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Corrected WBC for nRBCs and absolute counts

10 DOH Abu Dhabi (HAAD) lab licensing practice questions on Corrected WBC for nRBCs and absolute counts, part of Hematology. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Hard

    A 2-day-old baby born at 36 weeks is feeding poorly and has an unstable temperature. The analyzer, which has no nRBC channel, reports WBC 22.0 × 10^9/L. The 100-cell differential shows segmented neutrophils 40%, bands 14%, metamyelocytes 4%, myelocytes 2%, lymphocytes 30% and monocytes 10%, and 10 nRBCs are seen per 100 white cells. The neonatal unit treats an immature-to-total neutrophil (I:T) ratio above 0.2 as suggesting sepsis. Which pair of results is correct?

    1. ACorrected WBC 22.0 × 10^9/L; I:T ratio 0.33
    2. BCorrected WBC 19.8 × 10^9/L; I:T ratio 0.50
    3. CCorrected WBC 20.0 × 10^9/L; I:T ratio 0.33
    4. DCorrected WBC 20.0 × 10^9/L; I:T ratio 0.23
    Show answer

    Answer: C · Corrected WBC 20.0 × 10^9/L; I:T ratio 0.33

    The analyzer has counted the nRBCs as white cells, so the corrected WBC is 22.0 × 100 ÷ (100 + 10) = 20.0 × 10^9/L. The I:T ratio divides all immature neutrophils (bands, metamyelocytes and myelocytes: 14 + 4 + 2 = 20) by all neutrophils, mature and immature (40 + 20 = 60), giving 0.33, above the unit's 0.2 cut-off. Because the ratio uses percentages from the same differential, the nRBC correction does not change it. Using bands alone gives 0.23, dividing the immature cells by the segmented neutrophils alone gives 0.50, and subtracting 10% of the WBC instead of dividing by 1.10 gives 19.8.

    • A: 22.0 is the uncorrected count, which still includes the nRBCs.
    • B: Divide by 1.10 rather than subtract 10%, and divide by all neutrophils.
    • D: 0.23 uses bands alone; metamyelocytes and myelocytes are also immature.
  2. Question 2 · Hard

    An analyzer with a dedicated nRBC channel reports WBC 9.0 × 10^9/L, with nRBCs already excluded, and 20 nRBCs per 100 WBCs. A technologist applies the manual nRBC correction and reports 7.5 × 10^9/L. The reported value is:

    1. AFalsely high, because nRBCs are still in the WBC
    2. BCorrect, because 20 nRBCs per 100 WBCs need correcting
    3. CCorrect, provided the nRBCs were counted on the film
    4. DFalsely low, because the count was corrected twice
    Show answer

    Answer: D · Falsely low, because the count was corrected twice

    Analyzers with an nRBC channel identify nRBCs separately and remove them from the WBC before reporting it. Applying 100/(100 + 20) again lowers the true count of 9.0 to 7.5 × 10^9/L, an error of about 17%. Each laboratory must know which of its analyzers already report a corrected WBC. The manual formula is used only for counts that still include the nRBCs.

    • A: The nRBC channel had already removed the nRBCs from this WBC.
    • B: Correction is needed only when the WBC still includes the nRBCs; here it did not.
    • C: Where the nRBCs were counted does not matter; the WBC was already corrected.
  3. Question 3 · Hard

    A splenectomized 12-year-old with beta-thalassemia major has an uncorrected WBC of 24.0 × 10^9/L. The differential shows 40% neutrophils, and 60 nRBCs were seen per 100 WBCs. What is the absolute neutrophil count?

    1. A6.0 × 10^9/L
    2. B9.6 × 10^9/L
    3. C16.0 × 10^9/L
    4. D3.8 × 10^9/L
    Show answer

    Answer: A · 6.0 × 10^9/L

    First correct the WBC: 24.0 × 100 ÷ (100 + 60) = 15.0 × 10^9/L. Then ANC = 15.0 × 0.40 = 6.0 × 10^9/L. Using the uncorrected count would overstate the neutrophils by 60%. Splenectomized thalassemia patients often have very high nRBC counts, so this correction matters.

    • B: This uses the uncorrected WBC, which still includes the nRBCs.
    • C: Dividing by 60 instead of 160 inflates the count.
    • D: Removing 60% of the count overcorrects; there are 60 nRBCs per 100 WBCs, not 60% of all cells.
  4. Question 4 · Easy

    When nucleated red cells are counted per 100 white cells on the differential, the corrected WBC count is:

    1. AUncorrected WBC × nRBCs ÷ 100
    2. BUncorrected WBC − (nRBCs ÷ 100)
    3. CUncorrected WBC × (100 + nRBCs) ÷ 100
    4. DUncorrected WBC × 100 ÷ (100 + nRBCs)
    Show answer

    Answer: D · Uncorrected WBC × 100 ÷ (100 + nRBCs)

    Analyzers without an nRBC channel count nucleated red cells as white cells, so the WBC is too high. If n nRBCs are seen while 100 white cells are counted, only 100 of every 100 + n nucleated cells are true white cells. Multiplying the uncorrected count by 100/(100 + n) removes the nRBC share. The nRBC figure must be per 100 WBCs, not per 100 nucleated cells.

    • A: This gives a figure related to the nRBCs, not the true white cell count.
    • B: Subtracting n/100 mixes a ratio with a cell count and barely changes the result.
    • C: Multiplying by (100 + n)/100 raises the count instead of lowering it.
  5. Question 5 · Hard

    A clinic gives a chemotherapy dose only if the ANC (segmented plus band neutrophils) is at least 1.5 × 10^9/L. Today's WBC is 2.4 × 10^9/L: segmented neutrophils 52%, bands 8%, lymphocytes 30%, monocytes 10%. Which result and decision are correct?

    1. AANC 1.44 × 10^9/L; hold the dose
    2. BANC 1.68 × 10^9/L; give the dose
    3. CANC 1.25 × 10^9/L; hold the dose
    4. DANC 14.4 × 10^9/L; give the dose
    Show answer

    Answer: A · ANC 1.44 × 10^9/L; hold the dose

    ANC = WBC × (segmented + band neutrophils) = 2.4 × (0.52 + 0.08) = 2.4 × 0.60 = 1.44 × 10^9/L. This is just below 1.5, so the protocol says to hold the dose. Leaving out the bands gives 1.25, and adding the monocytes gives 1.68, which would wrongly allow treatment. Near a cut-off, the laboratory reports the ANC exactly and does not round it up.

    • B: Monocytes are not neutrophils and do not belong in the ANC.
    • C: This counts segmented neutrophils only; the protocol includes bands.
    • D: This is a decimal error: 60% of 2.4 is 1.44, not 14.4.
  6. Question 6 · Medium

    The uncorrected WBC is 15.6 × 10^9/L and the differential shows 30 nRBCs per 100 WBCs. What is the absolute nRBC count?

    1. A0.36 × 10^9/L
    2. B3.6 × 10^9/L
    3. C4.7 × 10^9/L
    4. D12.0 × 10^9/L

    Answer and explanation in LearnFree account

  7. Question 7 · Medium

    An analyzer that counts nRBCs as white cells reports WBC 22.4 × 10^9/L. The differential shows 12 nRBCs per 100 WBCs. What is the corrected WBC?

    1. A2.7 × 10^9/L
    2. B25.1 × 10^9/L
    3. C20.0 × 10^9/L
    4. D19.7 × 10^9/L

    Answer and explanation in LearnFree account

  8. Question 8 · Hard

    A patient with primary myelofibrosis has a leukoerythroblastic film. A 200-cell differential finds 50 nRBCs, and the analyzer WBC, which includes nRBCs, is 30.0 × 10^9/L. What is the corrected WBC?

    1. A37.5 × 10^9/L
    2. B24.0 × 10^9/L
    3. C20.0 × 10^9/L
    4. D22.5 × 10^9/L

    Answer and explanation in LearnFree account

  9. Question 9 · Medium

    An adult's WBC is 2.5 × 10^9/L with 30% neutrophils and 60% lymphocytes (reference: neutrophils 2.0-7.5, lymphocytes 1.0-4.0 × 10^9/L). What do these results show?

    1. AAbsolute neutropenia; lymphocytes normal
    2. BNormal absolute counts for both cells
    3. CAbsolute lymphocytosis; neutrophils normal
    4. DAbsolute neutropenia and lymphocytosis

    Answer and explanation in LearnFree account

  10. Question 10 · Medium

    On day 1 a patient's WBC is 12.0 × 10^9/L with 25% lymphocytes. On day 3 the WBC is 6.0 × 10^9/L with 50% lymphocytes. What happened to the absolute lymphocyte count?

    1. AIt stayed at 3.0 × 10^9/L
    2. BIt rose from 1.5 to 3.0 × 10^9/L
    3. CIt halved from 6.0 to 3.0 × 10^9/L
    4. DIt doubled from 3.0 to 6.0 × 10^9/L

    Answer and explanation in LearnFree account

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