Microbiology Quiz: Culture Based Identification
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Culture Based IdentificationQuestion 1 of 20

An isolate of a Gram-negative rod from a stool culture produces a red slant and yellow butt (K/A) with no gas or H2S production on Triple Sugar Iron (TSI) agar. The organism is also urease positive. Based on these two results, what is the most likely identity of the organism?

Salmonella enterica
Shigella sonnei
Escherichia coli
Yersinia enterocolitica
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Microbiology Quiz

Microbiology Quiz: Culture Based Identification

Practice Culture Based Identification in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Culture Based Identification, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An isolate of a Gram-negative rod from a stool culture produces a red slant and yellow butt (K/A) with no gas or H2S production on Triple Sugar Iron (TSI) agar. The organism is also urease positive. Based on these two results, what is the most likely identity of the organism?

  1. Salmonella enterica
  2. Shigella sonnei
  3. Escherichia coli
  4. Yersinia enterocolitica (correct answer)
Explanation: The TSI reaction K/A (alkaline/red slant, acid/yellow butt) indicates that the organism ferments glucose but not lactose or sucrose. The urease positive result is the key differentiator. Yersinia enterocolitica fits this profile (K/A on TSI, urease positive). Salmonella (A) is typically K/A with H2S and is urease negative. Shigella (B) is K/A but is urease negative. Escherichia coli (C) is A/A (acid/acid) as it ferments lactose and/or sucrose, and is urease negative.

Question 2

A laboratory isolates an Enterococcus species from a patient with endocarditis. To guide therapy, high-level aminoglycoside resistance (HLAR) screening is performed using agar plates containing high concentrations of gentamicin and streptomycin. No growth is observed on either plate. How should this result be interpreted in the context of antimicrobial susceptibility testing?

  1. The organism is susceptible to all aminoglycosides, and they can be used as monotherapy.
  2. The organism has intrinsic low-level resistance to aminoglycosides, so the test has no clinical value.
  3. The test is inconclusive and requires molecular methods to detect resistance genes.
  4. The absence of growth predicts synergy between a cell wall agent (like ampicillin or vancomycin) and an aminoglycoside. (correct answer)
Explanation: When you encounter questions about Enterococcus and aminoglycoside testing, remember that enterococci have unique resistance patterns that affect combination therapy strategies. Enterococcus species possess intrinsic low-level resistance to aminoglycosides due to poor drug uptake. However, when combined with cell wall-active agents like ampicillin or vancomycin, these agents disrupt the cell wall and allow aminoglycosides to penetrate effectively, creating bactericidal synergy. High-level aminoglycoside resistance (HLAR) screening specifically tests whether this synergy will work. The absence of growth on high-concentration gentamicin and streptomycin plates indicates the organism lacks HLAR mechanisms. This means the intrinsic resistance can still be overcome by combination therapy, predicting that synergy between a cell wall agent and an aminoglycoside will be effective. This is why answer D is correct. Answer A is wrong because aminoglycosides alone are never effective against enterococci due to intrinsic resistance—they require combination therapy. Answer B misinterprets the test's purpose; while enterococci do have intrinsic low-level resistance, HLAR testing specifically determines if synergy is possible, making it clinically valuable. Answer C is incorrect because the clear absence of growth provides definitive information—molecular methods aren't needed when phenotypic results are unambiguous. For microbiology exams, remember that Enterococcus antimicrobial testing often focuses on combination therapy potential rather than individual drug susceptibility. HLAR screening directly predicts synergy effectiveness, which is crucial for treating serious enterococcal infections like endocarditis.

Question 3

An automated identification system provides the following report for a Gram-negative rod isolated from a respiratory specimen from a cystic fibrosis patient:

Burkholderia cepacia complex - 95% probability Pandoraea apista - 4% probability

The laboratory has a policy to confirm this identification with a supplementary culture-based method due to the therapeutic and infection control implications.

Which selective medium would be most useful for confirming the identity as a member of the Burkholderia cepacia complex?

  1. Cystine-tellurite blood agar (CTBA).
  2. Cetrimide agar.
  3. Buffered charcoal yeast extract (BCYE) agar.
  4. Burkholderia cepacia selective agar (BCSA). (correct answer)
Explanation: When identifying Burkholderia cepacia complex in cystic fibrosis patients, confirmation is critical because these organisms are associated with severe complications and require strict infection control measures. The key is understanding that specialized selective media have been developed specifically for isolating and confirming particular bacterial groups. Burkholderia cepacia selective agar (BCSA) is specifically formulated to inhibit other respiratory flora while allowing B. cepacia complex organisms to grow and display their characteristic morphology and biochemical properties. This medium contains selective agents that suppress competing bacteria commonly found in respiratory specimens, making it the ideal confirmatory tool. Answer D is correct because BCSA provides both selectivity and the ability to observe distinctive colonial characteristics of B. cepacia complex. Answer A, cystine-tellurite blood agar (CTBA), is selective for Corynebacterium diphtheriae, not Burkholderia species. Answer B, cetrimide agar, selects for Pseudomonas aeruginosa and would not reliably differentiate B. cepacia complex from other gram-negative rods. Answer C, buffered charcoal yeast extract (BCYE) agar, is designed for Legionella species and lacks the selective properties needed for Burkholderia identification. For microbiology identification questions, remember that specialized selective media are named for their target organisms. When you see a question asking for confirmation of a specific bacterial complex, look for the medium that bears that organism's name—it's usually designed specifically for that purpose.

Question 4

A cerebrospinal fluid (CSF) specimen is received for bacterial culture. The direct Gram stain is negative. To ensure recovery of the most common potential pathogens, the specimen should be inoculated onto which combination of media and incubated under what conditions?

  1. Sheep blood agar and MacConkey agar, incubated aerobically.
  2. Sheep blood agar, chocolate agar, and a thioglycollate broth, with plates incubated in a CO2-enriched atmosphere. (correct answer)
  3. CNA agar and MacConkey agar, with plates incubated aerobically and anaerobically.
  4. Thayer-Martin agar and Hektoen Enteric agar, incubated in a microaerophilic atmosphere.
Explanation: CSF is a critical sterile site specimen. The most common agents of bacterial meningitis include fastidious organisms like Neisseria meningitidis and Haemophilus influenzae, which require chocolate agar and a CO2-enriched (capnophilic) atmosphere. Sheep blood agar is included for non-fastidious organisms like Streptococcus pneumoniae and Listeria. Thioglycollate broth serves as an enrichment backup, supporting the growth of low numbers of organisms and anaerobes. The other options are insufficient: (A) lacks chocolate agar for fastidious organisms. (C) is inappropriate for primary CSF culture. (D) includes media for specific pathogens not typically sought in a primary CSF workup.

Question 5

A CAMP test is set up to identify a beta-hemolytic Gram-positive coccus isolated from a vaginal swab of a pregnant woman. The isolate is streaked perpendicular to a streak of Staphylococcus aureus on a sheep blood agar plate. After incubation, a distinct arrowhead-shaped zone of enhanced hemolysis is observed where the two streaks meet. What biochemical principle does this positive result demonstrate?

  1. The isolate produces a DNase that synergizes with the staphylococcal hemolysin.
  2. The isolate produces hyaluronidase which enhances the diffusion of the staphylococcal toxin.
  3. The isolate hydrolyzes esculin in the presence of bile produced by the Staphylococcus aureus.
  4. The isolate's CAMP factor is a phospholipase that acts synergistically with the staphylococcal beta-hemolysin. (correct answer)
Explanation: When you encounter a CAMP test question, you're dealing with a classic diagnostic test for Group B Streptococcus (GBS), particularly important in obstetric settings since GBS can cause serious neonatal infections. The CAMP test works through a specific biochemical interaction. The suspected GBS isolate produces CAMP factor, a phospholipase that by itself causes minimal hemolysis. However, when Staphylococcus aureus is streaked nearby, it produces beta-hemolysin (sphingomyelinase). These two enzymes work synergistically—the CAMP factor enhances the activity of the staphylococcal beta-hemolysin, creating the characteristic arrowhead-shaped zone of enhanced hemolysis where their diffusion zones meet. Answer D correctly identifies this mechanism: CAMP factor is indeed a phospholipase that acts synergistically with staphylococcal beta-hemolysin. Answer A is incorrect because CAMP factor is not a DNase—it's a phospholipase. While some streptococci produce DNases, that's not the mechanism of the CAMP test. Answer B misidentifies the enzyme as hyaluronidase. Though some bacteria produce hyaluronidase as a spreading factor, this isn't involved in the CAMP reaction. Answer C confuses this with the bile-esculin test, which is used for enterococci identification, not the CAMP test mechanism. Remember: CAMP test = phospholipase + beta-hemolysin synergy. The distinctive arrowhead pattern is your visual cue that two different enzymes are working together, not just one enzyme acting alone.

Question 6

A synovial fluid specimen from a patient with acute septic arthritis is sent to the lab. A direct Gram stain reveals numerous neutrophils and intracellular Gram-negative diplococci. The specimen is inoculated onto a sheep blood agar (BAP) plate and a MacConkey (MAC) agar plate, which are then incubated at 35°C in ambient air. After 48 hours, no growth is observed on either plate. Which of the following is the most likely reason for the culture failure?

  1. The organisms observed in the Gram stain were non-viable due to prior antibiotic therapy.
  2. The specimen required anaerobic incubation conditions for the pathogen to grow.
  3. The organism is likely Neisseria gonorrhoeae, which requires an enriched medium and a capnophilic atmosphere. (correct answer)
  4. The Gram stain was misinterpreted, and the observed structures were stain precipitate or cellular debris.
Explanation: The clinical presentation and Gram stain morphology (Gram-negative diplococci in synovial fluid) are classic for Neisseria gonorrhoeae. This organism is fastidious and requires an enriched medium like chocolate agar and a capnophilic (CO2-enriched) atmosphere for growth. It will not grow on BAP or MAC in ambient air. While prior antibiotic therapy (A) or misinterpretation (D) are possibilities, the failure to provide appropriate growth conditions (C) is the most direct and probable laboratory-related cause for the discrepancy. The organism is not a strict anaerobe (B).

Question 7

A technologist is processing a stool specimen from a patient with a 5-day history of bloody diarrhea. The physician has specifically requested a workup for enterohemorrhagic E. coli (EHEC), Salmonella, Shigella, and Campylobacter.

Which combination of primary plating media is most appropriate to recover all the requested potential pathogens from this specimen?

  1. Sheep Blood Agar (BAP), Chocolate Agar (CHOC), and MacConkey Agar (MAC).
  2. MacConkey Agar (MAC), Hektoen Enteric (HE) Agar, and Thayer-Martin Agar.
  3. Sorbitol MacConkey (SMAC) Agar, Hektoen Enteric (HE) Agar, and Campy Blood Agar incubated microaerophilically. (correct answer)
  4. CNA Agar, MacConkey (MAC) Agar, and Thioglycollate Broth incubated anaerobically.
Explanation: To recover the specified pathogens, a specific media battery is required. Sorbitol MacConkey (SMAC) agar is used to screen for EHEC O157:H7, which typically does not ferment sorbitol and produces colorless colonies. Hektoen Enteric (HE) agar is selective and differential for Salmonella and Shigella. Campy Blood Agar is a selective medium for Campylobacter species and must be incubated in a microaerophilic atmosphere. The other options are incorrect: (A) is a standard respiratory or wound setup, not for enteric pathogens. (B) includes Thayer-Martin, which is for Neisseria. (D) includes CNA for Gram-positives and anaerobic broth, which are not primary for this specific request.

Question 8

A technologist inoculates a 0.001 mL calibrated loop of a clean-catch midstream urine specimen onto a sheep blood agar plate. After 24 hours of incubation, 65 colonies of a single morphotype are counted on the plate. What is the colony count in CFU/mL, and what is the clinical significance?

  1. 65 CFU/mL; represents contamination.
  2. 6,500 CFU/mL; likely represents contamination or colonization.
  3. 65,000 CFU/mL; potentially significant and warrants identification. (correct answer)
  4. 650,000 CFU/mL; represents a significant bacteriuria requiring full workup.
Explanation: The colony count is calculated by multiplying the number of colonies by the dilution factor of the loop. The dilution factor is the reciprocal of the loop volume (1 / 0.001 mL = 1000). Therefore, the count is 65 colonies * 1000 = 65,000 CFU/mL. A count in this range (typically >10,000 CFU/mL of a single uropathogen) is considered potentially significant and requires, at a minimum, identification of the organism. Counts below 10,000 CFU/mL are often considered contamination (A, B), while counts >100,000 CFU/mL are definitively significant (D). 65,000 CFU/mL falls into the intermediate category that requires further investigation.

Question 9

A culture from an expectorated sputum sample yields four different colony morphotypes on sheep blood agar, with counts ranging from 10,000 to 50,000 CFU/mL for each. The morphotypes resemble viridans streptococci, coagulase-negative staphylococci, Corynebacterium species, and Neisseria species. What is the most appropriate action and report for the technologist to take?

  1. Perform a full identification and susceptibility test on all four morphotypes.
  2. Quantify and report each organism, but perform susceptibility testing only on the Neisseria species.
  3. Report the result as 'Moderate growth of normal oropharyngeal flora.' and perform no further workup. (correct answer)
  4. Request a new, better-quality specimen, such as a bronchial wash, as the current sample is contaminated.
Explanation: Expectorated sputum is non-sterile and is expected to be contaminated with flora from the upper respiratory tract. The organisms described (viridans strep, CoNS, Corynebacterium, non-pathogenic Neisseria) are all components of normal oropharyngeal flora. In the absence of a predominant potential pathogen (like S. pneumoniae or H. influenzae), and with multiple morphotypes at moderate counts, the finding is indicative of contamination, not infection. Therefore, the most appropriate report is 'normal oropharyngeal flora'. Performing a full workup (A, B) is a waste of resources and could lead to inappropriate antibiotic treatment. Requesting a new specimen (D) might be an option, but reporting the current finding is the immediate correct action.

Question 10

An indole test is performed on a lactose-negative, motile, Gram-negative rod isolated from a wound. The organism is inoculated into tryptone broth, incubated, and then Kovac's reagent is added. No red ring forms at the surface. A control organism, E. coli, tested with the same reagents, shows a dark red ring. What can be concluded?

  1. The test is negative, and the organism does not produce the enzyme tryptophanase. (correct answer)
  2. The test is invalid because the reagent is expired, as indicated by the control.
  3. The test is positive, indicating the organism is likely a species of Proteus.
  4. The organism is likely an oxidizer, and the test is not appropriate for this group of bacteria.
Explanation: The indole test detects the production of the enzyme tryptophanase, which breaks down tryptophan into indole, pyruvic acid, and ammonia. Kovac's reagent reacts with indole to produce a red color. The absence of the red ring indicates a negative result, meaning the organism does not produce tryptophanase. The positive result with the E. coli control confirms that the media and reagents are working correctly, invalidating option (B). A positive test would be a red ring, making (C) incorrect. While many oxidizers are indole-negative, the test is still valid and performed on them; the conclusion is simply that the reaction is negative (D).

Question 11

A laboratory is performing quality control on a new batch of MacConkey agar. The Escherichia coli ATCC 25922 strain produces flat, dry, pink colonies. The Staphylococcus aureus ATCC 25923 strain shows pinpoint growth. The Proteus mirabilis ATCC 12453 strain produces colorless colonies. Which result indicates a problem with the medium's formulation?

  1. The growth of Escherichia coli as pink colonies.
  2. The pinpoint growth of Staphylococcus aureus. (correct answer)
  3. The growth of Proteus mirabilis as colorless colonies.
  4. The absence of swarming by Proteus mirabilis.
Explanation: MacConkey agar is designed to be selective, completely inhibiting the growth of most Gram-positive organisms like Staphylococcus aureus. The observation of any growth, even pinpoint colonies, indicates that the selective agents (bile salts and crystal violet) are not at the correct concentration or are defective. Therefore, the medium fails its selectivity quality control parameter. The results for E. coli (lactose fermenter, pink) (A) and P. mirabilis (non-lactose fermenter, colorless) (C) are correct differential reactions. The absence of swarming by Proteus (D) is also an expected outcome, as the medium is formulated to inhibit this.

Question 12

An organism isolated from a burn wound fails to grow on MacConkey agar but grows well on sheep blood agar, producing a characteristic fruity, grape-like odor and a metallic sheen. A Gram stain shows Gram-negative bacilli. An oxidase test is strongly positive. Which of the following biochemical characteristics is most likely associated with this organism?

  1. Oxidizes glucose but does not ferment it. (correct answer)
  2. Ferments glucose and lactose rapidly.
  3. Is indole positive and motile.
  4. Produces H2S on Triple Sugar Iron (TSI) agar.
Explanation: The combination of characteristics (Gram-negative bacillus, no growth on MAC, grape-like odor, metallic sheen, oxidase-positive) is classic for Pseudomonas aeruginosa. P. aeruginosa is a non-fermenter, meaning it metabolizes carbohydrates oxidatively rather than fermentatively. Therefore, it oxidizes glucose but does not ferment it. Rapid fermentation of glucose and lactose (B) is characteristic of coliforms like E. coli. Indole positivity (C) is characteristic of E. coli. H2S production (D) is characteristic of Salmonella, Proteus, and other species, but not P. aeruginosa.

Question 13

A urine culture is plated on a chromogenic agar designed for urinary pathogens. After 24-hour incubation, the plate shows >100,000 CFU/mL of smooth, pink-to-red colonies. According to the manufacturer's guide, this color indicates beta-galactosidase activity. The plate also shows about 10,000 CFU/mL of small, blue colonies, indicating beta-glucosidase activity. What is the correct interpretation and next step?

  1. The culture is mixed and likely contaminated; request a recollection.
  2. Perform a full workup and susceptibility testing on both the pink and blue colony types.
  3. Report >100,000 CFU/mL of a lactose-fermenting organism, likely E. coli, and perform a workup on the pink colonies only. (correct answer)
  4. Report the blue colonies as Enterococcus and perform a workup only if the patient is symptomatic.
Explanation: Quantitative culture guidelines are critical. The pink colonies are present in a clinically significant number (>100,000 CFU/mL), and their beta-galactosidase activity (lactose fermentation) suggests a common uropathogen like E. coli. This requires a full workup. The blue colonies (Enterococcus) are present at a low, likely insignificant count (10,000 CFU/mL) and represent probable colonization or contamination. Therefore, the primary focus should be on identifying and testing the predominant pathogen. Requesting a recollection (A) is unnecessary when a clear predominant pathogen is present. Working up both (B) is inefficient and not clinically indicated. Focusing only on the minor organism (D) is incorrect.

Question 14

An organism isolated from a burn wound fails to grow on MacConkey agar but grows well on sheep blood agar, producing a characteristic fruity, grape-like odor and a metallic sheen. A Gram stain shows Gram-negative bacilli. An oxidase test is strongly positive. Which of the following biochemical characteristics is most likely associated with this organism?

  1. Oxidizes glucose but does not ferment it. (correct answer)
  2. Ferments glucose and lactose rapidly.
  3. Is indole positive and motile.
  4. Produces H2S on Triple Sugar Iron (TSI) agar.
Explanation: The combination of characteristics (Gram-negative bacillus, no growth on MAC, grape-like odor, metallic sheen, oxidase-positive) is classic for Pseudomonas aeruginosa. P. aeruginosa is a non-fermenter, meaning it metabolizes carbohydrates oxidatively rather than fermentatively. Therefore, it oxidizes glucose but does not ferment it. Rapid fermentation of glucose and lactose (B) is characteristic of coliforms like E. coli. Indole positivity (C) is characteristic of E. coli. H2S production (D) is characteristic of Salmonella, Proteus, and other species, but not P. aeruginosa.

Question 15

An isolate of a Gram-negative rod from a stool culture produces a red slant and yellow butt (K/A) with no gas or H2S production on Triple Sugar Iron (TSI) agar. The organism is also urease positive. Based on these two results, what is the most likely identity of the organism?

  1. Salmonella enterica
  2. Shigella sonnei
  3. Escherichia coli
  4. Yersinia enterocolitica (correct answer)
Explanation: The TSI reaction K/A (alkaline/red slant, acid/yellow butt) indicates that the organism ferments glucose but not lactose or sucrose. The urease positive result is the key differentiator. Yersinia enterocolitica fits this profile (K/A on TSI, urease positive). Salmonella (A) is typically K/A with H2S and is urease negative. Shigella (B) is K/A but is urease negative. Escherichia coli (C) is A/A (acid/acid) as it ferments lactose and/or sucrose, and is urease negative.

Question 16

A CAMP test is set up to identify a beta-hemolytic Gram-positive coccus isolated from a vaginal swab of a pregnant woman. The isolate is streaked perpendicular to a streak of Staphylococcus aureus on a sheep blood agar plate. After incubation, a distinct arrowhead-shaped zone of enhanced hemolysis is observed where the two streaks meet. What biochemical principle does this positive result demonstrate?

  1. The isolate produces a DNase that synergizes with the staphylococcal hemolysin.
  2. The isolate produces hyaluronidase which enhances the diffusion of the staphylococcal toxin.
  3. The isolate hydrolyzes esculin in the presence of bile produced by the Staphylococcus aureus.
  4. The isolate's CAMP factor is a phospholipase that acts synergistically with the staphylococcal beta-hemolysin. (correct answer)
Explanation: When you encounter a CAMP test question, you're dealing with a classic diagnostic test for Group B Streptococcus (GBS), particularly important in obstetric settings since GBS can cause serious neonatal infections. The CAMP test works through a specific biochemical interaction. The suspected GBS isolate produces CAMP factor, a phospholipase that by itself causes minimal hemolysis. However, when Staphylococcus aureus is streaked nearby, it produces beta-hemolysin (sphingomyelinase). These two enzymes work synergistically—the CAMP factor enhances the activity of the staphylococcal beta-hemolysin, creating the characteristic arrowhead-shaped zone of enhanced hemolysis where their diffusion zones meet. Answer D correctly identifies this mechanism: CAMP factor is indeed a phospholipase that acts synergistically with staphylococcal beta-hemolysin. Answer A is incorrect because CAMP factor is not a DNase—it's a phospholipase. While some streptococci produce DNases, that's not the mechanism of the CAMP test. Answer B misidentifies the enzyme as hyaluronidase. Though some bacteria produce hyaluronidase as a spreading factor, this isn't involved in the CAMP reaction. Answer C confuses this with the bile-esculin test, which is used for enterococci identification, not the CAMP test mechanism. Remember: CAMP test = phospholipase + beta-hemolysin synergy. The distinctive arrowhead pattern is your visual cue that two different enzymes are working together, not just one enzyme acting alone.

Question 17

A technologist inoculates a 0.001 mL calibrated loop of a clean-catch midstream urine specimen onto a sheep blood agar plate. After 24 hours of incubation, 65 colonies of a single morphotype are counted on the plate. What is the colony count in CFU/mL, and what is the clinical significance?

  1. 65 CFU/mL; represents contamination.
  2. 6,500 CFU/mL; likely represents contamination or colonization.
  3. 65,000 CFU/mL; potentially significant and warrants identification. (correct answer)
  4. 650,000 CFU/mL; represents a significant bacteriuria requiring full workup.
Explanation: The colony count is calculated by multiplying the number of colonies by the dilution factor of the loop. The dilution factor is the reciprocal of the loop volume (1 / 0.001 mL = 1000). Therefore, the count is 65 colonies * 1000 = 65,000 CFU/mL. A count in this range (typically >10,000 CFU/mL of a single uropathogen) is considered potentially significant and requires, at a minimum, identification of the organism. Counts below 10,000 CFU/mL are often considered contamination (A, B), while counts >100,000 CFU/mL are definitively significant (D). 65,000 CFU/mL falls into the intermediate category that requires further investigation.

Question 18

A laboratory isolates an Enterococcus species from a patient with endocarditis. To guide therapy, high-level aminoglycoside resistance (HLAR) screening is performed using agar plates containing high concentrations of gentamicin and streptomycin. No growth is observed on either plate. How should this result be interpreted in the context of antimicrobial susceptibility testing?

  1. The organism is susceptible to all aminoglycosides, and they can be used as monotherapy.
  2. The organism has intrinsic low-level resistance to aminoglycosides, so the test has no clinical value.
  3. The test is inconclusive and requires molecular methods to detect resistance genes.
  4. The absence of growth predicts synergy between a cell wall agent (like ampicillin or vancomycin) and an aminoglycoside. (correct answer)
Explanation: When you encounter questions about Enterococcus and aminoglycoside testing, remember that enterococci have unique resistance patterns that affect combination therapy strategies. Enterococcus species possess intrinsic low-level resistance to aminoglycosides due to poor drug uptake. However, when combined with cell wall-active agents like ampicillin or vancomycin, these agents disrupt the cell wall and allow aminoglycosides to penetrate effectively, creating bactericidal synergy. High-level aminoglycoside resistance (HLAR) screening specifically tests whether this synergy will work. The absence of growth on high-concentration gentamicin and streptomycin plates indicates the organism lacks HLAR mechanisms. This means the intrinsic resistance can still be overcome by combination therapy, predicting that synergy between a cell wall agent and an aminoglycoside will be effective. This is why answer D is correct. Answer A is wrong because aminoglycosides alone are never effective against enterococci due to intrinsic resistance—they require combination therapy. Answer B misinterprets the test's purpose; while enterococci do have intrinsic low-level resistance, HLAR testing specifically determines if synergy is possible, making it clinically valuable. Answer C is incorrect because the clear absence of growth provides definitive information—molecular methods aren't needed when phenotypic results are unambiguous. For microbiology exams, remember that Enterococcus antimicrobial testing often focuses on combination therapy potential rather than individual drug susceptibility. HLAR screening directly predicts synergy effectiveness, which is crucial for treating serious enterococcal infections like endocarditis.

Question 19

A thin, spreading, beta-hemolytic colony with a rough, 'ground-glass' appearance is isolated from a blood culture. The Gram stain shows large, spore-forming, Gram-positive rods in chains. The isolate is catalase positive and motile. Which of the following is the most likely identification?

  1. Clostridium perfringens
  2. Bacillus anthracis
  3. Bacillus cereus (correct answer)
  4. Listeria monocytogenes
Explanation: The described features are characteristic of the Bacillus cereus group. Large Gram-positive rods, spore formation, catalase positivity, beta-hemolysis, and motility are all key features. This must be differentiated from Bacillus anthracis (B), which is morphologically similar but is characteristically non-hemolytic and non-motile. Clostridium perfringens (A) is an obligate anaerobe and is catalase negative. Listeria monocytogenes (D) is a Gram-positive rod but is smaller, non-spore-forming, and has a distinct tumbling motility at room temperature.

Question 20

An automated identification system provides the following report for a Gram-negative rod isolated from a respiratory specimen from a cystic fibrosis patient:

Burkholderia cepacia complex - 95% probability Pandoraea apista - 4% probability

The laboratory has a policy to confirm this identification with a supplementary culture-based method due to the therapeutic and infection control implications.

Which selective medium would be most useful for confirming the identity as a member of the Burkholderia cepacia complex?

  1. Cystine-tellurite blood agar (CTBA).
  2. Cetrimide agar.
  3. Buffered charcoal yeast extract (BCYE) agar.
  4. Burkholderia cepacia selective agar (BCSA). (correct answer)
Explanation: When identifying Burkholderia cepacia complex in cystic fibrosis patients, confirmation is critical because these organisms are associated with severe complications and require strict infection control measures. The key is understanding that specialized selective media have been developed specifically for isolating and confirming particular bacterial groups. Burkholderia cepacia selective agar (BCSA) is specifically formulated to inhibit other respiratory flora while allowing B. cepacia complex organisms to grow and display their characteristic morphology and biochemical properties. This medium contains selective agents that suppress competing bacteria commonly found in respiratory specimens, making it the ideal confirmatory tool. Answer D is correct because BCSA provides both selectivity and the ability to observe distinctive colonial characteristics of B. cepacia complex. Answer A, cystine-tellurite blood agar (CTBA), is selective for Corynebacterium diphtheriae, not Burkholderia species. Answer B, cetrimide agar, selects for Pseudomonas aeruginosa and would not reliably differentiate B. cepacia complex from other gram-negative rods. Answer C, buffered charcoal yeast extract (BCYE) agar, is designed for Legionella species and lacks the selective properties needed for Burkholderia identification. For microbiology identification questions, remember that specialized selective media are named for their target organisms. When you see a question asking for confirmation of a specific bacterial complex, look for the medium that bears that organism's name—it's usually designed specifically for that purpose.