Why Tissue Composition Matters
Mechanistic tissue-to-plasma partitioning models represent a tissue as a set of environments in which a drug may distribute. Composition measurements help describe the amounts of water, neutral lipid, phospholipid, and protein available in that tissue. The measurements are inputs describing the tissue, not measurements of the drug’s affinity for those components. A composition table must therefore be interpreted and converted consistently before it can be used in a prediction.
The earlier material on Kp definitions established that a tissue-to-plasma partition coefficient depends on the concentration basis and the compartments being compared. Here, the focus is narrower: how to identify tissue-composition inputs and calculate the corresponding phase volumes. These calculations do not by themselves produce a Kp value.
Composition data may be reported as grams per 100 g wet tissue, a mass fraction of wet tissue, grams per gram of tissue, a dry-weight fraction, or a volume fraction. These formats are not interchangeable without conversion. Always record the tissue basis, the units, and whether each reported value is a mass or a volume before doing arithmetic.
For a mass-based input, the conversion is straightforward. If a tissue sample has mass \(M\) and constituent \(i\) has mass fraction \(f_i\), then \(m_i=f_iM\), and the phase volume follows from \(V_i=m_i/\rho_i\). If the data are already volume fractions or volumes, dividing by density again would be an error.
Converting Composition to Phase Volumes
The four named inputs are water, neutral lipid, phospholipid, and protein. In a composition-based calculation, their masses must be defined on the same tissue basis and must represent non-overlapping categories. For instance, if a data source reports total lipid and a separate phospholipid amount, the definition of “total lipid” must be checked before treating the two as distinct inputs. Otherwise, phospholipid could be counted twice.
For each constituent, divide its mass by the density assigned to that constituent. The density should be stated with units, such as grams per millilitre. The following calculation also requires an explicit approximation: the reported mass and chosen density adequately represent the volume assigned to that constituent. If a model provides standard densities or conventions, use those consistently; do not combine densities from one convention with composition inputs from another without justification.
If the bulk tissue density \(\rho_{\mathrm{tissue}}\) and tissue mass \(M\) are known, the tissue’s bulk volume is \(V_{\mathrm{tissue}}=M/\rho_{\mathrm{tissue}}\). One may then calculate the difference between the bulk volume and the sum of the four assigned phase volumes. This difference is a bookkeeping residual. It is not automatically a fifth biological phase, nor should it be assigned to an additional constituent unless the model definition calls for that interpretation.
This calculation assumes that the tissue mass, constituent masses, component densities, and bulk density all refer to compatible conditions and the same tissue definition. It also treats the assigned volumes as additive for the purpose of bookkeeping. Real tissues are heterogeneous, and their measured constituents need not behave like four isolated containers whose volumes can be measured independently. A negative residual is a warning to investigate data definitions, units, densities, or overlap; it is not evidence of a physically negative compartment.
Confirm whether each composition value is a mass, mass fraction, or volume, and whether it is expressed per wet tissue, dry tissue, or another basis.
Choose the specified sample mass \(M\), or convert the reported per-unit-tissue values to the required mass.
For mass-based inputs, calculate each mass and divide by its stated density, keeping units explicit.
Compare the sum of phase volumes with bulk tissue volume when it is available. Interpret the difference as a residual unless its biological meaning is independently specified.
Worked Example: Phase Volumes in a Hypothetical 100 g Tissue Sample
Suppose a hypothetical tissue sample has mass \(100\) g. Its specified constituent masses are \(70.0\) g water, \(2.001\) g neutral lipid, \(1.50\) g phospholipid, and \(18.0\) g protein. For this example only, use densities of \(1.00\), \(0.900\), \(1.00\), and \(1.35\) g/mL, respectively. Suppose the bulk tissue density is \(1.03\) g/mL. Calculate the four phase volumes, their sum, and the residual.
The inputs are all masses for the same 100 g wet-tissue sample, so they can be converted directly. Water has volume \(70.0\text{ g}/(1.00\text{ g/mL})=70.0\) mL. Neutral lipid has volume \(2.001\text{ g}/(0.900\text{ g/mL})=2.22333\) mL. Phospholipid has volume \(1.50\text{ g}/(1.00\text{ g/mL})=1.50\) mL. Protein has volume \(18.0\text{ g}/(1.35\text{ g/mL})=13.3333\) mL.
Add the unrounded component volumes: \(70.0+2.001/0.900+1.50+18.0/1.35=87.0567\) mL, or \(87.1\) mL to three significant figures. The bulk tissue volume is \(100\text{ g}/(1.03\text{ g/mL})=97.0874\) mL. Thus the residual, calculated before rounding, is \(97.0874-87.0567=10.0307\) mL, or \(10.03\) mL to four significant figures.
The four specified constituent masses sum to \(91.501\) g, so they do not describe all of the sample’s mass. The \(10.03\) mL residual is the difference between bulk volume and the four assigned volumes; it is not obtained by treating the unlisted \(8.499\) g as though it had a known density. The calculation does not identify what accounts for the residual.
Scaling and Checking the Data Basis
When data are reported as fractions, scaling to a sample mass is necessary before using densities. For example, a water mass fraction of \(0.72\) means \(72\) g of water per \(100\) g tissue on the stated basis. It does not mean \(0.72\) mL of water per gram unless that is how the source explicitly reports the measurement. The distinction matters because the density conversion applies to mass, not to a number whose basis has not been established.
Worked Example: Scaling Hypothetical Composition Fractions
A hypothetical \(250\) g tissue sample has mass fractions \(0.72\) water, \(0.035\) neutral lipid, \(0.025\) phospholipid, and \(0.18\) protein. Use the same illustrative densities as above: \(1.00\), \(0.900\), \(1.00\), and \(1.35\) g/mL. The bulk density is \(1.05\) g/mL. Find each phase volume and the residual.
First calculate masses: water is \(0.72(250)=180\) g; neutral lipid is \(0.035(250)=8.75\) g; phospholipid is \(0.025(250)=6.25\) g; and protein is \(0.18(250)=45.0\) g. The phase volumes are \(180/1.00=180\) mL, \(8.75/0.900=9.72222\) mL, \(6.25/1.00=6.25\) mL, and \(45.0/1.35=33.3333\) mL, respectively.
Their unrounded sum is \(180+8.75/0.900+6.25+45.0/1.35=229.3056\) mL, which rounds to \(229.3\) mL. The bulk volume is \(250/1.05=238.0952\) mL. Therefore, \(V_{\mathrm{residual}}=238.0952-229.3056=8.78968\) mL, or \(8.790\) mL to four significant figures.
The four mass fractions sum to \(0.96\), corresponding to \(240\) g; the remaining \(10\) g is not assigned to these four inputs. The residual calculation uses volumes and bulk density, not a presumed density for that unassigned mass. This is why the residual cannot be calculated simply by converting the leftover mass with one of the four component densities.
Worked Example: Converting Per-Gram Wet-Tissue Values
A hypothetical composition table reports \(0.68\) g water, \(0.04\) g neutral lipid, \(0.03\) g phospholipid, and \(0.16\) g protein per gram of wet tissue. For a \(50\) g sample, use densities of \(1.00\), \(0.900\), \(1.00\), and \(1.35\) g/mL, respectively, and a bulk tissue density of \(1.02\) g/mL.
Multiply each reported amount by \(50\) g tissue: the constituent masses are \(34\) g water, \(2\) g neutral lipid, \(1.5\) g phospholipid, and \(8\) g protein. Converting these masses gives \(34/1.00=34.0\) mL water, \(2/0.900=2.22222\) mL neutral lipid, \(1.5/1.00=1.50\) mL phospholipid, and \(8/1.35=5.92593\) mL protein.
The four phase volumes sum to \(34+2/0.900+1.5+8/1.35=43.6481\) mL. Bulk tissue volume is \(50/1.02=49.0196\) mL, so the residual is \(49.0196-43.6481=5.37146\) mL. The reported per-gram values are mass amounts, not volumes; the density conversion is therefore required.
Diagnostics, Alternatives, and Limitations
Before accepting a set of phase volumes, check the calculation and the source data. The constituent values should share a tissue basis, and the mass fractions should be plausible as a set. A sum below one may be expected when other constituents are omitted; a sum above one usually indicates incompatible denominators, overlapping categories, or a conversion error. Compare the component-volume sum with bulk tissue volume if both can be calculated, but do not force the residual to zero by rescaling inputs unless the chosen method specifically requires that operation.
| Input format | Appropriate handling | Common error |
|---|---|---|
| Mass fraction of wet tissue | Multiply by the specified wet tissue mass, then divide by component density. | Treating a mass fraction as a volume fraction. |
| Mass per gram of tissue | Multiply by the required tissue mass, then divide by component density. | Using the per-gram value as a volume without conversion. |
| Mass per 100 g tissue | Scale to the sample mass if it differs from 100 g, then convert using density. | Applying the 100 g values unchanged to a different sample size. |
| Volume fraction or volume | Use as a volume input on its stated basis, after any needed scaling. | Dividing by density a second time. |
A composition-based approach is one way to parameterize tissue properties. An alternative is to use directly measured tissue-to-plasma partition coefficients when suitable measurements are available; those measurements answer a different input question and do not require reconstructing phase volumes in the same way. A further practical alternative is to use a curated composition table or a model’s prescribed tissue values rather than assemble values from multiple sources. Whichever route is chosen, tissue definition, measurement basis, and reference concentration conventions must remain compatible.
Composition tables are not universal constants. Tissue source, species, physiological state, sample handling, and analytical definitions can affect reported values. Some tables report wet-weight measurements, while others may use dry tissue or a different convention. The four named constituents also do not necessarily account for every tissue constituent or every volume contribution. A good residual check can reveal inconsistency, but it cannot establish the correct biology on its own.
Common Pitfall: Adding Rounded Values
A tempting shortcut is to round each calculated phase volume and then add the rounded numbers. In the first example, rounding the neutral-lipid and protein volumes early changes the displayed sum, and subtracting that rounded sum from a rounded bulk volume can change the reported residual. Keep full calculator precision through the summation and subtraction, then round the final results to a precision justified by the inputs. Also, do not interpret a positive residual as a measured “other” compartment: it is only the difference implied by the chosen data and assumptions.
Another frequent mistake is to mix wet-weight and dry-weight data as if they shared the same denominator. A dry-weight fraction cannot be multiplied by total wet tissue mass without first accounting for the water basis. Likewise, if a source already supplies volume fractions, density conversion is not appropriate. In each case, the correction is to return to the source definitions and write the units explicitly before calculating.
Check Your Understanding
Try each question, then compare your work with the concise solutions below.
- A \(40\) g hypothetical tissue sample contains \(0.60\) g water per gram of wet tissue. How much water mass is represented?
- A \(3.0\) g neutral-lipid amount has an illustrative density of \(0.90\) g/mL. What is its phase volume?
- Why should a reported volume fraction not be divided by density again?
- What does a positive residual between bulk tissue volume and the four calculated phase volumes establish, and what does it not establish?
Solutions. (1) The water mass is \(0.60\text{ g/g}\times40\text{ g}=24\) g. (2) The volume is \(3.0\text{ g}/(0.90\text{ g/mL})=3.3\) mL to two significant figures. (3) A volume fraction already describes volume on its stated basis; dividing by density would apply a mass-to-volume conversion to a quantity that is already a volume. (4) It establishes only that the specified bulk volume exceeds the sum of the assigned volumes under the stated inputs and assumptions. It does not identify the source of the difference or prove that it is a distinct biological phase.