MLT GlobeLaboratory Operations: Lab math – page 2
48 Laboratory Operations MCQs on Lab math with answers and explanations.
A serum creatinine is 1.0 mg/dL. Creatinine has a molecular weight of 113 g/mol. The SI result is approximately:
1.0 mg/dL = 10 mg/L; 10 ÷ 113 = 0.0884 mmol/L = 88.4 µmol/L. 8.8 µmol/L misses the dL-to-L factor of 10.
A total calcium is 10 mg/dL. Calcium has an atomic weight of 40 and a valence of 2. What is the result in mEq/L?
10 mg/dL = 100 mg/L; 100/40 = 2.5 mmol/L. mEq/L = mmol/L × valence = 2.5 × 2 = 5.0 mEq/L. 2.5 is the mmol/L value.
A high sample is diluted by mixing 1 part serum with 4 parts saline. The diluted sample reads 60 U/L. The result to report is:
1 part + 4 parts = 5 total parts, a 1:5 dilution. 60 × 5 = 300 U/L. Multiplying by 4 treats the ratio as the dilution.
How many grams of copper sulfate pentahydrate (CuSO4·5H2O, MW 249.7 g/mol) are needed to make 1 L of 0.1 mol/L CuSO4?
Grams = M × MW × L = 0.1 × 249.7 × 1 = 24.97 g. Using the anhydrous weight (159.6) would give 15.96 g, which is wrong for the hydrate.
How much 95% ethanol is needed to prepare 500 mL of 70% ethanol?
V1 = (70 × 500)/95 = 368 mL, made up to 500 mL with water. 350 mL is 70% of 500 and ignores that the stock is only 95%.
A 1 mol/L sodium hydroxide solution (MW 40 g/mol) expressed as a percent (w/v) solution is:
1 mol/L = 40 g/L = 4 g/100 mL = 4% w/v. 40% would require 40 g per 100 mL.
How many grams of H2SO4 (MW 98 g/mol) are required to prepare 1 L of a 1 N solution?
Equivalent weight = MW/valence = 98/2 = 49 g. For 1 N in 1 L, 49 g is needed. 98 g would make a 1 mol/L (2 N) solution.
A 24-hour urine volume is 1440 mL with a urine creatinine of 100 mg/dL. Plasma creatinine is 1.0 mg/dL. The uncorrected creatinine clearance is:
Clearance = (U × V)/P. V = 1440 mL/1440 min = 1.0 mL/min. (100 × 1.0)/1.0 = 100 mL/min. Using the total volume without converting to per minute is the common error.
A technologist must prepare 5.0 mL total of a 1:25 dilution of serum in saline. Which volumes are correct?
Sample volume = total volume ÷ dilution factor = 5.0/25 = 0.2 mL; add 4.8 mL saline to reach 5.0 mL. Adding 5.0 mL saline would give 1:26.
A blood urea nitrogen (BUN) result is 28 mg/dL. Each urea molecule contains 2 nitrogen atoms (28 g of nitrogen per mole of urea). What is the result as urea in mmol/L?
28 mg/dL = 280 mg/L; each mole of urea has 28 g of nitrogen, so 280 ÷ 28 = 10 mmol/L. Dividing by 56 would give 5 mmol/L in error.
A solution has an absorbance of 0.50. What is its percent transmittance (%T)?
A = 2 − log %T, so log %T = 1.5 and %T = 10^1.5 ≈ 31.6%. Absorbance and %T are logarithmically, not linearly, related, so 50% is wrong.
Serum results: sodium 140 mmol/L, glucose 180 mg/dL, BUN 28 mg/dL. Using 2 × Na + glucose/18 + BUN/2.8, the calculated osmolality is:
2 × 140 = 280; 180 ÷ 18 = 10; 28 ÷ 2.8 = 10; total = 300 mOsm/kg. Omitting either glucose or BUN term gives 290.
A fasting lipid profile shows total cholesterol 200 mg/dL, HDL cholesterol 50 mg/dL and triglycerides 150 mg/dL. Using the Friedewald equation, LDL cholesterol is:
LDL = TC − HDL − TG/5 = 200 − 50 − 30 = 120 mg/dL. Forgetting the VLDL term (TG/5) gives 150 mg/dL.
A serial dilution is made in which each tube is a 1:3 dilution of the previous tube. Tube 1 is a 1:3 dilution of serum. What is the dilution in tube 4?
Serial dilutions multiply: 1:3 × 3 × 3 × 3 = 1:81 in tube 4. Adding (3 × 4 = 12) instead of multiplying is a common error; 1:27 is tube 3.
A serum potassium is 4.0 mEq/L. Potassium has an atomic weight of 39 and a valence of 1. What is the result in mg/dL?
For a monovalent ion, 4.0 mEq/L = 4.0 mmol/L × 39 mg/mmol = 156 mg/L = 15.6 mg/dL. Reporting mg/L as mg/dL gives 156.
An automated WBC count is 12.0 × 10⁹/L. The differential shows 20 nucleated red blood cells per 100 WBCs. The corrected WBC count is:
Corrected WBC = observed WBC × 100 ÷ (100 + nRBC) = 12.0 × 100/120 = 10.0 × 10⁹/L. Subtracting 20% (9.6) is incorrect.
A patient has a reticulocyte count of 6.0% and a haematocrit of 20%. Using a normal haematocrit of 45%, the corrected reticulocyte count is:
Corrected retic = 6.0% × (20/45) = 2.67%, about 2.7%. Inverting the ratio (45/20) gives 13.5%.
9 g of glucose (molecular weight 180 g/mol) is dissolved in water to a final volume of 500 mL. What is the molarity?
9 ÷ 180 = 0.05 mol, and 0.05 mol ÷ 0.5 L = 0.1 mol/L.
What volume of a 1 mol/L stock solution is needed to prepare 100 mL of a 0.1 mol/L solution?
Using C1V1 = C2V2: 1 × V1 = 0.1 × 100, so V1 = 10 mL, made up to 100 mL.
A 1:20 dilution of blood is made using 0.05 mL of blood. How much diluent must be added?
Total volume = 0.05 × 20 = 1.0 mL, so diluent = 1.0 − 0.05 = 0.95 mL.