13.1 Percent Solutions, Hydration Calculations & Dilutions

13.1 Percent Solutions, Hydration Calculations & Dilutions

Quick Summary: Quantitative precision in reagent preparation is a fundamental prerequisite for diagnostic reproducibility in the histopathology laboratory. Minor mathematical errors in calculating solute mass, compensating for crystalline waters of hydration, adjusting for certified dye purity, or performing serial antibody titrations directly cause diagnostic staining failures, excessive non-specific background, or irreversible tissue destruction. The American Society for Clinical Pathology (ASCP) HTL examination evaluates mastery across four foundational mathematical domains: percent solutions (weight-in-volume and volume-in-volume), biological dye purity corrections, hydration state conversions (correcting for crystal lattice water), and dilution stoichiometry ($C_1 V_1 = C_2 V_2$ and serial antibody titration). Technologists must master both the theoretical chemical principles and practical laboratory execution of these protocols.

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1. The Metric System and SI Prefixes

The ASCP BOC content outline lists the metric system as the first item under Laboratory Mathematics, because every downstream calculation depends on carrying units correctly.

Conversions that appear on the examination

• 1 cm = 10 mm = 10,000 micrometers = 10,000,000 nm
• 1 micrometer = 1,000 nm, so a 5 micrometer paraffin section is 5,000 nm and a 70 nm ultrathin section is 0.07 micrometers
• 1 L = 1,000 mL, and for dilute aqueous solutions 1 mL of water = 1 g, which is why weight-in-volume percent works out cleanly
• 1 g = 1,000 mg = 1,000,000 micrograms
• Temperature conversions: degrees C = (degrees F - 32) x 5/9, so a 60 degrees C slide oven is 140 degrees F

[!IMPORTANT] Carry the units through every step of a calculation. A dilution problem worked in milliliters and answered in liters is the single most common arithmetic error on laboratory mathematics questions, and the numeric answer will usually still appear among the distractors.

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2. Mathematical Foundations & Volumetric Principles in Histotechnology

Reagent formulation in the clinical histology laboratory requires a rigorous understanding of metric units, dimensional analysis, and volumetric equipment accuracy. Solid reagents are measured by mass (grams, milligrams, or micrograms) using calibrated analytical balances, whereas liquid solvents and solutes are measured by volume (liters, milliliters, or microliters).

VOLUMETRIC GLASSWARE ACCURACY HIERARCHY:

[ Volumetric Flask (Class A) ] ──> HIGHEST ACCURACY (±0.05% to ±0.15%)
          │                        Calibrated To Contain (TC) at 20°C
          │                        Mandatory for molarity, normality, standard buffers
          ▼
[ Graduated Cylinder ]         ──> MODERATE ACCURACY (±0.5% to ±1.0%)
          │                        Calibrated To Deliver (TD) or To Contain (TC)
          │                        Used for general reagents, wash solutions, solvent mixtures
          ▼
[ Beaker / Erlenmeyer Flask ]  ──> LOW ACCURACY (±5.0%)
                                   Strictly for initial dissolution and mixing
                                   NEVER used for final volume calibration!

The "Quantity Sufficient" (q.s.) Principle

When preparing any liquid reagent containing a dissolved solid solute, the technologist must never measure the full target volume of solvent first and subsequently add the dry chemical. Solid solutes occupy physical volume upon dissolution. Adding 10 g of a salt to 100 mL of water results in a final volume significantly greater than 100 mL, thereby diluting the final reagent below its intended target concentration.

$\text{Correct Workflow:} \quad \text{Weigh Solute} \longrightarrow \text{Dissolve in } 60\% \text{ to } 80\% \text{ Total Solvent} \longrightarrow \text{Dilute q.s. to Meniscus in Volumetric Container}$

1. Dissolution: The dry solute is weighed on an analytical balance and transferred to a beaker containing approximately 60% to 80% of the total required solvent (typically deionized or distilled water).
2. Solvation: The mixture is stirred with a magnetic stir bar until all crystals have dissolved completely into solution.
3. Volumetric Transfer: The dissolved solution is quantitatively transferred into a calibrated volumetric flask or graduated cylinder, using a small stream of solvent to rinse residual solute from the beaker.
4. Quantity Sufficient (q.s.): Additional solvent is added slowly until the bottom of the liquid meniscus aligns precisely with the calibration ring at eye level.
5. Homogenization: The flask is stoppered and inverted 10 to 15 times to ensure complete homogeneity.

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3. Weight-in-Volume Percent Solutions (% w/v)

A weight-in-volume percent solution (% w/v) is defined as the number of grams of dry solid solute dissolved in a total volume of 100 mL of final solution. In clinical laboratory science, percent means "parts per hundred":

$1\% \text{ (w/v)} = \frac{1.0\text{ gram solute}}{100\text{ mL total solution}}$

General Formula for Weight-in-Volume Calculations

$\text{Mass of Solute (g)} = \left( \frac{\text{Desired Percentage}}{100} \right) \times \text{Total Volume Desired (mL)}$

$\text{Mass of Solute (g)} = \text{Desired Decimal Concentration} \times \text{Total Volume Desired (mL)}$

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Worked Calculation 1: Preparing 250 mL of 3% Periodic Acid

Periodic acid ($H_5IO_6$) is the critical primary oxidizing agent in the Periodic Acid-Schiff (PAS) stain. It selectively cleaves carbon-carbon bonds between adjacent 1,2-glycol groups to generate dialdehydes that react with Schiff reagent.

• Problem Statement: Calculate the mass of crystalline periodic acid required to prepare 250 mL of a 3.0% (w/v) aqueous working solution.
• Step 1: Identify Known Variables:
  • Desired Concentration = $3.0%$
  • Total Volume = $250\text{ mL}$
• Step 2: Apply the Formula: $\text{Mass (g)} = \left( \frac{3.0}{100} \right) \times 250\text{ mL} = 0.030 \times 250\text{ mL} = 7.50\text{ g}$
• Step 3: Laboratory Execution:
  1. Weigh exactly 7.50 g of ACS-grade crystalline periodic acid on a calibrated balance.
  2. Dissolve the crystals in approximately 175 mL of deionized water in a clean glass beaker.
  3. Quantitatively transfer the dissolved solution to a 250 mL volumetric flask.
  4. Add deionized water until the bottom of the meniscus touches the 250 mL calibration line.
  5. Stopper and invert to mix. Store in an amber glass bottle at 4°C to protect from light-induced degradation.

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Worked Calculation 2: Preparing 500 mL of 0.9% Physiological Saline

Normal physiological saline (0.9% NaCl w/v) is used extensively in frozen section tissue preparation, fresh specimen transport, and immunohistochemistry rinse protocols to maintain osmotic equilibrium.

• Problem Statement: Determine the mass of sodium chloride ($NaCl$) required to formulate 500 mL of 0.9% (w/v) physiological saline.
• Step 1: Identify Known Variables:
  • Desired Concentration = $0.9%$
  • Total Volume = $500\text{ mL}$
• Step 2: Apply the Formula: $\text{Mass (g)} = \left( \frac{0.9}{100} \right) \times 500\text{ mL} = 0.009 \times 500\text{ mL} = 4.50\text{ g}$
• Step 3: Dimensional Analysis Verification: $500\text{ mL} \times \left( \frac{0.9\text{ g } NaCl}{100\text{ mL solution}} \right) = 4.50\text{ g } NaCl$
• Step 4: Laboratory Execution: Weigh 4.50 g of high-purity $NaCl$, dissolve in ~350 mL of deionized water, and dilute q.s. to exactly 500 mL in a volumetric cylinder.

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4. Volume-in-Volume Percent Solutions (% v/v)

A volume-in-volume percent solution (% v/v) is defined as the volume of liquid solute (in mL) present per 100 mL of final solution. It is utilized when mixing two miscible liquids, such as concentrated liquid acids, commercial formalin, alcohols, or organic clearing solvents.

$\text{Volume of Liquid Solute (mL)} = \left( \frac{\text{Desired Percentage}}{100} \right) \times \text{Total Volume Desired (mL)}$

The Histological Formalin Nomenclature Convention

A critical trap on the ASCP HTL examination centers on the clinical definition of formalin versus formaldehyde:

FORMALIN VS FORMALDEHYDE TERMINOLOGY:

[ Pure Saturated Stock ] ──> 37% to 40% Formaldehyde Gas Dissolved in Water
                             CLINICALLY DESIGNATED AS: "100% Formalin"
                                       │
                                       ▼
[ Routine Fixative ]     ──> 10% Aqueous Dilution of Stock
                             CLINICALLY DESIGNATED AS: "10% Formalin"
                             ACTUAL CHEMICAL CONCENTRATION: ~3.7% to 4.0% Formaldehyde Gas

• Pure Formaldehyde: At room temperature, formaldehyde ($HCHO$) is a pungent gas. Water can dissolve a maximum of 37% to 40% formaldehyde gas by weight at room temperature.
• 100% Formalin: By long-standing histological convention, this commercial 37% to 40% saturated aqueous formaldehyde solution is designated as 100% formalin.
• 10% Formalin: Standard routine histological fixative is formulated by diluting 10 volumes of commercial stock formalin with 90 volumes of diluent (water or phosphate buffer). Therefore, a 10% formalin solution contains 3.7% to 4.0% pure formaldehyde gas.

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Worked Calculation 3: Preparing 1.0 Liter of 10% Formalin

• Problem Statement: Calculate the volume of concentrated 37%–40% stock formalin required to formulate 1,000 mL of 10% unbuffered formalin.
• Step 1: Identify Known Variables:
  • Desired Concentration = $10%$
  • Total Volume = $1,000\text{ mL}$
• Step 2: Apply the Formula: $\text{Volume of Stock Formalin (mL)} = \left( \frac{10}{100} \right) \times 1000\text{ mL} = 100\text{ mL}$
• Step 3: Diluent Calculation: $\text{Volume of Diluent (Water)} = 1000\text{ mL} - 100\text{ mL} = 900\text{ mL}$
• Step 4: Execution: Measure 100 mL of 37%–40% stock formalin in a graduated cylinder under a certified chemical fume hood. Add to 900 mL of distilled or deionized water in a dedicated chemical storage carboy.

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Worked Calculation 4: Preparing 70% Reagent Alcohol from Stock Alcohols

Histology processing and staining sequences rely on graded alcohol series (70%, 80%, 95%, 100%) for gentle dehydration and controlled rehydration.

Case A: Preparing 70% Alcohol from 95% Commercial Stock Ethanol

Commercial laboratory ethanol is routinely supplied as an azeotropic mixture of 95% ethanol and 5% water because simple distillation cannot exceed 95.6% purity.

• Conservation of Mass Equation: $C_1 \times V_1 = C_2 \times V_2$
  • $C_1 = 95%$ (Stock concentration)
  • $V_1 = \text{Unknown volume of stock needed}$
  • $C_2 = 70%$ (Target working concentration)
  • $V_2 = 1,000\text{ mL}$ (Target working volume)
• Solve for $V_1$: $95 \times V_1 = 70 \times 1000$ $V_1 = \frac{70,000}{95} = 736.84\text{ mL} \approx 736.8\text{ mL}$
• Execution: Measure 736.8 mL of 95% stock ethanol. Pour into a 1,000 mL graduated cylinder. Add deionized water until the liquid volume reaches exactly 1,000 mL.

Case B: Preparing 70% Alcohol from 100% Absolute Ethanol

• Equation: $100 \times V_1 = 70 \times 1000$
• Solve for $V_1$: $V_1 = \frac{70,000}{100} = 700.0\text{ mL}$
• Execution: Measure 700.0 mL of absolute ethanol. Dilute q.s. with deionized water to 1,000 mL.

[!NOTE] Volume Contraction Phenomenon: Mixing alcohol and water induces hydrogen bonding between hydroxyl groups and water dipoles, causing thermodynamic volume contraction (~3% volume reduction). Therefore, technologists must never simply mix 700 mL ethanol + 300 mL water; they must add water until the mixture reaches the 1,000 mL calibration line.

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5. Biological Stain Commission (BSC) Certification & Dye Content Purity Corrections

Commercial powdered biological dyes are rarely 100% pure chemical chromophores. During industrial organic synthesis, raw dyes crystallize with inorganic salts (such as sodium chloride, $NaCl$, or sodium sulfate, $Na_2SO_4$), dextrin fillers, residual moisture, and positional isomers. To ensure reproducible staining in diagnostic pathology, biological dyes are tested and certified by the Biological Stain Commission (BSC).

Every bottle of BSC-certified dye carries a certification label stating the specific dye content percentage (or dye purity assay) for that manufacturer lot (e.g., "Total Dye Content: 80%").

THE DYE CONTENT CORRECTION PRINCIPLE:

Commercial Dye Powder = [ Active Colored Chromophore ] + [ Inert Inorganic Salts (NaCl, Na2SO4) ]

If a protocol specifies 2.0 g of pure dye, and the powder contains only 80% active dye:
Weighing 2.0 g of powder delivers only 1.6 g of active dye -> 20% DEFICIT (Pale/Sub-optimal Staining!)

CORRECT APPROACH:
Technologist must weigh MORE bulk powder to achieve the specified active chromophore mass!

Universal Dye Content Correction Formulas

$\text{Actual Mass to Weigh (g)} = \frac{\text{Specified Mass of Pure Dye (g)}}{\text{Dye Content Fraction}} = \frac{\text{Specified Mass (g)}}{\text{Dye Content } \% / 100}$

$\text{Actual Mass to Weigh (g)} = \text{Specified Mass (g)} \times \left( \frac{100}{\text{Certified Dye Content } \%} \right)$

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Step-by-Step Worked Example: Preparing Toluidine Blue Staining Solution with Dye Purity Correction

• Problem Statement: A diagnostic protocol for mast cell metachromatic staining requires 2.00 g of pure Toluidine Blue O dissolved in 200 mL of buffer (a 1.0% w/v solution). The stock bottle in the laboratory is certified by the Biological Stain Commission with a total dye content of 82.0%. What mass of raw dye powder must be weighed to formulate this reagent correctly?
• Step 1: Identify Known Variables:
  • Specified Pure Dye Mass = $2.00\text{ g}$
  • Certified Dye Content = $82.0% = 0.820$
• Step 2: Apply the Dye Purity Formula: $\text{Actual Mass to Weigh} = \frac{2.00\text{ g}}{0.820} = 2.439\text{ g} \approx 2.44\text{ g}$
• Step 3: Verification: $\text{Active Dye Delivered} = 2.439\text{ g powder} \times 0.820 = 2.00\text{ g pure dye}$
• Step 4: Clinical Impact: If the technologist weighed only 2.00 g of the uncorrected bulk powder, the solution would deliver only $2.00 \times 0.82 = 1.64\text{ g}$ of active chromophore, creating an 18% deficit that results in weak metachromatic staining and faint mast cell granule visualization.

Lot-to-Lot Dye Adjustment Formula

When transitioning between two dye lots with differing certified percentages:

$\text{New Mass to Weigh} = \text{Old Mass} \times \left( \frac{\text{Old Lot Dye Content } \%}{\text{New Lot Dye Content } \%} \right)$

• Example: If an established protocol used 1.20 g of an old dye lot with 75% purity, and the new lot has 90% purity: $\text{New Mass} = 1.20\text{ g} \times \left( \frac{75}{90} \right) = 1.20 \times 0.8333 = 1.00\text{ g}$

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6. Hydration States & Molecular Weight Conversions

Many solid chemical reagents used in staining solutions, fixatives, and biological buffers crystallize with water molecules bound stoichiometrically within their crystal lattice. These bound water molecules are designated as waters of hydration or waters of crystallization.

HYDRATION STATES IN CHEMICAL CRYSTALS:

1. ANHYDROUS SALT:     [ Salt Cation ][ Salt Anion ]                     (Zero Lattice Water)
2. MONOHYDRATE:        [ Salt Cation ][ Salt Anion ] · [ H2O ]           (1 Mole Water / Mole Salt)
3. PENTAHYDRATE:       [ Salt Cation ][ Salt Anion ] · [ 5 H2O ]         (5 Moles Water / Mole Salt)
4. HEPTAHYDRATE:       [ Salt Cation ][ Salt Anion ] · [ 7 H2O ]         (7 Moles Water / Mole Salt)
5. DECAHYDRATE:        [ Salt Cation ][ Salt Anion ] · [ 10 H2O ]        (10 Moles Water / Mole Salt)

CRITICAL HISTOLOGY RULE:
Bound water increases the total molecular weight of the salt package.
Weighing an uncorrected hydrated salt when an anhydrous formula is specified results in
SEVERE UNDER-CONCENTRATION of the active chemical in the staining bath!

Universal Hydration Conversion Formulas

$\text{Mass of Hydrated Salt (g)} = \text{Specified Mass of Anhydrous Salt (g)} \times \left( \frac{\text{MW of Hydrated Salt}}{\text{MW of Anhydrous Salt}} \right)$

$\text{Mass of Anhydrous Salt (g)} = \text{Specified Mass of Hydrated Salt (g)} \times \left( \frac{\text{MW of Anhydrous Salt}}{\text{MW of Hydrated Salt}} \right)$

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Worked Calculation 5: Cupric Sulfate Hydration Conversion

Cupric sulfate is used as a mordant, tissue differentiator, and constituent in Bouin-Hollande fixative.

• Problem Statement: A histology protocol specifies 10.00 g of anhydrous cupric sulfate ($CuSO_4$, MW = 159.61 g/mol). The chemical shelf contains only cupric sulfate pentahydrate ($CuSO_4 \cdot 5H_2O$, MW = 249.68 g/mol). Calculate the mass of pentahydrate required.
• Step 1: Identify Molecular Weights:
  • $\text{MW Anhydrous } (CuSO_4) = 159.61\text{ g/mol}$
  • $\text{MW Pentahydrate } (CuSO_4 \cdot 5H_2O) = 249.68\text{ g/mol}$
• Step 2: Water Content Analysis: Bound crystal water constitutes $(5 \times 18.015) / 249.68 = 90.075 / 249.68 = 36.08%$ of the pentahydrate mass!
• Step 3: Apply the Hydration Formula: $\text{Mass of Pentahydrate} = 10.00\text{ g} \times \left( \frac{249.68}{159.61} \right) = 10.00 \times 1.5643 = 15.643\text{ g} \approx 15.64\text{ g}$
• Step 4: Chemical Verification:
  • Moles of $CuSO_4$ in 10.00 g anhydrous: $10.00 / 159.61 = 0.06265\text{ mol}$
  • Moles of $CuSO_4$ in 15.64 g pentahydrate: $15.643 / 249.68 = 0.06265\text{ mol}$
  • The active copper ion concentration is identical.

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Worked Calculation 6: Sodium Carbonate Hydration Conversion

Sodium carbonate is utilized in silver impregnation sensitizers and alkaline differentiation solutions.

• Problem Statement: A silver stain protocol specifies 5.00 g of anhydrous sodium carbonate ($Na_2CO_3$, MW = 105.99 g/mol). The inventory contains sodium carbonate monohydrate ($Na_2CO_3 \cdot H_2O$, MW = 124.00 g/mol) and sodium carbonate decahydrate ($Na_2CO_3 \cdot 10H_2O$, MW = 286.14 g/mol).
• Case A: Substituting Monohydrate: $\text{Mass Monohydrate} = 5.00\text{ g} \times \left( \frac{124.00}{105.99} \right) = 5.00 \times 1.1699 = 5.85\text{ g}$
• Case B: Substituting Decahydrate: $\text{Mass Decahydrate} = 5.00\text{ g} \times \left( \frac{286.14}{105.99} \right) = 5.00 \times 2.6997 = 13.50\text{ g}$
• Insight: In the decahydrate form, water makes up 63% of the total mass! Weighing 5.0 g of decahydrate instead of anhydrous provides only 1.85 g of active sodium carbonate, crippling the silver impregnation reaction.

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Worked Calculation 7: Dibasic Sodium Phosphate Hydration Conversion

• Problem Statement: A Gomori methenamine silver (GMS) buffer specifies 14.20 g of anhydrous dibasic sodium phosphate ($Na_2HPO_4$, MW = 141.96 g/mol). Only dibasic sodium phosphate heptahydrate ($Na_2HPO_4 \cdot 7H_2O$, MW = 268.07 g/mol) is available.
• Application: $\text{Mass Heptahydrate} = 14.20\text{ g} \times \left( \frac{268.07}{141.96} \right) = 14.20 \times 1.88835 = 26.81\text{ g}$

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7. Dilution Mathematics: $C_1 V_1 = C_2 V_2$ & IHC Serial Antibody Titration

Dilution is the process of reducing the concentration of a solute by adding solvent. Because the absolute quantity of solute remains constant during dilution, the relationship between initial and final states is governed by the conservation of mass:

$C_1 \times V_1 = C_2 \times V_2$

Where:

• $C_1 = \text{Initial (stock) concentration}$
• $V_1 = \text{Volume of stock solution required}$
• $C_2 = \text{Desired (final working) concentration}$
• $V_2 = \text{Final total working volume desired}$

Understanding Dilution Notation (1:D)

In clinical histotechnology and immunohistochemistry, a 1:D dilution signifies 1 part of concentrated solute per D total parts of working solution:

$\text{Dilution Ratio } 1:D \implies \frac{1\text{ volume stock solute}}{D\text{ total volumes working solution}}$ $\text{Volume of Diluent Required} = \text{Total Volume} - \text{Volume of Stock Solute}$ $\text{Dilution Factor (DF)} = \frac{C_1}{C_2} = \frac{V_2}{V_1} = D$

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Serial Dilutions in Immunohistochemistry (IHC)

Primary antibody optimization is required when validating a new antibody clone, qualifying a new lot number, or optimizing the signal-to-noise ratio in diagnostic immunohistochemistry. The goal of an antibody titration series is to identify the dilution that provides intense, crisp target chromogen deposition with zero non-specific background.

IMMUNOHISTOCHEMISTRY 2-FOLD SERIAL DILUTION SERIES (1:50 TO 1:800):

[ Stock Primary Ab ]
        │
        ▼ (20 µL Stock + 980 µL Diluent)
  [ Tube 1: 1:50 ] ──(Transfer 500 µL)──► [ Tube 2: 1:100 ] ──(Transfer 500 µL)──► [ Tube 3: 1:200 ]
  (Total: 1000 µL)                         + 500 µL Diluent                         + 500 µL Diluent
                                                  │                                        │
                                                  ▼ (Transfer 500 µL)                      ▼ (Transfer 500 µL)
                                          [ Tube 4: 1:400 ] ──(Transfer 500 µL)──► [ Tube 5: 1:800 ]
                                           + 500 µL Diluent                         + 500 µL Diluent

Direct Pipetting vs. Serial Dilution Mechanics

To prepare 1.0 mL (1,000 µL) of working solutions directly:

• 1:50: $V_{\text{antibody}} = 1000 / 50 = 20.0,\mu\text{L}; \quad V_{\text{diluent}} = 980.0,\mu\text{L}$
• 1:100: $V_{\text{antibody}} = 1000 / 100 = 10.0,\mu\text{L}; \quad V_{\text{diluent}} = 990.0,\mu\text{L}$
• 1:200: $V_{\text{antibody}} = 1000 / 200 = 5.0,\mu\text{L}; \quad V_{\text{diluent}} = 995.0,\mu\text{L}$
• 1:400: $V_{\text{antibody}} = 1000 / 400 = 2.5,\mu\text{L}; \quad V_{\text{diluent}} = 997.5,\mu\text{L}$
• 1:800: $V_{\text{antibody}} = 1000 / 800 = 1.25,\mu\text{L}; \quad V_{\text{diluent}} = 998.75,\mu\text{L}$

[!WARNING] The Micropipetting Precision Limit: Pipetting volumes below 2.0 µL introduces significant volumetric error (exceeding ±10% to ±15%) due to tip surface tension, liquid adhesion to plastic walls, and piston inaccuracies. Direct preparation of high dilutions (e.g., 1:800 requiring 1.25 µL) produces erratic antibody concentrations. Technologists must utilize two-fold serial dilutions where aliquots of $\ge 250,\mu\text{L}$ to $500,\mu\text{L}$ are transferred sequentially.

Two-Fold Serial Dilution Execution Protocol

1. Tube 1 (1:50): Dispense 980 µL of antibody diluent into a 1.5 mL microcentrifuge tube. Add 20.0 µL of stock antibody using calibrated micropipettor forward pipetting. Vortex gently.
2. Tubes 2 through 5: Dispense exactly 500 µL of antibody diluent into each tube.
3. Transfer 1: Aspirate 500 µL from Tube 1 (1:50), dispense into Tube 2, and mix thoroughly by pipetting up and down 5 times. Concentration in Tube 2 becomes $1:50 \times 1/2 = 1:100$.
4. Transfer 2: Aspirate 500 µL from Tube 2, dispense into Tube 3, and mix. Concentration becomes $1:100 \times 1/2 = 1:200$.
5. Transfer 3: Aspirate 500 µL from Tube 3, dispense into Tube 4, and mix. Concentration becomes $1:200 \times 1/2 = 1:400$.
6. Transfer 4: Aspirate 500 µL from Tube 4, dispense into Tube 5, and mix. Concentration becomes $1:400 \times 1/2 = 1:800$. Discard 500 µL from Tube 5 so all working tubes maintain equal 500 µL volumes.

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8. Practical Reagent Preparation Reference Table