Question 1 · Hard
A 61-year-old man with polycythemia vera, awaiting venesection, has a hematocrit of 0.66 L/L (66%). His coagulation screen, from a correctly filled tube holding 0.3 mL of 3.2% citrate for 2.7 mL of blood, gives PT 16.9 s (reference 11.0-13.5 s) and APTT 43 s (25-35 s), although he has never bled abnormally. A repeat sample will be taken by syringe, adding exactly 2.7 mL of blood to an adjusted tube. Using C = 0.00185 × V × (100 − Hct), with V the blood volume in mL and Hct in %, how should the tube be prepared?
- AKeep the full 0.30 mL of citrate unchanged
- BRemove 0.13 mL of citrate, leaving 0.17 mL
- CRemove 0.11 mL of citrate, leaving 0.19 mL
- DRemove 0.17 mL of citrate, leaving 0.13 mL
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Answer: B · Remove 0.13 mL of citrate, leaving 0.17 mL
C = 0.00185 × 2.7 × (100 − 66) = 0.00185 × 2.7 × 34 = 0.17 mL, so 0.30 − 0.17 = 0.13 mL of citrate must be removed. With a hematocrit of 0.66 L/L there is much less plasma for the same amount of citrate, so the excess citrate binds the calcium in the reagents and falsely prolongs the PT and APTT, which explains his first results. CLSI H21 advises this adjustment when the hematocrit is above 0.55 L/L. Removing 0.17 mL confuses the citrate to keep with the citrate to remove, and leaving 0.19 mL comes from using the 3.0 mL total tube volume instead of the 2.7 mL of blood.
- A: Above a hematocrit of 0.55 L/L, CLSI H21 advises reducing the citrate.
- C: This uses the 3.0 mL total tube volume instead of the 2.7 mL of blood.
- D: 0.17 mL is the citrate to keep, not the amount to remove.