Filtered load
F = GFR × unbound plasma concentration. If given total plasma concentration, multiply by the unbound fraction first. GFR is a volume per time; filtered load is an amount per time.
SKILLS · BIOLOGY & BIOCHEMISTRY
Renal mass balance
Track amounts per minute. A urine measurement tells you what leaves the kidney; it does not reveal every movement along the way.
01 · THE IDEA
Excretion is the result of filtration plus secretion minus reabsorption.
E = F + S − R; therefore E − F = S − R
All four terms are rates, such as mg/min. This balance assumes no production, metabolism, or accumulation of the substance in the kidney. For a protein-bound compound, assume only its unbound portion is filtered.
Excreted load
E = urine concentration × urine flow. A higher concentration alone does not establish a higher excretion rate: the flow may have changed too.
Net is a difference
E − F > 0 means net secretion; E − F < 0 means net reabsorption. Neither result separately identifies gross secretion and gross reabsorption.
CHANGE ONE PART OF THE MODEL
Different gross rates, the same net result
Filtered load stays at 4 mg/min and excretion at 7 mg/min. Explore several possible pairs of secretion and reabsorption that fit those measurements.
Gross secretion3 mg/min
Gross reabsorption0 mg/min
Net addition3 mg/min
4 + 3 − 0 = 7 mg/min excreted
The slider shows compatible possibilities, not additional measurements. The net result alone cannot choose between them.
02 · WORKED EXAMPLES
Follow the setup, then the numbers
1. From measurements to a net rate
GFR is 80 mL/min. Total plasma X is 0.25 mg/mL and its unbound fraction is 0.20. Urine X is 3.5 mg/mL at a flow of 2 mL/min.
- Unbound X = 0.25 × 0.20 = 0.05 mg/mL.
- Filtered load = 80 × 0.05 = 4 mg/min.
- Excretion = 3.5 × 2 = 7 mg/min.
- Net tubular addition = 7 − 4 = 3 mg/min.
Carry this forward: The net addition is 3 mg/min; gross secretion could be larger if reabsorption also occurs.
2. Infer a filtration rate from a marker
Marker M is freely filtered and neither secreted, reabsorbed, produced, nor metabolized. Plasma M is 0.04 mg/mL; urine M is 1.6 mg/mL at 2 mL/min.
- Marker excretion = 1.6 × 2 = 3.2 mg/min.
- Marker clearance = 3.2 mg/min ÷ 0.04 mg/mL = 80 mL/min.
- Under the stated marker assumptions, its clearance equals GFR.
Carry this forward: The special marker assumptions justify calling its clearance GFR. Do not make that substitution for every compound.
03 · TRY IT IN DIFFERENT FORMS
Practice in different forms
For each question, identify the requested quantity and write the setup before calculating. Use a hint if needed, then choose an answer to see the explanation.
Untimed learning practice. These responses stay on this page and reset when you leave or reload; they do not change your test results.
QUESTION 1 / 4 · B/B
Only unbound compound filters
GFR is 120 mL/min. Total plasma X is 0.20 mg/mL, of which 25% is unbound. Only unbound X filters. What is the filtered load?
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TAKE INTO A FRESH PASSAGE
Your setup checklist
- Convert concentration to an amount-per-time rate.
- Use unbound plasma concentration for filtration when the model requires it.
- Calculate E − F and interpret its sign.
- State separately what is known about net transport and what remains unknown about gross transport.
Before calculating, explain what each term represents and what the requested answer would mean.
Concept references
All scenarios and questions here are original learning exercises. These references support the underlying concepts; the simplified assumptions are stated within each example.