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MedExams

Urine

18 ASCP MLS practice questions on Urine, part of Microbiology. The first 18 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Hard

    A 21-year-old woman has had dysuria and frequency for 2 days. Urinalysis: leukocyte esterase positive, nitrite negative, WBC 20-30/HPF, few squamous epithelial cells. A clean-catch midstream urine is plated with a 0.001 mL calibrated loop. At 24 hours, sheep blood agar shows 60 white, nonhemolytic colonies of a single type, and MacConkey agar has no growth. The colonies are catalase-positive gram-positive cocci in clusters, slide and tube coagulase negative, with a 9 mm zone around a 5 µg novobiocin disk (resistant 16 mm or less). How should the culture be reported?

    1. AS. saprophyticus, 6 x 10^4 CFU/mL
    2. BMixed skin flora; recollect the specimen
    3. CCoagulase-negative staph, 6 x 10^3 CFU/mL
    4. DS. epidermidis, 6 x 10^4 CFU/mL, contaminant
    Show answer

    Answer: A · S. saprophyticus, 6 x 10^4 CFU/mL

    With a 0.001 mL loop each colony equals 1,000 CFU/mL, so 60 colonies are 6 x 10^4 CFU/mL of a single organism. Catalase-positive, coagulase-negative staphylococci that are resistant to novobiocin are S. saprophyticus, a common cause of cystitis in young, sexually active women. In a symptomatic young woman with pyuria, a pure culture at this count is significant. A negative nitrite is common, because S. saprophyticus seldom reduces nitrate to nitrite. Novobiocin-susceptible S. epidermidis, by contrast, is more often a contaminant.

    • B: A single colony type with pyuria is not mixed skin flora.
    • C: That count uses a 0.01 mL loop factor; a 0.001 mL loop multiplies by 1,000.
    • D: S. epidermidis is novobiocin susceptible; this isolate is resistant.
  2. Question 2 · Hard

    A 58-year-old woman receiving chemotherapy for acute myeloid leukemia has a fever of 38.7 °C and new dysuria. Her neutrophil count is 0.1 x 10^3/µL [0.1 x 10^9/L]. A clean-catch urine is tested within 30 minutes: leukocyte esterase negative, nitrite positive; sediment WBC 0-2/HPF, many bacteria, no squamous epithelial cells. Culture with a 0.001 mL loop grows 150 colonies of a single lactose-fermenting gram-negative rod, later identified as Klebsiella pneumoniae. Which interpretation best fits these results?

    1. AContaminated specimen; collect another
    2. BUTI; neutropenia explains the absent pyuria
    3. CBacterial growth from a testing delay
    4. DAsymptomatic bacteriuria; no report needed
    Show answer

    Answer: B · UTI; neutropenia explains the absent pyuria

    The culture shows more than 10^5 CFU/mL (150 colonies x 1,000) of a single uropathogen, and the positive nitrite fits gram-negative rods that reduce nitrate. The leukocyte esterase test and the sediment white cell count both depend on neutrophils, and a patient with almost no circulating neutrophils cannot produce pyuria. In neutropenic patients a urinary tract infection may therefore show bacteriuria without white cells, and fever with dysuria makes this a symptomatic infection. A pure culture without squamous cells argues against contamination, and testing within 30 minutes excludes growth during storage.

    • A: A pure culture with no squamous cells does not suggest contamination.
    • C: The urine was tested within 30 minutes, so growth during storage is excluded.
    • D: She has fever and dysuria, so this is a symptomatic infection.
  3. Question 3 · Medium

    A clean-catch urine from a 24-year-old woman with dysuria was plated with a 0.001 mL calibrated loop. The blood agar plate after overnight incubation is shown. What colony count should be reported?

    Figure: MedExams
    Blood agar plate with separate, evenly spread white colonies of one type and no confluent growth.
    1. A27,000 CFU/mL
    2. B27 CFU/mL
    3. C2,700 CFU/mL
    4. D270,000 CFU/mL
    Show answer

    Answer: A · 27,000 CFU/mL

    There are 27 colonies on the plate. A 0.001 mL loop delivers one thousandth of a millilitre, so each colony represents 1,000 CFU/mL: 27 x 1,000 = 27,000 CFU/mL (2.7 x 10^4 CFU/mL, or 2.7 x 10^7 CFU/L). Using the factor for a 0.01 mL loop (x 100) gives 2,700, and an extra factor of 10 gives 270,000. Count the whole plate, and dip the loop vertically so that it picks up the stated volume.

    • B: 27 CFU/mL ignores the volume the loop delivers.
    • C: 2,700 CFU/mL uses the factor for a 0.01 mL loop (x 100).
    • D: 270,000 CFU/mL multiplies by 10,000 instead of 1,000.
  4. Question 4 · Hard

    A straight-catheter urine from a 68-year-old woman with fever was plated with a 0.001 mL loop, and three colony types grew: large gray (A), smaller white (B) and pinpoint yellow (C), as shown. For catheter urine, the laboratory identifies and tests each organism at 10^4 CFU/mL or more if no more than two reach that level, and reports mixed growth with a request for a new specimen if three or more do. What should be done?

    Figure: MedExams
    Blood agar plate with separate colonies of three sizes: many large gray colonies, fewer smaller white colonies and a few pinpoint yellow colonies.
    1. AIdentify and test types A and B
    2. BReport no significant growth
    3. CIdentify and test all three types
    4. DReport mixed growth; request a new specimen
    Show answer

    Answer: A · Identify and test types A and B

    Counting gives 21 type A, 14 type B and 4 type C colonies. With the 0.001 mL loop each colony is 1,000 CFU/mL, so A is 21,000, B is 14,000 and C is 4,000 CFU/mL. Only A and B reach 10^4 CFU/mL, so by the stated policy both are identified and tested, and C is reported by count without workup. Calling C significant would wrongly trigger the mixed-growth rule, and using the 0.01 mL loop factor would put every count below 10^4.

    • B: A and B exceed 10^4 CFU/mL when the 0.001 mL loop factor (x 1,000) is used.
    • C: Type C, at 4,000 CFU/mL, is below the 10^4 cut-off.
    • D: Only two organisms reach 10^4 CFU/mL, so the mixed-growth rule does not apply.
  5. Question 5 · Hard

    Urinalysis: pH 8.5, many triple phosphate (struvite) crystals and a strong ammonia odor. Culture: more than 100,000 CFU/mL of a swarming, oxidase-negative, lactose-negative gram-negative rod. The bacterial enzyme that best explains the urinalysis findings is:

    1. ANitrate reductase
    2. BUrease
    3. CTryptophanase
    4. DPhenylalanine deaminase
    Show answer

    Answer: B · Urease

    Proteus mirabilis produces a strong urease that splits urea into ammonia and carbon dioxide. The ammonia raises the urine pH and gives the odor, and magnesium ammonium phosphate (struvite) then precipitates as crystals and can form staghorn stones. Swarming, a negative oxidase and no lactose fermentation fit Proteus. Stones can harbor the organism and make the infection hard to clear.

    • A: Nitrate reductase produces nitrite but does not make the urine alkaline.
    • C: Tryptophanase makes indole, and P. mirabilis is indole negative.
    • D: Phenylalanine deaminase is positive in Proteus, but urea, not phenylalanine, is the source of the ammonia in urine.
  6. Question 6 · Hard

    A suprapubic bladder aspirate from a febrile 2-month-old infant is plated with a 0.01 mL loop and grows 3 colonies of Escherichia coli. The correct count and interpretation are:

    1. A300 CFU/mL; significant growth
    2. B30 CFU/mL; below a significant level
    3. C300 CFU/mL; below a significant level
    4. D3,000 CFU/mL; significant growth

    Answer and explanation in LearnFree account

  7. Question 7 · Easy

    Which coagulase-negative staphylococcus is a common cause of cystitis in young, sexually active women?

    1. AStaphylococcus epidermidis
    2. BStaphylococcus haemolyticus
    3. CStaphylococcus saprophyticus
    4. DStaphylococcus lugdunensis

    Answer and explanation in LearnFree account

  8. Question 8 · Medium

    At her first prenatal visit at 12 weeks, a woman with no urinary symptoms has two consecutive clean-catch urines that each grow more than 100,000 CFU/mL of Escherichia coli. This finding is best classified as:

    1. AAcute pyelonephritis that needs admission
    2. BAsymptomatic bacteriuria that should be treated
    3. CAsymptomatic bacteriuria; no treatment needed
    4. DContamination that needs a new specimen

    Answer and explanation in LearnFree account

  9. Question 9 · Hard

    A clean-catch urine from a woman with dysuria is plated with a 0.001 mL loop. Next morning the blood agar plate has 35 colonies of Escherichia coli and 4 colonies of diphtheroids. Which colony count should be reported for E. coli?

    1. A35,000 CFU/mL
    2. B350,000 CFU/mL
    3. C3,500 CFU/mL
    4. D39,000 CFU/mL

    Answer and explanation in LearnFree account

  10. Question 10 · Medium

    On a Gram stain of uncentrifuged urine, at least one bacterium per oil-immersion field corresponds to a colony count of about:

    1. A1,000 CFU/mL
    2. B1,000,000 CFU/mL
    3. C10,000 CFU/mL
    4. D100,000 CFU/mL

    Answer and explanation in LearnFree account

  11. Question 11 · Easy

    For a routine clean-catch midstream urine culture, the calibrated loop most often used delivers:

    1. A0.001 mL
    2. B0.1 mL
    3. C0.01 mL
    4. D1.0 mL

    Answer and explanation in LearnFree account

  12. Question 12 · Hard

    A 78-year-old man with a long-term catheter has urine pH 8.0, many white cells and gram-positive rods on Gram stain. Routine culture is negative at 24 hours, but tiny colonies appear at 48 hours. They are catalase positive and strongly urease positive within minutes. The most likely organism is:

    1. ALactobacillus species
    2. BCorynebacterium striatum
    3. CProteus mirabilis
    4. DCorynebacterium urealyticum

    Answer and explanation in LearnFree account

  13. Question 13 · Medium

    A first-morning urine is positive for leukocyte esterase and negative for nitrite. The patient takes no vitamin C. Culture grows more than 100,000 CFU/mL of Enterococcus faecalis. The negative nitrite is best explained because enterococci:

    1. ANeed an alkaline urine to form nitrite
    2. BGrow too slowly to form nitrite overnight
    3. CReduce nitrite further to nitrogen gas
    4. DDo not reduce nitrate to nitrite

    Answer and explanation in LearnFree account

  14. Question 14 · Hard

    A 66-year-old man who had pulmonary tuberculosis years ago has persistent hematuria and pyuria with an acidic urine (pH 5.5). Repeated routine urine cultures show no growth at 48 hours. Which organism is most likely?

    1. AMycobacterium tuberculosis
    2. BEnterococcus faecalis
    3. CCorynebacterium urealyticum
    4. DProteus mirabilis

    Answer and explanation in LearnFree account

  15. Question 15 · Hard

    A clinic tested a urine right after collection: leukocyte esterase and nitrite negative, no white cells in the sediment. The specimen then sat at room temperature for 6 hours before plating, and three gram-negative rods each grew above 100,000 CFU/mL. What is the most likely explanation?

    1. AInfection with organisms that do not form nitrite
    2. BAntibiotic treatment before the specimen was taken
    3. CBacteria multiplied in the urine after collection
    4. DAn early infection that has not yet caused pyuria

    Answer and explanation in LearnFree account

  16. Question 16 · Medium

    A clean-catch urine from an outpatient grows a lactobacillus, a viridans streptococcus and a coagulase-negative staphylococcus, each at 10,000 to 20,000 CFU/mL, with none predominant. What is the best action?

    1. AReport the staphylococcus as the pathogen
    2. BIdentify all three and test susceptibility
    3. CReincubate the plates for 48 more hours
    4. DReport mixed flora; request a new specimen

    Answer and explanation in LearnFree account

  17. Question 17 · Medium

    A urine culture grows more than 100,000 CFU/mL of Escherichia coli, and the sediment shows many white cells and white cell casts. What do the casts add to the interpretation?

    1. AThe pyuria is from vaginal contamination
    2. BThe infection involves the kidney
    3. CThe infection is limited to the urethra
    4. DThe infection is limited to the bladder

    Answer and explanation in LearnFree account

  18. Question 18 · Medium

    A lactose-fermenting gram-negative rod from urine is indole positive and citrate negative. What is the most likely organism?

    1. AEscherichia coli
    2. BPseudomonas aeruginosa
    3. CKlebsiella pneumoniae
    4. DProteus mirabilis

    Answer and explanation in LearnFree account

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