# Calculations of Fluid Shifts and Osmolarities After Infusion

Calculations of Fluid Shifts and Osmolarities After Infusion of Hypertonic Saline If 2 liters of a hypertonic 3.0 % NaCl solution are infused into the extracellular fluid compartment of a 70 kg patient whose initial plasma osmolarity is 280 mOsml/l, what would be the intracellular and extracellular fluid volumes and osmolarities after reaching osmotic equilibrium?

Assuming that extracellular fluid volume is 20 % and intracellular fluid volume 40 % of the body weight. Initial Conditions Volume (liters)

Concentration (mOsm/l) Total (mOsm)

Extracellular fluid 14 280 3 920

Intracellular fluid 28 280

7 840 Total body fluid 42 280

11 760 2 L of 3 % NaCl = 30 g NaCl per liter. Because the molecular weight is 58.5 g/mol. This mean there is about 0.513 mole of NaCl per liter (30:58.5). For 2 liters of solution this would be 1.026 mole (2 x 0.516). Because 1 mole of NaCl equals 2 osmoles = 2 x 1.026 = 2.052 osmoles = 2 052 mOsm

Instantaneous Effect of Adding 2 Liters of 3.0 % NaCl Volume (liters) Extracellular fluid

16 Concentration (mOsm/l) 373

14 + 2 5972 : 16 Intracellular fluid

28 280 Total body fluid 44

no equilibrium Total (mOsm) 5972

3920 + 2052 7 840 13 812 Final osmolarity after reaching equilibrium must be: 13 812 : 44 = 313.9 mOsm/L

Effect of Adding 2 Liters of 3.0 % NaCl after Osmotic Equilibrium Volume (liters) Extracellular fluid

19.02 Concentration (mOsm/l) Total

(mOsm) 313.9 5 972 313.9

7 840 313.9 13 812

5 972:313.9 Intracellular fluid 24.98 7 840:313.9

Total body fluid 42 13 812:313.9 One can see that adding 2 liters of hypertonic NaCl solution causes more tha 5-liters increase in extracellular fluid volume (19.02 14 = 5.02), while decreasing intracellular fluid volume by

3 liters (24.98 28 = -3.02). Kidneys serve following functions: 1. 2. 3. 4.

5. 6. Excretion of metabolic waste products and foreign chemicals Regulation of water and electrolyte balance Regulation of body fluid osmolality Regulation of acid-base balance

Metabolism of hormones Gluconeogenesis GLOMERULRN KAPILRY Eferentn arteriola

Aferentn arteriola PGC Vazokonstrikce aferentn arterioly

PGC Vazokonstrikce eferentn arterioly PGC GLOMERULRN

FILTRACE PRTOK KRVE LEDVINAMI Autoregulation of Glomerular Filtration Rate and Renal Blood Flow

1. Myogenic Mechanism 2. Tubuloglomerular Feedback Use of Clearance Methods to Quantify Kidney Function Renal clearance of a substance is the volume of plasma that is completely cleared of the substance by the kidneys per unit time

1. 2. Tubular Processing of the Glomerular Filtrate

Urinary excretion = Glomerular Filtration Tubular reabsorption + Tubular Secretion The processes of glomerular filtration and tubular reabsorption are quantitatively very large relative to urinary excretion for many substances. Unlike glomerular filtration, tubular reabsorption is highly selective.

Filtration, Reabsorption and Excretion Rates of Different Substances by the Kidneys Amount Filtered Amount Reabsorbed

Amount Excreted % of Filtered Load Reabsorbed

Glucose (g/day) 180 180

0 100 Bicarbonate (mmol/day)

4 320 4 318 2 99.9

Sodium (mmol/day) 25 560 25 410

150 99.4 Chloride (mmol/day)

19 440 19 260 180

99.1 Potassium (mmol/day) 756

664 92 87.8 Creatinine

(g/day) 1.8 0 1.8

0 Comparison of sodium and water reabsorption along the tubule Tubular segment Proximal tubule

Descending thin limb of Henles loop Ascending thin limb and thick ascending limb of Henles loop Distal convoluted tubule Collecting-duct system Percent of filtered load reabsorbed (%)

Sodium Water 65 65 0

10 25 0

>24 (during dehydration) Renal Regulation of Potassium Balance 3 Na+ K+ ATP

2 K+ GLUT4 glukza Mineralokortikoidn receptor

Inzulnov receptor inzuln cAMP aldosteron

Summary of tubular potassium transport Tubular segment Normal- or high-potassium Low-potassium diet diet

Proximal tubule Reabsorption (60-80 %) Reabsorption (55 %)

Thick ascending limb Reabsorption (5-25 %) Reabsorption (30 %) Distal convoluted tubule

Secretion Reabsorption Cortical collecting duct (Principals cells)

Substantial secretion (15-180 %) 0 Cortical collecting duct

(Intecalated cells, type A) Reabsorption (10 %) Reabsorption (10 %) Medullary collecting duct

Reabsorption (5 %) Reabsorption (5 %) Lumen Na

+ Intersticiln Prostor ENaC

K+ K+ K+ ClHlavn Buka principal cell Sbracho Kanlku

3 Na+ ATP 2 K+ Lumen

Vmezeen buka intercalated cell Sbracho kanlku Intersticiln Prostor 3 Na+ ATP

H+ HCO3- ATP

H+ K+ K+ Cl- ATP

2 K+ Cl- Homeostatic Control of Potassium Secretion by the Cortical Collecting Duct (3 key factors)

1. Plasma concentration of potassium 2. Plasma levels of aldosterone 3. Delivery of sodium to the distal nephron Ad.1: The principal cells contains an isoform of Na-K-ATPase that is especially sensitive to increases in the concentrations in peritubular capillaries.

Ad.2: The luminal membrane pathway that allows potassium to exit the cell must be open and this is the function of aldosterone. Ad.3: With an increased delivery of sodium to the cortical collecting duct, more sodium

enters principal cells, and more potassium is secreted. Regulation of Extracellular Fluid Osmolarity and Sodium Concentration Obligatory urine volume 600 mOsm/day

= 0.5 L/day 1200 mOsm/L Pro nept moskou vodu? 1 L mosk vody = 1 200 mOsm = pjem 1 200 mOsm/L Organizmus se mus denn zbavit minimln 600 mOsm denn To znamen, e musme vylouit 1800 mOsm, co i pi tvorb maximln

koncentrovan moi (1200 mOsm/l) musme vylouit 1.5 L. Z toho vyplv, e mme minimln ztrtu 500 ml. Australian hopping mouse Notomys alexis Klokanomy spinifexov

Can concentrate urine to 10 000 mOsm/L Quantifying renal urine concentration and dilution: Free water and osmolar clearance Osmolar clearance (Cosm): this is the volume of plasma cleared of solutes each minute. Posm = plasma osmolarity, 300 mOsm/L Uosm = urine osmolarity, 600 mOsml/L

V = urine flow, 1 ml/min (0.001 L/min) Uosm x V Cosm = (2 ml/min) 600 x 0.001

= 0.6 mOsm/min = = 0.002 L/min

This P means that 2 ml 300 of plasma are being cleared solute each minute 300ofmOsm/L osm

Free-water clearance (CH20): is calculated as the difference between urine flow and C osm Uosm x V CH20 = V Cosm = V Posm CH20 = 1 ml/min 2 ml/min = - 1 ml/min When CH20 is positive, excess water is being excreted by the kidneys, when C H20 is negative excess solutes ,are being removed from the plasma by the kidneys and water is being conserved.

Thus, whenever urine osmolarity is greater than plasma osmolarity, free-water clearance is negative, indicating water conservation Estimating plasma osmolarity from plasma sodium concentration Posm = 2.1 x Plasma sodium concentration

Sodium ions and associated anions (bicarbonate and chloride) represents 94 % of the ECFV solutes. Glucose and urea contribue about 3 5 %. Renal Control of Acid-Base Balance 1.

There must be a balance between the production of H+ and the net removal of H+ from the body. 2. Precise H+ regulation is essential because the activities of almost all enzyme systems in the body are influenced by H+ concentration. 3. Na+ = 142 mmol/L, H+ = 0.00004 mmol/L 4. pH = -log [H+] = -log[0.00004] = 7.4 5. There are three primary systems that regulate the H+ concentration

in body fluids to prevent acidosis: A/ Chemical acid-base buffer systems of the body fluids B/ Lungs C/ Kidneys Buffering of Hydrogen Ions in the Body Fluids Buffer + H+

H Buffer Daily production of H+ = 80 mmol, Body fluid concentration = 0.00004 mmol/L Bicarbonate Buffer System

H+ + HCO3- H2CO3 Henderson-Hasselbalch Equation: HCO3pH = 6.1 + log 0.03 x pCO2

CO2 + H2O Phosphate Buffer System It plays a major role in buffering renal tubular fluid and intracellular fluid HPO42- + H+

H2PO4- Proteins: Important Intracellular Buffer H+ + Hemoblobin HHemoglobin

Approximately 60 to 70 percent of the total chemical buffering of the body fluids is inside the cells, and most of this results from the intracellular proteins. Henderson-Hasselbalch Equation: HCO3pH = 6.1 + log

0.03 x pCO2 The kidneys regulate extracellular fluid H+ concentrations thought three fundamental mechanisms: 1. Reabsorption of filtered HCO32. Secretion of H+ 3. Production of new HCO3Ad. 1. 180 L/day x 24 mmol/L = 4320 mmol of HCO3-

Proximal tubule, thick ascending loop of Henle, early distal tubule Late distal tubule and collectiong tubules Phosphate and Ammonia Buffers Minimal urine pH is 4.5, corresponding to an H+ concentration 0.03 mmol/L.

In order, to excrete the 80 mmol of nonvaletile acid formed each day, about 2667 liters of urine would have to be excreted if the H+ remained free in solution. 500 mmol/day of H+ must be sometimes excreted. Therefore, whenever an H+ secreted into the tubular lumen combines with a buffer other than, HCO3- the net effect is addition of a new HCO3to the blood.

A second buffer system in the tubular fluid that is even more important quantitatively than the phosphate buffer system is composed of ammonia (NH 3) and the ammonium ion (NH4+). Proximal tubule, thick ascending limb of the loop of Henle, distal tubule

Collecting duct Quantifying Renal Acid-Base Excretion Net acid excretion = NH+4 excretion + Urinary titrable acid bicarbonate excretion

The most important stimuli for increasing H+ secretion by the tubules are: 1. An increase in pCO2 of extracellular fluid. 2. An in H+ concentration in extracellular fluid.

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