BIOS242 Fundamentals of Microbiology with Lab - Exam 3 | Chamberlain University 2026-2027
BIOS242 Fundamentals of Microbiology with Lab
Exam 3 Study Guide
Chamberlain University | 2026-2027 Curriculum | 75 Questions | Graded A+
Verified by Experts | Latest 2026/2027 Update
Section 1: Microbial Genetics and Gene Expression (Q1-Q15)
Q1: During DNA replication in Escherichia coli, which enzyme is responsible for unwinding the double helix at the replication
fork by breaking hydrogen bonds between complementary nitrogenous bases?
A. DNA polymerase III
B. Helicase [CORRECT]
C. Ligase
D. Primase
Correct Answer: B
Rationale: Helicase is the enzyme that unwinds the DNA double helix at the replication fork by breaking hydrogen bonds between A-T and G-C
base pairs, creating single-stranded templates. DNA polymerase III (Option A) synthesizes new DNA strands in the 5-prime to 3-prime
direction. Ligase (Option C) joins Okazaki fragments on the lagging strand. Primase (Option D) synthesizes short RNA primers. Distinguishing
the roles of replication enzymes is a core BIOS242 competency.
Q2: A microbiology student observes that on the lagging strand during DNA replication, short DNA fragments are produced.
These fragments are later joined by an enzyme. What are these fragments called and which enzyme joins them?
A. Okazaki fragments; joined by DNA ligase [CORRECT]
B. Okazaki fragments; joined by DNA polymerase I
C. Primer fragments; joined by helicase
D. Fragments of Watson strand; joined by primase
Correct Answer: A
Rationale: The lagging strand is synthesized discontinuously as short segments called Okazaki fragments, each initiated by an RNA primer.
DNA ligase seals the nicks between adjacent Okazaki fragments by forming phosphodiester bonds, creating a continuous DNA strand. DNA
polymerase I (Option B) removes the RNA primers and replaces them with DNA but does not join fragments. Helicase (Option C) unwinds
DNA, and primase (Option D) synthesizes primers. Understanding lagging strand mechanics is essential in BIOS242.
Q3: A researcher is studying the lac operon in E. coli. In the presence of both lactose and glucose, which regulatory state BEST
describes the lac operon?
A. Fully active; both the repressor is inactive and catabolite activator protein (CAP) is bound to the promoter
B. Inactive; the repressor is bound to the operator and CAP is not bound
C. Partially active; the repressor is inactive due to allolactose but CAP is not bound due to high glucose (cAMP is
low) [CORRECT]
D. Fully repressed; the repressor is active and CAP is bound
Correct Answer: C
Rationale: When both lactose and glucose are present, lactose binds to the lac repressor, inactivating it and allowing transcription. However,
high glucose levels keep intracellular cAMP low, preventing CAP-cAMP binding to the promoter. Without CAP, transcription occurs at a very
low basal level, not full activation. Full activation requires both lactose presence (repressor inactive) AND low glucose (high cAMP, CAP
bound). Option A describes the state with lactose only. Option B describes the state with glucose only and no lactose. Option D is contradictory
since CAP binding promotes, not represses, transcription. The lac operon is a fundamental BIOS242 concept.
Page 1
,BIOS242 Fundamentals of Microbiology with Lab - Exam 3 | Chamberlain University 2026-2027
Q4: A point mutation changes a codon from GAA to GUA. The original codon GAA codes for glutamic acid, and GUA codes for
valine. What type of mutation has occurred?
A. Silent mutation
B. Missense mutation [CORRECT]
C. Nonsense mutation
D. Frameshift mutation
Correct Answer: B
Rationale: A missense mutation changes a codon to one that encodes a different amino acid. GAA (glutamic acid) to GUA (valine) results in a
single amino acid substitution in the protein. A silent mutation (Option A) would change the codon but not the encoded amino acid. A nonsense
mutation (Option C) would change the codon to a stop codon (UAA, UAG, or UGA), truncating the protein. A frameshift mutation (Option D)
results from insertions or deletions not in multiples of three. Identifying mutation types is a critical BIOS242 genetics competency.
Q5: A nursing student is reviewing transcription in prokaryotes. Which statement about prokaryotic transcription is CORRECT?
A. Prokaryotic mRNA requires intron splicing before translation
B. Prokaryotic RNA polymerase recognizes promoter sequences and transcribes mRNA that is immediately
available for translation without processing [CORRECT]
C. Prokaryotic transcription occurs in the nucleus
D. Prokaryotic mRNA has a 5-prime cap and 3-prime poly-A tail
Correct Answer: B
Rationale: In prokaryotes, RNA polymerase directly binds promoter sequences (such as the -10 and -35 regions) and transcribes mRNA that is
translated immediately without processing. Prokaryotic mRNA does not contain introns (Option A is incorrect; intron splicing is eukaryotic).
Prokaryotes lack a nucleus (Option C is incorrect). The 5-prime cap and 3-prime poly-A tail (Option D) are eukaryotic modifications. These
differences are fundamental to BIOS242.
Q6: During translation, a tRNA molecule carries the amino acid methionine and has the anticodon 3-prime-UAC-5-prime. Which
mRNA codon will this tRNA bind to?
A. 3-prime-UAC-5-prime
B. 5-prime-AUG-3-prime [CORRECT]
C. 5-prime-UAC-3-prime
D. 3-prime-CAU-5-prime
Correct Answer: B
Rationale: The tRNA anticodon 3-prime-UAC-5-prime pairs with the mRNA codon 5-prime-AUG-3-prime through antiparallel base pairing
(A-U, U-A, G-C). AUG is the start codon encoding methionine, consistent with this tRNA carrying methionine. Options A and C do not
represent correct antiparallel pairing. Option D is the reverse complement in wrong orientation. Codon-anticodon recognition is essential to
BIOS242.
Q7: A patient is infected with a bacterium that has developed resistance to multiple antibiotics through the acquisition of a
plasmid from another bacterial cell via a pilus. Which mechanism of genetic recombination is described?
A. Transformation
B. Transduction
C. Conjugation [CORRECT]
D. Transposition
Correct Answer: C
Rationale: Conjugation is the transfer of genetic material (typically plasmids) between bacterial cells through direct contact via a sex pilus.
The F plasmid encodes pilus formation and plasmid transfer. Transformation (Option A) involves uptake of free DNA from the environment.
Transduction (Option B) involves bacteriophage-mediated DNA transfer. Transposition (Option D) is movement of transposable elements
within a genome. Conjugation is the primary mechanism of plasmid-mediated multidrug resistance spread, a key BIOS242 concept.
Page 2
, BIOS242 Fundamentals of Microbiology with Lab - Exam 3 | Chamberlain University 2026-2027
Q8: A researcher exposes a bacterial culture to UV radiation, causing thymine dimers to form in the DNA. Which DNA repair
mechanism is MOST directly responsible for repairing this type of damage?
A. Mismatch repair
B. Nucleotide excision repair [CORRECT]
C. SOS repair
D. Proofreading by DNA polymerase
Correct Answer: B
Rationale: Nucleotide excision repair (NER) is the primary mechanism for repairing thymine dimers caused by UV radiation. An excinuclease
recognizes the distortion, removes a short segment containing the dimer, and DNA polymerase fills the gap. Mismatch repair (Option A)
corrects replication errors. SOS repair (Option C) is error-prone emergency repair. Proofreading (Option D) corrects misincorporated bases
during replication. DNA repair mechanisms are a core BIOS242 topic.
Q9: Which statement about the tryptophan (trp) operon is CORRECT?
A. It is an inducible operon that is normally off and activated by the presence of tryptophan
B. It is a repressible operon that is normally on and turned off when tryptophan binds to the repressor protein
[CORRECT]
C. It uses positive control exclusively through CAP-cAMP
D. It is always transcribed regardless of tryptophan levels
Correct Answer: B
Rationale: The trp operon is a repressible operon normally transcribed when tryptophan levels are low. When tryptophan is abundant, it acts
as a corepressor by binding to the inactive repressor, activating it. The activated repressor then binds the operator, blocking transcription.
Option A incorrectly calls it inducible. Option C is incorrect; it uses negative repressor control. Option D is false. Distinguishing inducible
from repressible operons is a critical BIOS242 competency.
Q10: A bacteriophage infects a bacterial cell and during the lytic cycle, it accidentally packages fragments of the bacterial
chromosome instead of its own viral DNA. When this phage infects another bacterium, it transfers these bacterial genes. Which
process is described?
A. Specialized transduction
B. Generalized transduction [CORRECT]
C. Conjugation
D. Transformation
Correct Answer: B
Rationale: Generalized transduction occurs during the lytic cycle when a phage mistakenly packages random fragments of bacterial DNA into
the phage capsid. Specialized transduction (Option A) occurs with temperate phages during improper prophage excision. Conjugation (Option
C) and transformation (Option D) are different horizontal gene transfer mechanisms. Distinguishing generalized from specialized transduction
is a key BIOS242 competency.
Q11: A frameshift mutation occurs when three nucleotides are inserted into a gene. What is the MOST likely effect on the
resulting protein?
A. The entire reading frame is altered, causing all downstream amino acids to change
B. The reading frame is preserved because the insertion is a multiple of three, resulting in the addition of one extra
amino acid without affecting downstream codons [CORRECT]
C. No change occurs because the genetic code is degenerate
D. A single amino acid is substituted
Correct Answer: B
Rationale: When insertions or deletions occur in multiples of three nucleotides, the reading frame is preserved. Only the codons at the
insertion site are affected, adding one extra amino acid. A frameshift (Option A) occurs when the insertion is NOT a multiple of three. Option C
describes silent mutations. Option D describes missense mutations. Understanding the reading frame concept is essential in BIOS242.
Page 3
BIOS242 Fundamentals of Microbiology with Lab
Exam 3 Study Guide
Chamberlain University | 2026-2027 Curriculum | 75 Questions | Graded A+
Verified by Experts | Latest 2026/2027 Update
Section 1: Microbial Genetics and Gene Expression (Q1-Q15)
Q1: During DNA replication in Escherichia coli, which enzyme is responsible for unwinding the double helix at the replication
fork by breaking hydrogen bonds between complementary nitrogenous bases?
A. DNA polymerase III
B. Helicase [CORRECT]
C. Ligase
D. Primase
Correct Answer: B
Rationale: Helicase is the enzyme that unwinds the DNA double helix at the replication fork by breaking hydrogen bonds between A-T and G-C
base pairs, creating single-stranded templates. DNA polymerase III (Option A) synthesizes new DNA strands in the 5-prime to 3-prime
direction. Ligase (Option C) joins Okazaki fragments on the lagging strand. Primase (Option D) synthesizes short RNA primers. Distinguishing
the roles of replication enzymes is a core BIOS242 competency.
Q2: A microbiology student observes that on the lagging strand during DNA replication, short DNA fragments are produced.
These fragments are later joined by an enzyme. What are these fragments called and which enzyme joins them?
A. Okazaki fragments; joined by DNA ligase [CORRECT]
B. Okazaki fragments; joined by DNA polymerase I
C. Primer fragments; joined by helicase
D. Fragments of Watson strand; joined by primase
Correct Answer: A
Rationale: The lagging strand is synthesized discontinuously as short segments called Okazaki fragments, each initiated by an RNA primer.
DNA ligase seals the nicks between adjacent Okazaki fragments by forming phosphodiester bonds, creating a continuous DNA strand. DNA
polymerase I (Option B) removes the RNA primers and replaces them with DNA but does not join fragments. Helicase (Option C) unwinds
DNA, and primase (Option D) synthesizes primers. Understanding lagging strand mechanics is essential in BIOS242.
Q3: A researcher is studying the lac operon in E. coli. In the presence of both lactose and glucose, which regulatory state BEST
describes the lac operon?
A. Fully active; both the repressor is inactive and catabolite activator protein (CAP) is bound to the promoter
B. Inactive; the repressor is bound to the operator and CAP is not bound
C. Partially active; the repressor is inactive due to allolactose but CAP is not bound due to high glucose (cAMP is
low) [CORRECT]
D. Fully repressed; the repressor is active and CAP is bound
Correct Answer: C
Rationale: When both lactose and glucose are present, lactose binds to the lac repressor, inactivating it and allowing transcription. However,
high glucose levels keep intracellular cAMP low, preventing CAP-cAMP binding to the promoter. Without CAP, transcription occurs at a very
low basal level, not full activation. Full activation requires both lactose presence (repressor inactive) AND low glucose (high cAMP, CAP
bound). Option A describes the state with lactose only. Option B describes the state with glucose only and no lactose. Option D is contradictory
since CAP binding promotes, not represses, transcription. The lac operon is a fundamental BIOS242 concept.
Page 1
,BIOS242 Fundamentals of Microbiology with Lab - Exam 3 | Chamberlain University 2026-2027
Q4: A point mutation changes a codon from GAA to GUA. The original codon GAA codes for glutamic acid, and GUA codes for
valine. What type of mutation has occurred?
A. Silent mutation
B. Missense mutation [CORRECT]
C. Nonsense mutation
D. Frameshift mutation
Correct Answer: B
Rationale: A missense mutation changes a codon to one that encodes a different amino acid. GAA (glutamic acid) to GUA (valine) results in a
single amino acid substitution in the protein. A silent mutation (Option A) would change the codon but not the encoded amino acid. A nonsense
mutation (Option C) would change the codon to a stop codon (UAA, UAG, or UGA), truncating the protein. A frameshift mutation (Option D)
results from insertions or deletions not in multiples of three. Identifying mutation types is a critical BIOS242 genetics competency.
Q5: A nursing student is reviewing transcription in prokaryotes. Which statement about prokaryotic transcription is CORRECT?
A. Prokaryotic mRNA requires intron splicing before translation
B. Prokaryotic RNA polymerase recognizes promoter sequences and transcribes mRNA that is immediately
available for translation without processing [CORRECT]
C. Prokaryotic transcription occurs in the nucleus
D. Prokaryotic mRNA has a 5-prime cap and 3-prime poly-A tail
Correct Answer: B
Rationale: In prokaryotes, RNA polymerase directly binds promoter sequences (such as the -10 and -35 regions) and transcribes mRNA that is
translated immediately without processing. Prokaryotic mRNA does not contain introns (Option A is incorrect; intron splicing is eukaryotic).
Prokaryotes lack a nucleus (Option C is incorrect). The 5-prime cap and 3-prime poly-A tail (Option D) are eukaryotic modifications. These
differences are fundamental to BIOS242.
Q6: During translation, a tRNA molecule carries the amino acid methionine and has the anticodon 3-prime-UAC-5-prime. Which
mRNA codon will this tRNA bind to?
A. 3-prime-UAC-5-prime
B. 5-prime-AUG-3-prime [CORRECT]
C. 5-prime-UAC-3-prime
D. 3-prime-CAU-5-prime
Correct Answer: B
Rationale: The tRNA anticodon 3-prime-UAC-5-prime pairs with the mRNA codon 5-prime-AUG-3-prime through antiparallel base pairing
(A-U, U-A, G-C). AUG is the start codon encoding methionine, consistent with this tRNA carrying methionine. Options A and C do not
represent correct antiparallel pairing. Option D is the reverse complement in wrong orientation. Codon-anticodon recognition is essential to
BIOS242.
Q7: A patient is infected with a bacterium that has developed resistance to multiple antibiotics through the acquisition of a
plasmid from another bacterial cell via a pilus. Which mechanism of genetic recombination is described?
A. Transformation
B. Transduction
C. Conjugation [CORRECT]
D. Transposition
Correct Answer: C
Rationale: Conjugation is the transfer of genetic material (typically plasmids) between bacterial cells through direct contact via a sex pilus.
The F plasmid encodes pilus formation and plasmid transfer. Transformation (Option A) involves uptake of free DNA from the environment.
Transduction (Option B) involves bacteriophage-mediated DNA transfer. Transposition (Option D) is movement of transposable elements
within a genome. Conjugation is the primary mechanism of plasmid-mediated multidrug resistance spread, a key BIOS242 concept.
Page 2
, BIOS242 Fundamentals of Microbiology with Lab - Exam 3 | Chamberlain University 2026-2027
Q8: A researcher exposes a bacterial culture to UV radiation, causing thymine dimers to form in the DNA. Which DNA repair
mechanism is MOST directly responsible for repairing this type of damage?
A. Mismatch repair
B. Nucleotide excision repair [CORRECT]
C. SOS repair
D. Proofreading by DNA polymerase
Correct Answer: B
Rationale: Nucleotide excision repair (NER) is the primary mechanism for repairing thymine dimers caused by UV radiation. An excinuclease
recognizes the distortion, removes a short segment containing the dimer, and DNA polymerase fills the gap. Mismatch repair (Option A)
corrects replication errors. SOS repair (Option C) is error-prone emergency repair. Proofreading (Option D) corrects misincorporated bases
during replication. DNA repair mechanisms are a core BIOS242 topic.
Q9: Which statement about the tryptophan (trp) operon is CORRECT?
A. It is an inducible operon that is normally off and activated by the presence of tryptophan
B. It is a repressible operon that is normally on and turned off when tryptophan binds to the repressor protein
[CORRECT]
C. It uses positive control exclusively through CAP-cAMP
D. It is always transcribed regardless of tryptophan levels
Correct Answer: B
Rationale: The trp operon is a repressible operon normally transcribed when tryptophan levels are low. When tryptophan is abundant, it acts
as a corepressor by binding to the inactive repressor, activating it. The activated repressor then binds the operator, blocking transcription.
Option A incorrectly calls it inducible. Option C is incorrect; it uses negative repressor control. Option D is false. Distinguishing inducible
from repressible operons is a critical BIOS242 competency.
Q10: A bacteriophage infects a bacterial cell and during the lytic cycle, it accidentally packages fragments of the bacterial
chromosome instead of its own viral DNA. When this phage infects another bacterium, it transfers these bacterial genes. Which
process is described?
A. Specialized transduction
B. Generalized transduction [CORRECT]
C. Conjugation
D. Transformation
Correct Answer: B
Rationale: Generalized transduction occurs during the lytic cycle when a phage mistakenly packages random fragments of bacterial DNA into
the phage capsid. Specialized transduction (Option A) occurs with temperate phages during improper prophage excision. Conjugation (Option
C) and transformation (Option D) are different horizontal gene transfer mechanisms. Distinguishing generalized from specialized transduction
is a key BIOS242 competency.
Q11: A frameshift mutation occurs when three nucleotides are inserted into a gene. What is the MOST likely effect on the
resulting protein?
A. The entire reading frame is altered, causing all downstream amino acids to change
B. The reading frame is preserved because the insertion is a multiple of three, resulting in the addition of one extra
amino acid without affecting downstream codons [CORRECT]
C. No change occurs because the genetic code is degenerate
D. A single amino acid is substituted
Correct Answer: B
Rationale: When insertions or deletions occur in multiples of three nucleotides, the reading frame is preserved. Only the codons at the
insertion site are affected, adding one extra amino acid. A frameshift (Option A) occurs when the insertion is NOT a multiple of three. Option C
describes silent mutations. Option D describes missense mutations. Understanding the reading frame concept is essential in BIOS242.
Page 3
