COT Microbiology 4 — Questions and Answers
Question 1: A microbiology technician is examining a culture that produces a green-blue pigment and has a fruity odor. The organism is gram-negative, non-fermentative, and grows at 42°C. Which organism is most likely responsible?
- Pseudomonas aeruginosa (Correct answer)
- Escherichia coli
- Klebsiella pneumoniae
- Acinetobacter baumannii
Correct answer: Pseudomonas aeruginosa
Pseudomonas aeruginosa characteristically produces pyocyanin (blue-green pigment) and has a grape-like or fruity odor due to aminoacetophenone. It is gram-negative, non-fermentative, oxidase-positive, and can grow at 42°C, which distinguishes it from other Pseudomonas species.
Question 2: In the Kirby-Bauer disk diffusion test, the zone of inhibition around an antibiotic disk is measured and compared to CLSI breakpoints. If a zone measures exactly at the intermediate breakpoint, how should the organism's susceptibility be reported?
- Susceptible
- Intermediate (Correct answer)
- Resistant
- Non-susceptible
Correct answer: Intermediate
When the zone of inhibition falls exactly at the intermediate breakpoint value, the organism is reported as Intermediate (I). This category suggests the drug may be clinically effective in body sites where it achieves high concentrations, but the outcome is uncertain in standard dosing regimens.
Question 3: A spinal fluid specimen is received in the microbiology laboratory. The Gram stain reveals gram-positive diplococci with a surrounding clear halo. Which additional test would BEST confirm the presumptive identification?
- Optochin susceptibility and bile solubility (Correct answer)
- Coagulase test
- Oxidase test
- Urease test
Correct answer: Optochin susceptibility and bile solubility
Gram-positive diplococci with a clear halo (capsule) in CSF are highly suspicious for Streptococcus pneumoniae. Optochin susceptibility (P disk) and bile solubility are the classic confirmatory tests for S. pneumoniae. Coagulase identifies Staphylococcus aureus, oxidase differentiates gram-negative organisms, and urease is used for organisms like Cryptococcus or Helicobacter.
Question 4: A 28-year-old patient presents with a painless ulcer on the genitalia. Dark-field microscopy of the lesion exudate reveals motile, corkscrew-shaped organisms with tight coils. Culture attempts on standard laboratory media are unsuccessful. What is the most likely diagnosis?
- Chancroid caused by Haemophilus ducreyi
- Primary syphilis caused by Treponema pallidum (Correct answer)
- Genital herpes caused by HSV-2
- Lymphogranuloma venereum caused by Chlamydia trachomatis
Correct answer: Primary syphilis caused by Treponema pallidum
Treponema pallidum, the causative agent of primary syphilis, presents as a painless chancre and is visualized as tightly coiled, corkscrew-shaped spirochetes on dark-field microscopy. Crucially, T. pallidum cannot be cultured on standard laboratory media. Chancroid (H. ducreyi) causes painful ulcers; HSV-2 causes vesicular lesions; and Chlamydia is an intracellular organism.
Question 5: During a urinary tract infection workup, a urine culture grows two distinct colony morphologies on blood agar after 24 hours. The laboratory protocol requires a minimum of how many colonies per milliliter (CFU/mL) to be considered clinically significant for a clean-catch midstream urine from a symptomatic female patient?
- 100,000 CFU/mL (10⁵) (Correct answer)
- 10,000 CFU/mL (10⁴)
- 1,000 CFU/mL (10³)
- 500 CFU/mL
Correct answer: 100,000 CFU/mL (10⁵)
The traditional threshold for a clinically significant UTI in a symptomatic female is ≥100,000 CFU/mL (10⁵) from a clean-catch midstream urine specimen. This criterion, established by Kass, helps distinguish true infection from contamination. Lower thresholds (10²–10⁴) may apply in specific scenarios such as catheterized specimens or symptomatic males, but 10⁵ remains the standard for symptomatic females with clean-catch specimens.
Question 6: A stool culture from a patient with bloody diarrhea grows a colorless colony on MacConkey agar that is sorbitol-negative. Further testing confirms the organism is O157:H7 serotype. Which virulence mechanism is PRIMARILY responsible for the life-threatening complication of hemolytic uremic syndrome (HUS)?
- Invasion of the intestinal epithelium causing direct tissue destruction
- Production of Shiga toxin that damages renal endothelial cells (Correct answer)
- Release of exotoxin A that inhibits protein synthesis in hepatocytes
- Formation of biofilm that obstructs renal tubules
Correct answer: Production of Shiga toxin that damages renal endothelial cells
E. coli O157:H7 produces Shiga toxins (Stx1 and Stx2), which are absorbed into the bloodstream and bind to Gb3 receptors on renal glomerular endothelial cells. The toxin inhibits protein synthesis by inactivating the 60S ribosomal subunit, causing endothelial cell death, platelet aggregation, and the triad of HUS: microangiopathic hemolytic anemia, thrombocytopenia, and acute renal failure.
COT Microbiology 4
A microbiology technician is examining a culture that produces a green-blue pigment and has a fruity odor.
The organism is gram-negative, non-fermentative, and grows at 42°C.
Which organism is most likely responsible?