Microbiology Practice Test 34
Microbiology NCLEX Practice Test
Microbiology is a key topic within the NCLEX test plan, located under Nursing Science → Clinical Foundations → Microbiology. This section explains pathogens, host defenses, and antimicrobial stewardship essential for infection control. Each test contains 50 questions designed to mirror the difficulty and variety of the real exam.
This is the 34th part of the Microbiology series. To explore all practice tests under this topic, use the “Back to Main Topic” button at the end of the page.
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Microbiology Practice Test 34
Which one of the following is an irregular, gram-positive rod?
- Bordetella pertussis
- Corynebacterium diphtheriae
- Burkholderia pseudomallei
- Streptococcus pyogenes
- Pneumocystis jirovecii
Explanation: Answer reason: This organism is classically a gram-positive, non-spore-forming, irregular rod due to its cell wall structure and morphology. In contrast, Bordetella pertussis and Burkholderia pseudomallei are gram-negative rods, and Streptococcus pyogenes is a gram-positive coccus in chains rather than a rod. Pneumocystis jirovecii is a fungal pathogen, not a bacterium, so it would not be described as a gram-positive rod.
Infection by which of the following results in the formation of Ghon complexes?
- Bordetella pertussis
- Corynebacterium diphtheriae
- Mycobacterium tuberculosis
- Streptococcus pyogenes
- Blastomyces dermatidis
Explanation: Answer reason: This forms after inhalation of the organism and initiation of a cell-mediated immune response, producing granulomatous inflammation and caseation. The other organisms listed cause different hallmark syndromes (e.g., pertussis with whooping cough, diphtheria with pseudomembrane) rather than a primary lung/lymph node granulomatous complex. The finding is therefore most specifically associated with primary infection by the tuberculosis bacillus.
All of the following are characteristic of Pseudomonas aeruginosa EXCEPT?
- Gram-positive cell wall.
- Resistance to many types of disinfectants and antibiotics.
- Growth in moist environments.
- Production of pyocyanin.
- Rod-shaped.
Explanation: Answer reason: Pseudomonas aeruginosa is a gram-negative, aerobic, oxidase-positive rod with an outer membrane and relatively thin peptidoglycan layer, so it does not have a gram-positive cell wall structure. It commonly thrives in moist environments (e.g., sinks, respiratory equipment, burn wounds) and is well known for intrinsic and acquired resistance mechanisms such as efflux pumps, low-permeability porins, and biofilm formation. It can produce characteristic pigments including pyocyanin, which contributes to blue-green coloration and virulence via oxidative injury. Its bacillary (rod-shaped) morphology is consistent with this organism, making the gram-positive cell wall statement the exception.
All of the following are characteristic of the Group A beta-hemolytic streptococci EXCEPT?
- Methicillin resistance.
- M proteins.
- The ability to damage cell membranes.
- The ability to dissolve blood clots.
- Hyaluronidase production.
Explanation: Answer reason: Group A beta-hemolytic streptococci (Streptococcus pyogenes) are classically identified by virulence factors such as M protein (antiphagocytic), streptolysins that damage host cell membranes, streptokinase that lyses fibrin clots, and hyaluronidase that facilitates tissue spread. Methicillin resistance is a hallmark concept tied to Staphylococcus aureus (MRSA) via altered penicillin-binding proteins, not to S. pyogenes. Clinically, GAS remains generally susceptible to penicillin, making beta-lactam resistance via “methicillin resistance” an incorrect characteristic. The other listed features align with well-known GAS enzymes/toxins used in pathogenesis and spread.
In which of the following respects is measles similar to German measles (rubella)?
- They have a similar type of rash.
- They are caused by the same virus.
- Encephalitis is a possible complication.
- Congenital complications may occur.
- They can be controlled by vaccination.
Explanation: Answer reason: Measles (rubeola) and rubella are distinct viral illnesses, but both are preventable through routine immunization programs. Both are targeted by the combined MMR vaccine, which provides population-level control via high coverage and herd immunity. The other choices are not consistently shared: the rash patterns and associated findings differ, they are caused by different viruses, and congenital complications are classically associated with rubella rather than measles. Encephalitis can occur with measles and is not a typical shared hallmark for rubella in standard exam framing.
The patient has a papular rash. Microscopic examination of skin scrapings reveals small eight-legged animals. The etiology is?
- Candida.
- Microsporum.
- Pseudomonas aeruginosa.
- Staphylococcus aureus.
- Sarcoptes.
Explanation: Answer reason: Eight-legged organisms seen on skin scrapings indicate an ectoparasitic arthropod (mite), not bacteria or fungi. Sarcoptes scabiei causes scabies, which classically presents with an intensely pruritic papular rash and mites/eggs/scybala identified on microscopic examination of scrapings. Candida and Microsporum are fungi and would show yeast/pseudohyphae or hyphae, respectively, rather than visible arthropods. Pseudomonas aeruginosa and Staphylococcus aureus are bacteria and would not appear as small eight-legged animals on microscopy.
The patient has vesicles and scabs over her forehead. Microscopic examination of skin scrapings shows gram-positive cocci in clusters. The etiology is?
- Candida.
- Microsporum.
- Pseudomonas aeruginosa.
- Staphylococcus aureus.
- Sarcoptes.
Explanation: Answer reason: Gram-positive cocci in clusters on microscopy is the classic morphologic pattern of staphylococci, most notably S. aureus. Vesicles and scabbing on the face are consistent with superficial pyogenic skin infection such as impetigo, where this organism is a leading cause. Candida and Microsporum are fungi and would not appear as gram-positive cocci in clusters on a Gram stain. Pseudomonas is a gram-negative rod, and Sarcoptes is a mite diagnosed by identifying mites/eggs rather than gram-positive cocci.
The patient has scaling skin on his fingers. Conidiospores are seen in microscopic examination of skin scrapings. The etiology is?
- Candida.
- Microsporum.
- Pseudomonas aeruginosa.
- Staphylococcus aureus.
- Mycobacterium ulcerans.
Explanation: Answer reason: Conidia (including conidiospores/arthroconidia) in skin scrapings point to a dermatophyte causing tinea of the hands (tinea manuum), which presents with dry, scaly skin on fingers/palms. Microsporum species are classic dermatophytes that form conidia and infect keratinized tissues (stratum corneum, hair, nails). Candida more often causes moist erythematous intertrigo/paronychia with budding yeast and pseudohyphae rather than conidia in scrapings. The bacterial options would not produce fungal conidiospores on microscopy of skin scrapings.
A patient is experiencing profuse greenish-yellow, foul-smelling discharge from her vagina. She is complaining of itching and irritation. What is the most likely treatment?
- Metronidazole
- Cephalosporins
- Acyclovir
- Miconazole
- No treatment is available.
Explanation: Answer reason: Profuse greenish-yellow, foul-smelling vaginal discharge with itching/irritation is most consistent with Trichomonas vaginalis infection, which is treated with a nitroimidazole. This directly supports use of metronidazole as standard therapy (often with partner treatment to prevent reinfection). By contrast, miconazole targets Candida (typically thick white “cottage cheese” discharge), and acyclovir targets HSV lesions rather than malodorous discharge.
Which one of the following statements about genital warts is FALSE?
- It is transmitted by direct contact.
- It is caused by papillomaviruses.
- It is always precancerous.
- It is treated by removing them.
- It can lead to cervical cancer in women.
Explanation: Answer reason: Genital warts are caused by human papillomavirus, most commonly low-risk types (e.g., HPV 6 and 11), which are associated with benign condyloma rather than malignant transformation. High-risk HPV types (e.g., 16 and 18) are the ones strongly linked to cervical dysplasia and cervical cancer, and these often do not present as exophytic warts. Transmission occurs via direct skin-to-skin sexual contact, and management commonly involves destruction/removal of visible lesions, although this does not guarantee eradication of the virus. Therefore, stating that genital warts are always precancerous is incorrect because many are due to low-risk HPV with minimal oncogenic potential.
Which of the following statements regarding a bacterium that is R+ is FALSE?
- It possesses a plasmid.
- R+ can be transferred to a cell of the same species.
- It is resistant to certain drugs and heavy metals.
- It is F+.
- R+ can be transferred to a different species.
Explanation: Answer reason: R plasmids (R factors) encode antibiotic and sometimes heavy-metal resistance and are extrachromosomal DNA elements that can be horizontally transferred between bacteria, including across species. Therefore, having a plasmid, being resistant to certain drugs/heavy metals, and being transferable within the same species or to a different species are all consistent with an R+ state. The fertility factor (F plasmid) specifically confers the conjugation machinery that defines an F+ donor, and an organism can be R+ without carrying the F factor. While some R plasmids can be conjugative or can integrate with transfer functions, R+ status alone does not imply F+.
According to the operon model, for the synthesis of an inducible enzyme to occur, the?
- End-product must not be in excess.
- Substrate must bind to the enzyme.
- Substrate must bind to the repressor.
- Repressor must bind to the operator.
- Repressor must not be synthesized.
Explanation: Answer reason: Inducible operons (e.g., the lac operon) are normally OFF because an active repressor sits on the operator and blocks transcription. When the inducer (typically the substrate or a derivative such as allolactose) is present, it binds the repressor and changes its conformation so it can no longer bind the operator, allowing RNA polymerase to transcribe the structural genes. This is the defining control step for induction and is distinct from substrate binding to the enzyme, which affects catalysis rather than gene expression. Options implying repressor binding to the operator would maintain repression, not permit enzyme synthesis.
An enzyme that makes covalent bonds between the sugar of one nucleotide and the phosphate groups in another nucleotide in DNA is?
- RNA polymerase.
- DNA ligase
- DNA helicase.
- Transposase.
- DNA polymerase.
Explanation: Answer reason: This is especially critical during DNA replication on the lagging strand, where Okazaki fragments must be joined into a continuous strand. DNA polymerase primarily adds nucleotides to an existing 3′-OH end but does not seal pre-existing breaks between fragments. Helicase unwinds the double helix, and RNA polymerase synthesizes RNA rather than joining DNA backbone breaks.
The term aerotolerant anaerobe refers to an organism that?
- Does not use oxygen but tolerates it.
- Is killed by oxygen.
- Tolerates normal atmospheric nitrogen gas levels.
- Requires less oxygen than is present in air.
- Requires more oxygen than is present in air.
Explanation: Answer reason: Aerotolerant anaerobes rely on anaerobic metabolism and do not use oxygen as a terminal electron acceptor, but they can survive in its presence. This implies they possess enough protective mechanisms (e.g., enzymes that detoxify reactive oxygen species) to avoid oxygen-mediated cellular damage. In contrast, obligate anaerobes are harmed or killed by oxygen because they lack these defenses. Options describing altered oxygen requirements refer instead to microaerophiles (lower than air) or organisms thriving at higher oxygen levels, not aerotolerant anaerobes.
Which of the following is an example of a metabolic activity that could be used to measure microbial growth?
- Standard plate count
- Glucose consumption
- Direct microscopic count
- Turbidity
- Most probable number (MPN)
Explanation: Answer reason: Monitoring depletion of a nutrient like glucose directly reflects cellular metabolism and typically correlates with increasing biomass/cell number in an active culture. In contrast, standard plate count, direct microscopic count, and MPN are enumeration methods (counting viable cells or estimating numbers) rather than metabolic readouts. Turbidity is an indirect measure of cell density based on light scattering, not a metabolic activity measure.
Which of the following is NOT a direct method to measure microbial growth?
- Direct microscopic count
- Standard plate count
- Filtration on a support membrane followed by incubation on medium
- Metabolic activity
- Most probable number (MPN)
Explanation: Answer reason: Metabolic activity assays instead infer growth indirectly by measuring byproducts or biochemical processes (e.g., CO2 production, O2 consumption, acid production, ATP), which may change without a proportional change in cell count. Because metabolism can vary with growth phase, nutrient availability, or stress, it is not a direct enumeration of organisms. In contrast, plate-based methods and MPN are centered on counting viable organisms capable of replication under the test conditions.
Which of the following is the best definition of generation time?
- The length of time needed for lag phase
- The length of time needed for a cell to divide
- The minimum rate of doubling
- The duration of log phase
- The time needed for nuclear division
Explanation: Answer reason: This makes it a time measurement describing how long it takes a single cell to divide into two daughter cells during exponential growth. Lag phase and log phase are growth-curve stages and their durations describe phases of adaptation or rapid growth, not the intrinsic division interval. Nuclear division is not an appropriate defining feature for bacteria (which lack a true nucleus) and does not capture population doubling time.
A urease test is used to identify Mycobacterium tuberculosis because?
- Urease is a sign of tuberculosis.
- M. tuberculosis produces urease.
- Urea accumulates during tuberculosis.
- Some bacteria reduce nitrate ion.
- M. bovis can cause tuberculosis.
Explanation: Answer reason: M. tuberculosis produces urease. Biochemical identification tests rely on detecting characteristic enzyme activities of organisms grown in culture. Urease-positive organisms hydrolyze urea into ammonia and carbon dioxide, raising the pH and causing a color change in the test medium. M. tuberculosis characteristically demonstrates urease activity, so a positive urease test supports its identification among mycobacteria. The nitrate reduction statement is a different biochemical test and does not explain why the urease test specifically is useful here.
Which of the following statements regarding the Entner-Doudoroff pathway is TRUE?
- It involves glycolysis.
- It involves the pentose phosphate pathway.
- NADH is generated.
- ATP is generated.
- NADH and ATP are generated.
Explanation: Answer reason: The Entner–Doudoroff pathway is an alternative bacterial glucose catabolic route that produces energy and reducing equivalents during carbohydrate breakdown. It yields a net ATP gain (classically 1 ATP per glucose) via substrate-level phosphorylation and also produces NADH during oxidation steps (along with NADPH earlier in the pathway). Therefore, the statement that both ATP and NADH are generated best captures the key energetic outputs of this pathway. Options stating only one product are incomplete, and describing it purely as “glycolysis” or “pentose phosphate” is misleading because it is a distinct pathway that shares intermediates/enzymes with those routes rather than being identical to them.
Which microscope is used to observe viruses and the internal structure of thinly sectioned cells?
- Transmission electron microscope
- Darkfield microscope
- Fluorescence microscope
- Brightfield microscope
- Scanning electron microscope
Explanation: Answer reason: Transmission electron microscopy passes electrons through ultrathin sections, producing high-resolution images of internal cellular ultrastructure (e.g., organelles, viral particles within cells). Scanning electron microscopy primarily shows surface topography rather than internal structures. Darkfield, fluorescence, and brightfield microscopes rely on light and are generally insufficient for direct visualization of most viruses and fine internal ultrastructure at the needed resolution.
Assume you stain Bacillus by applying malachite green with heat and then counterstaining with safranin. Through the microscope, the green structures are?
- Cell walls.
- Capsules.
- Endospores.
- Flagella.
- The answer cannot be determined.
Explanation: Answer reason: Malachite green with heat is the classic Schaeffer–Fulton endospore stain, in which heating drives the primary dye into the tough spore coat. After rinsing, vegetative cells decolorize while endospores retain the green dye because of their keratin-like, highly resistant layers. Safranin then counterstains the decolorized vegetative cells red/pink, creating the key contrast used to identify sporulation in Bacillus. Structures like capsules and flagella require different specialized stains and would not be expected to appear as the green elements in this protocol.
Which microscope is used to observe a specimen that emits light when illuminated with an ultraviolet light?
- Compound light microscope
- Phase-contrast microscope
- Darkfield microscope
- Fluorescence microscope
- Electron microscope
Explanation: Answer reason: This makes it ideal for organisms or structures that naturally fluoresce or are tagged with fluorescent dyes (e.g., immunofluorescence). In contrast, a compound light microscope relies on visible light transmission without fluorescence emission. Electron microscopy uses electron beams rather than UV-induced emission, so it does not fit the mechanism described.
This microscope produces an image of a light cell against a dark background; internal structures are NOT visible?
- Compound light microscope
- Phase-contrast microscope
- Darkfield microscope
- Fluorescence microscope
- Electron microscope
Explanation: Answer reason: This is especially useful for viewing thin, unstained, living organisms, but it does not provide clear visualization of internal cellular structures. In contrast, phase-contrast enhances differences in refractive index and is specifically used to visualize internal structures in living cells. Fluorescence requires fluorochromes and emits colored signals on a dark background rather than a simple light cell silhouette.
The following is true of quaternary ammonium compounds EXCEPT?
- They are non-toxic at lower concentrations.
- They are tasteless.
- They are effective when combined with soaps.
- They are stable.
- They may be an ingredient in mouthwash.
Explanation: Answer reason: Quaternary ammonium compounds (cationic detergents) are surface-active agents whose antimicrobial activity is reduced by anionic substances. Soaps and many detergents are anionic and can inactivate quats through chemical incompatibility, so combining them decreases effectiveness rather than improving it. In contrast, quats are generally stable, relatively low-toxicity at appropriate dilutions, and can be incorporated into some oral-care or disinfectant formulations depending on the specific compound and concentration. This makes the statement about being effective with soaps the exception.
Which of the following pairs of terms is mismatched?
- Bacteriostatic — kills vegetative bacterial cells
- Germicide — kills microbes
- Virucide — inactivates viruses
- Sterilant — destroys all living microorganisms
- Fungicide — kills yeasts and molds
Explanation: Answer reason: Killing vegetative bacterial cells is the defining feature of bactericidal activity, not bacteriostatic activity. The other pairings are consistent with standard definitions: germicides kill microorganisms broadly, virucides inactivate viruses (which are not truly “killed”), sterilants eliminate all forms of microbial life, and fungicides target yeasts and molds. Therefore, the mismatch is the statement attributing killing to a bacteriostatic agent.
Which of the following is a limitation of the autoclave?
- It requires an excessively long time to achieve sterilization.
- It cannot inactivate viruses.
- It cannot kill endospores.
- It cannot be used with heat-labile materials.
- It cannot be used with glassware.
Explanation: Answer reason: Autoclaving sterilizes by saturated steam under pressure at high temperature, so it is inherently unsuitable for items that are damaged or denatured by heat. This method reliably inactivates viruses and is specifically effective against bacterial endospores when proper time/temperature/pressure parameters are met. Sterilization cycles are generally efficient rather than “excessively long” compared with many low-temperature alternatives. Glassware is commonly autoclaved, so incompatibility with glass is not a true limitation.
Which of the following pairs is mismatched?
- Ionizing radiation — hydroxyl radicals
- Ozone — takes electrons from substances
- Plasma sterilization — free radicals
- Supercritical fluids — CO2
- Ultraviolet radiation — desiccation
Explanation: Answer reason: UV radiation (nonionizing) mainly damages microbial DNA by forming pyrimidine (thymine) dimers, leading to replication failure. Desiccation instead refers to dehydration/inhibition of microbial growth due to lack of water and is not the mechanism by which UV inactivates organisms. In contrast, ionizing radiation generating hydroxyl radicals, ozone acting as a strong oxidizer/electron acceptor, plasma sterilization producing reactive species, and supercritical CO2 use well-established mechanisms consistent with their pairings.
Which of the following treatments is the most effective for controlling microbial growth?
- 63°C for 30 minutes
- 72°C for 15 seconds
- 140°C for 4 seconds
- They are equivalent treatments.
- None of the answers is correct.
Explanation: Answer reason: Thermal control of microbial growth depends on achieving a sufficient time–temperature exposure to produce a comparable log reduction in organisms. These three heat regimens correspond to standard pasteurization/ultra-high-temperature approaches designed to deliver similar overall lethality (higher temperature requires much shorter time). As long as the target organism profile and product conditions match the intended pasteurization standard, the microbial reduction is considered equivalent across these validated combinations. The distractors focus on one specific regimen as “most effective,” but effectiveness here is defined by equivalent microbial kill rather than simply the highest temperature.
Which of the following is a nucleated, unicellular organism that, if you changed the incubation temperature, would form filaments with conidiospores?
- Ascomycete
- Cellular slime mold
- Euglenozoa
- Tapeworm
- Plasmodial slime mold
Explanation: Answer reason: The mold phase characteristically produces asexual conidia (conidiospores) on hyphae, which is a defining fungal reproductive feature. Ascomycetes include many clinically relevant dimorphic fungi (e.g., several systemic mycoses) that show this temperature-dependent morphology. Protozoa like euglenozoa and helminths such as tapeworms do not form hyphae or conidia, and slime molds are not true fungi and are not defined by conidiospore-producing filamentous phases.
Which of the following statements about helminths is FALSE?
- They are heterotrophic.
- They are multicellular animals.
- They have eukaryotic cells.
- All are parasites.
- Some have male and female reproductive organs in one animal.
Explanation: Answer reason: Helminths are eukaryotic, multicellular animals that obtain nutrients heterotrophically, and many are medically important as parasitic worms. However, “helminth” as a broad biological grouping includes numerous free-living species (especially among nematodes) that do not parasitize a host, making the statement incorrect. The other choices are consistent with helminths being animals with eukaryotic cells and requiring organic nutrients. In addition, several helminths are hermaphroditic (e.g., many cestodes and trematodes), supporting the statement about having both reproductive organs in one organism.
Which of the following statements about the oomycote algae is FALSE?
- They form hyphae.
- They produce zoospores in a sporangium.
- They cause plant diseases.
- They have chlorophyll.
- They reproduce sexually.
Explanation: Answer reason: Oomycetes are fungus-like water molds (stramenopiles) that obtain nutrients by absorption rather than photosynthesis, so they lack chlorophyll. They characteristically grow as filamentous hyphae and produce motile zoospores within sporangia, reflecting their affinity for aquatic environments. Many are important plant pathogens (e.g., downy mildews, Phytophthora species), making plant disease causation a true statement. They also have a sexual cycle producing oospores after gamete fusion, so sexual reproduction is also true.
All of the following are characteristic of the Platyhelminthes EXCEPT that they?
- Are hermaphroditic.
- Are dorsoventrally flattened.
- Have highly developed digestive and nervous systems.
- Can be divided into flukes and tapeworms.
- Are multicellular animals.
Explanation: Answer reason: Platyhelminths (flatworms) are acoelomate, dorsoventrally flattened helminths with relatively simple organ systems adapted to parasitism or free-living habitats. Their nervous system is primitive (ladder-like with ganglia), and their digestive system is not highly developed—many have an incomplete gut, and cestodes (tapeworms) lack a digestive tract entirely and absorb nutrients through the tegument. In contrast, being hermaphroditic, multicellular, and classifiable into flukes (trematodes) and tapeworms (cestodes) are standard characteristics taught for this phylum. Therefore the statement claiming advanced digestive and nervous systems is the exception.
Which of the following pairs is mismatched?
- Nematodes — complete digestive tract
- Cestodes — segmented body made of proglottids
- Trematodes — flukes
- Nematodes — many are free-living
- Cestodes — all are free-living
Explanation: Answer reason: Cestodes (tapeworms) are obligate parasitic flatworms that inhabit host intestines and lack a free-living adult lifestyle, so describing them as all free-living is incorrect. The other pairings are standard identifiers: cestodes are segmented into proglottids, trematodes are flukes, and nematodes characteristically have a complete digestive tract. While nematodes include both parasitic and many free-living species, cestodes do not, making this the mismatched pair.
Microorganisms are essential to our life. Each of the following is an example of a beneficial function of microorganisms EXCEPT?
- Alternative fuel production.
- Bioremediation.
- Gene therapy.
- Agriculture.
- Increased number of illnesses.
Explanation: Answer reason: The key principle is that many microbial activities are beneficial to humans and ecosystems, but pathogenic effects are harmful outcomes rather than benefits. Microorganisms can be used to generate biofuels through fermentation and other metabolic pathways, providing alternative fuel production. They also break down pollutants in bioremediation and support agriculture through soil nutrient cycling and symbiotic nitrogen fixation. An increase in illnesses reflects microbial pathogenicity and disease transmission, which is the exception to beneficial functions.
Viruses are not considered living organisms because they?
- Cannot reproduce by themselves.
- Are structurally very simple.
- Can only be visualized using an electron microscope.
- Are typically associated with disease.
- Are ubiquitous in nature.
Explanation: Answer reason: A core criterion for life is the ability to reproduce independently and carry out essential metabolic processes. Viruses are obligate intracellular parasites that lack the cellular machinery (eg, ribosomes and energy-generating systems) needed to replicate on their own, so they must hijack a host cell to produce new virions. Structural simplicity and requiring an electron microscope are descriptive features but do not define whether something is living. Association with disease and ubiquity also do not address the fundamental dependence of viruses on a host for replication.
Which of the following are found primarily in the intestines of humans?
- Gram-negative aerobic rods and cocci
- Aerobic, helical bacteria
- Facultatively anaerobic gram-negative rods
- Gram-positive cocci
- Endospore-forming rods
Explanation: Answer reason: Enteric gram-negative rods (e.g., many Enterobacterales) are classic facultative anaerobes and are common members of normal intestinal flora. Options emphasizing strictly aerobic organisms fit better with higher-oxygen niches such as skin or upper respiratory tract rather than the colon. Endospore-forming rods and gram-positive cocci can be present, but they are not the primary defining bacterial group of the intestinal microbiota compared with enteric facultative anaerobic gram-negative rods.
Which of the following pairs is NOT correctly matched?
- Elementary body — Escherichia
- Endospore — Bacillus
- Endospore — Clostridium
- Heterocyst — cyanobacteria
- Myxospore — gliding bacteria
Explanation: Answer reason: Escherichia coli does not form elementary bodies; it is a free-living Gram-negative rod that reproduces by binary fission without such a cycle. In contrast, Bacillus and Clostridium are classic endospore-formers, making those matches correct. Heterocysts are specialized nitrogen-fixing cells in some cyanobacteria, and myxospores are associated with myxobacteria (gliding bacteria), supporting the remaining pairings.
You have isolated a bacterium that grows in a medium containing only inorganic nutrients. Ammonia is oxidized to a nitrate ion. This bacterium is?
- Gram-negative.
- Using anaerobic respiration.
- A chemoautotroph.
- A photoautotroph.
- A photoheterotroph.
Explanation: Answer reason: Oxidation of an inorganic compound (ammonia) to derive energy is chemolithotrophy, and growth on only inorganic nutrients implies carbon is obtained from CO2 (autotrophy). Nitrifying bacteria (e.g., ammonia-oxidizers) use the energy released from ammonia oxidation to drive ATP generation and fix CO2, fitting the chemoautotroph category. The stem does not provide information about Gram stain or oxygen use, so those choices are not supported. Phototroph options are incorrect because the energy source described is chemical (ammonia), not light.
Which of the following is NOT a characteristic inherent of the non-endospore-forming gram-positive rods?
- Are aerotolerant
- Carry out fermentative metabolism
- Display a branched filamentous morphology
- Nonpathogenic
- Lack cell walls
Explanation: Answer reason: Non–endospore-forming gram-positive rods include organisms such as Corynebacterium, Listeria, and Actinomyces, which characteristically possess cell walls. Several are aerotolerant and commonly rely on fermentative metabolism, and some (e.g., Actinomyces) can show branching filamentous forms. A common distractor is “nonpathogenic,” because many non–spore-forming gram-positive rods can cause disease, but the only option that fundamentally contradicts the group’s defining structure is the lack of a cell wall.
Which hepatitis virus has only one serotype, stimulates antibodies that are thought to result in lifelong immunity, has an incubation period of approximately 28 days, and can unknowingly be passed by the host?
- Hepatitis A.
- Hepatitis B.
- Hepatitis C.
- Hepatitis D.
Explanation: Answer reason: HAV is a fecal–oral, non-enveloped RNA virus with essentially one serotype, so post-infection antibodies are typically protective long term. Its incubation period averages about 28 days (roughly 15–50 days), matching the timeline in the stem. People can shed virus in stool and be contagious before symptoms (including jaundice) appear, so transmission can occur unknowingly. In contrast, HBV/HCV have longer typical incubation windows and are blood/body-fluid transmitted, and HCV commonly leads to chronic infection rather than reliable lifelong immunity.
The primary victims of the influenza pandemic of 1918-1919 were?
- Members of the military.
- Infants and the elderly.
- Residents of Spain; therefore, it was known as the "Spanish Flu."
- Young adults.
- Residents of the U.S.
Explanation: Answer reason: The key epidemiologic feature of the 1918–1919 influenza pandemic was an unusually high mortality in healthy adults, producing a characteristic “W-shaped” age-specific death curve. This contrasts with typical seasonal influenza, where the highest risk is concentrated in infants and older adults. Proposed mechanisms include severe viral pneumonia and an exaggerated host immune response contributing to acute lung injury in robust immune systems. While outbreaks were prominent in military settings and many countries (including the U.S.), these describe exposure locations rather than the primary age group with excess deaths.
Which of the following is mismatched?
- Beef — E. coli O157:H7
- Custard and cream pies — Staphylococcus aureus
- Eggs — Trichinella spiralis
- Chicken — Campylobacter jejuni
- Oysters — Vibrio parahaemolyticus
Explanation: Answer reason: Trichinella spiralis is classically acquired from undercooked pork or wild game (e.g., bear), where encysted larvae are present in muscle tissue, not from eggs. In contrast, E. coli O157:H7 is linked to undercooked beef, Staphylococcus aureus toxin to cream-filled pastries, Campylobacter jejuni to poultry, and Vibrio species to raw/undercooked shellfish. Therefore the eggs pairing is the mismatched association.
A urine sample collected directly from the urinary bladder?
- Is sterile.
- Contains fewer than 100 pathogens/ml.
- Contains fewer than 10,000 pathogens/ml.
- Contains more than 100,000 pathogens/ml.
- Has leukocyte esterase.
Explanation: Answer reason: In a healthy individual, urine within the bladder is normally free of microorganisms because the upper urinary tract is not colonized and normal flow helps prevent ascent of bacteria. The presence of significant pathogen counts is expected in contaminated specimens (e.g., poor clean-catch technique) or true UTI, not in urine taken directly from the bladder under aseptic conditions. Colony-count thresholds (e.g., 10^5 CFU/mL) are used to interpret culture results and do not define baseline bladder urine. Leukocyte esterase indicates pyuria/inflammation and is not an expected property of normal bladder urine.
Which of the following pertains to an improperly canned food product on a store shelf that has a bad odor and putrefaction of the canned material?
- Mesophilic spoilage
- Flat sour spoilage
- Bacterial soft rot
- Thermophilic anaerobic spoilage
- Hydrogen sulfide spoilage
Explanation: Answer reason: Their metabolic activity commonly leads to proteolysis and putrefaction, producing foul odors and obvious decomposition. In contrast, flat sour spoilage is characterized by acid production without gas and typically lacks strong putrid odor, making it a common distractor. Thermophilic anaerobic spoilage is more associated with high-temperature conditions rather than ordinary shelf storage. Hydrogen sulfide spoilage can create sulfur odors, but the stem emphasizes general putrefaction and shelf-temperature growth conditions, which aligns best with mesophilic spoilage.
Microorganisms themselves are commercial products. Which of the following microbes is available as a product sold in retail stores?
- Bacillus thuringiensis
- Lactobacillus
- Rhizobium
- Saccharomyces
- All of these organisms can be found in various commercial products.
Explanation: Answer reason: The key principle is that many microbes are intentionally marketed for agriculture, food/fermentation, and health-related uses. Bacillus thuringiensis is sold as a biological insecticide for garden and crop pest control. Lactobacillus is commonly sold as probiotic supplements and is used in yogurt/fermented foods, while Saccharomyces is sold widely as baker’s/brewer’s yeast and also as probiotic preparations (e.g., S. boulardii). Rhizobium is commercially available as seed/soil inoculants to enhance legume nitrogen fixation, so each listed organism can be found in commercial products.
You are growing Bacillus subtilis, an aerobic bacterium that can also carry out fermentation when required, in a bioreactor and notice that the growth rate has slowed and the pH has decreased. What could you add?
- Glucose
- Lactose
- Peptides
- Oxygen
- Lactic acid
Explanation: Answer reason: Increasing aeration pushes metabolism back toward aerobic respiration, improving ATP yield and growth while reducing acid end-product buildup. Adding more carbohydrate substrate would typically worsen acidification by fueling fermentation under oxygen limitation. Adding lactic acid would further decrease pH and inhibit growth rather than restore it.
Which of the following is an advantage of using E. coli to make a human gene product?
- Endotoxin may be in the product.
- It does not secrete most proteins.
- Its genes are well known.
- It cannot process introns.
- Endotoxin may be in the product and it does not secrete most proteins.
Explanation: Answer reason: A major advantage of E. coli as an expression host is that it is genetically well-characterized, easy to manipulate, and has extensive tools (vectors, promoters, selectable markers) that make cloning and high-yield production straightforward. This predictability supports efficient insertion and control of recombinant human genes and scaling up manufacturing. By contrast, endotoxin contamination and poor secretion are disadvantages that complicate purification and safety. Inability to process introns is also a disadvantage because many human genes require intron removal, often necessitating cDNA rather than genomic DNA for expression.
If you have inserted a gene in the Ti plasmid, the next step in genetic engineering is?
- Transformation of E. coli with Ti plasmid.
- Splicing T DNA into a plasmid.
- Transformation of an animal cell.
- Inserting the Ti plasmid into Agrobacterium.
- Inserting the Ti plasmid into a plant cell.
Explanation: Answer reason: Plant genetic engineering with the Ti plasmid relies on Agrobacterium tumefaciens as the biological vector that naturally transfers T-DNA into plant cells. After the gene of interest is inserted into the Ti plasmid (recombinant plasmid construction), the plasmid must be introduced into Agrobacterium so the bacterium can mediate transfer into plant tissue. This step is essential because Ti plasmids function within Agrobacterium and their transfer machinery is provided by the bacterium. Options involving E. coli or animal cells do not match the Ti plasmid system, and direct delivery into a plant cell bypasses the standard Agrobacterium-mediated transfer mechanism.
Which of the following statements about archaea is FALSE?
- They are prokaryotes.
- They lack peptidoglycan in their cell walls.
- Some are thermoacidophiles; others are extreme halophiles.
- They evolved before bacteria.
- Some produce methane from carbon dioxide and hydrogen.
Explanation: Answer reason: Archaea are prokaryotic organisms that are distinct from bacteria in cell wall composition and molecular biology, and their evolutionary history is best described as divergence from a common ancestor rather than a proven “before” sequence. A key distinguishing feature is the absence of peptidoglycan (they may have pseudopeptidoglycan or protein S-layers), which supports the truth of the cell wall statement. Many archaea are extremophiles (e.g., thermoacidophiles and extreme halophiles), and methanogenesis from CO2 and H2 is a classic archaeal metabolic pathway. The “evolved before bacteria” claim overstates what can be concluded and is the false statement among the options.
Which of the following statements about members of the Kingdom Plantae is FALSE?
- They are multicellular.
- They are composed of eukaryotic cells.
- They undergo photosynthesis.
- They use organic carbon sources.
- They synthesize organic molecules.
Explanation: Answer reason: Members of Kingdom Plantae are primarily photoautotrophs, meaning they use carbon dioxide as their main carbon source and light as their energy source to build organic compounds. Using organic carbon sources is a hallmark of heterotrophs (e.g., animals, fungi, many bacteria), not typical plants. The other choices align with core plant characteristics: plants are multicellular eukaryotes and (in general classification) perform photosynthesis and generate organic molecules like carbohydrates. While a small number of plants can be partially heterotrophic (e.g., parasitic plants), the statement as a general feature of Plantae is false.
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