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Physiology · Blood: reticulocyte correction
Sample itemA patient has a haematocrit of 20% and a reticulocyte count of 6%. Taking the normal haematocrit as 45%, what is the corrected reticulocyte percentage (before any maturation-time adjustment)?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
A. 2.7%
Corrected reticulocyte count = observed reticulocyte percentage multiplied by patient haematocrit divided by normal haematocrit. Here 6 x 20/45 = 2.67, which rounds to 2.7%. The correction is needed because few red cells in the denominator inflate the raw percentage.
Option by option
A. 2.7% (correct)
Correct. 6 x 20/45 = 2.67%.
B. 6.0%
The raw percentage is uncorrected and overstates the marrow output in an anaemic patient.
C. 1.5%
The arithmetic does not produce this value; 20/45 is about 0.44, not 0.25.
D. 13.5%
This results from inverting the fraction (6 x 45/20), which wrongly inflates the count.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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