MCB 136 Review | Kidneys and Body Fluids

Review Answers | Body Fluid

Q injected = 10 mg; Q lost = 300 ml urine · 0.005 mg/ml = 1.5 mg
Therefore Q in body = 8.5 mg

Conc. after equilibration = 0.0005 mg/ml
Therefore, volume of inulin space = Q/conc = 8.5 mg/0.0005 mg/ml = 17,000 ml or 17 L

Inulin space very closely approximates ECF, which we would expect to be ~24% of body mass.
Therefore, 0.24 · body mass = 17L · (kg/L) body mass = 17kg/0.24 = 70.8 kg so 70 kg is the closest.

Insufficient data to calculate blood volume.
Plasma volume = plasma protein space = Q_inj (none lost in short time) / plasma conc
Therefore, plasma volume = 60 痢 / 0.02 痢/ml = 3000 ml or 3 L

Blood volume = plasma vol/(1-H) Hematocrit = 1 - plasma fraction = 1 - 0.6 = 0.4
Therefore, blood vol = 3L/(1-0.4) = 5 L
ISF = ECF - Plasma volume; ISF = 17 L - 3 L = 14 L
(a) Cr space: Q_inj = Q_equil or: 1 · 10⁶ cpm/ml · 2 ml = 2800 cpm/ml · Cr space
Therefore: Cr space = (1 · 10⁶ cpm/ml · 2 ml) / (2800 cpm/ml) = 714 ml (Total Blood Volume)

(b) Inulin space: (Inulin is in plasma, not cells, but data give cpm / ml whole blood.)
Q_inj = Q_equil or: 4 · 10⁶ cpm/ml · 2 ml = [2200 cpm/ml / (1-H)] · Inulin space
Therefore: Inulin space = (4 · 10⁶ cpm/ml · 2 ml) / (4400 cpm/ml) = 1818 ml (Extracellular Fluid Volume)

(c) Interstitial fluid volume = extracellular fluid volume - plasma volume
ISF = 1818 ml - [714 · (1-H)] = 1461 ml
0.9 g NaCl/100 ml solution = (0.9 g / 0.1 L) / (22.99 g/mol Na + 35.45 g/mol Cl) = 0.154 mol/L = 154 mM
Osmolarity: with full dissociation → 2 mOsm / mmol 154 mmol/L · 2 mOsm/mmol = 308 mOsm/L
10 mg Ca²⁺/100 ml = (10 mg / 0.1 L) / (40 mg/mmol) = 2.5 mM Ca²⁺ = 5.0 mEq/L (two charges / ion)

2.2 mg Mg²⁺/100 ml = (2.2 mg / 0.1 L) / (24.3 mg/mmol) = 0.905 mM Mg²⁺ = 1.81 mEq/L
Gibbs-Donnan effect: diffusible cation will have higher concentration on the side with the polyanionic, non-diffusible protein
a. no change
b. swell
c. no change
d. shrink
e. initially no change, but because urea is permeable to RBC membranes, the cells will swell and burst.
f. no change
g. glycerol is permeable; cells will swell and burst
h. shrink
i. initially cells will shrink, but because glycerol is permeable cells will eventually swell & burst.

Functional Anatomy & Glomerular Filtration Answers

Renal clearance is in effect the amount, or volume, of plasma that is completely cleared of a substance per unit time, and it is ascertained from the following expression: Clearance = [U]_x· V / [P]_x where [U]_x and [P]_x are the urine and plasma concentrations, respectively, of substance x, and V is urine flow in ml/min.

Clearance is expressed as vol per unit time, or usually ml/min.

The clearance of a substance gives us some empirical measurement of renal function, namely, how much the substance is cleared or removed from the blood per unit time. Clearance data, especially when compared with reference values (e.g., clearance of inulin) tells us whether the substance is secreted or reabsorbed by the kidney. However, it does not tell us anything of the precise mechanism nor the specific location of the reabsorptive or secretory process.
To measure total glomerular filtration rate (GFR), we use an innocuous substance that is small enough to freely pass the renal filter and not so small as to leak across cell membranes. The substance must also not be secreted or reabsorbed by the renal tubule. Inulin is a substance that meets these requirements. The principle used here is that Q_filt = Q_excr if no secretion or reabsorption of the substance occurs after the initial (glomerular) filtration, so that urine vol/min · concentration in urine must equal filtrate vol/min · concentration in filtrate (which equals concentration in plasma for a freely filtered substance).

SNGFR is the GFR for a single nephron. It is observed that SNGFR is sensitive to the rate of flow and composition of the fluid flowing past the macula densa in the same nephron.
Fick Principle: In the circulatory system, the amount of a given material, X, entering a tissue through an artery (A) would equal the amount leaving through a vein (V) plus or minus the amount consumed or produced in the tissue. The amounts of X passing particular points in the vessels would be the concentrations times the flow rates, or [A]_x· flow rate and [V]_x· flow rate.

Therefore, [A]_x· flow = ([V]_x· flow) + Q or Q = ([A]_x· flow) - ([V]_x· flow)

In the steady state, flow in is equal to flow out. Therefore, flow (ml/min) = Q (mg/min) / ([A]_x - [V]_x)

Renal Clearance: For renal plasma flow (RPF) from the Fick Principle, we could write: RPF = Q / ([A]_x - [V]_x)

where the concentration of substance X is measured in plasma from the renal artery and renal vein, and Q is the amount of substance cleared or removed from the blood.

For a substance that is entirely cleared from the blood in one passage through the kidney (substance is filtered and secreted, and is below concentration that would saturate secretory mechanisms),
1) the concentration in venous plasma is zero (V_x = 0), and
2) all of Q ends up in urine, where it can be represented as [U]_x·V

therefore, renal plasma flow = Q/[A]_x = [U]_x·V / [P]_x

Thus, for this very special case renal plasma flow and clearance are identical. PAH is a substance that is nearly ideal for RPF determination, since it is both freely filtered and (at low concentrations) completely removed from peritubular capillaries by active secretion. However, the small portion of efferent arteriolar blood that goes through the vasa recta instead of the peritubular capillaries retains its PAH, so the venous concentration only approaches zero and the measured clearance only approaches RPF.
See text, reader, and notes.
(a) Glucose reabsorption = amount filtered - amount excreted = (GFR · [P]_glu) - E_glu
= (125 ml/min · 4.1 mg/ml) - (190 mg/min) = 322.5 mg/min

(b) Max. rate of glucose reabsorption = 322.5 mg/min Since filtered glucose greatly exceeds reabsorbed glucose (i.e., much glucose is being excreted in urine), the reabsorption rate calculated in (a) is the maximum rate of reabsorption (T_m) for glucose.

(c) Urine flow will be considerably elevated because the loss of solute (glucose) into tubular fluid will retain a good deal of water by an osmotic effect. This is generally called solute diuresis.
(a) [P]_urea = E_urea / C_urea = 18 mg/min / 75 ml/min = 0.24 mg/ml

(b ) Urea in glomerular filtrate = [P]_urea · GFR = 30 mg/min
Urea excreted in urine = 18 mg/min, so fraction of filtered urea which is excreted = 18/30 = 0.6
1. 1. GFR = (U_inV) / P_in = (165)(2.2)/3 = 121 ml/min
  2. Amount that would be filtered if freely filtered = P_D · GFR.
    Actual dextran filtered = U_DV (0 reab. and 0 secr.).
    Therefore, fractional filtration = U_DV/(P_D · GFR) = (15 · 2.2) / (0.4 · 121) = 0.68
    ( Note: = (U_DV / P_D) / GFR = C_D/GFR )
  3. U_DsV / (P_Ds · GFR) = (6 · 2.2)/(0.9 · 121) = 0.12
2. Both dextrans are the same size; only difference is negative charge on sulfonated species. Barrier for filtration is negatively charged.
See reader (336, 347) and text (Ch 9).
A countercurrent exchanger is a means to conserve an energy gradient by CC exchange. No input of energy is involved, simply a means to conserve an existing potential energy pool. A countercurrent multiplier creates a gradient using some form of energy.
c. (D,J)
d.
e.
c.
f. (GNUIP)
e. (70)
e.
b. (L,H)
c. (37 mm Hg)

Regulation of Salt & Water Balance Answers

Review your text and reader.
ECV decreases, [Na] increases, [ADH] increases, water permeability of collecting duct increases, osmolarity of urine increases.
Diabetes mellitus. Increased blood glucose leads to increased [glucose] in tubules, which cannot reabsorb it all. This leads to increased osmotic content of tubular fluid compared to normal, and decrease in amount of water reabsorbed. There is increased urine production and loss of body water, increased thirst.
a. increase b. increase (both amiloride and furosemide are diuretics) c. decrease d. decrease
ADH will be increased in plasma.
Increased renin → increased angiotensins I & II → increased aldosterone → increased Na transport by distal tubule and collecting duct.
b.
Reduced blood pressure → increased renal vasoconstriction and reduced glomerular pressure → reduced filtration → reduced urine
Produces a hormone (ANF) that is released into blood to affect kidney function when blood volume is too large.
Aldosterone causes increased Na⁺ absorption by stimulating synthesis of Na⁺ channels and Na/K-ATPase (also a recently discovered faster action to increase open times of apical Na⁺ channels). ANF causes reduced Na⁺ and water absorption by increasing GFR and reducing renin and aldosterone; also directly inhibits Na⁺ absorption by collecting duct cells.
d.
d.
e. (P, C, H₂O)

Salt, water and pH Answers

Lower. The increased aldosterone causes positive sodium balance (i.e., increased plasma Na⁺), which reflexly inhibits renin secretion. Thus one observes a high plasma aldosterone and a low plasma renin, a strong tipoff to the presence of the disease, since in almost every other situation renin and aldosterone change in the same direction because the renin-angiotensin system is the primary control of aldosterone secretion.

E. All. The hypodynamic circulation of congestive heart failure causes impaired perfusion of the kidneys, which in turn causes increased secretion of renin. This also results in increased production of angiotensin II, which stimulates aldosterone secretion. Increased sympathetic activity to renal vasculature decreases renal plasma flow and glomerular filtration. The hypodynamic circulation also increases ADH release from the posterior pituitary. The action of the ADH is to increase permeability of the collecting ducts to water (thereby increasing water reabsorption).

Even though the bicarbonate/CO₂ buffer system is operating far from its pK (~6.1), it is extremely effective because of the volatile component of the system, CO₂, whose evolution through the lungs can be effectively regulated.

See notes and text.

Compensation raises pH.

H⁺ secretion refers to the transport of protons across a membrane; in the kidney, it would be secretion from proximal and distal tubular cells to the respective lumen. H⁺ excretion refers to that which is ultimately lost or excreted in the urine. You would have secretion but not excretion if, say, HCO₃^- were converted to CO₂ and H₂O by the H⁺ secretory process and not lost into the urine. Generally H⁺ excretion is measured as titratable acidity plus the NH₄⁺ in the urine.

	Normal	Hypothetical Metabolic Acidosis

Daily filtered load of HCO₃^-	3750 mEq/day	1500 mEq/day
H⁺ secretion: for HCO₃^-	3750 mEq/day	1500 mEq/day
For T.A. & NH₄⁺	30 mEq/day	300 mEq/day
Total H⁺ secretion	3780 mEq/day	1800 mEq/day
H⁺ excretion (T.A.+NH₄⁺)	30 mEq/day	300 mEq/day

7.4 = 6.1 + log [20/(33.3 · 0.03)] = 6.1 + log [20/1]
(a) After 10 mEq H⁺ are introduced into a closed system (liquid solution only), ~10 mmoles HCO₃^- are converted to 10 mmoles dissolved CO₂
[HCO₃^-] becomes 10 mM
pH = 6.1 + log[10/(1+10)]) = 6.1 + log[10/11] = 6.1 + log 0.909 = 6.1 - 0.04 = 6.06

(b) In a system equilibrated with an atmosphere having a constant PCO₂ of 33.3 torr,
[HCO₃^-] again becomes 10 mM, but CO₂ leaves liquid until equilibrated with 33.3 torr PCO₂.
pH = 6.1 + log[10/(33.3 · 0.03)] = 6.1 + 1.0 = 7.1

(c) After equilibration with an atmosphere having PCO₂ of 16.6 torr,
[HCO₃^-] again becomes 10 mM, but [CO₂] equilibrates with 16.6 torr PCO₂.
pH = 6.1 + log[10/(16.6 · 0.03)] = 6.1 + log 20 = 6.1 + 1.3 = 7.4 (compensated)

(d) a would be a system without any gas outlet (not physiological at all).
b compares to a body in which the lungs are not capable of compensation.
c compares to a physiological system where the lungs, after compensation via hyperventilation, bring the blood pH back toward normal.
f. (H⁺, I)
b.
b.
b. (D, D, I)
b. (I, I, I)