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Renal Elimination and Renal Transporters · Tutorial 433 of 799

Unbound Drug Filtration Across the Glomerulus

Learn to calculate the amount of drug filtered per unit time and distinguish clearance referenced to total plasma concentration from clearance referenced to unbound concentration.

Advanced 10 min read

What You'll Learn

  • Define the glomerular sieving coefficient relative to unbound plasma concentration.
  • Derive filtered drug flux as GFR multiplied by unbound concentration and the sieving coefficient.
  • Calculate filtration flux and clearance with consistent volume and concentration units.
  • Explain how plasma protein binding changes filtration clearance referenced to total plasma concentration.
  • Identify assumptions and diagnose disagreements between predicted filtration and observed renal elimination.

From Plasma Concentration to Filtered Drug Amount

Glomerular filtration is a route by which drug leaves the plasma and enters the tubular fluid. Its rate depends on how much drug is available in the plasma water, how much fluid the kidneys filter, and how readily the drug crosses the glomerular filtration barrier. These quantities must be kept distinct: the plasma concentration reported for a drug is often total concentration, whereas filtration is driven by unbound drug.

The calculation below isolates filtration. It does not, by itself, describe what happens to drug after it enters the tubule. Tubular reabsorption or secretion can alter the amount ultimately excreted in urine, so filtration flux should not be equated with net urinary excretion unless those other processes are absent.

Definition: Let \(C_p\) be total plasma drug concentration and \(f_{u,p}\) the unbound fraction in plasma. The unbound plasma concentration is \(C_{u,p}=f_{u,p}C_p\). The glomerular sieving coefficient, \(\theta_g\), is the ratio of drug concentration in newly formed filtrate to unbound drug concentration in plasma. Under this convention, \(\theta_g=1\) means unbound drug is freely filtered, while \(0\leq\theta_g<1\) means its concentration in filtrate is lower than its unbound plasma concentration. The glomerular filtration rate, GFR, is the volume of plasma water filtered per unit time.

This definition of \(\theta_g\) is relative to unbound plasma concentration. If a data source instead defines a sieving ratio relative to total plasma concentration, that ratio already incorporates the effect of protein binding; multiplying it by \(f_{u,p}\) again would count binding twice. Always check the denominator used for the reported coefficient.

Deriving Filtration Flux and Clearance

Filtration flux is the amount of drug entering the filtrate per unit time. By definition, each unit volume of filtrate contains drug at concentration \(C_{\mathrm{filtrate}}=\theta_g C_{u,p}\). Multiplying this concentration by the volume of filtrate produced per unit time gives the amount filtered per unit time.

$$ \dot A_{\mathrm{fil}} = \mathrm{GFR}\,C_{\mathrm{filtrate}} = \mathrm{GFR}\,\theta_g C_{u,p} = \mathrm{GFR}\,\theta_g f_{u,p}C_p $$

The units verify the result: volume per time multiplied by amount per volume gives amount per time. For example, if GFR is in L/min and concentration is in mg/L, \(\dot A_{\mathrm{fil}}\) is in mg/min. Clearance is the flux divided by the concentration used as its reference. Thus, filtration clearance referenced to total plasma concentration is \(\dot A_{\mathrm{fil}}/C_p\), whereas filtration clearance referenced to unbound plasma concentration is \(\dot A_{\mathrm{fil}}/C_{u,p}\).

$$ CL_{\mathrm{fil},p} = \frac{\dot A_{\mathrm{fil}}}{C_p} = \mathrm{GFR}\,f_{u,p}\theta_g \qquad CL_{\mathrm{fil},u} = \frac{\dot A_{\mathrm{fil}}}{C_{u,p}} = \mathrm{GFR}\,\theta_g $$

Both quantities have units of volume per time, but they answer different questions. \(CL_{\mathrm{fil},p}\) is the filtration flux expressed relative to total plasma concentration; \(CL_{\mathrm{fil},u}\) expresses the same flux relative to unbound plasma concentration. When binding is negligible, \(f_{u,p}\) is near 1 and the two clearances are similar. When binding is substantial, total-plasma-referenced filtration clearance is reduced in proportion to \(f_{u,p}\), provided the sieving coefficient is defined as above.

Assumptions: The calculation treats plasma concentration, \(f_{u,p}\), GFR, and \(\theta_g\) as applicable to the same time and physiological state. It assumes that the concentration in filtrate can be represented by the stated sieving relationship and that GFR is expressed in units consistent with concentration. Using \(C_{u,p}=f_{u,p}C_p\) also assumes the unbound fraction is known for the relevant concentration; if binding is concentration-dependent, \(f_{u,p}\) must be evaluated at that concentration. A constant filtration clearance additionally requires the relevant parameters not to change over the concentration range being considered.

Worked Calculations

Worked Example: A Freely Filtered, Mostly Unbound Drug

Consider a hypothetical drug with total plasma concentration \(C_p=2.00\ \mathrm{mg/L}\), unbound fraction \(f_{u,p}=0.950\), and sieving coefficient \(\theta_g=0.980\). Suppose GFR is \(120\ \mathrm{mL/min}\). Calculate its unbound concentration, filtration flux, and clearance referenced to total and unbound plasma concentration.

First, calculate unbound concentration:

$$ C_{u,p}=f_{u,p}C_p =(0.950)(2.00\ \mathrm{mg/L}) =1.90\ \mathrm{mg/L} $$

Convert GFR to L/min so it is consistent with concentration units: \(120\ \mathrm{mL/min}=0.120\ \mathrm{L/min}\). Then the filtration flux is:

$$ \dot A_{\mathrm{fil}} =(0.120\ \mathrm{L/min})(0.980)(1.90\ \mathrm{mg/L}) =0.22344\ \mathrm{mg/min} $$

To three significant figures, the filtered amount is \(0.223\ \mathrm{mg/min}\). The clearance referenced to total concentration is:

$$ CL_{\mathrm{fil},p} =\frac{0.22344\ \mathrm{mg/min}}{2.00\ \mathrm{mg/L}} =0.11172\ \mathrm{L/min} =112\ \mathrm{mL/min} $$

The clearance referenced to unbound concentration is:

$$ CL_{\mathrm{fil},u} =\frac{0.22344\ \mathrm{mg/min}}{1.90\ \mathrm{mg/L}} =0.1176\ \mathrm{L/min} =118\ \mathrm{mL/min} $$

These results also follow from \(\mathrm{GFR}\,f_{u,p}\theta_g\) and \(\mathrm{GFR}\,\theta_g\), respectively. The drug is close to freely filtered, but its total-plasma-referenced filtration clearance is slightly below GFR because both \(f_{u,p}\) and \(\theta_g\) are slightly below 1.

Worked Example: Strong Plasma Protein Binding

For a hypothetical compound, let \(C_p=10.0\ \mathrm{mg/L}\), \(f_{u,p}=0.0200\), \(\theta_g=1.00\), and GFR \(=105\ \mathrm{mL/min}\). Calculate filtration flux and both filtration clearances.

Only the unbound concentration enters the filtration expression:

$$ C_{u,p}=(0.0200)(10.0\ \mathrm{mg/L}) =0.200\ \mathrm{mg/L} $$

In L/min, GFR is \(0.105\ \mathrm{L/min}\). Therefore:

$$ \dot A_{\mathrm{fil}} =(0.105\ \mathrm{L/min})(1.00)(0.200\ \mathrm{mg/L}) =0.0210\ \mathrm{mg/min} $$

The total-plasma-referenced clearance is:

$$ CL_{\mathrm{fil},p} =\frac{0.0210\ \mathrm{mg/min}}{10.0\ \mathrm{mg/L}} =0.00210\ \mathrm{L/min} =2.10\ \mathrm{mL/min} $$

The unbound-plasma-referenced clearance is:

$$ CL_{\mathrm{fil},u} =\frac{0.0210\ \mathrm{mg/min}}{0.200\ \mathrm{mg/L}} =0.105\ \mathrm{L/min} =105\ \mathrm{mL/min} $$

The low total-plasma-referenced clearance does not mean that the filtration barrier strongly excludes the unbound drug: \(\theta_g=1.00\). It reflects the small unbound fraction. At the stated concentration, filtration removes \(0.0210\ \mathrm{mg}\) each minute. This calculation uses the specified unbound fraction; it does not require assuming that protein-bound molecules pass intact into filtrate.

Worked Example: Reduced Sieving of Unbound Drug

Suppose a hypothetical drug has \(C_p=4.00\ \mathrm{mg/L}\), \(f_{u,p}=0.700\), \(\theta_g=0.350\), and GFR \(=90.0\ \mathrm{mL/min}\). Calculate its filtered flux and clearance referenced to total plasma concentration. Then compare its flux with the result if \(\theta_g\) were 1.00 and all other values were unchanged.

The unbound concentration is:

$$ C_{u,p}=(0.700)(4.00\ \mathrm{mg/L}) =2.80\ \mathrm{mg/L} $$

Convert GFR to \(0.0900\ \mathrm{L/min}\). With the given sieving coefficient:

$$ \dot A_{\mathrm{fil}} =(0.0900\ \mathrm{L/min})(0.350)(2.80\ \mathrm{mg/L}) =0.0882\ \mathrm{mg/min} $$

The total-plasma-referenced filtration clearance is:

$$ CL_{\mathrm{fil},p} =\frac{0.0882\ \mathrm{mg/min}}{4.00\ \mathrm{mg/L}} =0.02205\ \mathrm{L/min} =22.1\ \mathrm{mL/min} $$

If \(\theta_g=1.00\), the flux would instead be:

$$ (0.0900\ \mathrm{L/min})(1.00)(2.80\ \mathrm{mg/L}) =0.252\ \mathrm{mg/min} $$

The stated sieving coefficient therefore reduces filtration flux to \(0.350\) of the freely filtered value, with unbound concentration and GFR held fixed. The total-plasma-referenced clearance is also proportional to \(\theta_g\); it is not determined by GFR and binding alone.

Interpreting the Calculation and Checking It

The expression for filtration flux is a direct consequence of a volume rate multiplied by a concentration. Its simple form is useful in a PBPK model because it makes the dependencies explicit: at fixed GFR and \(\theta_g\), doubling \(C_{u,p}\) doubles filtration flux; at fixed concentrations and \(\theta_g\), doubling GFR doubles it. These proportionalities follow from the equation and apply only while the relevant parameters remain unchanged.

A practical calculation should pass several checks. First, concentration in filtrate must be \(\theta_g C_{u,p}\) under the convention used here. Second, the GFR and concentration units must cancel to amount per time. Third, the calculated filtration clearance should equal flux divided by the concentration named in its subscript. For \(0\leq f_{u,p}\leq1\) and \(0\leq\theta_g\leq1\), the model implies \(CL_{\mathrm{fil},p}\leq\mathrm{GFR}\) and \(CL_{\mathrm{fil},u}\leq\mathrm{GFR}\). A result exceeding GFR under these definitions signals a unit, definition, or calculation problem.

Filtration is only one term in renal drug handling. The earlier material on renal drug mass balance distinguishes filtration from the net balance that determines urinary drug recovery. If observed renal clearance differs from the filtration-only prediction, that difference does not by itself identify a mechanism. Check the concentration basis, binding measurement, GFR, sieving-coefficient definition, and units first. If these are appropriate, processes beyond filtration or a mismatch between the model assumptions and the physiological state may contribute. The filtration equation alone cannot separate those possibilities.

Common pitfall: “The drug is 98% protein-bound, so 98% of its plasma concentration is filtered.” This reverses the role of binding: under the stated model, the unbound fraction is \(0.020\), so \(C_{u,p}=0.020C_p\), and filtration flux is \(\mathrm{GFR}\,\theta_g(0.020C_p)\). The correction is to multiply total concentration by the unbound fraction, then apply the sieving coefficient and GFR.

A second common error is to call \(\mathrm{GFR}\) the filtration clearance for every drug. That is only the unbound-referenced clearance when \(\theta_g=1\); the clearance referenced to total plasma concentration is \(\mathrm{GFR}\,f_{u,p}\theta_g\). A third error is to apply \(f_{u,p}\) when the provided sieving ratio is already defined relative to total plasma concentration. Establishing the coefficient's definition prevents this double adjustment.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A hypothetical drug has \(C_p=3.00\ \mathrm{mg/L}\), \(f_{u,p}=0.400\), \(\theta_g=0.800\), and GFR \(=100\ \mathrm{mL/min}\). What is its filtration flux?
  2. For the drug in question 1, calculate filtration clearance referenced to total plasma concentration and to unbound plasma concentration.
  3. If an unbound-referenced sieving coefficient is \(\theta_g=0.600\), what does this say about filtrate concentration relative to unbound plasma concentration?
  4. Why is it incorrect to multiply by \(f_{u,p}\) if a supplied sieving ratio is already defined as filtrate concentration divided by total plasma concentration?

Solutions. For question 1, \(C_{u,p}=(0.400)(3.00)=1.20\ \mathrm{mg/L}\), and GFR is \(0.100\ \mathrm{L/min}\). Thus \(\dot A_{\mathrm{fil}}=(0.100)(0.800)(1.20)=0.0960\ \mathrm{mg/min}\). For question 2, \(CL_{\mathrm{fil},p}=0.0960/3.00=0.0320\ \mathrm{L/min}=32.0\ \mathrm{mL/min}\); \(CL_{\mathrm{fil},u}=0.0960/1.20=0.0800\ \mathrm{L/min}=80.0\ \mathrm{mL/min}\). For question 3, filtrate concentration is \(0.600\) times unbound plasma concentration. For question 4, the total-referenced ratio already includes binding's effect, so multiplying by \(f_{u,p}\) again would apply that effect twice.