Amount filtered

The amount of substance A filtered per unit time is calculated from the plasma concentration and GFR.

Amount filtered [latex]= P \cdot GFR[/latex]

For example, say P = 130 mM and GFR = 122 ml/min.

Amount filtered [latex]= 130 mM \cdot 122 ml/min[/latex]

To calculate this value we need to be mindful of the units.  For this calculation we have mM (mmoles/l) and ml/min.  To get an accurate value we need to convert the ml/min into standard units, i.e. l/min, to match the mmoles/l.

So the actual calculation is 130 x (122 x10^-3) = 15.86 mmoles/min

Amount excreted

The amount of substance A excreted per unit time is calculated from the urine concentration and urine flow rate.

Amount excreted [latex]= U \cdot V[/latex]

For example, say U = 22 mM and V = 15 ml/min.

Amount  excreted [latex]= 22 mM \cdot 15 ml/min[/latex]

To calculate this value we need to be mindful of the units.  For this calculation we have mM (mmoles/l) and ml/min.  To get an accurate value we need to convert the ml/min into standard units, i.e. l/min, to match the mmoles/l.

So the actual calculation is 22 x (15 x10^-3) = 0.33 mmoles/min

Amount reabsorbed

The amount of substance A reabsorbed per unit time is the difference between the amount filtered and the amount excreted.

[latex]= (P \cdot GFR)  –  (U \cdot V)[/latex]

For the examples above:

The amount reabsorbed = 15.86 - 0.33 = 15.53 mmoles/min

If your amount reabsorbed is a negative number (i.e. the amount excreted is greater than the amount filtered) this tells you that the solute has been secreted by the nephron.