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A PROGRAMMED INTRODUCTION TO THE
QUANTITATION OF RENAL FUNCTION
by Nora Laiken, PhD
Many aspects of renal function can be quantitatively analyzed. In
fact, a firm understanding of the quantitation of renal function is an
essential prerequisite for the study of renal physiology and
pathophysiology. This lesson is designed to help you acquire such an
understanding and develop skills in solving typical renal function problems.
Table of Contents
A. Basic definitions and fundamental equations . . . . . . . . . 2
B. General excretion models and their quantitation. . . . . . . 7
C. Fractional excretion, fractional delivery, and
fractional reabsorption . . . . . . . . . . . . . 28
Revised, September 2008
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A. Basic definitions and fundamental equations.
1. The purpose of this section is to define several important terms used in the
quantitation of renal function and then to combine these terms to form a set
of fundamental equations that describe the basic functions of the kidneys.
2. PX represents the concentration of a substance X in plasma. Commonly
used units for PX include mg/ml, mg/dl (i.e., mg/100 ml; also called mg %),
mosmol/liter, mmol/liter (mM), and (for electrolytes) meq/liter. For
example, typical normal PX values for some important plasma constituents
are as follows:
PNa = 140 mM PG = 80 mg/dl (G=glucose, fasting value)
PK = 4 mM PCr= 1 mg/dl (Cr = creatinine)
PCl = 110 mM
PHCO3 = 24 mM
3. UX represents the concentration of a substance X in urine. Commonlyused units for UX, as for PX, are ____. Because UX for a given substance
can vary markedly depending upon the diet, fluid intake, and physiological
state of the individual, it is impossible to list meaningful normal values;
however, many examples will be encountered in the problems presented
throughout this lesson.
mg/ml, mg/dl,mosmol/liter,
mmol/liter (mM),
meq/liter
4. V represents the volume of urine excreted per unit time, i.e., the urine
flow rate.1 Commonly used units for V are ml/min or liters/day. Although
V can vary from less than 0.5 ml/min to greater than 15 ml/min, depending
upon the diet, fluid intake, and physiological state of the individual, a
typical normal value for V
is 1 ml/min or 1.5 liters/day.
5. GFR represents the glomerular filtration rate and is defined as the
volume of plasma filtered (i.e., the volume of plasma passing from the
glomerular capillaries into Bowmans space) per unit time. Commonly used
units for GFR are ml/min and liters/day. In humans, a typical normal GFR
is 125 ml/min or 180 liters/day.
__________1Frequently, the symbol V is used instead of V, or V is called the urine volume instead of the urine flow rate. However,
the symbol V and the term urine flow rate are preferable.
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6. RPF represents the renal plasma flow and is defined as the volume of
plasma flowing to the kidneys through the renal arteries per unit time.1
Commonly used units for RPF, as for GFR, are _____ . In humans, a
typical normal RPF is 625 ml/min or 900 liters/day.
ml/min, liters/day
7. FF represents the filtration fraction and is defined as the fraction of the
RPF that is filtered into Bowmans space. Thus
FF = GFR (A-1)
RPF
Using the typical normal values for GFR and RPF listed in questions 5 and
6, a typical normal FF is ____. 0.2
8. TX represents the rate of tubular transport and is defined as the amount
of a substance X transported across the tubular epithelium per unit
time. Commonly used units for TX are mg/min, mmol/min, or (for
electrolytes) meq/min. Note that the term TX is used foreitherthe rate ofreabsorption (i.e., the rate of transport from the _____ fluid to the _____
capillaries) orthe rate of secretion (i.e., the rate of transport from _____
capillaries to the _____ fluid); the direction of transport must be specified
when the parameter is used, if not obvious from the context (sometimes, TXrand TXs are used to represent the rate of tubular reabsorption and the rate of
tubular secretion, respectively). TX values for several substances will be
calculated in Section B.
tubular
peritubular
peritubular
tubular
__________1RPF should not be confused with RBF, the renal blood flow. RPF and RBF are related as follows:
RBF = RPF
1 - Hct
where Hct represents the hematocrit.
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9. Summary of basic terms used in the quantitation of renal function:
Definition Commonly Used Units Typical Normal Values
PX Concentration of X in plasma mg/ml, mg/dl, mmol/liter (mM),
mosmol/liter, meq/liter
PNa = 140 mM, PK= 4 mM,
PCl = 110 mM, PHCO3 = 24 mM
UX Concentration of X in urine mg/ml, mg/dl, mmol/liter (mM),mosmol/liter, meq/liter
Varies markedly with diet, fluidintake, physiological state
V Urine flow rate ml/min, liters/day Varies markedly with diet, fluid
intake, physiological state, but
typically 1 ml/min or 1.5
liters/day
GFR Glomerular filtration rate ml/min, liters/day 125 ml/min or 180 liters/day
RPF Renal plasma flow ml/min, liters/day 625 ml/min or 900 liters/day
FF Filtration fraction 0.2
TX Rate of tubular transport(reabsorption orsecretion)
mg/min, mmol/min, meq/min See Section B for examples
In the remainder of this section, the above terms will be combined to form a set
of fundamental equations that describe the basic functions of the kidneys.
10. For any substance X that is freely filtered from the glomerular capillaries
into Bowmans space,1 the amount filtered per unit time is given by
Amount filtered per unit time = PXGFR (A-2)
The amount filtered per unit time often is loosely referred to as the amount
filtered. However, in this lesson, the term filtered load will be used when
referring to the quantity PXGFR.
__________1A freely filtered substance is one whose concentration in the filtrate is equal to its concentration in plasma
([filtrate]/[plasma] ratio = 1.0). Freely filtered substances include any water-soluble molecule that has the following two
properties: (1) it has a molecular weight < 5000 Da; as the molecular weight increases above 5000 Da, the
[filtrate]/[plasma] ratio gradually decreases (for example, myoglobin {17,000 Da} has a [filtrate]/[plasma] ratio of 0.75,
while serum albumin {69,000 Da} has a [filtrate]/[plasma] ratio of 0.01); and (2) it is not significantly bound to plasma
proteins. In general, if a substance is said to be filtered, it can be assumed that the substance is freely filtered unless
otherwise noted.
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11. For any substance X, the amount excreted per unit time is given by1
Amount excreted per unit time = UXV (A-3)
The amount excreted per unit time often is loosely referred to as the
amount excreted. However, in this lesson, the termrate of excretion will
be used when referring to the quantity UXV.
12. The clearance of any substance X, CX, is defined as
CX = UXV (A-4)
PX
Note that this definition of CX is valid for all X exceptH2O and electrolyte-
free H2O (EFW). As the clearance concept is an important one in renal
physiology, it will be discussed in more detail in the following question.
13. You have learned that the rate of excretion of a substance X is given byUXV
. Note that UXV describes the renal excretion of X from the urines
perspective. For example, if UX = 80 mg/ml and V = 0.5 ml/min, from the
urines perspective the renal excretion of X is best described as _____ . But
from the plasmas perspective, it might be more relevant to ask what
volume of plasma supplied the X that was excreted. Since volume =
amount/concentration, then:
Volume of plasma needed per = Amount of X excreted per unit time
unit time to supply excreted X Plasma concentration of X
= UXV
PX
For example, if 40 mg of X is excreted per min (UXV = 40 mg/min) and PX
= 2 mg/ml, then the volume of plasma needed per min to supply the
excreted X is _____. Note that the volume of plasma that supplies the
excreted X is thereby stripped orcleared of its X, i.e.,
Volumeof plasma needed per = UXV = Volume of plasma cleared
unit time to supply excreted X PX of X per unit time
Thus, the quantity UXV/PXrepresents the volume of plasma cleared of X
per unit time and is called the clearance of X (CX; cf. equation A-4). Note
that since CX represents a volume per unit time, its units are _____.
UXV = 40 mg/min
20 ml/min
ml/min, liters/day
__________1It is important to distinguish between excretion (the elimination of a substance in urine) and
secretion (the transport or diffusion of a substance from the peritubular capillaries or
epithelial cells into the tubular fluid).
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14. To consolidate your understanding of the concepts presented in this section,
complete the following table. The data were obtained from a patient with
GFR = 120 ml/min and V= 1 ml/min.
Filtered Rate of
PX UX Load Excretion CX
PXGFR UXV CX
mmol/min ml/min
or
mg/min
Na+ 140 mmol/liter 70 mmol/liter 16.8 0.07 0.5
K+ 4 mmol/liter 35 mmol/liter 0.48 0.035
8.75
Cl- 110 mmol/liter 70 mmol/liter 13.2 0.07 0.64
HCO3- 25 mmol/liter 1.4 mmol/liter 3.0 0.00
1
0.056
Creatinine 1 mg/dl 125 mg/dl 1.2 1.25 125
Glucose 70 mg/dl 0 84 0 0
Urea1 30 mg/dl 1500 mg/dl 36 15 501Actually, PUrea and UUrea are most commonly expressed as blood urea nitrogen (BUN) and urine urea
nitrogen (UUN), respectively. Urea has a mw of 60 Da, of which 28 is nitrogen, i.e., on a weight basis,
urea is 2 nitrogen. Thus, PUrea = 30 mg/dl is equivalent to BUN 15 mg/dl, while UUrea = 1500
mg/dl is equivalent to UUN 750 mg/dl.
15. In Section B of this lesson, the handling of certain substances by the kidney
will be summarized using graphs that illustrate the effect of plasma
concentration (PX) on four of the important quantities defined in this
section. Specifically, for each of these substances, fourrenal function vs.
PX curves will be shown:
(1) Filtered load ( ____ ) vs. PX
(2) Rate of excretion ( ______ ) vs. PX
(3) Rate of tubular transport ( ____ ) vs. PX
(4) Clearance ( ____ ) vs. PX
Note that the filtered load, rate of excretion, and rate of tubular transport all
share the same units (e.g., ____ ). Thus, the first three curves generally are
plotted on a single graph. It also should be noted that since these curves
typically are plotted at constant GFR, the filtered load vs. PX curve is a
straight line with slope ____ .1
PXGFR
UXV
TX
CX
mg/min, mmol/min
GFR
__________1This may not be true if the substance is bound to plasma proteins, as the extent of binding will depend on the plasma
concentration of the substance. However, this effect of protein binding will not be considered here.
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B. General excretion models and their quantitation.
1. Different substances are handled by the kidney in different ways. On the
macroscopic level, five general excretion models are possible:
(1) No filtration or secretion(2) Filtration only; no reabsorption or secretion
(3) Filtration and reabsorption
(4) Filtration and secretion
(5) Filtration, reabsorption, and secretion
Each of these models will be described in this section. More specifically,
for each model equations for the rate of excretion and clearance will be
derived. In addition, renal function vs. PX curves will be shown for several
important substances.
2. No filtration or secretion. A substance that is neither filtered nor added to
the tubular fluid by secretion cannot be excreted. Therefore,
Rate of excretion = UXV = 0 (B-1)
and
CX = UXV = 0 (B-2)
PX
Such substances are either (i) too large to cross the filtration barrier (e.g.,
the [filtrate]/[plasma] ratio for serum albumin, mw 69,000 Da, is 0.01; see
footnote 1, p 4); or (ii) tightly bound to plasma proteins (e.g., unconjugated
bilirubin is virtually absent from tubular fluid and urine because of its tightbinding to serum albumin).
3. Filtrationonly, no reabsorption or secretion. For a substance that is
freely filtered1 but neither reabsorbed nor secreted
Rate of excretion = Filtered load
i.e.,
UXV = PXGFR (B-3)
Examples of such substances are inulin (a fructose polymer with mw 5000
Da) and mannitol (a hexose that is neither found in nor metabolized by the
body).__________1The term freely filtered is defined in footnote 1, p 4.
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4. You have learned that the clearance of a substance X, CX, is defined by the
equation CX = UXV/PX. However, for a substance that is freely filtered but
neither reabsorbed nor secreted, such as inulin (equation B-3)
UInV = PIn GFR
so that
CIn = UInV/PIn = PIn GFR/PIn
or
CIn = GFR (B-4)
Equation B-4 makes sense: if inulin is freely filtered but neither reabsorbed
nor secreted, the volume of plasma that is cleared of inulin per unit time
(CIn) will equal the volume of plasma that is filtered per unit time (GFR); no
inulin is addedto the tubular fluid by secretion (in which case the volume
of plasma cleared of inulin per unit time would be greaterthan the GFR)
and no inulin is removedfrom the tubular fluid by reabsorption and returned
to the plasma (in which case the volume of plasma cleared of inulin perunit time would be less than the GFR). In fact, in quantitative discussions of
renal function, GFR and CIn often are used interchangeably. Furthermore,
since CIn can be obtained from readily measurable parameters, _____ , this
means that GFR can be measured noninvasively.1
UIn, V, PIn
5. GFR can be equated to the ____ of any substance that is ____ ____ , but
neither _____ nor _____ , such as the polysaccharide ____ , i.e., GFR =
____ . However, this polysaccharide must be infused IV in order to obtain
the latter quantity and therefore is not used routinely to determine GFR in
humans. Instead, for clinical purposes, GFR is estimated using the
endogenous substance creatinine, as described below (see question 20).
clearance
freely filtered
reabsorbed
secreted
inulin
CIn
6. The renal handling of a substance that is freely filtered but neither reab-
sorbed nor secreted can be conveniently summarized with renal function vs.
PX curves (question A-15), using inulin as an example. The data needed to
construct these curves are obtained by completing the table below:
__________1In addition to being freely filtered but neither reabsorbed nor secreted, inulin is
physiologically inert (most importantly, it does not alter renal function) and is easily
measured. Any substance used for the determination of GFR must satisfy these criteria as
well as the criterion of being freely filtered but neither reabsorbed nor secreted.
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PIn UIn V
CIn (= GFR) Filtered Rate of
mg/dl mg/ml ml/min Load Excretion
50 60 1.0
100 160 0.75
150 225 0.8
200 240 1.0
150
100
50Clearance(ml/min)
10050 150 200
PIn (mg/dl)
200
100
50RateofE
xcretion
(mg/min)
10050 150 200
PIn (mg/dl)
150
FilteredLoad(mg/min)
CIn PInGFR UInV
ml/min mg/min mg/min
120 60 60
120 120 120
120 180 180
120 240 240
(1) Filtered load (= ____ ) vs. PIn. Since GFR (= CIn) is constant, this
plot is a ____ ____ with slope ____ (see question A-15).
(2) Rate of excretion (= ____ ) vs. PIn. Since the rate of excretion
equals the ____ ____ for a substance that is filtered only (equation
B-3), this curve is identical to the _____ curve.
(3) Clearance vs. PIn. Since CIn = GFR and GFR is constant, CIn is
independentof PIn.
PInGFR
straight line
GFR (120 ml/min)
UInV
filtered load
filtered load vs. PIn
7. Filtration and reabsorption. For a substance that is filtered and
reabsorbed
Rate of excretion = Filtered load Rate of tubular reabsorption
so that
UXV
= ____
____
Examples of such substances are glucose, amino acids, sodium, chloride,
and phosphate.
PXGFRTX
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8. For a substance that is filtered and reabsorbed
UXV = PXGFR TX (B-5)
Since CX = UXV/PX, this means that
CX = PXGFR TXPX
= GFR TX /PX
or CX = CIn TX /PX (B-6)
Note, therefore, that for a substance that is filtered and reabsorbed
CX < CIn (B-7)
This result is expected: if a substance is filtered and reabsorbed (i.e., ifsome
of the filtered substance is returned to the _____ ), a smaller volume of____ will be cleared per unit time compared to the volume cleared if a
substance is filtered only (so that none of the filtered substance is returned
to the _____ ).
plasmaplasma
plasma
9. While a detailed description of the reabsorption processes in the kidney is
beyond the scope of this lesson, it is important to note that for many
reabsorbed substances, the reabsorption process is Tm-limited, i.e., it has a
maximum rate Tm (Tm = transport maximum).1 Glucose, amino acids,
and phosphate are among the substances whose reabsorption is Tm-limited.
Furthermore, many important substances that exhibit Tm-limited
reabsorption (including glucose, amino acids, and phosphate) also exhibit
the so-called threshold phenomenon. This means that the affinity of thetransport mechanism for the substance is so high that the reabsorption of the
substance from the tubular fluid is virtually complete when the filtered load
does notexceed Tm, i.e., when PXGFR TmX, then the entire filtered load
is reabsorbed
TX = PXGFR (B-8)
so that
UXV = PXGFR TX = PXGFR PXGFR
i.e., UXV = 0 (B-9)
which means that
__________1It is important to note that the Tm for some substances is subject to physiological regulation, i.e., only under a given
set of physiological conditions can it be regarded as afixedmaximum rate of reabsorption. For example, the Tm for
phosphate is decreasedby parathyroid hormone (PTH).
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CX = UXV
= 0 (B-10)
PX
When the filtered load exceeds Tm (i.e., when the transport mechanism is
saturated), the excess load will be excreted. Thus, when PXGFR > TmX, the
rate of reabsorption TX is fixed at its maximum value TmX
TX = TmX (B-11)
and the remaining X is excreted
UXV= PXGFR TmX (B-12)
so that CX = UXV = PXGFR TmX
PX PX
= GFR TmX/ PX
or
CX = CIn TmX/ PX (B-13)
i.e., some _____ is cleared of X (CX is _____ than 0). However, because
some X is returned to the _____ by _____ , the volume of _____ cleared
per unit time is ____ than the volume of _____ filtered per unit time (CX is
_____ than CIn [GFR]). Note that it is possible to define a plasma
concentration PThX, the so-called renal threshold for X, at which the
substance first appears in the urine. Theoretically (see question 11 below),
PThX represents that plasma concentration at which the filtered load exactly
saturates the transport mechanism (for a given GFR)
PThXGFR = TmX
orPThX = TmX/GFR (B-14)
The normal plasma concentrations of glucose and amino acids are well
below their respective renal thresholds, so that these substances normally
are absentfrom urine (see question 10 below). In contrast, the normal
plasma phosphate concentration generally is above the renal threshold for
phosphate, so that some phosphate normally is excretedin the urine (see
question 12 below).1
plasma
greater
plasma
reabsorption
plasma
less
plasma
less
__________1Since PTH decreases Tm for phosphate (see footnote 1, p 10), it also decreases PTh for
phosphate.
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10. The renal handling of substances that exhibit Tm-limited reabsorption and
the threshold phenomenon can be conveniently summarized with renal
function vs. PX curves, using glucose as an example. The data needed to
construct these curves can be obtained by completing the table below.
During the time period that the data were obtained, CIn was noted to have a
constant value of 125 ml/min.
PG UG V Filtered Rate of Rate of CG
mg/dl mg/ml ml/min Load Excretion Tubular
Reabs
50 0 1.0
150 0 0.8
250 0 0.8
300 tr. 0.75
350 70 0.9
400 125 1.0
500 250 1.0
150
100
50Clearan
ce(ml/min)
200 400 600
PG
(mg/dl)
400
200
100RateofExcr
etion
(mg/min)
PG
(mg/dl)
300
FilteredL
oad(mg/min)
RateofTubularR
eabsorption
(mg/min)
Inulin
200 400 600
500
600
nl
PXGFR UXV
TX CG
mg/min mg/min mg/min ml/min
63 0 63 0
188 0 188 0
313 0 313 0
375 Tr. 375 tr.
438 63 375 18
500 125 375 31
625 250 375 50
(1) Filtered load vs. PG. Since GFR (= CIn) is constant, a ____ ____ is
obtained with slope ____ (see question A-15).
straight line
GFR
(2) Rate of excretion vs. PG.
(3) Rate of tubular reabsorption vs. PG.
Taken together, curves (2) and (3) illustrate both the Tm-limitednature of glucose reabsorption and the threshold phenomenon:
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For PG ____ mg/dl, the transport mechanism (with its high
____ for glucose) is able to reabsorb the entire filtered load,
i.e., TG = ____ and UGV = ____ (equations B-8 and B-9).
For PG > ____ mg/dl, the filtered load exceeds the maximum
rate of tubular reabsorption, TmG (= ____ ), and the excessfiltered load is excreted. PG = ____ mg/dl therefore represents
PThG, the ____ ____ for glucose at this GFR. Note that for PG >
PThG, the rate of excretion vs. PG curve parallels the ______
curve, i.e., it is a ____ ____ with slope ____ , as predicted by
equation B-12.
(4) Clearance vs. PG.
For PG PThG, CG = ____ (equation B-10).
For PG > PThG, CG increases, and if additional data were
available CG would asymptotically approach CIn (which has
been included on the graph as a reference line). This ispredicted by equation B-13 and makes sense: when PG is very
high, the filtered load PGGFR is so much greater than TmG that
the rate of excretion UGV is approximately equal to PGGFR, as
predicted by equation B-12 when PGGFR >>> TmG. Thus, at
very high PG, glucose behaves approximately like a substance
such as inulin that is ______ only, and CG approaches _____.
Note that when PG is in the normal range, indicated by the bracket below the
abscissa, PG < PThG, so that UG = ____ and CG = ____ (equations B-9 and B-
10).
300 mg/dl
affinity
PGGFR
0
300 mg/dl
375 mg/min300 mg/dl
renal threshold
filtered load vs. PG
straight line
GFR
0
filtered
CIn
0
0
11. It should be noted that the data and glucose titration curves presented in theprevious question are somewhat idealized, as the curves for rate of excretion
vs. PG, rate of tubular reabsorption vs. PG, and clearance vs. PG show an
abruptchange when the rate of tubular reabsorption reaches its maximum
value. Actual glucose titration curves show a more gradual change in this
region:
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150
100
50Clearance(ml/min)
200 400 600
PG (mg/dl)
400
200
100RateofExcretion
(mg/m
in)
PG (mg/dl)
300
FilteredLoad(mg/min)
RateofTubularReabsorption
(mg/min)
Inulin
200 400 600
500
600
This feature of renal titration curves is called splay.1 Note that because of splay, PThG (defined as the
value of PG at which glucose first appears in urine) is 200 mg/dl insteadof 300 mg/dl, even though
the transport mechanism does notbecome saturated (i.e., TG does notreach TmG) until PG 300
mg/dl (see question 10). PG = 300 mg/dl therefore only can be regarded as the theoretical renalthreshold for glucose.
__________1Two explanations for splay commonly are offered:
(1) Reaction of glucose with the Na+-glucose cotransporter. Using the symbol R for the Na+-glucose cotransporter,
the reaction between glucose and the cotransporter can be written as
GR G + R
and
K = [G] [R]
[GR]
where K is the dissociation constant for the glucose-cotransporter complex. Even if K is very small (i.e., if the
affinity of the cotransporter for glucose is very large), a finite concentration of glucose must be present in thetubular fluid in order to saturate the cotransporter. Hence, some glucose will be excreted before the theoretical
renal threshold (= TmG /GFR; see equation B-14) is reached.
(2) Nephron heterogeneity. Considerable heterogeneity in the filtration capacity of glomeruli and the reabsorption
capacity of renal tubules is believed to exist. For example, some nephrons may produce average volumes of
filtrate but have subnormal reabsorption capacities; these nephrons are likely to excrete glucose at a lower PG
than the average nephron. In contrast, other nephrons may produce small volumes of filtrate but have high
reabsorption capacities; these nephrons are likely to excrete glucose at a higher PG than the average nephron.
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12. To consolidate your understanding of Tm-limited
reabsorption and the threshold phenomenon, use the
data below to construct renal titration curves for
phosphate. During the time period that the data
were obtained, CIn was noted to have the constant
value of 120 ml/min.
PP UP V Filtered Rate of Rate of CP
mmol/l mmol/l ml/min Load Excretion Tubular
Reabs
0.2 0 0.8
0.4 0 0.75
0.6 tr. 1.0
0.7 7.5 0.8
0.8 17.3 0.75
1.0 25 1.0
1.2 40 1.1
1.5 66.7 1.2
2.0 140 1.0
PPGFR UPV TP CP
mmol/min mmol/min mmol/min ml/min
0.024 0 0.024 0
0.048 0 0.048 0
0.072 tr. 0.072 tr.
0.084 0.006 0.078 8.6
0.096 0.013 0.083 16
0.120 0.025 0.095 25
0.144 0.044 0.1 37
0.180 0.080 0.1 53
0.240 0.140 0.1 70
150
100
50Clearance(ml/min)
1.0 2.0
PP (mmol/liter)
0.2
0.1
RateofExcretion
(mmol/min)
PP (mmol/liter)
FilteredLoad(mmol/min)
RateofTubularReabsorption
(mmol/min)
Inulin
1.0 2.0
0.3
nl
150
100
50Clearance(ml/min)
1.0 2.0
PP (mmol/liter)
0.2
0.1
RateofExcretion
(mmol/min)
PP (mmol/liter)
FilteredLoad(mmol/min)
RateofTubularReabsorption
(mmol/min)
Inulin
1.0 2.0
0.3
nl
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The basic features of the curves are similar to those for glucose (questions
10 and 11). PThP, the plasma phosphate concentration at which phosphate
first appears in the ____ , is approximately ____ mmol/liter, while TmP , the
_____ , is approximately ____ . Like the glucose titration curves in
question 11, and in contrast to the idealized curves in question 10, the
changes in the region at which the rate of tubular reabsorption reaches its
maximum value are gradual, i.e., the curves exhibit ____ ; thus, phosphateappears in the ____before the rate of tubular reabsorption reaches TmP.
Note that when PP is in the normal range, indicated by the bracket below the
abscissa, PP > PThP, so that UPV > ____ and CP > ____ (equations B-12 and
B-13).
urine
0.6 mmol/liter
transport maximum
for phosphate
0.1 mmol/minsplay
urine
0
0
13. Filtration and secretion. For a substance that is filtered and secreted
Rate of excretion = Filtered load + Rate of tubular secretion
i.e.,1
UXV
= ____ + ____
Examples of such substances are organic acids (e.g., penicillin, para-aminohippuric acid [PAH], furosemide, uric acid) and organic bases (e.g.,
creatinine, cimetidine, trimethoprim).2
PXGFR
TX
14. For a substance that is filtered and secreted
UXV = PXGFR + TX (B-15)
Since CX = UXV/PX, this means that
CX = PXGFR + TX
PX
= GFR + TX /PX
or
CX = CIn + TX /PX (B-16)
__________1Note that TX is used to represent the rate of tubular reabsorption (question 7) as well as the rate of tubular secretion (see
question A-8). The direction of transport must be specified when the parameter is used, if not obvious from the context.2It should be noted that many substances that are secreted are partially bound to plasma proteins and therefore are not
freely filtered (see footnote 1, p 4). For such substances, the filtered load is more accurately given by PXGFRF, where F
is the fraction of the substance in plasma that is freely filterable (i.e., unbound). However, since the purpose of this
lesson is to provide a general introduction to the quantitation of renal function, the effects of protein binding will not be
considered further here.
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Thus, for a substance that is filtered and secreted
CX > CIn (B-17)
This result is expected: if a substance is filtered and secreted (i.e., if some of
the substance that was notfiltered from the _____ capillaries into _____
space leaves the _____ capillaries and is added to the _____ _____ ), alarger volume of ____ will be cleared per unit time compared to the
volume cleared if a substance is filtered only.
glomerular
Bowmansperitubular
tubular fluid
plasma
15. A detailed description of the secretory processes in the kidney is beyond the
scope of this lesson. However, it should be noted that certain secreted
substances exhibit Tm-limited secretion, a phenomenon that is completely
analogous to Tm-limited reabsorption; such substances have a maximum
rate of secretion, Tm (Tm = ____ ____ ).1 PAH and creatinine are
examples of substances whose secretion is Tm-limited. Some secreted
substances, most notably PAH, also exhibit a phenomenon that is somewhat
analogous to the threshold phenomenon for reabsorption (question 9).Specifically, for a substance like PAH, the affinity of the transport
mechanism for the substance is so high that essentially all of the substance
in the peritubular capillaries is secreted into the tubular fluid as long as the
rate of delivery of the substance to the peritubular capillaries does not
exceed Tm. As a first approximation (see question 16 below), any part of
the RPF that does not get filtered enters the efferent arterioles and
peritubular capillaries, so that the rate of plasma flow through the
peritubular capillaries can be equated to RPF GFR and the rate of delivery
of a substance to the peritubular capillaries can be equated to PX(RPF
GFR). Thus, for a substance like PAH, as long as PX(RPF GFR) does not
exceed Tm, the entire amount delivered to the peritubular capillaries is
secreted, i.e.,
TX = PX (RPF GFR) (B-18)
Therefore
UXV = PXGFR + TX = PXGFR + PX (RPF GFR)
= PXRPF (B-19)
and
CX = UXV = RPF (B-20)
PX
i.e., the entire RPF is cleared of X. Equation B-20 makes sense: ifessentially all of the X delivered to the peritubular capillaries is secreted,
transport maximum
__________1Note that the symbol Tm is used to represent the transport maximum for reabsorption (question 9)
as well as the transport maximum for secretion.
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any plasma that is not cleared of X by filtration will be cleared of X by
secretion. In contrast, if PX(RPF GFR) exceeds Tm, so that the transport
mechanism is ____
TX = TmX (B-21)
so that
UXV = PXGFR + TmX (B-22)and1
CX = CIn + TmX/PX (B-23)
saturated
Because some X is secreted, the volume of _____ cleared of X per unit
time is ____ than the volume of _____ filtered per unit time (CX is _____
than CIn [GFR]). However, because some X remains in the peritubular
capillaries, the RPF is notcompletely cleared of X (CX is _____ than
RPF).
plasma
greater
plasma
greater
less
16. In the previous question, it was noted that the ____ of the transport
mechanism for substances such as PAH is so high that essentially all of the
substance in the ____ ____ is secreted into the ____ ____ as long as thetransport mechanism is not ____ . It was shown that under such conditions
CX = RPF (equation B-20), or, considering PAH as an example
CPAH = RPF (B-24)
As noted above, equation B-24 makes sense: if essentiallyall of the PAH in
the ____ ____ is secreted into the ____ ____ . any plasma that is not
cleared of PAH by filtration should be cleared of PAH by _____ , so
that the entire _____ is cleared of PAH. According to equation B-24, it
should be possible to determine the RPF noninvasively, from readily
measurable parameters ( _____ ). However, in actuality the RPF is not
completely cleared of PAH, even when the transport mechanism is not _____ , i.e., equation B-24 should be written as CPAH RPF. This is
because 10% of the RPF supplies portions of the kidney that cannot
remove PAH (e.g., the renal medulla [via the vasa recta capillaries], capsule,
and pelvis, the perirenal fat), i.e., 10% of the RPF does notflow through
the glomerular and/or peritubular capillaries and therefore cannotbe
cleared of PAH by filtration and secretion. The rate of plasma flow to
regions of the kidney that can remove PAH often is termed the effective
renal plasma flow (ERPF). Thus, the rate of plasma flow through the
peritubular capillaries is most accurately expressed as ERPF GFR, i.e.,
when the rate of PAH delivery does not exceed Tm (compare to equations
B-18 to B-20)
affinity
peritubular capstubular fluid
saturated
peritubular caps
tubular fluid
secretion
RPF
UPAH, V, PPAH
saturated
__________1While it is possible to define a plasma concentration at which the rate of delivery to the peritubular capillaries exactly
saturates the transport mechanism (at a given GFR and RPF), in analogy to PThX (equation B-14), this seldom is done, as
the term renal threshold refers specifically to the plasma concentration at which a substance first appears in the urine.
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TPAH = PPAH (ERPF GFR) (B-25)
so that
UPAHV = PPAH ERPF (B-26)
and
CPAH = ERPF (B-27)
In other words, while CPAH provides a good estimate of the RPF, it actuallymeasures the ERPF. For clinical purposes, the approximate RPF values
obtained by CPAH measurements are quite useful. If an accurate
determination of RPF is needed, the Fick principle can be used, as discussed
in the following question.
17. Because the rate of plasma flow through the peritubular capillaries is most
accurately expressed as ERPF GFR, where ERPF represents the ____
RPF, CPAH provides an accurate measurement of ERPF (as long as the
transport mechanism is not ____ ), but only an estimate of RPF. While
such an estimate is quite useful for clinical purposes, it is possible to obtain
an accurate RPF value with the Fick principle
RPF = UPAH V (B-28)
PPAH PRV,PAH
where PRV,PAH represents the concentration of PAH in the renal vein, so that
PPAH PRV,PAH represents the renal arterio-venous plasma concentration
difference for PAH.1 Thus, since ERPF = CPAH = UPAHV/PPAH
ERPF = RPF (PPAH PRV,PAH) (B-29)
PPAH
or
ERPF = RPF EPAH (B-30)
where EPAH represents the extraction ratio for PAH, defined as (PPAH
PRV,PAH)/PPAH. When the transport mechanism is not saturated, EPAH 0.90.2
effective
saturated
__________1Note that the renal vein must be catheterized to obtain PRV,PAH . Also note that many substances
other than PAH can be used when determining RPF with the Fick principle.2If PAH were completely cleared or extracted by the kidneys, PRV,PAH would equal 0 (i.e., PAH
would be completely absent from the renal vein) and EPAH would equal 1.0 .
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18. The renal handling of substances that exhibit Tm-limited secretion and that are almost completely
removed from the peritubular capillaries if the transport mechanism is not saturated can be
conveniently summarized with renal function vs. PX curves, using PAH as an example. The data
needed to construct these curves can be obtained by completing the table below. Data also are
included to allow the rate of PAH delivery to the peritubular capillaries to be calculated. During the
time period that the data were obtained, CIn was noted to have a constant value of 120 ml/min.
[Important note : Answers are provided at the bottom of the page, notin an answer column along theright margin, and the renal function vs. PX curves are shown on p 21.]
PPAH UPAH V Filtered Rate of Rate of CPAH ERPF Rate of PAH
mg/dl mg/ml ml/min Load Excretion Tubular Delivery to
Secretion Perit Caps
4 40 0.6
8 60 0.8
12 72 1.0
16 120 0.8
18 102 1.0 - - 1 - - 1
20 130 0.8 - - - -
24 136 0.8 - - - -
30 116 1.0 - - - -1The ERPF can be equated to CPAHonly when the transport mechanism is notsaturated. Hence, ERPF and the rate of
PAH delivery to the peritubular capillaries cannot be calculated from the data provided when PPAH > 16 mg/dl.
_________________________________________________________________________________Answers
Filtered Rate of Rate of CPAH ERPF Rate of PAH
Load Excretion Tubular ml/min ml/min Delivery to
mg/min mg/min Secretion (= CPAH) Perit Caps
mg/min mg/min
PPAH(ERPF-GFR)
4.8 24 19 600 600 19
9.6 48 38 600 600 38
14 72 58 600 600 58
19 96 77 600 600 77
22 102 80 567 - - - -
24 104 80 520 - - - -
29 109 80 454 - - - -
36 116 80 387 - - - -
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600
400
200Clearance(m
l/min)
PPAH (mg/dl)
80
40
RateofExcretion
(mg/min)
PPAH (mg/dl)
FilteredLoad(m
g/min)
RateofTubularSecr
etion
(mg/min)
Inulin
10 20
120
30 10 20 30
The important features of these renal function vs PX curves are as follows:
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(1) Filtered load vs. PPAH. Since GFR (= CIn) is constant, a ____ ____
is obtained with slope ____ .
(2) Rate of excretion vs. PPAH.
(3) Rate of tubular secretion vs. PPAH.
Taken together, curves (2) and (3) illustrate the Tm-limited nature
of PAH secretion and the ability of the transport mechanism toremove essentially all of the PAH delivered to the _____ capillaries
if it is not saturated:
For PPAH _____mg/dl, the transport mechanism (with its high
____ for PAH) is able to secrete essentially all of the PAH
delivered to the _____ capillaries. Thus, TPAH (calculated from
the data as follows: _____ ) equals the rate of PAH delivery to
the _____ capillaries (calculated from the data as follows:
_____ ).
For PPAH > _____ mg/dl, the rate of PAH delivery to the _____
capillaries exceeds the maximum rate of tubular secretion,
TmPAH ( = ____ ). PPAH ____ mg/dl therefore represents theplasma PAH concentration at which the transport mechanismis
saturated at this GFR and ERPF. Note that for PPAH > ____
mg/dl, the rate of excretion vs. PPAH curve parallels the _____
curve, i.e., it is a _____ _____ with slope ____ , as predicted
by equation B-22.
(4) Clearance vs. PPAH.
For PPAH ____ mg/dl, CPAH = ERPF (equation B-27).
For PPAH > ____ mg/dl, CPAH decreases, and if additional data
were available CPAH would asymptotically approach CIn (which
has been included on the graph as a reference line). This ispredicted by equation B-23 and makes sense: when PPAH is
very high, the filtered load PPAHGFR is so much greater than
TmPAH that the rate of excretion UPAHV is approximately equal
to PPAHGFR, as predicted by equation B-22 when PPAH >>>
TmPAH. Thus, at very high PPAH, PAH behaves approximately
like a substance such as inulin that is _____ only, i.e., CPAHapproaches _____.
It should be noted that the above curves exhibit a gradual change as the rate
of tubular secretion reaches its maximum level. An analogous gradual
change was seen in the renal titration curves for glucose (question 11) and
phosphate (question 12) and is called ____ . The latter term also can beapplied to the PAH curves.1
straight line
GFR
peritubular
16 mg/dl
affinity
peritubular
UPAHV PPAHGFR
peritubular
PPAH(ERPF GFR)
16 mg/dl
peritubular
80 mg/min
16 mg/dl
filtered load vs.
PPAH
straight line
GFR
16 mg/dl
16 mg/dl
filtered
CIn
splay
__________1The explanations offered for the splay in the PAH curves are analogous to those presented in
footnote 1, p 14 for the splay in the glucose titration curves.
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19. To consolidate your understanding of Tm-limited secretion and the removal
of secreted substances from the peritubular capillaries, use the data below to
construct renal titration curves for diodrast, a contrast agent used in
radiology whose renal handling is similar to PAH. During the time period
that the data were obtained, CIn was noted to have a constant value of 125
ml/min. [Important note: Answers are provided at the bottom ofpage 24,notin an answer column along the right margin.]
PD UD V Filtered Rate of Rate of CD ERPF Rate of D
mg/dl mg/ml ml/min Load Excretion Tubular Delivery to
Secretion Perit Caps
4 25 1.0
6 37.5 1.0
8 62.5 0.8
10 60.5 1.0 - - 1 - - 1
12 74.4 0.9 - - - -
16 77 1.0 - - - -
20 102.5 0.8 - - - -
30 94.5 1.0 - - - -1The ERPF can be equated to CDonly when the transport mechanism is notsaturated. Hence, ERPF and the rate of
diodrast delivery to the peritubular capillaries cannot be calculated from the data provided when PD > 8 mg/dl.
600
400
200Clearance(ml/min)
PD (mg/dl)
80
40
RateofExcretion
(mg/min)
PD (mg/dl)
FilteredLoad(mg/min)
RateofTubularSecretion
(mg/min)
Inulin
10 20 30 10 20 30
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The basic features of the curves are similar to those for PAH (question 18).
TmD, the ______ , is approximately ____ , and the transport mechanism
becomes saturated at PD ____ mg/dl. That the transport mechanism is
able to remove essentially all of the diodrast from the ____ ____ when the
transport mechanism is not _____ is shown by the fact that TD (calculated
from the data as follows: _____ ) equals the rate of diodrast delivery to the
_____ _____ (calculated from the data as follows: _____ ). Like the PAHtitration curves, the changes in the region where the rate of tubular secretion
reaches its maximum level are gradual, i.e., the curves exhibit ____.
transport maximum
for diodrast
57 mg/min
10 mg/dl
peritubular caps
saturated
UDV PDGFRperitubular caps
PD(ERPF GFR)
Splay
20. As noted in question 13, creatinine is an organic base that is filtered and
secreted. However, normally in humans the amount secreted is only about
10 to 15% of the amount filtered. This means that as an approximation
Rate of excretion Filtered load
or
UCrV PCrGFR (B-31)
________________________________________________________________Answers for p 23 Filtered Rate of Rate of CD ERPF Rate of D
Load Excretion Tubular ml/min ml/min Delivery to
mg/min mg/min Secretion Perit Caps
mg/min mg/min
5 25 20 625 625 20
7.5 37.5 30 625 625 30
10 50 40 625 625 40
12.5 60.5 48 605 - - - -
15 67 52 558 - - - -
20 77 57 481 - - - -
25 82 57 410 - - - -
37.5 94.5 57 315 - - - -
600
400
200Clearance
(ml/min)
PD
(mg/dl)
80
40
RateofExcretion
(mg/min)
PD
(mg/dl)
FilteredLoa
d(mg/min)
RateofTubularS
ecretion
(mg/min)
Inulin
10 20 30 10 20 30
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20. (cont.)
Hence, as an approximation, creatinine behaves like a substance that is
filtered only, such as the polysaccharide ____ , and its clearance can be
used to determine the GFR (see question 4 above):
CCr = UCrV GFR (B-32)
PCr
In fact, for clinical purposes, creatinine is preferable to the polysaccharide
____ , even though the resulting GFR value will be an approximation.1
This is because creatinine is an endogenous molecule that is produced from
muscle protein metabolism at a relatively constant rate, whereas the
polysaccharide ____ must be _____ in order to determine GFR (see
question 5 above).
inulin
inulin
inulin
infused IV
21. At this point, it may be helpful to briefly summarize the use of clearance
measurements for the determination of hemodynamic parameters in the
kidney:
Most Accurate Useful ApproximationDetermination for Clinical Purposes
GFR CIn CCr
RPF Fick principle CPAH
Furthermore, recalling that the filtration fraction, FF, is defined as
GFR/RPF (question A-7):
FF CInRPF from Fick
CCr
CPAH
__________1Because a small amount of creatinine is secreted in humans, the GFR will be slightly overestimated when determined
from CCr. However, the overestimate is at least partly corrected because PCrvalues, as measured in clinical laboratories,
tend to be too high (since the colorimetric methods for the determination of PCrmeasure other chromagens as well).
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22. Filtration, reabsorption, and secretion. For a substance that is filtered,
reabsorbed, and secreted
Rate of excretion = Filtered load Rate of tubular reabsorption
+ Rate of tubular secretion
For quantitative purposes, a substance that is filtered, reabsorbed, andsecreted usually is treated according to whether there is netreabsorption or
netsecretion:
If there is net reabsorption (i.e., if the rate of excretion is ____ than
the filtered load), the substance is treated as if it were filtered and
reabsorbed (see question 7 above), even though secretion also has
occurred. Thus, the rate of excretion can be described by the
equation _____ (with TX referring to the netrate of ____ ____),
and the clearance of the substance is ____ than CIn.
If there is net secretion (i.e., if the rate of excretion is ____ than the
filtered load), the substance is treated as if it were filtered and
secreted (see question 13 above), even though reabsorption also
has occurred. Thus, the rate of excretion can be described by the
equation _____ (with TX referring to the netrate of ____ ____),
and the clearance of the substance is ____ than CIn.
Examples of substances that are filtered, reabsorbed, and secreted include
potassium, urea, and uric acid.Note that while bicarbonate normally is filtered andreabsorbed, under conditions of extreme alkalosis it can be filtered, reabsorbed, and secreted(although in the latter case, there is netbicarbonate reabsorption).
less
UXV = PXGFR
TX
tubular
reabsorption
less
greater
UXV = PXGFR +
TX
tubular secretion
greater
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23. Summary of general excretion models:
Model Examples Rate of Excretion Clearance Other Remarks
No filtration or
secretion
Plasma proteins,
substances that are
tightly bound to
plasma proteins
UXV = 0 CX = 0
Filtration only Inulin, mannitol UXV = PXGFR CX = GFR CIn used for non-
invasive measure-
ment of GFR
Filtration and
reabsorption
Special case: Tm-
limited reabsorption,
threshold phenomenon
Glucose, amino
acids, sodium,
chloride, phosphate
Glucose, amino
acids, phosphate
UXV = PXGFR TX
If PXGFR TmX,
UXV = 0
If PXGFR > TmX,
UXV
=PXGFR
TmX
CX = CIn TX /PX
(i.e., CX < CIn)
If PXGFR TmX,
CX = 0
If PXGFR > TmX,
CX = CIn
TmX/PX
Filtration and secretion
Special case: Tm-
limited secretion with
complete clearance
from peritubular caps
when transport
mechanism is notsaturated
Organic acids,
organic bases
PAH
UXV = PXGFR + TX
If rate of delivery to
perit caps TmX,
UXV PXRPF
= PXERPF
If rate of delivery to
perit caps > TmX ,
UXV=PXGFR+TmX
CX = CIn + TX /PX
(i.e., CX > CIn)
If rate of delivery to
perit caps TmX,
CX RPF
= ERPF
If rate of delivery to
perit caps > TmX,
CX=CIn + TmX /PX
Because Cr secr is
minor, CCris used
for noninvasive
estimation of GFR
CPAH is used for
noninvasive mea-
surement of ERPF
and noninvasive
estimation of RPF
Filtration, reabsorption,
and secretion
Potassium, urea,
uric acid
Ifnetreabsorption,
same as for filtration
and reabsorption
Ifnetsecretion, same
as for filtration and
secretion
Ifnetreabsorption,
same as for filtration
and reabsorption
Ifnetsecretion, same
as for filtration and
secretion
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C. Fractional excretion, fractional delivery, and fractional reabsorption.
1. The fractional excretion, fractional delivery, and fractional reabsorption are
quantities with numerous applications in renal physiology. This section will
define these quantities and introduce the equations used in their calculation.
2. The fractional excretion(FE) can be defined as the fraction of a filteredsubstance that is excreted in the urine
FE = amount excreted
amount filtered
Dividing both numerator and denominator by time
FE = amount excreted per unit time
amount filtered per unit time
= rate of excretion
filtered loadi.e.,
FEX = ______
Snce CX = UXV/PX, FEX also can be written as
FEX = ______
Furthermore, since GFR = CIn and GFR CCr, additional equations for
FEX are
FEX = ______
FEX ______
UXV
PXGFR
CX
GFR
CX/CIn
CX/CCr
3. The fractional excretion can be calculated using any of the following
equations:
FEX = UXV (C-1)
PXGFR
FEX = CX (C-2)
GFR
FEX = CX (C-3)
CIn
FEX CX (C-4)
CCr
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Two additional equations for fractional excretion also are important.
Substituting CIn = UInV/PIn for GFR in equation C-1 gives
FEX = UX/PX (C-5)
UIn/PIn
and using CCr = UCrV/PCr as an approximation for GFR in equation C-1gives
FEX UX/PX (C-6)
UCr/PCr
Thus, equations C-5 and C-6 allow FEX to be calculated withoutmeasuring
____. Note that the quantity UX/PXby itselftells nothing about how much of
the filtered X is excreted because the value of UX/PX will vary depending on
the amount of water that is reabsorbed. Dividing UX/PX by UIn/PIn (or
UCr/PCr) corrects for water reabsorption (see below).
V
4. Of the equations for FEX, those in which U/P concentration ratios appear are
of particular interest (equations C-5 and C-6). One feature is that theyallow FEX to be calculated withoutmeasuring ____ , as noted above. An
additional feature can be appreciated with the aid of a terminology
switch: instead of using the term fractional excretion (FEX), we can use the
alternative term fractional delivery (FDX) because the fraction of the
filtered X that is excreted in the urine also represents the fraction of the
filtered X that is delivered to the urine. Thus, FDX can be substituted for
FEX in equations C-5 and C-6. Now, imagine moving from the urine up into
the tubule, sampling tubular fluid instead of urine. We could then substitute
tubular fluid concentration values (TF) for urine concentration values (U) in
equations C-5 and C-6:
FDX = TFX/PX (C-7)TFIn /PIn
and
FDX TFX/PX (C-8)
TFCr /PCr
These equations can be used to calculate the fraction of the filtered X that is
delivered to the point in the tubule from which the tubular fluid sample
was obtained (in experimental studies of renal function, such samples are obtained
by micropuncture). Note that the quantity TFX/PXby itselftells nothing about
how much of the filtered X is delivered to some point in the tubule
because the value of TFX/PX will vary depending on the amount of water
that is reabsorbed. Dividing TFX/PX by TFIn/PIn (or TFCr/PCr) corrects forwater reabsorption (see below).
V
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5. The fractional reabsorption (FR) can be defined as the fraction of a
filtered substance that has been reabsorbed. Once the fractional excretion
has been determined (by any equation in the C-1 to C-6 set), the fractional
reabsorption can be calculated easily, since
FRX = 1 FEX (C-9)
Similarly, once the fraction of a filtered substance that has been delivered
to some point in the tubule has been determined (using equation C-7 or C-
8), the fraction of the filtered substance reabsorbed from the tubular fluid
during its journey from Bowmans space to that point is given by
FRX = 1 FDX (C-10)
6. Summary of equations for fractional excretion, fractional delivery, and
fractional reabsorption:
Fractional Excretion
(FEX)
Fractional Reabsorption
(FRX)
UXV
PXGFR
CX
GFR
CX
CIn
CX
CIn
UX /PX
UIn/PIn
UX /PX
UCr/PCr
1 FEX
Fractional Delivery to
Point of Sample
(FDX)
Fraction Reabsorbed up to
Point of Sample
(FRX)
TFX /PX
TFIn/PIn
TFX /PX
TFCr/PCr
1 FDX
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7. To consolidate your understanding of fractional excretion, fractional
delivery, and fractional reabsorption, complete the following table:
Inulin Potassium
Low K+ Diet Normal Diet Vegetarian Diet
PX
TFX
end of proximal tubule
macula densa
UX
0.5 mg/ml
1.5 mg/ml
6 mg/ml
50 mg/ml
4 mM
4 mM
4.8 mM
20 mM
4 mM
4 mM
4.8 mM
80 mM
4 mM
4 mM
4.8 mM
640 mM
Fractional delivery to
point of sample
end of proximal tubule
macula densa
Fractional excretion
Fraction reabsorbed up
to point of sample
end of proximal tubule
macula densa
Fractional reabsorption - - 1
1.0
1.0
1.0
0
0
0
0.33
0.1
0.05
0.67
0.9
0.95
0.33
0.1
0.2
0.67
0.9
0.8
0.33
0.1
1.6
0.67
0.9
-- 1
1Since the fractional excretion exceeds 1.0, the amount excreted is greater than the
amount filtered, i.e., net secretion has occurred.
8. Equations for the fractional excretion, fractional delivery, and fractional
reabsorption of a substance now have been derived. However, it should be
noted that the fractional excretion, fractional delivery, and fractional
reabsorption of water also are important quantities, particularly when the
discussing the urinary concentration and dilution mechanisms. The
equations for water (which have many similar features to the equations in
the table on p 30; see question 12 below) are derived in the following
questions.
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9. The fractional excretion of water (FEH2O) can be defined as the fraction of
filtered water that is excreted
FEH2O = amount excreted
amount filtered
Dividing both numerator and denominator by time
FEH2O = amount excreted per unit time
amount filtered per unit time
= V (C-11)
GFR
A more useful equation for FEH2O is obtained by substituting CIn = UInV/PIn
for GFR
FEH2O = 1 (C-12)
UIn/PIn
Equation C-12 makes sense: since inulin is filtered but neither reabsorbed
nor secreted, the difference between UIn and PIn can be attributed exclusively
to the reabsorption of water. For example, if U In = 10PIn, ___ % of the
filtered water must have been reabsorbed, ____ % is excreted; this intuitive
result also is predicted by equation C-12. Another very useful equation for
FEH2O is obtained by substituting CCr = UCrV/PCras an approximation for
GFR in equation C-11
FEH2O 1 (C-13)
UCr/PCr
Note that equations C-12 and C-13 allow FEH2O to be calculated without
measuring ____ . Another important feature of equations C-12 and C-13 is
discussed in the following question.
90%
10%
V
10. Of the equations for FEH2O, those in which U/P ratios appear (equations C-
12 and C-13) are particularly useful. One feature is that they allow FEH2O to
be calculated withoutmeasuring ____ , as noted above. An additional
feature can be appreciated with the aid of the same terminology switch
that was introduced when FEX was discussed above (question 4): instead of
using the term fractional excretion of water (FEH2O), we can use the
alternative term fractional deliveryof water (FDH2O)because the fraction
of the filtered water that is excreted in the urine also represents the fractionof the filtered water that is delivered to the urine. Thus, FDH2O can be
substituted for FEH2O in equations C-12 and C-13. Now, imagine moving
from the urine up into the tubule, sampling tubular fluid instead of urine.
We could then substitute tubular fluid concentration values (TF) for urine
concentration values (U) in equations C-12 and C-13
V
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FDH2O = 1 (C-14)
TFIn/PIn
and
FDH2O 1 (C-15)
TFCr/PCr
Thus, the ratios TFIn/PIn and TFCr/PCr are very useful in discussions of the
urinary concentration and dilution mechanisms, since the fraction of filtered
water delivered to various points in the tubule are important in such
discussions (see question 14 below).
11. The fractional reabsorption of water (FRH2O) can be defined as the
fraction of filtered water that has been reabsorbed. Once the fractional
excretion of water has been determined (by any equation in the C-11 to C-
13 set), the fractional reabsorption can be calculated easily, since
FRH2O = 1 FEH2O (C-16)
Similarly, once the fraction of filtered water that has been delivered to
some point in the tubule has been determined (using equation C-14 or C-
15), the fraction of filtered water reabsorbed from the tubular fluid during
its journey from Bowmans space to that point is given by
FRH2O = 1 FDH2O (C-17)
12. Summary of equations for fractional excretion, fractional delivery, and
fractional reabsorption of water:
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Fractional Excretion of Water
(FEH2O)
Fractional Reabsorption of Water
(FRH2O)
V
GFR
1UIn/PIn
1
UCr/PCr
1 FEH2O
Fractional Delivery of Water
to Point of Sample
(FDH2O)
Fraction of Water Reabsorbed
up to Point of Sample
(FRH2O)
1
TFIn/PIn
1TFCr/PCr
1 FDH2O
13. Having derived the above equations for FEH2O, FDH2O, and FRH2O, it should
be noted that the same equations also could be obtained from the equations
for FEX, FDX, and FRX (question 6) by setting UX/PX (or TFX/PX) equal to
1.0 . This substitution is valid because UH2O/PH2O and TFH2O/PH2Odo equal
1.0: the concentration of water is 55 mol/liter in plasma, urine, and tubular
fluid.
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14. To consolidate your understanding of FEH2O, FDH2O, and FRH2O, complete
the following table:UCSD Medical Student UCSD Medical Student
during a Santa Ana after a Post-OP
Condition Beer Binge
PIn
TFIn
end of proximal tubule
tip of papilla
macula densa
end of cortical CD
middle of medullary CD
UIn
0.1 mg/ml
0.3 mg/ml
1.2 mg/ml
1.2 mg/ml
2.0 mg/ml
10 mg/ml
20 mg/ml
0.1 mg/ml
0.25 mg/ml
0.8 mg/ml
0.8 mg/ml
0.8 ml/ml
0.8 mg/ml
0.8 mg/ml
Fraction of filtered H2O delivered
to point of sample
end of proximal tubule
tip of papilla
macula densa
end of cortical CD
middle of medullary CD
FEH2O
Fraction of filtered H2O reabsorbed
up to point of sample
end of proximal tubule
tip of papilla
macula densa
end of cortical CD
middle of medullary CD
FRH2O
0.33
0.083
0.083
0.05
0.01
0.005
0.67
0.917
0.917
0.95
0.99
0.995
0.4
0.125
0.125
0.125
0.125
0.125
0.6
0.875
0.875
0.875
0.875
0.875
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