Principles · Types · Isolated Tissue Methods · Multiple-Point Assays
Definition, why bioassays are needed, types, and ideal properties
Bioassay (biological assay) is a method of finding out how strong (potent) a drug or chemical is by looking at its effect on living cells, tissues, or whole animals and comparing it with a standard drug of known strength.
Teaching definition: Bioassay is the estimation of the relative potency of an active principle in a test preparation by comparing its effect with that of a standard preparation on a suitable living system.
Based on nature of response
Based on method (key for practicals)
| Method | Subtypes | Use |
|---|---|---|
| Direct (simple) | Matching · Bracketing | Quick estimation; limited accuracy |
| Graphical | Interpolation method | More objective; uses full dose–response curve |
| Multiple-point | Three-point (2+1) · Four-point (2+2) | Statistical evaluation; highest accuracy |
Whole animal and isolated tissue models, Tyrode solution, organ bath assembly, and drug dilutions
| Drug | Animal | Response Measured |
|---|---|---|
| Noradrenaline | Cat (spinal) | Rise in blood pressure |
| Digitalis | Guinea pig | Death due to cardiac toxicity (quantal) |
| Vasopressin | Rat | Reduction in urine volume (antidiuretic effect) |
| Insulin | Mice | Hypoglycaemic convulsions |
| d-Tubocurarine | Rabbit | Head drop due to skeletal muscle paralysis |
| Oxytocin | Rabbit | Milk ejection |
| Anabolic steroid | Castrated male rat | Increase in weight of levator ani muscle |
Whole animal models are closer to clinical situations but are expensive, time-consuming, and have ethical constraints.
| Drug | Tissue Preparation | Response |
|---|---|---|
| Acetylcholine | Frog rectus abdominis muscle | Contraction |
| Histamine | Guinea pig ileum | Contraction |
| Adrenaline / Noradrenaline | Rat colon | Relaxation |
| Oxytocin | Rat uterus | Contraction |
| 5-Hydroxytryptamine (5-HT) | Rat stomach fundus | Contraction |
Isolated tissue preparations are suitable for graded dose–response curves and for matching, bracketing, interpolation, and multiple-point assays.
To keep an isolated tissue alive in an organ bath, it is immersed in physiological salt solution (PSS) that mimics extracellular fluid. The solution is oxygenated and maintained at 30–37°C.
| Component | Amount (g/L) | Main Role |
|---|---|---|
| NaCl | 8.0 | Provides sodium and chloride; main contributor to osmolarity |
| NaHCO₃ | 1.0 | Buffer to maintain pH |
| Glucose | 1.0 | Energy source for the tissue |
| KCl | 0.2 | Provides potassium for resting membrane potential |
| CaCl₂ | 0.2 | Calcium required for contraction |
| MgCl₂ | 0.1 | Cofactor and membrane stabilizer |
| NaH₂PO₄ | 0.05 | Additional buffer and phosphate source |
Unit conversions
10 mg/mL1 mg/mL0.1 mg/mLSerial dilutions (1:10)
Matching bioassay and bracketing bioassay — principles, procedure, advantages, and limitations
Principle
A fixed dose of the standard drug (S) is selected and its response is recorded. Doses of the test preparation (T) are adjusted until a response is obtained that is visually equal to the response of the standard. From the doses that give equal responses, the relative potency of the test is calculated.
Example: 2 mL of test = same effect as 0.5 mL of 1 mg/mL standard → test concentration = 0.25 mg/mL
Advantages
Limitations
Principle
A convenient test dose (T) is chosen first. Standard doses are adjusted until two of them — S1 and S2 — produce responses that just fall below and above the response of the test. The test response is thus "bracketed" between two standard responses.
From these three responses, the potency of the test solution is computed by proportion or by a percentage response calculation (e.g. calculating the percentage position of T between S1 and S2).
Recording pattern
Advantages
Limitations
Log dose–response curve of the standard used to read off the equivalent dose of the test preparation
Prepare several graded doses of the standard drug and record their responses on the tissue.
Plot log dose (X-axis) vs response (Y-axis) for the standard. The middle portion is nearly straight.
Administer one or more doses of the test solution and record their responses on the same tissue.
For each test response, locate the same level on the Y-axis, move horizontally to meet the standard curve, then move down to the X-axis to read the equivalent standard dose.
Compare the actual test dose with the equivalent standard dose to calculate the concentration or potency of the test.
Three-point (10 steps) and four-point assays with Latin square randomization and statistical evaluation
Uses two standard doses (S1, S2) and one test dose (T), all lying on the linear part of the log dose–response curve.
Example dose–response data (from standard curve)
| Dose (µg) | Response (cm) | Log dose | % Response |
|---|---|---|---|
| 2 | 1.0 | 0.30 | 13.3% |
| 4 | 1.75 | 0.60 | 23.3% |
| 8 | 3.0 | 0.90 | 40.0% |
| 16 | 4.25 | 1.20 | 56.6% ← S1 |
| 32 | 5.75 | 1.50 | 76.6% ← S2 |
| 64 | 7.5 | 1.80 | 100% |
Give increasing doses of the standard drug (e.g. 2, 4, 8, 16, 32, 64 µg) to the tissue. Measure response (height of contraction) for each dose.
Measure response height for each concentration using the kymograph recording or digital read-out in centimetres or arbitrary units.
Prepare a table of dose, response (cm), log dose, and percent response (taking maximum response as 100%).
Draw log dose (X-axis) vs % response (Y-axis). The middle portion (20–80% response) is almost straight and is used for selecting S1 and S2.
Identify S1, S2, and T. S1 and S2 responses: 25–75% of maximum; S2 = double S1. T: response between S1 and S2.
Repeat responses of S1, S2, and T until T reliably falls between S1 and S2. Note responses until they stabilise.
Apply Latin square randomization: Cycle 1 → S1–S2–T · Cycle 2 → S2–T–S1 · Cycle 3 → T–S1–S2. Each dose appears once in each position.
Record each solution across 3 periods and calculate mean responses for S1, S2, and T (denoted S̄1, S̄2, T̄).
Apply the formula for concentration of unknown (see below) using mean response heights.
Substitute values and compute the answer — the concentration of the test solution.
Example — Mean response table (Step 8)
| Solution | Period 1 (cm) | Period 2 (cm) | Period 3 (cm) | Mean response (cm) |
|---|---|---|---|---|
| S1 (16 µg/mL) | 3.0 | 3.0 | 3.0 | 3.0 |
| S2 (32 µg/mL) | 4.25 | 4.25 | 4.25 | 4.25 |
| T (test) | 3.25 | 3.25 | 3.25 | 3.25 |
Formula — concentration of unknown (Step 9)
log(Ct) = log(S1) + [log(S2/S1)] × [(T̄ − S̄1) / (S̄2 − S̄1)]
S1 = lower standard dose · S2 = higher standard dose · T = dose of test solutionS̄1, S̄2, T̄ = mean responses to S1, S2, and T respectivelyWorked example (Step 10)
Comparison: Three-point vs Four-point
| Feature | Three-Point Assay | Four-Point Assay |
|---|---|---|
| Standard doses | 2 (S1, S2) | 2 (S1, S2) |
| Test doses | 1 (T) | 2 (T1, T2) |
| Total responses (typical) | 9 (three cycles) | 16 (four cycles) |
| Accuracy | Good | Higher than three-point |
| Time taken | Moderate | Longer |
| Statistical evaluation | Possible | Better error estimation |
Causes of variability, how to reduce them, and the role of bioassays in pharmacology