Why Correct a Composition-Based Kp?
A composition-based tissue-to-plasma partition coefficient, \(K_p\), predicts the ratio of total drug concentrations in tissue and plasma at equilibrium. The prediction depends not only on how strongly a drug binds to tissue constituents, but also on how much aqueous volume is present in each compartment. A correction that uses unbound fractions without accounting for these different water fractions can therefore give the wrong total-concentration ratio.
The Berezhkovskiy correction makes that water-volume factor explicit. Under its equilibrium assumptions, it expresses \(K_p\) as the ratio of plasma to tissue compartment unbound fractions, multiplied by the tissue-to-plasma water-fraction ratio. The correction is an algebraic relationship: it does not replace the composition information or the assumptions used to obtain the underlying binding and distribution quantities.
This definition distinguishes two related quantities that are easy to confuse. The ratio of free aqueous concentration to total concentration is \(C_f/C=1/F\). The compartment unbound fraction is instead \(f_u=f_wC_f/C=f_w/F\), because the free aqueous contribution per unit bulk volume is \(f_wC_f\). These quantities are equal only when the water fraction is one.
Deriving the Berezhkovskiy Correction
The correction follows directly from the definitions. First assume the drug has reached equilibrium and that the free aqueous concentrations in tissue and plasma are equal. This is a useful approximation when the freely diffusible species is at equilibrium across the compartments and there is no active transport or other process maintaining a free-concentration gradient. Then \(C_{f,t}/C_{f,p}=1\), so the total-concentration ratio is the ratio of the capacity factors.
By definition, \(K_p=C_t/C_p=(F_tC_{f,t})/(F_pC_{f,p})\).
If \(C_{f,t}=C_{f,p}\), their ratio cancels, leaving \(K_p=F_t/F_p\).
Since \(f_{u,t}=f_{w,t}/F_t\) and \(f_{u,p}=f_{w,p}/F_p\), rearrangement gives \(F_t=f_{w,t}/f_{u,t}\) and \(F_p=f_{w,p}/f_{u,p}\).
Substituting those expressions into \(F_t/F_p\) gives the tissue-to-plasma water-fraction ratio multiplied by the plasma-to-tissue unbound-fraction ratio.
The order of the terms matters: tissue water fraction is in the numerator and plasma water fraction in the denominator, while plasma unbound fraction is in the numerator and tissue unbound fraction in the denominator. If the free aqueous concentrations are not equal, the more general relationship is \(K_p=(F_t/F_p)(C_{f,t}/C_{f,p})\). The correction alone does not account for a free-concentration gradient.
Worked Calculations
Worked Example: Calculating Kp from Capacity Factors
Hypothetical data. Suppose tissue has a water fraction of \(0.70\), and its total capacity factor is \(F_t=2.00\). Plasma has a water fraction of \(0.93\), and \(F_p=1.20\). Assume equal free aqueous concentrations at equilibrium. Calculate \(K_p\), then calculate both compartment unbound fractions and verify the corrected equation.
The direct capacity-factor calculation is
Thus, the predicted total tissue concentration is 1.667 times the total plasma concentration, when both are expressed on their stated bulk-volume bases. The compartment unbound fractions are
Using the correction gives
As a basis check, \(C_{f,t}/C_t=1/F_t=0.500\), whereas the compartment unbound fraction is \(0.350\). The difference is expected: the free aqueous contribution per unit tissue volume is \(0.70C_{f,t}\), not \(C_{f,t}\).
Worked Example: Correcting an Unbound-Fraction Ratio
Hypothetical data. A composition-based assessment provides a tissue water fraction of \(0.60\), a plasma water fraction of \(0.90\), a tissue compartment unbound fraction of \(0.25\), and a plasma compartment unbound fraction of \(0.75\). Assume these fractions use the definition above and that free aqueous concentrations are equal. What is the corrected \(K_p\), and how would omitting the water correction affect the answer?
First calculate the uncorrected unbound-fraction ratio:
This is not yet the total-concentration \(K_p\). The water-fraction multiplier is \(0.60/0.90=0.6667\), so
The same result follows by recovering the capacity factors: \(F_t=0.60/0.25=2.40\) and \(F_p=0.90/0.75=1.20\), giving \(F_t/F_p=2.40/1.20=2.000\). Omitting the water correction would report \(3.00\), an overestimate in this example because the tissue water fraction is smaller than the plasma water fraction.
Worked Example: Reconstructing the Correction from Composition Terms
Hypothetical data. Suppose tissue capacity per unit free aqueous concentration is composed of a water contribution of \(0.60\), a neutral-lipid contribution of \(0.40\), and a binding contribution of \(1.50\). Plasma capacity consists of a water contribution of \(0.92\) and a protein-binding contribution of \(0.58\). Assume the listed terms are additive capacity contributions on compatible concentration bases, and assume equal free aqueous concentrations.
The total capacity factors are \(F_t=0.60+0.40+1.50=2.50\) and \(F_p=0.92+0.58=1.50\). The direct prediction is
The compartment unbound fractions implied by these capacities are \(f_{u,t}=0.60/2.50=0.240\) and \(f_{u,p}=0.92/1.50=0.6133\). Applying the Berezhkovskiy form,
The two calculations agree because they use the same capacities, water fractions, and equilibrium assumption. The phase contributions in this example are illustrative model inputs, not measurements from a real drug or tissue.
Assumptions, Checks, and Failure Modes
The correction is useful only when its inputs have compatible meanings and its assumptions are plausible. Check the following before interpreting a result:
- Equilibrium: The ratio concerns equilibrium concentrations. A tissue sampled before equilibration can have a different measured tissue-to-plasma ratio.
- Free aqueous concentration: The simple formula assumes \(C_{f,t}=C_{f,p}\). If pH-dependent speciation or another mechanism produces different effective free concentrations, the omitted ratio \(C_{f,t}/C_{f,p}\) matters.
- Consistent bases: Tissue and plasma total concentrations, water fractions, and unbound fractions must refer to the same defined compartments and compatible volume bases.
- Correct definition of unbound fraction: Use \(f_u=f_w/F\) for the compartment unbound fraction in this correction. Do not substitute \(1/F\) for it unless the water fraction is one.
- Capacity assumptions: The \(F\) values must adequately represent aqueous drug and binding contributions. If composition terms are incomplete or binding is not adequately described, algebraic correction cannot repair the underlying prediction.
A useful numerical check is to reconstruct capacity factors from the supplied fractions: \(F_t=f_{w,t}/f_{u,t}\) and \(F_p=f_{w,p}/f_{u,p}\). Under the stated definitions, \(F_t\) must be at least \(f_{w,t}\), and \(F_p\) must be at least \(f_{w,p}\), when additional contributions to total concentration are nonnegative. A computed value below the corresponding water fraction signals inconsistent inputs, definitions, or units. Also verify that \(K_p\) is positive and that the water multiplier has not been inverted.
How the Correction Relates to Other Approaches
A direct composition-based calculation sums appropriate phase-capacity contributions to obtain \(F_t\) and \(F_p\), then calculates \(K_p=F_t/F_p\) under equal free aqueous concentrations. The Berezhkovskiy correction instead expresses those same capacity factors through water fractions and compartment unbound fractions. If both routes use consistent inputs and assumptions, they are algebraically equivalent, as the worked examples show.
This distinction is practical rather than a claim that one route is universally more accurate. Direct capacity calculations make phase contributions visible; the correction can be convenient when unbound fractions and water fractions are the available quantities. A measured equilibrium \(K_p\) is a different kind of information: it can be used to assess a prediction, but it does not by itself establish that the model assumptions or proposed tissue binding mechanism are correct. The Schmitt method is another composition-based prediction framework; choosing between frameworks requires attention to their inputs and assumptions rather than applying the correction as if it were an independent source of new binding data.
The correction has important limits. It does not account by itself for nonequilibrium distribution, active transport, a barrier-maintained concentration gradient, missing binding processes, or uncertainty in the composition and unbound-fraction inputs. For ionizable drugs, the effective free concentrations can depend on the local chemical environment; if they differ between tissue and plasma, the general expression with \(C_{f,t}/C_{f,p}\) must be considered rather than silently assuming that this ratio is one. Thus, a numerically precise corrected value is not necessarily a biologically reliable prediction.
Check Your Understanding
Answer each question using the definitions and assumptions stated above. Brief solutions follow the questions.
- A tissue has \(f_{w,t}=0.72\) and \(F_t=2.40\). What are \(C_{f,t}/C_t\) and the compartment unbound fraction?
- Given \(f_{w,t}=0.64\), \(f_{w,p}=0.90\), \(f_{u,t}=0.20\), and \(f_{u,p}=0.60\), calculate corrected \(K_p\) assuming equal free aqueous concentrations.
- For the inputs in question 2, recover \(F_t\) and \(F_p\), then verify \(K_p\) from their ratio.
- What additional factor is required if the free aqueous concentrations are not equal?
- A calculation gives \(F_t=0.50\) when \(f_{w,t}=0.70\), with nonnegative binding contributions. What should be checked?
Solutions. (1) \(C_{f,t}/C_t=1/F_t=1/2.40=0.4167\). The compartment unbound fraction is \(f_{u,t}=f_{w,t}/F_t=0.72/2.40=0.300\); these are not the same. (2) \(K_p=(0.64/0.90)(0.60/0.20)=0.7111\times3.00=2.133\). (3) \(F_t=0.64/0.20=3.20\), \(F_p=0.90/0.60=1.50\), and \(F_t/F_p=3.20/1.50=2.133\), which agrees. (4) Multiply \(F_t/F_p\) by \(C_{f,t}/C_{f,p}\); the short correction formula assumes that ratio equals one. (5) Check the definitions, units, and input consistency: under the stated additive-capacity assumptions, total capacity cannot be less than the aqueous contribution \(f_{w,t}=0.70\).