Free practice final
Microbiology practice final (free, with answers)
An intro microbiology final typically covers bacterial structure and growth, microbial metabolism and genetics, sterilization and antibiotics, viruses, and how the host fights infection. These 15 questions include the calculations that often appear, such as generation time, plate counts, and D-values. Try each question yourself before opening the answer.
Topics
- Cell Structure
- Microbial Growth
- Microbial Metabolism
- Microbial Genetics
- Control of Microbial Growth
- Antimicrobial Drugs
- Viruses
- Pathogenesis and Immunity
15 questions
Try each one before you open its answer.
1. In the Gram stain, why do Gram-negative cells lose the crystal violet–iodine complex during the alcohol decolorizing step?
- A Their peptidoglycan layer is thin, and alcohol disrupts their outer membrane so the dye complex washes out
- B They have no cell wall at all
- C Alcohol dissolves their thick peptidoglycan layer
- D Teichoic acids in their wall repel the dye
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Answer: A. Their peptidoglycan layer is thin, and alcohol disrupts their outer membrane so the dye complex washes out
Gram-positive cells have a thick peptidoglycan layer that traps the dye complex when alcohol dehydrates it. Gram-negative cells have only a thin layer under a lipid outer membrane, so the complex escapes and the cells take up the safranin counterstain. Teichoic acids are found in Gram-positive walls.
2. A culture in exponential growth goes from 1.0 × 10³ cells to 6.4 × 10⁴ cells in 3 hours. What is its generation time?
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Answer: 30 minutes
The population grew 64-fold, and 64 = 2⁶, so there were 6 generations. 180 minutes ÷ 6 generations = 30 minutes per generation.
3. In a tube of thioglycolate broth, where O₂ is present near the top and absent at the bottom, a bacterium grows throughout the tube but most densely near the top. How is it best classified?
- A Obligate aerobe
- B Aerotolerant anaerobe
- C Microaerophile
- D Facultative anaerobe
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Answer: D. Facultative anaerobe
Growth all the way down shows it can grow without O₂, and the heavier growth at the top shows it grows better with O₂, because aerobic respiration yields more ATP. An aerotolerant anaerobe would grow evenly, since it does not use O₂.
4. During fermentation, what is the final electron acceptor for the electrons carried by NADH?
- A O₂
- B Nitrate (NO₃⁻)
- C An organic molecule made from the substrate, such as pyruvate
- D Sulfate (SO₄²⁻)
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Answer: C. An organic molecule made from the substrate, such as pyruvate
Fermentation regenerates NAD⁺ by handing electrons to an organic molecule, for example reducing pyruvate to lactate. O₂ is the acceptor in aerobic respiration, and nitrate or sulfate serve in anaerobic respiration.
5. A gene passes from one bacterial strain to another. The transfer still happens when the strains are separated by a filter that blocks cells but not DNA or viruses, and it is stopped by adding DNase to the medium. Which mechanism is responsible?
- A Transformation
- B Transduction
- C Conjugation
- D Transposition
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Answer: A. Transformation
In transformation, cells take up naked DNA from the medium, which DNase destroys. Transduced DNA is protected inside a phage capsid, and conjugation needs direct cell-to-cell contact, which the filter prevents.
6. You spread 0.1 mL of a 10⁻⁵ dilution of a culture on a plate and count 150 colonies. What was the concentration of the original culture?
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Answer: 1.5 × 10⁸ CFU/mL
CFU/mL = colonies ÷ (volume plated × dilution) = 150 ÷ (0.1 mL × 10⁻⁵) = 1.5 × 10⁸ CFU/mL.
7. E. coli growing on glycerol, with no glucose or lactose, has a lacI mutation that makes the lac repressor unable to bind the operator. What happens to lac operon expression?
- A It is never expressed, even if lactose is added
- B It is expressed only when lactose is present
- C It is expressed even though no lactose is present
- D It is expressed only when glucose is added
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Answer: C. It is expressed even though no lactose is present
Normally the repressor sits on the operator until lactose, as allolactose, pulls it off. A repressor that cannot bind the operator leaves the operon on all the time (constitutive expression), and with no glucose present nothing else holds it down.
8. A bacterium near a hydrothermal vent gets its energy by oxidizing H₂S and builds its cell material from CO₂. What is its nutritional type?
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Answer: Chemoautotroph (chemolithoautotroph)
It gets energy from chemical oxidation rather than light, so it is a chemotroph, and its electron donor is inorganic (H₂S), so it is a lithotroph. Fixing CO₂ as its carbon source makes it an autotroph.
9. Which method is best for removing bacteria from a heat-sensitive antibiotic solution without damaging the drug?
- A Autoclaving at 121 °C for 15 minutes
- B Pasteurization
- C Dry heat at 170 °C for 2 hours
- D Passing it through a 0.22 μm membrane filter
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Answer: D. Passing it through a 0.22 μm membrane filter
A 0.22 μm filter physically holds back bacteria without heating the solution. Autoclaving and dry heat would destroy a heat-sensitive drug, and pasteurization only reduces microbes rather than removing them.
10. At 121 °C, the D-value for a spore suspension is 1.5 minutes. How long must it be held at 121 °C to reduce the population from 1 × 10⁸ to 1 × 10² spores?
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Answer: 9 minutes
The D-value is the time to kill 90% of the population, or one log reduction. Going from 10⁸ to 10² is 6 log reductions, and 6 × 1.5 min = 9 min.
11. Mycoplasma pneumoniae has no cell wall. Which of these antibiotics would you expect to be ineffective against it?
- A Azithromycin (a macrolide)
- B Penicillin (a β-lactam)
- C Doxycycline (a tetracycline)
- D Levofloxacin (a fluoroquinolone)
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Answer: B. Penicillin (a β-lactam)
β-lactams kill bacteria by blocking the enzymes that cross-link peptidoglycan, so they have no target in a cell without a wall. The other three act on ribosomes or on DNA gyrase and topoisomerase IV, which Mycoplasma still has.
12. Which enzyme must a negative-sense, single-stranded RNA virus carry inside its virion to start an infection?
- A Reverse transcriptase
- B RNA-dependent RNA polymerase
- C DNA-dependent DNA polymerase
- D Integrase
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Answer: B. RNA-dependent RNA polymerase
A negative-sense genome cannot be translated directly; it must first be copied into positive-sense mRNA. The host cell has no polymerase that can transcribe the viral RNA genome into mRNA, so the virus brings its own RNA-dependent RNA polymerase.
13. In the lysogenic cycle, what is the term for a phage genome that has become integrated into the host bacterium's chromosome?
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Answer: Prophage
The integrated prophage is copied along with the bacterial chromosome each time the cell divides. A signal such as DNA damage can make it excise and enter the lytic cycle.
14. Fever and septic shock in Gram-negative infections are often triggered by endotoxin. What is endotoxin?
- A A protein toxin secreted by actively growing cells
- B Lipopolysaccharide (LPS) of the outer membrane, whose lipid A portion is toxic
- C Teichoic acid from the peptidoglycan layer
- D A capsule polysaccharide that blocks phagocytosis
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Answer: B. Lipopolysaccharide (LPS) of the outer membrane, whose lipid A portion is toxic
Endotoxin is part of the Gram-negative outer membrane and is released mainly when cells lyse. Its lipid A portion triggers a strong inflammatory response that can cause fever and shock. Secreted protein toxins are exotoxins.
15. Cytotoxic (CD8⁺) T cells recognize virus-infected cells by binding viral peptides displayed on which molecules?
- A MHC class II molecules
- B Antibodies bound to Fc receptors
- C MHC class I molecules
- D Toll-like receptors
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Answer: C. MHC class I molecules
Nearly all nucleated cells display peptides from proteins made inside them on MHC class I, and CD8 binds MHC class I. MHC class II, found on antigen-presenting cells, presents to CD4⁺ helper T cells instead.