Solute and Water Transport in Peritoneal Dialysis: A Case-Based Primer

Special Article Solute and Water Transport in Peritoneal Dialysis: A Case-Based Primer Ramesh Khanna, MD Peritoneal dialysis (PD) is an effective therapy for patients with end-stage kidney disease. Dialysis solutions containing physiologic concentrations of electrolytes and base, as well as glucose often at supraphysiologic concentrations, are infused into the peritoneal cavity for solute and water exchange, and the patient’s own peritoneal membrane is used for dialysis. The peritoneal membrane is dominated by small pores, which allow transport of water and small-molecular-size solutes, including electrolytes, by way of both diffusion and convection. Through small pores, diffusion allows the movement of solutes from the high-concentration compartment to a lower-concentration region. Also, through small pores, water and solutes move together by convection in response to an osmotic force. The glucose in the dialysis solution generates osmotic force to drive convection. In addition to small pores, the peritoneal membrane contains a specialized water channel, aquaporin 1, which is also present in capillaries of the peritoneal membrane. These specialized water channels, which are upregulated by glucose, allow water transport without solute (free water) in response to the osmotic force induced by glucose in the PD solution. During a PD exchange, net loss or gain of electrolytes and base is determined by both their gradient between capillary blood and dialysis solution and the net ultrafiltration volume. Developing a PD prescription, including the amount of glucose used, and changing the prescription in response to dietary changes and/or loss of residual kidney function requires a sound understanding of the peritoneal physiology. The case studies presented here help solidify the basic elements of PD prescription and how the PD prescription should be altered in response to changing clinical situations. Am J Kidney Dis. -(-):---. ª 2016 by the National Kidney Foundation, Inc. INDEX WORDS: Peritoneal dialysis (PD); peritoneal physiology; peritoneal membrane; peritoneum; pores; diffusive transport; convective transport; solute transfer; water channel; aquaporin 1; PD exchange; ultrafiltration; electrolyes; icodextrin; dwell time; end-stage renal disease; review.

Ramesh Khanna, MD, was the J. Michael Lazarus Distinguished Award recipient at the 2016 National Kidney Foundation Spring Clinical Meetings, together with Karl Nolph, MD, who was awarded this distinction posthumously. This award was established to honor Dr Lazarus for his major contributions to the clinical science and care of dialysis patients and to recognize individuals whose research has yielded novel insights related to renal replacement therapy.

INTRODUCTION Overview Peritoneal dialysis (PD) is an effective treatment for patients with chronic kidney failure. PD uses a patient’s own peritoneal membrane, across which fluids and solutes such as creatinine, urea, electrolytes, glucose, and many other unmeasured uremic toxins are exchanged between the capillary blood and PD solution. The PD solution is infused into the peritoneal cavity through a permanently implanted silicone rubber tube, the PD catheter. After dwelling in the cavity for a predetermined period, the PD solution is drained out and fresh PD solution is reinstilled. Such exchanges are performed regularly throughout the day, either with manual exchanges as in continuous ambulatory PD (CAPD) or with the help of a machine as in automated PD (also referred to as continuous cycling PD [CCPD]). Experience with PD therapy Am J Kidney Dis. 2016;-(-):---

over the past 3 to 4 decades has shown that 5-year patient survival is comparable to in-center hemodialysis. Importantly, PD costs much less than hemodialysis in most parts of the world and can readily be performed at home without the need for complicated machines. Critically, successful PD requires a hygienic environment and attention to detail to minimize the risk for peritonitis. Peritoneum The fraction of the peritoneal membrane actively involved in the PD exchange is largely unknown. Nevertheless, it is apparent that the surface area of peritoneum involved in the transport process is not a fixed quantity. Therefore, it is more meaningful to measure it functionally. Consequently, physiologists have suggested estimating the proportion of functional From the Division of Nephrology, Department of Medicine, University of Missouri-Columbia, Columbia, MO. Received May 24, 2016. Accepted in revised form November 27, 2016. Address correspondence to Ramesh Khanna, MD, Division of Nephrology, Department of Medicine, University of MissouriColumbia, Columbia, MO 65212. E-mail: khannar@health. missouri.edu  2016 by the National Kidney Foundation, Inc. 0272-6386 http://dx.doi.org/10.1053/j.ajkd.2016.11.007 1

Ramesh Khanna

pore area in relation to the effective diffusion path length of the membrane. This calculation takes into account capillary surface area and the distance between capillary and dialysis solution/mesothelial contact.1-5 The capillary wall (to a great extent), the interstitium, the mesothelial cell layer (to a lesser extent), and the unstirred dialysate fluid layer at the mesothelium (to a minor extent) all offer resistance to the free transport of water and solutes during dialysis. Pores in the Peritoneal Membrane The peritoneal membrane contains predominantly small pores of 40 to 50 Å and rare large pores of 150 Å. Specialized water channels, ultrasmall pores or aquaporin 1, are present in the mesothelial cells and the capillaries.6-9 In the presence of glucose, these channels are upregulated.10 Aquaporin 1 channels allow transport of solute-free water in response to a crystalloid-induced osmotic pressure. In clinical practice, such water movement is termed free water transport. Small pores and aquaporin 1 play major roles in solute and water transport. Because of their paucity, large pores do not contribute significantly to solute and water transport. Macromolecules such as albumin and larger proteins are transported through large pores at a very low rate and do not reach equilibrium between capillary blood and dialysis solution even after a 24-hour PD solution dwell, although their concentration in the PD solution is extremely low. Nevertheless, such small losses of protein and albumin result in lower serum protein and albumin levels in PD patients. Diffusive and Convective Transport Solute and water are transported through the peritoneal membrane by diffusion and convection. Small pores allow the transport of small-molecular-size solutes, including electrolytes, by way of both diffusion and convection. During the diffusion process, the solute concentration gradient between 2 compartments (capillary blood and PD solution in the peritoneal cavity) that are separated by the semipermeable peritoneal membrane allows solute to move from a higher to a lower concentration compartment. The rate of transfer by diffusion of a solute is determined by the diffusive permeability (ratio of free diffusion coefficient and the distance of diffusion) of the peritoneum to that particular solute and the surface area available for transport. Convective solute transport occurs in conjunction with water transport from one compartment to the other. In PD, the glucose in the dialysis solution in the peritoneal cavity generates crystalloid osmotic pressure, which, along with transmembrane hydrostatic pressure, promotes convective water and solute transport from the capillary blood to the dialysis 2

solution. The effectiveness of glucose in promoting water and solute transport during convection is determined by the resistance offered by the peritoneal membrane to glucose transport from the peritoneal cavity to the capillaries. In physiology, this resistance is expressed as the osmotic reflection coefficient. For the peritoneal membrane, the reflection coefficients reported for small solutes vary greatly, depending on the measurement method,11 and differ for different solutes and pores. By definition, the reflection coefficient is 1.0 for a solute with complete resistance, such that the solute cannot cross the membrane, and 0 when the membrane offers no resistance to solute movement. The osmotic reflection coefficient for small osmotic solutes including glucose is very low, reportedly between 0.02 and 0.05 when calculated by assessing the transperitoneal ultrafiltration rate in response to the solute concentration rate.11 However, other studies have noted reflection coefficients that are significantly greater, between 0.2 and 0.6, based on assessing the transperitoneal solute transport rate during a hypertonic exchange.12,13 During a peritoneal dialysis exchange, the reflection coefficient for glucose is the mean of its reflection coefficient for all pores, including small and large pores and aquaporin 1 channels. The sieving coefficient is determined by the ratio of the solute concentration in the filtrate divided by its concentration in plasma in the absence of diffusion. The value, 1.0 minus the sieving coefficient, is an approximate estimate of the reflection coefficient for a given solute; this is called the rejection coefficient. Solute Transfer Dialysis solution contains physiologic concentrations of electrolytes, including sodium, calcium, magnesium, and bicarbonate. Consequently, diffusion plays an inconsequential role in the removal of these electrolytes from the body during peritoneal dialysis. These electrolytes may enter the body by diffusion from the dialysate into the body based on their concentration gradient. Accordingly, ultrafiltration is the dominant way of removing electrolytes from the body. During a PD exchange, both small pores and aquaporin 1 contribute somewhat equally to water transport.3,14,15 Because water transport through aquaporin 1 is solute free, sodium concentration decreases in the dialysate during the first 2 to 3 hours of a PD exchange because the high volume of free water, generated through aquaporin 1 in the early phase of an exchange in response to glucose-induced osmotic pressure, dilutes the sodium in the dialysis solution. Due to free water loss from the blood compartment, serum sodium concentration increases. This phenomenon, in which Am J Kidney Dis. 2016;-(-):---

Solute and Water Transport in PD

sodium is left behind in the blood compartment while water enters the dialysate compartment, is called sodium sieving. In the later phase of an exchange, when the dialysate glucose concentration declines, resulting in a decline in ultrafiltration rate, diffusion of sodium from the capillary blood tends to equilibrate with that of the dialysate sodium; such sodium equilibration typically occurs with longer duration dwell exchanges, as often is the case with CAPD. In summary, the large volume of hypotonic ultrafiltrate that occurs in the early phase of an exchange dilutes the sodium in the dialysate. In the later phase of this exchange, when much of the glucose may have been absorbed into the blood compartment and very little ultrafiltration is occurring, sodium diffuses from the blood to the dialysate to equilibrate the sodium concentration. The longer the dwell, the more likely that sodium will be in equilibrium between the capillary blood and dialysate.16 The clinical implication of this observation is that short dwell exchanges using hypertonic dialysis solution will remove more water than sodium from the extracellular fluid compartment. This can lead to hypernatremia.17 Calcium and magnesium transport though the peritoneum is similar to sodium transport. The standard PD solution calcium concentration is slightly higher than that of plasma. Consequently, calcium diffuses into capillary blood from the dialysis solution. However, if calcium removal is desired, lower calcium dialysis solutions can be used; these lower calcium solutions are typical in the United States. Studies have documented positive mass transfer of calcium and magnesium during CAPD exchanges using isotonic exchanges18 and approach zero balance with hypertonic solutions as convective transport of these electrolytes neutralizes the diffusive transport.18,19 PD solution typically is free of potassium. Therefore, diffusion is the predominant way of removing potassium during peritoneal dialysis. Convective removal of potassium is inconsequential because the potassium concentration in blood is low. In other words, assuming the plasma potassium level is 4 mEq/L, 1 L of ultrafiltration through small pores can remove only 4 mEq of potassium. Traditionally, PD solution does not contain bicarbonate, but rather contains lactate as the buffer. Consequently, bicarbonate concentration in the PD solution is zero. Therefore, bicarbonate transport from the blood to PD solution during dialysis exchanges occurs through diffusion and additional losses of bicarbonate from the blood compartment occurring via convective clearance during ultrafiltration. However, bicarbonate losses during PD exchanges are offset by lactate diffusion from the PD solution into blood followed by conversion in the body to bicarbonate. Bicarbonate loss through convection Am J Kidney Dis. 2016;-(-):---

when ultrafiltration rates are high tends to be substantial, especially with use of hypertonic dialysis solution exchanges. With a lactate concentration in the PD solution of 40 mmol/L, total alkali gain is virtually identical to total acid production (54.2 vs 52.4 mEq/d), suggesting that PD patients will be in neutral acid-base balance.20 However, it should be noted that PD patients with liberal fluid intake who therefore require high ultrafiltration volumes using hypertonic solutions will have higher bicarbonate losses through convection and therefore be at higher risk for chronic metabolic acidosis. Ultrafiltration In the United States, 2 types of dialysis solutions currently are available for inducing ultrafiltration during the PD: dextrose (glucose)-based solutions and those containing icodextrin, a high-molecular-weight polymer (16.8 kDa) of glucose derived from corn starch that is poorly absorbed from the peritoneal cavity. Table 1 shows the essential characteristics of dextrose and icodextrin formulations. It is reported that average icodextrin absorption during an 8-hour PD exchange is 19%,21 much lower than would be seen for glucose, and when using a 7.5% icodextrin solution, average ultrafiltration rate is 1.4 to 2.3 mL/min.22 Glucose in the PD fluid generates crystalloid osmotic force. This osmotic force coupled with transmembrane hydrostatic pressure generates a substantial amount of transcapillary ultrafiltration. Water is transported from the capillary blood to dialysate compartment through both aquaporin 1 and small pores in response to osmotic forces generated by the glucose. A large fraction of this ultrafiltrate subsequently is reabsorbed via back filtration in venules Table 1. Dextrose and Icodextrin Dialysis Solution Compositions

Dextrose, g/dL Icodextrin, g/dL Sodium, mEq/L Chloride, mEq/L Calcium, mEq/L Magnesium, mEq/L Lactate, mEq/L Osmolarity, mOsm/kg pH

Icodextrin

Dextrose

—

1.5, 2.5, 4.25

7.5 132 96 2.5 0.5 40 282-286 5.2

132 96 2.5 0.5 40 346, 395, 485 5.2

—

Note: Icodextrin peritoneal dialysis solution contains 7.5% icodextrin as the primary osmotic agent. Additional components of the icodextrin formulation are detailed on this table. The electrolyte composition and buffer (lactate) are identical in both icodextrin and dextrose formulations. Icodextrin is essentially iso-osmolar with respect to plasma (w285 mOsm/L). In contrast, dextrose-containing solutions are hyperosmolar relative to plasma; osmolarities range between 346 (for 1.5% dextrose) and 485 mOsm/kg (4.25% dextrose). 3

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and capillaries, whereas a smaller fraction is reabsorbed through the lymphatics located in the under surface of the right diaphragm. Net ultrafiltration, assessed by quantifying 24-hour dialysate drain minus fill volume, is only a fraction of the total amount of ultrafiltration that was generated at the capillary level. Back absorption of fluid into capillaries follows Starling forces, whereas lymphatic absorption is convective and driven by the negative force created by expansion of the chest wall during inspiration. Icodextrin, being a macromolecule, induces colloidal osmosis.23 Because macromolecules are unable to cross capillary walls, the oncotic pressure generated by icodextrin tends to pull fluid in its direction. This process is similar to oncotic pressure in the blood compartment. Oncotic pressure is generated in the interior of blood vessels by the presence of large quantities of albumin. Because albumin is unable to escape through the capillary wall, oncotic pressure in the capillary interior tends to draw fluid in its direction. Icodextrin, being a macromolecule, induces fluid transport at the capillary level in its direction even in an isotonic or hypotonic state. Consequently, dialysis solution containing icodextrin draws fluid out of capillary blood despite low osmotic force, through ubiquitous small pores (Table 2). In contrast, icodextrin solution fails to draw water via aquaporin 1 due to low numbers relative to electrolytes and extremely low osmotic force compared to glucose. Also in contrast to glucose, icodextrin stays in the PD fluid for long periods due to extremely slow metabolism and poor absorption from the peritoneal cavity. Consequently, icodextrin solution continues to draw fluid into the peritoneal compartment for a longer period than glucose-based solution and does not cause sodium sieving. Equations Because 3.8% dextrose and 7.5% icodextrin PD solutions induce almost similar 8-hour ultrafiltration volumes during PD exchanges, these 2 solutions are used here for comparison and calculations. The average ultrafiltration rate can be calculated using equation 1 (Box 1). A major determinant for average ultrafiltration rate per exchange with glucose is the rate of glucose absorption and the speed of decline of the glucose gradient. The osmotic pressure generated by the glucose and icodextrin for small pores and for aquaporin are depicted

in equations 2 and 3 (Box 1). The osmotic reflection coefficient of an osmotic agent determines its effectiveness. As mentioned, for solutes totally impermeable, the reflection coefficient is 1, and for solutes that pass through with no hindrance, it is 0. For solutes with a reflection coefficient of 1, each 1 mOsm exerts an osmotic pressure of 19.3 mm Hg according to Van’t Hoff’s law.24,25 Box 1 also includes equations for osmotic gradients at the start of an exchange with 3.8% dextrose and 7.5% icodextrin solutions and for osmotic pressures as a function of the reflection coefficient for 3.8% glucose and 7.5% icodextrin solutions for both small pores and aquaporin. Using these equations, a summary of differences between 3.8% glucose and 7.5% icodextrin PD solution characteristics for small pores and aquaporin is depicted in Table 2.

CLINICAL APPLICATION OF TRANSPORT THEORY The characterization in the previous section of glucose and icodextrin PD solutions with reference to osmotic pressure exerted by glucose and icodextrin allows us to answer several clinical questions relative to their clinical use. Tables 1 and 2 guide the approach to these questions. Why does 3.8% glucose solution, and not 7.5% icodextrin, induce free water transport across aquaporin 1? Refer to Table 1 and equation 1 to understand how the numbers cited below are derived. The reflection coefficients for glucose and icodextrin are the same, both being 1.0 for aquaporin 1 channels. Notably, the effective gradient (gradient 3 reflection coefficient) for glucose at the time of initiation of the exchange is very high (209 mOsm/L), as opposed to icodextrin, which at 4.7 mOsm/L appears inconsequential. Accordingly, the initial osmotic pressure for glucose (effective osmotic gradient 3 reflection coefficient 3 19.3 5 209 mOsm/L 3 1 3 19.3 mm Hg/mOsm/L 5 4,034 mm Hg/mOsm/L) is very high compared to that for icodextrin (4.7 mOsm/L 3 1 3 19.3 mm Hg/ mOsm/L 5 90.7 mm Hg/mOsm/L). This high free water transport with glucose-based solution can occur in the setting of very low (2%) pore area. Because of inconsequential free water transport with icodextrin through aquaporin 1, no sodium sieving occurs with icodextrin; this is in contrast to free water transport

Table 2. Summary of Differences Between 3.8% Glucose and 7.5% Icodextrin Characteristics for Small Pores and Aquaporin 1 Pore Type

Osmotic Agent

Initial Gradient, mOsm/L

Reflection Coefficient

Osmotic Pressure,a mm Hg/mOsm/L

Pore Area, %

Small pores Small pores Aquaporin 1 Aquaporin 1

3.86% glucose 7.5% icodextrin 3.86% glucose 7.5% icodextrin

209 4.7 209 4.7

0.05 1.0 1.0 1.0

201.7 90.7 4,034 90.7

95 95 2 2

Gradient 3 reflection coefficient 3 19.3 5 osmotic pressure mm Hg/mOsm/L.

a

4

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Solute and Water Transport in PD Box 1. Equations Equation 1: Average UF rate during a PD exchange (mL/min) 5 [average osmotic pressure + net transmembrane hydrostatic pressure 2 net oncotic pressure (mm Hg)] 3 [hydraulic conductivity (cm/min/mm Hg)] 3 [total effective pore area (cm2)] Osmotic Pressurea Equation 2: Osmotic pressure with glucose solution (mm Hg) 5 [0.05 (glucose D 2 glucose P) + 0.05 (Na D 2 Na P) + 0.0 (urea D 2 urea P) + 0.05 (Cl D 2 Cl P) + etc] 3 [19.3 mm Hg/mOsm/L]b Equation 3: Osmotic pressure with icodextrin solution (mm Hg) 5 [1.0 (icodextrin D 2 icodextrin P) + 0.05 (Na D 2 Na P) + 0.0 (urea D 2 urea P) + 0.05 (Cl D 2 Cl P) + etc] 3 [19.3 mm Hg/mOsm/L]b Osmotic Gradients at Exchange Initiation for 3.86% Glucose and 7.5% Icodextrin 3.86% dextrose 5 (38,600 mg/L)/(180 mg/mOsm) 5 214 mOsm/L in dialysate glucose D 2 glucose P 5 (214 mOsm/L 2 5 mOsm/L) 5 209 mOsm/L gradient (assuming a plasma glucose concentration of 90 mg/dL) 7.5% icodextrin 5 75,000 mg/L/16,000 mg/mOsm 5 4.7 mOsm/L in dialysate icodextrin D 2 icodextrin P 5 4.7 2 0 5 4.7 mOsm/L gradient Initial Gradients 3 Reflection Coefficients Small Pores Glucose: 209 3 0.05 5 10.45 effective mOsm/L, where 0.05 is the small pore reflection coefficient for glucose Icodextrin: 4.7 3 1.0 5 4.7 effective mOsm/L, where 1.0 is the small pore reflection coefficient for icodextrin Aquaporins Glucose: 209 3 1.0 5 209 effective mOsm/L, where 1.0 is the AQP1 reflection coefficient for glucose Icodextrin: 4.7 3 1.0 5 4.7 effective mOsm/L, where 1.0 is the AQP1 reflection coefficient for icodextrin Abbreviations: AQP1, aquaporin 1; D, dialysis solution; P, plasma; PD, peritoneal dialysis; UF, ultrafiltration. a The equations shown are for small pores, where it is assumed that the reflection coefficient for glucose, Na, and Cl are 0.05 and for icodextrin is 1.0. For corresponding equations for AQP1, for both glucose and icodextrin solutions, use the same equations as for small pores but use a reflection coefficient of 1.0 for both glucose and icodextrin. b Solutes with a reflection coefficient of 1, every 1 mOsmol exerts an osmotic pressure of 19.3 mm Hg according to Van’t Hoff’s law.24,25

through aquaporin 1 associated with the glucose PD solution. Because of inconsequential free water transport with icodextrin, no dilution of sodium in the dialysate occurs in the early phase of an exchange with icodextrin PD solution. Therefore, the difference in total ultrafiltration volume between glucose and icodextrin solutions under a standardized condition is a good reflection of free water transport volume occurring with glucose. Why does icodextrin yield minimal ultrafiltration through aquaporin 1? Although icodextrin’s reflection coefficient is high and similar to that of glucose (1.0), the effective Am J Kidney Dis. 2016;-(-):---

gradient is extremely low (4.7 mOsm/L). Consequently, as shown previously, the initial osmotic pressure is low (90.7 mm Hg/mOsm/L), and pore area also is low due to a lack of aquaporin 1 upregulation. Why does icodextrin solution yield ultrafiltration even with osmolality lower than plasma? Because the magnitude of the effective osmotic gradient related to icodextrin is greater than the combined negative effective gradients for other solutes, there is a positive effective osmotic gradient. This is a consequence of the very low reflection coefficients for solutes such as sodium, chloride, urea, etc. For example, if the icodextrin dialysate to plasma osmolar 5

Ramesh Khanna

gradient [D-P gradient] is 15 and the gradients for glucose, sodium, chloride, and urea are 25, 25, 25, and 230, respectively, effective osmotic gradient could be estimated as follows, incorporating the gradients for these osmoles and their reflection coefficients:

of these case examples, we ignore the Gibbs-Donnan effect on either side of the peritoneal membrane. Refer to Table 2 for the composition of electrolytes in the dextrose dialysis solutions.

5 icodextrin 1 glucose 1 sodium 1 chloride

A 36-year-old man treated with CAPD (4 3 2 L with daily ultrafiltration [UF] volume 5 0.5 L) and daily urine sodium (Na) losses of 66 mEq has a usual serum Na level of 132. He becomes anuric after exposure to intravenous contrast, aminoglycosides, and excessive use of ibuprofen for arthritis.

1 urea osmotic pressures 5ð15 3 1:0Þ 1 ð25 3 0:05Þ 1 ð25 3 0:05Þ 1 ð25 3 0:05Þ 1 ð230 3 0Þ 5 520:75 5 14:25 mOsm=L This positive osmotic gradient would generate an osmotic pressure (osmotic gradient 3 19.3) of 82.025 mm Hg/mOsm/L due to icodextrin, which is sufficient to yield enough sustained slow ultrafiltration. Although the PD solution may become hypo-osmotic, why does it still yield ultrafiltration late in a hypertonic glucose exchange? Late in a hypertonic exchange, the D-P glucose gradient may be positive, while D-P sodium, chloride, and urea are negative. For example, if they are respectively 110, 24, 24, and 210 mOsm/L, net dialysate osmolality would be w8 mOsm/L lower than plasma osmolality. Incorporating the reflection coefficients, the effective osmotic gradients 3 reflection coefficients 5 glucose (10 3 0.05) 1 (28 3 0.05) 1 (210 3 0) 5 10.1 mOsm/L. This positive osmotic gradient due to glucose remains sufficient to yield a continued low rate of ultrafiltration. The phenomenon of hypo-osmolar ultrafiltration (Fig 1) has been previously explained in our work (see Fig 7 in Nolph et al26). In that publication, we clarified that: The low electrolyte ultrafiltrate helps to shorten the time to crystalloid equilibrium and probably explains the overshoot where the dialysate becomes hypo-osmolar. Glucose equilibrium was not observed during the six hours. Clinical studies have also reported glucose dysequilibrium well past the time when dialysate osmolality decreases to serum osmolality. Some transcapillary ultrafiltration appears to persist during the hypo-osmolar phase. Osmotic pressure is not balanced at osmotic equilibrium, and dysequilibrium of electrolytes and glucose persists. Since the osmotic pressure asserted by a solute is a function of the molal concentration and reflection coefficient, the relatively higher reflection coefficient for glucose may generate osmotic pressure favoring transcapillary ultrafiltration even when the dialysate has become slightly hypo-osmolar.26(p225)

CASE STUDIES The following case studies can help clarify some of the physiology discussed previously. For the purpose 6

Case 1

Question What changes in the CAPD treatment will maintain the same serum Na level, assuming no change in intake or stool losses? Answer With total infused dialysate volume of 8 L, net UF volume of 0.5 L, and a dialysate Na concentration of 132 mEq/L, Na removal by CAPD 5 [(serum Na 2 dialysis solution Na) 3 8 L infusion volume] 1 (0.5 L UF volume 3 132) 5 [(132 2 132) 3 8] 1 [0.5 3 132] 5 66 mEq. After becoming anuric, at a serum Na level of 132 mEq/L, diffusion removal will still be 0 regardless of exchange volume or number; that is: [(132 2 132) 3 instilled volume] 5 0 no matter the amount of the instilled volume. Therefore, increasing Na removal by 66 mEq with CAPD must occur by UF (convection), and to maintain prior total Na clearance, (132 3 daily UF volume) must equal 66 mEq of previous Na CAPD removal plus 66 mEq additional clearance (that was excreted through urine), for a total of 132 mEq. Thus, 132 3 UF volume 5 132; reorganizing this equation to solve for UF volume, UF volume equals 1 L. Accordingly, the new prescription must increase UF volume to 1 L such that an additional 66 mEq of sodium could be excreted. If the CAPD prescription were to be left unchanged (ie, no increase in UF volume), the serum Na concentration will increase. The final steady-state serum Na level can also be calculated: [(serum Na 2 dialysis solution Na) 3 instilled volume] 1 [serum Na 3 UF volume] 5 66 1 66 5 132. Thus, [(serum Na 2 132) 3 8] 1 [serum Na 3 0.5] 5 132 mEq of daily CAPD Na removal. Rearranging this equation to solve for serum Na results in a serum Na value of 139.8 mEq/L, indicating that there would be w70 mEq of Na removed by convection and 64 mEq of Na removed by diffusion. Am J Kidney Dis. 2016;-(-):---

Solute and Water Transport in PD

3. Calculate the daily Na and water retention after 20 days when serum Na level stabilized at 127 mEq/L (assume that diet and potassium [K] intake are unchanged for all). Answers Assume that 1 month ago, total-body water, V 5 0.58 3 60 kg 5 35 L. Total-body Na 5 135 mEq/L 3 35 L 5 4,725 mEq.

Figure 1. Mean transcapillary ultrafiltration (UF) rate as a function of dwell time in studies using a 15% dextrose dialysate solution. The rate corresponds to the overall mean lymphatic reabsorption rate, and milestones related to dialysate volume and composition characteristics are indicated. Abbreviation: IP, intraperitoneal. Reproduced with permission of the International Society of Nephrology from Nolph et al.26

Take-Home Points

 When serum Na level equals dialysis solution Na concentration, clearance of Na by CAPD is convective  Any increases in Na removal when serum Na level remains equal to dialysis solution Na concentration must be accomplished by increasing the total UF  Removal of Na in CAPD by diffusion requires serum Na level to be above the dialysis solution Na concentration Case 2 A 56-year-old anuric man had a stable serum Na level (135 mEq/L) and weight (60 kg) while performing CAPD with 4 3 2.5 L exchanges for 2 years. Over a 20-day period, his serum Na level decreased to 127 mEq/L, his weight increased to 70 kg, and pitting ankle edema became evident. His weight then continued to increase at 0.5 kg/d and edema worsened, but Na level stabilized at 127 mEq/L. His daily UF volume remained stable at 1 L/d. In the first 2 years, daily dialysate Na losses repeatedly measured 140.5 mEq and UF volume was always near 1 L. Given his weight gain and stable UF volume, it is apparent that fluid intake has increased by 0.5 L/d in recent weeks. Na intake appears unchanged from dietary assessment. Questions

1. Calculate the changes in total-body water and Na. 2. Calculate the daily Na and water retention over the initial 20 days of weight gain. Am J Kidney Dis. 2016;-(-):---

1. Current total-body water would be 5 35 L 1 10 L 5 45 L and total-body Na 5 127 3 45 5 5,715, for a net gain of 5,715 2 4,725 5 990 mEq (assumed to be all Na). 2. Net water gain over 20 days was 10 L (0.5 L/d) and Na gain was 990 mEq (49.5 mEq/d), representing hyponatric expansion of body water. 3. Na removal and presumably intake in the stable early years was near 140.5 mEq (plus stool and other losses—assume negligible for calculations). When serum Na level stabilized at 127 mEq/L, daily Na removal in each 1 L of UF volume would be near 127 mEq, but daily Na uptake from dialysis solution by back diffusion into blood would be approximately equal to 10 L (infused volume) 3 (132 mEq/L 2 127 mEq/L) 5 50 mEq. Therefore, net Na removal would be 127 mEq 2 50 mEq 5 77 mEq/d. Daily net Na balance would be 5 140.5 mEq (in the diet) 2 77 mEq (net out in dialysate) 5 163.5 mEq/d, retained with 0.5 L of water. Take-Home Points

 Increasing water intake in CAPD initially causes hyponatric expansion and hyponatremia  The resultant hyponatremia leads to Na absorption from PD solution into the blood by diffusion  Serum Na level stabilizes when the daily positive Na balance (in mEq)/daily water gain (in L) 5 serum Na (mEq/L)  Thereafter, increased water intake yields isonatric expansion with increasing edema and stable serum Na level  Reducing fluid intake by 500 mL/d would return the patient to baseline Case 3 A 31-year-old man treated with CAPD for 3 years presents with a serum Na level of 120 mEq/L. His weight is stable. A peritoneal equilibration test shows average transport; this has remained stable since initiating PD therapy. Urine output has declined from 1.5 to 0.5 L/d over the past year. The patient uses 4 3 2.5 L exchanges daily, and he is increasingly using 2.5% and, occasionally, 4.25% exchanges to maintain weight. The 7

Ramesh Khanna

patient has always adhered to a 2-g (88-mEq) Na diet. Total weekly Kt/V urea is 2.5, and normalized protein nitrogen appearance (nPNA) is 1.2. Questions

1. Assuming that fluid and Na intakes have remained essentially constant, what is the likely cause of the hyponatremia? 2. If serum Na is steady at 120 mEq/L what is the total daily Na loss in urine, stool, and dialysate? 3. Estimate the dialysate Na losses. Answers

1. Na removal per liter of UF volume is greater than per liter of urine for most patients. When hyponatremia develops, dialysis losses of Na diminish. Loss of urine output resulted in shifting Na loss from the kidney to UF. The patient adjusted his prescription to maintain weight through increasing UF. Consequently, higher UF resulted higher Na losses and hyponatremia developed. 2. The total loss of Na in this steady state of Na of 120 mEq/L should be w88 mEq/d (consistent with being at steady state in the presence of a 2-g Na intake diet). 3. At a stable serum Na level of 120 mEq/L, Na losses by dialysis would approach zero (12 mEq/L uptake by diffusion from the dialysate into the blood 3 10 L 5 120 mEq/d Na gain) because the gain in Na due to diffusion from the dialysate into blood is balanced by the fact that each 1 L of UF volume contains 120 mEq of Na. Accordingly, daily Na losses would be dictated by the daily fluid intake and UF volume. Take-Home Points

 Na loss per liter of UF volume is greater than urine losses  Loss of urine output will result in hyponatremia if fluid and Na intake remain unchanged  A new steady state will be achieved that is appropriate for Na and fluid intake Case 4 A 62-year-old anuric man treated with CAPD for 3 years has had a stable weight (60 kg) and serum Na level (132 mEq/L). He uses 4 3 2.5 L of PD solution daily and has 11-L drain volumes on daily collections for a net UF volume of 1 L. Na intake has been estimated from diet to be near 3 g (132 mEq) per day, which is the amount removed in 1 L of UF volume at a serum Na level of 132 mEq/L (assuming negligible Na sieving with long-dwell CAPD exchanges). Removal of Na by diffusion would be expected to be near zero because the PD solutions and serum both have an Na concentration of 132 mEq/L. 8

The patient moves into a boarding house where the cook adds salt generously to the meals. His daily Na intake increases to 5 g (220 mEq). The patient continues the same degree of fluid restriction despite increasing thirst. His weight remains unchanged. For this example, assume that V 5 35 L (58% of body weight) and also assume that V and total-body K have not changed. Questions 1. Calculate the total increase in body Na and the new serum Na level once a new steady state is reached. How many mEq of the daily Na removal will be by diffusion and how many by convection in the new steady state? 2. What are the new volumes of extracellular fluid (ECF) and intracellular fluid (ICF) assuming that they were previously 11.6 and 23.4 L? Answers

1. With new daily intake of 5 g of dietary Na, the daily net Na removal by PD must be near 5 g or 220 mEq (neglecting stool and other losses) to maintain steady state. His UF volume is unchanged at 1 L, as suggested by the fact that his weight has remained unchanged. Thus: ½1 L of UF volume 3 serum Na 1 ½ðserum Na 2 132Þ 3 10 L 5 220 mEq 1x 1 ½10 3 ðx 2 132Þ 5 220 1x 1 10x 2 1; 320 5 220 11x 5 1; 540 x 5 140 mEq=L 5 new serum Na 1 L 3 140 mEq=L 5 removal by UF 5 140 mEq 10 3 ð140 mEq=L 2 132 mEq=LÞ 5 removal by diffusion 5 10 3 8 mEq=L 5 80 mEq Consequently, 140 mEq of Na would be removed through UF (convection) and 80 mEq would be removed by diffusion. 2. Total-body cation was 132 mEq/L 3 35 L, or 4,620 mEq, and is now 140 3 35 L, or 4,900 mEq, for a gain of 280 mEq (assumed that all cations in the extracellular space are Na). ICF K 5 132 mEq/L 3 23.4 5 3,089 mEq. Accordingly, the new ICF volume must be (3,089 mEq)/(140 mEq/L) 5 22 L and ECF is 35 L 2 22 L 5 13 L for an ECF gain of 13.0 L 2 11.6 L 5 1.4 L (assume that all cations in the intracellular space are K). Am J Kidney Dis. 2016;-(-):---

Solute and Water Transport in PD

Take-Home Point

 Isolated increases in Na intake in anuric CAPD patients result in an increasing serum Na level until dialysate Na removal increases to achieve Na balance. However, if greater thirst causes increased water intake, indefinite isonatric expansion and increasing edema could occur Case 5 A 43-year-old anuric woman has been on CAPD therapy with 4 3 2.5 L exchanges daily for 1 year. Serum K level has been steady at 4.0 mEq/L on a daily diet estimated on several occasions to contain 60 mEq of K. Weight is steady at a daily UF volume of 1.5 L. The patient eliminates many K-containing foods from her diet and K intake decreases to 30 mEq/ d without a change in Na or water balance. For all cases dealing with K, although measured serum K level is 0.5-mEq/L greater than plasma K level and true serum K level due to inevitable low-grade hemolysis with clotting and serum separation, for ease of understanding concepts, assume that measured serum K and plasma K values are identical. Also, assume that plasma water and Gibbs-Donnan corrections cancel one another. Also, assume that all cations in the intracellular space are K. Questions 1. Estimate dialysis losses of K by diffusion and convection, as well as nondialysis K losses prior to the dietary change. 2. Estimate the dialysis removal of K by diffusion and convection in the new steady state and give the new steady serum K level. Assume that nondialysis K losses are minimally changed. Answers 1. Infusion dialysis solution contains no K. Therefore, the diffusive K gradient would be serum K less dialysate K, or 4.0 mEq/L 2 0.0 mEq/L 5 4.0 mEq/L. The diffusive removal of K would be 4.0 mEq/L 3 10 L infused volume per day 5 40 mEq/d. The 1.5-L UF volume (convective clearance) would remove close to 6.0 mEq/d of K (1.5 L UF volume 3 dialysate K 4.0 mEq/L). Therefore, total daily dialysis K removal would be 46 mEq (40 mEq through diffusion and 6 mEq through UF [convection]). Given that steady state is present, nondialysis losses of K, mainly stool losses, must be 60 mEq/d 2 46 mEq/d 5 14 mEq/d. 2. Given current daily stool K loss of 14 mEq/d, with a new daily K intake of 30 mEq, dialysis must remove 30 mEq 2 14 mEq 5 16 mEq if stool losses remained unchanged. Am J Kidney Dis. 2016;-(-):---

Thus, (10 L infused volume 3 x mEq/L plasma K) 1 (1.5 L 3 x mEq/L plasma K) 5 16 mEq. This equation can be rewritten as: x mEq=L plasma K 5 16 mEq=11:5 L 5 1:4 mEq=L Accordingly, if nothing else has changed, the serum K level would be near 1.4 mEq/L. If the serum K level is higher than this, it implies that stool losses have declined. Take-Home Points

 K losses by diffusion and convection can be easily estimated from serum and dialysate K concentrations  If K intake is known, stool losses can be estimated  If dietary K and urine and stool K losses are known, serum K level can be predicted in CAPD patients Case 6 A 43-year-old anuric man treated with CAPD for 2 years has had a serum K level near 6.0 mEq/L on repeated occasions. He finally admits to using a K-containing salt substitute. He has been using 4 3 2.5 L exchanges daily and achieves daily UF of 1.5 L. The patient stops the salt substitute and serum K level decreases to 4.0 mEq/L. There were no changes in diet, fluid intake, dietary Na intake, UF volume, or body weight. Question Calculate the daily basal dietary K intake and the K intake due to the salt substitute. Answer Total K removal by dialysis when serum K level was 6.0 mEq/L could be estimated as [plasma K (mEq/L) 3 UF (L/d)] 1 {10 L/d infused volume 3 [plasma K (mEq/L) 2 dialysis solution K (mEq/L)]}. This is [6.0 mEq/L 3 1.5 L] 1 [10 L 3 (6.0 mEq/L – 0 mEq/L)] 5 69 mEq. Note that convective (through UF) and diffusive K removals are 9 and 60 mEq, respectively. Dialysis K removal when serum K level was 4.0 mEq/L could be estimated as [plasma K (mEq/L) 3 UF (L/d)] 1 {10 L infused volume 3 [plasma K (mEq/L) 2 dialysis solution K (mEq/L)]}. This is [4.0 mEq/L 3 1.5 L] 1 [10 L 3 (4.0 mEq/L 2 0 mEq/ L)] 5 46 mEq, representing 6 mEq by convection (UF) and 40 mEq by diffusion. Assuming that these dialysis removals plus stool losses represent daily K intake during steady states and neglecting any changes in stool losses, K intake as salt substitute would have been 69 2 46 5 23 mEq/d. 9

Ramesh Khanna

Take-Home Points

 Isolated changes in K intake in anuric CAPD patients result in new steady states when dialysis removals adjust to achieve balance  Dialysis removal of K in CAPD is mainly by diffusion, which is mainly a function of plasma K (note: infusion dialysis solution contains no K) Case 7 A 59-year-old anuric man on CAPD therapy for 2 years has a stable serum K level of 4.5 mEq/L. He uses 4 3 2.5 L exchanges daily with a daily UF volume of 2 L. Questions

1. Estimate his daily K intake minus stool losses of K. 2. If he increases his dietary K intake by 30 mEq/d, calculate the new steady-state serum K level (assume that stool losses, Na balance, and water balance are unchanged).

Case 8 A 36-year-old patient treated with CAPD (4 3 2 L daily) averages 1 L of UF volume and urinary losses of 20 mEq of K daily; serum K level is maintained at 4.0 mEq/L. The patient becomes anuric after exposure to intravenous contrast and ibuprofen. Questions

1. Calculate the change in serum K level assuming that K intake and stool losses stay constant. 2. What changes in the CAPD prescription would maintain serum K level at 4.0 mEq/L, assuming that K intake and stool losses stay constant? Answers

1. Initial K removal through CAPD: 5 ½serum K ðmEq=LÞ 3 UF ðL=dÞ 1 ½10 L infused volume 3 ðserum K 2 dialysis solution KÞ

Answers

1. Equation for calculating K removal in dialysate: 5 ½plasma K ðmEq=LÞ 3 UF ðL=dÞ 1 f10 L infused volume 3 ½plasma K ðmEq=LÞ 2 dialysis solution K ðmEq=LÞg 5 ð2 L=d 3 4:5 mEq=LÞ 1 ð10 L=d 3 4:5 mEq=LÞ 5 54 mEq=d In a steady state, this should represent dietary intake minus stool losses. Accordingly, dietary K intake is 5 54 mEq 1 stool losses. 2. Adding 30 mEq of K to the diet with no change in stool losses would increase daily removal K through dialysis in the new steady state to 84 mEq/d: 84 mEq=d 5 ð10 L=d 3 plasma KÞ 1ð2:0 L=d ultrafiltration 3 plasma KÞ Rearranging the equation: plasma K 5 ð84 mEq=dÞ=ð12 L=dÞ 5 7 mEq=L Therefore, assuming no change in stool K losses, the new plasma K level would be 7.0 mEq/L. Take-Home Points

 Increases in K intake in anuric CAPD patients allow predictable increases in serum K level if urine and stool K losses are unchanged 10

 In the absence of change in PD prescription, increases in K intake increases dialysis K losses mainly through diffusion

5 ð4:0 mEq=L 3 1:0 L=d UFÞ 1 ð4:0 mEq=L 3 8:0 L=dÞ 5 36 mEq=d Total PD and urine K losses were 36 mEq/d 1 20 mEq/d 5 56 mEq/d. Stool losses are unknown but constant. With anuria and no change in the CAPD prescription, then (serum K 3 1 L UF volume) 1 (serum K 3 8 L infused volume) must equal 56 mEq. New serum K 5 (56 mEq/L)/(9L/d) 5 6:2 mEq/L. The net increase in serum K level would be 2.2 mEq/L. 2. To keep serum K level at 4.0 mEq/L following loss of residual kidney function, dialysis removal of K must equal 56 mEq/d. This means that serum K 3 daily drain volume must equal 56 mEq of K. Daily drain volume 5 56 mEq/4.0 mEq/L 5 14.0 L. Regimens that could meet this requirement include four 3-L exchanges plus 2 L of daily UF volume; this regimen may not be possible for many individuals. Alternatively, dietary reduction of 15 to 20 mEq/d of K may be necessary. CCPD may be another option with 10 L of nightly volume and a daytime 2.0-L exchange. Total daily UF volume in this scenario would need to be 2.0 L. For a CCPD regimen with 11 L of nightly volume and a daytime 2-L exchange, daily total UF volume could decrease to 1 L. Note that for each additional 4 mEq of K removal by PD (at a serum K of 4.0 mEq/L), an increase of 1 L in drain volume is needed. Am J Kidney Dis. 2016;-(-):---

Solute and Water Transport in PD

Take-Home Points

 In an anuric PD patient, at a serum K level of 4.0 mEq/L, each 1 L of drain volume removes w4.0 mEq of K  With loss of residual kidney function, compensation for urinary K losses of 20 mEq/d would require 5.0 L of additional PD drain volume to maintain a serum K level of 4.0 mEq/L, assuming that daily dietary K intake remains unchanged  Residual urine volume is worth protecting for a number of reasons including K excretion Case 9 A 28-year-old anuric man treated with CAPD uses 4 3 2 L exchanges daily, with a daily UF volume of 1.0 L. Serum K level is 4.0 mEq/L. His PD prescription, changed to meet adequacy targets and clinical evidence of fluid retention, becomes 4 exchanges of 2.5 L daily with a daily UF volume of 1.5 L. Question

1. Calculate the new serum K level. Assume that dietary intake including K intake and K stool losses do not change. Answer

1. Initial daily K removal 5 (serum K 4.0 mEq/L 3 8 L/d infused volume) 1 (serum K 4.0 mEq/L 3 1 L/d ultrafiltration) 5 36 mEq/d After the prescription change, K removal must be 36 mEq daily at steady state and therefore: ðserum K 3 10 L infused volumeÞ 1 ðserum K 3 1:5 L UF volumeÞ 5 36 mEq=L 11:5 L 3 serum K 5 36 mEq=d Rearranging the equation, 36 mEq/L/11.5 L 5 3.24 mEq/L. The new serum K level would be 3.2 mEq/L. To adjust for this, the patient should be encouraged to increase dietary K intake to bring the serum K to a safer level. Take-Home Points

 Increases in exchange volumes and/or increases in daily UF will increase K removal  Serum K level will decrease to a lower steady-state value unless intakes increase

SUMMARY Understanding the peritoneal membrane structure and solute and water transport physiology is critical in understanding the various changes that occur during Am J Kidney Dis. 2016;-(-):---

the PD. The peritoneal membrane is dominated by small pores, which allow transport of smallmolecular-size solutes including electrolytes by way of both diffusion and convection. Specialized water channels, aquaporin 1, are present in mesothelial cells and capillaries. These channels allow free water transport in response to an osmotic gradient. The solute reflection coefficient is a major determinant of ultrafiltration rate during a PD exchange. PD solution contains physiologic concentrations of electrolytes and base. During a PD exchange, net loss or gain of electrolytes and base is determined by their gradient between capillary blood and dialysis solution and the volume of net ultrafiltration. Understanding the concept of crystalloid and colloid osmosis helps explain the variation in solute and water transport between glucose- and icodextrincontaining PD solutions. Dialysis prescription and change in response to dietary intake and loss of residual kidney function require a sound understanding of the peritoneal physiology. The case studies presented here help solidify the basic elements of PD prescription and how logical alteration of PD prescription should be done in response to changes.

ACKNOWLEDGEMENTS The author acknowledges the contribution of the late Karl D. Nolph, MD, in mentoring and conceiving the case studies. Support: None. Financial Disclosure: The author declares that he has no relevant financial interests.

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