Clinical Protein and Enzyme Analysis Practice Quiz (58 questions)

1. A laboratory needs glassware for prolonged heating and sterilization, but it may also encounter strong alkali. Which comparison between borosilicate and aluminosilicate glass is accurate?

A. Aluminosilicate is six times stronger, but borosilicate is more resistant to scratches and alkali.

B. Both types resist scratches and alkali equally, but only aluminosilicate tolerates prolonged heating.

C. Borosilicate tolerates prolonged heating, while aluminosilicate is stronger and more resistant to scratches and alkali.

D. Borosilicate is more resistant to scratches and alkali, while aluminosilicate tolerates prolonged heating better.

Answer: (C)

Explanation

Borosilicate tolerates prolonged heating and sterilization, whereas aluminosilicate is six times stronger and better resists scratches and alkali attack. The tempting reverse comparison assigns aluminosilicate’s scratch and alkali resistance to borosilicate.

2. A technician needs to deliver a measured fraction of a pipette’s capacity. Which pipette design is suited to this task?

A. A graduated pipette, because its single calibration mark defines one fixed delivery volume.

B. A volumetric transfer pipette, because it has subdivisions for selecting different volumes.

C. A volumetric transfer pipette, because its single mark indicates fractions of its capacity.

D. A graduated pipette, because its subdivisions allow fractions of its capacity to be measured and delivered.

Answer: (D)

Explanation

A graduated pipette has subdivisions for measuring and delivering fractions of its capacity. A transfer pipette has a single calibration mark and is used for a fixed volume, rather than for selecting fractions.

3. A sample is viscous and difficult to pipette. Which micropipette mechanism is recommended for handling it?

A. A positive-displacement micropipette, because an air cushion separates its piston from the sample.

B. A positive-displacement micropipette, because its piston contacts the liquid directly.

C. An air-displacement micropipette, because an air cushion separates its piston from the sample.

D. An air-displacement micropipette, because its piston contacts the liquid directly.

Answer: (B)

Explanation

Positive-displacement micropipettes place the piston in direct contact with the liquid and are recommended for viscous, dense, or problematic samples. Air-displacement models use an air cushion between the piston and sample.

4. A procedure requires maximum reagent purity and minimal interference. Which reagent grade is the best fit?

A. ACS analytical grade, because it is intended for procedures requiring the lowest possible interference.

B. ACS analytical grade, because it is suitable for most analytical procedures.

C. Ultrapure spectrograde or nanograde, because it is intended for maximum purity and minimal interference.

D. Ultrapure spectrograde or nanograde, because it is the routine choice for most analytical procedures.

Answer: (C)

Explanation

Ultrapure spectrograde or nanograde reagents are used when maximum purity and minimal interference are required. ACS analytical grade suits most procedures, but it is not the specified choice when those requirements are especially stringent.

5. A laboratory needs water for preparing standards and carrying out a procedure requiring high accuracy and precision. Which type should it select?

A. Type I, which is the purest and is used for standards and high-accuracy procedures.

B. Type II, which is the purest and is intended for standards requiring high precision.

C. Type II, which is used for standards requiring the highest accuracy and precision.

D. Type III, which is the purest and is intended for accurate reagent preparation.

Answer: (A)

Explanation

Type I is the purest reagent water and is used for standards and procedures requiring high accuracy and precision. Type II suits most laboratory procedures, but it is not identified as the highest-purity choice.

6. After a body-fluid spill, which disinfection procedure meets the specified concentration and contact-time requirements?

A. Apply 5.25% sodium hypochlorite or 10% bleach diluted 1:10, and maintain contact for at least 10 minutes.

B. Apply 10% sodium hypochlorite or 5.25% bleach diluted 1:10, and maintain contact for at least 20 minutes.

C. Apply 5.25% sodium hypochlorite or undiluted 10% bleach, and maintain contact for at least 20 minutes.

D. Apply 5.25% sodium hypochlorite or 10% bleach diluted 1:10, and maintain contact for at least 20 minutes.

Answer: (D)

Explanation

The procedure calls for 5.25% sodium hypochlorite or 10% bleach diluted 1:10, with at least 20 minutes of contact. A shorter contact period does not meet the stated minimum.

7. Which description correctly distinguishes Class II and Class III biological safety cabinets?

A. Class II cabinets are gas-tight, negative-pressure enclosures with glove ports; Class III cabinets use two HEPA filters to protect the worker and work.

B. Class II cabinets use two HEPA filters and protect the worker and work; Class III cabinets are gas-tight, negative-pressure enclosures with glove ports.

C. Class II cabinets protect the work but not the worker; Class III cabinets are open-front enclosures with two HEPA filters.

D. Class II cabinets use two HEPA filters to protect the worker; Class III cabinets use glove ports but operate without negative pressure.

Answer: (B)

Explanation

Class II cabinets use two HEPA filters and protect both the worker and the work, while Class III cabinets are gas-tight negative-pressure enclosures with glove ports. The reversed description incorrectly assigns Class III’s enclosure design to Class II.

8. A solute has a molecular weight of 60 and a valence of 3. What is its equivalent weight, and what does normality count per liter?

A. The equivalent weight is 180, and normality counts moles of solute per liter of solution.

B. The equivalent weight is 20, and normality counts moles of solute per liter of solution.

C. The equivalent weight is 20, and normality counts equivalents of solute per liter of solution.

D. The equivalent weight is 63, and normality counts equivalents of solute per liter of solution.

Answer: (C)

Explanation

Equivalent weight is molecular weight divided by valence, giving $60/3=20$, and normality is equivalents per liter. The plausible alternative that describes moles per liter is the definition of molarity, not normality.

9. A technician dilutes 10 mL of a stock solution to a final volume of 100 mL. If the stock concentration is 2 mol/L, what is the final concentration?

A. 20 mol/L, found by applying $C_1V_1=C_2V_2$.

B. 0.2 mol/L, found by applying $C_1V_1=C_2V_2$.

C. 2 mol/L, because dilution leaves the concentration unchanged.

D. 0.02 mol/L, found by dividing the final volume by the stock concentration.

Answer: (B)

Explanation

Using $C_1V_1=C_2V_2$ gives $C_2=(2\times10)/100=0.2$ mol/L. A concentration of 2 mol/L would ignore the increase in volume during dilution.

10. For a normally distributed data set, approximately what proportion of observations falls within two standard deviations of the mean?

A. About 95%, with this proportion corresponding to two standard deviations.

B. About 99%, with this proportion corresponding to two standard deviations.

C. About 95%, with this proportion corresponding to three standard deviations.

D. About 68%, with this proportion corresponding to one standard deviation.

Answer: (A)

Explanation

Approximately 95% of observations in a normally distributed data set fall within two standard deviations of the mean. The 68% figure applies to one standard deviation, while approximately 99% applies to three.

11. A method produces results that cluster closely together but are consistently far from the true value. Which description best fits these results?

A. They are accurate but imprecise.

B. They are neither accurate nor precise.

C. They are both accurate and precise.

D. They are precise but inaccurate.

Answer: (D)

Explanation

Precision describes close agreement among repeated results, while accuracy describes closeness to the true value. Results can be tightly clustered and still be inaccurate when the cluster lies far from the truth.

12. A control system shows unpredictable scatter around its target, while another consistently reads too high. Which error classification applies, respectively?

A. Both errors affect accuracy in the same way.

B. Systematic error affects precision; random error affects accuracy.

C. Random error affects accuracy; systematic error affects precision.

D. Random error affects precision; systematic error affects accuracy.

Answer: (D)

Explanation

Unpredictable variation is random error and reduces precision, whereas a consistent one-direction bias is systematic error and affects accuracy. The consistent high readings are not random scatter; they indicate systematic bias.

13. Control values gradually rise across the mean over six consecutive days. What pattern does this represent?

A. A trend, because values gradually move across the mean.

B. A shift, because values change over consecutive days.

C. A trend, because the values remain on one side of the mean.

D. A shift, because the values remain on one side of the mean.

Answer: (A)

Explanation

A gradual increase or decrease across the mean over six consecutive days is a trend. A shift instead involves an abrupt change followed by values remaining on one side of the mean for six consecutive days.

14. When a blood-pressure cuff is used to assist venipuncture, which combination follows the recommended preanalytical limits?

A. Apply the tourniquet for less than one minute and inflate the cuff to 60–80 mm Hg.

B. Apply the tourniquet for about two minutes and inflate the cuff to 40–60 mm Hg.

C. Apply the tourniquet for about two minutes and inflate the cuff to 60–80 mm Hg.

D. Apply the tourniquet for less than one minute and inflate the cuff to 40–60 mm Hg.

Answer: (D)

Explanation

The tourniquet should be applied for less than one minute, and the cuff should be inflated to 40–60 mm Hg. A longer tourniquet application exceeds the stated limit even if the cuff pressure is correct.

15. Which sequence gives the correct evacuated-tube order of draw?

A. Sterile culture, light blue, red, plasma separator, serum separator, green, lavender, gray.

B. Sterile culture, light blue, red, serum separator, plasma separator, green, lavender, gray.

C. Light blue, sterile culture, red, serum separator, plasma separator, lavender, green, gray.

D. Sterile culture, red, light blue, serum separator, green, plasma separator, lavender, gray.

Answer: (B)

Explanation

The order begins with the sterile culture tube, followed by light blue, red, serum separator, plasma separator, green, lavender, and gray. The sequence with plasma separator before serum separator reverses their specified positions.

16. A specimen is hemolyzed. Which pattern of analyte changes is expected?

A. Both intracellular analytes and lipase may fall, while hexokinase glucose may rise.

B. Many intracellular analytes, including potassium, may rise, while lipase and hexokinase glucose may fall.

C. Both intracellular analytes and lipase may rise, while hexokinase glucose may fall.

D. Many intracellular analytes, including potassium, may fall, while lipase and hexokinase glucose may rise.

Answer: (B)

Explanation

Hemolysis can increase intracellular analytes such as potassium and decrease lipase and glucose measured by hexokinase. The reverse pattern for intracellular analytes and lipase does not match the stated effects.

17. In a spectrophotometric assay, concentration and path length both increase while molar absorptivity stays constant. What happens to absorbance under the Beer-Lambert law?

A. Absorbance depends on concentration but not on path length.

B. Absorbance increases because it is proportional to both concentration and path length.

C. Absorbance decreases because greater path length allows more light to reach the detector.

D. Absorbance stays constant because molar absorptivity has not changed.

Answer: (B)

Explanation

The Beer-Lambert law is $A=abc$, so absorbance is directly proportional to both path length and concentration when molar absorptivity is constant. The unchanged molar absorptivity does not prevent absorbance from rising as the other factors increase.

18. A laboratory needs to monitor whether an assay continues to perform as expected after calibration. Which material serves that purpose?

A. A control with target values used to monitor assay performance.

B. A reference cuvet used to compensate for source-intensity variation.

C. A standard with known concentration used to calibrate or quantify the assay.

D. A blank used to subtract absorbance not due to the analyte.

Answer: (A)

Explanation

A control has target values and is used to monitor assay performance. A standard has known purity and concentration for calibration or quantitation, so it serves a different purpose.

19. An optical method detects light scattered by particles in a suspension rather than light blocked by them. Which method is being used?

A. Nephelometry, which measures blocked light.

B. Turbidimetry, which measures scattered light.

C. Nephelometry, which measures scattered light.

D. Turbidimetry, which measures transmitted light.

Answer: (C)

Explanation

Nephelometry measures light scattered by particles in a suspension. Turbidimetry instead measures light blocked by the suspension.

20. A diagnostic test correctly identifies 72 of 80 people who have a disease. What is its diagnostic sensitivity?

A. 90%, calculated as true negatives divided by true negatives plus false positives

B. 80%, calculated as true positives divided by all people tested

C. 10%, calculated as false negatives divided by true positives plus false negatives

D. 90%, calculated as true positives divided by true positives plus false negatives

Answer: (D)

Explanation

Sensitivity is the proportion of people with the disease who test positive, so it is calculated as $\frac{TP}{TP+FN}\times100$. Specificity instead uses true negatives and false positives, so it does not measure this test’s ability to detect affected people.

21. A patient receives a positive test result. Which measure describes the probability that the patient has the disease?

A. Diagnostic sensitivity

B. Negative predictive value

C. Diagnostic specificity

D. Positive predictive value

Answer: (D)

Explanation

Positive predictive value describes the probability of disease given a positive result. Negative predictive value instead describes the probability of absence of disease given a negative result.

22. Which feature distinguishes fluorescence polarization immunoassay (FPIA) from microparticle capture enzyme immunoassay (MEIA)?

A. FPIA uses microparticles and a fluorogenic substrate, whereas MEIA uses polarized light without separation

B. FPIA is applied to high-molecular-weight substances, whereas MEIA is applied to low-molecular-weight substances

C. FPIA requires separation of bound and free components, whereas MEIA proceeds without a separation step

D. FPIA is homogeneous and requires no separation, whereas MEIA is heterogeneous and requires separation

Answer: (D)

Explanation

FPIA is a homogeneous method that uses polarized light to excite a fluorescent label and does not require separation; MEIA is heterogeneous and requires separation. The claim that FPIA requires separation reverses the central distinction between these techniques.

23. Which mass spectrometry method is the reference method for detecting inborn errors of metabolism?

A. Gas chromatography–mass spectrometry

B. MALDI-TOF mass spectrometry

C. Isotope dilution mass spectrometry

D. Tandem mass spectrometry

Answer: (D)

Explanation

Tandem mass spectrometry is a reference method for detecting inborn errors of metabolism. Gas chromatography–mass spectrometry is instead the gold standard for drug analysis.

24. A laboratory needs batch testing, random access, and STAT testing using separate containers for each sample. Which analysis format best fits these needs?

A. Discrete analysis

B. Continuous-flow analysis

C. Microparticle capture analysis

D. Centrifugal analysis

Answer: (A)

Explanation

Discrete analysis uses separate containers and supports batch, random-access, and STAT testing. Centrifugal analysis can support batch work, but its stated limitation is low throughput and it does not provide the same combination of testing modes.

25. After a meal, which hormone promotes cellular glucose uptake and glycogenesis while lowering plasma glucose?

A. Glucagon, produced by pancreatic islet alpha cells

B. Insulin, produced by pancreatic islet beta cells

C. Cortisol, produced by the adrenal cortex

D. Epinephrine, produced by the adrenal medulla

Answer: (B)

Explanation

Insulin is produced by pancreatic islet beta cells and lowers plasma glucose by promoting cellular uptake and glycogenesis, among other actions. Glucagon is a plausible contrast, but it raises glucose by promoting gluconeogenesis and glycogenolysis.

26. During fasting, which pancreatic hormone promotes gluconeogenesis and glycogenolysis while inhibiting glycolysis?

A. Glucagon, produced by pancreatic islet alpha cells

B. Thyroxine, which promotes intestinal glucose absorption

C. Insulin, produced by pancreatic islet beta cells

D. Growth hormone, which inhibits glycolysis

Answer: (A)

Explanation

Glucagon is produced by pancreatic islet alpha cells and promotes gluconeogenesis and glycogenolysis while inhibiting glycolysis. Insulin is the key opposing hormone, promoting glucose uptake and inhibiting glycogen breakdown.

27. A plasma glucose result is 38 mg/dL. How should this value be classified using the stated hypoglycemia thresholds?

A. It is above the hypoglycemia cutoff but below the panic-value threshold

B. It meets the hypoglycemia cutoff but does not reach the panic-value threshold

C. It is below the hypoglycemia cutoff and also below the panic-value threshold

D. It is below the hypoglycemia cutoff but above the panic-value threshold

Answer: (C)

Explanation

Hypoglycemia is defined as plasma glucose below 50 mg/dL, and the panic-value threshold is below 40 mg/dL, so 38 mg/dL is below both. A value between 40 and 50 mg/dL would meet the hypoglycemia definition without falling below the panic threshold.

28. Which set of findings fulfills Whipple’s triad?

A. Symptoms of hypoglycemia, increased insulin, and decreased β-hydroxybutyrate

B. Symptoms after a meal, low plasma glucose, and increased proinsulin

C. Symptoms of hypoglycemia, low plasma glucose, and symptom relief after glucose correction

D. Low plasma glucose, increased C-peptide, and decreased β-hydroxybutyrate

Answer: (C)

Explanation

Whipple’s triad consists of hypoglycemic symptoms, a low plasma glucose measurement, and relief of symptoms after glucose correction. Increased insulin, proinsulin, and C-peptide with decreased β-hydroxybutyrate are additional findings relevant to insulinoma diagnosis, not the triad itself.

29. Which pattern distinguishes type 1 diabetes from type 2 diabetes?

A. Type 1 involves absolute insulin deficiency from beta-cell destruction, while type 2 involves insulin resistance with progressive insulin deficiency.

B. Type 1 is glucose intolerance arising during pregnancy, while type 2 is autoimmune beta-cell destruction.

C. Type 1 involves reduced insulin sensitivity without beta-cell loss, while type 2 involves a fixed absence of insulin production.

D. Type 1 involves insulin resistance with progressive insulin deficiency, while type 2 involves absolute deficiency from beta-cell destruction.

Answer: (A)

Explanation

Type 1 diabetes results from beta-cell destruction and absolute insulin deficiency, whereas type 2 combines insulin resistance with progressive insulin deficiency. The second pattern reverses the defining distinction between the two types.

30. A patient has classic symptoms of diabetes and a random plasma glucose of 205 mg/dL. Which interpretation is consistent with a diagnostic threshold?

A. The result meets the fasting-glucose threshold because it exceeds 126 mg/dL, regardless of symptoms.

B. The result meets the HbA1c threshold because a glucose value above 200 mg/dL predicts HbA1c above 6.5%.

C. The result meets the random-glucose threshold because it exceeds 200 mg/dL and symptoms are present.

D. The result does not meet a diagnostic threshold unless a 2-hour glucose exceeds 200 mg/dL.

Answer: (C)

Explanation

Random plasma glucose above 200 mg/dL in the presence of diabetes symptoms meets a stated diagnostic threshold, subject to the specified confirmation requirements. A fasting threshold is a separate criterion and cannot be applied to a random sample.

31. Which testing sequence is used to diagnose gestational diabetes in the described two-step approach?

A. A 100 g, 1-hour OGTT is used for screening, followed by a 50 g, 3-hour OGTT for diagnosis.

B. A 50 g, 1-hour OGTT is used for screening, followed by a 100 g, 3-hour OGTT for diagnosis.

C. A 75 g, 2-hour OGTT is used for screening, followed by a 100 g, 1-hour OGTT for diagnosis.

D. A 50 g, 3-hour OGTT is used for screening, followed by a 75 g, 2-hour OGTT for diagnosis.

Answer: (B)

Explanation

The screening test uses a 50 g glucose load measured at 1 hour, and diagnosis uses a 100 g, 3-hour OGTT, with at least two values meeting or exceeding the specified cutoffs. The 75 g, 2-hour test is a distinct OGTT protocol, not the stated gestational diagnostic sequence.

32. A fasting plasma glucose is 112 mg/dL. Which interpretation fits the stated prediabetes range and fasting preparation?

A. It is within the prediabetes range of 140–149 mg/dL, with fasting defined as 8–14 hours without food.

B. It is within the prediabetes range of 100–125 mg/dL, with fasting defined as at least 8 hours without food.

C. It is within the diabetes range of at least 126 mg/dL, with fasting defined as at least 6 hours without food.

D. It is below the prediabetes range of 100–125 mg/dL, which applies to a 2-hour glucose measurement.

Answer: (B)

Explanation

A fasting value of 112 mg/dL falls within the stated prediabetes interval of 100–125 mg/dL, and fasting requires at least 8 hours without food. The 140–149 mg/dL interval applies to the 2-hour glucose measure, not fasting glucose.

33. A glucose assay gives a falsely increased result when a sample contains a reducing substance other than glucose. Which method distinction best explains this finding?

A. Enzymatic methods are less specific than copper-reduction methods, so they produce the false increase.

B. Copper-reduction methods can respond to any reducing agent, whereas hexokinase is the reference method and avoids those interferences.

C. Hexokinase responds to any reducing agent, whereas copper-reduction methods are the reference method for glucose.

D. Copper-reduction methods measure only glucose, whereas hexokinase is vulnerable to unrelated reducing substances.

Answer: (B)

Explanation

Copper-reduction assays are nonspecific because reducing agents other than glucose can produce increased readings, while hexokinase is the reference method and is not subject to those described interferences. The claim that copper-reduction methods are glucose-specific reverses the central distinction.

34. A child presents with hypoglycemia, ketonuria, lactic acidosis, dyslipidemia, and hyperuricemia. Which enzyme deficiency is characteristic of Von Gierke disease?

A. Glucose-6-phosphatase deficiency

B. Glucose-6-phosphate dehydrogenase deficiency

C. Lysosomal acid alpha-glucosidase deficiency

D. Hepatic glycogen phosphorylase deficiency

Answer: (A)

Explanation

Von Gierke disease, or glycogen storage disease type Ia, is caused by glucose-6-phosphatase deficiency and has the listed metabolic features. Glucose-6-phosphate dehydrogenase deficiency is a different enzyme defect and does not define this glycogen storage disease.

35. Which apolipoprotein acts as a cofactor for lipoprotein lipase (LPL)?

A. Apo C-II

B. Apo B-100

C. Apo C-III

D. Apo A-I

Answer: (A)

Explanation

Apo C-II serves as a cofactor for LPL, supporting lipoprotein metabolism. Apo C-III has the contrasting role of inhibiting LPL.

36. After a meal, which particles carry dietary triglycerides from the intestine, while liver-produced endogenous triglycerides are carried in which particles?

A. Chylomicrons carry dietary triglycerides, while VLDL carries endogenous triglycerides.

B. HDL carries dietary triglycerides, while LDL carries endogenous triglycerides.

C. LDL carries dietary triglycerides, while HDL carries endogenous triglycerides.

D. VLDL carries dietary triglycerides, while chylomicrons carry endogenous triglycerides.

Answer: (A)

Explanation

Chylomicrons transport dietary triglycerides absorbed from the intestine, whereas VLDL transports triglycerides produced by the liver. Reversing these origins confuses the principal distinction between the two particles.

37. A lipid profile shows high cholesterol and high triglycerides, with IDL and β-VLDL implicated. Which Fredrickson type matches this pattern?

A. Type 2b

B. Type 3

C. Type 4

D. Type 5

Answer: (B)

Explanation

Fredrickson type 3 is associated with IDL and β-VLDL and elevated cholesterol and triglycerides. Type 2b also raises both lipids, but it involves LDL and VLDL rather than IDL and β-VLDL.

38. A patient is scheduled for a lipid profile and a separate total cholesterol test. Which preparation advice is appropriate?

A. Fast for cholesterol testing alone; the lipid profile does not require fasting.

B. Fast for both tests because cholesterol measurement requires fasting.

C. Avoid food for 24 hours before both tests to ensure accurate results.

D. Fast for 12–14 hours before the lipid profile; fasting is not required for cholesterol testing alone.

Answer: (D)

Explanation

A lipid profile requires 12–14 hours of fasting, while cholesterol testing alone does not. Requiring fasting for both tests incorrectly applies the lipid-profile preparation to cholesterol testing by itself.

39. A patient's triglyceride concentration is 460 mg/dL. Which approach is appropriate for determining LDL cholesterol?

A. Use the Friedewald equation with a triglyceride divisor of 5 because the calculation remains reliable at this level.

B. Use direct LDL measurement because the Friedewald calculation is unreliable above 400 mg/dL triglycerides.

C. Use the Friedewald equation with a triglyceride divisor of 6.5 because that is its standard divisor in mg/dL.

D. Estimate LDL by dividing total cholesterol by the triglyceride concentration.

Answer: (B)

Explanation

The Friedewald calculation is unreliable when triglycerides exceed 400 mg/dL, so direct LDL measurement is appropriate at 460 mg/dL. The divisor of 5 is used in the equation when calculation is reliable, but it does not resolve this limitation.

40. A laboratory result gives BUN as 10 mg/dL. Using the stated conversion, what is the corresponding urea concentration?

A. 12.14 mg/dL, calculated by adding 2.14 to BUN.

B. 21.4 mg/dL, calculated by multiplying BUN by 2.14.

C. 10 mg/dL, because BUN and urea have the same numerical value.

D. 4.7 mg/dL, calculated by dividing BUN by 2.14.

Answer: (B)

Explanation

Urea is calculated by multiplying BUN by 2.14, so 10 mg/dL BUN corresponds to 21.4 mg/dL urea. Dividing by 2.14 reverses the conversion and instead converts urea toward BUN.

41. Which comparison of urea and creatinine handling by the kidneys is accurate?

A. Urea and creatinine are both freely filtered and substantially reabsorbed in the tubules.

B. Urea is freely filtered and 40–60% reabsorbed, whereas creatinine is filtered, not reabsorbed, and minimally secreted.

C. Urea is freely filtered and not reabsorbed, whereas creatinine is filtered and substantially reabsorbed.

D. Urea is not filtered and is secreted, whereas creatinine is freely filtered and substantially reabsorbed.

Answer: (B)

Explanation

Urea is freely filtered, with 40–60% reabsorbed, while creatinine is filtered, not reabsorbed, and minimally secreted. The claim that creatinine is substantially reabsorbed incorrectly assigns it the key tubular behavior of urea.

42. A protein consists of multiple subunits held together by noncovalent forces. Which level of protein structure does this describe?

A. Quaternary structure, which arises from interactions among multiple protein subunits.

B. Secondary structure, which includes α-helices and β-pleated sheets stabilized by hydrogen bonds.

C. Primary structure, which specifies the amino acid types and their sequence.

D. Tertiary structure, which describes the three-dimensional conformation of one protein molecule.

Answer: (A)

Explanation

Quaternary structure describes the interactions among multiple protein subunits held together by noncovalent forces. Tertiary structure concerns a molecule’s three-dimensional conformation, rather than the assembly of multiple subunits.

43. During serum protein electrophoresis at pH 8.6, in which direction do the proteins migrate?

A. Away from both electrodes, because the proteins have no net charge at this pH.

B. From the sample wells near the cathode toward the positive anode, because the proteins carry a net positive charge.

C. From the sample wells near the anode toward the cathode, because the proteins carry a net negative charge.

D. From the sample wells near the cathode toward the positive anode, because the proteins carry a net negative charge.

Answer: (D)

Explanation

At pH 8.6, serum proteins carry a net negative charge and migrate from the wells near the cathode toward the positive anode. A positive charge would drive migration toward the cathode, not the anode.

44. Which statement correctly characterizes pre-albumin, also known as transthyretin?

A. It inhibits proteolytic enzymes and is the most abundant protein in the α1 band.

B. It transports primarily T4 and vitamin A associated with retinol-binding protein and is a positive acute-phase reactant.

C. It transports primarily T4 and vitamin A associated with retinol-binding protein and is a negative acute-phase reactant.

D. It binds free hemoglobin to preserve iron and is a positive acute-phase reactant.

Answer: (C)

Explanation

Pre-albumin transports primarily T4 and vitamin A associated with retinol-binding protein, and it is a negative acute-phase reactant. The positive acute-phase classification is a common confusion with proteins such as α1-antitrypsin.

45. A patient has markedly decreased serum albumin. Which consequence can follow from this change?

A. Reduced plasma oncotic pressure, promoting an exudative effusion through inflammation.

B. Reduced plasma oncotic pressure, promoting edema and transudative effusion.

C. Increased plasma oncotic pressure, promoting edema and transudative effusion.

D. Increased plasma oncotic pressure, preventing fluid movement into tissues.

Answer: (B)

Explanation

Decreased albumin lowers plasma oncotic pressure and can lead to edema and transudative effusion. An exudative effusion is not the stated consequence of reduced oncotic pressure; it is associated with a different mechanism.

46. Which protein is the most abundant in the α1 band and inhibits proteolytic enzymes such as elastase?

A. α2-macroglobulin, a protease inhibitor that rises markedly in nephrotic syndrome.

B. α1-antitrypsin, a positive acute-phase reactant that inhibits several proteolytic enzymes.

C. Haptoglobin, a positive acute-phase reactant that binds free hemoglobin and preserves iron.

D. Pre-albumin, a negative acute-phase reactant that transports T4 and vitamin A.

Answer: (B)

Explanation

α1-antitrypsin is the most abundant α1-band protein, is a positive acute-phase reactant, and inhibits enzymes including elastase. Haptoglobin is also a positive acute-phase reactant, but its key role is binding free hemoglobin.

47. What happens to haptoglobin concentration during intravascular hemolysis, and why?

A. It decreases as haptoglobin–hemoglobin complexes are removed by the liver.

B. It remains unchanged because haptoglobin does not interact with free hemoglobin.

C. It decreases because the liver removes unbound haptoglobin before it binds hemoglobin.

D. It increases as it binds free hemoglobin and preserves iron in circulation.

Answer: (A)

Explanation

Haptoglobin binds free hemoglobin, and its concentration decreases during intravascular hemolysis as the resulting complexes are removed by the liver. Although haptoglobin is a positive acute-phase reactant, that property does not prevent this decrease during hemolysis.

48. Which condition can contribute to hypoproteinemia?

A. Dehydration causing concentrated plasma proteins

B. Increased synthesis of immunoglobulins

C. Reduced hepatic protein synthesis

D. Increased intestinal absorption of dietary protein

Answer: (C)

Explanation

Reduced hepatic synthesis can lower circulating protein levels, as can protein loss, malnutrition, increased catabolism, or immunodeficiency. Dehydration instead causes relative hyperproteinemia by concentrating proteins in the blood.

49. A serum protein electrophoresis result shows a sharp M spike caused by an absolute increase in one immunoglobulin. Which disorder pattern does this indicate?

A. Hypoproteinemia from malnutrition

B. Monoclonal gammopathy

C. Relative hyperproteinemia from dehydration

D. Polyclonal gammopathy

Answer: (B)

Explanation

A monoclonal gammopathy involves a sharp increase in one immunoglobulin, commonly IgG or IgM. Polyclonal gammopathy instead produces a diffuse increase in the gamma region.

50. Which collection method is used to quantify total urine protein?

A. A timed two-hour urine specimen

B. A timed 24-hour urine specimen

C. A single random urine specimen

D. A first-morning urine dipstick reading

Answer: (B)

Explanation

Total urine protein is quantified using a 24-hour urine specimen, with a reference value below 100 mg per day. A dipstick reading is not the specified collection method for quantifying total daily protein.

51. A patient with diabetes has increased urinary albumin that is not detected by a dipstick. What is this finding called, and why is it clinically important?

A. Bilirubinuria, which signals early diabetic nephropathy

B. Microalbuminuria, which can precede diabetic nephropathy

C. Proteinuria, which confirms established nephrotic syndrome

D. Hematuria, which can precede diabetic nephropathy

Answer: (B)

Explanation

Microalbuminuria is increased urinary albumin below dipstick detection and can precede diabetic nephropathy. Nephrotic syndrome is a different condition and is not established by this specific finding.

52. Why is myoglobin more useful for ruling out than establishing myocardial infarction?

A. It rises late and remains elevated for several weeks

B. It is cardiac-specific but rises after troponin

C. It rises early but lacks cardiac specificity

D. It is released by ventricles in response to pressure overload

Answer: (C)

Explanation

Myoglobin rises earlier than troponin after myocardial infarction, but it is also found in skeletal muscle and therefore lacks cardiac specificity. Cardiac troponins are more suitable for establishing cardiac injury because they are cardiac-specific.

53. Which marker is considered the gold standard for acute coronary syndrome because it is cardiac-specific and remains elevated?

A. Creatine kinase from skeletal muscle

B. Myoglobin

C. B-type natriuretic peptide

D. Cardiac troponin

Answer: (D)

Explanation

Cardiac troponins are the gold-standard markers for acute coronary syndrome because they are cardiac-specific and remain elevated. Myoglobin can rise earlier, but its lack of cardiac specificity makes it less suitable for confirming myocardial injury.

54. A patient with congestive heart failure has volume expansion and increased cardiac pressure. Which protein is released by the ventricles in response to these changes?

A. B-type natriuretic peptide

B. Cardiac troponin T

C. Myoglobin

D. Cardiac troponin I

Answer: (A)

Explanation

BNP is secreted by myocardial ventricles in response to volume expansion and pressure overload, and its levels increase in congestive heart failure. Troponins are markers of cardiac injury rather than this ventricular response to loading.

55. How does the Kjeldahl method estimate protein content?

A. It measures ultraviolet absorption and applies a factor of 6.25 to aromatic amino acids

B. It measures peptide-bond color and applies a factor of 16, assuming protein is 6.25% nitrogen

C. It measures nitrogen and applies a factor of 6.25, assuming nitrogen is 16% of protein mass

D. It measures copper binding and assumes each protein contains 16% peptide bonds

Answer: (C)

Explanation

The Kjeldahl method estimates protein from nitrogen content, assuming nitrogen accounts for 16% of protein weight and using a conversion factor of 6.25. Measuring peptide-bond color describes the biuret method, not Kjeldahl analysis.

56. What chemical interaction produces the colored signal in the biuret method?

A. Bilirubin binds albumin in alkaline medium, forming a violet complex

B. Nitrogen reacts with acid to form a colored peptide complex

C. Aromatic amino acids absorb ultraviolet light near 280 nm

D. Cupric ions bind peptide bonds in alkaline medium, forming a violet or purple complex

Answer: (D)

Explanation

In the biuret method, cupric ions bind peptide bonds in an alkaline medium to form a violet or purple complex whose intensity tracks protein concentration. Ultraviolet absorption near 280 nm instead reflects aromatic amino acids.

57. Which function is performed by the liver?

A. Detoxifying drugs and ammonia while synthesizing proteins and coagulation factors

B. Producing cardiac troponins while storing water-soluble vitamins

C. Filtering plasma proteins into urine while conjugating unconjugated bilirubin

D. Converting peptide bonds into urobilinogen while synthesizing myoglobin

Answer: (A)

Explanation

The liver detoxifies substances including drugs and ammonia, and it synthesizes or metabolizes proteins and coagulation factors. Cardiac troponins are cardiac markers, not products of the liver's listed functions.

58. Which sequence correctly describes bilirubin metabolism from red-cell breakdown to intestinal products?

A. Red-cell breakdown forms urobilinogen, which the liver converts to unconjugated bilirubin before bile secretion

B. Hemoglobin releases heme, which becomes biliverdin and albumin-bound unconjugated bilirubin; the liver conjugates it before it enters bile and the intestine

C. Hemoglobin releases bilirubin, which becomes biliverdin and binds albumin; the kidney conjugates it before it enters the intestine

D. Heme becomes conjugated bilirubin in blood, which binds albumin before the liver converts it to biliverdin

Answer: (B)

Explanation

Red-cell destruction releases hemoglobin; heme becomes biliverdin and then albumin-bound unconjugated bilirubin, which hepatocytes conjugate before secretion into bile and delivery to the intestine. The kidney does not perform the conjugation step; intestinal bacteria generate urobilinogen and stool pigments.