1. Ask Yourself! Can it be filtered? Is it reabsorbed? Is it secreted? What factors regulate the amount filtered, reabsorbed, and secreted? –Size –Permeability (charge) –Surface area

2. 5L/min CO 1.25L/min RBF 660 ml/min RPF 125 ml/min GFR Filtration fraction: GFR/RPF Urine formation 2ml/min

3. RBF (renal blood flow) ml/min RBF = ¼ CO RBF = ¼ (5L/min) = 1.25L/min= 1250ml/min Turns out, it really is a little less than this (some vasoconstriction, etc.) so we will use 1200ml/min 1.25-1.2L/min RBF

4. Effective Renal Blood Flow (ERBF; ml/min) Only about 90% of the RBF actually reaches the nephrons at the glomerulus, therefore; this number indicates the estimated, or effective, blood flow that actually reaches the nephrons. ERBF = 0.9(RBF) ERBF = 0.9(1200ml/min) = 1080 ml/min 1.08L/min

5. Renal Plasma Flow (RPF in ml/min) Since it is the plasma that is actually filtered in the kidney, it is important to know the amount of plasma that actually reaches the kidneys which is represented by this number. RPF = RBF(1 – HCT) Normal HCT =.45 (men) RPF = 1200 ml/min(1-.45) = 660 ml/min RPF ≈ 660ml/min (Normal value used in lab) 660 ml/min

6. Effective Renal Plasma Flow (ERPF in ml/min) As in the blood flow, only about 90% of the plasma that flows through the renal artery actually reaches the nephrons. This figure represents the estimated amount of plasma that reaches the glomerulus. ERPF = 0.9(RPF) ERPF = 0.9(660 ml/min) = 594 ml/min ERPF ≈ 594 ml/min (Normal value used in lab) 594 ml/min

7. Glomerular Filtration Rate (GFR in ml/min) This figure refers to the amount of plasma that is actually filtered into Bowman’s capsule. About 20% of the plasma that reaches the glomerulus is filtered but the lab manual figures this number by reducing the amount per day to amount per minute. About 180 L of plasma is filtered daily. GFR = 180 L/day =.125 L/min = 125 ml/min 1440 min/day GFR ≈ 125 ml/min (Normal value used in lab) 125 ml/min GFR Filtration fraction: GFR/RPF

8. Filtered Load (aka-”tubular load”) The total amount of any non- protein or non-protein bound substance filtered into bowman’s capsule. Each solute is independently calculated! Filtration Load or Tubular Load s = (GFR) x [ substance] plasma Example: glucose: 180 L/day x 1g/L Compare what is filtered to what is actually found in the urine…..tells you what is reabsorbed or secreted (think about this).

9. Excretion in mg/min This will determine the amount of a substance that is excreted in the urine in a specific time frame considering filtration, secretion, & reabsorption. (again, each substance/solute is independent) Excretion s = filtration s + secretion s - reabsorption s Be ready to rearrange the above equation to solve for other values! Excretion = (Urine Flow Rate)(Urine Concentration s ) This can also be used for excretion

10. Clearance Renal clearance of a substance is the volume of plasma completely cleared of a substance per min by the kidneys. Clearance is a general concept that describes the rate at which substances are removed (cleared) from the plasma— units ml/min.

11. Renal Clearance Value (RCV in ml/min) Renal clearance of a substance is the volume of plasma completely cleared of a substance per min. RCVs = U s x R P s RCV s x P s = U s x R Where: RCVs = clearance of substance S P s = plasma conc. of substance S U s = urine conc. of substance S R = urine flow rate

12. How long would it take to clear a substance that is only filtered? This is ERPF (594ml) arriving in the glomerulus with a substance dissolved in the plasma This is GFR (125) with some of the substance entering the glomerular capsule (filtered load) This is the 469ml of plasma that did not get filtered with the substance dissolved in it leaving the glomerulus to leave the kidney and recirculates through the body.

13. Production of Glomerular Ultrafiltrate Excretion rate = Filtration rate Secretion rate Reabsorption rate A substance that is neither reabsorbed nor secreted can be used to determine the amount of filtrate produced (GFR) We use inulin or creatinine Filtration Excretion +-

14. Inulin Small polysaccharide (energy storage in some plants) Not digested Not reabsorbed or secreted Amount of inulin excreted / min = amount of inulin filtered / min

15. Renal Clearance of Inulin: Calculating GFR 1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P i ) from Jung et al. (1992) Filtration load = GFR X P i (mg/min) (ml/min) (mg/ml)

16. 1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P i ) 3. Obtain urine sample and determine rate of inulin excretion Excretion per minute = R X U i (mg/min) (ml/min) (mg/ml) R = Urine flow rate U i = inulin concentration in urine Renal Clearance of Inulin: Calculating GFR

17. 1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P) 3. Obtain urine sample and determine rate of inulin excretion 4. Solve for GFR For inulin... Thus, Quantity filtered per minute (mg/min) = Quantity excreted per minute (mg/min) = GFR X P (ml/min) (mg/ml) R X U (ml/min) (mg/ml) GFR = R X U i = RCV i PiPi R = urine flow rate U i = inulin concentration in urine P i = inulin concentration in plasma Renal Clearance of Inulin: Calculating GFR

18. Example A doctor suspects her patient has kidney damage and has inulin infused into his vein. She then recovered 30 mg/ml of inulin is his blood plasma and 0.5 mg/ml of inulin in his urine. If the rate of urine formation was 2 ml/min, what is the GFR of the patient? GFR = R X U i PiPi GFR = 2 ml/min x 30 mg/ml = 120 ml/min 0.5 mg/ml Stages of Kidney Disease StageKidney damageGFR (ml/min) 1Little (e.g. some proteinuria)90 or above 2Mild60 to 89 3Moderate30 to 59 4Severe15 to 29 5Kidney failureLess than 15 Can we use creatinine? What are its advantages? Disadvantages? Yes!Endogenousslight over-estimate of GFR

19. Theoretically, if a substance is completely cleared from the plasma, its clearance rate would equal effective renal plasma flow. Use of Clearance to Estimate Renal Plasma Flow Cx = renal plasma flow Paraminohippuric acid (PAH) is freely filtered and secreted and is virtually completely cleared from the renal plasma Excretion rate = Filtration rate Secretion rate Reabsorption rate +-

20. Measuring Effective Renal Plasma Flow [RA] PAH x ERPF [RV] PAH x ERPF [U] PAH x R Amount of PAH entering the kidney = the amount leaving the kidney [RA] PAH x ERPF =( [RV] PAH x ERPF) + ([U] PAH x R) Solve for ERPF ERPF = [U] PAH x R [RA] PAH - [RV] PAH Because PAH is filtered and completely secreted, [RV] PAH is nearly zero! SO.........

21. Use of PAH Clearance to Estimate Renal Plasma Flow 1. amount enter kidney = ERPF x P PAH 3. ERPF x P pah = U PAH x R ERPF = U PAH x R P PAH ERPF = Clearance PAH 2. amount entered = amount excreted ~ Copyright © 2006 by Elsevier, Inc.

22. Clearances of Different Substances Clearance of inulin (C in ) = GFR if Cx < Cin: indicates reabsorption of x Clearance of PAH (C pah ) ~ effective renal plasma flow SubstanceClearance (ml/min inulin (used to estimate GFR)125 Creatinine (used to estimate GFR) 140 (slight secretion) PAH(used for ERPF/ RPF/ RBF)594 glucose 0 sodium 0.9 urea 70 Clearance creatinine(C creat ) ~ 140 (used to estimate GFR) if Cx > Cin: indicates secretion of x

23. Use the “long” excretion equation for calculation of tubular reabsorption Reabsorption = Filtration -Excretion Filt s = GFR x Ps (Ps = Plasma conc of s) Excret s = Us x R Us = Urine conc of s R = urine flow rate (when Excret s < Filt s)

24. Use the “long” excretion equation for calculation of tubular secretion (when Excret s > Filt s) Secretion = Excretion - Filtration Filt s = GFR x Ps Excret s = Us x R