A 61-year-old man with JAK2-positive polycythemia vera has been treated with venesection every 4 to 6 weeks for 2 years, aiming for a hematocrit below 45%, and he takes only low-dose aspirin. At diagnosis his MCV was 89 fL. Today's results: RBC 6.80 × 10^6/µL (6.80 × 10^12/L), hemoglobin 13.6 g/dL (136 g/L), hematocrit 43.5% (0.435 L/L); WBC and platelets are mildly raised. Calculate the red cell indices from these values. The results are best explained by:

Iron deficiency from venesection

Coexisting beta-thalassemia trait

Normocytic cells with good control

Relapse that needs more venesection

A sample from an adult with known iron deficiency gives Hb 7.8 g/dL (78 g/L), Hct 29% (0.29 L/L), MCV 66 fL and MCHC 27 g/dL. The film shows hypochromic microcytes without clumps, and the plasma is clear. The rule of three is not met. What should the technologist do?

Report the results; hypochromia explains the gap

Recollect because the sample has clotted

Warm the sample to 37 °C and rerun it

Replace the plasma with saline and rerun

Each pair of results below comes from a sample with normocytic, normochromic cells on the film. Which pair breaks the rule of three and should be checked before release?

Hb 11.0 g/dL (110 g/L), Hct 39% (0.39 L/L)

Hb 12.5 g/dL (125 g/L), Hct 36% (0.36 L/L)

Hb 14.0 g/dL (140 g/L), Hct 42% (0.42 L/L)

Hb 15.0 g/dL (150 g/L), Hct 47% (0.47 L/L)

Hematocrit and red cell indices

11 ASCP MLS practice questions on Hematocrit and red cell indices, part of Hematology. 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 61-year-old man with JAK2-positive polycythemia vera has been treated with venesection every 4 to 6 weeks for 2 years, aiming for a hematocrit below 45%, and he takes only low-dose aspirin. At diagnosis his MCV was 89 fL. Today's results: RBC 6.80 × 10^6/µL (6.80 × 10^12/L), hemoglobin 13.6 g/dL (136 g/L), hematocrit 43.5% (0.435 L/L); WBC and platelets are mildly raised. Calculate the red cell indices from these values. The results are best explained by:
1. ACoexisting beta-thalassemia trait
2. BIron deficiency from venesection
3. CNormocytic cells with good control
4. DRelapse that needs more venesection
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Answer: B · Iron deficiency from venesection
MCV = Hct ÷ RBC × 10 = 43.5 ÷ 6.80 × 10 = 64 fL; MCH = Hb ÷ RBC × 10 = 13.6 ÷ 6.80 × 10 = 20.0 pg; MCHC = Hb ÷ Hct × 100 = 31.3 g/dL. The cells are microcytic and hypochromic, and the rule of three no longer holds (13.6 × 3 = 40.8, below the hematocrit of 43.5). Each venesection removes about 200 to 250 mg of iron, and the resulting iron deficiency is expected, because it limits red cell production while the marrow keeps making many small cells. Thalassemia trait is unlikely because his MCV was normal at diagnosis. A hematocrit below the 45% target means the disease is controlled, not relapsing.
- A: Thalassemia trait is lifelong, but his MCV was 89 fL at diagnosis.
- C: The calculated MCV is 64 fL and the MCH 20 pg, which are microcytic and hypochromic.
- D: The hematocrit is below the 45% target; the high RBC count reflects many small cells.
Question 2 · Medium
The table shows a CBC run before and after the sample's plasma was replaced with an equal volume of diluent. Which interference best explains the first results?
Figure: MedExams
1. ASpherocytosis
2. BIn vitro hemolysis
3. CCold agglutinins
4. DLipemia
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Answer: D · Lipemia
In the first run the hemoglobin is too high for the RBC count and hematocrit, so MCH (37.8 pg) and MCHC (42.0 g/dL) are far above their reference intervals while the RBC count and MCV are normal. Replacing the plasma brings hemoglobin, MCH and MCHC down to 12.6 g/dL, 29.9 pg and 33.2 g/dL, so the error came from the plasma. Turbid, lipemic plasma adds to the absorbance read in the hemoglobin channel but does not change the cell counts. An MCHC well above 36 g/dL is a signal to look for an interference like this before release.
- A: Spherocytes raise MCHC only slightly, and plasma replacement would not change it.
- B: In vitro hemolysis lowers the RBC count and hematocrit.
- C: Cold agglutinins lower the RBC count and raise the MCV, and warming corrects them.
Question 3 · Hard
A sample from an adult with known iron deficiency gives Hb 7.8 g/dL (78 g/L), Hct 29% (0.29 L/L), MCV 66 fL and MCHC 27 g/dL. The film shows hypochromic microcytes without clumps, and the plasma is clear. The rule of three is not met. What should the technologist do?
1. AReport the results; hypochromia explains the gap
2. BRecollect because the sample has clotted
3. CWarm the sample to 37 °C and rerun it
4. DReplace the plasma with saline and rerun
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Answer: A · Report the results; hypochromia explains the gap
The rule of three assumes a normal MCHC; when red cells carry clearly less hemoglobin than normal, the hematocrit is high relative to the hemoglobin and the rule fails by design. Here the low MCHC matches the hypochromic cells on the film, there are no clumps and the plasma is clear. The results are internally consistent and can be reported. Warming and plasma replacement correct cold agglutinins and lipemia, which raise the MCHC rather than lower it.
- B: A clotted sample gives falsely low counts and visible clots, not a low MCHC that matches the film.
- C: Warming corrects cold agglutinins, which raise the MCHC and show clumps on the film.
- D: Saline replacement corrects lipemia, which falsely raises hemoglobin and MCHC.
Question 4 · Medium
Each pair of results below comes from a sample with normocytic, normochromic cells on the film. Which pair breaks the rule of three and should be checked before release?
1. AHb 12.5 g/dL (125 g/L), Hct 36% (0.36 L/L)
2. BHb 11.0 g/dL (110 g/L), Hct 39% (0.39 L/L)
3. CHb 14.0 g/dL (140 g/L), Hct 42% (0.42 L/L)
4. DHb 15.0 g/dL (150 g/L), Hct 47% (0.47 L/L)
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Answer: B · Hb 11.0 g/dL (110 g/L), Hct 39% (0.39 L/L)
Hemoglobin × 3 predicts the hematocrit within ±3 percentage points. For Hb 11.0 g/dL the expected hematocrit is 30% to 36%, so 39% falls outside, and the implied MCHC of 11.0 ÷ 39 × 100 = 28 g/dL does not fit cells that look normochromic. The other pairs all fall within ±3 of Hb × 3. A failed check prompts review of the specimen, the film and the analyzer before the CBC is reported.
- A: 12.5 × 3 = 37.5, and 36% is within ±3.
- C: 14.0 × 3 = 42, an exact match.
- D: 15.0 × 3 = 45, and 47% is within ±3.
Question 5 · Easy
In a sample with normocytic, normochromic red cells, the rule of three predicts that the hematocrit equals:
1. AMCHC × 3, ± 3%
2. BRBC count × 3, ± 3%
3. CHemoglobin ÷ 3, ± 3%
4. DHemoglobin × 3, ± 3%
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Answer: D · Hemoglobin × 3, ± 3%
For normal-sized, normally filled red cells, the hemoglobin (g/dL) is about three times the RBC count (× 10^6/µL), and the hematocrit (%) is about three times the hemoglobin, within ±3. For example, a hemoglobin of 14.0 g/dL (140 g/L) predicts a hematocrit of 39% to 45%. A result outside this window in a normocytic sample suggests a specimen or analyzer problem and is checked before release. The rule does not hold for clearly hypochromic or otherwise abnormal red cells.
- A: MCHC is itself calculated from hemoglobin and hematocrit; tripling it means nothing.
- B: RBC count × 3 predicts the hemoglobin, not the hematocrit.
- C: The hematocrit is about three times the hemoglobin, not one third of it.
Question 6 · Hard
A CBC gives RBC 4.00 × 10^6/µL (4.00 × 10^12/L), hemoglobin 8.4 g/dL (84 g/L) and hematocrit 28% (0.28 L/L). The red cells are best classified as:
1. ANormocytic, hypochromic
2. BMicrocytic, normochromic
3. CMicrocytic, hypochromic
4. DNormocytic, normochromic
Answer and explanation in LearnFree account
Question 7 · Hard
An analyzer reports a red cell volume standard deviation of 12.6 fL and an MCV of 70 fL for a woman with fatigue. What is the RDW-CV?
1. A5.6%
2. B18.0%
3. C1.8%
4. D12.6%
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Question 8 · Medium
A CBC shows RBC 4.40 × 10^6/µL (4.40 × 10^12/L), hemoglobin 13.2 g/dL (132 g/L) and hematocrit 40% (0.40 L/L). What is the MCH?
1. A33 pg
2. B3.0 pg
3. C30 pg
4. D91 pg
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Question 9 · Medium
In a patient with sickle cell anemia, the spun microhematocrit reads several points higher than the analyzer's calculated hematocrit. The tube was spun and read correctly. The most likely reason is:
1. AMore plasma trapped between the rigid sickled cells
2. BDehydration concentrating the patient's blood
3. CHemolysis of fragile sickled cells during spinning
4. DExcess EDTA shrinking the red cells in the tube
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Question 10 · Hard
A 70-year-old man has RBC 2.20 × 10^6/µL (2.20 × 10^12/L), hemoglobin 8.1 g/dL (81 g/L) and hematocrit 25% (0.25 L/L). His MCH is raised. The best explanation is:
1. ACold agglutinins falsely lowering the RBC count
2. BLipemia falsely raising the hemoglobin value
3. CLarger cells, each holding more hemoglobin
4. DHyperchromic cells with an MCHC above 36 g/dL
Answer and explanation in LearnFree account
Question 11 · Easy
Hematocrit 36% and RBC count 4.0 × 10^12/L. What is the MCV?
1. A111 fL
2. B144 fL
3. C90 fL
4. D9 fL
Answer and explanation in LearnFree account

Hematocrit and red cell indices

The table shows a CBC run before and after the sample's plasma was replaced with an equal volume of diluent. Which interference best explains the first results?

A sample from an adult with known iron deficiency gives Hb 7.8 g/dL (78 g/L), Hct 29% (0.29 L/L), MCV 66 fL and MCHC 27 g/dL. The film shows hypochromic microcytes without clumps, and the plasma is clear. The rule of three is not met. What should the technologist do?

Each pair of results below comes from a sample with normocytic, normochromic cells on the film. Which pair breaks the rule of three and should be checked before release?

In a sample with normocytic, normochromic red cells, the rule of three predicts that the hematocrit equals:

A CBC gives RBC 4.00 × 10^6/µL (4.00 × 10^12/L), hemoglobin 8.4 g/dL (84 g/L) and hematocrit 28% (0.28 L/L). The red cells are best classified as:

An analyzer reports a red cell volume standard deviation of 12.6 fL and an MCV of 70 fL for a woman with fatigue. What is the RDW-CV?

A CBC shows RBC 4.40 × 10^6/µL (4.40 × 10^12/L), hemoglobin 13.2 g/dL (132 g/L) and hematocrit 40% (0.40 L/L). What is the MCH?

In a patient with sickle cell anemia, the spun microhematocrit reads several points higher than the analyzer's calculated hematocrit. The tube was spun and read correctly. The most likely reason is:

A 70-year-old man has RBC 2.20 × 10^6/µL (2.20 × 10^12/L), hemoglobin 8.1 g/dL (81 g/L) and hematocrit 25% (0.25 L/L). His MCH is raised. The best explanation is:

Hematocrit 36% and RBC count 4.0 × 10^12/L. What is the MCV?