13.4 Laboratory Mathematics: Dilutions, Molarity, Normality, Standard Curves & Diagnostic Statistics

13.4 Laboratory Mathematics: Dilutions, Molarity, Normality, Standard Curves & Diagnostic Statistics

[!NOTE] Laboratory Mathematics is a named sub-area of the Laboratory Operations domain (V.C in the BOC content outline), covering concentration/volume/dilutions, molarity and normality, standard curves, mean/median/mode/confidence intervals, and sensitivity/specificity/predictive value. Separately, the content guideline lists seven calculations examinees are expected to know outright. Arithmetic errors are the cheapest points to lose on this examination and the easiest to prevent.

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The BOC's Explicit Calculation List

Beyond that list, the anion gap, osmolality and the osmolal gap, corrected calcium, the CK-MB relative index, the delta ratio, and the coefficient of variation all appear routinely in scenario items, and each is worked through in its own chapter.

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Concentration, Volume, and Dilutions

The Convention That Trips People Up

In clinical laboratory usage, a 1:10 dilution means one part specimen diluted to a total of ten parts — one part specimen plus nine parts diluent. The dilution factor is 10, and the diluted result must be multiplied by 10 to recover the original concentration.

$\text{Dilution Factor} = \frac{\text{Total Volume}}{\text{Volume of Specimen}}$

$\text{Reported Result} = \text{Measured (Diluted) Result} \times \text{Dilution Factor}$

Worked example. A serum glucose exceeds the analytical measurement range. The technologist prepares a 1:5 dilution by adding 0.10 mL of serum to 0.40 mL of saline (total 0.50 mL) and the analyzer reads 340 mg/dL.

$\text{Reported Glucose} = 340 \times 5 = 1{,}700\ \text{mg/dL}$

The result must be released with a comment documenting the dilution, and the concentration must fall inside the clinically reportable range established for that assay (Section 13.2).

Serial and Compound Dilutions

A serial dilution repeats the same dilution step through a row of tubes. Four tubes each carrying a 1:2 dilution give final dilutions of 1:2, 1:4, 1:8, and 1:16. The general form is:

$\text{Final Dilution} = (\text{Dilution per tube})^{n}$

A compound dilution multiplies unlike steps: a 1:10 dilution followed by a 1:50 dilution of that product yields a final dilution of 1:500 and a dilution factor of 500.

Preparing Solutions from a Concentrated Stock

$C_1 V_1 = C_2 V_2$

Worked example. Prepare 500 mL of 0.100 mol/L hydrochloric acid from a 2.00 mol/L stock:

$V_1 = \frac{C_2 V_2}{C_1} = \frac{0.100 \times 500}{2.00} = 25.0\ \text{mL}$

Measure 25.0 mL of stock into a volumetric flask and bring to a final volume of 500 mL — quantity sufficient to volume, never "add 500 mL of water."

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Molarity, Normality, and Percent Solutions

$N = M \times \text{valence}$

Converting Conventional Units to SI Units

Because the examination presents results in both conventional and SI units, the conversion between mg/dL and mmol/L must be automatic. Multiply by 10 to move from deciliters to liters, then divide by the molecular weight:

$\text{mmol/L} = \frac{\text{mg/dL} \times 10}{\text{Molecular Weight}}$

Worked example. A total calcium of 10.0 mg/dL (atomic weight 40.08):

$\frac{10.0 \times 10}{40.08} = 2.5\ \text{mmol/L}$

which is exactly why the BOC's SI calcium interval (2.2 to 2.6 mmol/L) corresponds to the conventional 8.6 to 10.2 mg/dL.

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Standard Curves and Beer's Law

The Law Itself

$A = \varepsilon b c$

where $A$ is absorbance, $\varepsilon$ is the molar absorptivity (L per mol per cm), $b$ is the light path in centimeters, and $c$ is concentration in mol/L. Absorbance and transmittance are related logarithmically:

$A = 2 - \log_{10}(\%T)$

Worked example. A solution transmits 25% of the incident light:

$A = 2 - \log_{10}(25) = 2 - 1.398 = 0.602$

Calculating an Unknown Against a Single Standard

When the calibration is linear and passes through the origin, concentration is directly proportional to absorbance:

$C_{\text{unknown}} = \frac{A_{\text{unknown}}}{A_{\text{standard}}} \times C_{\text{standard}}$

Worked example. A 200 mg/dL standard reads an absorbance of 0.400; the patient specimen reads 0.260:

$C = \frac{0.260}{0.400} \times 200 = 130\ \text{mg/dL}$

Enzyme Activity from a Rate of Absorbance Change

Coupled NAD(H) enzyme assays are read at 340 nm, where NADH has a molar absorptivity of 6,220 L per mol per cm. Activity in international units per liter (micromoles of substrate converted per minute per liter) is:

$\text{U/L} = \frac{\Delta A/\text{min}}{\varepsilon \times b} \times \frac{TV}{SV} \times 10^{6}$

where $TV$ is the total reaction volume and $SV$ is the sample volume.

Worked example. An ALT reaction gives $\Delta A/\text{min}$ = 0.010 with a 1.00 cm path, a total reaction volume of 1.00 mL, and a sample volume of 0.020 mL:

$\text{U/L} = \frac{0.010}{6220 \times 1.00} \times \frac{1.00}{0.020} \times 10^{6} = (1.608 \times 10^{-6}) \times 50 \times 10^{6} \approx 80.4\ \text{U/L}$

When the Curve Is Not a Straight Line

A multi-point calibration curve is required whenever the relationship deviates from linearity — competitive immunoassays are typically fitted with a logit-log or four-parameter logistic model, and enzyme-immunoassay curves flatten at both extremes. A single-standard calculation applied to a nonlinear assay is a systematic error, not a shortcut. Stray light (Section 12.1) also produces negative deviation from Beer's law at high absorbance, imposing a ceiling on the usable curve.

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Descriptive Statistics

$\text{CV}\% = \frac{\text{SD}}{\bar{x}} \times 100$

Worked example. Five quality-control glucose values: 88, 94, 98, 98, 122 mg/dL. The sum is 500, so the mean is 100 mg/dL; ranked, the middle value gives a median of 98 mg/dL; the most frequent value gives a mode of 98 mg/dL. The mean sits above both because the single 122 mg/dL outlier pulls it upward — this is exactly why the median is preferred for skewed analyte distributions.

Confidence Interval versus Reference Interval

These two are constantly confused, and the distinction is examinable.

$\text{SEM} = \frac{\text{SD}}{\sqrt{n}} \qquad \text{95\% CI of the mean} = \bar{x} \pm 1.96 \times \text{SEM}$

• A confidence interval describes the precision with which a parameter (usually the mean) has been estimated. It narrows as $n$ increases.
• A reference interval describes the spread of individual values in a healthy population — the central 95%, estimated as the mean plus or minus 1.96 SD for Gaussian data, or as the 2.5th to 97.5th percentiles for non-Gaussian data (Section 1.2). It does not narrow as $n$ increases.

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Diagnostic Performance: Sensitivity, Specificity, and Predictive Value

Every diagnostic-performance calculation comes from one 2 x 2 table.

$\text{Sensitivity} = \frac{TP}{TP + FN} \times 100 \qquad \text{Specificity} = \frac{TN}{TN + FP} \times 100$

$\text{PPV} = \frac{TP}{TP + FP} \times 100 \qquad \text{NPV} = \frac{TN}{TN + FN} \times 100$

Read the denominators, because that is the whole distinction: sensitivity and specificity divide by the disease columns and are therefore intrinsic properties of the assay, independent of how common the disease is. Predictive values divide by the test-result rows and therefore change with prevalence.

Worked Example: Prevalence Drives Predictive Value

A cardiac biomarker has 95% sensitivity and 90% specificity.

Chest-pain unit, disease prevalence 10%, n = 1,000:

• Diseased = 100: TP = 95, FN = 5
• Non-diseased = 900: FP = 90, TN = 810

$\text{PPV} = \frac{95}{95 + 90} \times 100 = 51.4\% \qquad \text{NPV} = \frac{810}{810 + 5} \times 100 = 99.4\%$

Asymptomatic screening population, disease prevalence 1%, n = 10,000:

• Diseased = 100: TP = 95, FN = 5
• Non-diseased = 9,900: FP = 990, TN = 8,910

$\text{PPV} = \frac{95}{95 + 990} \times 100 = 8.8\%$

The assay did not change; only the population did. At 1% prevalence roughly eleven out of twelve positive results are false. This is the quantitative argument against indiscriminate screening with a good-but-imperfect test, and it is why a "highly sensitive" test can still generate mostly false positives.

• A highly sensitive test with few false negatives is best for ruling out disease when negative.
• A highly specific test with few false positives is best for confirming disease when positive — the logic behind screening drugs of abuse by immunoassay and confirming by GC-MS or LC-MS/MS (Section 11.3).
• Efficiency is the overall proportion correctly classified: (TP + TN) divided by the total.

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Clearance and Timed-Urine Calculations

Any timed collection converts a concentration into a quantity excreted per unit time. Watch the units: 1 dL equals 100 mL.

$\text{Analyte Excreted per 24 h} = \text{Urine Concentration} \times \text{Total Volume}$

Worked example. A 24-hour urine of 2,400 mL contains protein at 24 mg/dL. Because 2,400 mL equals 24 dL:

$24\ \text{mg/dL} \times 24\ \text{dL} = 576\ \text{mg/24 h}$

Creatinine clearance (Section 5.3) is the same idea normalized to plasma concentration and expressed per minute:

$\text{CrCl (mL/min)} = \frac{U_{\text{cr}} \times V}{P_{\text{cr}} \times 1440} \times \frac{1.73}{\text{BSA}}$

Worked example. Urine creatinine 110 mg/dL, 24-hour volume 1,440 mL, serum creatinine 1.1 mg/dL:

$\text{CrCl} = \frac{110 \times 1440}{1.1 \times 1440} = \frac{110}{1.1} = 100\ \text{mL/min}$

Body-surface-area normalization for a patient with a BSA of 2.00 square meters:

$100 \times \frac{1.73}{2.00} = 86.5\ \text{mL/min per 1.73 m}^2$

The dominant error in creatinine clearance is not arithmetic — it is an incomplete or over-collected timed specimen. Total urine creatinine excretion (roughly 15 to 25 mg/kg per 24 hours in men and 10 to 20 mg/kg per 24 hours in women, and stable for a given individual) is used to judge whether the collection was adequate before the clearance is reported at all.