A 46-year-old woman with no symptoms and no medicines has blood drawn at a community clinic at 14:00 into a serum separator tube for a routine panel. The tube is left unspun at room temperature, reaches the laboratory at 10:00 the next day and is centrifuged then. Results: glucose 41 mg/dL [2.3 mmol/L], potassium 6.2 mmol/L, phosphate 5.8 mg/dL [1.87 mmol/L] (reference 2.5-4.5 mg/dL), calcium 9.4 mg/dL [2.35 mmol/L] (reference 8.6-10.2 mg/dL), ALP 72 U/L (reference 40-130 U/L), creatinine 0.9 mg/dL [80 µmol/L]. The hemolysis index is 0 and the serum looks clear. The most likely explanation and the correct action are:

Delayed separation; recollect and spin within 2 hours

EDTA carryover; recollect using the right draw order

In vitro hemolysis; recollect with a larger needle

Insulinoma; measure insulin and C-peptide

Which collection approach stops glycolysis almost at once, giving the most accurate glucose for diagnosis?

Unspun lithium heparin gel tube

Some glucose oxidase reagents contain mutarotase. Its role is to:

Convert alpha-D-glucose to the beta form

Remove hydrogen peroxide from the sample

Destroy ascorbic acid in the sample

Phosphorylate glucose before oxidation

Carbohydrates: test procedures and interferences

11 ASCP MLS practice questions on Carbohydrates: 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 46-year-old woman with no symptoms and no medicines has blood drawn at a community clinic at 14:00 into a serum separator tube for a routine panel. The tube is left unspun at room temperature, reaches the laboratory at 10:00 the next day and is centrifuged then. Results: glucose 41 mg/dL [2.3 mmol/L], potassium 6.2 mmol/L, phosphate 5.8 mg/dL [1.87 mmol/L] (reference 2.5-4.5 mg/dL), calcium 9.4 mg/dL [2.35 mmol/L] (reference 8.6-10.2 mg/dL), ALP 72 U/L (reference 40-130 U/L), creatinine 0.9 mg/dL [80 µmol/L]. The hemolysis index is 0 and the serum looks clear. The most likely explanation and the correct action are:
1. AEDTA carryover; recollect using the right draw order
2. BIn vitro hemolysis; recollect with a larger needle
3. CInsulinoma; measure insulin and C-peptide
4. DDelayed separation; recollect and spin within 2 hours
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Answer: D · Delayed separation; recollect and spin within 2 hours
The tube stood on the cells for about 20 hours. The cells kept using glucose, and once their energy ran out the sodium-potassium pump failed, so potassium leaked into the serum; phosphate rose as cell organic phosphates were broken down. A normal calcium and ALP exclude EDTA contamination, which chelates both, and a hemolysis index of 0 excludes in vitro hemolysis. A well patient with low glucose plus high potassium and phosphate has a specimen problem, not hypoglycemia. CLSI GP44 limits contact of serum or plasma with cells to 2 hours, so the sample should be recollected and spun promptly, with glucose in a fluoride tube if a delay is expected.
- A: EDTA lowers calcium and ALP, and both are normal here.
- B: The hemolysis index is 0; potassium leaked from intact cells.
- C: A symptom-free patient with high potassium and phosphate points to a handling error, not true hypoglycemia.
Question 2 · Easy
Which glycolytic enzyme does sodium fluoride inhibit to preserve glucose in a blood sample?
1. AEnolase
2. BLactate dehydrogenase
3. CHexokinase
4. DGlucose oxidase
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Answer: A · Enolase
Fluoride inhibits enolase, which acts late in glycolysis, so the blood cells stop consuming glucose. It is usually paired with potassium oxalate as the anticoagulant in gray-top tubes. Because enolase acts late, glucose can still fall during the first one to two hours after collection. Prompt separation of plasma remains good practice.
- B: Lactate dehydrogenase converts pyruvate to lactate; fluoride does not target it.
- C: Hexokinase acts at the first step of glycolysis and is not the target of fluoride.
- D: Glucose oxidase is a reagent enzyme, not part of glycolysis in blood cells.
Question 3 · Medium
Which collection approach stops glycolysis almost at once, giving the most accurate glucose for diagnosis?
1. AUnspun lithium heparin gel tube
2. BCitrate-buffered fluoride tube
3. CPlain sodium fluoride tube
4. DEDTA tube at room temperature
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Answer: B · Citrate-buffered fluoride tube
Fluoride blocks enolase, a late step, so glucose still falls by several percent in the first hour or two until it takes full effect. Acidifying the sample with citrate buffer inhibits the early enzymes of glycolysis and stops glucose loss at once. Guidelines accept this, or placing the tube in an ice-water slurry and separating the plasma within about 30 minutes. Early glycolytic loss can make a diabetic fasting value look normal.
- A: Cells left in contact with plasma keep consuming glucose.
- C: With fluoride alone, glucose still falls during the first one to two hours.
- D: EDTA does not stop glycolysis, so glucose keeps falling.
Question 4 · Easy
Some glucose oxidase reagents contain mutarotase. Its role is to:
1. ARemove hydrogen peroxide from the sample
2. BConvert alpha-D-glucose to the beta form
3. CDestroy ascorbic acid in the sample
4. DPhosphorylate glucose before oxidation
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Answer: B · Convert alpha-D-glucose to the beta form
In solution, glucose exists as about 36% alpha and 64% beta anomers. Glucose oxidase reacts only with beta-D-glucose, so mutarotase speeds the conversion of the alpha form and lets the reaction finish quickly. Without it, a short kinetic reading could underestimate glucose. The hexokinase method does not need this step.
- A: Hydrogen peroxide is the product the peroxidase step needs; mutarotase does not remove it.
- C: Ascorbate oxidase, not mutarotase, is used to remove ascorbic acid.
- D: Phosphorylation is the hexokinase reaction, not part of the glucose oxidase method.
Question 5 · Easy
The hexokinase glucose method is read at 340 nm. The rise in absorbance comes from the formation of:
1. AA quinoneimine dye
2. BHydrogen peroxide
3. CNADPH
4. DGlucose-6-phosphate
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Answer: C · NADPH
Hexokinase uses ATP to phosphorylate glucose to glucose-6-phosphate. Glucose-6-phosphate dehydrogenase then oxidizes glucose-6-phosphate and reduces NADP+ to NADPH, which absorbs at 340 nm in proportion to the glucose present. Some reagents use a bacterial enzyme with NAD+, forming NADH, which absorbs at the same wavelength. The coupled reaction is highly specific, which is why hexokinase is the basis of the reference method.
- A: A quinoneimine dye is the colored product of the peroxidase (Trinder) step.
- B: Hydrogen peroxide is a product of the glucose oxidase reaction.
- D: Glucose-6-phosphate is an intermediate and does not absorb at 340 nm.
Question 6 · Medium
In the glucose oxidase-peroxidase (Trinder) method, a high ascorbic acid level in the sample tends to give:
1. AFalsely high results, by forming extra dye
2. BFalsely high results, by activating glucose oxidase
3. CNo change, because peroxidase is specific
4. DFalsely low results, by using up peroxide
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Question 7 · Hard
A patient on peritoneal dialysis with icodextrin has a bedside meter glucose of 260 mg/dL (14.4 mmol/L) and no symptoms. The laboratory plasma glucose by hexokinase is 92 mg/dL (5.1 mmol/L). Which meter enzyme best explains the difference?
1. AHexokinase
2. BGlucose oxidase
3. CGlucose dehydrogenase with FAD
4. DGlucose dehydrogenase with PQQ
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Question 8 · Medium
With a glucose oxidase test strip, which factor tends to make a capillary meter glucose read falsely low?
1. ALow hematocrit
2. BSugar on the fingertip
3. CLow blood oxygen
4. DHigh hematocrit
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Question 9 · Hard
A glucose standard of 100 mg/dL (5.55 mmol/L) gives an absorbance of 0.200. A patient's plasma, diluted 1 part with 4 parts saline because it exceeded the linear range, gives an absorbance of 0.300. What is the patient's glucose?
1. A750 mg/dL (41.6 mmol/L)
2. B600 mg/dL (33.3 mmol/L)
3. C333 mg/dL (18.5 mmol/L)
4. D150 mg/dL (8.3 mmol/L)
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Question 10 · Medium
With a normal hematocrit, whole blood glucose reads roughly 10 to 15 percent lower than plasma glucose from the same sample. The main reason is:
1. AHemoglobin inhibits the glucose enzymes
2. BRed cells hold less water than plasma
3. CCentrifugation concentrates the plasma
4. DRed cells consume glucose during the test
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Question 11 · Easy
Blood for glucose is collected in a plain tube and left unseparated at room temperature for 3 hours. What happens to the glucose result?
1. AIt does not change
2. BIt is falsely low
3. CIt is falsely high
4. DIt rises, then returns to normal
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Carbohydrates: test procedures and interferences

Which glycolytic enzyme does sodium fluoride inhibit to preserve glucose in a blood sample?

Which collection approach stops glycolysis almost at once, giving the most accurate glucose for diagnosis?

Some glucose oxidase reagents contain mutarotase. Its role is to:

The hexokinase glucose method is read at 340 nm. The rise in absorbance comes from the formation of:

In the glucose oxidase-peroxidase (Trinder) method, a high ascorbic acid level in the sample tends to give:

A patient on peritoneal dialysis with icodextrin has a bedside meter glucose of 260 mg/dL (14.4 mmol/L) and no symptoms. The laboratory plasma glucose by hexokinase is 92 mg/dL (5.1 mmol/L). Which meter enzyme best explains the difference?

With a glucose oxidase test strip, which factor tends to make a capillary meter glucose read falsely low?

A glucose standard of 100 mg/dL (5.55 mmol/L) gives an absorbance of 0.200. A patient's plasma, diluted 1 part with 4 parts saline because it exceeded the linear range, gives an absorbance of 0.300. What is the patient's glucose?

With a normal hematocrit, whole blood glucose reads roughly 10 to 15 percent lower than plasma glucose from the same sample. The main reason is:

Blood for glucose is collected in a plain tube and left unseparated at room temperature for 3 hours. What happens to the glucose result?