Formulas

A spinal fluid that is slightly hazy is briefly examined microscopically. The technologist performing the count decides to make a 1:10 dilution using 30 µL of sample. What volume of diluent should be used?

A simple 1:10 dilution, as this is the case, uses the formula: sample volume/(sample volume + diluent volume) = 1/10 or sample volume x 10 = sample volume + diluent volume 30 µL x 10 = 30 µL + diluent volume diluent volume = 300 µL - 30 µL Diluent volume = 270 µL So, to make a 1:10 dilution, 270 µL of diluent should be added to 30 µL of sample, .

A manual white blood cell count was performed by the hematology technologist. The cell counts for both sides were 152 and 164 respectively. All nine large squares were counted on each side. The dilution for this kit was pre-measured at 1:100. What should the technologist report as the white cell count?

Calculation: Cells Counted (in this case the average of both sides) X dilution factor (in this case 100) / # of squares counted (in this case 9) X area of each square (1 mm2) X 0.1 mm (depth factor) So, in this problem: 158 x 100 / 9 x 1 x 0.1 mm = 17555.55/mm3 (can be converted to 17.5 x 109/L*) *There are 1,000,000 mm3 in a liter (L). So 17555.55 X 1,000,000 = 17.5 x 109/L

Determine the molarity of a solution of HCl that is contained in a Class A 1-L volumetric flask filled to the calibration mark. The content label reads 32 g HCl. The gmw of HCl=36.46 g/mol. 0.88 mol/L

Molarity (M) is often expressed in units of moles per liter (mol/L). Remember that 1 mol of a substance is equal to the gram molecular weight (gmw) of that substance. First, you need to first determine which units are needed for the final answer. In this case, the final units needed are moles per liter (mol/L). Next, you need to look at the units we have, which are grams and liters. HCl may be expressed as moles and grams. The gmw of HCl is calculated to equal 36.46 g/mol. Rearrange the equation so that grams can be canceled out and the remaining units are mol/L. The equation is: (32 g HCl / L) X (1 mol/ 36.46 g HCl) = 0.88 mol/L The correct answer is 0.88 mol/L

Which of the following formulas would you use to calculate absolute cell counts?

The absolute cell count is the percentage of a single leukocyte type multiplied by the total WBC count.

What is the corrected WBC if a differential count shows 20 nucleated erythrocytes per 100 leukocytes and the uncorrected leukocyte count is 12.0 X 109

The formula for correcting a white blood cell count in the presence of more than 5 nucleated red blood cells per 100 white blood cells is: (Uncorrected WBC countx100) /(# NRBCs per 100 WBCs + 100) (12.0 X 109/L X 100) / (20 + 100) = 10.0 X 109/L

Using the formula for creatinine clearance, calculate the clearance based on the values given below: Urine creatinine: 100 mg/dL Plasma creatinine: 1.9 mg/dL Urine volume: 1375 mL/24hrs Surface Area: 1.2 m2 Round to 1 decimal place.

The formula for creatinine clearance (CrCl) is: UV/P x 1.73/SA Where U is the urine creatinine in mg/dL, V is urine volume in mL/min, P is plasma creatinine in mg/dL, and SA is surface area of the patient in m2. Setting up this problem, the first thing that needs to be done is to correct units. CrCl is reported in mL/min and mL/min/m2 if correcting for body surface area. 1375 mL/24 hrs is converted to 0.95 mL/min by dividing by 1440 (hours in a day times minutes in a day) and can then be entered into the formula: (100 mg/dL)(0.95mL/min)/(1.9 mg/dL) x 1.73/1.2 m2 Here, you can see the mg/dL cancel out and the units you are left with are mL/min/m2. Now, you just calculate and you will get 72 mL/min/m2.

Recovery studies will show whether a method is able to accurately measure an analyte. In a recovery experiment, a small aliquot of concentrated analyte is added (spiked) into a patient sample (matrix) and then measured by the method being evaluated. If all the precautions were observed when preparing the samples, what would be the recovery percentage for the following example (to the nearest whole number)? baseline = 7.50 mg/dL; patient sample = 8.35 mg/dL (added analyte = 0.95 mg/dL); recovered = 0.85 mg/dL

The purpose of performing a recovery study is to determine how much of the analyte can be detected (recovered) in the presence of all the other compounds in the matrix (type of patient sample). % recovered = [ (value measured in spiked sample - value measured in baseline) / concentration added ] X 100 For this study, [ (8.35 mg/dL - 7.50 mg/dL) / 0.95 mg/dL ] X 100 = ( 0.85 / 0.95 ) X 100 = 89%

A sample of cerebrospinal fluid is diluted 1:100; the standard 9 squares of a hemocytometer are counted on each side for a total of 18 large squares. Side 1-- 186 nucleated cells counted Side 2-- 184 nucleated cells counted Total nucleated cells = 370 Using the standard hemocytometer formula shown below, what is the nucleated cell count per microliter (µL)? Cells/µL = number of cells counted x dilution / number of larger squares x 0.1

Using the formula on the right, Cells/µL = 370 x 100 / 18 x 0.1 Cells/µL = 37000 / 1.8 Cells/µL = 20556 or 2.06 x 104