7.5 - Urinalysis and Body Fluids Problem-Solving

Given the following dry reagent strip urinalysis results, select the most appropriate course of action:

pH = 8.0, Protein = 1+, Glucose = Neg, Blood = Neg, Ketone = Neg, Nitrite = Neg, Bilirubin = Neg

A. Report the results, assuming acceptable quality control

B. Check pH with a ph meter before reporting

C. Perform a turbidimetric protein test, instead of the dipstick protein test, and report

D. Request a new specimen

C. Perform a turbidimetric protein test, instead of the dipstick protein test, and report

• Reasoning: Alkaline urine (pH ≥8.0) can cause false-positive protein results on dipsticks (protein error of indicators depends on pH-sensitive color change).

• Next step: Confirm protein with a quantitative/turbidimetric method (e.g., SSA test) when pH is high or if results seem inconsistent.

• Key takeaway:

  • Alkaline urine → confirmatory testing required

  • Always correlate pH with protein results before reporting to avoid false positives.

Given the following urinalysis results, select the most appropriate course of action:

pH: 8.0, Protein = Trace, Glucose = Neg, Ketone = Small, Blood = Neg, Nitrite = Neg

Micro: RBCs = 0-2, WBCs = 20-50, Bacteria = Large, Crystals = Small, CaCO3

A. Call for a new specimen because urine was contaminated in vitro

B. Recheck pH because CaCO3 does not occur at alkaline pH

C. No indication of error is present; results indicate a UTI

D. Report all results except bacteria because the nitrite was neg

C. No indication of error is present; results indicate a UTI

• Reasoning: Elevated WBCs (20–50/hpf) and bacteria (large) indicate infection; nitrite-negative does not exclude UTI (some pathogens are non-nitrate reducers, e.g., Enterococcus).

• Context: pH 8.0 is common in infection due to urease-producing bacteria (Proteus, Klebsiella), which alkalinize urine.

• Key takeaway:

  • UTI pattern: ↑WBCs + bacteria → infection even if nitrite is negative.

  • CaCO₃ crystals are normal in alkaline urine and not an error indicator.

A 6-mL peds urine sample is processed for UA. The sediment is prepared by centrifuging all the urine remaining after performing biochemical tests. The following results are obtained: SG: 1.015, Blood: Large, Leukocytes: Moderate, Protein: 2+, RBCs: 5-10, WBCs: 5-10. Select the most appropriate course of action:

A. Report the results; blood & protein correlate with micro results

B. Report biochemical results only; request a new sample for the micro exam

C. Request a new sample & report as QNS

B. Report biochemical results only; request a new sample for the micro exam

• Reasoning: Only 6 mL of urine was available — insufficient volume (QNS) for proper biochemical and microscopictesting; sediment reliability compromised.

• Context: A valid microscopic exam requires adequate volume (≈10–12 mL) to ensure accurate sediment concentration.

• Key takeaway:

  • QNS = biochemical results valid, micro unreliable → request recollection.

  • Always note when specimen volume affects sediment accuracy.

Given the following urinalysis results, select the most appropriate course of action:

pH: 6.5, Protein: Neg, Glucose: Neg, Ketone: Trace, Blood: Neg, Bilirubin: Neg

Micro Findings: Mucus = small, Ammonium Urate = large

A. Recheck urine pH

B. Report these results, assuming acceptable quality control

C. Repeat the dry reagent strip tests to confirm the ketone result

D. Request a new sample and repeat the urinalysis

A. Recheck urine pH

• Reasoning: Ammonium urate crystals form only in alkaline or neutral urine, not at pH 6.5 (acidic) — the pH result conflicts with microscopic findings.

• Context: pH error may result from old or improperly stored specimens (bacterial urea breakdown raises pH).

• Key takeaway:

  • Crystals must correlate with pH — discordance → recheck.

  • Ammonium urate → alkaline urine indicator; verify dipstick accuracy.

Given the following urinalysis results, select the most appropriate first course of action:

pH: 6.0, Protein: Neg, Glucose: Neg, Ketone: Neg, Blood: Neg, Bilirubin: Neg

Other findings - Color: intense yellow, transparency: clear

Micro: Crystals, Bilirubin granules: small

A. Repeat the dry reagent strip test for bilirubin

B. Request a new sample

C. Recheck the pH

D. Perform a test for urinary urobilinogen

A. Repeat the dry reagent strip test for bilirubin

• Reasoning: Bilirubin granules seen microscopically but negative dipstick bilirubin suggests a false-negative pad reaction (bilirubin unstable, oxidizes to biliverdin, or strip degraded by light exposure).

• Context: Bilirubin in urine is pathologic (prehepatic → normal; hepatic/posthepatic → positive). Verification is essential before interpreting.

• Key takeaway:

  • Microscopic bilirubin + negative strip → repeat bilirubin pad.

  • Confirm before proceeding to other tests (e.g., urobilinogen).

A biochemical profile gives the following results:

Creatinine: 1.4 mg/dL, BUN: 35 mg/dL, K: 5.5 mmol/L

All other results are normal, and all tests are in control. Urine from the patient has an osmolality of 975 mOsm/kg. Select the most appropriate course of action.

A. Check for hemolysis

B. Repeat the BUN, and report only if elevated

C. Repeat the serum creatinine, and report only if elevated

D. Report these results

D. Report these results

• Reasoning: Elevated BUN (35 mg/dL) and creatinine (1.4 mg/dL) with high urine osmolality (975 mOsm/kg) indicate true renal concentration ability intact, not specimen error.

• Context: BUN:Cr ratio ≈ 25:1 → consistent with prerenal azotemia (e.g., dehydration); K⁺ 5.5 mmol/L aligns with decreased renal perfusion.

• Key takeaway:

  • No QC or specimen error indicated → report.

  • Pattern fits prerenal azotemia, not analytical interference.

• BUN: 7–18 mg/dL → ↑35 mg/dL = elevated

• Creatinine: 0.6–1.2 mg/dL → ↑1.4 mg/dL = mildly elevated

• BUN:Creatinine ratio: 10–20:1 → ≈25:1 = prerenal pattern (e.g., dehydration)

• Potassium (K⁺): 3.5–5.0 mmol/L → ↑5.5 mmol/L = mild hyperkalemia

• Urine osmolality: 300–900 mOsm/kg → 975 mOsm/kg = concentrated, indicates intact tubular function

A 2 p.m. urinalysis shows trace glucose on the dry reagent strip test. Fasting blood glucose drawn 8 hours earlier is 100 mg/dL. No other results are abnormal. Elect the most appropriate course of action.

A. Repeat the urine glucose, and report if positive

B. Perform a test for reducing sugars, and report the result

C. Perform a quantitative urine glucose; report as trace if greater than 100 mg/dL

D. Request a new urine specimen

A. Repeat the urine glucose, and report if positive

• Reasoning: A trace glucose finding with a normal blood glucose may reflect a transient or borderline strip reaction(e.g., contamination, temperature, timing error). Repeating the dipstick verifies the result before further workup.

• Context: The glucose oxidase pad is specific for glucose, so confirmatory “reducing sugar” testing is unnecessary unless the result persists or occurs in infants.

• Key takeaway:

  • First step: always repeat questionable dipstick results to confirm.

  • Only escalate to reducing sugar or quantitative testing if repeat is still positive or clinical context warrants (e.g., pediatric patient).

Following a transfusion reaction, urine from a patient gives positive test results for blood and protein. The SG is 1.015. No RBCs or WBCs are seen in the microscopic examination. These results:

A. Indicate renal injury induced by transfusion reaction

B. Support the finding of an extravascular transfusion reaction

C. Support the finding of an intravascular transfusion reaction

D. Rule out a transfusion reaction caused by RBC incompatibility

C. Support the finding of an intravascular transfusion reaction

• Reasoning: Positive blood + no RBCs microscopically = free hemoglobin (hemoglobinuria) from RBC lysis, not intact cells.

• Context: Seen in intravascular hemolysis (e.g., ABO incompatibility) causing hemoglobinemia → hemoglobinuria → transient proteinuria.

• Key takeaway:

  • Dipstick “blood” = Hb/Mb or RBCs; absence of RBCs → hemoglobinuria.

  • Confirms intravascular, not extravascular, transfusion reaction.

A urine sample taken after a suspected transfusion reaction has a positive test result for blood, but intact RBCs are not seen on microscopic examination. Which test result would rule out an intravascular hemolytic transfusion reaction?

A. Negative urine urobilinogen

B. Serum unconjugated bilirubin below 1.0 mg/dL

C. Serum potassium below 6.0 mmol/L

D. Normal plasma haptoglobin

D. Normal plasma haptoglobin

• Reasoning: In intravascular hemolysis, free hemoglobin binds haptoglobin, causing low plasma haptoglobin. A normal value rules it out.

• Context: Positive dipstick for blood without RBCs = hemoglobinuria, but confirmatory plasma tests (↓haptoglobin, ↑K⁺, ↑LDH, ↑unconjugated bilirubin) establish diagnosis.

• Key takeaway:

  • Normal haptoglobin → no significant intravascular hemolysis.

  • Always correlate urine hemoglobin with plasma hemolysis markers.

Given the following urinalysis results, select the most appropriate course for action:

pH: 5.0, Protein: Neg, Glucose: 1,000 mg/dL, Blood: Neg, Bilirubin: Neg, Ketone: Moderate, SSA Protein: 1+

A. Report the SSA protein test result instead of the dry reagent strip test result

B. Call for a list of medications administered to the patient

C. Perform a quantitative urinary albumin

D. Perform a test for Microalbuminuria

B. Call for a list of medications administered to the patient

• Reasoning: SSA protein 1+ with dipstick negative suggests interference or non-albumin proteins (e.g., drugs, contrast media, antibiotics like penicillin or sulfonamides).

• Context: SSA test detects all proteins (albumin + globulins + Bence-Jones + drugs), while dipstick detects albumin only.

• Key takeaway:

  • Discordant SSA vs dipstick → check medications before further testing.

  • Confirm true proteinuria only after ruling out drug interference.

Urinalysis results from a 35-year-old woman are as follows:

SG: 1.015, pH: 7.5, Protein: Trace, Glucose: Small, Ketone: Neg, Blood: Neg, Leukocytes: Moderate

Microscopic Findings - RBCs: 5-10/HPF, WBCs: 25-50/HPF

Select the most appropriate course of action

A. Recheck the blood reaction; if negative, look for budding yeast

B. Repeat the WBC count

C. Report all results except that for blood

D. Request the list of medications used

A. Recheck the blood reaction; if negative, look for budding yeast

• Reasoning: WBCs↑ (25–50/HPF) and glucose present suggest possible UTI or yeast infection; RBCs on micro but negative dipstick blood indicates possible interference or misidentification (yeast can mimic RBCs).

• Context: Yeast infections are common in women, especially with glycosuria, and can produce false RBC appearancemicroscopically.

• Key takeaway:

  • Mismatch between dipstick blood and micro RBCs → recheck blood pad, then examine for yeast.

  • Yeast + WBCs + glucose = possible candiduria.

A routine urinalysis gives the following results:

pH: 6.5, Protein: Neg, Blood: Neg, Glucose: Trace, Ketone: Neg

Micro findings - Blood Casts: 5-10/LPF, Mucus: small, Crystals: large, amorphous

These results are most likely explained by:

A. False-negative blood reaction

B. False-negative protein reaction

C. Pseudocasts of urate mistaken for true casts

D. Mucus mistaken for casts

C. Pseudocasts of urate mistaken for true casts

• Reasoning: Blood casts require true hematuria and proteinuria—both are negative here, so casts seen are likely amorphous urate aggregates mimicking casts.

• Context: Amorphous urates form in acidic to neutral urine (pH ≤7.0) and may appear cylindrical, resembling casts under low power.

• Key takeaway:

  • No protein + no blood → casts not physiologic.

  • Likely pseudocasts (e.g., urates, mucus) misinterpreted microscopically.

When examining a urinary sediment under 400x magnification, the medical laboratory scientist (MLS) noted many RBCs to have cytoplasmic blebs and an irregular distribution of the hemoglobin. This phenomenon is most often caused by:

A. Intravascular hemolytic anemia

B. Glomerular disease

C. Hypotonic or alkaline urine

D. Severe dehydration

B. Glomerular disease

• Reasoning: RBCs in dilute or alkaline urine lose membrane integrity—water influx causes swelling, ghosting, and bleb formation with uneven hemoglobin distribution.

• Context: These are crenated or ghost RBCs, reflecting urine osmolarity/pH effects, not disease pathology.

• Key takeaway:

  • RBC morphology in urine ≠ blood morphology.

  • Hypotonic or alkaline urine → RBC lysis or distortion.

A urine specimen is dark orange and turns brown after storage in the refrigerator overnight. The MLS requests a new specimen. The second specimen is bright orange and is tested immediately. Which test result would differ between the two specimens?

A. Ketone

B. Leukocyte esterase

C. Urobilinogen

D. Nitrite

C. Urobilinogen

• Reasoning: Urobilinogen is unstable and oxidizes to urobilin on standing, especially with light or refrigeration—causing darkening of urine and decreased reactivity on retesting.

• Context: Fresh specimen remains bright orange with detectable urobilinogen; old sample → oxidized, false-low/negative result.

• Key takeaway:

  • Urobilinogen degrades on standing → test promptly.

  • Color change (orange → brown) signals oxidation to urobilin.

A patient's random urine sample consistently contains a trace of protein but no casts, cells, or other biochemical abnormality. The first voided morning sample is consistently negative for protein. These findings can be explained by:

A. Normal diurnal variation in protein loss

B. Early glomerulonephritis

C. Orthostatic or postural albuminuria

D. Microalbuminuria

C. Orthostatic or postural albuminuria

• Reasoning: Proteinuria present in random (daytime) urine but absent in first-morning sample indicates orthostatic albuminuria — benign, posture-related increase in albumin excretion.

• Context: Common in young adults, due to increased renal venous pressure when upright; disappears after recumbency overnight.

• Key takeaway:

  • Daytime-only proteinuria = orthostatic.

  • Morning negative sample confirms benign cause.

A urine sample with a pH of 8.0 and a specific gravity of 1.005 had a small positive blood reaction but is negative for protein, and no RBCs are present in the microscopic examination of urinary sediment. What best explains these findings?

A. High pH and low SG caused a false-positive blood reaction

B. The blood reaction and protein reaction are discrepant

C. Hemoglobin is present without intact RBCs because of hemolysis

D. An error was made in the microscopic examination

C. Hemoglobin is present without intact RBCs because of hemolysis

• Reasoning: Positive blood pad + no RBCs microscopically = free hemoglobin (hemoglobinuria) or myoglobin from lysis or muscle injury.

• Context: Alkaline, dilute urine (pH 8.0, SG 1.005) promotes RBC lysis, leaving only soluble hemoglobin to react with the pad.

• Key takeaway:

  • Dipstick “blood” = detects Hb/Mb peroxidase activity.

  • RBCs absent → hemoglobinuria from lysed cells.

A urine sample has a negative blood reaction and 5 to 10 cells per high-power field that resemble RBCs. What is the best course of action?

A. Mix a drop of sediment with 1 drop of WBC counting fluid and re-examine

B. Report the results without further testing

C. Repeat the blood test, and if negative, report the results

D. If the leukocyte esterase test is positive, report the cells as WBCs

A. Mix a drop of sediment with 1 drops of of WBC counting fluid and re-examine

• Reasoning: Cells resembling RBCs but with negative blood pad suggest yeast, oil droplets, or WBCs—WBC counting fluid (acetic acid) lyses RBCs but not other cells, clarifying identity.

• Context: This quick differential helps distinguish RBCs from look-alikes without repeating the full analysis.

• Key takeaway:

  • Acetic acid test = confirm RBC identity.

  • RBCs lyse → not RBCs → likely yeast or WBCs.

A toluidine blue chamber count on CSF gives the following values:

CSF Counts - WBCs: 10x10^6/L, RBCs: 1,000x10^6/L

Peripheral Blood Counts - WBCs: 5x10^9/L, RBCs: 5x10^12/L

After correcting the WBC count in CSF, the MLS should next:

A. Report the WBC count as 9 x 10^6/L without additional testing

B. Report the WBC count and number of PMNs identified by the chamber count

C. Perform a differential on a direct smear on the CSF

D. Concentrate CSF using a cytocentrifuge and perform a differential

D. Concentrate CSF using a cytocentrifuge and perform a differential

• Correction: Blood WBC/RBC = 5×10⁹ / 5×10¹² = 0.001 → contaminant WBC = 0.001 × 1,000×10⁶ = 1×10⁶/L → corrected CSF WBC = 10 − 1 = 9×10⁶/L.

• Next step: CSF cell ID requires a cytospin differential (best morphology at low counts); direct smear/chamber PMN % is inadequate.

• Key takeaway: After correction for a traumatic tap, report corrected WBC and perform cytospin diff for clinical interpretation.

A blood-tainted pleural fluid is submitted for culture. Which test result would be most conclusive in classifying the fluid as an exudate?

A. Test - LD fluid/serum, Result - 0.65

B. Test - Total Protein, Result - 3.2 g/dL

C. RBC Count - 10,000/uL

D. WBC Count - 1500/uL

A. Test - LD fluid/serum, Result - 0.65

• Reasoning: By Light’s criteria, a pleural fluid is an exudate if LD ratio > 0.6, protein ratio > 0.5, or fluid LD > ⅔ upper limit of serum LD.

• Context: Exudates = increased vascular permeability (infection, malignancy); transudates = systemic pressure/oncotic imbalance (CHF, cirrhosis).

• Key takeaway:

  • LD ratio > 0.6 → exudate confirmed.

  • Protein 3.2 g/dL or cell counts alone can’t classify reliably.

A pleural fluid submitted to the laboratory is milky in appearance. Which test would be most useful in differentiating between a chylous and pseudochylous effusion?

A. Fluid to serum triglyceride ratio

B. Fluid WBC count

C. Fluid total protein

D. Fluid:serum LD ratio

A. Fluid to serum triglyceride ratio

• Reasoning: Chylous effusion = high triglycerides (>110 mg/dL) from lymphatic leakage (chyle); pseudochylous = cholesterol-rich from chronic inflammation.

• Context: Both appear milky, but only chylous contains true chylomicrons (confirmed by triglyceride level or Sudan stain).

• Key takeaway:

  • Triglyceride ↑ → chylous; cholesterol ↑ → pseudochylous.

  • Triglyceride testing = definitive differentiation.

A CSF sample from an 8-year-old child with a fever of unknown origin was tested for glucose, total protein, lactate, and IgG index. Glucose was 180 mg/dL, but all other results were within the reference range. The CSF WBC count was 9x 106/L, and the RBC count was 10 x 106/L. The differential showed 50% lymphocytes, 35% monocytes, 10% macrophages, 3% neutrophils, and 2% neuroectodermal cells. What is the most likely cause of these results?

A. Aseptic meningitis

B. Traumatic tap

C. Subarachnoid hemorrhage

D. Hyperglycemia

D. Hyperglycemia

• Reasoning: CSF glucose normally equals ~60–70% of plasma glucose; it rises only when blood glucose is elevated.Normal protein, lactate, and cell counts exclude infection or hemorrhage.

• Context: CSF glucose reflects serum glucose, not CNS metabolism. Infection (aseptic/bacterial meningitis) would decrease CSF glucose and elevate WBCs/protein.

• Key takeaway:

  • High CSF glucose → systemic hyperglycemia.

  • CSF glucose never elevated intrinsically.

A WBC count and diff performed on ascites fluid gave a WBC count of 20,000/uL with 90% macrophages. The gross appearance of the fluid was described as "thick & bloody." It was noted that several clusters of these cells were observed and that the majority of the cells contained many vacuoles resembling paper-punch holes. What do the observations suggest?

A. Malignant mesothelial cells were counted as macrophages

B. Adenocarcinoma from a metastatic site

C. Lymphoma infiltrating peritoneal cavity

A. Malignant mesothelial cells were counted as macrophages

• Reasoning: Clusters of vacuolated “paper-punch” cells suggest reactive or malignant mesothelial cells, which can mimic macrophages on morphology.

• Context: True macrophages are usually single, not clustered, and have uniform cytoplasm; mesothelial cells form clusters/sheets with vacuolated borders.

• Key takeaway:

  • Clustered, vacuolated cells → mesothelial origin.

  • Misidentification can falsely elevate “macrophage” count.

Given the following data for creatinine clearance, select the most appropriate course of action.

Volume: 2.8 L/day; Surface Area: 1.73 m2; urine creatinine: 100 mg/dL; serum creatinine: 1.2 mg/dL

A. Report a creatinine clearance of 162 mL/min

B. Repeat the urine creatinine; results point to a dilution error

C. Request a new 24-hour urine sample

D. Request the patient's age and gender

C. Request a new 24-hour urine sample

• Reasoning: The calculated creatinine clearance (≈162 mL/min) is physiologically too high, suggesting collection timing or volume error, not an analytical issue.

• Context: Proper 24-hr collection is critical—missed or excess urine causes major clearance errors. Normal adult range ≈ 90–130 mL/min (1.73 m²).

• Key takeaway:

  • Implausible clearance → recollect sample.

  • Always verify timing, completeness, and preservation of 24-hr urine before reporting.

An elevated amylase is obtained on a stat serum collected at 8pm. An amylase performed a 8am that morning was within normal limits. The MLS also noted that urine amylase was measured at 6pm. Select the best course of action

A. Repeat stat amylase; report only if in normal limits

B. Repeat both morning & afternoon serum amylase, & report only if they agree

C. Request new specimen; don't report results of stat sample

D. Review amylase result on 6 pm urine sample; if elevated, report stat amylase

D. Review amylase result on 6 pm urine sample; if elevated, report stat amylase

• Reasoning: Urine amylase parallels serum amylase but stays elevated longer; checking the urine result confirms whether the 8 pm serum rise reflects true pancreatic activity rather than specimen or analytical error.

• Context: Sudden rise after normal morning value may indicate acute pancreatitis onset or post-collection delay artifact—urine amylase helps differentiate.

• Key takeaway:

  • Concordant elevated urine amylase → true rise → report.

  • Discrepant or normal urine → re-evaluate spec

Results of an FLM study from a patient with diabetes mellitus are as follows:

L/S Ratio: 2.0, Phophatidyl Glycerol: Negative, Creatinine: 2.5 mg/dL

Given these results, the MLS should:

A. Report the result and recommend repeating the L/S ratio in 24 hours

B. Perform scanning spectrophotometry on the fluid to determine if blood is present

C. Repeat the L/S ratio after 4 hours and report those results

D. Report results as invalid

A. Report the result and recommend repeating the L/S ratio in 24 hours

• Reasoning: The L/S ratio of 2.0 suggests possible maturity, but absence of phosphatidylglycerol (PG) makes the result borderline, especially in diabetic pregnancies where lung surfactant development is delayed.

• Context: PG appears last in fetal lung maturation and provides confirmation of maturity when present.

• Key takeaway:

  • Borderline or discordant FLM → repeat in 24 hours.

  • In diabetics, PG detection is essential before confirming maturity.

A 24-hour urine sample from an adult submitted for Catecholamines gives a result of 140 ug/day (upper reference limit 150 ug/day). The 24-hour urine creatinine level is 0.6 g/day. Select the best course of action.

A. Check the urine pH to verify that it is less than 2.0

B. Report the result in ug Catecholamines per milligram of creatinine

C. Request a new 24-hour urine sample

D. Measure the VMA, and report the Catecholamine result only if elevated

C. Request a new 24-hour urine sample

• Reasoning: The urine creatinine (0.6 g/day) is too low for a complete 24-hour collection (normal ≈ 1.0–1.8 g/day for adults), indicating an incomplete or under-collected sample.

• Context: Catecholamine excretion is time- and volume-dependent; incomplete collection invalidates quantitative interpretation.

• Key takeaway:

  • Low creatinine = incomplete 24-hr urine → recollect.

  • Always verify collection adequacy before reporting timed urine analytes.

A sperm motility test was performed & 200 sperm were evaluated in each of two duplicates. The first sample showed progressive movement in 50% and nonprogressive movement in 35%, and 15% were immotile. The second showed progressive movement in 35% and nonprogressive movement in 35%, and 30% were immotile. What is the best course of action?

A. Report the average of the two values for progressive movement

B. Report the higher of the two values

C. Repeat the motility test

D. Call for a new specimen

C. Repeat the motility test

• Reasoning: Duplicate motility results are discordant (progressive 50% vs 35%; total 85% vs 70%) — exceeds typical ≤10% allowable difference for duplicate semen analyses.

• Key takeaway: When duplicate counts disagree beyond limits, repeat motility on the same specimen (new fields/observers if needed) rather than averaging or picking the higher value.

A quantitative serum hCG is ordered on a male patient. The technologist should:

A. Perform the test and report the result

B. Request that the order be cancelled

C. Perform the test and report the result if negative

D. Perform the test and report the result only if greater than 25 IU/L

A. Perform the test and report the result

• Reasoning: hCG testing can be clinically relevant in males—especially to evaluate testicular germ cell tumors, trophoblastic disease, or ectopic hCG production.

• Context: The test should not be canceled based on gender; interpretation depends on clinical context, not patient sex.

• Key takeaway:

  • Always perform and report ordered tests if properly requested.

  • Elevated hCG in males may indicate testicular cancer or tumor marker elevation.

When testing for drugs of abuse in urine, which of the following test results indicate dilution and would be cause for rejecting the sample?

A. Temperature at sample submission 92F

B. SG 1.002; creatinine 15 mg/dL

C. pH 5.8; temperature 94F

D. SG 1.012; creatinine 25 mg/dL

B. SG 1.002; creatinine 15 mg/dL

• Reasoning: Dilute urine is defined by specific gravity <1.003 and/or creatinine <20 mg/dL — indicating possible water loading or adulteration.

• Context: Valid specimen criteria: temp 90–100°F, SG 1.003–1.030, creatinine ≥20 mg/dL, pH 4.5–9.0.

• Key takeaway:

  • SG <1.003 + creatinine <20 → dilution.

  • Such samples should be rejected and recollected under supervision.

A urine specimen has an SG of 1.025 & is strongly positive for nitrite. All other dry reagent strip test results are normal, and the micro exam was unremarkable, showing no WBCs or bacteria. The urine sample was submitted as part of a pre-employment physical exam that also includes drug testing. Which most likely caused these results?

A. A viral infection of the kidney

B. A urinary tract infection in an immunosuppressed person

C. An adulterated urine specimen

D. Error in reading the nitrite pad

C. An adulterated urine specimen

• Reasoning: Strong nitrite with no WBCs/bacteria on micro = discordant for UTI → suggests oxidant/nitrite adulterant used to mask drugs.

• Context/Action: In pre-employment screens, check specimen validity (oxidants, nitrite, pH, creatinine, SG) and recollect under observation.

• Key takeaway: Nitrite pad ↑ without pyuria/bacteriuria → think adulteration, not infection.

A CSF sample submitted for cell counts has a visible clot. What is the best course of action?

A. Count RBCs and WBCs manually after diluting the fluid with normal saline

B. Tease the cells out of the clot before counting, then dilute with WBC counting fluid

C. Request a new sample

D. Perform a WBC count without correction

C. Request a new sample

• Reasoning: A visible clot in CSF indicates high protein or fibrinogen, often due to traumatic tap or contamination, making cell counts invalid—cells are trapped in the clot.

• Context: Proper CSF should not clot; normal fibrinogen is absent. Clot formation suggests blood contamination or blockage of flow.

• Key takeaway:

  • Clotted CSF → reject and recollect.

  • Do not attempt manual or diluted counts—results will be unreliable.

Total hemolytic complement and glucose are ordered on a synovial fluid sample that is too viscous to pipet. What is the best course of action?

A. Dilute the sample in saline

B. Add 1 mg/mL hyaluronidase to the sample, and incubate at room temperature for 30 minutes

C. Warm the sample to 65C for 10 minutes

D. Request a new specimen

B. Add 1 mg/mL hyaluronidase to the sample, and incubate at room temperature for 30 minutes

• Reasoning: Synovial fluid viscosity is due to hyaluronic acid; this must be broken down before testing to ensure accurate pipetting and analyte recovery.

• Context: Hyaluronidase enzyme digests hyaluronic acid, making the sample fluid enough for testing without altering key chemistry values.

• Key takeaway:

  • Viscous synovial fluid → treat with hyaluronidase.

  • Ensures valid measurements for glucose, protein, complement, etc.

A CSF cytosine smear shows many smudge cells and macrophages with torn cell membranes. What most likely caused this problem?

A. Failure to add albumin to the cytospin cup

B. Failure to collect the CSF in EDTA

C. Centrifuge speed too low

D. Improper alignment

A. Failure to add albumin to the cytospin cup

• Reasoning: CSF has low protein, so cells are fragile and rupture easily during cytocentrifugation unless albumin (e.g., 1% BSA or serum) is added as a stabilizer.

• Context: Albumin coats and cushions cells, preventing smudge cells and membrane tearing.

• Key takeaway:

  • Add albumin to fragile fluids (CSF, serous) before cytospin.

  • Prevents cell distortion and artifact formation.

An automated electronic blood cell counter was used to count RBCs and WBCs in a turbid pleural fluid sample. The WBC count was 5 x 10^10/L (50,000/uL) and the RBC count was 5.5 x 10^10/L (55,000/uL). What is the significance of the RBC count?

A. The RBC count is not significant and should be reported as 5,000/uL

B. The RBC count should be reported as determined by the analyzer

C. A manual RBC count should be performed

D. A manual RBC and WBC count should be performed and reported instead

A. The RBC count is not significant and should be reported as 5,000/uL

• Reasoning: Automated electronic counters often overcount noncellular debris or protein aggregates in turbid serous fluids, yielding falsely high RBC counts.

• Context: A pleural fluid RBC of 5 × 10¹⁰/L (50,000/µL) is too high for such a sample and likely spurious; report only a rounded, clinically insignificant estimate to reflect minor blood contamination.

• Key takeaway:

  • Body fluid RBC counts >5,000/µL generally indicate contamination or trauma.

  • When analyzer output is unreliable, qualitative reporting (e.g., “~5,000/µL, not significant”) is appropriate.

A lamellar body count was performed on an amniotic fluid sample that was slightly pink within 1 hour of specimen collection. The sample was stored at 4C prior to analysis. The result was 25,00/uL, classified as intermediate risk of RDS. The physician waited 24 hours and collected a new sample that was counted within 2 hours of collection on the same instrument. The LCB count of the new sample was 14,00/uL and the patient was reclassified as high risk for delivery Which statement best explains these results?

A) Loss of lamellar bodies occurred in the second sample because of storage.

B) Blood caused a falsely elevated result for the first sample

C) The fetal status changed in 24 hours because of respiratory illness

D) The difference in counts is the result of day to day physiological and instrument variance.

B. Blood caused a falsely elevated result for the first sample

• Reasoning: Pink (bloody) amniotic fluid introduces platelets that are similar in size to lamellar bodies, falsely increasing LBC on particle counters.

• Context: LBC is relatively stable with short 4 °C storage; a drop from ~25k → ~14k/µL far exceeds normal analytic/physiologic variation and is best explained by platelet interference in the first sample.

• Key takeaway: Avoid blood contamination for LBC—platelets ↑ count → false maturity.