UA 7.5 Urinalysis and Body Fluids Problem-Solving

Question 1

1. 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

Answer 1

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

Highly buffered alkaline urine may cause a false-positive result on the dry reagent strip protein test by titrating the acid buffer on the reagent pad. Protein should be measured by an alternate method that is not subject to positive interference by highly buffered alkaline urine

Question 2

2. 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
   Microscopic findings:
   RBCs = 0–2/HPF
   WBCs = 20–50/HPF
   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 test was negative

Answer 2

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

Question 3

3. SITUATION: A 6-mL pediatric urine sample is processed for routine urinalysis in the usual manner. The sediment is prepared by centrifuging all of the urine remaining after performing the biochemical tests. The following results are obtained:
   SG = 1.015
   Blood = Large
   Leukocytes = Moderate
   Protein = 2+
   RBCs: 5–10/HPF
   WBCs: 5–10/HPF
   Select the most appropriate course of action.

A. Report these results; blood and protein correlate with microscopic results
B. Report biochemical results only; request a new sample for the microscopic examination
C. Request a new sample and report as quantity not sufficient (QNS)
D. Recentrifuge the supernatant and repeat the microscopic examination

Answer 3

B. Report biochemical results only; request a new sample for the microscopic examination

Question 4

4. 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
   Microscopic 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

Answer 4

A. Recheck urine pH

Question 5

5. 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
   Microscopic: 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

Answer 5

A. Repeat the dry reagent strip test for bilirubin

Question 6

6. 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 normal
C. Repeat the serum creatinine, and report only if elevated
D. Report these results

Answer 6

D. Report these results

Patients with prerenal failure usually have a BUN:creatinine ratio greater than 20:1. Reduced renal blood flow causes increased urea reabsorption and high urine osmolality. Patients are usually hypertensive and show fluid retention and hyperkalemia

Question 7

7. 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. Select 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

Answer 7

A. Repeat the urine glucose, and report if positive

Question 8

8. 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

Answer 8

C. Support the finding of an intravascular transfusion reaction

RBCs usually remain intact at a SG of 1.015. The absence of RBCs, WBCs, and casts points to hemoglobinuria caused by intravascular hemolysis rather than glomerular injury. A positive protein reaction will occur if sufficient hemoglobin is present.

Question 9

9. 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

Answer 9

D. Normal plasma haptoglobin

The plasma free hemoglobin will be increased immediately after a hemolytic transfusion reaction, and the haptoglobin will be decreased. The hemoglobin will be eliminated by the kidneys, but the haptoglobin will remain low or undetectable for 2 to 3 days. Normal urine urobilinogen and serum unconjugated bilirubin help in ruling out extravascular hemolysis. Pretransfusion potassium is needed to evaluate the contribution of hemolysis to the post-transfusion result

Question 10

10. Given the following urinalysis results, select the most appropriate course of 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

Answer 10

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

The combination of glucose and ketone positivity of urine points to a patient with insulin-dependent diabetes. A false-positive SSA test result is likely if tolbutamide (Orinase) has been administered

Question 11

11. 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

Answer 11

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

Question 12

12. A routine urinalysis gives the following results:
    pH =6.5
    Protein = Neg
    Blood = Neg
    Glucose= Trace
    Ketone = Neg
    Microscopic findings:
    Blood casts: 5–10/LFP
    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

Answer 12

C. Pseudocasts of urate mistaken for true casts

Question 13

13. SITUATION: When examining a urinary sediment under 400× 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

Answer 13

B. Glomerular disease

Question 14

14. SITUATION: 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

Answer 14

C. Urobilinogen

Question 15

15. 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

Answer 15

C. Orthostatic or postural albuminuria

Question 16

16. 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

Answer 16

C. Hemoglobin is present without intact RBCs because of hemolysis

Question 17

17. 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

Answer 17

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

Question 18

18. A toluidine blue chamber count on CSF gives the following values:
    CSF Counts
    WBCs 10 × 10^6/L
    RBCs 1,000 × 10^6/L
    Peripheral Blood Counts
    WBCs 5 × 10^9/L
    RBCs 5 × 10^12/L
    After correcting the WBC count in CSF, the MLS should next:

A. Report the WBC count as 9 × 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 of the CSF
D. Concentrate CSF using a cytocentrifuge and perform a differential

Answer 18

D. Concentrate CSF using a cytocentrifuge and perform a differential

Question 19

19. 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. Test: RBC count, Result: 10,000/µL
D. Test: WBC count, Result: 1500/µL

Answer 19

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

A traumatic tap makes classification of fluids difficult on the basis of cell counts and protein. The values reported for protein, RBCs, and WBCs can occur in either an exudate or bloody transudate, but the LD ratio is significant

Question 20

20. 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

Answer 20

A. Fluid to serum triglyceride ratio

Question 21

21. 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 9 × 10^6/L, and the RBC count was 10 × 106^6/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

Answer 21

D. Hyperglycemia

CSF glucose is approximately 60% of the plasma glucose but may be somewhat lower in a person with diabetes. The reference range is approximately 40 to 70 mg/dL. A CSF glucose level above 70 mg/dL is caused by a high plasma glucose that equilibrated with CSF. Therefore, hyperglycorrhachia is caused by hyperglycemia. The WBC count in a child between 5 and 12 years of age is 0 to 10 × 106/L (0–10/µL). The normal RBC count and protein rule out subarachnoid hemorrhage and traumatic tap. Although aseptic meningitis cannot be ruled out conclusively, it is unlikely given a normal WBC count and IgG index.

Question 22

22. A WBC count and differential performed on ascites fluid gave a WBC count of 20,000/µL with 90% macrophages. The gross appearance of the fluid was described by the MLS as “thick and bloody.” It was noted on the report 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 above suggest?

A. Malignant mesothelial cells were counted as macrophages
B. Adenocarcinoma from a metastatic site
C. Lymphoma infiltrating the peritoneal cavity
D. Nodular sclerosing type Hodgkin disease

Answer 22

A. Malignant mesothelial cells were counted as macrophages

Question 23

23. 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

Answer 23

C. Request a new 24-hour urine sample

A calculated clearance in excess of 140 mL/min is greater than the upper physiological limit. The high volume per day suggests addition of H2 O, or urine that should have been voided and discarded at the start of sample collection. The result should be considered invalid.

Question 24

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

A. Repeat the stat amylase; report only if within normal limits
B. Repeat both the morning and afternoon serum amylase, and report only if they agree
C. Request a new specimen; do not report results of the stat sample
D. Review the amylase result on the 6 p.m. urine sample; if elevated, report the stat amylase

Answer 24

D. Review the amylase result on the 6 p.m. urine sample; if elevated, report the stat amylase

Serum amylase peaks 2 to 10 hours after an episode of acute pancreatitis, and this may have caused the elevated serum amylase at 8 p.m. Urinary amylase parallels serum amylase; therefore, a positive urine test at 6 p.m. makes sample collection error unlikely

Question 25

25. Results of an FLM study from a patient with diabetes mellitus are as follows:
    L/S ratio = 2.0; Phosphatidyl glycerol = Neg; 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

Answer 25

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

In patients with poorly controlled diabetes, lung maturity may be delayed and an L/S ratio of 2:1 may be associated with respiratory distress syndrome. A positive PG spot correlates with an L/S ratio of 2:1 or higher and rules out a falsely increased result caused by blood contamination. However, the appearance of PG in amniotic fluid is often delayed in diabetes. The best course of action is to wait an additional 24 hours and perform another L/S ratio on a fresh sample of amniotic fluid because an L/S ratio of 3:1 would indicate a high probability of lung maturity.

Question 26

26. A 24-hour urine sample from an adult submitted for catecholamines gives a result of 140 µg/day (upper reference limit 150 µg/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 µg 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

Answer 26

C. Request a new 24-hour urine sample

Urine creatinine of less than 0.8 g/day indicates incomplete sample collection. The patient’s daily catecholamine excretion would be misinterpreted from this result.

Question 27

27. A sperm motility test was performed and 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

Answer 27

C. Repeat the motility test

Sperm motility should be performed in duplicate and 200 cells per sample in several fields should be evaluated for movement. Agreement between duplicates must be within a specified percentage based on the category with the highest percentage. In this case, progressive movement is highest at 50%, and the difference between replicates should be no greater than 10%

Question 28

28. 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

Answer 28

A. Perform the test and report the result

Question 29

29. SITUATION: A lamellar body count (LBC) was performed on an amniotic fluid sample that was slightly pink within 1 hour of specimen collection. The sample was stored at 4°C prior to analysis. The result was 25,000/µL, 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 14,000/µL 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

Answer 29

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

Question 30

30. 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 92°F
B. SG 1.002; creatinine 15 mg/dL
C. pH 5.8; temperature 94°C
D. SG 1.012; creatinine 25 mg/dL

Answer 30

B. SG 1.002; creatinine 15 mg/dL

Question 31

31. SITUATION: A urine specimen has an SG of 1.025 and is strongly positive for nitrite. All other dry reagent strip test results are normal, and the microscopic examination was unremarkable, showing no WBCs or bacteria. The urine sample was submitted as part of a pre-employment physical examination 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 caused by poor reflectometer calibration

Answer 31

C. An adulterated urine specimen

Question 32

32. 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

Answer 32

C. Request a new sample

Question 33

33. 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 65°C for 10 minutes
D. Request a new specimen

Answer 33

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

Question 34

34. A CSF cytospin 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

Answer 34

A. Failure to add albumin to the cytospin cup

Question 35

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

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

Answer 35

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

When cell counters that perform CBCs are used, the WBCs are not lysed in the RBC bath and would be counted as RBCs. In this case, the empyemic fluid would cause the RBC count to be erroneously elevated, and this should be corrected before reporting by subtracting the WBC count from the RBC count.