Examples

1. A typical adult human ingests of 8-10 g about NaCl each day. There is no metabolic production of this inorganic compound, so for a steady state to be maintained (as is further discussed in Chapter 29) NaCl must be excreted (by the sum of urinary, sweat, and fecal routes) at the same rate of 8-10 g per day.

2. Water is typically ingested at a rate of 1-2 L/ day with about 150-250 mL/day added from metabolism of various substrates such as glucose. Water loss, which occurs via the same routes as for NaCl, must equal this daily intake.

3. Little urea is ingested in the diet, but it is synthesized in the liver at a rate of about 30 g/day or 20 mg/min from the HCO3~ and NH4+ produced by the degradation of amino acids. Thus, for a steady state to exist, urea must be eliminated at the same rate, and this occurs primarily via urinary excretion.

Examples

1. If the urea concentration in the arterial plasma [expressed in the usual clinical units of blood urea nitrogen concentration (BUN)]; see later discussion in this chapter] is present in arterial plasma at 25 mg/dL, and the total rate of plasma flow into the kidneys via the two renal arteries is 690 mL/min, the input rate of urea is 172.5 mg/min (25 mg/dL • 6.9 dL/min). The normal rate of urea excretion is about 20 mg/min. Therefore, the difference between the rate of urea flow into the kidney (via the renal artery) and the rate of its excretion into the urine must be equal to the rate of urea return to the plasma via the renal vein and lymphatics, or approximately 152.5 mg/min.

2. In the case of water, the mass flow may be calculated as for any substance. However, because water is by far the main body constituent, its ''concentration'' is virtually the same in all biological fluids, and its mass flow is proportional to the volume flow. Thus, the rate of delivery of water to the kidney is equal to the rate of plasma flow, or about 690 mL/min (for convenience, ignoring the volume occupied by macromole-cules such as proteins). If the UF is 1 mL/min, then the sum of the renal venous and lymphatic flow must be 689 mL/min.

that flow into the nephron must flow out. There are two routes of inflow into the nephron: the filtered mass flow and contributions due to secretion along the nephron. Water enters the nephron only by filtration. on the other hand, solutes may not only be filtered at the glomerulus but also secreted into various nephron segments by epithelial transport processes that are described in the two following chapters. Solutes and water that enter the nephron by these routes must leave the nephron in equal amounts by two routes: the mass flow excreted in the final urine and absorption across the epithelium of the various nephron segments.

FIGURE 2 Mass balance relations for the kidney. In the steady state, the output of any substance from the kidney via the urine, renal vein, and lymphatic flows must be equal to the rate of input via the renal artery. For simplicity, this neglects any production or metabolism of the substance within the kidney, which is negligible for most substances.

FIGURE 2 Mass balance relations for the kidney. In the steady state, the output of any substance from the kidney via the urine, renal vein, and lymphatic flows must be equal to the rate of input via the renal artery. For simplicity, this neglects any production or metabolism of the substance within the kidney, which is negligible for most substances.

FIGURE 3 Mass balance relations along the nephron. The rate of excretion must equal the rate of filtration plus the rate of secretion minus the rate of reabsorption.

Figure 3 shows diagrammatically the operation of all 2 million nephrons together as if they were one giant nephron. For any given substance, inflow to the nephron must equal outflow, or, in terms of the rates of these processes:

Filtration + secretion = excretion + reabsorption. (4)

If this equation is rearranged, the rate of excretion of a substance can be calculated directly as:

Excretion = filtration — reabsorption + secretion. (5)

Equation [5] is the fundamental mass balance equation that must be appreciated in understanding the components that regulate the final excretion of water and solutes by the kidney, as shown in Fig. 3. A solute is filtered at the glomerulus at a certain rate. This may be augmented by secretion and diminished by reabsorption, resulting in the final rate of excretion. This view does not specify where the reabsorptive and secretory processes occur. Some solutes may be reabsorbed in one region of the nephron and secreted in another; some solutes are only reabsorbed and not secreted, and some are neither reabsorbed nor secreted.

When a solute is reabsorbed across the epithelium of the nephron, it is taken up into the peritubular capillaries; this subtracts from its rate of excretion in the urine and adds to its outflow from the kidney via the renal vein. When a substance is secreted, it is transported by the epithelium from the renal interstitium into the tubular fluid, whereas the interstitial concentration of the substance is maintained by passive diffusion of the solute out of the peritubular capillaries. Thus, secretion of a solute enhances its rate of excretion via the urine and reduces its rate of outflow from the kidney via the renal vein. These processes are shown diagram-matically in Fig. 4.

As for all mass flow processes, the rates of filtration and excretion are equal to the product of the volume flow

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FIGURE 4 Schematic diagram of solute movements between the nephron and renal capillary networks. An ultrafiltrate of plasma is formed by filtration out of the glomerular capillaries. The amount of a given substance that is filtered can be reduced by reabsorption along the nephron, which returns solute through the renal interstitium to the peritubular capillaries and, thus, to the systemic circulation. Secretion adds to the amount of the substance excreted by moving the substance through the renal interstitium from the peritubular capillaries into the lumen of the nephron and, thus, reduces the amount of the substance that returns to the systemic circulation via the renal venous plasma.

FIGURE 4 Schematic diagram of solute movements between the nephron and renal capillary networks. An ultrafiltrate of plasma is formed by filtration out of the glomerular capillaries. The amount of a given substance that is filtered can be reduced by reabsorption along the nephron, which returns solute through the renal interstitium to the peritubular capillaries and, thus, to the systemic circulation. Secretion adds to the amount of the substance excreted by moving the substance through the renal interstitium from the peritubular capillaries into the lumen of the nephron and, thus, reduces the amount of the substance that returns to the systemic circulation via the renal venous plasma.

rate and the solute concentration in that volume (see Eq. [1]). In the case of glomerular filtration, the volume flow is the glomerular filtration rate (GFR). For most substances that are freely filtered and not bound to plasma proteins, their concentration in the glomerular ultrafiltrate will be equal to their plasma concentration Px, where x stands for any solute meeting these criteria. The filtration rate of the substance x can be calculated as

Filtration rate = GFR Px.

Similarly, the excretion rate for a given substance x can be calculated from the rate of UF and the concentration of the substance in the urine (Ux):

These calculations become important when one attempts to determine quantitatively the effectiveness of the kidney in filtering and excreting substances, as illustrated by the determination of the glomerular rate.

Determination of the Glomerular Filtration Rate

The GFR is an important clinical index of the extent and rate of progression of renal disease. In most renal diseases, there is a progressive loss of filtering glomeruli due to sclerosis and destruction. Consequently, the GFR falls as the disease progresses. The clinician tries to obtain an indication of the GFR first to determine the extent of renal damage that has already occurred and then, at intervals, to determine the rate at which the disease is abating or progressing.

Ideally, one would like to be able to measure the GFR directly. This could be accomplished if one could measure the rate of excretion of a substance that is neither reabsorbed nor secreted. If such a substance were freely filtered, the rate of its excretion would be precisely equal to the rate of its filtration (see Eq. [5] and Fig. 5). one substance that meets the criteria is inulin. Inulin is a polyfructose molecule with a molecular weight of 5500 Da. The molecular size of inulin is sufficiently small that it is freely filtered at the glomerulus. However, inulin is too large to be reabsorbed passively by the nephron, and there are no active transport processes that reabsorb or secrete inulin. This relationship can be expressed quantitatively as follows, using Eqs. [6] and [7]:

Rate of inulin filtration = rate of inulin excretion GFR • Pin = UF • Uin ;

FIGURE 5 The rate of inulin excretion is equal to the rate of inulin filtration at the glomerulus because inulin is neither secreted nor reabsorbed.

where the subscript "in" refers to inulin. Rearranging gives:

The right-hand side of Eq. [9] is referred to as the clearance of inulin. Conceptually, it represents the milliliters per minute of plasma that flow through the kidney from which inulin is completely removed. In the case of this substance, which is neither reabsorbed nor secreted, this clearance equals the GFR.

Clearance Concept

Plasma clearance, although a complete abstraction, is widely used to describe the effectiveness with which the kidney excretes various substances, and the student should study carefully the following discussion and quantitative relationships to understand the concept. Consider the relationship between the flows of fluid and their inulin concentrations in Fig. 6. In this example, it is assumed that the inulin concentration in the plasma is 1 mg/dL. Of the approximately 130 mL/min of fluid that are filtered at the glomerulus, more than 99%, or about 129 mL/min, is normally reabsorbed, resulting in a UF of 1 mL/min. Because inulin is neither reabsorbed nor secreted, its concentration in the tubular fluid rises along the nephron until it reaches a maximum concentration 130-fold higher than in the plasma: 130 mg/dL. This figure is determined by rearranging Eq. [9] to give Uin = (GFR . Pin)/UF.

On the other hand, the reabsorbed fluid that is returned to the peritubular circulation contains no inulin and, thus, the inulin concentration of the renal venous plasma is diluted. The concentration of inulin in the renal venous plasma can be calculated for the example in Fig. 6 by dividing the amount of inulin flowing out of the kidney via the renal vein by the volume flow rate. As seen in Fig. 6, if we neglect the small amount of inulin that leaves the kidney by the lymphatics, the amount of inulin flowing out of the kidney by the renal vein will be equal to the amount flowing in by the renal artery minus the amount that is excreted. The urea inflow via the renal arteries is 6.9 dL/ min • 1 mg/dL = 6.9 mg/min, and the amount that is filtered (and therefore, excreted) is 0.01 dL/min • 130 mg/dL = 1.3 mg/min. Thus, the amount of inulin leaving the kidney via the renal vein will be (6.9 — 1.3) = 5.6 mg/min. The renal venous plasma flow equals the arterial inflow minus the UF, or 689 mL/min (6.89 dL/ min). Thus, the inulin concentration in the renal venous blood will be (5.6 mg/min)/(6.89 dL/min) = 0.81 mg/ dL. This flow would be equivalent to a combined flow of 560 mL/min containing 1 mg/dL of inulin (equivalent to the flow through the efferent arterioles after filtration)

Renal Flow
FIGURE 6 The clearance of inulin is equivalent to the renal plasma flow from which inulin has been completely removed plus the urine flow rate (UF). See text for explanation.

plus a flow of 129 mL/min containing no inulin (equivalent to the reabsorbed fluid).

These considerations lead directly to the clearance concept. The clearance of inulin is equal to the flow of renal venous blood that can be considered to be free (cleared) of inulin plus the UF that is lost: 129 mL/min + 1 mL/min =130 mL/min. This is equal to the clearance of inulin calculated using Eq. [9]. Thus, in general, the clearance of any substance (Cx) from the plasma is:

If the urine and plasma concentrations in Eq. [10] are in the same units (e.g., mg/dL or mmol/L), the clearance has the same units as the urine volume flow rate. Thus, clearances are traditionally reported in units of milliliters per minute.

A substance that is reabsorbed along the nephron has a clearance that is less than the GFR because some of the filtered amount returns to the peritubular capillaries and adds to the amount flowing out of the renal vein. Alternatively, any substance that is secreted along the nephron has a clearance that is higher than the GFR because the peritubular blood flow loses some of the solute due to secretion, resulting in a lesser amount of solute flowing out of the kidney via the renal vein.

Practical Indicators of the Glomerular Filtration Rate

The importance of inulin, a substance that is neither reabsorbed nor secreted, is that it allows the GFR to be determined from three parameters that are measured relatively easily: UF, the plasma concentration of inulin, and the urine concentration of inulin. The only problem is that inulin does not occur naturally in the mammalian body and, therefore, to measure its clearance, inulin must be infused intravenously to produce readily measurable plasma concentrations.

Because of its large size, inulin distributes only in the extracellular space. To obtain a plasma inulin concentration of 1 mg/dL, as in the preceding example, if the extracellular fluid volume is 15 L, one has to infuse a priming dose of 150 mg. However, because of the renal clearance of the inulin, its plasma concentration immediately begins to fall. To maintain a constant plasma inulin concentration while the clearance is being measured, inulin has to be constantly infused at the same rate at which it is being lost in the urine, for example, at 1.3 mg/min in the preceding example. Obviously, the measurement of an inulin clearance can be conducted only in the hospital, where intravenous infusions are possible. In practical terms, this is seldom done in a patient except as part of a research protocol when the best estimates of GFR are required. Other substances that are neither reabsorbed nor secreted can be used clinically, but, as with inulin, they must be infused because they are not normally present in the body.

To avoid the complications of intravenous infusion, it is preferable to measure the clearance of an endogenous substance that approximates the GFR. Historically, physicians sometimes measured the clearance of urea as an approximation. However, urea is reabsorbed to a substantial degree along the nephron, and its rate of reabsorption is quite dependent on the UF. Consequently, the urea clearance not only underestimates the true GFR substantially, but it is dependent on the UF.

A better approximation of the GFR is obtained by measuring the clearance of creatinine. Creatinine is produced in the body by the metabolism of creatine and phosphocreatine derived from the breakdown of muscle. Muscle tissue is constantly being broken down and rebuilt in the body, and meats (i.e., animal muscle) provide an additional source in the usual Western diet. Almost no creatinine is directly ingested; therefore, in order for the body to maintain a constant creatinine content, the rate of creatinine excretion must equal its rate of metabolic production, which in a 70-kg man is somewhat less than 2 g/day. Creatinine is actively secreted into the urine in the late proximal nephron, but at a relatively slow rate in comparison with its rate of glomerular filtration. Consequently, the rate of creatinine excretion is normally only about 10% higher than its rate of filtration, and the creatinine clearance provides a useful measure, albeit a slight overestimate, of the GFR.

In practical terms, the rate of creatinine production and, thus, its rate of excretion, varies in the course of a day. Production increases about 1 hr after a meal and wanes during the night. Therefore, in order to estimate the GFR of a kidney patient, the nephrologist will usually have the patient collect his or her entire urine volume over a 24-hr period. The creatinine concentration in that urine is an average of the urine concentration over a 24-hr period, the so-called clearance period. A blood sample is usually drawn at some point during or just at the end of the clearance period, when the patient comes to the physician's office with his or her sample of urine. Knowing both the urine and plasma creatinine concentrations, one needs only the average UF rate during the clearance period in order to calculate the creatinine clearance according to Eq. [10]. The average UF rate is simply the quotient of the total urine volume and the total minutes over which it was collected, with the final answer given in units of milliliters per minute.

However, in following a patient over a long period of chronic renal failure, the physician may only occasionally measure the creatinine clearance. Instead, the physician may more conveniently follow the plasma creatinine concentration (Pcr) as an index of the GFR. The reason that this concentration reflects the GFR can be understood from the following analysis of the mass balance concepts developed in this chapter.

As noted earlier, creatinine is produced by metabolism at a rate of ~2 g/day. To maintain mass balance of creatinine, the body must match this rate of production by the excretion rate. However, the creatinine excretion rate is approximately equal to the rate of creatinine filtration, if one neglects the small contribution of secretion. Thus, from Eq. [8]:

where the subscript "cr" indicates creatinine. Actually, because of the small amount of creatinine secreted, the rate of creatinine filtration is closer to 1.3 mg/min. Using this approximation and rearranging the equation, one can see that the plasma creatinine concentration is inversely proportional to the GFR:

Therefore, for a normal GFR of 130 mL/min, one would predict that the plasma creatinine concentration would be (1.3 mg/min)/(1.3 dL/min) = 1.0 mg/dL. (As in the case of many plasma nonelectrolytes, the creatinine concentration is given in clinical units of milligrams per decaliter.) Thus, the equation predicts what is regarded as the normal plasma creatinine concentration rate of 0.8-1.2 mg/dL. Using Eq. [12], we can examine what happens to the plasma creatinine concentration when the GFR is progressively decreased by renal disease.

If the GFR were halved to 65 mL/min by a loss of functioning nephrons, Eq. [12] predicts that the plasma concentration of creatinine would double to 2 mg/dL. The increase in plasma creatinine concentration would occur merely because of the necessity of the rate of filtration to increase to match the rate of creatinine production. With GFR reduced by half, Eq. [6] indicates that the same rate of creatinine filtration can be maintained only by a doubling of the creatinine concentration in the plasma. If the GFR fell progressively over a period of years, as in chronic renal failure, the plasma creatinine concentration would slowly rise, so that at any given time the rate of creatinine excretion would equal the rate of production. There would be only imperceptible transient periods in which the rate of excretion lagged behind the rate of production resulting in a rise in the plasma creatinine concentration.

It is again most important to realize that an elevated Pcr does not indicate that the kidneys are not excreting a normal amount of creatinine. When a physician sees a normal patient or one with renal damage, either patient is usually approximately in balance with respect to creatinine, that is, he or she is excreting it at the rate at which it is being produced. The problem is that the patient with the damaged kidney requires a higher plasma creatinine concentration to maintain this normal rate of excretion, which is determined primarily by filtration.

If the loss of functioning nephrons progresses in the patient, the creatinine concentration will rise in inverse proportion to the fall in GFR. According to Eq. [12], when the GFR is 25% of normal, the creatinine concentration will be 4 mg/dL; at a GFR of 12.5%, 8 mg/dL, etc. Thus, as illustrated in Fig. 7, the plasma creatinine concentration rises geometrically as the GFR falls.

Because the rate of urea excretion is also primarily dependent on the GFR, measurements of plasma urea concentration are also followed. The urea concentration is traditionally reported in the rather obscure clinical units of blood urea nitrogen (BUN, mg/dL). The BUN actually is the concentration of the nitrogen atoms associated with urea. Because there are two nitrogen

FIGURE 7 Change in the plasma creatinine concentration (Pcr) as a function of the glomerular filtration rate (GFR). In this example, the rate of creatinine production, which is assumed to remain constant, was ~1.8 g/day, or 1.3 mg/min.

atoms per urea molecule, a 1.0 mmol/L urea concentration is equivalent to a BUN of 2.8 mg/dL. The normal BUN is in the range of 10-20 mg/dL, which is, therefore, equivalent to a urea concentration of 3.6-7.1 mmol/L.

Note from Fig. 7 that even a doubling of the normal plasma creatinine concentration indicates a significant fall in the GFR to one-half normal, but the diagnosis of renal damage should never be based on a single such

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