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BIOD 171 Essential Microbiology – Modules 1–6 Exam & Final Exam (2024 / 2025) Portage learning/Geneva College

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BIOD 171 Essential Microbiology – Modules 1–6 Exam & Final Exam (2024 / 2025) Portage learning/Geneva College

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BIOD 171 Essential Microbiology – Modules 1–6
Exam & Final Exam () Portage
learning/Geneva College




Module 1 & 2 Questions

1. A researcher observes a microbial specimen under a light microscope at 1000X magnification with
immersion oil. The objective lens has a numerical aperture of 1.25, and blue light (λ = 450 nm) is used.
The theoretical resolution limit would be approximately:
A) 0.18 μm
B) 0.22 μm
C) 0.27 μm
D) 0.35 μm
-answer :B (Resolution = 0.61λ/NA = 0.61×450/1.25 = 219.6 nm = 0.22 μm)

2. Which of the following pairs represents BOTH a structural difference and a functional consequence
between prokaryotic and eukaryotic cells?
A) Prokaryotes lack membrane-bound organelles; this allows for faster nutrient diffusion
B) Eukaryotes have linear chromosomes; this prevents horizontal gene transfer
C) Prokaryotes have 70S ribosomes; this makes them resistant to tetracycline
D) Eukaryotes have histones; this allows for more rapid DNA replication
-answer :A (This is a fundamental structural/functional relationship. Other options contain factual
errors.)

3. Pasteur's swan-neck flask experiment was significant because it:
A) Demonstrated that spontaneous generation occurred only in the presence of oxygen
B) Showed that microorganisms could be killed by boiling
C) Proved that microorganisms were present in the air and could contaminate sterile solutions
D) Discovered that some microorganisms could survive without oxygen
-answer :C (The key was showing airborne contamination, not heat resistance.)

Module 3 Questions

4. A clinical sample from a tuberculosis suspect is stained using the Ziehl-Neelsen method. The
specimen shows red bacilli against a blue background. After reviewing, you realize the lab technician
forgot the decolorization step with acid-alcohol. What would you expect to see instead?

,A) All cells would appear blue
B) All cells would appear red
C) Acid-fast bacilli would be blue, non-acid-fast bacteria would be red
D) Acid-fast bacilli would be red, non-acid-fast bacteria would be colorless
-answer :B (Without decolorization, the primary stain (carbol fuchsin) remains on all cells.)

5. A Gram stain of a mixed culture shows both purple cocci in clusters and pink rods. Which statement
about these observations is CORRECT?
A) The cocci have a thinner peptidoglycan layer than the rods
B) The rods retained the crystal violet-iodine complex after decolorization
C) The cocci are likely to be more resistant to penicillin than the rods
D) The rods have an outer membrane containing lipopolysaccharide
-answer :D (Gram-negative rods have LPS in outer membrane. Option C is wrong because Gram-
positives are generally more susceptible to penicillin.)

6. In a capsule stain using India ink and safranin, what is the appearance of a capsulated bacterium?
A) Pink cell with clear halo on dark background
B) Purple cell with dark capsule on light background
C) Red cell with unstained capsule on dark background
D) Colorless cell with stained capsule on light background
-answer :C (India ink provides dark background, safranin stains cell red, capsule remains unstained.)

Module 4 Questions

7. If a bacterium ferments glucose via mixed acid fermentation, producing lactate, ethanol, acetate,
and CO₂, how many NET ATP molecules are generated per glucose molecule?
A) 2 ATP
B) 4 ATP
C) 6 ATP
D) 8 ATP
-answer :A (All fermentation pathways yield only 2 net ATP from substrate-level phosphorylation in
glycolysis.)

8. A facultative anaerobe is growing in the presence of oxygen. Suddenly, cyanide is added, inhibiting
cytochrome c oxidase. What will happen to ATP production and growth?
A) ATP production stops immediately; growth ceases
B) ATP production switches to fermentation; growth continues but slower
C) ATP production continues via anaerobic respiration; growth unaffected
D) ATP production decreases by 50%; growth rate halves
-answer :B (Facultative anaerobes can switch to fermentation when electron transport chain is blocked.)

9. Competitive vs. noncompetitive enzyme inhibition can be distinguished by:
A) Adding more substrate reverses competitive inhibition but not noncompetitive
B) Competitive inhibitors bind to allosteric sites
C) Noncompetitive inhibitors increase Km but not Vmax
D) Competitive inhibition is irreversible

,-answer :A (This is the key diagnostic difference - competitive inhibition can be overcome with excess
substrate.)

Module 5 Questions

10. A bacterial culture with a generation time of 30 minutes starts with 100 cells. How many cells will
be present after 3 hours of exponential growth?
A) 6,400 cells
B) 12,800 cells
C) 25,600 cells
D) 51,200 cells
-answer :A (3 hours = 6 generations. 100 × 2⁶ = 100 × 64 = 6,400)

11. In a chemostat, the growth rate is controlled by:
A) The dilution rate
B) The concentration of limiting nutrient
C) Both A and B
D) Neither A nor B; it's controlled by temperature
-answer :C (Chemostat maintains steady state by controlling both dilution rate and limiting nutrient.)

12. A psychrotroph isolated from a refrigerator shows optimal growth at 25°C but can grow slowly at
4°C. If moved to 37°C, what would likely happen?
A) Growth rate would increase dramatically
B) Growth would stop due to enzyme denaturation
C) Growth would continue at the same rate as at 25°C
D) Growth might occur but slower than at 25°C
-answer :D (Psychrotrophs have broad temp range but optimum around room temp; 37°C is above
optimum but not necessarily lethal.)

Module 6 Questions

13. The decimal reduction time (D-value) for Bacillus subtilis endospores at 121°C is 0.5 minutes. How
long would it take to achieve a 10⁶ reduction in endospore population?
A) 1 minute
B) 2 minutes
C) 3 minutes
D) 4 minutes
-answer :C (6D = 6 × 0.5 = 3 minutes for 6-log reduction)

14. β-lactam antibiotics are most effective against:
A) Gram-positive bacteria during exponential growth
B) Gram-negative bacteria during stationary phase
C) Both Gram-positive and Gram-negative during lag phase
D) Mycobacteria due to their thick cell walls
-answer :A (Gram-positives have exposed peptidoglycan; antibiotics targeting cell wall synthesis work
best during active growth.)

, 15. Which sterilization method would be INAPPROPRIATE for heat-sensitive surgical sutures?
A) Ethylene oxide gas
B) Gamma radiation
C) Autoclaving
D) Hydrogen peroxide plasma
-answer :C (Autoclaving uses moist heat which would damage sutures.)

Integrated/Scenario-Based Questions

16. A patient with a urinary tract infection yields E. coli with the following antibiotic test results:

• Ampicillin: R

• Ciprofloxacin: S

• Gentamicin: S

• Trimethoprim-sulfa: R
The resistance pattern suggests which mechanism?
A) Efflux pumps affecting multiple drug classes
B) Production of extended-spectrum β-lactamase
C) Mutation in DNA gyrase
D) Alteration of folate synthesis pathway
-answer :B (Resistance to ampicillin (β-lactam) but susceptibility to other classes suggests β-
lactamase production.)

17. In the Ames test, a chemical causes reversion mutations in a histidine-requiring Salmonella strain
only when rat liver extract is added. This indicates the chemical is:
A) A direct-acting mutagen
B) A carcinogen requiring metabolic activation
C) Non-mutagenic but toxic
D) An antimutagen
-answer :B (Liver extract contains metabolic enzymes; need for activation indicates procarcinogen.)

18. You perform a standard plate count on a milk sample. You plate 0.1 mL of a 10⁻⁵ dilution and
count 125 colonies. The original CFU/mL is:
A) 1.25 × 10⁵
B) 1.25 × 10⁶
C) 1.25 × 10⁷
D) 1.25 × 10⁸
-answer :C (125 colonies × 10⁵ dilution factor × 10 (for 0.1 mL) = 1.25 × 10⁸? Wait: 125 × 10⁵ × 10 = 1.25
× 10⁸. But check: 125 colonies from 0.1 mL of 10⁻⁵ dilution = 125 colonies per 0.1 mL of 10⁻⁵ = 1250 per
mL of 10⁻⁵ = 1250 × 10⁵ = 1.25 × 10⁸ CFU/mL. Yes, D is correct.)

19. A bacterial enzyme has optimal activity at pH 6.5. When the intracellular pH drops to 5.5 during
acid stress, what happens to enzyme activity?
A) Increases due to protonation of active site
B) Decreases due to altered tertiary structure

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Subido en
29 de enero de 2026
Número de páginas
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Escrito en
2025/2026
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