7.5 Urinalysis and Body Fluids Problem-Solving

Question 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

Question 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

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 supernant and repeat the microscopic examination

Answer 3

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

This discrepancy between the blood reaction and RBC count resulted from spinning less than 12 mL of urine. When volume is below 12 mL, the sample should be diluted with saline to 12 mL before concentrating. Results are multiplied by the dilution (12 mL/mL urine) to give the correct range.

Question 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

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. Request 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. Request the dry reagent strip test for bilirubin

Question 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

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 at 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

Urine glucose concentration is dependent on blood glucose concentration at the time urine is formed. The postprandial glucose (2 pm) level exceedded the renal threshold, resulting in trace glycosuria. Tests for reducing sugars are not used to confirm a positive urine glucose test result.

Question 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

A urine sample taken after a suspected transfusion reaction has a positive test results 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

Question 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
Ketone = Moderate
Bilirubin = Neg
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 likley if tolbutamide (Orinase) has been administered.

Question 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

A nonhemolyzed trace may have been overlooked and the blood test should repeated. A false-negative result (e.g., megadoes of vitamin C) rarely occurs. Yeast cells often accompany pyuria and glycosuria and are easily mistaken for RBCs.

Question 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

SITUATION: 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

Answer 13

B. Glomerular disease

When RBCs pass through the damaged endothelial wall of the glomerulus, they become distorted, and such cells are described as dysmorphic in appearance. They are characterized by uneven distribution of hemoglobin, cytoplasmic blebs, and an asymmetrical membrane distinct from crenation. The cytoplasma may be extruded from the cell and may aggregate at the membrane giving the cell a wavy appearance. A predominance of dysmorphic RBCs in the microscopic examination points to glomerular bleeding as opposed to hematuria from other causes. Intravascular hemolytic anemia causes hemoglobinuria, rather than hematuria. RBCs lyse in hypotonic and alkaline urine. Severe dehydration is not a cause of hematuria.

Question 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

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

Protein and other constituents of urine will often be highest in the first morning void. A normal first-voided sample makes glomerular disease highly unlikely. Orthostatic albuminuria is a benign condition sometimes seen in adolescents who are tall and have a bent posture that puts pressure on the kidneys. The quantitiy of albumin excreted into the urine is small. Diagnosis is made by demonstrating a positive test after a person is erect for several hours, and the absence of proteinuria when the person is recombent. Microalbuminuria seen in persons with diabetes is usually accompanied by a positive test result for urinary glucose.

Question 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

RBCs will lyse in alkaline or dilute urine, and this sample displays both. The blood test is sensitive to as little is 0.015 mg/dL hemoglobin, and the protein test is significantly less sensitive. As a result, trace to small positive blood and negative protein are commonly encountered.

Question 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

When 5 to 10 RBCs/HPFs are seen, the blood test should show a nonhemolyzed trace. Therefore, it is likely that the cells are not RBCs. RBCs are easily confused with nonbudding yeast and may also be mistaken for WBCs. RBCs will lyse in dilute acetic acid, but yeast and WBCs will not. If a yeast infection is present, then the leukocyte esterase test result will likely be positive; therefore, the leukocyte esterase test cannot be used to determine the identity of the cells. The Sternheimer-Malbin stain can be used to differentiate WBCs from RBCs and yeast.

Question 18

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

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

Peripheral Blood Counts
- WBCs: 5 x 10^9/L
- RBCs: 5 x 10^12/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 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

A differential is performed by using CSF concentrate on all neonatal samples and whenever the WBC count is greater than 5uL. A toluidine blue chamber count of PMNs is not sufficiently sensitive to detect neutrophilic pleocytosis.

Question 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/uL
D. Test: WBC count, Result: 1,500/u:

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 WBC can occur in either an excudate or bloody transudate, but the LD ratio is significant.

Question 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

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 x 10^6/L, and the RBC count was 10 x 10^6/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 x 10^6/L (0-10/uL). 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

A WBC count and differential performed on ascites fluid gave a WBC count of 20,000/uL 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

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

Volume = 2.8 L/day
surface area= 1.73 m^2
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 H2O, or urine that should haven been voided and discarded at the start of sample collection. The result should be considered invalid.

Question 24

An elevated amylase is obtained on a stat serum collected at 8pm. An amylase perform at 8am that morning was within normal limits. The MLS also noted that urine amylase was measured at 6pm. 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 6pm urine sample; if elevated, report the stat amylase

Answer 24

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

Question 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 fasely 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

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

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

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

Question 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

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 4C prior to analysis. The result was 25,000/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 14,000/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

Answer 29

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

Lamellar bodies are small particles containing pulmonary surfactants that are made by type II pneumocytes, and their number in amniotic fluid increases as the concentration of phospholipids increases. They are abou the same size as platelets and are counted in the platelet channel of cell counters. If the amnotic fluid sample is contaminated with blood, platelets will falsely raise the LBC. AMniotic fluid samples for LBC are stable for several days when stored at 4C. However, cutoffs for FLM need to be established by each laboratory because there are significant differences in LBCs between different counters.

Question 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
D. SG 1.012; creatinine 25 mg/dL

Answer 30

B. SG 1.002; creatinine 15 mg/dL

Tampering with a sample submitted for abuse substance testing can be either by dilution or substituition. Substance Abuse and Mental Health Services Administration (SAMHSA) certified workplace drug testing laboratories are required to test for both, and reject samples based on SAMHSA cutoffs. A specimen is too dilute for testing if the creainine is less than 20 mg/dL and SG below 1.003. A sample is considered substituted if creatinine is less than 5.0 mg/dL and SG less than 1.002. Values in A and C for pH and temperature are within acceptable limits.

Question 31

SITUATION: 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 reflectormeter calibration

Answer 31

C. An adulterated urine specimen

Urine validity testing for drugs of abuse includes tests for nitrite, glutathione, pyridinium dichromate, and peroxide in addition to pH, SG, and creatinine. These substances are known to cause negative interference in the enzyme-multiplied immunoassay technique (EMIT test). A viral infection of the kidney would be associated with high numbers of renal tubular epithelial cells and leukocytes. An infection in an immunosuppresed person would still produce urinary WBCs. Although laboratory error is possible, a false-positive result caused by a reflectometer error would be suspected if the test pad were negative when reading it maunally.

Question 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 dilution 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

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

Joint fluid is too viscous to pipet accurately and cannot be diluted accurately. Complement is heat labile, and total hemolytic complement is destroyed when the sample is heated to 56C for 5 minutes. Joint fluid is difficult to collect, and a new sample is likely to have the same problem.

Question 34

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 cytosine 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 cytosine cup

Question 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 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

Answer 35

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

Electronic cell counters are validated for body fluid cell counts with specific minimum detection limits. For most counters, this is 50 to 200 total nucleated cells (TNCs)/uL and 10,000 RBC/uL. Specialized counters using fluorescent and image analysis can achieve low end sensitivities for both TNC and RBC counts of 0 to 1/uL. Because the RBCs are lysed in the WBC bath, the WBC count represents the number of nucleated cells present. 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.