Microbiology Quiz: Multi Step Diagnostics
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Multi Step DiagnosticsQuestion 1 of 20

A culture from a cat bite wound grows a fastidious Gram-negative coccobacillus identified as Pasteurella multocida. The physician is considering treatment with ampicillin. The laboratory performs a nitrocefin-based beta-lactamase test on the isolate, which yields a positive result (a color change from yellow to red).

What is the most important clinical implication of this specific test result for this isolate?

Ampicillin is an appropriate therapy, as this test has low specificity for Pasteurella.
The organism is resistant to ampicillin; a beta-lactamase inhibitor combination should be used.
The test is likely a false positive and susceptibility testing via MIC should be performed.
The organism is resistant to all penicillins and cephalosporins.
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Microbiology Quiz

Microbiology Quiz: Multi Step Diagnostics

Practice Multi Step Diagnostics 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 Multi Step Diagnostics, 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

A culture from a cat bite wound grows a fastidious Gram-negative coccobacillus identified as Pasteurella multocida. The physician is considering treatment with ampicillin. The laboratory performs a nitrocefin-based beta-lactamase test on the isolate, which yields a positive result (a color change from yellow to red).

What is the most important clinical implication of this specific test result for this isolate?

  1. Ampicillin is an appropriate therapy, as this test has low specificity for Pasteurella.
  2. The organism is resistant to ampicillin; a beta-lactamase inhibitor combination should be used. (correct answer)
  3. The test is likely a false positive and susceptibility testing via MIC should be performed.
  4. The organism is resistant to all penicillins and cephalosporins.
Explanation: A positive beta-lactamase test indicates the organism produces an enzyme that hydrolyzes the beta-lactam ring of penicillin-class antibiotics like ampicillin, rendering them ineffective. While not all Pasteurella produce this enzyme, this particular isolate does. Therefore, ampicillin monotherapy would likely fail. A combination agent, such as amoxicillin-clavulanate, which includes a beta-lactamase inhibitor, would be a more appropriate choice.

Question 2

A sterile body fluid culture grows Gram-positive cocci in chains. An automated identification system reports the isolate as Staphylococcus lugdunensis with 99% confidence. The technologist notes the discrepancy between the Gram stain morphology and the final identification.

What is the most crucial and immediate manual test to perform to resolve this discrepancy?

  1. Perform a manual catalase test on the isolate. (correct answer)
  2. Report Staphylococcus lugdunensis based on the high confidence score.
  3. Request a new specimen due to potential pre-analytical error.
  4. Perform a PYR test to confirm the identification.
Explanation: There is a fundamental conflict between the observed morphology (chains, suggesting Streptococcus/Enterococcus) and the automated identification (Staphylococcus). The most basic, rapid, and critical test to differentiate these two genera is the catalase test. Staphylococcus species are catalase-positive, while Streptococcus and Enterococcus species are catalase-negative. This single test will immediately clarify which identification path is correct before proceeding with further tests or reporting a potentially erroneous result.

Question 3

A cerebrospinal fluid (CSF) sample is collected from a patient with clinical signs of acute meningitis. A Gram stain performed on the cytocentrifuged specimen is negative for organisms. The sample is cultured, and after 48 hours of incubation, no growth is observed. However, a multiplex PCR panel performed directly on the CSF specimen is positive for Neisseria meningitidis DNA.

What is the most probable explanation for the discrepancy between the culture and PCR results?

  1. The PCR result is a false positive due to contamination in the laboratory.
  2. The patient received empiric antibiotic therapy prior to the lumbar puncture. (correct answer)
  3. The organism is a rare, non-culturable variant of Neisseria meningitidis.
  4. The Gram stain was performed incorrectly, leading to a false-negative result.
Explanation: PCR detects nucleic acid, which can be present even if the bacteria are no longer viable. Administration of antibiotics before specimen collection can kill bacteria, rendering the culture negative. However, the bacterial DNA remains and can be detected by the highly sensitive PCR method. While other options are theoretically possible, prior antibiotic administration is the most common clinical reason for culture-negative, PCR-positive results in bacterial meningitis cases.

Question 4

A virology lab performs a plaque reduction neutralization test (PRNT) to determine the titer of neutralizing antibodies in a patient's serum. Serial dilutions of serum are incubated with virus and then added to cell monolayers. The control wells (virus, no serum) develop an average of 200 plaques. The results for the serum dilutions are as follows: 1:20 dilution yields 10 plaques; 1:40 dilution yields 45 plaques; 1:80 dilution yields 95 plaques; 1:160 dilution yields 160 plaques.

The PRNT₅₀ titer is defined as the reciprocal of the highest serum dilution that reduces the plaque count by 50% or more. What is the PRNT₅₀ titer of this serum sample?

  1. 40
  2. 80 (correct answer)
  3. 100
  4. 200
Explanation: First, calculate the 50% reduction endpoint. The control has 200 plaques, so a 50% reduction means a plaque count of 100 or less. Second, examine the dilutions to find the highest (most dilute) sample that meets this criterion. The 1:20 dilution (10 plaques), 1:40 dilution (45 plaques), and 1:80 dilution (95 plaques) all reduce the count by at least 50%. The 1:160 dilution (160 plaques) does not. The titer is the reciprocal of the highest dilution that works, which is 1:80. Therefore, the titer is 80.

Question 5

A sputum sample is processed for mycobacterial investigation. The direct acid-fast stain (Ziehl-Neelsen) is positive, showing pink bacilli. The specimen is decontaminated and inoculated onto both a Löwenstein-Jensen (LJ) solid medium slant and into a liquid-based automated culture system. The liquid culture system flags positive for growth after 10 days. At the same time point, the LJ slant shows no visible colonies.

What is the best explanation for this difference in culture results?

  1. The initial acid-fast stain was a false positive.
  2. The organism is a non-tuberculous mycobacterium that cannot grow on LJ medium.
  3. Liquid culture media generally support more rapid detection of mycobacterial growth than solid media. (correct answer)
  4. The LJ slant was likely contaminated with an inhibitor during processing.
Explanation: This scenario represents a typical workflow in mycobacteriology. Liquid culture systems, such as the mycobacteria growth indicator tube (MGIT) system, are known to reduce the time to detection for Mycobacterium tuberculosis complex compared to traditional solid media like LJ agar. It is common for liquid cultures to be positive in 1-3 weeks, while growth on LJ may take 3-8 weeks to become visible. The positive liquid culture confirms the validity of the acid-fast stain.

Question 6

An isolate of Staphylococcus aureus from a wound infection undergoes antimicrobial susceptibility testing. The initial automated report shows the organism is resistant to erythromycin but susceptible to clindamycin. Due to this pattern, a D-test is performed by placing an erythromycin disk and a clindamycin disk in close proximity on a Mueller-Hinton agar plate.

The D-test shows a distinct flattening of the clindamycin inhibition zone adjacent to the erythromycin disk. How should the laboratory report the clindamycin susceptibility result?

  1. Report clindamycin as Susceptible, as per the initial automated result.
  2. Report clindamycin as Intermediate and recommend alternative therapies.
  3. Report clindamycin as Resistant due to inducible resistance. (correct answer)
  4. Request a new specimen as the test results are considered indeterminate.
Explanation: The flattening of the clindamycin zone in the shape of a 'D' indicates inducible resistance, mediated by the erm gene. This gene confers resistance to erythromycin and can be induced to confer resistance to clindamycin. Although the organism appears susceptible to clindamycin in vitro without the presence of an inducer (erythromycin), treatment with clindamycin could fail due to in vivo induction of resistance. Therefore, the result must be edited to report clindamycin as Resistant.

Question 7

A patient presents with a fever and maculopapular rash. Serological testing for a specific viral infection is ordered. The acute serum sample, taken on day 3 of illness, shows an IgM titer of 1:20 and an IgG titer of 1:32. A convalescent serum sample is drawn 3 weeks later.

Which of the following convalescent serum results would provide the most definitive evidence of a recent, primary infection?

  1. IgM < 1:10, IgG 1:32
  2. IgM 1:40, IgG 1:512 (correct answer)
  3. IgM 1:20, IgG 1:64
  4. IgM < 1:10, IgG 1:1024
Explanation: The gold standard for serological diagnosis of a recent primary infection is the demonstration of seroconversion, defined as at least a four-fold rise in IgG titer between acute and convalescent samples. In this case, the acute IgG is 1:32. A four-fold rise would be 1:128 or greater. Option B shows a rise from 1:32 to 1:512 (a 16-fold rise) and a persistent or rising IgM titer, which is classic for a primary infection in the convalescent phase. Option D shows a sufficient IgG rise, but the disappearance of IgM makes B a slightly better representation of a 3-week convalescent sample.

Question 8

An aspirate from a deep abdominal abscess is Gram stained, revealing a mix of Gram-positive cocci and Gram-negative rods. The specimen is plated on aerobic blood agar, MacConkey agar, and anaerobic blood agar. After 48 hours, the aerobic plates show no growth, but the anaerobic plate shows heavy growth of multiple colony types.

What is the most appropriate next action for the microbiologist to take?

  1. Report 'No growth' based on the negative aerobic plates.
  2. Re-incubate the aerobic plates for an additional 72 hours.
  3. Subculture individual colony morphotypes from the anaerobic plate for further identification. (correct answer)
  4. Perform an acid-fast stain on the original specimen to look for mycobacteria.
Explanation: The diagnostic findings point to an infection with obligate anaerobes. The specimen source (deep abscess) is classic for anaerobic infections. The Gram stain showed mixed flora. Crucially, growth occurred only on the plate incubated anaerobically. This confirms the presence of obligate anaerobes. The next logical step is to isolate the different bacteria growing on the anaerobic plate by subculturing them to new media to obtain pure cultures for identification and susceptibility testing.

Question 9

A Gram-negative, facultatively anaerobic rod isolated from a wound is oxidase-negative. To further characterize its metabolism, it is inoculated into MR-VP broth. After incubation, the Methyl Red (MR) test is positive (bright red), and the Voges-Proskauer (VP) test is negative (no color change).

These results indicate that the organism ferments glucose primarily via which metabolic pathway?

  1. The 2,3-butanediol pathway
  2. The pentose phosphate pathway
  3. The Entner-Doudoroff pathway
  4. The mixed acid fermentation pathway (correct answer)
Explanation: When you encounter MR-VP test results in microbiology, you're looking at how bacteria ferment glucose and what end products they produce. These classic differential tests help distinguish between closely related Gram-negative rods, particularly within the Enterobacteriaceae family. The mixed acid fermentation pathway produces multiple acidic end products including acetic acid, lactic acid, succinic acid, and formic acid. This pathway generates enough acid to lower the pH significantly, which is exactly what the Methyl Red test detects. A positive MR test (bright red color) confirms substantial acid production and low pH. Additionally, mixed acid fermentation does not produce acetoin or 2,3-butanediol as major end products, explaining why the VP test remains negative. Choice A is incorrect because the 2,3-butanediol pathway would produce a positive VP test, as this pathway generates acetoin and 2,3-butanediol - the compounds the Voges-Proskauer test specifically detects. Choice B is wrong because the pentose phosphate pathway is primarily involved in generating NADPH and ribose-5-phosphate, not fermentation end products that would affect MR-VP results. Choice D is incorrect because the Entner-Doudoroff pathway is an alternative glucose breakdown pathway found mainly in certain Gram-positive bacteria and some pseudomonads, not the typical fermentation pathway being tested here. Remember this pattern: MR positive + VP negative = mixed acid fermentation, while MR negative + VP positive = 2,3-butanediol fermentation. This distinction is crucial for identifying enterobacteria like E. coli versus Enterobacter species.

Question 10

A throat swab from a child with pharyngitis is cultured on a blood agar plate. The plate shows small colonies surrounded by a wide zone of complete hemolysis. A Gram stain of the colonies reveals Gram-positive cocci in chains, and a catalase test is negative. A bacitracin susceptibility disk placed on the plate shows a distinct zone of inhibition. A PYR test performed on the isolate is also positive.

The combination of these test results provides a definitive identification of which organism?

  1. Streptococcus pyogenes (Group A) (correct answer)
  2. Streptococcus agalactiae (Group B)
  3. Enterococcus faecalis
  4. Streptococcus pneumoniae
Explanation: This is a classic multi-step identification algorithm. Catalase-negative, Gram-positive cocci are streptococci. Beta-hemolysis narrows the possibilities. Bacitracin susceptibility is a presumptive test for S. pyogenes. The positive PYR (pyrrolidonyl arylamidase) test confirms the identification, as S. pyogenes is PYR-positive, while other beta-hemolytic streptococci like S. agalactiae are negative. Enterococcus can be PYR-positive but is not typically beta-hemolytic and is not the primary target in a pharyngitis workup.

Question 11

A patient with suspected sepsis has two sets of blood cultures drawn from two different venipuncture sites. After 24 hours, the aerobic bottle from the first set flags positive. A Gram stain shows Gram-positive cocci in clusters. The remaining three bottles (aerobic and anaerobic from set 2, anaerobic from set 1) remain negative after 5 days. The isolate from the single positive bottle is identified as coagulase-negative Staphylococcus.

What is the most likely clinical interpretation of these findings?

  1. The patient has a true bacteremia caused by coagulase-negative Staphylococcus.
  2. The result represents specimen contamination during phlebotomy. (correct answer)
  3. The second blood culture set was drawn after the patient started effective antibiotic therapy.
  4. The anaerobic bottles contained an inhibitor that prevented the organism's growth.
Explanation: The interpretation of blood cultures requires analyzing the organism identity and the pattern of positivity. Coagulase-negative staphylococci are common skin flora and frequent contaminants. A true bloodstream infection is typically characterized by growth in multiple bottles across multiple sets. The finding of a common skin commensal in only one of four bottles is the classic pattern for contamination during specimen collection.

Question 12

A cerebrospinal fluid (CSF) sample is collected from a patient with clinical signs of acute meningitis. A Gram stain performed on the cytocentrifuged specimen is negative for organisms. The sample is cultured, and after 48 hours of incubation, no growth is observed. However, a multiplex PCR panel performed directly on the CSF specimen is positive for Neisseria meningitidis DNA.

What is the most probable explanation for the discrepancy between the culture and PCR results?

  1. The PCR result is a false positive due to contamination in the laboratory.
  2. The patient received empiric antibiotic therapy prior to the lumbar puncture. (correct answer)
  3. The organism is a rare, non-culturable variant of Neisseria meningitidis.
  4. The Gram stain was performed incorrectly, leading to a false-negative result.
Explanation: PCR detects nucleic acid, which can be present even if the bacteria are no longer viable. Administration of antibiotics before specimen collection can kill bacteria, rendering the culture negative. However, the bacterial DNA remains and can be detected by the highly sensitive PCR method. While other options are theoretically possible, prior antibiotic administration is the most common clinical reason for culture-negative, PCR-positive results in bacterial meningitis cases.

Question 13

A patient with suspected sepsis has two sets of blood cultures drawn from two different venipuncture sites. After 24 hours, the aerobic bottle from the first set flags positive. A Gram stain shows Gram-positive cocci in clusters. The remaining three bottles (aerobic and anaerobic from set 2, anaerobic from set 1) remain negative after 5 days. The isolate from the single positive bottle is identified as coagulase-negative Staphylococcus.

What is the most likely clinical interpretation of these findings?

  1. The patient has a true bacteremia caused by coagulase-negative Staphylococcus.
  2. The result represents specimen contamination during phlebotomy. (correct answer)
  3. The second blood culture set was drawn after the patient started effective antibiotic therapy.
  4. The anaerobic bottles contained an inhibitor that prevented the organism's growth.
Explanation: The interpretation of blood cultures requires analyzing the organism identity and the pattern of positivity. Coagulase-negative staphylococci are common skin flora and frequent contaminants. A true bloodstream infection is typically characterized by growth in multiple bottles across multiple sets. The finding of a common skin commensal in only one of four bottles is the classic pattern for contamination during specimen collection.

Question 14

A throat swab from a child with pharyngitis is cultured on a blood agar plate. The plate shows small colonies surrounded by a wide zone of complete hemolysis. A Gram stain of the colonies reveals Gram-positive cocci in chains, and a catalase test is negative. A bacitracin susceptibility disk placed on the plate shows a distinct zone of inhibition. A PYR test performed on the isolate is also positive.

The combination of these test results provides a definitive identification of which organism?

  1. Streptococcus pyogenes (Group A) (correct answer)
  2. Streptococcus agalactiae (Group B)
  3. Enterococcus faecalis
  4. Streptococcus pneumoniae
Explanation: This is a classic multi-step identification algorithm. Catalase-negative, Gram-positive cocci are streptococci. Beta-hemolysis narrows the possibilities. Bacitracin susceptibility is a presumptive test for S. pyogenes. The positive PYR (pyrrolidonyl arylamidase) test confirms the identification, as S. pyogenes is PYR-positive, while other beta-hemolytic streptococci like S. agalactiae are negative. Enterococcus can be PYR-positive but is not typically beta-hemolytic and is not the primary target in a pharyngitis workup.

Question 15

A biopsy of a skin lesion from a gardener is cultured for fungi. On Sabouraud dextrose agar at 25°C, a mold with delicate hyphae and ovoid conidia arranged in a 'daisy-head' pattern at the apex of conidiophores is observed. A subculture of this mold onto a rich medium incubated at 37°C results in the growth of small, cigar-shaped budding yeast cells.

This combination of thermal dimorphism and distinct microscopic morphologies is characteristic of which fungus?

  1. Coccidioides immitis
  2. Histoplasma capsulatum
  3. Aspergillus fumigatus
  4. Sporothrix schenckii (correct answer)
Explanation: This question requires integrating multiple pieces of information. The conversion from a mold form at 25°C to a yeast form at 37°C defines a dimorphic fungus. The specific morphology of the mold phase ('daisy-head' conidia) and the yeast phase ('cigar-shaped' bodies), combined with the clinical history of a gardener (linking to 'rose gardener's disease'), are all classic features of Sporothrix schenckii.

Question 16

A liquid stool sample from an immunocompromised patient with watery diarrhea is examined. A wet mount is unremarkable. A trichrome stained smear is prepared, which reveals several spherical, pink-staining structures approximately 5 µm in diameter. No distinct internal morphology is visible in these structures.

To confirm the identity of the suspected pathogen, which specialized stain should be performed next?

  1. Giemsa stain
  2. Calcofluor white stain
  3. Iron hematoxylin stain
  4. Modified acid-fast stain (correct answer)
Explanation: When you encounter an immunocompromised patient with diarrhea and see small, pink-staining spherical structures on trichrome stain, you're likely dealing with Cryptosporidium or Cyclospora - both important opportunistic parasites that cause prolonged diarrhea in immunocompromised hosts. The key diagnostic clue here is that trichrome stain shows these organisms but reveals no internal detail. This is characteristic of these coccidian parasites, which have acid-fast properties in their oocyst walls that make them resistant to standard staining techniques. Answer D is correct because the modified acid-fast stain is the gold standard for identifying Cryptosporidium and Cyclospora. This stain exploits the acid-fast nature of their oocyst walls, making them appear bright pink-red against a blue background, allowing definitive identification. Answer A (Giemsa) is used primarily for blood parasites like Plasmodium and some protozoa, but won't adequately stain these intestinal coccidian oocysts. Answer B (Calcofluor white) is a fungal stain that binds chitin and cellulose in fungal cell walls - irrelevant for parasitic protozoa. Answer C (Iron hematoxylin) is excellent for demonstrating internal morphology of parasites like Entamoeba, but these coccidian oocysts require the acid-fast technique for proper visualization. Study tip: Remember the triad for immunocompromised patients with diarrhea: small pink spheres on trichrome + no internal detail visible + need for acid-fast confirmation = Cryptosporidium or Cyclospora. This pattern appears frequently on microbiology exams.

Question 17

A sterile body fluid culture grows Gram-positive cocci in chains. An automated identification system reports the isolate as Staphylococcus lugdunensis with 99% confidence. The technologist notes the discrepancy between the Gram stain morphology and the final identification.

What is the most crucial and immediate manual test to perform to resolve this discrepancy?

  1. Perform a manual catalase test on the isolate. (correct answer)
  2. Report Staphylococcus lugdunensis based on the high confidence score.
  3. Request a new specimen due to potential pre-analytical error.
  4. Perform a PYR test to confirm the identification.
Explanation: There is a fundamental conflict between the observed morphology (chains, suggesting Streptococcus/Enterococcus) and the automated identification (Staphylococcus). The most basic, rapid, and critical test to differentiate these two genera is the catalase test. Staphylococcus species are catalase-positive, while Streptococcus and Enterococcus species are catalase-negative. This single test will immediately clarify which identification path is correct before proceeding with further tests or reporting a potentially erroneous result.

Question 18

An isolate from a cystic fibrosis patient's sputum is identified by two different methods. MALDI-TOF mass spectrometry provides a high-confidence identification of Burkholderia cepacia complex. In parallel, an automated biochemical identification panel identifies the organism as Stenotrophomonas maltophilia. Both are non-fermenting, Gram-negative rods.

Which single, rapid biochemical test would be most effective in resolving this discrepancy?

  1. Motility test
  2. Urease test
  3. Oxidase test (correct answer)
  4. DNAse test
Explanation: When two reliable identification systems provide conflicting results, a simple, rapid test that differentiates the two possibilities is needed. A key difference between these two organisms is their oxidase reaction. Burkholderia cepacia complex is typically oxidase-positive, whereas Stenotrophomonas maltophilia is characteristically oxidase-negative. This test provides a quick and reliable way to determine which identification is more likely correct.

Question 19

A Gram-negative, facultatively anaerobic rod isolated from a wound is oxidase-negative. To further characterize its metabolism, it is inoculated into MR-VP broth. After incubation, the Methyl Red (MR) test is positive (bright red), and the Voges-Proskauer (VP) test is negative (no color change).

These results indicate that the organism ferments glucose primarily via which metabolic pathway?

  1. The 2,3-butanediol pathway
  2. The pentose phosphate pathway
  3. The Entner-Doudoroff pathway
  4. The mixed acid fermentation pathway (correct answer)
Explanation: When you encounter MR-VP test results in microbiology, you're looking at how bacteria ferment glucose and what end products they produce. These classic differential tests help distinguish between closely related Gram-negative rods, particularly within the Enterobacteriaceae family. The mixed acid fermentation pathway produces multiple acidic end products including acetic acid, lactic acid, succinic acid, and formic acid. This pathway generates enough acid to lower the pH significantly, which is exactly what the Methyl Red test detects. A positive MR test (bright red color) confirms substantial acid production and low pH. Additionally, mixed acid fermentation does not produce acetoin or 2,3-butanediol as major end products, explaining why the VP test remains negative. Choice A is incorrect because the 2,3-butanediol pathway would produce a positive VP test, as this pathway generates acetoin and 2,3-butanediol - the compounds the Voges-Proskauer test specifically detects. Choice B is wrong because the pentose phosphate pathway is primarily involved in generating NADPH and ribose-5-phosphate, not fermentation end products that would affect MR-VP results. Choice D is incorrect because the Entner-Doudoroff pathway is an alternative glucose breakdown pathway found mainly in certain Gram-positive bacteria and some pseudomonads, not the typical fermentation pathway being tested here. Remember this pattern: MR positive + VP negative = mixed acid fermentation, while MR negative + VP positive = 2,3-butanediol fermentation. This distinction is crucial for identifying enterobacteria like E. coli versus Enterobacter species.

Question 20

An aspirate from a deep abdominal abscess is Gram stained, revealing a mix of Gram-positive cocci and Gram-negative rods. The specimen is plated on aerobic blood agar, MacConkey agar, and anaerobic blood agar. After 48 hours, the aerobic plates show no growth, but the anaerobic plate shows heavy growth of multiple colony types.

What is the most appropriate next action for the microbiologist to take?

  1. Report 'No growth' based on the negative aerobic plates.
  2. Re-incubate the aerobic plates for an additional 72 hours.
  3. Subculture individual colony morphotypes from the anaerobic plate for further identification. (correct answer)
  4. Perform an acid-fast stain on the original specimen to look for mycobacteria.
Explanation: The diagnostic findings point to an infection with obligate anaerobes. The specimen source (deep abscess) is classic for anaerobic infections. The Gram stain showed mixed flora. Crucially, growth occurred only on the plate incubated anaerobically. This confirms the presence of obligate anaerobes. The next logical step is to isolate the different bacteria growing on the anaerobic plate by subculturing them to new media to obtain pure cultures for identification and susceptibility testing.