Renal Function PDF Notes

WSUSOM Medical Physiology Rossi-Renal Physiology Page 1 of 22 Clearance and Measurement of Renal Function Clearance and Measurement of Renal Function Learning Objectives: 1. Clearance (renal clearance) A. Correctly define the concept of renal clearance o...

WSUSOM Medical Physiology Rossi-Renal Physiology Page 1 of 22 Clearance and Measurement of Renal Function Clearance and Measurement of Renal Function Learning Objectives: 1. Clearance (renal clearance) A. Correctly define the concept of renal clearance of any solute B. Calculate the clearance of any solute C. Develop a clear conceptual understanding of the utility of clearance measurements in assessing renal functional parameters such as a. glomerular filtration rate b. renal plasma flow c. Understand the underlying premises that permit these assessments 2. Know what may be used to measure glomerular filtration rate (GFR) and why and be able to calculate GFR 3. Distinguish total and effective renal plasma flow (RPF) and do the calculations that permit the assessment of each 4. Understand the source of creatinine, its handling by the kidney and the rationale for the use of the clearance of creatinine as a marker of GFR 5. Compare the use of the clearances of creatinine and inulin as indices of glomerular filtration rate 6. Apply measurements of GFR to decision making WSUSOM Medical Physiology Rossi-Renal Physiology Page 2 of 22 Clearance and Measurement of Renal Function Renal Clearance Definition: The renal clearance of a substance is the VOLUME of plasma from which that substance is completely cleared (removed) by the kidney per unit of TIME. Hence, the units of clearance are VOLUME TIME Cx = Ux * V Ax This is a VERY VERY IMPORTANT equation. It is used both in physiology and in clinical applications for assessment of parameters of renal function. Cx = clearance of solute “x” One may calculate the renal clearance of any substance that can be measured in the plasma and in the urine. Clearance principles are used to measure kidney function clinically and are applied daily in practice. WSUSOM Medical Physiology Rossi-Renal Physiology Page 3 of 22 Clearance and Measurement of Renal Function Excretory Rate of X is Ux * V = (GFR * Ax) - Rx + Sx Renal Clearance of X is Ux* V = GFR _ Rx + Sx Ax Ax Ax The first equation gives the EXCRETORY RATE of substance “x” which is the net result of filtration, reabsorption, and secretion. The second equation is that for renal clearance of x, Cx, which by definition is (Ux*V) / Ax. Note that the numerator (on the left) is the EXCRETORY RATE. If you divide both sides of the first equation by Ax, you get the formula for clearance on the left and the expression noted on the right. All the terms are now in volume/time. Let’s see what this implies with inulin, PAH and glucose... Inulin Clearance For inulin, Rin = 0 and Sin = 0, so Cin = Uin * V = GFR Ain Note that the clearance of inulin is CONSTANT over the whole range of plasma concentrations of inulin, Ain. The EXCRETORY RATE (Uin* V) of inulin is not constant, but its RENAL CLEARANCE ([Uin * V]/Ain) is constant. That is, plotting (Uin * V)/Ain vs Ain is constant. The SLOPE of the inulin excretory rate is constant (see earlier slide). Compare with the graph of Uin* V vs Ain in which the slope is the GFR. The slope in THAT graph is the Cin. WSUSOM Medical Physiology Rossi-Renal Physiology Page 4 of 22 Clearance and Measurement of Renal Function This slide illustrates the clearance of inulin. Note this is lighter green. (see animated step-wise slides from class on Canvas) Consider the following 1. Each rectangle is 100 ml of plasma water. Inulin is green (filled boxes). So there are 500 ml of plasma going into the glomerulus with inulin dissolved in it. 2. At the glomerulus, 100 ml of plasma gets filtered and this will have inulin in it at the SAME concentration as in plasma water. 3. Now that leaves 400 ml of plasma going into the efferent arteriole and that plasma has inulin in it just as before. (The glomerulus does not extract the inulin from the plasma. The inulin in the plasma that is filtered will be filtered with that plasma, but the inulin in the plasma that is left behind is left behind.) 4. The 100 ml of filtrate (tubular fluid) with its inulin then enters the tubules where a. The tubules reabsorb the water…99% (proximal ® descending loop ® collecting duct) back into the peritubular capillaries but NOT the inulin (see the rectangle is clear, no filled). b. The inulin is not reabsorbed and not secreted so remains in the tubules in the tiny bit of fluid that is left to become the final urine (~1% of all that is filtered). 5. So now if we look at the inulin content of the plasma water in the renal vein, it looks as if there were 400 ml of plasma with inulin at the original concentration and 99 ml of plasma with NO inulin...that is 99 ml of plasma has been TOTALLY CLEARED of its inulin. 6. Now you know that fluid does not act in compartments but that the inulin dissolves in the whole plasma but it is AS IF 99 ml of plasma has been cleared of inulin. Since inulin is ONLY filtered, the CLEARANCE of inulin = GFR. 7. 99 ml is close enough to 100 ml given the errors of urine collection and measuring the inulin in the plasma and urine. WSUSOM Medical Physiology Rossi-Renal Physiology Page 5 of 22 Clearance and Measurement of Renal Function Glucose Clearance For glucose Sglucose = 0 (glucose is not secreted) Cglucose = Uglucose * V = GFR - Rglucose Aglucose Aglucose At low Aglucose, all the glucose that is filtered is reabsorbed from the tubular fluid, hence, no glucose appears in the urine and NO glucose is “cleared” from the plasma. The clearance Cglucose is zero at these concentrations of Aglucose. At a certain point (threshold), the filtered glucose exceeds the transport capacity of the tubule to reabsorb the glucose (Tm), and glucose begins to appear in the urine (is excreted). The rate at which glucose appears in the urine is equal to the rate at which it is filtered, hence curve #3 approaches that for curve #1. Put another way, at high Aglucose values, Rglucose /Aglucose approaches zero and Cglucose approaches the GFR. Compare with the graph of Uglucose* V vs Aglucose in which the slope is parallel to that of the GFR. WSUSOM Medical Physiology Rossi-Renal Physiology Page 6 of 22 Clearance and Measurement of Renal Function This slide illustrates the clearance of glucose. Consider the following 1. At the glomerulus, 100 ml of plasma gets filtered and this will have glucose in it at the SAME concentration as in plasma water. 2. Now that leaves 400 ml of plasma going into the efferent arteriole and that plasma has glucose in it just as before. The glucose in the plasma that is filtered will be filtered with that plasma, but the glucose in the plasma that is left behind is left behind in the efferent arteriole at the same concentration. 3. The 100 ml of filtrate (tubular fluid) with its glucose then enters the tubules where a. In the case of GLUCOSE, the tubules reabsorb the water back into the peritubular capillaries as in the previous slide BUT b. GLUCOSE is reabsorbed by the proximal tubule. c. In normal individuals with normal plasma glucose, ALL the glucose is reabsorbed. 4. So now if we look at the glucose content of the plasma water in the renal vein, it looks as if there is 400 ml of plasma with glucose at the original concentration and 99 ml of plasma with glucose back in it. 5. Thus, in the case of glucose, it appears that NONE of the plasma water has been CLEARED of glucose. 6. The RENAL CLEARANCE of GLUCOSE (in this normal individual) = 0. (Note that this does not take into account that some of the glucose is actually utilized by the kidney for its own metabolic needs to generate the ATP for all the transport functions we will learn about shortly. The kidney also uses free fatty acids for energy not just glucose.) WSUSOM Medical Physiology Rossi-Renal Physiology Page 7 of 22 Clearance and Measurement of Renal Function PAH Clearance For PAH, Rpah = 0 Cpah = Upah * V = GFR + Spah Apah Apah At low Apah, all (or nearly so) the PAH is “cleared” from the plasma by a combination of filtration and secretion, hence Cpah is high. (We will later discuss how this can be used to assess effective renal plasma flow) As Apah increases, Spah increases to a constant maximum value (the tubular transport maximum Tm discussed later). As the Apah increases, the value for Apah exceeds the tubular maximum transport capacity by so much that the amount secreted is dwarfed by the amount of PAH filtered. Remember that filtration increases as the concentration of PAH increases and has no maximum limit. At high Apah values, Spah/Apah approaches zero and Cpah approaches the GFR Compare with the graph of Upah* V vs Apah in which the slope is very steep initially then is parallel to that of the GFR. WSUSOM Medical Physiology Rossi-Renal Physiology Page 8 of 22 Clearance and Measurement of Renal Function This slide illustrates the clearance of PAH. Consider the following 1. At the glomerulus, 100 ml of plasma gets filtered and this will have PAH in it at the SAME concentration as in plasma water. 2. Now that leaves 400 ml of plasma going into the efferent arteriole and that plasma has PAH in it just as before. The plasma water that is filtered has PAH in it at the same concentration as in plasma water and this 100 ml of filtrate (tubular fluid) with its PAH then enters the tubules where a. The tubules still reabsorb the water…99% (proximal ® descending loop ® collecting duct) back into the peritubular capillaries. b. The PAH that was filtered STAYS in the tubules since it cannot be reabsorbed…AND c. The PAH that remained in the plasma water of the efferent arteriole now enters the peritubular capillaries where at the proximal tubule the PAH is now SECRETED into the proximal tubule…leaving NO PAH in the peritubular capillaries. 3. So now if we look at the PAH content of the plasma water in the renal vein, there is NO PAH left in the plasma that comes from the peritubular capillaries…that is ALL of the plasma has been TOTALLY CLEARED of its PAH content. 4. Thus, all the PAH that entered the afferent arteriole has ended up in the urine (very concentrated in ~ 1 ml of urine) and 499 ml of the plasma water has been cleared of PAH. As a result the RENAL CLEARANCE of PAH = the EFFECTIVE renal plasma flow, that is, the plasma water that goes through the nephrons (not the part that goes to fat and fascia, etc. In this example we are not talking of plasma from the renal artery but already that plasma going to the afferent arterioles.) WSUSOM Medical Physiology Rossi-Renal Physiology Page 9 of 22 Clearance and Measurement of Renal Function Important Definitions bear repeating Renal Clearance: The renal clearance of a substance is the VOLUME of plasma from which that substance is completely cleared (removed) by the kidney per unit of TIME. PAH Clearance Observe that in this case (where the concentration of PAH is very, very low), the tubules have the capacity to secrete all of the PAH that is left in the efferent arteriole. As a result, the venous plasma coming from the tubules has no PAH in it. The plasma water has been completely CLEARED of PAH. Thus, at very low concentrations of PAH, the volume of plasma cleared of PAH is the volume of plasma that went through the nephrons! WSUSOM Medical Physiology Rossi-Renal Physiology Page 10 of 22 Clearance and Measurement of Renal Function At low Apah, all (or nearly so) the PAH is “cleared” from the plasma by a combination of filtration and secretion, hence Cpah is high. (We will later discuss how this can be used to assess effective renal plasma flow) As Apah increases, Spah increases to a constant maximum value (the tubular transport maximum Tm discussed later). As the Apah increases, the value for Apah exceeds the tubular maximum transport capacity by so much that the amount secreted is dwarfed by the amount of PAH filtered. Remember that filtration increases as the concentration of PAH increases and has no maximum limit. At high Apah values, Spah/Apah approaches zero and Cpah approaches the GFR. Compare with the graph of Upah* V vs Apah in which the slope is very steep initially then is parallel to that of the GFR. Definitions Clearance: The clearance of a substance is the VOLUME of plasma from which that substance is completely removed per unit of TIME. Tm: The Tm is the MAXIMAL transport rate of a substance by the renal tubules (secretion OR reabsorption). WSUSOM Medical Physiology Rossi-Renal Physiology Page 11 of 22 Clearance and Measurement of Renal Function Clinical Question 58 yo woman with long standing but stable chronic kidney disease due to lupus nephritis. She now develops multiple cysts in her kidneys, but the one on the left appears malignant (arrow). She wants to know what will happen if they take out the kidney with the tumor. Will she need dialysis if she has her kidney with the tumor removed? How can you use your knowledge of clearance of inulin and PAH (and creatinine) to help you counsel her in her decision making? Think about this question as you go through the next few slides…then we will use what we have learned to help answer her questions. The implications for clinical practice and the use of clearance concepts become very important and useful. Renal Hemodynamics Total Renal Blood Flow (total RBF): that blood which enters the kidney by the renal artery and perfuses kidney tissue as well as perirenal structures Effective Renal Blood Flow (effective RBF): that blood that perfuses the functional renal mass (ie, nephrons) » 90% of the total RBF WSUSOM Medical Physiology Rossi-Renal Physiology Page 12 of 22 Clearance and Measurement of Renal Function Methods for measuring effective RPF and RBF: 1. Effective RPF: At low Apah, Spah is much lower than the Tm for PAH secretion. Thus, as plasma flows through the glomeruli, PAH is filtered (» 20%). The PAH that is left in the peritubular capillaries (» 80%) is then fully secreted into the tubules…thus virtually no PAH remains in the plasma leaving the nephrons. Under these conditions, the clearance of PAH measures the effective RPF. Cpah = Upah* V = effective RPF Apah 2. Effective RBF: the effective RBF can be calculated from the effective RPF if the hematocrit (Hct) is known (expressed as a fraction, not as %) e.g, If the Hct is 40%, then plasma volume = 1 - 0.4 = 0.6 effective RBF = effective RPF 1 - Hct Effective RPF Thus, at low plasma concentrations of PAH, all the PAH is cleared from the plasma that goes through the nephrons (see below) in one pass. At higher plasma concentrations of PAH, the tubules cannot secrete all the PAH. Hence, the clearance of PAH would underestimate the value for effective RPF. Remember that the higher the plasma concentration of PAH the closer its clearance (slope of the excretory rate) is to the GFR (curve #2). Curve #1 is the clearance of inulin, the GFR. WSUSOM Medical Physiology Rossi-Renal Physiology Page 13 of 22 Clearance and Measurement of Renal Function Total RPF and RBF Blood to perirenal filtrate has 100 mg PAH in 100 ml structures…not through the nephrons 440 ml/min plasma (containing 440 mg PAH) leaves by the efferent arteriole to perfuse the tubules tubules secrete ALL PAH tubules reabsorb 99 mL filtered water V = 1 mL/min UPAH =100+440 = 540 mg/mL renal venous plasma flow = total RPF - V = 599 ml/min RVPAH = 60mg/599 mL PAH = 0.1 mg/mL Recall that total RPF is the flow to the nephrons PLUS the flow to perirenal structures. Apply to the diagram above: 1. Total RPF=600 ml/min and 600 mg of PAH are dissolved in the plasma. Thus, Apah = 1 mg/ml 2. Of the total RPF, 60 ml/min (containing 60 mg of PAH) bypasses the nephrons and 540 ml/min (containing 540 mg of PAH) perfuses the nephrons (effective RPF) 3. If GFR = 100 ml/min, the filtrate contains 100 mg PAH. Thus, 440 ml/min containing 440 mg of PAH perfuse the peritubular capillaries. 4. The tubules secrete all the PAH in the peritubular capillary plasma (440 mg) and they also reabsorb 99 ml of the filtered water. Thus, the final urine flow V = 1 ml/min and contains all the PAH that was filtered (100 mg) plus the PAH that was secreted (440 mg). The urine PAH Upah = 540 mg/ml 5. Renal venous plasma flow = total RPF - V = 599 ml/min. The renal venous (RV) blood contains the 60 mg of PAH that bypassed the nephrons. Therefore, Rvpah = 60 mg/599 ml, or roughly 0.1 mg/ml. WSUSOM Medical Physiology Rossi-Renal Physiology Page 14 of 22 Clearance and Measurement of Renal Function Effective RPF and RBF calculated Thus, given the values APAH = 1 mg/ mL; UPAH = 540 mg/mL;V = 1 ml/min; Hct = 40% effective RPF = CPAH = UPAH * V = 540 mg/mL * 1 mL/min = 540 mL/min APAH 1 mg/mL effective RBF = CPAH (1-Hct) = 540 mL/min/(1- 0.4) = 540 mL/min /0.6 = 900 mL/min Note that the EFFECTIVE renal plasma flow is that plasma which actually passes thru the glomeruli and peritubular capillaries. Hence, it is available to filtration and secretion. The Cpah only measures this effective RPF. TOTAL renal plasma (or blood) flow refers to all the plasma going into the main renal artery. Some of this plasma (typically ~10% but sometimes more), goes to other structures: perirenal fat, fascial tissue, upper ureter, etc. This plasma is NOT available for filtration and secretion. Can we measure this total RPF? Yes. To assess total renal plasma flow with PAH requires measurement of PAH in the renal vein (RV). Be sure you are clear on the distinction between total and effective RPF and RBF. Total RPF and RBF Total RPF = UPAH * V___ APAH - RVPAH Total RBF = total RPF 1 - Hct Total RPF = 540 mg/mL * 1 ml/min = 600 ml/min 1 mg/mL - 0.1 mg/mL For those who want to know how these formulae were derived here they are in the box… (Do not memorize.) WSUSOM Medical Physiology Rossi-Renal Physiology Page 15 of 22 Clearance and Measurement of Renal Function What if [PAH] is higher than can be completely secreted in one pass thru the kidney? filtrate has 200,000 mg PAH in 100 ml 350 ml/min plasma (containing 700,000 mg PAH) perfuse the tubules If tubular transport maximizes at 500 mg/min (we are using an example) tubules secrete maximum possible PAH or 500 mg/min Tubules will also overall reabsorb 99 mL filtered water V = 1 mL/min UPAH =200,000+500 = 200,500 mg/mL renal venous plasma flow = total RPF - V = 499 ml/min RVPAH = 799,500 mg/499 mL = 1,602 mg/mL In this example, the concentration of PAH in the Low concentration PAH plasma is very high. Cpah = eRPF Note that when the concentration of PAH exceeds the ability of the tubule to secrete the PAH that was not filtered, then the clearance of PAH will NOT equal the effective RPF but will approach the value of GFR Hi concentration PAH Cpah approaches GFR WSUSOM Medical Physiology Rossi-Renal Physiology Page 16 of 22 Clearance and Measurement of Renal Function If APAH is high, then CPAH approaches GFR Given, APAH = 2,000 mg/ mL UPAH = 200,500 mg/mL V = 1 ml/min Hct = 40% CPAH = UPAH * V = 200,500 mg/mL * 1 mL/min = 100.25 mL/min Thus, if PAH concentration is high and exceeds the secretory capacity of the tubules to eliminate it in ONE pass thru the kidney, all the PAH will NOT be “cleared” and the calculated CPAH approaches the GFR (box) and will NOT be an accurate measure of effective RPF! Thus, CPAH = effective RPF if and only if APAH is below the concentration at which ALL PAH entering the peritubular capillaries can be secreted (arrow). Glomerular Filtration Rate Excretion rate Ux * V = (GFR * Ax) - Rx + Sx Clearance Cx = Ux* V = GFR _ Rx + Sx Ax Ax Ax The formulas above give the arithmetic definitions of excretion and clearance. Be sure you are clear on the CONCEPTUAL difference between excretion and clearance that the formulas reflect: Excretion is the AMOUNT excreted in the urine per unit time: UNITS are amount/time Clearance is the VOLUME of plasma cleared of a substance per unit time: UNITS are volume/time Cinulin = GFR GFR = Uinulin * V Ainulin Sneaky question: Inulin is freely filtered at the glomerulus, not secreted or reabsorbed. If the plasma concentration of inulin is 10 mg/ml and the glomerulus filters 20% of the plasma and the GFR is 100 ml/min, - What is the concentration of inulin in the efferent arteriole? - How much inulin has been filtered in 1 min? WSUSOM Medical Physiology Rossi-Renal Physiology Page 17 of 22 Clearance and Measurement of Renal Function There are three common ways of measuring/estimating GFR: Cinulin = GFR. Inulin is filtered and neither reabsorbed (Rinulin = 0) or secreted (S inulin = 0), so Uinulin * V = GFR * Ainuin rearranging the formula GFR = Uinulin * V Ainulin Cinulin is the standard for measuring GFR, but is used less in human studies now (being replaced by a substance called iothalamate that is handled similarly to inulin by the kidney). Inulin Clearance Clearance of Inulin Thus, the clearance of inulin equals the GFR. = Cin = GFR = UinV Ain Creatinine Clearance Clearance of creatinine closely approximates the GFR. WSUSOM Medical Physiology Rossi-Renal Physiology Page 18 of 22 Clearance and Measurement of Renal Function Ccr » GFR. Creatinine (Cr) is produced as a result of creatine metabolism in muscle. Creatinine is released into the blood continuously in proportion to muscle mass. In the steady state, the amount of creatinine released into the blood equals the amount of creatinine excreted into the urine. Creatinine is filtered and not reabsorbed (Rcr = 0). A small amount is secreted, Scr (about 10%, see slide above). Despite creatinine secretion, the clearance of creatinine usually approximates the GFR. (The approximation is even closer since the lab measurement of plasma creatinine, Acr is usually overestimated by »10%.) Unless the GFR is very very low as in advanced renal failure OR when a steady state is not present as in acute renal failure, the Ccr gives a “good enough” estimate of the GFR. Ccr has the advantage that nothing needs to be injected. Thus, the GFR of an individual can be evaluated by obtaining a timed sample of urine and of plasma. (Under most clinical situations venous blood is drawn and serum not plasma is used. This is not perfectly precise but gives a good enough estimate given all the other variables that are also not controlled for: proper collection of urine, complete emptying of the bladder, proper timing of the collections, etc.) In the steady state, Fick’s Principle applies if muscle mass is constant then production of creatinine will be constant the only way to eliminate creatinine from the body is in the urine production rate creatinine = excretion rate creatinine excretion rate of creatinine (Ucreatinine * V) will be also be constant And, the way creatinine gets into the urine is by filtration (ignoring secretion), then the filtered load of creatinine is GFR * ACr and is also constant Thus, GFR * ACr = UCr * V Or GFR = UCr * V ACr Recall production of creatinine depends on muscle mass. From day to day, this is stable in the normal person…exceptions include loss of lean body mass (emaciated cancer patients), muscle necrosis (as in a car accident), body building with gain of muscle mass, amputation of lower limbs, etc. Since Ucr * V = GFR * Acr + Scr constant = GFR * Acr Constant = GFR Acr WSUSOM Medical Physiology Rossi-Renal Physiology Page 19 of 22 Clearance and Measurement of Renal Function Thus, even without known the urinary excretion rate of creatinine, if the muscle mass remains the same, the GFR will be INVERSELY proportional to the plasma [creatinine]. E.g., doubling the Acr indicates GFR decreased by ½ E.g., If a person has a GFR of 100 ml/min with a plasma Cr = 1.0, and then develops a kidney disease and kidney function decreases to a certain point and stabilizes. If plasma Cr at this new steady state, is 4.0, then the new GFR would be 1/4th what is was before, or 25 ml/min. Filtration Fraction Another assessment that can be made from these clearance values is the filtration fraction which is a measure of the proportion of the effective renal plasma flow that gets filtered, that is, GFR vs eRPF. Thus, filtration fraction = Cin_ = GFR Cpah eRPF Usually, filtration fraction is ~20% in a normal human. This may vary depending on relative vascular resistances of the afferent and efferent arterioles or intraglomerular factors. (These will be discussed in the next lecture.) Back to our patient 58 yo woman has long standing but stable chronic kidney disease due to lupus nephritis. She now develops multiple cysts in her kidneys, but the one on the left appears malignant (arrow). She wants to know what will happen if they take out the kidney with the tumor….will she need dialysis, etc.? So what are you going to do? How will you know if taking out her kidney will put her on dialysis? How will you reassure her? Or can you? WSUSOM Medical Physiology Rossi-Renal Physiology Page 20 of 22 Clearance and Measurement of Renal Function The Patient Serum creatinine before lupus kidney disease = 0.6 mg/dL At that point, her Ccr = 100 ml/min (@ GFR) After lupus nephritis = 1.8 mg/dL and stable Her new GFR @ 33 ml/min How can you test this to be sure? o Have her collect a timed urine (24 hr) for creatinine and draw her blood level of creatinine and measure her clearance. If her Ccr is now 33 ml/min, what will happen if you remove her left kidney? o How much kidney function will she lose? o Will she end up on dialysis? o What is her life expectancy on dialysis vs with the tumor? What can you do to check and be sure? o Renal Scan: with PAH (eRPF) and technetium (GFR) § In the best case scenario, that is, each kidney contributes 50% function? § What if the left kidney contributes 70% and the right 30%? Are there alternative treatments to surgery? When to use/not use creatinine clearance as a measure of GFR GFR = Ccr = Ucr* V Acr Ccr » GFR. Creatinine (Cr) is produced as a result of creatine metabolism from muscle. Creatinine is released into the blood continuously in proportion to muscle mass. In the steady state, the amount of creatinine released into the blood equals the amount of creatinine excreted into the urine. Using Ccr as a measure of GFR is based on the assumption the person is in a steady state (stable). Fick’s Principle applies: creatinine in = creatinine out. Even when the kidney is diseased and GFR is decreased, the amount of creatinine excreted into the urine (Ucr *V) in a steady state will be the same as when the GFR was normal. BUT in order for this to happen the plasma creatinine, Acr, will need to be higher. Acr got higher because in the NON steady state the creatinine could not be excreted, so accumulated in the plasma UNTIL the product GFR * Acr again = Ucr * V. WSUSOM Medical Physiology Rossi-Renal Physiology Page 21 of 22 Clearance and Measurement of Renal Function Creatinine is filtered and not reabsorbed (Rcr = 0). A small amount is secreted, Scr (about 10%). Despite creatinine secretion, the clearance of creatinine usually approximates the GFR. (The approximation is even closer since the lab measurement of plasma creatinine, Acr is usually overestimated by »10%). Unless the GFR is very very low as in advanced renal failure OR when a steady state is not present as in acute kidney injury, the Ccr gives a “good enough” estimate of the GFR. Ccr has the advantage that nothing needs to be injected. Thus, the GFR of an individual can be evaluated by obtaining a timed sample of urine and of plasma. Follow in the slide. When a disease or insult occurs to the kidney and the GFR drops (the non- steady state, which may be hours, days, months, years), the excretion of creatinine DOES go down. Since creatinine cannot then be excreted properly, it will accumulate in the plasma. The plasma concentration goes up. (Remember that the muscle mass has not changed so the muscles are still making the same amount of creatine which is being converted to creatinine.) Once the plasma creatinine rises to a level where the filtered load of creatinine, GFR * Acr again = the amount of creatinine made by the muscles, the urinary excretion of creatinine goes back to previous level…so that again the amount of creatinine excreted = amount of creatinine made by muscles. Note that if the GFR stays low, the UcrV is back to its previous level BUT at the expense of a higher Acr. This is why we can use the Acr as an index of GFR (recall the horrible formulas I told you NOT to memorize. They are based on this principle). Of course if the person loses/gains muscle mass, then this relationship will be more complex (e.g., amputation, muscle wasting in cancer, etc.). This is also why it does not make sense to do a creatinine clearance during the non-steady state. Clinical Question (also seen on Step 1) 28 year old man donates his kidney to his brother. His right kidney is removed. Which of the following indices would be expected to decrease in the donor after full recovery? A. Creatinine clearance B. Creatinine production C. Daily excretion of Na D. Plasma creatinine concentration E. Renal excretion of creatinine WSUSOM Medical Physiology Rossi-Renal Physiology Page 22 of 22 Clearance and Measurement of Renal Function If you understand the principle that creatinine excretion in the steady state is the same regardless of GFR, you should be able to answer this question (directly off old USMLE). Note that the question may “couch” things in different language such as “steady state” becomes “full recovery” or “stable.” It means he is in a steady state. (Under most clinical situations venous blood is drawn and serum not plasma is used. This is not perfectly precise but gives a good enough estimate given all the other variables that are also not controlled for: proper collection of urine, complete emptying of the bladder, proper timing of the collections, etc.)