PROBLEM SET 7.1 FLUID VOLUMES, GLOMERULAR FILTRATION AND CLEARANCE

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1 PROBLEM SET 7.1 FLUID VOLUMES, GLOMERULAR FILTRATION AND CLEARANCE ANSWER KEY 1. The time course of decay of plasma [inulin] shown in Fig can be simultaneously used to determine the ECF volume and the GFR. The disappearance of inulin is given by the first-order equation dn inulin / dt = - GFR P inulin where N inulin is the number of moles of inulin in the body and P inulin is its plasma concentration. The plasma concentration is the total number of moles of inulin in the body divided by its volume of distribution, the ECF: P inulin = N inulin /ECF Combining these two equations, we get d N inulin /dt = - GFR/ECF N inulin This equation describes a first-order decay curve. A. Separate variables and integrate the first-order decay equation between the definite limits of time t=0 and t=t, corresponding to N inulin = N 0 and N t. Separation of variables gives d N inulin / N inulin = -GFR/ECF dt (1) Definite integration gives Ln N t /N 0 = -GFR/ECF t (2) Taking the exponent of both sides gives N t = N 0 e -GFR/ECF t (3) B. Two hours after establishing steady-state in a 70 kg person, P inulin falls from the steady-state value of 20 mg% at steady-state to 7.2 mg%. Calculate GFR/ECF from the first-order decay equation. First we recognize that P inulin = N inulin / ECF so that P t / P 0 = N t / N 0 and we can use Eqn. (2) above: Ln 7.2/20 = -GFR/ECF 2 hr GFR/ECF = hr -1 = min -1 7.PS1.1

2 C. The total urine collected during 8 hours was 500 ml, and the average inulin concentration was 560 mg%. Calculate the ECF. The steady-state P inulin was 20 mg dl -1. The total amount of inulin in the ECF at that time was 500 ml x 560 mg dl -1 = 2800 mg. Thus the ECF is calculated as ECF = Amount/ concentration = 2800 mg /20 mg dl -1 = 14 L D. From B and C, calculate the GFR. Plugging in from B and C, we get GFR = m in -1 x 14 L = ml min A woman weighing 60 kg is given 10 mg in Evan s Blue dye intravenously. Ten minutes later a blood sample was obtained from another vein and colorimetric analysis of the plasma shows 0.4 mg% of the dye. Assume that the administered dye was evenly distributed in the plasma compartment by the end of the 10 min and that no dye was lost from the plasma during this period. A. Calculate the woman s plasma volume. The plasma volume is calculated as Volume = Amount /Concentration = 10 mg / 0.4 mg dl -1 = 25 dl = 2.5 L B. If the woman s hematocrit is 0.40, what is her total blood volume? Here P = (1-0.4) B; if P = 2.5 L, then B = 4.17 L 3. The empirical fit to the oncotic pressure due to albumin is given by Landis and Pappenheimer (Handbook of Physiology, Vol. 2, sec 2, pp , 1963) as albumin = 2.8 C C C 3 where albumin is in units of mm Hg and C is in units of g% (g per 100 ml of plasma). A. Calculate albumin s contribution to plasma oncotic pressure at the afferent arteriole when [album in] = 4 g%. This is a plug and chug. The equation is albumin = 2.8 C C C 3 and C = 4; The result gives albumin = 2.8 x x x 64 = = 14.8 mm Hg 7.PS1.2

3 B. Assume the filtration fraction is 0.2 and that the sieving coefficient for albumin = 0. What is the concentration of albumin at the venule end of the glomerular capillary? At the venule end the concentration will increase to C/ 0.8 = 5 mg% C. Calculate albumin s contribution to the plasma oncotic pressure at the efferent arteriolar end of the glomerulus At the venule end the osmotic pressure due to albumin would be albumin = 2.8 x x x 125 = = 20 mm Hg Thus, the osmotic pressure increases much faster than linearly. 4. Table 7.PS1.1 lists the diffusion coefficients of a variety of proteins. These were determined in water at 25 C at a viscosity of 1 x 10-3 Pa s, where Pa is Pascal = 1 N m -2. A. Calculate the Stokes radii for the proteins using the Stokes Einstein equation: D = kt/ 6 a s where D is the diffusion coefficient, k is Boltzmann s constant (=1.38 x joule molecule -1 K -1 ), is the viscosity and a s is the Stokes radius. Table 7.PS1.1 Diffusion coefficients and M r for a variety of proteins Protein Molecular Weight D x 10 7 (cm 2 s -1 ) Stokes Radius, a s milk lipase 6, Metallothionein 9, Cytochrome C 12, Ribonuclease 12, Myoglobin 16, Chymotrypsinogen 23, Carbonic anhydrase 30, Peroxidase II 44, Album in 68, Lactoperoxidase 92, Aldolase 149, The Stokes radii were calculated by a s = [1.38 x joules molecule -1 K -1 x 298 K / 6 x x 1 x 10-3 N m -2 x D x 10-4 m 2 cm -2 ] 7.PS1.3

4 a s = 2.18 x / D where D is in units of cm 2 s -1 ; using the numerical values in the table below a s = 2.18 x 10-8 /D where a s is given in m Table 7.PS1.1 Diffusion coefficients and M r for a variety of proteins Protein Molecular Weight D x 10 7 (cm 2 s -1 ) Stokes Radius, a s milk lipase 6, nm Metallothionein 9, nm Cytochrome C 12, nm Ribonuclease 12, nm Myoglobin 16, nm Chymotrypsinogen 23, nm Carbonic anhydrase 30, nm Peroxidase II 44, nm Album in 68, nm Lactoperoxidase 92, nm Aldolase 149, nm B. Based on these calculations and Fig.7.3.6, what is the approximate molecular weight cut-off of the kidney glomerulus? (This is the molecular weight that is 50% retained by the membrane). The molecular weight cut-off is the point at which = 0.5; from Fig this is difficult to determine; extrapolation of the curve gives = 0.5 below a s = 2.0 nm, suggesting a cut-off of about 18, Given that the average glomerular capillary pressure is about 55 mm Hg, plasma oncotic pressure is about 28 mm Hg, the hydrostatic pressure within Bowman s space is 20 mm Hg, ultrafiltrate oncotic pressure is 0 mm Hg, the GFR is 120 ml min -1, and each kidney weighs 125 g, define a filtration coefficient and find its magnitude. The equation here is GFR = K f [ P - ] (1) so we can define the filtration coefficient operationally as K f = GFR/ [ P - ] (2) Here GFR = 120 ml m in -1 and P - = 55 mm Hg - 20 mm Hg - 28 mm Hg = 7 mm Hg. 7.PS1.4

5 Plugging into Eq. (2) we get K f = 120 m L m in -1 / 7 mm Hg = 17.1 ml min -1 mm Hg The following test results were obtained over a 24-hour period: urine volume = 1.4 L urine [inulin] = 100 mg% urine [urea] = 220 m mol L -1 urine [PAH] = 70 mg ml -1 plasma [inulin] = 1 mg% plasma [urea] = 5 m mol L -1 plasma [PAH] = 0.2 mg ml -1 hematocrit = 0.40 Calculate: A. C inulin, C urea, C PAH The clearances are all calculated as Q U U x / P x. Plugging in, we get C inulin = 1.4 L/1440 min x 1 mg ml -1 /.01 mg ml -1 = 97.2 ml min -1 C urea = 1.4 L/1440 min x 220 m mol L -1 / 5 m mol L -1 = 42.8 ml min -1 C PAH = 1.4 L/1440 min x 70 mg ml -1 / 0.2 mg L -1 = ml min -1 B. ERPF The ERPF, effective renal plasma flow, is the clearance of PAH, given above as ml min -1 C. The rate of urea filtration The rate of urea filtration is just C inulin x P urea = 97.2 m L m in -1 x 5 mmol L -1 = mmol min -1. D. The rate of urea excretion The rate of urea excretion is just Q U x U urea = 1.4 L/1440 min x 220 m mol L -1 = mmol min -1 E. The rate of urea reabsorption The rate of urea reabsorption is the difference between the rate filtered and the rate excreted: Rate of urea reabsorption = m mol min mmol min -1 = mmol min -1 F. The rate of PAH filtration The rate of PAH filtration is C inulin x P PAH = 97.2 m L m in -1 x 0.2 mg ml -1 = mg min -1 G. The rate of PAH excretion The rate of PAH excretion is Q U x U PAH = ml m in -1 x 70 mg ml -1 = mg min -1 7.PS1.5

6 H. The rate of PAH filtration secretion (assum ing no PAH is reabsorbed) is Rate of PAH excreted - rate of PAH filtered = mg min mg min -1 = mg min The following data were obtained from a research animal: urine flow rate: 1.5 ml m in -1 urine [inulin]: 150 mg% plasma [inulin]: 1.5 mg% urine [A]: plasma [A]: 18 mg% 1.8 mg% urine [B]: 200 mg% plasma [B]: 1.0 mg % urine [C]: plasma [C]: 20 mg% 0.2 mg% A. Based on this inform ation, postulate how the kidney handles substances A, B and C and justify your answer. B. Propose experiment(s) to test your postulates. We can calculate the clearance of each substance by taking the ratio of Q U U x /P x. The clearances are: C inulin = 150 ml m in -1 C A = 15 ml m in -1 C B = 300 m L m in -1 C C = 150 m L m in -1 Reasonable hypotheses are that substance A is either not freely filtered or it is reabsorbed; substance B is secreted; substance C is freely filtered but neither secreted nor reabsorbed. Substance A has a lower clearance than inulin, which may be because it isn t filtered or because it is reabsorbed. Substance B must be secreted, but it is not certain that it is freely filtered. Substance C has the same clearance as inulin, but that may be coincidental. These postulates can be tested most simply by varying the plasma concentration and examining the clearance as a function of P X. If A is reabsorbed, we should be able to saturate the transport m echanism and determine a Tm. Sim ilarly with B; if it is secreted, we should be able to saturate its transport mechanism. 8. A person has TBW = 42L and ECF = 14 L and a plasma osmolarity of 300 mosm. After losing 4 g of NaCl and 2 L in sweat, what is the new ECF and ICF volume and osmolarity? The total osmoles in the body in the beginning is 42 L x 300 mosm = mosmoles. This is distributed as 28 L x 300 mosm = 8400 mosmoles in the ICF and 14 x 300 mosm =4200 mosmoles in the ECF. Losing 4g of NaCl is equivalent to 4g / 58.4 g mol -1 = 68.5 m oles of NaCl This is 2 osmole mol -1 x 68.5 = 137 mosm oles. 7.PS1.6

7 Thus, the total osmoles in the body following the loss is mosmoles mosmoles = mosm oles. The total volume in which this load of osmolytes is distributed is now 40 L (42L - 2 L lost). The new osm olarity therefore should be about osmoles / 40 L = mosm Nearly all of the NaCl lost came from the ECF. Thus the osmoles in the ECF is 4200 mosmole -137 mosmole = 4063 mosmole ; The ECF volume is thus 4063 mosmole / mosm = L The ICF is calculated as 8400 mosmol / mosm = L 9. A person has TBW = 50 L and ECF = 16 L and an original osm olarity of 295 mosm. The person eats a meal that contains 6 g of NaCl and 1 g K Cl and 0.5 L of water. Assume it is all absorbed and distributed. Calculate the estimated new ICF and ECF volume and osmolarity. The original total osmolarity is 50 L x 295 mosm = m osmole Ingesting 6 g NaCl adds 2 osmoles mol -1 x 6 g / 58.4 g mol NaCl -1 = mosmol to the total; 1 g K Cl adds 2 osmoles mol -1 x 1 g / g mol KCl -1 = 26.8 mosmol to the total. The final osmolytes in the body after the m eal is absorbed and distributed is = mosmole The final volume is 50 L L = 50.5 L The final osmolarity is thus mosmol / 50.5 L = mosm The ICF osmolytes are 34 L x 295 mosm mosmol = mosmol So the ICF volume is mosmoles / mosm = L The ECF osmolytes are 16 L x 295 mosm mosmol = mosmol So the ECF volume is mosmoles / mosm = L 7.PS1.7

8 10. One model of the glomerular membrane is a microporous membrane in which right cylindrical pores penetrate all the way through the membrane. Assume that the pores have a length of 50 nm and a raius of 3.5 nm. The viscosity of plasma is Pa s. The average hydrostatic pressure in the glomerulus is 60 mm Hg, hydrostatic pressure in Bowman s space is 20 mm Hg and the average oncotic pressure of glomerular capillary blood is 28 mm Hg. A. Calculate the flow through a single pore assuming laminar flow (use the Poiseuille flow equation). The flow is given as Q V = a 4 / 8 P/ x where a is the radius of the pore, is the viscosity of the medium, P is the pressure difference on the two sides of the pore and x is the length of the pore. The pressure difference is P = P GC - P BS - BS = 60 mm Hg - 20 mm Hg - 28 mm Hg = 12 m m Hg. Since 1 mm Hg = Pa, P = 12 mm Hg x Pa mm Hg -1 = Pa Inserting in the remaining variables: a = 3.5 x 10-9 m, = Pa s and x = 50 x 10-9 m, we have Q V = x 1.5 x m 4 / Pa s x Pa / 50 x 10-9 m = 9.42 x m 3 s -1 This can be converted to ml m in -1 by multiplying by 10 6 cm 3 m -3 and 60 s min -1 : 9.42 x m 3 s -1 x 10 6 cm 3 m -3 x 60 s min -1 = 5.65 x ml min -1 B. How many pores would there have to be to produce a normal GFR? The normal GFR = 120 ml m in -1 Dividing by the flow per pore, we get the number of pores: 120 m L m in -1 / 5.65 x ml min -1 pore -1 = 2.12 x pores C. If the total aggregate area of the kidneys for filtration is 1.5 m 2, what is the density of the pores (number of pores per unit area) The pore density is 2.12 x pores / 1.5 m 2 = x pores m -2 D. What fraction of the area is present as pores? The fraction of the area is the number of pores x the cross-sectional area per pore / total area. The area per pore is a 2 = x (3.5 x 10-9 m) 2 = 3.85 x m 2 Fractional area = x pores x 3.85 x m 2 / m 2 = This calculation suggests that some 5% of the area of the kidney is the interior of tiny pores through which glomerular ultrafiltrate forms. The actual number is in fact close to 5%. 7.PS1.8

9 11. Assume that myoglobin is a spherical protein with a Stoke s radius of about 1.9 nm and that the pores in the renal corpuscle have a radius of 3.5 nm. A. Give an estimate of, the reflection coefficient, for myoglobin. Recall here that if the protein hits the rim, it is assumed to be reflected back into the plasma side of the membrane. The formula for the reflection coefficient given the radius of the pore and the solute was given in Eqn. (2.7.21) as = 1 - (1 - a s / a) 2 where a s is the radius of the spherical protein and a is the radius of the pore. Inserting these values, we get = 1 - (1-1.9/3.5) 2 = B. Calculate the sieving coefficient for myoglobin, assuming that the only restriction for myoglobin filtration is at the pore entrance. The sieving coefficient would be 1 - = C. In rhabdom yolysis, m uscle mem branes leak m yoglobin into the blood. What do you think happens to that myoglobin, based on your answer to A and B? Some of it is filtered. It overwhelms the degradative mechanisms for filtered protein, and becomes highly concentrated in the distal neprhon, sometimes causing blockage of flow. 12. The following test results were obtained over a 24-hour period: urine volume = 1.2 L urine [inulin] = 110 mg% urine [creatinine] = 170 mg% plasma [inulin] = 0.8 mg% plasma [creatinine] = 1.2 mg % hematocrit = 0.40 A. Calculate the clearance of inulin. The clearance of inulin is C inulin = Q U U inulin / P inulin Q U = 1.2 L day -1 / 24 hr day -1 x 60 min hr -1 = m L m in -1 C inulin = ml m in -1 x 110 mg% / 0.8 mg% = ml min -1 B. Calculate the clearance of creatinine. C creatinine = Q U U creatinine / P creatinine = ml m in -1 x 170 mg% /1.2 mg% = ml min -1 C. What is the GFR? The GFR is the inulin clearance, ml min -1 D. Creatinine is an endogenous by-product of muscle metabolism, originating from creatine. Assuming a steady-state, estimate the daily production of creatinine. 7.PS1.9

10 At steady-state, the rate of creatinine production must be equal to the rate of creatinine elimination from the body: Q synthesis = Q excretion = Q U U creatinine = 1.2 L day -1 x 170 mg % = 1200 ml day -1 x 170 mg / 100 ml = 2.04 g day -1 E. If the filtration fraction is 0.18, estimate the effective renal plasma flow. FF = GFR / RPF = ml m in -1 / RPF = 0.18 RPF = ml m in -1 The ERPF is usually about 90% of the RPF: ERPF = 573 ml min The following test results were obtained over a 24-hour period: urine volume = 1.3 L urine [inulin] = 133 mg% urine [glucose] = 0 mg% plasma [inulin] = 0.8 mg% plasma [glucose] = 90 mg % hematocrit = 0.40 A. Determine the GFR. The GFR is calculated as the clearance of inulin: GFR = C inulin = Q U U inulin / P inulin Here Q U = 1.3 L day -1 / 24 hr day -1 x 60 min hr -1 = ml m in -1 C inulin = ml m in -1 x 133 mg% / 0.8 mg% = ml min -1 B. Calculate the clearance of glucose. The clearance of glucose is zero, because U glucose = 0 C. Estimate the daily filtered load of glucose. This is how much glucose is filtered per day. The daily filtered load of glucose is glucose x GFR x P glucose = 1.0 x ml m in -1 x 90 mg % = m g m in -1 x 60 min hr -1 x 24 hr day -1 = g day -1 D. How much glucose is excreted per day? The daily excretion of glucose is zero. E. How much glucose is reabsorbed from the ultrafiltrate, per day? The entire filtered load, which is g day -1 7.PS1.10

11 14. Suppose you injected someone with deuterated water, 2 H 2 O. At som e tim e, the m arker water would be evenly distributed among all exchangeable fluid compartments. The 2 H 2 O is filtered through the glomerulus, and then reabsorbed. Assume that the total body water is 42 L and ECF is 14 L. The GFR is 120 m L m in -1 and urine flow is a constant 1 ml m in -1. Assume that ingested fluids replenish the urinary losses on a continuous basis. A. How long would the 2 H 2 O remain in the body? Derive an equation for plasma [ 2 H 2 O] as a function of time. The disappearance of D 2 O due to exrection in the urine is given as dn D2O /dt = - Q U P D2O And the plasma concentration of D 2 O is P D2O = N D2O / TBW Combining these two equations, we have dn D2O / N D2O = - Q U / TBW dt Definite integration from time t=0 to time t=t gives ln N D2O (t) / N D2O (0) = - Q U / TBW t Taking the exponent of both sides, we get N D2O (t) = N D2O (0) e -Qu/TBW t Since the number of moles of D 2 O is proportional to its concentration, we have P D2O (t) = N D2O (t) / TBW = N D2O (0) / TBW e -Qu/TBW t = P D2O (0) e -Qu/TBW t B. What would the half-life of 2 H 2 O be in the plasma? The half-life is the time it takes for P D2O (t 1/2 ) to be ½ P D2O (0) In this case, we have 1 / 2 = e -Qu/TBW t 1/2 Taking the ln of both sides: -ln 2 = -Qu/TBW t 1/2 t 1/2 = TBW / Qu x ln 2 = 42 L / 1 ml m in -1 x = min Suppose you inject the sam e person with inulin and that a peak [inulin] was obtained some minutes later. C. Derive an equation for plasma [inulin] as a function of time. This is identical in form to the equation for [ 2 H 2 O] except that the excretion rate is set by the GFR and not the flow of urine, Q U, and the volume of distribution is the ECF and not the TBW: 7.PS1.11

12 P inulin (t) = P inulin (0) e -GFR/ECF t D. What would the half-life of inulin be in the plasma? The half-life of inulin is t 1/2 = ECF/GFR x ln 2 = 14 L / 120 m L m in -1 x = min E. How does this compare to B? The half-life of inulin is much shorter than the half-life of 2 H 2 O, because the clearance of inulin is much higher, and its volume of distribution is smaller. 15. A person weighs 105 kg. His total body water is 48 L. A. Calculate his lean body mass. Eqn. (7.1.5) Gives LBM = TBW /0.73 = 48 L x 1 kg L -1 / 0.73 = kg B. Calculate the per cent of body fat in excess of the lean body mass. The per cent of body fat is calculated as (total mass - LBM) / total mass = (105 kg kg) / 105 kg = = 37.3% C. W ould you say that this person is obese? W hy or why not? This person would be classified as obese. Usually, body fat in excess of 20% of body weight is classified as obese. 16. The Stoke s radius for water is 0.1 nm; urea s radius is 0.16 nm and the radii for glucose and sucrose are 0.36 nm and 0.44 nm, respectively. The radius for inulin, a fructose polymer with a molecular weight of 5.5 KDa, is about 1.6 nm. Assume the radius of the pores in the glomerulus is 3.5 nm. Hint: the area available to the solvent is NOT the dimensions of the pore. The reflection coefficient is a parameter of area available to solute relative to water. A. Calculate the reflection coefficient for each of the solutes given above, on the assumption that the available area for the solutes is (a - a s ) 2, where a is the radius of the pore and a s is the radius of the solute. The formula for the reflection coefficient given the radius of the pore and the solute was given in Eqn. (2.7.21) as = 1 - (1 - a s / a) 2 where a s is the radius of the spherical protein and a is the radius of the pore. However, what we really mean by radius of the pore is the radius of the pore available to water. So we should use a - a H2O 7.PS1.12

13 as the actual radius of the pore, because a reflection coefficient for water has no meaning: reflection coefficients are calculated relative to water. Similarly, we should use a s - a H2O as a measure of the size of the solute. Inserting these values, we get urea = 1 - (1-0.06/3.4) 2 = glucose = 1 - (1-0.26/3.4) 2 = sucrose = 1 - (1-0.34/3.4) 2 = 0.19 inulin = 1 - (1-1.5/3.4) 2 = B. Based on this reasoning, estimate the sieving coefficient,, for these small molecules. The sieving coefficient can be estimated as 1-. These would be urea = = glucose = = sucrose = = inulin = = C. Does your calculation support the idea that inulin is freely filtered? W hy or W hy not? The calculation does NOT support the idea that inulin is freely filtered. According to these calculations, there should be significant retardation of the inulin at the pores. Thus, either the dimensions of the equivalent pores is wrong, or the sizes of the m olecules are wrong, or there is something else going on that is not incorporated into the model of the kidney. 7.PS1.13

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