UA 7.5 Urinalysis and Body Fluids Problem-Solving Flashcards
- 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
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
- 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
C. No indication of error is present; results indicate a UTI
- 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
B. Report biochemical results only; request a new sample for the microscopic examination
- 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
A. Recheck urine pH
- 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
A. Repeat the dry reagent strip test for bilirubin
- 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
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
- 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
A. Repeat the urine glucose, and report if positive
- 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
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.
- 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
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
- 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
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
- 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
- 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
C. Pseudocasts of urate mistaken for true casts
- 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
B. Glomerular disease
- 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
C. Urobilinogen
- 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
- 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
- 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 drop of WBC counting fluid and re-examine
- 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
D. Concentrate CSF using a cytocentrifuge and perform a differential
- 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
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
- 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
- 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
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.
- 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
A. Malignant mesothelial cells were counted as macrophages
- 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
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.
- 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
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
