate the two strands of the DNA double helix at the replication fork. DNA Gyrase
(topoisomerase) relieves supercoiling tension ahead of the fork, DNA Polymerase III synthesizes the
new strand, and Primase synthesizes RNA primers.
2. During DNA replication in Escherichia coli, Okazaki fragments are joined together into
a continuous strand by which enzyme?
A) DNA Polymerase I
B) RNA Polymerase
C) DNA Ligase
D) Topoisomerase IV
Answer: C
Rationale: DNA ligase catalyzes the formation of phosphodiester bonds between adjacent Okazaki
fragments on the lagging strand after DNA Polymerase I has removed the RNA primers and replaced
them with DNA nucleotides.
3. What is the role of the promoter region in bacterial transcription?
A) It marks the site where DNA replication begins.
, B) It codes for the ribosomal binding site on mRNA.
C) It serves as the recognition and binding site for RNA polymerase via the sigma factor.
D) It signals the termination of mRNA synthesis.
Answer: C
Rationale: The promoter is a specific DNA sequence located upstream of a gene where RNA
polymerase (facilitated by the sigma factor in prokaryotes) binds to initiate transcription. It is not
translated into protein and does not serve as a replication origin.
4. A point mutation that results in the substitution of one amino acid for another in the
resulting polypeptide chain is classified as a:
A) Silent mutation
B) Nonsense mutation
C) Missense mutation
D) Frameshift mutation
Answer: C
Rationale: A missense mutation is a nucleotide substitution that alters a codon so that it codes for a
different amino acid. A silent mutation alters the codon without changing the amino acid, while a
nonsense mutation creates a premature stop codon.
5. In the lac operon of E. coli, what occurs when lactose is present and glucose is
absent?
A) The repressor binds to the operator, blocking transcription.
B) Allolactose binds the repressor, inactivating it, while elevated cAMP allows CAP to bind the
promoter, leading to high transcription.
C) CAP is inactivated by high levels of cAMP, preventing RNA polymerase binding.
D) Transcription is completely shut down due to catabolite repression.
Answer: B
Rationale: When lactose is present, allolactose inactivates the repressor protein so it releases the
operator. Low glucose elevates intracellular cAMP, forming a cAMP-CAP complex that binds the
promoter and recruits RNA polymerase for maximal transcription.
6. The transfer of naked DNA from a donor cell to a competent recipient cell in
environment is known as:
A) Conjugation
B) Generalized Transduction
C) Specialized Transduction
D) Transformation
Answer: D
, Rationale: Transformation involves the uptake of free, naked extracellular DNA by a physiologically
competent bacterial cell. Conjugation requires cell-to-cell contact via a pilus, and transduction
requires a bacteriophage vector.
7. Which mechanism of horizontal gene transfer specifically requires a bacteriophage
(bacterial virus) to transport bacterial genes from one host to another?
A) Transduction
B) Transformation
C) Conjugation
D) Transposition
Answer: A
Rationale: Transduction is the transfer of bacterial DNA from a donor to a recipient cell mediated by
a bacteriophage during either the lytic cycle (generalized) or lysogenic cycle (specialized).
8. An Hfr (High Frequency of Recombination) bacterial cell is characterized by:
A) Carrying an autonomous F plasmid separate from the chromosome.
B) Having an F plasmid integrated directly into its bacterial chromosome.
C) Lacking the genes necessary to produce a sex pilus.
D) Containing multiple antibiotic-resistant R plasmids.
Answer: B
Rationale: An Hfr cell has integrated its F (fertility) plasmid into the circular bacterial chromosome via
homologous recombination, allowing high-frequency transfer of chromosomal genes during
conjugation.
9. Which Ames test outcome indicates that a chemical compound is a potential mutagen
and carcinogen?
A) No bacterial colonies grow on histidine-deficient media.
B) A significantly higher number of revertant Salmonella colonies grow on histidine-deficient
media compared to the control.
C) The bacterial colonies require ampicillin to survive on nutrient agar.
D) The test strain loses its cell wall and converts into an L-form.
Answer: B
Rationale: The Ames test utilizes a histidine-auxotrophic strain of Salmonella enterica. If a compound
induces a high rate of back-mutations (reversion to His+), allowing colonies to grow without added
histidine, the chemical is mutagenic.
10. Thymine dimers are covalent linkages between adjacent pyrimidine bases caused
