A 64-year-old man with a left ventricular assist device (LVAD) is monitored for pump thrombosis, which raises LD. Three samples over 2 days, including one drawn by an experienced phlebotomist through a straight needle, all have pink plasma and a free hemoglobin of about 180 mg/dL [1.8 g/L]. The laboratory's rule suppresses LD when free hemoglobin exceeds 50 mg/dL [0.5 g/L]. Latest results: LD 1,480 U/L (reference 125-220), potassium 4.3 mmol/L, AST 64 U/L (below 40), haptoglobin below 10 mg/dL [0.1 g/L], reticulocytes 4.2%. What should the laboratory do?

Report LD with a comment on in vivo hemolysis

Report LD as falsely raised by in vitro hemolysis

Suppress LD and request another recollection

Apply a hemolysis correction factor to LD

Alkaline phosphatase measured on EDTA plasma is far lower than on serum from the same draw. The reason is that EDTA:

Chelates the zinc and magnesium ALP needs

Lowers the reaction pH below its optimum

Precipitates the bone isoenzyme of ALP

A 72-year-old woman without chest pain has total CK 160 U/L and CK-MB by immunoinhibition 90 U/L. Troponin is normal on two samples. The most likely explanation is:

Macro-CK interfering with the method

Enzymes: test procedures and interferences

11 ASCP MLS practice questions on Enzymes: test procedures and interferences, part of Chemistry. The first 11 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Hard
A 64-year-old man with a left ventricular assist device (LVAD) is monitored for pump thrombosis, which raises LD. Three samples over 2 days, including one drawn by an experienced phlebotomist through a straight needle, all have pink plasma and a free hemoglobin of about 180 mg/dL [1.8 g/L]. The laboratory's rule suppresses LD when free hemoglobin exceeds 50 mg/dL [0.5 g/L]. Latest results: LD 1,480 U/L (reference 125-220), potassium 4.3 mmol/L, AST 64 U/L (below 40), haptoglobin below 10 mg/dL [0.1 g/L], reticulocytes 4.2%. What should the laboratory do?
1. AReport LD as falsely raised by in vitro hemolysis
2. BReport LD with a comment on in vivo hemolysis
3. CSuppress LD and request another recollection
4. DApply a hemolysis correction factor to LD
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Answer: B · Report LD with a comment on in vivo hemolysis
Hemolysis in the tube releases red cell potassium and LD together, so in vitro hemolysis usually raises potassium. Here potassium is normal, haptoglobin is absent, reticulocytes are raised and three carefully drawn samples look the same, so the hemolysis is happening in the circulation (in vivo), as is common with LVADs. Recollecting cannot remove it, and the raised LD is the patient's true value, which is exactly what the clinicians need to detect pump thrombosis. The laboratory should report the LD with a comment that the sample shows in vivo hemolysis, rather than apply a correction formula. Rules that suppress hemolyzed results need an exception for this situation.
- A: Normal potassium, absent haptoglobin and identical draws show the hemolysis happened in the patient.
- C: Three careful draws gave the same result; another will too, and the clinical answer is lost.
- D: Correction formulas are unreliable and not recommended; this LD is real circulating LD.
Question 2 · Medium
Alkaline phosphatase measured on EDTA plasma is far lower than on serum from the same draw. The reason is that EDTA:
1. AChelates the zinc and magnesium ALP needs
2. BLowers the reaction pH below its optimum
3. CCauses hemolysis in the sample
4. DPrecipitates the bone isoenzyme of ALP
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Answer: A · Chelates the zinc and magnesium ALP needs
ALP is a zinc metalloenzyme that is activated by magnesium, and EDTA binds both ions, so little activity remains. Citrate and oxalate also chelate divalent cations and are unsuitable for ALP. Serum or heparin plasma is the right sample. A very low ALP with a very low calcium and a high potassium suggests EDTA contamination, for example from a wrong order of draw.
- B: The reagent is strongly buffered at an alkaline pH of about 10.4.
- C: EDTA does not hemolyze red cells.
- D: EDTA does not precipitate ALP isoenzymes.
Question 3 · Hard
A 72-year-old woman without chest pain has total CK 160 U/L and CK-MB by immunoinhibition 90 U/L. Troponin is normal on two samples. The most likely explanation is:
1. AAcute myocardial infarction
2. BRecent skeletal muscle injury
3. CMacro-CK interfering with the method
4. DHypothyroid myopathy
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Answer: C · Macro-CK interfering with the method
Immunoinhibition blocks M subunits with anti-M antibody, measures the remaining B-subunit activity and doubles it to report CK-MB. CK-BB, macro-CK (usually CK-BB bound to immunoglobulin, or oligomeric mitochondrial CK) and adenylate kinase are not inhibited, so they give a falsely high CK-MB, here over half of the total. CK-MB from heart muscle seldom exceeds about 25% of total CK, and the normal troponins argue against infarction. A CK-MB mass immunoassay or CK electrophoresis resolves the question.
- A: Troponin is normal twice, and the CK-MB fraction is implausibly high for heart muscle.
- B: Skeletal muscle releases mainly CK-MM, which the anti-M antibody inhibits.
- D: Hypothyroid myopathy raises CK-MM, which this method inhibits.
Question 4 · Hard
An LD method follows NADH at 340 nm in a 1 cm cuvette (molar absorptivity 6.22 x 10^3 L/mol/cm). 50 µL of serum is added to 1.00 mL of reagent, and absorbance changes by 0.150 over 3 minutes at a steady rate. What is the LD activity?
1. A8 U/L
2. B506 U/L
3. C161 U/L
4. D169 U/L
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Answer: D · 169 U/L
Activity (U/L) = (ΔA/min x total volume x 10^6) / (ε x path length x sample volume). Here ΔA/min = 0.150/3 = 0.050 and total volume = 1.00 + 0.05 = 1.05 mL, so U/L = 0.050 x (1.05/0.05) x 10^6 / 6,220 = 169 U/L. The factor 10^6 converts mol to µmol, and the volume ratio converts the activity in the cuvette back to serum. Leaving the sample out of the total volume, or skipping the per-minute step, gives the usual wrong answers.
- A: 8 U/L leaves out the total-to-sample volume ratio.
- B: 506 U/L uses the 3-minute change as if it were per minute.
- C: 161 U/L leaves the 50 µL of sample out of the total volume.
Question 5 · Easy
Many enzyme methods follow the formation or use of NADH. At which wavelength is NADH read?
1. A340 nm
2. B405 nm
3. C540 nm
4. D260 nm
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Answer: A · 340 nm
NADH absorbs strongly at 340 nm while NAD+ does not, so the conversion between the two can be followed directly. ALT, AST, LD and CK methods all rely on this, either directly or through a coupled indicator reaction. The rate of absorbance change per minute is proportional to enzyme activity. The molar absorptivity of NADH at 340 nm, 6.22 x 10^3 L/mol/cm, converts that rate into units.
- B: 405 nm is used for p-nitrophenol (ALP) and p-nitroaniline (GGT) products.
- C: 540 nm is used for the biuret and cyanmethemoglobin methods.
- D: Both NAD+ and NADH absorb at 260 nm, so it cannot tell them apart.
Question 6 · Easy
A serum ALT of 1 U/L means that each liter of serum, under the stated assay conditions, converts:
1. A1 µmol of substrate per hour
2. B1 µmol of substrate per minute
3. C1 mmol of substrate per minute
4. D1 mol of substrate per second
Answer and explanation in LearnFree account
Question 7 · Easy
A serum sample from a difficult draw is visibly hemolyzed. Which enzyme result will be falsely raised the most?
1. AAlanine aminotransferase
2. BLactate dehydrogenase
3. CAlkaline phosphatase
4. DGamma-glutamyltransferase
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Question 8 · Hard
An ALT method follows the fall in NADH absorbance at 340 nm. One sample reads 1.20 at 0 s, 0.42 at 30 s, 0.18 at 60 s, 0.17 at 90 s and 0.17 at 120 s. The analyzer flags a low result. What should be done?
1. AReport the result, since the final rate is steady
2. BDilute the sample with diluent and repeat the assay
3. CExtend the reading window and report the mean
4. DRepeat with a larger volume of the same sample
Answer and explanation in LearnFree account
Question 9 · Medium
An incubator fault makes an AST method designed for 37 °C run at 32 °C. Patient results will be:
1. AFalsely low, because activity falls as temperature drops
2. BFalsely high, as substrate depletion is slower
3. CFalsely high, as the enzyme is more stable
4. DUnchanged, as zero-order kinetics still apply
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Question 10 · Medium
Hemolysis can falsely raise CK activity, although red cells contain essentially no CK. Which red cell enzyme causes this?
1. AGlucose-6-phosphate dehydrogenase
2. BAdenylate kinase
3. CHexokinase
4. DLactate dehydrogenase
Answer and explanation in LearnFree account
Question 11 · Medium
Why is a continuous-monitoring (kinetic) method preferred to a two-point fixed-time method for enzyme activity?
1. AIt measures enzyme mass instead of activity
2. BIt removes the need for temperature control
3. CIt needs no substrate excess in the reagent
4. DIt confirms that the rate stays linear throughout
Answer and explanation in LearnFree account

Enzymes: test procedures and interferences

Alkaline phosphatase measured on EDTA plasma is far lower than on serum from the same draw. The reason is that EDTA:

A 72-year-old woman without chest pain has total CK 160 U/L and CK-MB by immunoinhibition 90 U/L. Troponin is normal on two samples. The most likely explanation is:

An LD method follows NADH at 340 nm in a 1 cm cuvette (molar absorptivity 6.22 x 10^3 L/mol/cm). 50 µL of serum is added to 1.00 mL of reagent, and absorbance changes by 0.150 over 3 minutes at a steady rate. What is the LD activity?

Many enzyme methods follow the formation or use of NADH. At which wavelength is NADH read?

A serum ALT of 1 U/L means that each liter of serum, under the stated assay conditions, converts:

A serum sample from a difficult draw is visibly hemolyzed. Which enzyme result will be falsely raised the most?

An ALT method follows the fall in NADH absorbance at 340 nm. One sample reads 1.20 at 0 s, 0.42 at 30 s, 0.18 at 60 s, 0.17 at 90 s and 0.17 at 120 s. The analyzer flags a low result. What should be done?

An incubator fault makes an AST method designed for 37 °C run at 32 °C. Patient results will be:

Hemolysis can falsely raise CK activity, although red cells contain essentially no CK. Which red cell enzyme causes this?

Why is a continuous-monitoring (kinetic) method preferred to a two-point fixed-time method for enzyme activity?