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| Completely excreted by the kidneys in one pass, way of measuring RPF. |
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| Mass balance equation in the kidneys |
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Arterial content=Venous content + urine content
RPFa x Pa= RPFv x Pv + U x V |
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Signifies the minimal amount of arterial RPF required to produce the observed solute content in urine. This assumes that there is no venous return of solute.
High clearance signifies high urine excretion. Clearance is usually an imaginary small volume. In reality, the kidneys get a much larger blood supply than necessary to remove all of the solute.
Clearance= UV/Pa |
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Excreted completely in a single pass through the kidney, not reabsorbed into venous system.
Therefore, clearance equals RPF. PAH is a good way of measuring RPF. |
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| Creatinine and inulin are removed from the blood only by glomerular filtration. Therefore, clearance equals GFR. |
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| Filtration + secretion - reabsorption |
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| Concentration of the solute in the afferent arterioles is the same as in Bowman's Space. |
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| Fractional excretion (FE)= |
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Urinary excretion/ filtered load
UV/P x GFR
since clearance= UV/P
FE=C/GFR |
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| Fractional reabsorption (FR) |
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| Since solute must be reabsorbed or excreted, FR=1-FE |
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| Relation between clearance, GFR, and fractional excretion |
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1. When C=GFR, then FE=1.0
2. When there is excretion by filtration and net secretion
C>GFR, then FE>1.0
3. Excretion by filtration and net reabsorption
C4. No urinary excretion (for proteins or glucose)
C and FE=0 |
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| Time independent calculation of FE |
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