Body Fluids Bruce M

Faculty version with model answers

Body Fluids Bruce M. Koeppen, M.D., Ph.D. University of Connecticut Health Center

Introduction

By weight, water comprises approximately 60% of the adult human body (this percentage varies with the amount of adipose tissue (as the amount of adipose tissue increases the percentage of body weight attributed to water decreases). This total body water (TBW) is contained within two major compartments; the intracellular fluid (ICF – 40% of body weight) and the extracellular fluid (ECF – 20% of body weight). Fluid within the ECF is further subdivided into plasma and the fluid surrounding cells (interstitial fluid).

Water readily moves between these various fluid compartments. Water movement between the ICF and ECF is driven by differences in osmotic pressure, whereas movement between the vascular compartment (i.e., plasma) and the interstitial fluid compartment is driven by Starling forces across the wall of the capillaries.

In this conference we will examine the volumes and composition of the various body fluid compartments, and how fluid shifts occur under a number of conditions.

Recommended Reading:

Pg. 1 - 15: Renal Physiology, 3rd ed., Koeppen & Stanton, C.V. Mosby, 2001.

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Capillary Fluid Movement

Fluid movement across the wall of a capillary is driven by the sum of the Starling forces; two of which favor (Pc and Πis) and two of which oppose (Pis and Πc) fluid movement out of the capillaries. • Pc: hydrostatic pressure in the capillary • Pis: hydrostatic pressure in the interstitium • Πc: oncotic pressure of proteins in the capillary • Πis: oncotic pressure of proteins in the interstitium

Fluid movement across the capillary can be quantitated as:

Fluid Flow = Kf [(Pc - Pis) - σ(Πc-Πis)]

Where: Kf is the filtration coefficient and takes into consideration the intrinsic permeability of the capillary wall to fluid movement (also called the hydraulic conductivity), as well as the surface area available for fluid flow; and σ is the reflection coefficient for proteins, which varies from 0 (i.e., the capillary wall is freely permeable to proteins) to 1 (i.e., the capillary wall is impermeable to proteins).

Fluid Movement Between the ICF and ECF

Water can move between the ICF and ECF under a number of important clinical situations. The effects of these shifts on the volumes of the body fluid compartment can be approximated using the following principles:

©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -1-

• The volumes of the various body fluid compartments can be estimated in the normal adult as:

Total Body Water (TBW) = 0.6 x body weight Extracellular Fluid Volume (ECF) = 0.2 x body weight Intracellular Fluid Volume (ICF) = 0.4 x body weight Plasma Volume (P) = 0.25 x ECF volume Interstitial Fluid Volume (IF) = 0.75 x ECF volume

• All exchange of water and solutes with the external environment occur through the ECF (e.g., intravenous infusion and intake or loss via the gastrointestinal tract). Changes in the ICF are secondary to fluid shifts between the ECF and ICF. Fluid shifts only occur if the perturbation of the ECF alters its osmolality. Note: the osmolality of the ECF can be quickly estimated by doubling the plasma [Na+], because Na+ and its attendant anions are the major osmoles of the ECF. • Except for brief periods of seconds to minutes, the ICF and ECF are in osmotic equilibrium. A measurement of plasma osmolality will provide a measure of both ECF and ICF osmolality. • For the sake of simplification, it can be assumed that equilibration between the ICF and ECF occurs only by movement of water, and not by movement of osmotically active solutes. • Conservation of mass must be maintained, especially when considering either addition of water and/or solutes to the body or their excretion from the body.

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1. Use the following Starling forces for questions 1A and 1B.

Arterial End Venous End Pc 35 mmHg 10 mmHg Pis - 3 mmHg -3 mmHg Πc 26 mmHg 26 mmHg Πis 5 mmHg 5 mmHg

A. On the following graph, plot: (Pc - Pis) and σ(Πc - Πis). Assume that plasma proteins cannot cross the capillary wall (i.e., that σ = 1).

30 Pc-Pis mmHg

20 σ(Πc-Πis)

©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -2-

0 Arterial Venous End End

Note that proteins cannot exert an oncotic pressure if they freely cross the capillary wall. Accordingly, the reflection coefficient (σ) takes into account the permeability of the capillary to proteins. A capillary that is essentially impermeable to proteins (e.g., glomerular capillary of the kidney) will have σ = 1 as in this example. If the capillary is freely permeable to protein (e.g., liver sinusoid) σ = 0.

What can you conclude from this graph about the movement of fluid into and/or out of the capillary?

The Starling forces are such that there will be net movement of fluid out of the lumen of the capillary at the arterial end, and net movement into the capillary lumen at the venous end.

B. Replot (Pc - Pis) and σ(Πc - Πis) on the following graph assuming that the capillary wall has a significant permeability to plasma proteins (i.e., σ = 0.4).

30 Pc-Pis mmHg 20

10 σ(Πc-Πis)

0 Arterial Venous End End

This clearly emphasizes the importance of considering the permeability of the capillary to proteins. With a significant permeability (i.e., σ = 0.4) there is a much reduced effect of protein on the movement of fluid across the capillary. In the extreme where σ = 0, as in the liver sinusoids, the only driving forces for movement of fluid across the capillary wall would be the capillary and interstitial hydrostatic pressures.

What can you conclude from this graph about the movement of fluid into and/or out of the capillary under this condition?

As noted above, because σ is less than 1, the protein within the capillary lumen and interstitium will not exert the full oncotic pressure (in this example it is reduced by 60%). As a result of this, fluid leaves

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the capillary lumen along its entire length. This overall net movement of fluid into the interstitium will result in formation of lymph.

Recent studies indicate that σ in muscle capillaries is in the range of 0.8 - 0.9, and given the estimated Starling forces, fluid is probably filtered along the entire length of the capillary. In liver sinusoids, σ is near 0, and fluid is filtered through out the sinusoid driven by hydrostatic pressure only.

2. Edema is the abnormal accumulation of fluid in the interstitial fluid compartment. Edema can be localized or generalized. The formation of edema requires an alteration in the Starling forces across the capillary wall, or a change in the permeability of the capillary wall such that there is increased movement of fluid out of the capillary into the interstitium. As the terms imply, localized edema is confined to a particular portion of the body or vascular bed, whereas generalized edema results from increased movement of fluid out of the capillaries into the interstitum in vascular beds through out the body. Give an example of localized edema, and an example of generalized edema. Explain the pathogenesis of each in terms of the Starling forces across the capillary wall.

Localized edema: A bee sting is an example of localized edema. In the case of the bee string, inflammatory and vasoactive mediators (e.g., histamine, cytokines, etc.) at the sting site increase both capillary permeability and cause local vasodilation. This increases permeability of the capillary wall to protein (i.e., reduces σ), and increases capillary hydrostatic pressure. The net effect of these two changes is illustrated below (compare to graph shown in 1A). These changes in capillary permeability and capillary hydrostatic pressure will result in more fluid moving out of the capillary with localized edema formation.

40 Pc-Pis

30 mmHg 20

10 Πc-Πis 0 Arterial Venous End End

Generalized edema: The classic example of generalized edema is that seen with congestive heart failure. Hydrostatic pressure within the capillary lumen is an important driving force for the movement of fluid out of the capillary; this pressure is determined by precapillary and postcapillary resistances. Most capillary beds have a well developed precapillary sphincter that helps maintain a relatively constant capillary hydrostatic pressure in the face of changes in arterial pressure. In most capillary beds there is not an effective postcapillary sphincter (the exception is the glomerulus of the kidney). Therefore changes in venous pressure can have important effects on capillary hydrostatic pressure. With congestive heart failure there is an increase in venous volume and pressure. The increased venous pressure is transmitted down to the level of the capillary resulting in an increase in capillary hydrostatic pressure. This increased pressure will in turn cause excess accumulation of fluid in the interstitial

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space (i.e., edema). Because the change in venous pressure effects the Starling forces in capillaries throughout the body the formation of edema is generalized. Edema is most easily detected in the lower extremities, this occurs because gravity acting on the column of blood in the veins further increases the hydrostatic pressure in the capillaries.

The following graph depicts the changes in Starling forces that would be seen in this situation. In this example σ = 1(compare to graph shown in 1A).

30 Pc-Pis mmHg

20 Πc-Πis

0 Arterial Venous 3. Patients with generalized edema are frequently treated with low salt diets, or diuretics. Based End End on what you know about generalized edema formation, what is the rationale for these forms of therapy? How will this therapy affect the Starling forces across the capillary wall?

As fluid moves out of the vascular compartment into the interstitial compartment vascular volume will decrease. Because of the compliance of the interstitial space, approximately 3 liters of excess fluid must accumulate before edema is clinically detectable. Edema is detected by physical examination. In ambulatory patients the edema accumulates in the lower limbs (e.g., ankles, pretibial area). In bedridden patients edema can also be detected in the small of the back (presacral) and around the eyes (periorbital). Edema in the lungs (pulmonary edema) is detected by ascultation of the chest with a stethoscope. Since the plasma volume is only 3 - 4 liters, vascular collapse would occur if some mechanism did not operate to replenish the plasma volume as fluid moves out of the capillaries into the interstitium. The kidneys respond to the decreased plasma volume by retaining NaCl and water. Therefore, because the interstitial fluid compartment must increase in volume by approximately 3 liters before edema can be detected on physical examination, there must be NaCl and water retention by the kidneys to maintain plasma volume in the face of the shift of fluid out of the vasculature. Thus clinically detectable generalized edema will only occur if there is both an alteration in Starling forces and NaCl retention by the kidneys. Possible therapies for minimizing edema formation are directed at reducing or preventing NaCl and water retention by the kidneys. These strategies include restriction of dietary NaCl and preventing renal NaCl reabsorption with diuretic agents.

By preventing renal NaCl and water retention vascular volume will decrease as described above, and thus hydrostatic pressure within the capillary will decrease. This decrease in capillary hydrostatic pressure will reduce the net driving force for fluid movement out of the capillary, and thereby reduce interstitial fluid accumulation (i.e., edema).

4. A healthy volunteer (body weight = 50 kg) is infused with 1 L of a 5% dextrose solution. This solution is essentially iso-osmotic to plasma. What would be the immediate and long-term

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(several hour) effects of this infusion on the following parameters? Assume an initial plasma [Na+] of 145 mEq/L, and for simplicity no urine output.

Immediate Effect: ECF volume: ____11____ L ICF volume: ____20____ L Plasma [Na+]: ___132____ mEq/L

This is an isotonic solution, and will therefore remain confined initially to the ECF. The ECF volume will increase by 1 L, and ICF volume will be unchanged (initial volumes being 10 L (ECF) and 20 L (ICF), respectively). Plasma [Na+] will decrease because of the addition of 1 L of Na+-free solution to the ECF. The new plasma [Na+] is calculated by:

Initial ECF Na+ content = 145 mEq/L x 10 L = 1,450 mEq New [Na+] = 1,450 mEq/11 L = 132 mEq/L

Long-term Effect: ECF volume: ____10.33__ L ICF volume: ____20.67__ L Plasma [Na+]: ____141___ mEq/L

Long-term, the dextrose will be metabolized to CO2 and H2O. Thus, infusion of a dextrose solution is equivalent to infusion of solute-free water. After metabolism and equilibration over several hours, the 1 L of infused fluid will distribute into both the ICF and ECF proportional to the ratio of their volumes (2/3 into the ICF and 1/3 into the ECF).

ICF volume = 20 L + 0.67 L = 20.67 L ECF volume = 10 L + 0.3 L = 10.3 L New [Na+] = 1,450 mEq/10.3 L = 141 mEq/L

Based on these effects of the 5% dextrose solution on the volumes and composition of the body fluids, how would this solution be used clinically?

As noted above a dextrose solution is equivalent to solute-free water. Therefore, these solutions would be used when the patient had lost solute-free water and body fluid osmolality is elevated (ie., hypernatremia).

5. A second healthy volunteer (body weight = 50 kg) is infused with 1 L of a 0.9% NaCl solution. This solution is essentially iso-osmotic to plasma. What would be the immediate and long- term (several hour) effects of this infusion on the following parameters? Assume an initial plasma [Na+] of 145 mEq/L, and for simplicity no urine output.

Immediate Effect: ECF volume: ____11____ L ICF volume: ____20____ L Plasma [Na+]: ___145____ mEq/L

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A 0.9% NaCl solution is isotonic saline. Therefore the entire infused volume will remain in the ECF. In this example the ECF will increase by 1 L, and the ICF will not change. One approach to this problem is to calculate the amount of Na+ in the infused volume, and then determine the effect on the plasma [Na+]. For example:

0.9% NaCl = 154 mEq/L Amount of infused Na+ = 1 L x 154 mEq/L = 154 mEq New ECF Na+ content = 1,450 mEq + 154 mEq = 1,604 mEq New plasma [Na+] = 1,604 mEq/11 L = 146 mEq/L

However, for simplicity we assume that there will be no appreciable change in the plasma [Na+], as indeed there is not. It should be emphasized that infusion of isotonic saline into an individual who has a normal plasma [Na+] will result in an increase in the volume of the ECF (by the infused amount) and no change in the [Na+].

Long-term Effect: ECF volume: ____11____ L ICF volume: ____20____ L Plasma [Na+]: ___145____ mEq/L

Because the isotonic saline does not change the osmolality of the ECF, there will be no fluid shift, and the entire infused volume will remain in the ECF until eventually excreted by the kidneys.

Based on these effects of the NaCl solution on the volumes and composition of the body fluids, how would this solution be used clinically?

Because isotonic saline will stay within the ECF, it is used clinically to treat a patient who is hypovolemic due to loss of isotonic fluid, as often occurs with diarrhea (see question #6).

6. A child (body weight = 30 kg) develops gastroenteritis with vomiting and diarrhea. Over a 2 day period he loses 2 kg of weight. His plasma [Na+] was initially 140 mEq/L, and is unchanged. Calculate the following:

Initial After Day 2 Total Body Water (L) 18 L 16 L ECF Volume (L) 6 L 4 L ICF Volume (L) 12 L 12 L Total Body Osmoles (mosmoles) 5,040 mOsm 4,480 mOsm ECF Osmoles (mosmoles) 1,680 mOsm 1,120 mOsm ICF Osmoles (mosmoles) 3,360 mOsm 3,360 mOsm

Points to highlight:

1. It is assumed that the 2 kg weight loss reflects only the loss of body fluids (ie., 2 L) and not tissue mass. Given the short time period involved, this is a reasonable assumption. 2. Because the plasma [Na+] has not changed this is isotonic loss of fluid. Consequently, all of the fluid loss will occur from the ECF (there is no osmotic gradient to cause fluid to leave the ICF). This a simplification, but again reasonable. 3. All of the osmole loss will occur from the ECF, essentially as NaCl. This is also a simplification because K+ is typically lost in the stool during diarrhea. Because K+ is the major intracellular cation, there is in fact lose of solute from both the ECF and ICF. However, the loss is greatest from the ECF, and the lose from the ICF is ignored. 4. Note that the quick and easy way to estimate plasma osmolality is to double the plasma [Na+]. This reflects the fact that Na+ and its attendant anions is the major determinate of plasma osmolality. A more accurate estimate of plasma osmolality can be obtained by also taking into account the plasma glucose and blood urea nitrogen (BUN) concentrations.

Posm = 2 x [Na+] + [glucose] / 18 + [BUN] / 2.8

Where the concentrations of glucose and urea are expressed in units of mg/dL. However, neither glucose (especially if insulin levels are adequate) nor urea are effective osmoles (i.e., they are highly permeable across cell membranes and therefore do not play a major role in shifting water between the ECF and ICF). Therefore, the osmolality calculated from 2 x [Na+] is the best estimate of effective osmolality of the ECF.

6. Calculations: Initial Conditions Total body water = 30 kg x 0.6 = 18 L ECF volume = 18 L x 1/3 = 6 L ICF volume = 18 L - 6 L = 12 L Total body osmoles = 140 x 2 x 18 L = 5,040 mOsmoles ECF osmoles = 6 L x 2 x 140 = 1,680 mOsmoles ICF osmoles = 5,040 - 1,680 = 3,360 mOsmoles

Day 2 Total body water = 18 L - 2 L = 16 L ECF volume = 6 L - 2 L = 4 L ICF volume = 12 L Total body osmoles = 140 x 2 x 16 L = 4,480 mOsmoles ECF osmoles = 4 L x 2 x 140 = 1,120 mOsmoles ICF osmoles = 4,480 - 1,120 = 3,360 mOsmoles

7. Another child (body weight = 30 kg) also develops gastroenteritis with vomiting and diarrhea. Over a 2 day period he also loses 2 kg of weight. However, his plasma [Na+], which was initially 140 mEq/L is now 130 mEq/L following the vomiting and diarrhea. Calculate the following: Hint: Calculate the total solute loss from the body, and for simplicity assume all of this solute lose represents Na+ and its anions from the ECF.

Initial After Day 2 Total Body Water (L) 18 L 16 L

ECF Volume (L) 6 L 3.1 L ICF Volume (L) 12 L 12.9 L Total Body Osmoles (mosmoles) 5,040 mOsm 4,160 mOsm

ECF Osmoles (mosmoles) 1,680 mOsm 800 mOsm ICF Osmoles (mosmoles) 3,360 mOsm 3,360 mOsm

Points to highlight: 1. Compared to the child in question #6, this child is in much more serious condition. This child has lost 2 L of fluid with excess NaCl from the ECF. As a result he has become hyponatremic. 2. For simplicity this problem has assumed that the only solute lost was NaCl from the ECF (see also question #6). This is not entirely accurate since K+, which is a major intracellular electrolyte, is commonly lost with diarrhea. Accordingly, both ECF and ICF solute loss occurs. However, the loss of solute from the ECF is greater. 3. Because of the hyponatremia there is now a driving force to cause water to move from the ECF to the ICF. This movement will further reduce the volume of the ECF (of which the vascular volume is a component). 4. Calculations: Initial Conditions Total body water = 30 kg x 0.6 = 18 L ECF volume = 18 L x 1/3 = 6 L ICF volume = 18 L - 6 L = 12 L Total body osmoles = 140 x 2 x 18 L = 5,040 mOsmoles ECF osmoles = 6 L x 2 x 140 = 1,680 mOsmoles ICF osmoles = 5,040 - 1,680 = 3,360 mOsmoles

Day 2 Total body water = 18 L - 2 L = 16 L

ECF volume = 800 mOsm/ 260 mOsm/kg H2O = 3.1 L ICF volume = 16 L – 3.1 L = 12.9 L Total body osmoles = 130 x 2 x 16 L = 4,160 mOsmoles ECF osmoles = 4,160 - 3,360 = 800 mOsmoles ICF osmoles = 3,360 mOsmoles

8. Compare the fluid and electrolyte treatment strategies that would be used to restore the volume and composition of the body fluids in the children in problems #6 and #7.

Both children need 2 L of water and some NaCl. For the child in problem #6 this can be accomplished by infusing 2 L of isotonic saline (0.9% NaCl). However, the child in problem #7 needs additional NaCl. Strictly speaking, this child could be treated with an infusion of an appropriate combination of isotonic and hypertonic saline. However, in practice only isotonic saline is administered, and over time the kidneys reabsorb sufficient quantities of Na+ to correct the Na+ deficit.

9. Two normal individuals (body weight = 60 kg) excrete the following urine over the same time period.

Subject A: 1 L of urine with an osmolality of 1,000 mOsm/kg H2O Subject B: 4 L of urine with an osmolality of 400 mOsm/kg H2O

If both individuals have no fluid intake, what will be their plasma osmolality? Hint: Assume both individuals have an initial plasma [Na+] of 145 mEq/L and thus a plasma osmolality of

approximately 290 mOsm/kg H2O.

Subject A: 270 mOsm/kg H2O

Subject B: 276 mOsm/kg H2O

Both individuals start out with the same total body water (36 L) and total body osmoles (10,440) + assuming a plasma [Na ] of 145 mEq/L and a plasma osmoality of 290 mOsm/kg H2O. Subject A loses 1 L of total body water and 1,000 osmoles of total body solute, resulting in a new plasma osmolality of:

Posm = (10,440 mOsm - 1,000 mOsm)/ 35 L = 270 mOsm/kg H2O

Subject B loses 4 L of total body water and 1,600 mOsm of total body osmoles, resulting in a new plasma osmolality of:

Posm = (10,440 mOsm - 1,600 mOsm)/ 32 L = 276 mOsm/kg H2O

It should be emphasized that hypotonic urine excretion, without appropriate replacement of solute-free water, will lead to an increase in the osmolality of the body fluids and the development of hypernatremia (e.g., as occurs in diabetes insipidus). Whereas hypertonic urine excretion, without a reduction in solute-free water ingestion, will lead to decrease in the osmolality of the body fluids and the development of hyponatremia (e.g., as occurs in the syndrome of inappropriate secretion of ADH; SIADH).