Page 43 - Fluid, Electrolyte, and Acid-Base Disorders in Small Animal Practice

32 APPLIED PHYSIOLOGY

Autoregulatory range 90%ofitisremovedinonepassthroughthekidneys.Itisnot
600 80 metabolized or excreted by any other organ. Thus, it
approximately meets the preceding assumptions and RPF
70
500 GFR 60 ¼ U PAH V/P PAH . Now it can be seen that the clearance of
Renal blood flow (ml/min) 400 RBF 50 Glomerular filtration rate (ml/min) a concentration much below the tubular transport maxi-
PAH is an estimate of RPF. When PAH is infused during
a clearance study, it is essential that P PAH be maintained at
mum (T max ) for PAH. If not, P VX cannot be neglected.
40
300
Somebloodflowsthroughregionsofthekidneysthatdo
30
notremovePAH(e.g.,renalcapsule,perirenalfat,andrenal
200
effective RPF is more appropriately used when speaking of
100 20 pelvis), and as a result, P VX is not really zero. Thus, the term
10 PAH clearance. Furthermore, only 90% of PAH is removed
from the blood during a single pass through the kidneys.
0 50 100 150 200 250
80 180 This also contributes to the fact that P VX for PAH is not
Renal arterial blood pressure (mm Hg) really zero. A closer approximation of RPF can be deter-
Figure 2-7 Autoregulation of renal blood flow and glomerular mined by sampling renal arterial and venous blood and
filtration rate. (Drawing by Tim Vojt.) measuring their respective PAH concentrations. The
extraction ratio for PAH is then determined:

The afferent arteriolar constriction causes SNGFR to
E X ¼ðP AX P VX Þ=P AX
decrease, thus decreasing filtration and minimizing
NaCl loss in that nephron. This effect occurs locally A more accurate calculation of RPF is then:
in the region of the juxtaglomerular interstitium.
Tubuloglomerular feedback represents a fine control that
RPF ¼ C PAH =E PAH
operates with a 10- to 12-second delay.
RPF ¼ U PAH V=P PAH E PAH
MEASUREMENT OF RENAL BLOOD
FLOW AND RENAL PLASMA FLOW The extraction ratio for PAH is 0.9 because approxi-
52
Consider the following mass balance equation : mately 90% of it is removed from the blood in a single pass
Amount entering the kidneys ¼ amount leaving the through the kidneys. Notice that if we substitute the
kidneys equation for E X into the preceding equation, we get
RPF ¼ U X V/(P AX P VX ), which is the same equation
P AX RPF A ¼ P VX RPF V þ U X V as derived before for RPF.
Another way to determine RPF is by use of the Fick
P AX RPF A P VX RPF V ¼ U X V principle, which states that the amount of a substance
(V) removed by an organ is equal to the blood flow to
where P AX is the renal arterial plasma concentration of x, the organ (Q) times the arteriovenous concentration
RPF A isthearterialrenalplasmaflow(RPF),P VX istherenal difference of the substance in question (C A C V ):
venousplasma concentration ofx, RPF V is the venousRPF,
U X is the urine concentration of x, and V is the urine flow. V ¼ QðC A C V Þ
If we ignore the slight difference between renal arterial
and venous plasma flow (with probably less than 1% Q ¼ V=ðC A C V Þ
error), the equation is simplified to:
Using the kidneys as an example and equating the amount
ðP AX P VX ÞRPF ¼ U X V of the substance removed to the amount excreted (U X V):

RPF ¼ U X V=ðP AX P VX Þ
RPF ¼ U X V=ðP AX P VX Þ
If we choose a substance that is completely removed from Note that this equation is identical to that derived before
the blood in one pass through the kidneys, P VX is zero using the mass balance principle.
and RPF ¼ U X V/P AX . If the substance x is not If the hematocrit is known, RBF can be calculated
metabolized and is not excreted by any organ other than from the RPF by using the following equation:
the kidneys, its concentration in any peripheral vessel
equals P AX . Thus, RPF ¼ U X V/P X . RPF
PAH is filtered by the glomeruli and secreted by the RBF ¼
ð1 hematocritÞ
peritubularcapillariesintothetubulessothatapproximately