1. (For use only by Loyola Stritch School of Medicine Students)
Body Fluids
Ivana Kuo, Ph.D.
(For use only by Loyola Stritch School of Medicine Students)
Of the 5 problems, I want you to focus particular attention on understanding the concepts behind questions 1, 2 and 5 for the renal block.

2. Case conference
There are 5 problems in this set, each dealing with a different aspect of body fluids:
1. Dilution principle
2. Darrow Yannet diagrams and the 6 challenges to body fluid homeostasis
3. Anion gap
4. The calculation of fractional reabsorption
5. Diuretics.

3. Note:
Of the 5 problems, I want you to focus particular attention on understanding the concepts behind questions 1, 2 and 5 for the renal block.
You will re-visit the anion gap (3) with Dr. Mignery and Dr. Pak.
Fractional reabsorption is of interest (4), and you should be able to calculate these values, but I will not be asking math questions in the exam

4. 1. Dilution Principle
A man weighing 75 kg is given 100 μCi of 35S-sulfate (radioactive- sulfate or radio-sulfate in micro-curries) and 99.8g of D2O. Allowing sufficient time for the substances to equilibrate in their respective compartments, blood samples are drawn after 20 minutes and 2 hours and treated to obtain protein-free filtrates of plasma.
From the results below, calculate the total body water, extracellular fluid volume, and intracellular fluid volume.

5. Dilution principle
Volume = (Amount given – Amount lost) / equilibrium concentration
35S-sulfate concentration = μCi / mL (after 20 min)
D2O concentration = 0.2 g D2O / 100 mL (after 2 hrs.)
Urinary loss of 35S-sulfate = 4% in 20 minutes
Urinary loss of D2O = 0.4% in 2 hours
TBW (L) = (99.8g – 0.004•99.8g) / (0.2g/100 mL) • L/1000 mL = 49.7 L
ECF (L) = (100μCi – .04•100μCi) / (0.0064μCi/mL) • L/1000 mL = 15 L
ICF (L) = TBW – ECF = 49.7 L – 15.0 L = 34.7 L
% Body Water = 100 • TBW / BW = 100 • (49.7 L / 75 kg) • (L/kg) = 66.3%

6. To solve this problem, revise lecture 37
Dilution principle
Volume = (Amount given – Amount lost) / equilibrium concentration
35S-sulfate concentration = μCi / mL (after 20 min)
D2O concentration = 0.2 g D2O / 100 mL (after 2 hrs.)
Urinary loss of 35S-sulfate = 4% in 20 minutes
Urinary loss of D2O = 0.4% in 2 hours
To solve this problem, revise lecture 37

7. Body Fluids Total body water is approximately 60% of body weight
High in newborn & adult males.
Low in adult females & obese.
Total body water (TBW) is distributed into two compartments:
Intracellular fluid (ICF) 2/3 TBW
Extracellular fluid (ECF)  1/3 TBW
ECF is composed of interstitial fluid & plasma.
Physiology Fifth Edition. Costanzo, © 2014, Elsevier

8. Body Fluids Intracellular fluid (ICF) 2/3 TBW
Major cations are K+ & Mg2+
Major anions are protein & organic phosphates
Extracellular fluid (ECF)  1/3 TBW
Major cation is Na+
Major anions are Cl- & HCO3-
rule
Physiology Fifth Edition. Costanzo, © 2014, Elsevier

9. (assuming a 70 kg human)
What does this information tell you?
If you know the body weight, you can ESTIMATE the total body water (TBW, ie: 60% of BW), as well as the ICF (ie: 40% of BW) and the ECF (20% of ECF).
The 60:40:20 rule is a pretty good approximation for healthy people, but of course, there will be variations to this.
In the clinic, to TRULY know what the TBW and the breakdown of the fluid compartments, you will have to MEASURE each of these. This can be achieved by using tracers that only enter certain water compartments (see next slide)

10. Body Fluid compartments
D2O (duterated, or heavy water) will be a proxy for all the water compartments
Other markers include radiolabeled sodium. These substances are high in the ECF space
We will be using TBW-ECF to calculate the ICF in our problem

11. Measurement of fluid volumes by the dilution method
To solve the problem, you need to know the below equation:
V space = (Amtx given – Amtx lost) / equilibrium [x] in space
Measure TBW with D2O
Measure ECF water with radiosodium or radiosulfate
Measure P water with Evans Blue (T-1824)
Calculate IC water = TBW – EC water
Calculate IS water = EC water – P water

12. Dilution principle
Volume = (Amount given – Amount lost) / equilibrium concentration
35S-sulfate concentration = μCi / mL (after 20 min)
D2O concentration = 0.2 g D2O / 100 mL (after 2 hrs.)
Urinary loss of 35S-sulfate = 4% in 20 minutes
Urinary loss of D2O = 0.4% in 2 hours
TBW (L) = (99.8g – 0.004•99.8g) / (0.2g/100 mL) • L/1000 mL = 49.7 L
ECF (L) = (100μCi – .04•100μCi) / (0.0064μCi/mL) • L/1000 mL = 15 L
ICF (L) = TBW – ECF = 49.7 L – 15.0 L = 34.7 L
% Body Water = 100 • TBW / BW = 100 • (49.7 L / 75 kg) • (L/kg) = 66.3%
THUS- in this case, the person in this example has a higher % water compared to the estimate of 60%

13. 2. Darrow-Yannet Diagrams
Darrow-Yannet diagrams display the relationships between intracellular and extracellular fluid compartment volumes and osmolarities during changes in states of hydration and osmolarity. These diagrams can be conveniently used to qualitatively understand fluid volume shifts and osmolarity changes during six classic challenges to homeostasis.

14. Darrow- Yannet Diagram

15. Six challenges to homeostasis of body fluids
Volume expansion (gain of body weight)
Hyperosmotic volume expansion
Isosmotic volume expansion
Hyposmotic volume expansion
Volume contraction (loss of body weight)
Hyperosmotic volume contraction
Isomotic volume contraction
Hyposmotic volume contraction

16. Solve fluid balance problems in 3 steps
Construct the Darrow-Yannet diagram
Draw the disturbance THIS ONLY HAPPENS TO THE ECF
Equilibrate the ICF and the ECF by moving water from hypotonic to hypertonic THE WATER SHIFT MAY BE ECF ICF, or from ICF to ECF

17. Hyperosmotic volume expansion (gain of volume >300 mOsm/L)
Examples
Gulping in sea water while swimming
High NaCl infusion (3% or 5% NaCl)
A patient with a head-injury patient is given a hypertonic solution intravenously in order to shrink brain cells, thereby helping to relieve intracranial pressure.
Here, the blue dotted lines point to the disturbance.
The Blue arrows point to the displacements.

18. The red line represents the body’s shift in water.
NOTE: the goal is to BALANCE the osmolarity in the ICF and the ECF
The Red arrows represent the changes that are made between the disturbed state (remember: disturbance only happens to the ECF!!)
The YELLOW ARROW points to a GAIN in fluids from START TO FINISH (in this example, that’s happening to the ECF)
The GREEN ARROW points to a LOSS in fluids from START to FINISH (in this example, that’s happening to the ICF)

19. COLOUR CODES FOR THE NEXT FEW SLIDES
The original state is always the black line
In these examples, the ICF = ICV (ie: fluid and volume are being used interchangeably). Similarly, the ECF = ECV
The DISTURBANCE is drawn with a BLUE DOTTED LINE and blue arrows
The CORRECTION is drawn with a RED SOLID LINE and red arrows
The OVERALL SHIFT (from the black line to the red line; ie: start to finish) in fluid is drawn with a GREEN ARROW to represent a GAIN in fluid or a YELLOW ARROW to represent a LOSS in fluid.

20. Isosmotic volume expansion (gain of volume = 300 mOsm/L)
Examples
Physiological saline
Receiving blood (blood transfusion)
THERE IS NO CHANGE IN OSMOLARITY, thus there is no change in fluids between the ICF and the ECF
NOTE: for the purposes here, ICF = ICV; ECF= ECV

21. Hyposmotic volume expansion (gain of volume < 300 mOsm/L)
Examples
Drink pure water.
SIADH (syndrome of inappropriate ADH) excess ADH.
Administration of physiological saline intravenously (hypotonic to patient) to a severely dehydrated patient.
The BLUE arrow and blue dotted lines represents the challenge
The Red line and red arrows represents the correction (and shifts in fluid and overall change in osmolarity)
The green arrows point to the overall EXPANSION of fluids from beginning to end

22. Isosmotic volume contraction (loss of volume = 300 mOsm/L)
Examples
Diarrhea.
Donating Blood.
Hemorrhage.
THERE IS NO CHANGE IN OSMOLARITY, thus there is no change in fluids between the ICF and the ECF
Yellow arrow points to the overall contraction

23. Hyperosmotic volume contraction
You lost hypotonic fluid which makes your body hypertonic (loss of volume <300mOsm)
Examples
Sweating (severe exercise)
Water deprivation
Hyperventilation
The BLUE arrow and blue dotted lines represents the challenge
The Red line and red arrows represents the correction (and shifts in fluid and overall change in osmolarity)
The yellow arrows point to the overall CONTRACTION (decrease) of fluids from beginning to end

24. Hyposmotic volume contraction
You lost hypertonic fluid which makes your body hypotonic (loss of volume > 300 mOsm/L)
Examples
Producing hyperosmotic urine
Adrenal insufficiency
(less Aldosterone which means decrease Na+ reabsorption).
The BLUE arrow and blue dotted lines represents the challenge
The Red line and red arrows represents the correction (and shifts in fluid and overall change in osmolarity)
The green arrow points to the gain of fluids from beginning to end
The yellow arrow points to the loss of fluid from beginning to end

25. Question 3: Anion Gap
The anion gap is the difference between the measured cations (positively charged ions) and the measured anions (negatively charged ions) in plasma or urine.
The magnitude of this difference (i.e., "gap") in the serum is often calculated in medicine when attempting to identify the cause of metabolic acidosis, a lower than normal pH in the blood.
If the gap is greater than normal, then high anion gap metabolic acidosis is diagnosed. The term "anion gap" usually implies "serum anion gap“.

26. Anion Gap [A-] = Na – Cl- – HCO3-
Anion Gap Equation

27. Anion Gap
RTA type 2 is a type of renal metabolic acidosis. Note that although the Anion gap is the same as the normal condition, the complement HCO3- ions is now much smaller. HCO3- acts as a base, thus, with decreased base, and buffering capacity, the patient will experience metabolic acidosis. The OPPOSITE is true with vomiting- you’re increasing your HCO3- complement, thus, you will be in alkalosis

28. Anion Gap 15 >15 pH Problem Anion Gap Renal / GI Other Organs Renal
If there’s a pH problem in the patient, you can diagnose what type it is based on the anion gap, using the chart to the left
PLEASE ALSO SEE DR. MIGNERY’s lectures
YES
Anion Gap 15
>15
Renal / GI
Other Organs CL
HCO3
Renal GI

29. Question 4: Fractional Reabsorption
It is possible to compute the fractional reabsorption of any filtered substance without necessitating the collection of urine.
In spite of this, it is still good practice for you to be able to calculate the fractional reabsorption if you were given the values from urine and the plasma.
Fx is the algebraic derivation for the fractional reabsorption of any substance x.
Note that % reabsorption of X (%RX) = 100 • FX.
For this problem, you are comparing the fractional reabsorption of K to that of creatinine, and you are making the following assumption:
Creatinine is being cleared ONLY by filtration, and there is no secretion.

30. Calculate the fractional reabsorption of K+ and percent reabsorption of K+ from the following laboratory values sampled from the arterial blood and urine.
UK+ = 70 meq/L
PK+ = 4.5 meq/L
UCR = 150 mg/dL
PCR = 1.3 mg/dL

31. We first need these formulas:
Formula for the clearance of creatinine:
Formula for the clearance of Potassium:
Where C is clearance
U is urinary concentration
V. Is the urinary flow rate
P is the plasma concentration
CCR= UCRV
PCR
CK= UKV PK
To get the fractional reabsorption of K, we first need to calculate the fractional CLEARANCE of K, relative to the Clearance of CR
FCK=CK
CCRc
Where FCK is the fractional clearance of K
UCRV
PCR
THUS: FCK=CK =
CCRc
UKV PK
UCRV
PCR
X PK
UKV
We can then rearrange this equation:
FCK=
Then you can cross out the V
FCCR= UCRPK
PCRUK
Then the FRACTIONAL REABSORPTION is equal to 1- the FRACTIONAL CLEARANCE
UCRPK
PCRUK
FK = 1-
Where FK is the fractional reabsorption of K

32. %RK+ = 100 • FK+ = 100 • 0.865 = 86.5% reabsorption of K+
Because FRACTIONAL REABSORPTION is equal to 1- the FRACTIONAL CLEARANCE
UCRPK
PCRUK
FK = 1-
FK+ = 1 – (70 / 4.5) / (150 / 1.3) = 0.865
%RK+ = 100 • FK+ = 100 • = 86.5% reabsorption of K+

33. Is the computed fractional reabsorption of potassium above consistent with a high K+ diet, a normal K+ diet, or a low K+ diet? Why?
K+ is filtered, reabsorbed, secreted and excreted.
K+ reabsorption is set by potassium dietary intake. The more the potassium in the diet, the less the potassium reabsorption.
Since a normal potassium reabsorption is 85%, the computed %RK+ represents a normal K+ diet.

34. Question 5: Diuretics a b d
e,f,g c

35. Concepts to take away from this section:
Although there are 7 classes of diuretics listed here, you should focus on the following:
1. Ask yourself WHY are non-Na targeting diuretics used in the proximal tubule. Answer: You don’t want to target the N reabsorption DIRECTLY because 67% of the Na reabsorption occurs there
2. Be able to distinguish between the K wasting and the K sparing diuretics. For the K wasting diuretics understand why, in particular, the loop diuretics and the thiazides are K wasting. I have included, here, the two extra slides from lecture 40 that explain this. Make sure you go back and review how K is secreted (lecture 40) from the principal cells.

36. Why K wasting?
Loop diuretics & thiazide diuretics inhibit Na “upstream” of K secretion sites
WHAT DOES THIS THEN DO TO THE PRINCIPAL CELLS DOWNSTREAM in the late Distal tubule and collecting duct?
1). Principal cells: Na uptake, as well as more activity by Na/K/ATPase. K+ pumping into cells , increasing driving force for K secretion from the principal
2). Increased flow rate; luminal K is diluted, thus increasing driving force for K secretion from the principal cells
For loop diuretics: there is direct inhibition of the Na/K/Cl-co-transport as well as the two reasons given above
New Slide

37. Why K sparing?
These diuretics act on the actions of the cells that are affected by ALDOSTERONE sensitive cells; i.e. on the principal cells
The K-sparing diuretics inhibit Na transport in the principal cells
Na/K/ATPase activity is downregulated
 K secretion is decreased
New Slide

38. A. Mannitol (hexan-1,2,3,4,5,6-hexol, C6H8(OH)6)
osmotic diuretic
non-reabsorbable polysaccharide
holds water in the tubules osmotically
weak diuretic
K+ wasting
Mannitol is a diuretic for the same reason that excess glucose results in diuresis- ie: holds water
B. Acetazolamide (Diamox)
Inhibits carbonic anhydrase
Inhibits bicarbonate reabsorption in proximal tubule
Decreases sodium reabsorption in collecting duct
Weak diuretic
K+ wasting 

39. D. Hydrochlorothiazide (Thiazide)
C. Furosemide (Lasix)
Inhibits salt reabsorption in thick ascending loop of Henle
Overwhelms distal tubule and collecting duct with sodium
Sodium reabsorption and excretion both increase
Strong diuretic
K+ wasting
Can induce hypokalemic metabolic alkalosis
Must give potassium salt supplements
D. Hydrochlorothiazide (Thiazide)
Inhibits sodium/chloride transport in distal tubule
Diuretic competes for chloride sites on transporters
Sodium load in distal tubule increases
Weak diuretic

40. F & G E E. Spironolactone (Aldactone)
Competes with aldosterone in DT and CD
Decreases sodium reabsorption
Weak diuretic
K+ sparing
F. Amiloride (Midamor)
Blocks epithelial sodium channels on luminal side of CD
G. Triamterene (Dyrenium)
F & G E