Numerous tests can assess renal function. Some are applicable only in animal experiments. N33-5 Others are useful in clinical settings and fall into two general categories:
1. Modern imaging techniques provide outstanding macroscopic views of renal blood flow, filtration, and excretory function.
2. Measurements of so-called renal clearance of various substances evaluate the ability of the kidneys to handle solutes and water.
In Vitro Preparations for Studying Renal Function in the Research Laboratory
Contributed by Gerhard Giebisch, Erich Windhager
Clearance methods treat the whole kidney as a “black box.” Thus, using clearance methods, it is very difficult to determine which nephron segments are responsible for which transport processes. It is also impossible to determine which nephrons are responsible for overall urinary excretion. To learn how single nephrons function, and also to understand how individual segments of the nephron contribute to overall nephron function, renal physiologists have developed a series of invasive techniques for studying the function of renal cells in the research laboratory.
It is possible, under microscopic observation, to insert a sharpened micropipette into tubule segments at or near the surface of the kidney or into exposed parts of the renal papilla. In these cases, one can collect fluid from identified tubule segments, as illustrated in Figure 33-9A for a superficially located proximal tubule. The investigator collects tubule fluid at a rate that is comparable to the flow rate in the undisturbed nephron and then analyzes this fluid for the solute of interest.
Stationary or Stopped-Flow Microperfusion
In the stationary or stopped-flow microperfusion approach, the investigator deposits an aqueous solution between two oil columns inside the tubule lumen and periodically withdraws samples for analysis (see Fig. 33-9B). An advantage of this technique is that one can choose the initial composition of the fluid deposited in the lumen.
The continuous microperfusion approach is similar to the stationary approach described above in that a pair of oil blocks isolate the artificial perfusion fluid from the tubule fluid normally present (see Fig. 33-9C). A difference is that one continuously infuses fluid by means of a carefully calibrated microperfusion pump and then collects the perfusate in a micropipette located at the distal end of the perfused tubule segment. In addition, it is possible to perfuse the peritubular capillary network by means of separate micropipettes. This approach permits one to evaluate the effects of changes in the peritubular environment on tubule transport. The simultaneous perfusion of the lumen and peritubular capillaries affords excellent control over the environment of the tubule cells at both the apical and basolateral surfaces.
Perfusion of Isolated Tubules
It is possible to microdissect a single segment from almost any part of the nephron and perfuse the single isolated tubule on the stage of a microscope (see Fig. 33-9D). The isolated perfused tubule technique offers two distinct advantages over the in vivo micropuncture and microperfusion techniques discussed above. First, one can study tubules from any segment of the nephron, not just superficial tubule segments. Second, the tubules can be isolated from nephron populations (juxtamedullary as well as superficial) located anywhere in the kidney. This consideration is important because of the heterogeneity that exists among nephrons. Thus, the isolated perfused tubule technique offers the ultimate in control over the environment of tubule cells.
In addition to the above four preparations, which permit one to study a defined tubule segment from a single nephron, several other preparations are available for studying cellular and subcellular events. This more narrowly focused approach is achieved at the expense of disrupting the normal tubule and vascular organization of the kidney.
Suspensions of Tubules or Single Cells
Suspensions of tubules or single cells have enough cell mass to permit the determination of cellular metabolism under a variety of experimental conditions. Although this approach can also be used to study solute transport, a disadvantage is that, because the cells have lost their normal polarity, it is impossible to assign function to either the apical or basolateral membranes.
Tubule Cells in Culture
A preparation of tubule cells in culture offers the advantage of preserving the sidedness of the apical and basolateral membranes. On the other hand, one must be aware that cells in culture—whether primary cultures or continuous cell lines—may acquire a phenotype different from that of tubule cells in vivo.
A membrane vesicle preparation is an extension of the single-cell suspension technique but takes the isolation a step further. When one disrupts tubule cells, the fragments of their plasma membranes spontaneously take the form of membrane vesicles—small, irregularly shaped volumes that are completely surrounded by a piece of cell membrane, complete with its lipids and proteins. It is possible to purify membrane vesicles into fractions that consist mainly of luminal (“brush-border”) cell membranes or basolateral cell membranes. These vesicles maintain many of the transport functions of tubule cells in vivo and can be examined in ways that allow one to study specific transport systems in vesicles derived primarily from the apical versus basolateral membrane (see Fig. 5-12). Two important caveats regarding this method should be recognized. First, vesicles may exhibit large variations in their passive leakiness via pathways that are not normally present in tubule cells in vivo. Second, many of the normal cytoplasmic constituents are absent from the vesicle preparation, so that signal-transduction mechanisms may not function normally.
This subchapter focuses on clearance measurements, which compare the rate at which the glomeruli filter a substance (water or a solute) with the rate at which the kidneys excrete it into the urine. By measuring the difference between the amounts filtered and excreted for a particular substance, we can estimate the net amount reabsorbed or secreted by the renal tubules and can thus gain insight into the three basic functions of the kidney: glomerular filtration, tubule reabsorption, and tubule secretion. Although widely used, clearance methods have the inherent limitation that they measure overall nephron function. This function is “overall” in two different senses. First, clearance sums many individual transport operations occurring sequentially along a nephron. Second, clearance sums the output of all 2 million nephrons in parallel. Hence, clearance cannot provide information on precise sites and mechanisms of transport. Such information can, however, come from studies of individual nephrons, tubule cells, or cell membranes. One can also apply the clearance concept to other problems, such as clearance of bile by the liver (see p. 963) or clearance of hormones from the blood (see p. 1021).
The clearance of a solute is the virtual volume of plasma that would be totally cleared of a solute in a given time
All solutes excreted into the urine ultimately come from the blood plasma perfusing the kidneys. Thus, the rate at which the kidney excretes a solute into the urine equals the rate at which the solute disappears from the plasma, provided the kidney does not produce, consume, or store the solute. Imagine that, in 1 minute, 700 mL of plasma flow through the kidneys. This plasma contains 0.7 L × 142 mM or ~100 mmol of Na+. Of this Na+, the kidneys remove and excrete into the urine only a tiny amount, ~0.14 mmol. In principle, these 0.14 mmol of Na+ could have come from only 1 mL of plasma, had all Na+ ions been removed (i.e., cleared) from this volume. The clearance of a solute is defined as the virtual volume of blood plasma (per unit time) needed to supply the amount of solute that appears in the urine. Thus, in our example, Na+ clearance was 1 mL/min, even though 700 mL of plasma flowed through the kidneys.
Renal clearance methods are based on the principle of mass balance and the special anatomy of the kidney (Fig. 33-7). For any solute (X) that the kidney does not synthesize, degrade, or accumulate, the only route of entry to the kidney is the renal artery, and the only two routes of exit are the renal vein and the ureter. Because the input of X equals the output of X,
PX,a and PX,v are plasma concentrations of X in the renal artery and renal vein, respectively. RPFa and RPFv are rates of renal plasma flow (RPF) in the renal artery and vein, respectively. UX is the concentration of X in urine. is urine flow (the overdot represents the time derivative of volume). The product UX ⋅ is the urinary excretion rate, the amount of X excreted in urine per unit time.
FIGURE 33-7 Solute mass balance in the kidney.
In developing the concept of renal clearance, we transform Equation 33-1 in two ways, both based on the assumption that the kidneys clear all X from an incoming volume of arterial plasma. First, we replace RPFa with the inflow of the virtual volume—the clearance of X (CX)—that provides just that amount of X that appears in the urine. Second, we assign the virtual venous output a value of zero. Thus, Equation 33-1 becomes
Solving for clearance yields
This is the classic clearance equation that describes the virtual volume of plasma that would be totally cleared of a solute in a given time (Table 33-2, section A). N33-6 We need to know only three parameters to compute the clearance of a solute X:
1. The concentration of X in the urine (UX)
2. The volume of urine formed in a given time ()
3. The concentration of X in systemic blood plasma (PX), which is the same as PX,a in Equation 33-1
A. Clearance of any substance X
CX may vary between zero, for a substance that does not appear in the urine (e.g., glucose), and ~700 mL/min (i.e., total RPF) for a substance (e.g., PAH) that is totally removed from the blood in a single pass through the kidney.
B. Clearance of PAH approximates RPF (at low plasma PAH concentration)
C. Clearance of inulin (In) equals GFR
Symbols of the Clearance Equation
Contributed by Erich Windhager, Gerhard Giebisch, Emile Boulpaep, Walter Boron
In Equation 33-3, we present the clearance equation* as
Here C is the clearance, X is the solute, UX is the concentration of X in the urine, is the flow of urine (volume per unit time), and PX is the concentration of X in the plasma. Most clinical texts replace the symbol simply with V. We use to distinguish an absolute volume of urine from the rate of urine production.
This use of symbols is very different from elsewhere in the text and, indeed, from elsewhere in the discipline of physiology. A cellular physiologist might write the same equation in the form
Indeed, an advantage of writing the clearance equation in its totally unofficial form (Equation NE 33-2) is that it avoids confusing PX with the standard symbol for the permeability to X.
*The following mnemonic is attributed to John M. Russell: Imagine that you are in Brussels on a sunny day, gazing at the famous fountain that features the statue of a little boy urinating into a pool of water. The sunlight (which contains ultraviolet or “UV” rays) is obviously above the stream of urine (“pee”) … yielding “UV over P.”
Together, the three basic functions of the kidney—glomerular filtration, tubule reabsorption, and tubule secretion—determine the renal clearance of a solute. In the special case in which the kidneys completely clear X from plasma during a single passage through the kidneys (PX,v = 0 in Equation 33-1), the renal clearance of X equals RPFa in Equation 33-1. Because para-aminohippurate (PAH) is just such a special solute, its clearance is a good estimate of RPFa, which we simplify to RPF (see Table 33-2, section B). We discuss RPF on pages 749–750.
For all solutes that do not behave like PAH, the renal venous plasma still contains some X. Thus, the virtual volume cleared of X in a given time is less than the total RPF. For most solutes, then, clearance describes a virtual volume of plasma that would be totally cleared of a solute, whereas in reality a much larger volume of plasma is partially cleared of the solute.
We can use a clearance approach to estimate another important renal parameter: glomerular filtration rate (GFR), which is the volume of fluid filtered into Bowman's capsule per unit time. Imagine a solute X that fulfills two criteria. First, X is freely filtered (i.e., concentration of X in Bowman's space is the same as that in blood plasma). Second, the tubules do not absorb, secrete, synthesize, degrade, or accumulate X. Thus, the amount of X that appears in the urine per unit time (UX ⋅ ) is the same as the amount of X that the glomerulus filters per unit time (PX ⋅ GFR):
The input to Bowman's space is also known as the filtered solute load and is generally given in mmol/min (or mg/min). Rearranging Equation 33-4, we have
Equation 33-5 is in exactly the same form as the classic clearance equation (see Equation 33-3). In other words, GFR is CX if X has the required properties. As discussed on pages 739–741, inulin is just such a solute (see Table 33-2, section C).
A solute's urinary excretion is the algebraic sum of its filtered load, reabsorption by tubules, and secretion by tubules
The homeostasis of body fluids critically depends on the ability of the kidneys to determine the amount of a given solute that they excrete into the urine. Renal excretion rate (EX) depends on three factors (Fig. 33-8):
1. The rate of filtration of X (FX), known as the filtered load (FX = GFR ⋅ PX)
2. The rate of reabsorption of X (RX) by the tubules
3. The rate of secretion of X (SX) by the tubules
FIGURE 33-8 Factors contributing to the net urinary excretion of a substance.
This interrelationship is expressed quantitatively by
For some substances (e.g., inulin), no reabsorption or secretion occurs. For most substances, either reabsorption or secretion determines the amount present in the final urine. However, for some substances, both reabsorption and secretion determine excretion.
If a solute is only reabsorbed, but not secreted, we can rearrange Equation 33-6 to obtain the rate of reabsorption:
Conversely, if a solute is only secreted, but not reabsorbed, the rate of secretion is
In applying Equations 33-7 and 33-8, we must keep in mind two important limitations. First, to estimate the rate at which a substance appears in the filtrate—the filtered load—from the product GFR ⋅ PX, we assume that PX is the freely filterable concentration of X. Indeed, many substances (particularly univalent electrolytes, urea, glucose, and amino acids) are freely filterable. However, if the solute (e.g., Ca2+, phosphate, Mg2+, and PAH) binds to protein, then it will not be freely filterable. For such solutes it is necessary to measure plasma binding and correct for the nonfilterable fraction of the solute. Second, for us to apply the mass balance equation (see Equation 33-6), the kidney must not synthesize, degrade, or accumulate the solute. An example of a solute that is synthesized by the kidney is ammonium. Examples of solutes degraded by the kidney include glutamine and glutamate (which are deaminated to yield ammonium) as well as several other amino acids and monocarboxylic and dicarboxylic acids.
When the kidney both reabsorbs and secretes a substance, clearance data are inadequate to describe renal handling. For example, if the proximal tubule completely reabsorbed a solute that a later segment then secreted, clearance data alone would imply that only filtration and some reabsorption occurred. We would have no reason to implicate secretion. Complex combinations of reabsorption and secretion occur with K+, uric acid, and urea.
Another useful parameter for gauging how the kidney handles a freely filtered solute is the fractional excretion (FE), which is the ratio of the amount excreted in the urine (UX ⋅ ) to the filtered load (PX ⋅ GFR):
According to Equation 33-3, however, the term (UX ⋅ /PX) is simply the clearance of X (CX):
As discussed on pages 739–741, one estimates GFR by measuring the clearance of inulin (CIn). Thus, the FE of a freely filterable solute is the same as the clearance ratio
Microscopic techniques make it possible to measure single-nephron rates of filtration, absorption, and secretion
Because clearance methods treat the kidney as a “black box,” reflecting the activity of many single nephrons and nephron segments, it is very difficult to determine which nephron segments are responsible for which transport processes. It is also impossible to determine which nephrons are responsible for overall urinary excretion. To learn how single nephrons function, and to understand how individual nephron segments contribute to overall nephron function, physiologists developed a series of invasive techniques for studying renal cells in the research laboratory (Fig. 33-9). N33-5
FIGURE 33-9 Methods for studying renal function in the research laboratory.
To apply the concept of clearance to a single nephron site, one uses the free-flow micropuncture approach (see Fig. 33-9A) and measures the concentration of the solute in the tubule fluid at that site (TFX), volume flow at that site (i.e., the collection rate), and plasma concentration (PX). By analogy with the macroscopic clearance equation, we can write a clearance equation for a single nephron:
Compared with Equation 33-3, here TFX replaces UX and “Volume collection rate” replaces . We can use this basic equation to compute the amount of fluid that a single nephron filters, as well as the amounts of fluid and solutes that a single tubule segment handles.
If X in Equation 33-12 is a marker for GFR (e.g., inulin, or In), one can calculate single-nephron GFR (SNGFR) using an equation that is similar to that in Table 33-2, section C, to calculate total GFR:
Using the numeric values for rat kidney shown in Figure 33-10, we can use Equation 33-13 to compute the SNGFR:
FIGURE 33-10 Measurement of SNGFR. Data are for the rat.
Handling of Water by Tubule Segments in a Single Nephron
We also can use the same information that we used to compute SNGFR to calculate the rate of water reabsorption between the glomerulus and the micropuncture site. The fraction of filtered water remaining at the micropuncture site is
Substituting the expression for SNGFR (see Equation 33-13) into Equation 33-15 yields
Thus, to know the fraction of filtered water remaining at the collection site, we do not need to know the collection rate, only the concentrations of inulin in blood plasma and at the collection site. In the example of Figure 33-10, in which the tubule reabsorbed two thirds of the fluid, the fraction of filtered water remaining at the sampling site is (1 mg/mL)/(3 mg/mL) or ~0.33.
The fraction of filtered water reabsorbed is 1 minus the fraction of filtered water remaining:
In our example, the fraction of filtered water reabsorbed is 1 − 0.33 or ~0.67.
Handling of Solutes by Tubule Segments in a Single Nephron
We can use the same concepts of single-nephron clearance to quantitate the reabsorption or secretion of any solute along the tubule. The first step is to estimate the fraction of the filtered solute remaining at the puncture site. This parameter—the fractional solute delivery—is the ratio of the amount of solute appearing at the micropuncture site to the amount of solute filtered at the glomerulus (i.e., single-nephron filtered load):
The numerator is the product of volume collection rate and tubule solute concentration (TFX), and the denominator is the product of SNGFR and the plasma solute concentration (PX):
In Equations 33-15 and 33-16, we saw that the ratio (volume collection rate/SNGFR) is (PIn/TFIn). Making this substitution in Equation 33-19, we obtain an alternative expression for fractional solute excretion:
The advantage of Equation 33-20 over Equation 33-19 is that no measurements of collection rates are required.
Equation 33-20 is important for understanding the transport of a solute along the nephron. If (TFX/PX)/(TFIn/PIn) is >1, secretion has occurred. Merely observing that the solute concentration in tubule fluid increases along the length of the nephron does not necessarily mean that the tubule secreted the solute; the concentration would also increase if water were reabsorbed. We can conclude that secretion has occurred only if the solute concentration in tubule fluid, relative to its concentration in filtrate, exceeds the concentration ratio of inulin. If (TFX/PX)/(TFIn/PIn) is <1, reabsorption has occurred.
Rather than referring to the fraction of the solute excreted (i.e., remaining) up to the point of the micropuncture site, we can also refer to the fractional solute reabsorption at the puncture site. By analogy to Equation 33-17 for water, this parameter for a solute is merely 1 minus the fractional solute excretion:
Thus, applying the principles of clearance and mass balance to a single nephron, we can calculate SNGFR as well as the fractional reabsorption of water and solutes at the micropuncture site.