1. RENAL FUNCTION IN THE ICU S. ESFANDIARI MD, SURGICAL INTENSIVE CARE UNIT

3. Case Study 86 year Male 40 Kg had been admitted to ICU for G.I. Bleeding last 12 hr. BUN = 65 Creat 1.6 Urine Output = 30 cc/h BP = 140/90 C.I. 3.4L/min PCWP = 28 Medication Lasix 40 mg. Q/ d + Renal dose Dopa Urine Creat ( 2H)= 105 mg/L Urine Na = 60 Urine OS= 400 GFR=? < 15 > =35 = 19 =125 =60 DIAG A= ATN B=PUMP FAILURE C= NORMAL RENAL FUNC D= INSU DATA FOR DIAG

4. RENAL CLEARANCE/GFR Practical issues regarding creatinine, clearance in SICU A. Estimate Creatinine Clearance CL Cr = (140 - age) x lean BW in Kg 72 x S Cr Female = Estimate Value x .85

5. Case Study Calculate G.F.R V.S Measured Creatinine Clearance Estimate GFR 140 - 86 x 40 = 19 72 x 1.6

6. Case Study GFR = Cr C 2h = 105 x .5 = 32 1.6 GFR = 32 is normal for this patient age, Diagnosis = normal kidney function; High BUN is due to GI bleeding

10. RENAL CLEARANCE/GFR (cont) 24 Creatinine Clearance or 2 Hours = CL Cr = Ucr x UV(cc/min) S cr needs volume of urine (cc/min) urine concentration of creatinine ( lab reports total amount of creatinine per vol of urine) serum concentration of creatinine

13. Case Study 4 (cont) Ccr = 140 - Age x B.W. = 140 - 43 x 105 = 110 72 x Pcr 72 x 1.4 Female correction = Ccr x .85 = 110 x .85 = 94

14. Case Study 4 ESTIMATED VS MEASURED CREATININE CLEARANCE 43 year old female; weight 105 kg; evaluated for possible renal disease. Plasma creatinine 1.4/DL Urine creatinine 62/DL Urine volume 24 hours 1080 cc Consider zosyn dose

15. case 4 C cr measured 24 hour urinary creatinine clearance Ccr = Ucr x Uv min = 62x .75 = 33 Pcr 1.4

16. Interpretation= GFR Dialysis <10 >6 End stage Aggressive hydration 40-60 2-6 Severe hypovolemia Diuretics/dialysis 10-20 2-4 ATN None,protection strategy 20-30 2-4 C.Renal insufficiency None 30-80 1-2 Age >60 None 80-120 1-2 Normal TX GFR Serum Cr condition

20. Case Study 2 (cont) FENa = UNa x Pcr x 100 = 65 x 1. 8 x 100 = 6% PNa x Ucr 130 x 15 2h urinary creatinine clearance Ccr = Ucr x UV (min) = 15 x 4 = 33 Pcr 1.8

21. CASE 2 ATN FENA 65 LOW GFR=33

22. Fractional excretion of Na Ratio of the Na excreted < 1% to Na reabsorbed 99%

26. FENA FENA= Una + x S cr x 100 S na X Ucr

29. Case Study 5 70 year old woman 80 kg in very good health admitted in ICU after surgical removal of her spleen. She has been hypotensive and on pressors in PACU for a period of 6 hours prior to ICU admission: BP 110/70 BUN = 20 Creat 1.6/L urine output 30cc/h CVP = 12 Lasix drip, 2 mg/h + Dopa renal dose Urine Na = 65 Urine osm = 320 Urine Creatinine 48 mg/L SN a=135 DIAGNOSIS A= ATN B= SIADH C= HYPOVOLEMIA D=INSU DATA FOR DIAGNOSIS

30. Case Study 5 (cont) Estimated Creatinine Clearance from Cock Croft - Gault Ccr = (140-age) x BW = 140 - 30 x 80 = 76 72 x PCr 72 x 1.6 Correction for Female .85 = 65

31. Case 5 Urinary Cr clearance 2 h Ccr= Ucr x UV (min) = 48 x .5 = 15 Pcr 1.6

32. Dx=ATN

39. Mechanism of Rhabdomyolosis Release of myoglobin, which is 25% of HG size rapidly filtering by glomerlus, precipitate as an acid ferrihematin in proximal tubes, hypovolemia, urine pH<6 low urine output enhancing the process

49. Case Study 7 (cont) FENA UNa - 111  Sr cr - 2 SNa - 160  Ucr - 15 x 100 = 9%

51. Case Study 45 year old male with Cirrhosis and large acites admitted in I.C.U.for hypotension and low urine out put Patient urinary output 20 cc/h C.I. 5L/min PCWP = 20 BP = 130/70 P Na= 130 U Na = 5 meq/l Urine osmo = 580 Pcr 2.6 BUN = 60 meq/l Urine Cr = 92

55. Case 3 DX= HRS LOW GFR =11 LOW FENA=.1 H/O LIVER DISEASE

58. Hepato-Renal

59. URINARY INDICES IN HEPATO-RENAL SYNDROME

61. case U Cr=80 UOS=750 UNA=65

68. Free Water Clearance In a patient with prerenal oliguria, the urine flow may be 0.5 ml/min, serum osmolarity 300 mOsm/liter, and urine osmolarity 600 mOsm/liter: Cl OSM = (600)(0.5) = 1.0 ml/min (300) Cl H2O = urine flow (ml/min) _ Cl OSM = 0.5 - 1.0 = -0.5 ml/min

69. Free Water Clearance Cl H2O = Urine Volume/min - Cl OSM Example: Serum OSM = 300 MSO Urine OSM = 600 MSO Urine Value 120 cc/h Cl OSM = Uosm x UV = 600 x 2 = 4 S OSM 300 Cl H2O = UV = Cl OSM = 2- 4 = - 2 

70. Tests to Evaluate Tubular Dysfunction Osmolar Clearance CL OSM = Uosm x Uv SERUM OSO Refers to total number of osmotically active solute particles instead of single substance: If CL OSM in equal with urine volume means no solute, would be cleared by kidney.

71. Free Water Clearance Cl OSM = (0.5)(200) = 0.36 ml/min (280) Cl H2O = 0.5 - 0.36 = 0.14 ml/min . IF FREE WATER Cr POSITIVE MEANS LACK OF CONCENTRATION CAPABILTY

72. FILTERATION, EXCRETION, REABSOBTION ELCTROLYTES

73. Free Water Clearance Free-water clearance (Cl H2O ) is defined as urine volume per minute minus osmolar clearance and is normally negative. Isosthenuria (urine osmolarity the same as plasma) is one of the earliest and most consistent functional characteristics of ATN As urine osmolality falls, osmolar clearance also falls and free-water clearance becomes less negative. If urine osmolality falls below that of serum, free-water clearance becomes positive, a fall in osmolar clearance may occur, even before a fall in Cl cr or serum Cr . in some patients as ATN is developing.