Faculty version with model answers Clearance Bruce M. Koeppen, M.D., Ph.D. University of Connecticut Health Center 1. Inulin is used in an experiment to measure the glomerular filtration rate (GFR). Inulin is continuously infused to achieve a steady-state concentration in the plasma of 1.0 mg/dL. Urine is collected over a 10 hour period. The total volume of urine is 1.5 L, and the urinary concentration of inulin is 440 mg/L. What is the GFR, as determined from the inulin clearance? Inulin Clearance = GFR = _____110____ ml/min Calculations: Urine flow rate = 1,500 ml/600 min = 2.5 ml/min Urine [inulin] = 440 mg/L = 0.44 mg/ml Plasma [inulin] = 1.0 mg/dL = 0.01 mg/ml Inulin clearance = GFR = (2.5 ml/min x 0.44 mg/ml)/0.01 mg/ml = 110 ml/min 2. In clinical situations the GFR of a patient is usually determined by measuring the clearance of creatinine. Like inulin, creatinine is freely filtered at the glomerulus. However, in contrast to inulin, a small amount is also secreted (approximately 10% of all creatinine excreted in the urine reflects this secretory component). Therefore creatinine is not an ideal marker for GFR. Despite this error, creatinine is used to measure GFR because it is produced endogenously by skeletal muscle, and does not require an infusion. Therefore it is easy to do a creatinine clearance, because the patient can do it at home by simply collecting their urine (usually over a 24 hr period), and having a single blood sample obtained to measure the plasma [creatinine]. What would be the estimated GFR of a patient with the following values for plasma and urine [creatinine] and urine volume. Plasma [creatinine] = 1.5 mg/dL Urine volume = 1.5 L/24 hours Urine [creatinine] = 1.44 gm/L Creatinine Clearance?~ GFR = _____100____ ml/min Calculation: Urine flow rate = 1,500 ml/1440 min = 1.04 ml/min Urine [creatinine] = 1.44 gm/L = 1.44 mg/ml Plasma [creatinine] = 1.5 mg/dL = 0.015 mg/ml Creatinine Clearance ≈ GFR = (1.04 ml/min x 1.44 mg/ml) / 0.015 mg/ml = 100 ml/min 3. One important consequence of renal disease is loss of nephrons. As nephrons are lost the GFR will decrease (remember that GFR is the sum of the filtration rates of all the nephrons in the kidneys). Therefore, as renal disease progresses the GFR declines, and the plasma [creatinine] increases. The relationship between GFR and plasma [creatinine] is illustrated below. ©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -1- 10 8 Plasm a 6 [ creat inine] (mg/ dL) 4 2 20 40 60 80 100 120 GFR (ml/ min) How do you explain the shape of this relationship, and what limitations does this present when trying to estimate GFR from a single plasma creatinine value? This curve reflects the fact that to maintain steady-state balance, excretion of creatinine by the kidneys must equal its production by skeletal muscle, and remain constant (this assumes production rate is constant). Because the excretion rate of creatinine essentially equals the creatinine filtered load (ignoring the secretory components of creatinine excretion), the following relationship applies: Ucr x V = GFR x Pcr Since Ucr x V is constant (ie., steady-state balance), the plasma concentration of creatinine (Pcr) must increase as GFR declines. Note that when GFR first begins to decline with renal disease, the change in plasma [creatinine] is quite small. In fact an individual can lose 1/3 to 1/2 of renal function and still have a plasma [creatinine] that is only slightly elevated above the normal reference value. Similarly, large changes in plasma [creatinine] occur with advanced renal disease even when only a small fraction of additional nephrons are lost. Thus predicting a precise GFR value from a single plasma [creatinine] determination is relatively inaccurate. However, monitoring the change in plasma [creatinine] over time is a useful way to monitor progression (or resolution) of renal disease. 4. A 25 year old man weighing 60 kg has a plasma [creatinine] of 1.4 mg/dL (nl = 0.8 - 1.4 mg/dL). A 24 hour urine collection is done to determine his creatinine clearance, and thereby estimate his GFR. The following data are obtained: Urine [creatinine] = 833 mg/L Urine volume = 1,080 ml/24 hour What is his creatinine clearance? Creatinine clearance = _____44.6____ ml/min Calculation: Urine flow rate = 1,080 ml/1440 min = 0.75 ml/min Urine [creatinine] = 833 mg/L = 0.833 mg/ml Plasma [creatinine] = 1.4 mg/dL = 0.014 mg/ml ©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -2- Creatinine Clearance ~ GFR = (0.75 ml/min x 0.833 mg/ml) / 0.014 mg/ml = 44.6 ml/min If the normal range of GFR for adult males is 90 - 140 ml/min, what would you conclude about the renal function of this man? The value of 44.6 ml/min for a creatinine clearance would seem to indicate a significant reduction in renal function (perhaps 50%). While his plasma [creatinine] is within the normal range (upper limit of normal), it is difficult to interpret because there is no baseline value. For example, it may have been 0.8 mg/dL previous to this current determination. This emphasizes the point that early on in renal disease small increases in plasma [creatinine] can occur with large initial declines in renal function. What error could lead to the calculation of a creatinine clearance that does not reflect the actual GFR? How could you determine if such an error occurred in this man, without having to repeat the 24 hour urine collection? (Hint: Muscle creatinine production is relatively constant and proportional to muscle mass. In adult males, the creatinine production rate is 20 -25 mg/kg body weight/day, and in adult females it is 15 - 20 mg/kg body weight/day). A falsely low creatinine clearance will be calculated if the urine collection was incomplete (ie., collected for less than a 24 hr period). Because creatinine production is relatively constant and proportional to muscle mass a simple calculation of the total amount of creatinine excreted in the urine will allow you to determine if the collection was complete. For example, this man weighs 60 kg, and his creatinine production should be in the range of 1,200 - 1,500 mg/24 hr. This should equal his creatinine excretion rate. However, his excretion rate is only about 900 mg/24 hr (1,080 ml x 0.833 mg/ml). Therefore, it appears that the low value for creatinine clearance may not a reflection of decreased renal function, but the result of an incomplete urine collection. Note: there is some reason to believe that perhaps this man does have some degree of decreased renal function. By taking the expected creatinine excretion rate of 1,200 - 1,500 mg/24 hr, a predicted creatinine clearance of 60 - 75 ml/min can be calculated. Unfortunately, because it is not know exactly how much urine was not collected, an accurate measure can only be obtained after the man does an appropriate 24 hour urine collection. 5. The renal handling of Na+ and K+ and urea are examined in two individuals. Both individuals eat the same diet, and are in steady-state balance. However, they have different glomerular filtration rates. Subject A has a GFR of 170 L/day, while the GFR of subject B is 50 L/day. The following data are obtained. Na+ Intake K+ Intake Urea Production GFR Urine Flow Subject mEq/day mEq/day gm/day L/day L/day A 100 50 8.5 170 1 B 100 50 8.5 50 1 Serum [Na+] Serum [K+] Serum BUN* GFR Subject mEq/L mEq/L mg/dL L/day A 145 3.5 10 170 ©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -3- B 145 3.5 34 50 *BUN = blood urea nitrogen Assume that the kidneys represent the only excretion route for Na+, K+ and urea. Calculate the following: Subject A Subject B Clearance Na+ (L/day) ____0.69___ ____0.69___ K+ (L/day) ____14.3___ ____14.3___ Urea (L/day) _____85____ _____25____ Filtered Load Na+ (mEq/day) ___24,650__ ____7,250___ K+ (mEq/day) _____595___ _____175___ Urea (mg/day) ___17,000__ __17,000___ Fractional Excretion Na+ (%) _____0.4____ _____1.4____ K+ (%) _____8.4____ ____28.6____ Urea (%) _____50____ _____50____ Tubular Reabsorption Na+ (mEq/day) __24,550 _ ___7,150___ K+ (mEq/day) ___545 __ _____125___ Urea (mg/day) ___8,500 __ ____8,500___ The following formulae are used: Clearance = excretion rate / plasma concentration Filtered Load = GFR x serum concentration Fractional Excretion = excretion rate / filtered load x 100 Tubular reabsorption = filtered load - excretion rate Calculations subject A: Clearance Na+: 100 mEq/d / 145 mEq/L = 0.69 L/day K+ : 50 mEq/d / 3.5 mEq/L = 14.3 L/day ©Bruce M. Koeppen, M.D., Ph.D., University of Connecticut Health Center -4- Urea: 8,500 mg/d / 100 mg/L = 85 L/day Filtered Load Na+: 170 L/d x 145 mEq/L = 24,650 mEq/day K+: 170 L/d x 3.5 mEq/L = 595 mEq/day Urea: 170 L/d x 100 mg/L = 17,000 mg/day Fractional Excretion Na+: 100 mEq/d / 24,650 mEq/d x 100 = 0.4% K+: 50 mEq/d / 595 mEq/d x 100 = 8.4% Urea: 8,500 mg/d / 17,000 mg/day x 100 = 50% Tubular Reabsorption Na+: 24,650 mEq/d - 100 mEq/d = 24,550 mEq/day K+: 595 mEq/d - 50 mEq/d = 545 mEq/day Urea: 17,000 mg/d - 8,500 mg/d = 8,500 mg/day Calculations subject B: Clearance Na+: 100 mEq/d / 145 mEq/L = 0.69 L/day K+ : 50 mEq/d / 3.5 mEq/L = 14.3 L/day Urea: 8,500 mg/d / 340 mg/L = 25 L/day Filtered Load Na+: 50 L/d x 145 mEq/L = 7,250 mEq/day K+: 50 L/d x 3.5 mEq/L = 175 mEq/day Urea: 500 L/d x 340 mg/L = 17,000 mg/day Fractional Excretion Na+: 100 mEq/d / 7,250 mEq/d x 100 = 1.4% K+: 50 mEq/d / 175 mEq/d x 100 = 28.6% Urea: 8,500 mg/d / 17,000 mg/day x 100 = 50% Tubular Reabsorption Na+: 7,250 mEq/d - 100 mEq/d = 7,150 mEq/day K+: 175 mEq/d - 50 mEq/d = 125 mEq/day Urea: 17,000 mg/d - 8,500 mg/d = 8,500 mg/day Based on the above data, explain each of the following: Note: questions A-D are all related, the point of each is to understand how Na+, K+ and urea are handled by the kidney, and how this can be discerned from the above calculations.