BIOL 3200 Final Exam Actual Exam V1 | BIOL 3200 General
Microbiology (BIOL 3200 Final Exam) | Auburn University
1. Which of the following components is unique to the cell walls of Gram-positive bacteria,
providing structural rigidity and contributing to the negative charge of the cell surface?
A. Lipopolysaccharide
B. Periplasm
C. Teichoic acids
D. Porins
Answer: C
Rationale: Teichoic acids are polyalcohol phosphates found covalently linked to
peptidoglycan in Gram-positive bacteria. They play a crucial role in maintaining cell wall
structure and binding divalent cations like magnesium. Gram-negative bacteria lack these
structures, instead possessing an outer membrane with lipopolysaccharides.
2. During the process of bacterial chemotaxis, how does a bacterium change its direction of
movement toward an attractant?
A. By increasing the frequency of tumbles
B. By utilizing cilia to steer toward the chemical gradient
C. By reversing the rotation of all flagella to clockwise simultaneously
D. By decreasing the frequency of tumbles and lengthening runs
Answer: D
Rationale: Bacteria move via a series of runs and tumbles controlled by flagellar rotation.
When moving toward an attractant, the biased random walk mechanism reduces the
frequency of tumbles, which are caused by clockwise rotation. This results in longer runs in
the direction of the favorable chemical gradient.
3. Which microscopy technique is best suited for viewing the internal ultrastructure of a viral
particle at the highest possible resolution?
A. Phase-contrast microscopy
B. Transmission electron microscopy (TEM)
C. Scanning electron microscopy (SEM)
D. Confocal scanning laser microscopy
Answer: B
,Rationale: Transmission electron microscopy uses an electron beam that passes through a
thin specimen section to image internal details. It offers significantly higher resolution than
light microscopy because the wavelength of electrons is much shorter than that of photons.
SEM is primarily used for surface topography rather than internal ultrastructure.
4. In the context of microbial growth, what occurs during the ‘stationary phase’ of a batch
culture?
A. The population is doubling at a constant maximum rate
B. The rate of cell division equals the rate of cell death
C. Cells are adjusting to new medium and synthesizing enzymes
D. The number of viable cells decreases exponentially
Answer: B
Rationale: Stationary phase is reached when nutrient limitation or waste accumulation
prevents further net growth. During this period, the growth rate and death rate are in
equilibrium, leading to a stable population count. This phase often triggers the expression
of survival genes and secondary metabolite production.
5. An organism that uses light as an energy source and organic compounds as a carbon source
is classified as a:
A. Photoheterotroph
B. Chemoautotroph
C. Photoautotroph
D. Chemoheterotroph
Answer: A
Rationale: Photoheterotrophs derive their energy from light but cannot fix carbon dioxide,
so they rely on organic compounds for carbon. This metabolic strategy is seen in organisms
like purple non-sulfur bacteria. In contrast, photoautotrophs can use light and CO2 to
synthesize all needed organic matter.
6. What is the primary function of the enzyme DNA gyrase (Topoisomerase II) during DNA
replication?
A. Joining Okazaki fragments together
B. Unwinding the double helix at the replication fork
C. Synthesizing RNA primers
D. Relieving torsional stress and supercoiling ahead of the fork
Answer: D
, Rationale: As DNA helicase unwinds the double helix, positive supercoiling builds up
ahead of the replication fork. DNA gyrase introduces negative supercoils or removes
positive ones to prevent the DNA from becoming too tightly knotted to proceed. This
enzyme is a major target for fluoroquinolone antibiotics like ciprofloxacin.
7. Which mechanism of horizontal gene transfer involves the uptake of ‘naked’ DNA directly
from the environment by a competent cell?
A. Conjugation
B. Transformation
C. Transduction
D. Transposition
Answer: B
Rationale: Transformation was famously demonstrated by Frederick Griffith in his
experiments with Streptococcus pneumoniae. It requires the recipient cell to be in a
physiological state known as competence to allow DNA to cross the cell membrane. Once
inside, the DNA can be integrated into the host genome via homologous recombination.
8. In the lac operon of Escherichia coli, what happens when both glucose and lactose are
present in the growth medium?
A. The operon is transcribed at high levels
B. Transcription is low due to catabolite repression (lack of cAMP-CAP complex)
C. The repressor binds the operator and prevents transcription
D. Lactose is metabolized first because it is a more efficient energy source
Answer: B
Rationale: E. coli exhibits diauxic growth where glucose is used preferentially over other
sugars. When glucose is present, cAMP levels are low, preventing the Catabolite Activator
Protein (CAP) from binding to the promoter. Consequently, even if lactose is present to
remove the repressor, transcription remains at a basal, low level until glucose is exhausted.
9. Which of the following describes a ‘missense’ mutation?
A. A change in the DNA sequence that results in a different amino acid
B. A change in the DNA sequence that results in a premature stop codon
C. A change in the DNA sequence that does not alter the amino acid sequence
D. An insertion or deletion that shifts the reading frame
Answer: A
Microbiology (BIOL 3200 Final Exam) | Auburn University
1. Which of the following components is unique to the cell walls of Gram-positive bacteria,
providing structural rigidity and contributing to the negative charge of the cell surface?
A. Lipopolysaccharide
B. Periplasm
C. Teichoic acids
D. Porins
Answer: C
Rationale: Teichoic acids are polyalcohol phosphates found covalently linked to
peptidoglycan in Gram-positive bacteria. They play a crucial role in maintaining cell wall
structure and binding divalent cations like magnesium. Gram-negative bacteria lack these
structures, instead possessing an outer membrane with lipopolysaccharides.
2. During the process of bacterial chemotaxis, how does a bacterium change its direction of
movement toward an attractant?
A. By increasing the frequency of tumbles
B. By utilizing cilia to steer toward the chemical gradient
C. By reversing the rotation of all flagella to clockwise simultaneously
D. By decreasing the frequency of tumbles and lengthening runs
Answer: D
Rationale: Bacteria move via a series of runs and tumbles controlled by flagellar rotation.
When moving toward an attractant, the biased random walk mechanism reduces the
frequency of tumbles, which are caused by clockwise rotation. This results in longer runs in
the direction of the favorable chemical gradient.
3. Which microscopy technique is best suited for viewing the internal ultrastructure of a viral
particle at the highest possible resolution?
A. Phase-contrast microscopy
B. Transmission electron microscopy (TEM)
C. Scanning electron microscopy (SEM)
D. Confocal scanning laser microscopy
Answer: B
,Rationale: Transmission electron microscopy uses an electron beam that passes through a
thin specimen section to image internal details. It offers significantly higher resolution than
light microscopy because the wavelength of electrons is much shorter than that of photons.
SEM is primarily used for surface topography rather than internal ultrastructure.
4. In the context of microbial growth, what occurs during the ‘stationary phase’ of a batch
culture?
A. The population is doubling at a constant maximum rate
B. The rate of cell division equals the rate of cell death
C. Cells are adjusting to new medium and synthesizing enzymes
D. The number of viable cells decreases exponentially
Answer: B
Rationale: Stationary phase is reached when nutrient limitation or waste accumulation
prevents further net growth. During this period, the growth rate and death rate are in
equilibrium, leading to a stable population count. This phase often triggers the expression
of survival genes and secondary metabolite production.
5. An organism that uses light as an energy source and organic compounds as a carbon source
is classified as a:
A. Photoheterotroph
B. Chemoautotroph
C. Photoautotroph
D. Chemoheterotroph
Answer: A
Rationale: Photoheterotrophs derive their energy from light but cannot fix carbon dioxide,
so they rely on organic compounds for carbon. This metabolic strategy is seen in organisms
like purple non-sulfur bacteria. In contrast, photoautotrophs can use light and CO2 to
synthesize all needed organic matter.
6. What is the primary function of the enzyme DNA gyrase (Topoisomerase II) during DNA
replication?
A. Joining Okazaki fragments together
B. Unwinding the double helix at the replication fork
C. Synthesizing RNA primers
D. Relieving torsional stress and supercoiling ahead of the fork
Answer: D
, Rationale: As DNA helicase unwinds the double helix, positive supercoiling builds up
ahead of the replication fork. DNA gyrase introduces negative supercoils or removes
positive ones to prevent the DNA from becoming too tightly knotted to proceed. This
enzyme is a major target for fluoroquinolone antibiotics like ciprofloxacin.
7. Which mechanism of horizontal gene transfer involves the uptake of ‘naked’ DNA directly
from the environment by a competent cell?
A. Conjugation
B. Transformation
C. Transduction
D. Transposition
Answer: B
Rationale: Transformation was famously demonstrated by Frederick Griffith in his
experiments with Streptococcus pneumoniae. It requires the recipient cell to be in a
physiological state known as competence to allow DNA to cross the cell membrane. Once
inside, the DNA can be integrated into the host genome via homologous recombination.
8. In the lac operon of Escherichia coli, what happens when both glucose and lactose are
present in the growth medium?
A. The operon is transcribed at high levels
B. Transcription is low due to catabolite repression (lack of cAMP-CAP complex)
C. The repressor binds the operator and prevents transcription
D. Lactose is metabolized first because it is a more efficient energy source
Answer: B
Rationale: E. coli exhibits diauxic growth where glucose is used preferentially over other
sugars. When glucose is present, cAMP levels are low, preventing the Catabolite Activator
Protein (CAP) from binding to the promoter. Consequently, even if lactose is present to
remove the repressor, transcription remains at a basal, low level until glucose is exhausted.
9. Which of the following describes a ‘missense’ mutation?
A. A change in the DNA sequence that results in a different amino acid
B. A change in the DNA sequence that results in a premature stop codon
C. A change in the DNA sequence that does not alter the amino acid sequence
D. An insertion or deletion that shifts the reading frame
Answer: A
