Nca Clinical Laboratory Sciences.pdf

11 Clinicai Chemistry Chapter Author

Nancy Brunzel

CLT Review Questions 1. The turbid appearance of lipemic serum is due to an elevated level of A. B. C. D.

cholesterol HDL phospholipid triglycerides

The answer is D. Lipemia is dueto the presence of large fat-containing micelles. Both chylomicrons and VLDL particles have large enough diameters to scatter light, giving the specimen a turbid appearance. Both particles contain a high proportion of triglycerides. (Kaplan and Pesce, pp. 427, 661) 2. A trough blood sample for routine therapeutic drug monitoring is usually obtained A. just after a dose is administered B. just before the next scheduled dose e. at the calculated peak time after a dose D. one half-life after a dose is administered

The answer is B. lndividuals differ markedly in their rates of clearance of drugs due to differences in absorption, distribution, metabolism, and excretion. Factors such as age, liver and kidney status, protein binding, and the presence of other drugs influence serum drug leveis. The trough levei, obtained immediately before administering a dose, is frequently used to monitor serum drug concentrations and adjust drug dosage regimens. (Kaplan and Pesce, p. 1087) 3. To quantitate urinary vanillylmandelic acid (VMA), which of the following specimens should be collected? A. B. C. D.

24-hour urine collection First moming urine sample Random, mid-stream clean catch Two-hour urine collection following the midday meal, i.e., 2:00-4:00 pm

The answer is A. Vanillylmandelic acid (VMA) is a metabolite of the catecholamines: epinephrine, norepinephrine, and dopamine. Dueto the potential effects of exercise, hydration, and body metabolism on excretion rates, a 24-hour collection is the specimen of choice. (Kaplan and Pesce, p. 1117; Bishop et al., p. 417) 1

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Clinicai Chemistry

4. Situation: It is 8:00 am and you are working in the Specimen Receiving and Processing Area of the laboratory. The following specimens are received. Which requires intervention before proceeding with processing and testing? Test requested

A. Alkaline phosphatase B. Glucose e. Blood gases D. Electrolytes (Na & K)

Specimen received

EDTA sodium fluoride heparin, in ice slush lithium heparin

Time collected

7:30 am 6:10 am 7:35 am 5:50 am

The answer is A. Alkaline phosphatase, as with many enzymes, should not be collected in EDTA or oxalate because these collection tubes chelate ions necessary for enzyme activity. Glucose levels in sodium fluoride specimens are stable at room temperature for 24 hours. Blood gas specimens collected anaerobically in heparin and stored in ice water slush are stable for 30 minutes after collection. Sodium and potassium leveis in plasma are stable for 1 week at room temperature. (Lehmann, pp. 47, 79, 140, 161) S. A single tube of cerebrospinal fluid is received in the laboratory and the fol-

lowing tests requested: total protein, albumin, IgG quantitation, microbial culture, Gram stain, leukocyte count, and differential cell count. The specimen should be sent to the various laboratories in which order? A. B. C. D.

Chemistry lab, hematology lab, microbiology lab Hematology lab, chemistry lab, microbiology lab Microbiology lab, hematology lab, chemistry lab Hematology lab, microbiology lab, chemistry lab

The answer is C. The microbial culture should be performed first to ensure that sterility of the specimen is not compromised. Usually, cell counts are performed next, followed by chemical testing. With low volume specimens, the physician often prioritizes the remaining tests requested. (Brunzel, p. 368)

6. Which of the following components determines the wavelength of light that will pass through the sample cuvette in a spectrophotometer? A. B. C. D.

Detector Light source Potentiometer Monochromator

The answer is D. The monochromator isolates the desired wavelength of light (monochromatic light) and excludes that of other wavelengths. (Burtis and Ashwood, pp. 62-64)

7. Which of the following specimens usually requires concentration before analysis? A. B. C. D.

Urine for osmolality Amniotic fluid for L/S ratio Plasma for making a protein-free filtrate Cerebrospinal fluid for protein electrophoresis

CLT Review Questions

The answer is D. Because the protein content of cerebrospinal fluid is usually low, it must be concentrated 80- to 100-fold before electrophoresis. This is commonly achieved using commercial concentrator systems. (Brunzel, pp. 376- 377; Lehmann, p. 32)

8. You need to prepare a 1 to 4 (1 :4 or 1/4) dilution of a serum specimen using saline before analysis. Which of the following pipetting steps would result in this dilution? A. B. C. D.

Pipet 0.5 mL serum, then Pipet 1.0 mL serum, then Pipet 1.5 mL serum, then Pipet 2.0 mL serum, then

add 1.0 mL saline add 2.0 mL saline add 4.5 mL saline add 5.0 mL saline

The answer is C. A 1 to 4 dilution means that of a total volume of 4 parts, 1 part is sample. The sample volume in choice C is 1.5 mL (1 part) and the total volume of the núxture is 6.0 mL or 4 parts; hence a 1 to 4 dilution of the sarnple is made. Choices A and B are 1:3 dilutions and choice D is a 1:2.5 dilution. (Bishop et al., pp. 22- 23; Lehmann, p. 1213) 9. Which of the following situations indicates that an instrument problem is present and requires further investigation? A. B. C. D.

A series of high anion gaps A series of high glucose results A series of low urine osmolality results A series of low urine specific gravity results

The answer is A. The anion gap is a useful quality control tool for an instrument that performs electrolyte measurements (Na+, K+, c1- , HC03 - ). Consistently abnormal gaps (either increased or decreased) in a series of samples can indicate a problem in one of the electrolyte measurements and requires further investigation before results are reported. (Bishop et al., pp. 274-275; Lehmann, pp. 138-139) 10. The major components of a spectrophotometer are represented in the following diagram. Which component determines the factor "b" in Beer's law, A = abc?

C

D

+

+

·~ A. B. C. D.

Component A Component B Component C Component D

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Clinicai Chemistry

The answer is C. The sample cuvette determines the length of the light path through the sample (i.e., the value of "b" in Beer's law). Typically, sample cuvettes with a light path of 1 cm are used. ln so doing, the path length (b) and molar absorptivity (a) for a particular analyte become a constant at a given wavelength. ln which case, the absorbance (A) observed is proportional to the analyte concentration (c). (Bishop et ai., pp. 98-101)

11. Samples for calcium analysis by atomic absorption spectrophotometry should be diluted with a lanthanum solution because lanthanum ions A. B. C. D.

blank for variations in flame temperature blank for variations in lamp intensity ernit light used as the internai standard enhance dissociation of calcium phosphate

The answer is D. Because of the requirement of a cool flame in atornic absorption spectrophotometry, some calcium salts are not broken into their component atoms; calcium phosphate is ao example. The electrons in these anion-bound calcium atoms are then unable to absorb the incident wavelength of light and are not measured. Lanthanum binds the sarne anions more tightly than does calcium, which releases calcium from these salts. This allows the calcium electrons to achieve their ground state, absorb light energy, and be measured. (Kaplan and Pesce, pp. 96, 549) 12. Ata pH of 7.4, which of the enzymes listed catalyzes the following reaction?

pyruvate + NADH ~ lactate + NAD A. B. C. D.

Lactate oxidase Lactate dehydrogenase Pyruvate kinase Pyruvate decarboxylase

The answer is B. The International Union of Biochernistry (IUB) assigns a systematic name to each enzyme, which defines the substrate, the reaction catalyzed, and the coenzyme involved, if any. ln addition, IUB assigns a "recommended name" that is shorter, trivial, and more useable. None of the following groups are present in the reaction: oxygen, carboxyl group, phosphate group, ATP, ADP, etc. Therefore, the enzyme for this reaction is not an oxidase, carboxylase, or kinase and choices A, C, and D are elirninated. (Bishop et al., pp.

208-209, 218-219) 13. Chromatographic separation of a rnixture of solutes is based on A. B. C. D.

variable solubilities of solutes in the mobile and stationary phases spectral differences of solutes in the mobile and stationary phases selective degradation of solutes by the mobile and stationary phases differential distribution of solutes between the mobile and stationary phases

The answer is D. Chromatography is a collective term referring to the process of separating a mixture of solutes by differential distribution of the solutes between two phases. One phase, the solvent, is mobile and carries solutes with it as it

CLT Review Questions

passes over or through the stationary phase. The solutes interact to differing degrees with the fixed or stationary phase thereby causing separation. Thus, solutes that differ in their attraction to the stationary phase can be separated from one another. (Kaplan and Pesce, p. 107) 14. ln a coupled enzymatic method for measuring serum cholesterol, the color change observed during the indicator reaction is dependent upon the generation of

A. B. C. D.

ATP NAD oxygen hydrogen peroxide

The answer is D. The enzymatic method for measuring total cholesterol incubates serum with cholesterol esterase (to release free cholesterol). Subsequently, cholesterol oxidase oxidizes the cholesterol producing hydrogen peroxide. ln the presence of peroxidase, the hydrogen peroxide produced oxidizes the reduced dye to forma colored product. (Kaplan and Pesce, pp. 672-674) 15. pH 8.6 is used for serum protein electrophoresis so that

A. all serum proteins will have a net negative charge B. all serum proteins will have a net positive charge e. electroendosmosis is avoided D. heat production is minimized

The answer is A. Proteins are ampholytes whose terminal arnino and carboxyl groups, as well as ionizable side groups on component arnino acids, change their charges with change in pH. At a pH higher than the pK of these ionizable groups, dissociable hydrogen ions are lost to the medium resulting in no charge on each arnino group anda negative charge on each carboxyl group. The net charge on the protein therefore becomes negative. The amount of heat produced and the buffer migration (electroendosmosis) that occur are determined in large part by the concentration ofthe buffer. (Kaplan and Pesce, p. 201; Bishop et al., pp. 189- 192)

16. When quantitating serum protein using the biuret reaction, the biuret reagent is reacting with A. peptide bonds in proteins B. tyrosine residues in proteins e. free arnino groups in proteins D. ammonia released from proteins

The answer is A. ln an alkaline solution, the cuprous ions of the biuret reagent form coordinate bonds with the carbonyl groups of peptide bonds. This association results in the characteristic blue colored complex. (Bishop et al., pp. 186-187) 17. Which one of the following protein fractions, when separated in serum by electrophoresis on cellulose acetate, contains a single protein?

A. Albumin B. Alpha 1-globulin

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C. Alphaz-globulin D. Beta-globulin The answer is A. The large peak of albumin seen on a serum electropherogram is virtually pure albumin. Because electrophoresis on cellulose acetate separates proteins according to their net charges, the other peaks seen are mixtures of the proteins that share approximately the sarne net charge. (Bishop et al., pp. 189- 192)

18. Which of the following statements about the hexokinase reaction for serum glucose quantitation is true? A. B. C. D.

The amount of hydrogen peroxide produced is measured During the reaction cupric ions are reduced to cuprous ions The reaction generates a green condensation product with o-toluidine The coupled indicator reaction generates NADPH from glucose-6-phosphate

The answer is D. Hexokinase catalyzes the phosphorylation of several monosaccharides using ATP as the phosphate donor and produces the corresponding sugar-6-phosphate. However, the indicator reaction is specific for the substrate, glucose-6-phosphate. During this reaction, glucose-6-phosphate dehydrogenase (G-6-PD) catalyzes the reaction of glucose-6-phosphate and NADP to form 6phosphogluconate and NADPH. It is the high specificity of the indicator reaction that prevents interference from other monosaccharides. (Bishop et al., pp. 306-308; Lehmann, p. 52)

19. Turbidimetric assays for serum lipase measure the A. B. C. D.

amount of bile acid produced amount of titratable acid produced rate of production of NADH rate of degradation of triglyceride micelles

The answer is D. Lipase acts at the surface of triglyceride micelles, hydrolyzing terminal fatty acids from glycerol. As the micelles become smaller, they scatter less light and the substrate suspension becomes less turbid. The rate of clearing of turbidity reflects the amount of lipase activity. (Kaplan and Pesce, pp. 568- 569) 20. Osmolality measurements determine the A. activity of ions per kilogram of solvent B. grams of dissolved solutes per kilogram of solvent e. moles of dissolved solutes per kilogram of solvent D. equivalents of dissolved solutes per kilogram of solvent The answer is C. Osmolality is defined as the number of moles of particles per kilogram of water. lt is a measure of a solution' s concentration and it is irrelevant whether the particles are ions or nonionized solutes. Therefore, regardless of molecular weight, size, or charge each dissolved solute contributes equally to the osmolality value. (Kaplan and Pesce, pp. 271-272)

CLT Review Questions

21. Which of the following formulas can be used to calculate serum osmolality? A. B. C. D.

2.5 X Na+ Na++ K+ + c1- + C02 content (1.86 X Na+)+ (1118 X glucose)+ (112.8 X BUN) + 9 (Na++ K+) - (CI- + HC03 - )

The answer is C. This formula, using routine serum determinations of Na, glucose, and BUN, can be utilized to estimate serum osmolality. Because the body maintains electrical neutrality, each serum cation is balanced by an anion. Sodium is the major serum cation; therefore, two (or more accurately, 1.86) X Na (in mmol/L) accounts for most ions present. Glucose and BUN are also major contributors to serum osmolality. Dividing glucose (in mg/dL) by 18 and BUN (in mg/dL) by 2.8 converts these concentrations to mmol/L. The remaining unmeasured solutes that contribute to the osmolality in serum normally amount to 9 mrnol/L. The calculated osmolality is used to determine the osmolal gap by subtracting this calculated osmolality value from the actual or measured value. An abnormally high osmolal gap can be due to the ingestion of a volatile substance such as ethanol, methanol, or ethylene glycol. (Bishop et ai. , pp. 257-258; Kaplan and Pesce, pp. 459-461) 22. The following chemical reaction is used to measure the activity of which of the following serum enzymes?

coo1

H

- e - NH2 1 CH2

1

coo-

+

cooc=o 1

.--+ _

coo+

1

H - C - NH2

1

1

CH2

CH2

CH2

1

1

1 CH2

1

cooA. B. C. D.

coo1 c=o

1 CH2

coo-

1

coo-

Alkaline phosphatase Aspartate aminotransferase Gamma glutamyltransferase Lactate dehydrogenase

The answer is B. The International Union of Biochemistry (IUB) assigns a systematic name to each enzyme, which defines the substrate, the reaction catalyzed, and the coenzyme involved, if any. ln addition, IUB assigns a "recommended name" that is shorter, trivial, and more useable. The transfer of an amino group from one molecule (aspartate) to another (alpha-ketoglutarate) occurs in this reaction (i.e., amino transfer). ln addition, neither lactate, a glutamyl residue, nor phosphate is present eliminating choices A, C, and D. (Bishop et al., p. 220) 23. When using an automated instrument, the amount of carryover between consecutive samples is not affected by A. rinsing the probe between samples B. separating consecutive samples in a tubing by air segments

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C. using a separate reaction chamber for each sample D. using a serum blank The answer is D. Carryover is the percent error produced by interaction or crosscontarrúnation between adjacent samples. All techniques that rinse the components that touch adjacent samples or that increase the physical separation between adjacent samples decrease carryover. (Burtis and Ashwood, pp. 213, 219; Bishop et al., pp. 137-141) 24. When perforrrúng a thin layer chromatography procedure, the solvent front moved 10.0 centimeters. The substance of interest moved 4.0 centimeters. What is the Rr for the substance of interest? A. 0.25 B. 0.40 e. 2.5 D. 4.0 The answer is B. The retention time or Rr of a substance or compound is defined as the distance of spot rrúgration divided by the distance of the mobile phase rrúgration. (Burtis and Ashwood, p. 111)

25. Calculate the corrected creatinine clearance using these data obtained from a person with a 1.73 m 2 body surface area:

Serum creatinine: Urine creatinine: Urine volume:

1.3 mg/dL 2.4 mg/mL 1000 mL/24 hour

A. 119 mL/rrún B. 128 mL/rrún e. 139 mL/min D. 167 mL/rrún The answer is B. The formula for calculating creatinine clearance (CCr) is: CCr = U • V X 1.73 m2 P SA

CCr = creatinine clearance in mL/rrúnute U = urine creatinine P =plasma/serum creatinine V =urine volume in mL per minute SA =patient's body surface area

When using this formula, it is required that the urine and serum creatinine concentrations be in the sarne units so that the units cancel. Therefore, the urine concentration must be converted to 240 mg/dL before using the formula. The urine volume (V) must also be converted from mL per 24 hours to mL per minutes as follows: urine volume (mL) 24 hour

x

1 hour 60 rrúnutes

= y in mL/minute

Because creatinine excretion varies with muscle mass (i.e., body surface area), comparison to a reference range requires that the clearance be corrected for the individual's body surface area in square meters. (Brunzel, pp. 106-109)

CLT Review Questions

26. A serum sample is diluted 1 to 3 (1:3) before analysis and the following results obtained: Total protein Albumin

- 4.1 g/dL - 1.5 g/dL

Which total protein concentration should be reported? A. B. C. D.

4.1 g/dL 8.2 g/dL 12.3 g/dL 16.4 g/dL

The answer is C. The dilution factor for the dilution is 3; hence 3 times 4.1 is 12.3 g/dL. (Bishop et al., pp. 22-23) 27. Which of the following formulas is an accurate rearrangement of Beer's

law, when using a calibration constant (K)? A. B. C. D.

c=A!K b =A• K A= c/K K = a• c A

The answer is A. When an assay follows Beer's law, a calibration constant (K) can be derived and used to calculate the concentration of the analyte in unknown samples. The factor "K" is determined from the analyte absorptivity (a) and the instrument path length (b). With this substitution and subsequent rearrangement of Beer's law, the expression e= A/K is obtained. (Burtis and Ashwood, pp. 59-60)

28. A patient with biliary obstruction has a serum bilirubin assay performed. The bilirubin results determined using the Jendrassik-Grof method are as follows: Total bilirubin: 0.8 mg/dL Conjugated bilirubin: 1.0 mg/dL The clinicai laboratory technician does not report the results. The results obtained are most consistent with A. B. C. D.

a technical error occurring during analysis insufficient accelerator added to the total bilirubin reaction a reduced reaction time for the conjugated bilirubin reaction excess diazo reagent added to the conjugated bilirubin reaction

The answer is A. The conjugated fraction of bilirubin cannot exceed the total amount of bilirubin in a sample. This indicates a technical error during sample analysis. Each of the reaction conditions presented in choices B, C, and D will not cause the results obtained. (Bishop et al., pp. 386-388)

29. ln an adult, a blood glucose leve! of 35 mg/dL is A. normal B. dangerously low e. dangerously high D. physiologically impossible

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The answer is B. True hypoglycemia of this magnitude can cause neurologic symptoms and may result in irreversible damage. A very low serum glucose value also may be an artifact caused by cellular metabolism or bacterial contamination if serum is not separated from cells promptly. (Kaplan and Pesce, pp. 634-635) 30. Measurements of urinary human chorionic gonadotropin (HCG) in men can be used to A. diagnose hypogonadism B. detect testicular tumors e. assess pituitary function D. detect excessive estrogen secretion The answer is B. ln addition to the detection of pregnancy, HCG is frequently used to detect and monitor germ cell tumors of the testis and ovary. HCG levels are also used to diagnose and monitor gestational trophoblastic disease such as hydatidiform mole, gestational choriocarcinoma, and placental-site trophoblastic tumor. (Bishop et al., pp. 490-491) 31. Xanthochromic cerebrospinal fluid is an indicator of A. bacterial meningitis B. increased pressure of cerebrospinal fluid e. increased protein concentration in cerebrospinal fluid D. cerebral hemorrhage The answer is D. Xanthochromia in spinal fluid is yellow pigmentation caused by the presence of bilirubin. The bilirubin results from breakdown of heme released from erythrocytes after bleeding into the brain or spinal column such as occurs in cerebral hemorrhage. (Brunzel, pp. 368-370; Bishop et al., p. 534) 32. A creatinine clearance result below the normal reference range most likely indicates a decrease in A. hepatic blood flow B. hepatic creatinine synthesis e. renal blood flow D. renal glomerular filtration The answer is D. A creatinine clearance test determines the volume (mL) of plasma cleared of creatinine per minute. Creatinine readily passes the glomerular filtration barrier of functioning glomeruli and is excreted in the urine. No tubular reabsorption and negligible tubular secretion of creatinine occurs in the nephrons. Assuming no preanalytical or analytical errors, when a creatinine clearance is below the appropriate reference interval, it indicates a decrease in functioning glomeruli, i.e., glomerular filtration. (Brunzel, pp. 106-109) 33. A physician suspects that a patient has Cushing's syndrome. Based on this information, which of the following tests would assist in this diagnosis? A. Cortisol levei B. Vanillylmandelic acid level

CLT Review Questions

C. Thyroid-stimulating hormone level D. 24-hour creatinine clearance The answer is A. Both Cushing's syndrome and Cushing's disease are associated with excessive leveis of cortisol. Hypercortisolism can be dueto tumors of the pituitary, adrenal glands, or ectopic ACTH-secreting tumors. Exogenous administration of glucocorticoids or ACTH can also result in symptoms of Cushing's syndrome. (Bishop et al., pp. 414-415) 34. Which of the glucose tolerances curves in the following figure meet NDDG (National Diabetes Data Group) criteria for the diagnosis of diabetes mellitus?

400

e 350

8

300

~

00

e

1 250

5 IJ

::::1

"iiO

200

m 150

.!! c:i..

so 30

60

90

120

150

180

Minutes

A. B. C. D.

Curves 1 and 2 Curves 1 and 4 Curves 3 and 4 Only curve 4

The answer is C. NDDG criteria for the diagnosis of diabetes mellitus include either (1) fasting serum glucose levei greater than 140 mg/dL on more than one occasion, or (2) two or more serum samples with glucose leveis greater than 200 mg/dL following a meal. Curve 3 meets the latter criterion and curve 4 meets both criteria. (Burtis and Ashwood, pp. 439-441; Bishop et al., p. 307) 35. Identify the results that are not in electrolyte balance. (Results are m mrnol/L.) Na+

K+

c1-

C02 content

A. 125 B. 135 c. 145 D. 150

4.5 3.5 4.0 5.0

100 95 90 110

10 28 15 30

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The answer is C. Electrolyte balance is determined by calculating the anion gap (AG) using the formula: (Na + K) - (Cl + C02 content) or (Na) - (Cl + C0 2 content). The latter calculation predominates because potassium can be significantly increased in samples that are hemolyzed. The AG or difference obtained reflects the presence of anions that are not included in the equation. The reference interval for the anion gap (without K) is 8 to 16 mmol/L; with K, the reference interval is 10 to 20 mrnol/L. (Lehmann, pp. 138-139) 36. Which set of results is consistent with a diagnosis of respiratory acidosis? Arterial pH

A. Decreased B. Decreased e. Increased D. lncreased

Arterial pC02

Decreased Increased Decreased Increased

The answer is B. Respiratory acidosis is characterized by a decreased blood pH caused by an excess of C02 , i.e., an absolute excess of carbonic acid. This occurs when the lungs are not able to adequately expel C02 . The carbonic acid (H2 C0 3 ) concentration can be calculated from the pC02 as follows: 0.03 X pC02 = H 2 C0 3 . When there is ao excess of carbonic acid relative to the bicarbonate concentration in the blood, the pH will decrease. Note that to evaluate acid-base disorders, the p0 2 is not needed. (Lehmann, p.163; Bishop et al., pp. 241-243)

37. Review the following serum test results: Creatinine Cholesterol Glucose Urea Uric acid

2.5 mg/dL 220 mg/dL 110 mg/dL 40 mg/dL 6.9 mg/dL

(0.7-1.5 mg/dL) (< 220 mg/dL) (70- 110 mg/dL) (8-26 mg/dL) (2.5- 7 .O mg/dL)

These results are most consistent with A. compromised renal function B. impaired glucose metabolism e. diagnosis of gouty arthritis D. increased risk for coronary artery disease The answer is A. Urea, a product of protein metabolism, and creatinine, a byproduct of muscle metabolism, are dependent on the kidney for elimination from the body. Hence, elevated serum values are associated with renal disease or loss of renal function. Whereas elevations in urea can also occur due to nonrenal factors, these conditions will not demonstrate the magnitude of creatinine increase also present in these results without compromised renal function. (Lehmann, pp. 155-1 56) 38. The amniotic fluid from a 40-year-old female is tested and the following results obtained:

Test

Result

Reference Interval

L/S ratio: Alpha-fetoprotein:

1.6 normal

> 2.0 normal

CLT Review Questions

Based on these results, if the fetus were delivered today it could have A . Down's syndrome B. an open neural tube defect e. respiratory distress syndrome D. an increased risk of renal failure The answer is C. An L/S ratio less than 2.0 is associated with increased risk of respiratory distress syndrome in the neonate. A decrease in alpha-fetoprotein (AFP) is associated with Down's syndrome; whereas, an increase is associated with open neural tube defects. Fetal renal function cannot be assessed from the results provided. (Burtis and Ashwood, pp. 906-917) 39. A plasma sample was analyzed using ion selective electrodes (ISE) and the following electrolyte results obtained: Na+ K+

c1HC03 -

= 140 mrnol/L = 14.0 mmol/L = 112 mmol/L = 18 mmol/L

These results are consistent with a specimen that A. B. C. D.

is severely lipernic is slightly hemolyzed was collected in an EDTA tube needs to be diluted and reanalyzed

The answer is C. The result that is of most concern is the potassium of 14.0 mmol/L, which is incompatible with life. Potassium EDTA is a commonly used anticoagulant tube for the collection of blood plasma. Occasionally, an EDTA plasma specimen is inadvertently analyzed for electrolytes, as in this case. Lipernia can affect electrolyte results but would cause a false decrease in electrolyte values if an indirect ISE method was used. Slight hemolysis is not correct because of the magnitude of increase in the potassium result. Note that the potassium value increases only ~ 0.6% with the release of 10 mg/dL hemoglobin from erythrocytes. Lastly, a potassium of 14.0 mmol/L is within the linear range of most potassium methodologies, elirninating the need to dilute this sample. (Burtis and Ashwood, pp. 37-38)

40. A patient suffering from an acute hemolytic episode has a total bilirubin of 2.2 mg/dL (0.2-1.1 mg/dL). Based on this information, which of the following conjugated and unconjugated bilirubin results would you expect? (Reference values are in parentheses.) Conjugated (0-0.2 mg/dL) A. 0.1

B. 0.5 e. LO D. 1.6

Unconjugated (O - 1.1 mg/dL) 2.1 1.7 1.2 0 .6

The answer is A. Acute hemolytic anemia can result in prehepatic jaundice characterized by increased serum bilirubin due to a process that precedes bilirubin processing by the liver. Despite a healthy liver, its ability to rapidly process the

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additional unconjugated bilirubin presented is compromised. This is reflected by an increased unconjugated bilirubin fraction with a normal conjugated fraction. (Kaplan and Pesce, pp. 511-512) 41. Which of the following urine specimens is most useful when screening for glucosuria? A. First morning specimen B. 2-hour postprandial specimen e. 24-hour urine specimen D. Midstream clean catch specimen

The answer is B. A 2-hour postprandial specimen collected after a meal is an ideal specimen to screen for glucosuria. ln a healthy individual, glucose does not appear in the urine because it is reabsorbed by the renal tubules. However, in an uncontrolled diabetic individual, the amount of glucose presented to the tubules exceeds their renal capacity for reabsorption. Because the arnount of glucose in the urine ultrafiltrate is dependent on the glucose concentration in the plasma, specimens collected on these individuais during a fasting state may not detect any glucose. (Brunzel, pp. 55, 172-174) 42. What is the longest time that a urine specimen can remain at room temperature before it is no longer considered acceptable for analysis? A. 1 hour B. 2 hours e. 3 hours D. 4 hours

The answer is B. Due to the changes that can take place in unpreserved urine, specimens should be analyzed within 2 hours of collection or precautions taken, such as refrigerate the specimen or add an appropriate preservative. (Brunzel, pp. 58-59; NCCLS, p. 4) 43. Which of the following situations requires corrective action before pro-

ceeding with specimen testing? A. B. C. D.

Room temperature of the laboratory is 25ºC Refractometer result obtained using Type 1 water is 1.000 Temperature of the refrigerator that stores the QC materiais reads 2ºC Reagent strip protein result is trace when using negative control material

The answer is D. Quality control materials are used to assess whether a test is perforrning properly, i.e., they monitor analytical performance. A negative control material should produce negative results when analyzed. A trace protein indicates a change in performance and is not acceptable. Hence, intervention is required to identify and correct the cause of the false-positive result before analyzing any patient samples. (Brunzel, pp. 3~0, 148- 152) 44. The urinalysis reagent strips from four different bottles are evaluated using

the current quality control (QC) materiais. Which bottle of reagent strips is acceptable for use today?

CLT Review Questions

A. B. C. D.

Bottle #1: Bottle #2: Bottle #3: Bottle #4:

expiration expiration expiration expiration

date is today; QC acceptable date was yesterday; QC acceptable date is next week; QC not acceptable date is tomorrow; QC not acceptable

The answer is A. Quality control materials monitor the performance of a test, i.e., each of the tests on the reagent strip. Acceptable QC results must be obtained before the reagent strips can be used to test patient samples. Reagent strips, as with other testing systems (e.g., kits), can be used until their expiration date is passed or exceeded. ln the case of bottle #1, that would be the next day. (Brunzel, pp. 39-41)

45. Which protein test is not able to detect immunoglobulin light chains (i.e., Bence Jones proteins) in urine? A. B. C. D.

Immunoelectrophoresis Reagent strip protein test Sulfosalicylic acid (SSA) precipitation test Protein coagulates between 40 to 60ºC

The answer is B. Immunoglobulin light chains will not be detected by commercial reagent strip protein tests based on the "protein error of indicators." These reagent strip tests are most sensitive for albumin. Originally, immunoglobulin light chains (or Bence Jones proteins) were recognized by their unique solubility characteristics, i.e., coagulates between 40 to 60ºC and redissolves at lOOºC. Sulfosalicylic acid is a general protein-precipitating agent and will precipitate ali proteins, including immunoglobulin light chains. Immunoelectrophoresis is a definitive method for identifying specific types of immunoglobulin polypeptides. (Brunzel, pp. 169-172)

46. When drugs containing free sulfhydryl groups are excreted in the urine, they can cause false-positive results for which reagent strip test? A. B. C. D.

Blood Ketones Leukocyte esterase Nitrite

The answer is B. Compounds that contain free sulfhydryl groups react with sodium nitroprusside (nitroferricyanide) to produce a false-positive reagent teststrip test for ketones. Examples of drugs containing free sulfhydryl groups include MESNA, a chemotherapy agent; penicillamine, a chelating agent; and captopril, an antihypertensive drug. (Brunzel, p. 179)

47. Which of the following sets of urinalysis results, physical appearance and specific gravity (SG), is physiologically possible and indicates a concentrated urine?

SG A. 1.000 B. 1.015 e. 1.030 D. 1.050

Color colorless brown yellow amber

15

16

1.

Clinicai Chemistry

The answer is C. The range of physiologically possible urine specific gravity is 1.002 to - 1.040. Both choice A, a SG equal to that of pure water, and D the body is unable to produce. Specific gravity values greater than 1.040 can be observed in urine that is contaminated with x-ray contrast dye. Urine color can provide a crude indicator ofurine concentration. Typically, urine is a shade ofyellow with the depth of the color indicating a more concentrated specimen, i.e., a pale yellow sample is less concentrated than a dark yellow urine. (Brunzel, pp. 121-122, 128- 129) 48. A urinalysis is performed on a specimen from a patient with active diabetes insipidus. Which of the following laboratory data sets is most consistent with this diagnosis? SG (refractometer) A. 1.005 B. 1.020 e. 1.005 D. 1.020

Glucose

Urine Volume/day

negative negative positive positive

3100 mL 2500 mL 3500 mL 2000 mL

The answer is A. Diabetes insipidus, a metabolic disorder, is characterized by polyuria (excretion of;::: 3 L/d) and polydipsia. Dueto either defective production of arginine vasopressin (antidiuretic hormone) or lack of renal response to the hormone, these individuals are unable to reabsorb water in their renal tubules. Hence, they produce copious amounts of dilute (low SG) urine. ln contrast, individuals with uncontrolled diabetes mellitus could produce large volumes of high specific gravity urine dueto the presence of glucose. (Brunzel, pp. 102, 273, 300)

49. The following results are obtained on a fresh randomly collected urine spec1men: Parameter

Result

Confirmatory test

Result

pH SG Blood Protein Glucose Ketone Bilirubin Urobilinogen Nitrite Leukocyte esterase

5.5 1.025 trace 30 mg/dL 100 mg/dL negative negative 1.0 mg/dL negative positive

Microscopic Exam:

2-5 RBCs per high power field 5-1 O WBCs per high power field

Refractometer

1.027

Clinitest

positive

Ictotest

positive

Which of the following statements regarding these results is true? A. B. C. D.

An abnormal amount of bilirubin is present in the urine Ascorbic acid is causing the positive leukocyte esterase The blood present is causing the protein result to be positive Radiographic contrast media (x-ray dye) is causing the high specific gravity

The answer is A. The Ictotest is more sensitive for bilirubin than the reagent strip test. Hence, it is possible to have a positive Ictotest but a negative reagent strip

CLT Review Questions

test. Ascorbic acid does not have any effect on the leukocyte esterase test of any reagent strip (see manufacturer's product insert). The reagent strip protein test is primarily sensitive to albumin. Hence, a trace blood result, which according to the reagent strip manufacturer corresponds to - 0.2 to 0.6 mg/dL hemoglobin, is insufficient to cause a positive test. The specific gravity of 1.025 is within the reference range. Radiographic contrast media in urine causes specific gravity results to exceed those physiologically possible, i.e., > 1.040. (Brunzel, pp. 154, 156, 171, 183- 184) 50. Which of the following sediment components is the best indicator that an infection or inflarnmatory process is taking place in the kidney and not in the lower urinary tract? A. RBCs

B. Bacteria WBC casts D. Granular casts

e.

The answer is C. Because casts are only formed in the kidney, the components entrapped within them also originate there. White blood cells respond to infective and inflammatory processes by localizing in the affected tissue. Therefore, WBC casts are indicative of an infection or inflammatory process occurring in the kidney. Blood cells can pathologically enter the urinary tract at any point. The presence of bacteria is useful in diagnosing a urinary tract infection; however, their presence does not indicate at which levei of the urinary tract the infection resides, i.e., is the infection in the urethra, bladder, renal pelvis, or renal interstitium? (Brunzel, pp. 235-236, 254) 51. Urine sediment that contains red blood cells, red blood cell casts, and protein is characteristic of A. bladder infection B. Fanconi syndrome e. nephrotic syndrome D. acute glomerulonephritis The answer is D. ln acute glomerulonephritis, glomerular inflamrnation and injury alters the permeability of the glomerular filtration barrier such that red blood cells and plasma proteins, particularly albumin, enter the renal tubules. As casts form in the distal tubules, the red blood cells present are incorporated into the cast matrix. (Brunzel, pp. 234- 235, 277) 52. Hyaline casts are found in increased numbers in the urine sediment A. B. C. D.

when the urine is alkaline following strenuous exercise when examined using bright light whenever an abnormal amount of protein is present

The answer is B. The number of hyaline casts in urine sediment increases following exercise and is not considered pathologic, particularly when they are the only abnormality present. ln these cases, temporary and minor dehydration enhances stagnation of the ultrafiltrate and increased cast formation. (Brunzel, pp. 229-233; Strasinger, p. 88)

•

17

18

1.

Clinicai Chemistry

53. Which type of microscopy would best aid in the differentiation of red blood cells from the form of monohydrate calcium oxalate crystals that resemble RBCs? A. B. C. D.

Darkfield microscopy Phase-contrast microscopy Polarizing microscopy Interference-contrast microscopy

The answer is C. Both the monohydrate and dihydrate forms of calcium oxalate demonstrate birefringence when using polarizing microscopy. ln contrast, cells, such as blood cells, epithelial cells, bacteria, etc., are not capable of refracting light; hence, they are not birefringent. This fact enables the differentiation of crystalline entities from look-alike cellular components in urine sediment. (Brunzel, pp. 15-16, 246-247) 54. During the microscopic exarnination of a urine sediment, a clinicai laboratory technician observes an entity approximately the size of a white blood cell that demonstrates a "flitting or jerky motion." The entity observed is most likely a A. bacterium B. glitter cell e. pinworm D. trichomonad

The answer is D. Trichomonads exhibit a characteristic flitting or jerky motion when observed alive in fresh urine sediment. They are protozoan flagellates with anterior and posterior flagella and an undulating membrane. However, urine is not an ideal medium for their survival and trichomonads can rapidly die, losing their motility. Dead trichomonads round up and can become impossible to differentiate from white blood cells. Their identification relies predominantly on their characteristic motion or the observation of flagella and an undulating membrane. (Brunzel, pp. 258-259) 55. Which of the following urinalysis findings include contradictory results that should not be reported? A. B. C. D.

pH 5.0, small blood, cystine crystals pH 6.0, bilirubin positive, ammonium biurate crystals pH 7 .O, protein trace, caleium oxalate crystals pH 7.5, nitrite positive, amorphous phosphates

The answer is B. Ammonium biurate crystals are found in alkaline or neutral urine (::=: pH 7.0) and will dissolve upon heating or the addition of acetic acid. Hence, there is a conflict in the reported pH and the crystal identified. Ammonium biurate crystals can resemble some forms of sulfonamide drug crystals, and it is the pH that assists in the proper identification of these similar appearing crystals. No conflicting results are present in the other choices. (Brunzel, p. 252) 56. The following crystals are observed in a randornly collected urine specimen with a pH of 7.0.

CLT Review Questions

The identity of these crystals is most likely A. cystine B . uric acid e. triple phosphate D. ammonium biurate The answer is C. Both triple phosphate and ammonium biurate crystals can be present in urine with an alkaline pH. However, ammonium biurate crystals have a yellow to brown coloration and are primarily spherical in shape. The most common forms of triple phosphate crystals are a 3- to 6-sided prism, frequently described as "coffin lids." Cystine and uric acid crystals are present only in acidic urines. (Brunzel, pp. 243- 252) 57. When present in the urine, which of the following substances could cause false-positive reagent strip blood results? A. B. C. D.

Ascorbic acid Myoglobin Free-sulfhydryl drug X-ray contrast media

The answer is B. The reagent strip test for blood is based on the pseudoperoxidase activity of the heme moiety. Myoglobin is also a heme-containing protein capable of oxidizing the chromogen on the reagent strip pad. (Brunzel, pp. 154, 158-162) 58. ln which of the glucose tolerances shown in the figure would you expect to find concurrent glycosuria? 400

e

350

~

300

..e 1

z~o

~

"

200

~

I~

... ~

100 ~

30

60

90

120

Minutes (Tietz, 3rd., p . 252)

150

180

19

20

1.

Clinicai Chemistry

A. B. C. D.

Curves 1 and 2 Curves 1 and 3 Curves 3 and 4 Only curve 4

The answer is C. The normal renal threshold for glucose is a plasma level of 160 to 180 mg/dL. There is a limited amount of reabsorption mechanism in the proximal convoluted tubules. At blood glucose levels higher than the renal threshold, the limited reabsorption allows excretion of the excess glucose in the urine. Both curves 3 and 4 exceed this renal threshold value. Individuals with renal disease, which includes many diabetic patients, may have even lower renal thresholds for glucose. (Brunzel, p. 172)

59. A physician wants a urinalysis performed on a midstream clean catch urine specimen to evaluate whether a woman has a urinary tract infection. Which of the following urinalysis results suggests that a new specimen should be collected? A. B. C. D.

Chemical exam: blood positive Chemical exam: nitrite negative Physical exam: pale yellow, clear Microscopic exam: many squamous epithelial cells

The answer is D. The presence of many squamous epithelial cells in a specimen indicates that it is not a midstream collection. The epithelial cells originate from the vaginal and perineal areas and indicate that the urine can also be contaminated with the normal flora (bacteria) that also resides in these areas. (Brunzel, pp. 56-57, 225- 226)

60. The following results are obtained on a fresh, randomly collected urine specimen: Parameter

Result

pH SG Blood Protein Glucose Ketone Bilirubin Urobi linogen Nitrite Leukocyte esterase

5.5 1.025

Microscopic Exam:

5-1 O RBCs per high power field 2-5 WBCs per high power field 2-5 fatty casts 0-2 waxy casts

small 500 mg/dL negative negative negative 1.0 mg/dL negative negative

Whích of the following disorders is most consistent with these results? A. B. C. D.

Cystitis Nephrotic syndrome Acute pyelonephritis Acute glomerulonephritis

CLS Review Questions

21

The answer is B. Nephrotic syndrome is characterized by heavy proteinuria. This results from changes in the permeability of the glomerular filtration barrier (GFB). Consequently, serum lipids are also able to pass the GFB into the urine and can appear in the urine sediment as free fat globules, in casts, or in cells called oval fat bodies. (Brunzel, pp. 238, 282-283)

CLS Review Questions 1. Which of the following analytes can deteriorate if a blood specimen is exposed to light? A. B. C. D.

Bilirubin Calcium Cholesterol Glucose

The answer is A. Bilirubin degrades upon exposure to light, fluorescent as well as indirect or direct sunlight. Specimens should be stored in the dark and refrigerated unless analyzed within three hours. (Bishop et ai., pp. 386-388) 2. Which of the following collection tubes is the specimen of choice for the deterrnination of glycated hemoglobin? A. B. C. D.

Citrate tube EDTA tube Red/marble top tube Ammonium heparin tube

The answer is B. Whole blood collected in EDTA is the preferred specimen. Most methods require subsequent preparation of a hemolysate using a celllysing agent. Whole blood specimens can be stored up to 5 days at 2-5ºC. (Lehmann, p. 50; Kaplan and Pesce, pp. 635-637) 3. Which of the following serum samples is satisfactory for alkaline phosphatase measurement? A. B. C. D.

Sample frozen ovemight Sample refrigerated for 1 hour Sample with moderate hemolysis Sample maintained at room temperature for 2 hours

The answer is A. Alkaline phosphatase in serum specimens is stable at room temperature for up to 4 hours. Significant increases in alkaline phosphatase activity will be obtained if samples are refrigerated or frozen. It is postulated that low temperatures cause complexes between ALP and lipoproteins to dissociate. Slight hemolysis is acceptable; however, more severely hemolyzed specimens should be recollected. (Kaplan and Pesce, pp. 521- 522) 4. A clinicai laboratory scientist (CLS) is summoned to the emergency room to draw blood on Jane Doe. At the sarne time, numerous injury victims begin to arrive by ambulances from a multiple car accident. The CLS does

'

22

1.

Clinicai Chemistry

not want to disturb the now busy ER staff. According to the roam board, Jane Doe is in Ward D. The CLS goes to Ward D and collects blood from the only woman present. She is unable to verify that it is actually Jane Doe because the patient does not speak English and does not have an identification bracelet. However, the woman nods approvingly when asked if her name is Jane Doe. The CLS labels the blood specimen as "Jane Doe" and retums to the laboratory. What should be done next? A. Proceed with testing; woman positively identified herself B. Specimen should not be used; positive identification of patient was never made C. Specimen should not be used until an ER staff member comes to the laboratory to sign a waiver verifying patient's identity D. Proceed with testing; "Jane" is a female name and she was the only female present in Ward D, so it must be her

The answer is B. A patient must be positively identified before their blood is collected and labeled as such. If a patient is unable to identify themselves, then another health care worker (nurse, physician, etc.), family member, or visitar must verify their identity. Incorrectly identified samples can produce life-threatening consequences when results are reported on the wrong patient. (Bishop et al., p. 41)

5. Which of the following changes occurs in a serum specimen that is maintained at 4ºC for 8 hours? A. B. C. D.

Amylase activity decreases Alkaline phosphatase activity increases Lactate dehydrogenase activity increases Alanine aminotransferase activity decreases

The answer is B. Alkaline phosphatase (ALP) activity falsely increases during storage. This phenomenon has been attributed to the dissociation of complexes formed between ALP and lipoproteins at low temperatures. ln contrast, lactate dehydrogenase activity decreases with refrigeration and samples should be maintained at room temperature. Alanine aminotransferase (ALT) and amylase activity is stable in refrigerated samples for 7 days or several months, respectively. (Bishop et al., pp. 221 , 223; Kaplan and Pesce, pp. 568, 610)

6. Which of the following measurements usually takes place at room temperature? A. B. C. D.

p0 2 by blood gas analyzer Sodium by potentiometry Ionized calcium by ion selective electrode Creatinine by kinetic alkaline picrate method

The answer is D. The kinetic alkaline picrate or Jaffe method is a spectrophotometric method based on the differential rate of colar formation of noncreatinine chromogens versus creatinine. This reaction can be performed at room temperature. ln contrast, blood gas analysis and potentiometric methods, such as sodium and ionized calcium analysis, require that measurements occur in a controlled environment usually maintained at 37ºC. (Kaplan and Pesce, pp. 281- 285)

CLS Review Questions

7. A centrifuge is loaded with patient samples and turned on. It begins to vibrate and dance across the table top. Which of the following statements best accounts for this observation? A. B. C. D.

The brake was left "ON" The brushes need to be replaced The rate of acceleration was too high The tubes are not balanced in the carriers

The answer is D. The balancing of specimen tubes in carriers is critica! for proper functioning of a centrifuge. When carriers are not properly balanced, a centrifuge vibrates and can do so quite vigorously, such that it will move across a flat surface. Newer centrifuges will automatically decelerate and shut down when carriers are improperly balanced. (Kaplan and Pesce, pp. 24-25) 8. Which of the following
10. Which of the following enzymes catalyzes the conversion of glucose to hydrogen peroxide and gluconic acid? A. B. C. D.

Peroxidase Hexokinase Glucose oxidase Glucose-6-phosphate dehydrogenase

The answer is C. Glucose oxidase catalyzes the oxidation of glucose by oxygen to form gluconic acid and hydrogen peroxide. Glucose is quantitated by deter-

23

24

1.

Clinical Chemistry

mining the amount of oxygen consumed in the reaction (amperometry) or by using a second "indicator" reaction to measure the amount of hydrogen peroxide produced. (Bishop et al., p. 307; Kaplan and Pesce, pp. 634-635) 11. Which of the following constituents has the greatest effect on serum osmolality? A. Glucose

B. Protein Sodium D. Urea

e.

The answer is C. Osmolality is a measure of a solution's concentration based on the number of solutes present per kilogram of solvent. Each solute regardless of molecular weight contributes equally to the osmolality. The major osmotic solutes in serum are sodium, chloride, glucose, and urea. The concentration of sodium far exceeds that of any other serum solute. Typical solute leveis in normal serum include: sodium - 140 mmol/L; glucose - 5.5 mmol/L (or 100 mg/dL); urea - 2.5 mmol/L (15 mg/dL). ln addition, the molar amount of the various serum proteins is very low. (Bishop et al., pp. 256-258) 12. When measuring serum bilirubin, the purpose of adding caffeine-sodium benzoate or dyphylline to the reaction mixture is to A. accelerate the reaction with unconjugated bilirubin

B. stop the reaction by destroying excess diazo reagent C. enable azobilirubin formation with conjugated bílirubin D. shift the wavelength absorbed by azobilirubin for increased sensitivity The answer is A. Unconjugated bilirubin reacts very slowly with the aqueous diazotizing colar reagent used in bilirubin assays. However, in the presence of an accelerating reagent, such as caffeine-sodium benzoate (Jendrassik-Grof method) or dyphylline (modified Jendrassik-Grof method), unconjugated bilirubin's solubility is enhanced and it can readily participate in the intended reaction to form azobilirubin. (Bishop et al., pp. 386- 388; Kaplan and Pesce, pp. 523-527) 13. When iontophoresis is used to collect sweat for chloride analysis, pilocarpine is used to A. B. C. D.

clean the skin area complex with chloride complete the circuit induce sweat secretion

The answer is D. Pilocarpine is driven into the skin surface by iontophoresis (the migration of ions induced by direct current). lt stimulates the production of sweat, which is subsequently collected on preweighed filter paper or gauze for the analysis of chloride or sodium. (Bishop et al., pp. 477-478)

14. When using atomic absorption spectrophotometry for calcium quantitation, lanthanum or strontium is routinely added to each sample to

CLS Review Questions

A. minimize matrix interferences B. prevent ionization of calcium atoms e. avoid interference from phosphate in the sample D. reduce fluctuations from the hollow-cathode lamp

The answer is C. The atoms produced by the fuel-rich flame used in atornic absorption spectrophotometry (AAS) are in their ground state and readily available for light absorption. However, one problem with AAS due to the cooler flame temperature is the inability of the flame to dissociate samples into free atorns. For example, phosphate can interfere with calcium analysis in specimens by the formation of calcium phosphate. ln order to prevent this interference, lanthanum or strontium, both of which form stable complexes with phosphate, is added prior to analysis. (Bishop et al., pp. 103- 105)

15. Ion-selective electrodes compare the voltage (potential) of the measuring electrode to the A. B. C. D.

reference voltage resistivity of the sample conductivity of the sample current required to establish the voltage

The answer is A. Ion-selective electrodes are potentiometric measurements based on the potential (voltage) difference that develops between two electrodes under conditions of zero current. The reference electrode provides a stable, easily reproducible half-cell potential (which does not change) while the indicating or measuring electrode produces a half-cell potential dependent on the "activity" of the analyte being measured. The potential difference between that develops between the measuring and reference electrode correlates to concentration of analyte in the specimen. (Kaplan and Pesce, pp. 278-281)

16. Which of the following statements best describes the principle of p0 2 measurement used in blood gas analyzers? A. B. C. D.

H+ are generated by a reaction at the electrode surface The amount of 0 2 oxidized to hydrogen peroxide is measured The number of electrons used to reduce 0 2 is measured A voltage between the measuring half-cell and a reference half-cell is determined

The answer is C. The electrode for p02 measurements is based on arnperometry. ln amperometry, a current is measured while a fixed potential is applied. ln the p02 electrode, a known stable voltage (potential) is maintained between an anode and a platinum cathode. As oxygen diffuses through a membrane it is reduced at the cathode by electrons furnished by the anode, i.e., 0 2 + 2H+ + 2e- -+ H 20 2 . The amount of current (electron flow) is measured and expressed as the p02 in mmHg. (Kaplan and Pesce, pp. 277, 287-288) 17. Which of the following methods is not used to quantitate serum albumin? A. Nephelornetry B. Electrophoresis at pH 8.6

25

r

26

1.

Clinicai Chemistry

C. Sulfosalicylic acid (SSA) precipitation test D . Dye-binding method using bromcresol green The answer is C. Bromcresol green, under appropriate conditions of pH and ionic strength, binds specifically to albumin. This shifts the wavelength of light absorbed by the dye. Electrophoresis of serum proteins results in a virtually pure band of albumin, in contrast to the other bands that are mixtures of proteins. Nephelornetry is specific for an individual protein by virtue of antigen-antibody recognition. Sulfosalicylic acid is a general protein precipitant and is not specific for albumin. (Kaplan and Pesce, pp. S 18-519) 18. If a moderately hemolyzed serum specimen is used for protein electrophoresis, which of the following protein fractions will be elevated? A. B. C. D.

Albumin Alpha1 -globulin Beta-globulin Gamma-globulin

The answer is C. Hemolysis causes an increase primarily in í3-globulins, the fraction where free hemoglobin migrates. However, a small amount of a hemoglobin-haptoglobin complex can also appear between the a 2 and fractions. (Burtis and Ashwood, p. 131) 19. LDL cholesterol can be estimated using the Friedewald formula: LDL =Total cholesterol - HDL -

triglyceride

s

This calculation should not be used when the A. B. C. D.

HDL cholesterol is greater than 40 mg/dL triglyceride levei is greater than 400 mg/dL plasma shows no visible evidence of lipemia total cholesterol is elevated based on the age and sex of the patient

The answer is B. The formula estimates cholesterol contained in LDL particles by subtracting cholesterol in other lipoprotein particles from total cholesterol. An essential assumption is that 20% (1/5 ) of VLDL particles is cholesterol and that measured triglyceride accurately estimates the amount of VLDL. When the triglyceride result is excessively high (> 400 mg/dL), this assumption is not valid. (Burtis and Ashwood, p. 488) 20. Using the following figure of "drug concentration versus time after oral dose," calculate this drug's half-life in the circulation. ~

~

4

~ 3

= b"' =

o ;::

2

~

8

e

1

Q

o~~~~~~~~~-

0123

4

56

Time after oral dose (h)

CLS Review Questions

A. 0.5 h B. 1.5 h e. 2.5 h D. 4h The answer is B . The biologic half-life of a drug is the length of time required for the blood concentration to decrease by one-half. ln the figure, the blood concentration is approximately 4 µg/mL at 2 h and approximately 2 µg/mL at 3.5 h; thus the concentration has decreased by one-half in 3.5 - 2 = 1.5 h. (Burtis and Ashwood, pp. 608-610)

21. The following data are obtained using a cholesterol method:

Sampie

Absorbance

150 mg/dL standard Normal control Abnormal control Patient

0.200 0.200 0.400 0.500

(mean: 150 mg/dL; lSD = 15) (mean: 275 mg/dL; lSD = 25)

The patient's cholesterol result should be reported as A. B. C. D.

60 mg/dL 150 mg/dL 375 mg/dL 500 mg/dL

The answer is C. Beer's law states that the absorbance of a solution is directly related to its concentration. Rearrangement yields the formula: Cu A)A. X C. where the subscript u denotes the unknown values and s denotes the standard values. (Kaplan and Pesce, pp. 38-39)

=

22. The following results are obtained from a 28-year-old diabetic patient:

Analyte Na+ K+ c1Glucose: BUN: Serum osmolality:

Result

Reference Interval

140 mmol/L 3.8 mmol/L 101 mmol/L 215 mg/dL 25 mg/dL 328 mOsmol/kg

(136-145) (3.5-5.0) (99-109) (70-105) (10-20) (275-295)

Based on this data, what is the patient's osmolal gap? A. B. C. D.

8 mOsmol/kg 27 mOsmol/kg 48 mOsmol/kg 52 mOsmol/kg

The answer is B. A common formula used to calculate osmolality, in mOsmol/kg, is Serum Osmolality =2 X Na+ (mmol/L) +

Glucose (mg/dL) BUN (mg/dL) + ---'----'"'----~ 18 2.8

The osmolal gap (OG) is then calculated as follows: Osmolal gap (mOsmol/kg) = measured osmolality - calculated osmolality (Kaplan & Pesce, pp. 271-272)

27

28

1.

Clinicai Chemistry

23. Using the following data, calculate the corrected creatinine clearance. Serum creatinine: Urine creatinine: Urine volume: Body surface area:

1.8 mg/dL 2.7 mg/mL 640 mL/24h 1.25 m 2

A. 41 mL/min B. 67 mL/min e. 92 mL/min D. 132 mL/min

The answer is C. The formula for calculating a corrected creatinine clearance (Ccr) is: Ccr = creatinine clearance in mL/minute U = urine creatinine P = plasma/serum creatinine V =urine volume in mL per minute SA =patient's body surface area

C _ U •V X 1.73 m 2 cr- p SA

When using this formula, it is required that the urine and serum creatinine concentrations be in the sarne units so that the units cancel. Therefore, the urine concentration must be converted to 270 mg/dL before using the formula. The urine volume (V) must also be converted from mL per 24 hours to mL per minutes as follows: urine volume (mL) 24 hour

X

1 hour 60 minutes

=

V in mL/minute

Because creatinine excretion varies with muscle mass (i.e., body surface area), comparison to a reference interval requires that the clearance be corrected for the individual's body surface area in square meters. (Brunzel, pp. 106-109) 24. An aspartate arninotransferase (AST) result obtained on a serum specimen diluted 1 to 3 (1 :3) is 42 U/L. Which of the following results should be reported?

A. B. C. D.

84 U/L 126 U/L 168 U/L 210 U/L

The answer is B. The dilution factor is 3; therefore, the undiluted sample has a concentration of 3 X 42 = 126 U/L. (Bishop et al., pp. 22-23) 25. A serum specimen is being analyzed for the activity of an enzyme and the following kinetic data obtained:

Time (min)

Absorbance

o

0.020 0.200 0.315 0.395 0.435 0.480

1 2 3 4 5

Select the statement that best summarizes these results.

CLS Review Questions

A. Readings are satisfactory; calculate the enzyme result B. Substrate depletion; repeat the assay using a serum dilution C. The 0- 3 min readings are satisfactory; use these for enzyme result calculation D. The 3- 5 min readings are satisfactory; use these for enzyme result calculation

The answer is B. The rate of change of absorbance (change in absorbance per minute) is not constant for any of the data given. This indicates substrate exhaustion where there is insufficient substrate present for all of the enzyme present to be continuously active during the analysis. Use of a serum dilution will allow sufficient substrate for zero-arder kinetics and still produce a measurable change in absorbance. (Kaplan and Pesce, pp. 1065-1068) 26. A serum protein electrophoresis is performed and unexpected results obtained. Ali protein bands migrated further than usual and the bands were further apart from each other. Which of the following would accdunt for the results obtained? A. B. C. D.

The support medium has deteriorated A higher voltage was used during analysis The wrong buffer was used; ionic strength higher than required The wrong buffer was used; pH was significantly lower than required

The answer is B. A higher voltage will cause proteins to migrate faster and to separate more from each other. A buffer of a higher ionic strength, as well as one with a lower pH, would reduce the distance of protein migration. A deteriorated support medium would not produce the changes observed. (Kaplan and Pesce, pp. 207- 208) 27. The normal ratio of bicarbonate ion to carbonic acid in arterial blood is A. 0.03:1 B. 1:1.8 e. 20:1 D. 6.1:7.4

The answer is C. The Henderson-Hasselbalch equation defines the ratio of base to acid that is required for a given pH. At normal arterial pH the ratio of concentrations of bicarbonate ion to carbonic acid is 20: 1. The pKa of this whole blood buffer system at 37QC is 6.1. (Bishop et al., p. 241) 28. Increased serum uric acid is found in each the following conditions except A. B. C. D.

gout hypothyroidism Lesch-Nyhan syndrome renal failure

The answer is B. Thyroid hormones have no specific effect on formation or elimination of uric acid. Gout is the disease caused by deposition of excessive uric acid in body spaces, e.g., joints. Lesch-Nyhan syndrome is a rare inborn errar of metabolism in which the salvage enzyme of purine catabolism is deficient. This ~esults in excessive production of the purine catabolite uric acid.

29

30

1.

Clinicai Chemistry

Renal failure results in inability to clear the blood of waste products including uric acid. (Bishop et al., pp. 348-350) "

1

29. Which of the following is nota criterion for the diagnosis of diabetes mellitus?

A. B. C. D.

A fasting glucose ;:::: 126 mg/dL A serum glucose ::; 40 mg/dL within 3h after an oral dose of glucose Classic symptoms and a plasma glucose ;::: 200 mg/dL at anytime A serum glucose ;:::: 200 mg/dL at 2h after an oral glucose dose

The answer is B. The diagnostic criteria for diabetes mellitus was revised by the American Diabetes Association and published in Diabetes Care 1997; 20:1183-1201. Currently, all of the situations described except choice B, a low serum glucose, are diagnostic of diabetes mellitus. (Burtis and Ashwood, p. 439) 30. If LDL receptors are nonfunctional dueto disease, the plasma levei of which lipid would increase the most? A. B. C. D.

Fatty acids Cholesterol Cholesterol esters Triglycerides

The answer is B. LDL is the major carrier of cholesterol and is considered an atherogenic lipoprotein. Because approximately 50% of LDL lipid is cholesterol, an increase in plasma LDL will increase the total cholesterol. LDL is cleared from the plasma by the li ver. The apolipoprotein B moiety of LDL binds to specific hepatocyte receptors, the receptors are subsequently intemalized, and the LDL is catabolized. When these receptors are not present or non-functional, plasma total cholesterol leveis are increased. (Kaplan and Pesce, pp. 651-652)

31. Which of the following serum protein electrophoresis pattems is most typical of the nephrotic syndrome?

albumin A. ! ! B. ! C. t t D. normal

alpha1 !

alpha2

beta

t

! ! t

t

! t

normal !

normal

t

gamma

t ! t t

The answer is A. ln the nephrotic syndrome, an increased permeability of the glomerular membrane allows proteins, particularly alburnin due to its high plasma concentration, to be lost in the urine in large quantities. As a result, the concentration of the remaining large MW proteins (e.g., armacroglobulin) "appear" to be present in an increased concentration. (Lehmann, p. 40) 32. Which analyte is most likely to be elevated in a specimen drawn 2 hours after an uncomplicated myocardial infarction?

A. B. C. D.

CK-MB (CK-2) Myoglobin Troponin I Troponin T

CLS Review Questions

The answer is B. Following an acute MI, myoglobin is the earliest cardiac marker. lt begins to rise within 2-3 h and peaks at 6-9 h. For comparison, CKMB begins to rise at 4-6 h and usually peaks within 12-24 h; the troponins, I and T, begin to rise at 4-8 h and usually peak within 18 h. (Lehmann, p. 104) 33. Which of the following serum results correlates best with the rapid cell tumover associated with chemotherapy treatment regimens? A. Creatinine of 2.5 mg/dL B. Potassium of 5.0 mmol/L C. Urea nitrogen of 30 mg/dL D. Uric acid of 10.0 mg/dL

The answer is D. Chemotherapy destroys cells releasing their DNA and RNA, which must be processed. The purine bases, adenine and guanine, derived from DNA and RNA are degraded and their ring structures are converted to uric acid for excretion in the urine. (Anderson and Cockayne, pp. 371-372) 34. Which set of serum electrolyte results (in mmol/L) is most likely obtained from a serum with an elevated lactate level? Na+

K+

c1-

HCQ3 -

A. 125

4.5 3.5 4.0 5.0

100 95 90 110

10 28 15 30

B. 135 e. 145 D. 150

The answer is C. As lactic acid is produced it dissociates into H+ ions and lactate anions. Because the body maintains electrical neutrality, as lactate accumulates, another anion is eliminated. ln the case of excessive production of lactic acid, bicarbonate is used to neutralize and eliminate the H+ ions produced. The end result is a metabolic acidosis with an increased anion gap. (Anderson and Cockayne,pp.421-423) 35. The following arterial blood gas results are obtained:

7.28 53 mmHg 75 mmHg 26 mmol/L These results correlate best with a patient experiencing A. metabolic acidosis B. metabolic alkalosis

C. respiratory acidosis D. respiratory alkalosis The answer is C. Based on the pH alone, this patient is experiencing acidosis. Next, the pC02 exceeds the "normal" reference range ( ~32-48 mmHg) indicating a respiratory component. The normal bicarbonate value (22-26 mmol/L) indicates that compensation has not yet occurred. (Lehmann, p.163; Kaplan and Pesce,pp.471-475)

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36. A patient with intermittent hypertension has an elevated value for urinary catecholamine metabolites (e.g., vanillylmandelic acid [VMA]). This result may indicate A. hyperaldosteronism

B. hypercortisolism C. idiopathic hypertension

D. pheochromocytoma The answer is D. Pheochromocytoma is a catecholamine-secreting tumor of the adrenal medulla. Epinephrine and norepinephrine are catecholamines. Most pheochromocytomas produce a mixture of the two. The release of catecholamines by the tumor accounts for the patient's hypertension. Catecholamine metabolites, including VMA, are excreted in the urine. (Kaplan and Pesce, pp. 924, 930-932; Anderson and Cockayne, p. 550)

37. Blood from a newborn has low thyroxine (T4 ) and elevated thyroid-stimulating hormone (TSH) compared to reference ranges for that age. These results are most consistent with A. congenital hypopituitarism

B. congenital primary hypothyroidism C. congenital secondary hypothyroidism

D . a normal response to pregnancy-induced changes in maternal thyroid function The answer is B. Production ofT4 by the thyroid gland has a negative feedback relationship with thyroid-stimulating hormone (TSH) produced by the anterior pituitary gland. Congenital abnormalities that prevent adequate production of T4 result in a high level of TSH through this feedback loop. The elevated T4 values seen in maternal serum are an artifact caused by an estrogen-induced increase in synthesis of thyroxine-binding globulin. (Bishop et al., p. 440; Anderson and Cockayne, pp. 512-513)

38. ln which of the following situations will the patient have a "normal" urine levei of human chorionic gonadotropin (HCG)?

A. B. C. D.

Ten days following a spontaneous abortion During the 101h week of a normal pregnancy During a molar pregnancy (hydatidiform mole) During the third trimester of a normal pregnancy

The answer is A. HCG is produced by trophoblastic tissue, absorbed into the maternal plasma, and then excreted in the urine. Loss of trophoblastic tissue, as in spontaneous abortion, results in rapid urinary clearance of the hormone and values that are less than expected for the presumed period of gestation. Trophoblastic tumors, such as molar pregnancy, are associated with elevated values in the absence of pregnancy. ln a normal pregnancy, HCG rises in maternal blood soon after implantation of the fertilized ovum and doubles approximately every two days during the first trimester. A very slow decline then occurs through the rest of the gestation period. (Bishop et al., pp. 422, 490-491; Anderson and Cockayne, pp. 659-660)

CLS Review Questions

39. The following results are obtained from a patient whose admission diagnosis is biliary obstruction:

Serum: Urine:

Test

Result

Conjugated bilirubin Total bilirubin Bilirubin Urobilinogen

Increased Increased Positive lncreased

Which of the results obtained is inconsistent with the admission diagnosis? A. B. C. D.

Serum conjugated bilirubin Serum total bilirubin Urinary bilirubin Urinary urobilinogen

The answer is D. Urobilinogen is formed in the intestinal lumen by bacterial action on bile. Approximately 20% of the urobilinogen is reabsorbed into the portal blood, removed by the liver, and re-excreted into the bile. Two to five percent of the urobilinogen escapes hepatic remova! and is excreted by the kidneys in the urine. A patient with obstructive liver disease excretes less bilirubin into the intestine; hence, less urobilinogen is formed and ultimately the urine urobilinogen decreases. ln contrast, urine bilirubin, the water-soluble conjugated form, increases because of regurgitation from the bile canaliculi into the blood due to the biliary obstruction. (Brunzel, pp. 180-183; Anderson and Cockayne, pp. 284-286) 40. Blood and cerebrospinal fluid (CSF) samples were collected from a patient within 30 minutes of each other. Which set of glucose results indicates possible bacterial meningitis?

Blood A. 60 mg/dL B. 100 mg/dL C. 200 mg/dL D. 200 mg/dL

CSF

40 mg/dL 60 mg/dL 30 mg/dL 120 mg/dL

The answer is C. ln the absence of bacteria or increased numbers of leukocytes, the glucose concentration in CSF should be 60 to 80% of the concurrent concentration in blood. (Bishop et al., p. 307) 41. Which of the following enzymes provides the best indication of obstructive liver disease, i.e., cholestasis?

A. B. C. D.

Amylase Alkaline phosphatase Aspartate aminotransferase Lactate dehydrogenase

The answer is B. Amylase, an enzyme of pancreatic origin is unaffected by li ver obstruction. Each of the remaining choices are enzymes present in liver tissue. However, only the synthesis of alkaline phosphatase (ALP), which is localized in cell membranes, is induced due to biliary obstruction. Both aspartate arninotransferase (AST) and lactate dehydrogenase (LD) are cytoplasmic enzymes released upon cell damage or death. Hence, depending on the extent and dura-

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tion of the obstructive disease process, normal values can still be obtained for LD and AST. If the disease process causes hepatic cell damage or death, increases in LD and AST will also be observed. (Burtis and Ashwood, pp. 366-367; Kaplan and Pesce, pp. 515- 516) 42. ln a cerebrospinal fluid (CSF) sample, which of the following proteins is quantitated to assess the permeability of the blood/brain barrier?

A Albumin B. IgG C. Transferrin D. Prealbumin The answer is A Albumin is usually employed as the reference protein for permeability because it is not synthesized to any extent in the CNS. ln a CSF sample with no blood contamination, the albumin present comes from the plasma by passing the blood/brain barrier. An increase in the perrneability of the blood/brain barrier to plasma proteins can be dueto high intracranial pressure (e.g., brain tumor, intracerebral hemorrhage) or due to an inflammatory process (e.g., bacterial or virai meningitis). (Bishop et al., pp. 197- 198; Brunzel, pp. 375-376) 43. A maternity patient complains of dysuria during a monthly visit with her doctor. The physician suspects a lower urinary tract infection (e.g., cystitis) and requests a urinalysis and urine culture. Which type of urine specimen should be collected from this patient?

A B. C. D.

Random Catheterized First moming Midstream clean catch

The answer is D. The specimen type (random, first morning, etc.) is notas important as the collection technique used in obtaining the specimen. A midstream clean catch will eliminate potential bacterial contamination from the perineum and vulva and is the specimen of choice for microbial culture in uncatheterized patients. A catheterized specimen would also be acceptable; however, in this case the pregnant female is unlikely to be catheterized. (Brunzel, pp. 53- 57) 44. Urine preservation by refrigeration can cause A. decreased pH due to glycolysis B. increased bacteria dueto proliferation e. photo-oxidation of bilirubin to biliverdin D. increased turbidity dueto precipitate of solutes The answer is D. Refrigeration can induce the precipitation of urine solutes. This amorphous and crystalline material will cause a reduction in the visual clarity of the specimen. (Brunzel, pp. 58-60) 45. Four calibration solutions were evaluated for use as the daily calibration check for the refractometer at the physiological "upper reference range" for urine specific gravity (SG). Which calibration solution should be selected?

CLS Review Questions

A. B. C. D.

Calibrator A - SG 1.015 Calibrator B - SG 1.025 Calibrator C - SG 1.035 Calibrator D - SG 1.055

The answer is C. Urine SG can vary from 1.002 to 1.035 depending on an individual's hydration. Values below or above this range are physiologically impossible and require further investigation. Radiographic contrast media in urine can typically produce SG values greater than 1.040 and adulteration of urine with water can produce values near 1.000. (Brunzel, pp. 153-156) 46. All of the following are acceptable preparations of urine sediment for rnicroscopic examination except A. filling a chamber of a commercial standardized slide B. over-filling the chamber of a commercial standardized slide C. transferring, using a disposable pipette, several drops of sediment onto a rnicroscope slide and coverslipping D. transferring, using a calibrated automatic pipette, 20 µL of sediment onto a microscope slide and coverslipping

The answer is C. A standardized slide should be used for rnicroscopic examinations because of the enhanced reproducibility of results. If unavailable, an exact amount of sediment quantitatively delivered onto a slide and coverslipped can be used. Of particular note is that the coverslip size and weight should be stated in the procedure and used at all times because the weight and size of the coverslip will affect the depth (i.e., volume) of urine sediment viewed. (Brunzel, pp. 210-211) 47. Which of these sugars cannot be detected in urine using the copper reduction test? A. B. C. D.

Fructose Galactose Arabinose Sucrose

The answer is D. The copper reduction test detects carbohydrates by the reducing power of their free aldehyde groups. Sucrose is a disaccharide that has no free aldehyde group and does not produce the yellow-orange salts of oxidized copper. Sucrose is not absorbed or produced by the body. It only appears in urine as an artifact. (Brunzel, p. 176) 48. Which of the following tests is not used to assess the kidney's ability to concentrate the urine? A. B. C. D.

pH Refractive index Osmolality Specific gravity

The answer is A. Refractive índex, osmolality, and specific gravity are methods of measurements used to assess the concentration of dissolved solutes in urine. Urine pH, a measure of hydrogen ion concentration, reflects the diet and the body's regulation of its acid-base balance. (Brunzel, pp. 128- 136)

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Clinica! Chemistry

49. Ali of the following substances can affect the detection of urine glucose using regent strips except A. B. C. D.

ascorbic acid bleach free-sulfhydryl drugs galactose

The answer is D. All urine reagent strips utilize the enzyme, glucose oxidase, which is specific for glucose; hence, no other sugars (e.g., galactose) can react. ln contrast, false-negative urine glucose can result from cellular or bacterial glycolysis if an unpreserved urine specimen is left at room temperature for a prolonged period of time. Another cause for a false-negative glucose is excessive amounts of ascorbic acid (;::: 50 mg/dL) in the urine. Note, however, that Chemstrip reagent strips (Boehringer Mannheim Corp, lndianapolis, IN) are NOT affected by the presence of ascorbic acid due to a patented iodate scavenger pad that eliminates this interference. ln contrast, urine contaminated with bleach will produce a false-positive glucose. (Brunzel, pp. 58-59, 189-190; Strasinger, pp.

60--61) 50. Review the following urinalysis results:

Macroscopic Exam: Parameter pH SG Blood Protein Glucose Ketone Bilirubin Urobilinogen Nitrite Leukocyte esterase Microscopic Exam:

Result 5.5 l.020 small 30 mg/dL negative negative negative 1.0 mg/dL negative negative 5-1 O RBCs per high power field 0-2 WBCs per high power field 0-2 granular casts few transitional epithelial cells few sulfonamide crystals

Which result(s) requires additional action before reporting? A. Blood vs. RBCs; test for ascorbic acid B. Sulfonamide crystals; perform confirmatory test and check patient medications C. Granular casts; should see larger amount of protein, perform protein precipitation test D. pH and crystal identification; sulfonamide crystals precipitate in alkaline urine The answer is B . Sulfonamide crystals should be confirmed before reporting by performing a chemical test and by ensuring that the individual is actually taking a sulfonamide-containing medication. (Brunzel, p. 249)

CLS Review Questions

51. Which of the following crystals can be present in an alkaline urine? A. B. C. D.

Cystine Calcium oxalate Triple phosphate Sodium urate

The answer is C. Tripie phosphate is magnesium-ammonium phosphate, a salt that can precipitate in urine with an alkaline pH. At an acid pH, the salt dissociates into its soluble component ians. The other crystals listed are observed in their crystalline forros in neutral or acid urine. (Brunzel, pp. 241-253; Strasinger, pp. 92-95) 52. The presence of waxy casts in urine sediment and a fixed urine specific gravity of 1.010 correlates best with A. cystitis B. glomerulonephritis e. acute pyelonephritis D. renal failure

The answer is D. Casts are formed in the distal and collecting tubules of the nephron. Therefore, cystitis, an infection of the lower urinary tract (e.g., the urinary bladder), does not induce cast formation. The remaining conditions can all present with a variety of casts. However, renal failure is characterized by the presence of waxy and broad casts. ln addition, the fixed specific gravity indicates the inability of the nephrons to selectively reabsorb and secrete solutes, a hallmark of renal failure. (Brunzel, pp. 287-292) 53. Urine sediment that contains free fat globules and fatty casts is characteristic of A. a bladder infection B. the Fanconi syndrome C. the nephrotic syndrome D. acute glomerulonephritis The answer is C. ln the nephrotic syndrome, increased permeability of the glomerular filtration barrier allows the passage of plasma proteins, particularly albumin, and lipids into the urine. The lipids that are able to cross the filtration barrier can get enmeshed in casts that are forming in the distal tubules. Note that the lipid can be either cholesterol or neutral fats (triglyceride) and can be free floating, in casts, or in cells called oval fat bodies. (Brunzel, pp. 238, 255-256, 277) 54. Which of the following statements about finely granular casts in urine sediment is true? A. They are indicative of end-stage renal disease B. They are frequently seen when women have trichomoniasis C. Numerous finely granular casts are diagnostic for acute glomerulonephritis D. A few finely granular casts can be seen in urine from normal healthy individuals

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Clinicai Chemistry

The answer is D. The granules in finely granular casts are the by-products of protein metabolism excreted by renal tubular epithelial cells. Hence, they are seen in urine from normal healthy individuais. Waxy casts are usually observed with end-stage renal disease; whereas in acute glomerulonephritis, red blood cell and hemoglobin casts are considered pathognomonic. Vaginal trichomoniasis and urine cast formation are unrelated processes. (Brunzel, p. 237) 55. When examining "unstained" urine sediment, cellular detail is best when observed using A. B. C. D.

bright-field microscopy darkfield microscopy phase-contrast microscopy polarizing microscopy

The answer is A. Bright-field microscopy, with optimized adjustments, enables detailed viewing of unstained urine sediment components. ln contrast, phasecontrast microscopy can produce bright haloes that reduce the visualization of detail and dimension. (Brunzel, pp. 13-17) 56. The following urinalysis results are obtained on a urine specimen: Reagent strip blood: Microscopic exam:

negative 10--15 RBCs per high-power field

Which of the following statements best explains these results? A. B. C. D.

The microscopically identified RBCs are really yeast Myoglobin is causing a false-negative reagent strip blood test Ascorbic acid is causing a false-negative reagent strip blood test The microscopically identified RBCs are really monohydrate calcium oxalate crystals

The answer is C. Ascorbic acid is a reducing substance that can cause the reagent strip blood test to be falsely negative with some reagent strips, i.e., Multistix. Note that Chemstrip reagent strips are not affected by urine ascorbic acid because of an iodate scavenger pad on their blood and glucose reaction pads. Even though the reagent strip tests are less sensitive to intracellular hemoglobin than they are to free hemoglobin, they should be positive in the presence of this number of RBCs. The identification of the cells can be confirmed by addition of weak acetic acid that will lyse RBCs but not yeast or WBCs. Monohydrate calcium oxalate crystals can be readily differentiated from RBCs using polarizing microscopy. (Brunzel, pp. 158-162, 188-190, 218- 219) 57. Review the following urinalysis results:

Macroscopic Exam: Parameter pH SG Blood Protein Glucose Ketone Bilirubin Urobilinogen Nitrite Leukocyte esterase

Result 6.0 1.020 moderate 500 mg/dL negative negative negative 1.0 mg/dL negative negative

CLS Review Questions

Microscopic Exam: 10-25 RBCs per high-power field; dysmorphic forms present 2-5 WBCs per high-power field 2-5 hyaline casts 2-5 RBC casts 0-2 granular casts few uric acid crystals

These results are most consistent with a diagnosis of A. gout B. glomerulonephritis e. pyelonephritis D. urinary-tract obstruction The answer is B. Glomerulonephritis is an inflammation of the glomeruli that results in proteinuria and hematuria. Classic urine microscopic findings include: increased RBCs, often dysmorphic; increased WBCs, RBC and hemoglobin casts; granular casts and occasionally WBC and renal cell casts. (Brunzel, pp. 276-278, Strasinger, p. 32) 58. The following urinalysis results are obtained:

Glucose by reagent strip: negative Ketones by reagent strip: positive These results are most consistent with A. starvation B. polydipsia C. diabetes mellitus D. diabetes insipidus The answer is A. Ketones (acetoacetate, B-hydroxybutyrate, and acetone) are produced when the liver must oxidize fatty acids due to limited carbohydrate availability. As a result, large amounts of acetyl CoA are formed that exceed the Krebs cycle capacity, and the liver mitochondria begin active ketogenesis to reduce the levels of acetyl CoA. Note that this process will occur in any state where insufficient carbohydrates are available for cellular energy requirements. (Brunzel, pp. 177-178; Strasinger, pp. 62-63) 59. Each day, two laboratories perform an inter-laboratory urinalysis correlation. After analysis is completed in lab A, a urine specimen is selected and sent to lab B. Today, the physical and chemical examinations correlate but the microscopic results do not. Review the results obtained.

RBCs/hpf WBCs/hpf

LabA

LabB

5-10 0-2

25- 50 5-10

Which of the following situations could account for this discrepancy? A. B. C. D.

Lab B concentrated a smaller volume of specimen Lab A centrifuged the specimen twice as long as Lab B Lab B centrifuged the specimen at a lower speed than Lab A Lab A used the brake on the centrifuge when processing the specimen

The answer is D. If the brake is used after the centrifugation of urine, urine sediment components can become resuspended leading to falsely low or decreased '

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numbers of sediment entities, i.e., RBCs, WBCs. The other options describe situations that would result in higher micrascopic results obtained by Lab A compared to Lab B. (Brunzel, pp. 39-41, 417) 60. Which of the following is an "initial" step in a protocol for troubleshooting an automated method when quality contrai results are not acceptable? A. Repeat analysis of QC materiais using a fresh aliquot or different vial B. Recalibrate the instrument, then reanalyze contrais and patient samples C. Change reagents, then repeat analysis using a new lot number of QC materiais D. Perform periodic maintenance, then recalibrate and reanalyze contrais and patient samples The answer is A. Actions necessary to bring a system back into contrai vary. Traubleshooting should always occur in a stepwise fashion and documentation must be maintained, e.g., in an "out-of-contral" log book. The easiest and first step when QC results are not acceptable is to simply repeat the analysis using a fresh aliquot of the QC material or open a new bottle of the sarne lot number. Contrais can deteriorate or become contaminated while in use. Instrument recalibration, changing reagents, and performing periodic maintenance may have to be performed to get a system back into contrai; however, these are usually not "initial" steps taken to investigate and resolve unacceptable QC results. (Kaplan and Pesce, pp. 394-395)

References Anderson SC, Cockayne A. Clinica! Chemistry. Philadelphia: WB Saunders Company, 1993. Bishop ML, Duben-Engelkirk JL, Fody EP (eds). Clinica! Chemistry: Principies, Procedures, Correlations (3rd ed). Philadelphia: Lippincott, 1996. Brunzel NA. Fundamentais of Urine and Body Fluid Analysis. Philadelphia: WB Saunders Company, 1994. Burtis CA, Ashwood ER (eds). Fundamentais of Clinica[ Chemistry (5th ed). Philadelphia: WB Saunders Company, 2001. Kaplan LA, Pesce AJ. Clinicai Chemistry: Theory, Analysis, and Correlation (3rd ed). St. Louis: Mosby, 1996. Lehmann CA (ed). Saunders Manual of Clinicai Laboratory Science. Philadelphia: WB Saunders Company, 1998. Strasinger SK. Urinalysis and Body Fluids (3rd ed). Philadelphia: FA Davis, 1994.

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