Page 165 - Color_Atlas_of_Physiology_5th_Ed._-_A._Despopoulos_2003

Glomerular Filtration and Clearance Indicators present in the plasma are used to
measure GFR. They must have the following
The glomerular filtration rate (GFR) is the total properties:
volume of fluid filtered by the glomeruli per — They must be freely filterable
unit time. It is normally about 120 mL/min per — Their filtered amount must not change due
2
1.73 m of body surface area, equivalent to to resorption or secretion in the tubule
around 180 L/day. Accordingly, the volume of — They must not be metabolized in the kidney
exchangeable extracellular fluid of the whole — They must not alter renal function
body (ca. 17 L) enters the renal tubules about Inulin, which must be infused intravenously,
Kidneys, Salt, and Water Balance lute H 2O excretion (= urine output/time = V U) is (! A) is calculated as the plasma concentra-
10 times a day. About 99% of the GFR returns to
fulfills these requirements. Endogenous crea-
the extracellular compartment by tubular re-
tinine (normally present in blood) can also be
used with certain limitations.
absorption. The mean fractional excretion of
The amount of indicator filtered over time
H 2O is therefore about 1% of the GFR, and abso-
.
tion of the indicator (P In, in g/L or mol/L) times
about 1 to 2 L per day. (The filtration of dis-
the GFR in L/min. The same amount of indica-
solved substances is described on p. 154).
tor/time appears in the urine (conditions 2 and
The GFR makes up about 20% of renal
.
3; see above) and is calculated as V U (in L/min),
plasma flow, RPF (! p. 150). The filtration frac-
tion (FF) is defined as the ratio of GFR/RPF. The
times the indicator conc. in urine (U In, in g/L or
.
.
peptide hormone that increases efferent arte-
VU ! U In
[L/min] (! A).
GFR !
riolar resistance (R e) while lowering afferent
P In
7 filtration fraction is increased by atriopeptin, a mol/L, resp.), i.e. P In ! GFR = V U ! U In, or: [7.8]
The expression on the right of Eq. 7.8 repre-
arteriolar resistance (R a). This raises the effec-
tive filtration pressure in the glomerular capil- sents clearance, regardless of which substance
laries without significantly changing the over- is being investigated. Therefore, the inulin or
all resistance in the renal circulation. creatinine clearance represents the GFR. (Al-
The effective filtration pressure (P eff) is the though the plasma concentration of creat-
driving “force” for filtration. P eff is the glomer- inine, P cr, rises as the GFR falls, P cr alone is a
ular capillary pressure (P cap ! 48 mmHg) quite unreliable measure of GFR.)
minus the pressure in Bowman’s capsule (P Bow Clearance can also be regarded as the
! 13 mmHg) and the oncotic pressure in completely indicator-free (or cleared) plasma
plasma (π cap = 25 to 35 mmHg): volume flowing through the kidney per unit
P eff ! P cap –P Bow –π cap [7.6] time. Fractional excretion (FE) is the ratio of
P eff at the arterial end of the capillaries equals clearance of a given substance X to inulin
48–13–25 = 10 mmHg. Because of the high fil- clearance (C X/C In) and defines which fraction
tration fraction, the plasma protein concentra- of the filtered quantity of X was excreted (cf.
tion and, therefore, π cap values along the glo- p. 154). FE " 1 if the substance is removed from
+
–
merular capillaries increase (! p. 378) and P eff the tubule by reabsorption (e.g. Na , Cl , amino
decreases. (The mean effective filtration pres- acids, glucose, etc.; ! B1), and FE # 1 if the
sure, P eff, is therefore used in Eq. 7.7.) Thus, fil- substance is subject to filtration plus tubular
tration ceases (near distal end of capillary) secretion (! B2). For PAH (! p. 150), tubular
when π cap rises to about 35 mmHg, decreasing secretion is so effective that FE PAH ! 5 (500%).
P eff to zero (filtration equilibrium). The absolute rate of reabsorption or secre-
GFR is the product of P eff (mean for all glo- tion of a freely filterable substance X (mol/
meruli), the glomerular filtration area A (de- min) is calculated as the difference between
pendent on the number of intact glomeruli), the filtered amount/time (GFR · P X) and the ex-
.
and the water permeability k of the glomerular creted amount/time (V U · U X), where a positive
filter. The ultrafiltration coefficient K f is used to result means net reabsorption and a negative
represent A · k. This yields net secretion. (For inulin, the result would be
GFR ! P eff ! K f. [7.7] zero.)
152
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