Title MOD11-4-Tubular Function 1 (Renal 3).pdf ✅

PHYSIO ● PHYSIOLOGY Tubular Function 1 (Renal 3) TRANS 4 V MODULE 11 ictor Mendoza, MD, MSC, FPCP, FPCC, FPSP March 3, 2023 LECTURE OUTLINE I Tubular Function A. Functions of the Kidney B. Basic Processes of the Nephron 1. Filtration 2. Reabsorption/Resorption 3. Secretion C. Excretion D. Renal Clearance 1. Inulin Clearance 2. Glucose Clearance 3. Urea Clearance 4. Penicillin Clearance II. Renal Transport Processes in the Tubules A. Mechanisms III. Reabsorption A. Active Transport of Sodium B. Secondary Active Transport C. Saturation of Renal Transport 1. Transport Maximum 2. Effect of PG on Glucose Clearance LECTURE OBJECTIVES 1. Discuss the function of the tubules: a. Proximal Tubule b. Loop of Henle → Thin descending → Thin ascending → Thick ascending c. Distal Convoluted Tubule & Connecting tubule d. Collecting ducts → Cortical collecting duct → Medullary collecting duct → Papillary collecting duct I. TUBULAR FUNCTION A. FUNCTIONS OF THE KIDNEY ● Excretion of metabolic waste products and foreign chemicals ● Regulation of water and electrolyte balances ● Regulation of body fluid osmolality and electrolyte concentrations ● Regulation of arterial pressure ● Regulation of acid-base balance ● Secretion, metabolism, and excretion of hormones ● Gluconeogenesis B. BASIC PROCESSES OF THE NEPHRON (FOR URINE FORMATION) Figure 1. 3 processes in the body for urine formation. Dr. Victor Mendoza’s lecture, timestamp 00:58 1 FILTRATION See previous lecture / trans, discussed by Dr. Victor Mendoza 2 REABSORPTION /RESORPTION MOVEMENT of water and solutes From TUBULAR LUMEN (TL) to PERITUBULAR CAPILLARIES (PC) towards back to plasma From TL → PC Opposite of secretion 3 SECRETION TRANSFER of materials from PERITUBULAR CAPILLARIES towards LUMEN From PC → TL Figure 2. Urinary Excretion. Victor Mendoza Lecture Video Part 1 Group 1A, 2A, & 3A | Tubular Function 1 (Renal 3) 1
C. EXCRETION ● Also called “Urine Formation” Excretion = filtration – reabsorption + secretion 1. Excretion rate is dependent on a Filtration Rate b Whether the substance is reabsorbed, secreted, or both as it passes through the tubule D. RENAL CLEARANCE ● The rate at which a solute disappears from the body by excretion or metabolism ● If substance is cleared only by renal excretion: ○ Clearance is expressed as the volume of plasma passing through the kidneys that has been totally cleared of that solute in a given period time ○ ↑ renal clearance = ↑ plasma that is cleared of the substance (substance eliminated) from the body ○ Unit: ml/min a Inulin Clearance b Glucose Clearance c Urea Clearance d Penicillin Clearance 1. Inulin Clearance ● A plant product that is filtered but neither reabsorbed, secreted nor metabolized by the kidneys. ● Used to determine clearance rate, and therefore, nephron function ○ Suppose the plasma concentration of inulin is 4 units/100 mL, and the glomerular filtration rate (GFR) is 100mL/min. Figure 3. Inulin Clearance. Victor Mendoza Lecture Video Part 1 INULIN CLEARANCE a Inulin molecules enter the glomerulus through the afferent arteriole. b The plasma concentration of inulin is 4 mL/min, and it is filtered. Therefore, in 100 mL of filtrate, there are 4 units of inulin. It then passes through the tubules. c Upon passing through the tubules, none of the inulin is absorbed. The inulin molecules are 100% excreted. However, it is reabsorbed back towards peritubular capillaries. d Therefore, 100 mL/min of plasma is cleared of inulin as it passes through the kidneys. The inulin clearance is 100 mL/min. 2. Glucose Clearance Figure 4. Glucose Clearance. Victor Mendoza Lecture Video Part 1 GLUCOSE CLEARANCE a The hypothetical concentration of 4 units of Glucose per 100 mL of filtrate is the same concentration as the plasma. b The filtrate passes through the tubules. c All 4 units of glucose are reabsorbed together with the 100 mL of filtrate. Therefore, 100 mL of filtrate and 100% of glucose are reabsorbed (no glucose is excreted). UREA CLEARANCE a Again, we have 4 units of plasma hypothetically per 100 mL. Therefore, if urea is filtered the filtrate concentration is also 4 units per 100 mL. b As it passes through the renal tubules, 50% of urea is reabsorbed and 50% is excreted. This means 2 units per 100 mL of urea are excreted. Thus, the urea clearance is 50 mL/min Group 1A, 2A, & 3A | Tubular Function 1 (Renal 3) 2
Figure 5. Urea Clearance. Victor Mendoza Lecture Video Part 1 Figure 6. Urea Clearance. Victor Mendoza Lecture Video Part 1 Plasma Concentration = Filtrate Concentration ● If Plasma concentration is 4/100ml, then Filtrate concentration is also 4/100ml. ● If 50% of the substances reabsorbed as it passes through the tubules. Therefore, the concentration in the reabsorbed substance is 2 units per 100 mL ● If we divide 100mL that has been reabsorbed into 2 containers: ○ Container A: all substances that reabsorbed (2 units) ■ If we make it container A have a concentration of the substance similar to the plasma, every 2 units will have 50 mL ○ Container B: 50mL has no concentration of substance ■ Clearance is 50 mL/min 3. Penicillin Clearance Figure 7. Penicillin Clearance. Victor Mendoza Lecture Video Part 1 PENICILLIN CLEARANCE a The hypothetical concentration of 4 units of Penicillin per 100 mL of filtrate. b It is filtered and move toward the tubules c None of them is reabsorbed but some additional penicillin is secreted d The half of 100 mL is secreted towards the tubules, adding the amount of filtrate that has been absorbed without the substance + 2 units/100mL remain in the container. e Clearance: 100+50 = 150 ml/min 4. Equation for Renal Clearance ● Where ○ x = substance ○ Cx = clearance of substance ○ Ux = urine concentration of substance ○ Px = plasma concentration of substance ○ V = urine flow rate (ml/min) ● Amount of substance excreted = (filtered - reabsorbed + secreted) UxV = GFR x Px ± Tx ● The amount that is actually excreted (UxV) is the product of glomerular filtration rate (GFR) and the plasma concentration (Px) of the particular substance, to which you either add (if (if secreted) or subtract (if reabsorbed) Group 1A, 2A, & 3A | Tubular Function 1 (Renal 3) 3
COMPARISON OF CLEARANCE OF A SUBSTANCE WITH CLEARANCE OF INULIN Figure 8. Clearance of known substances. Victor Mendoza Lecture Video Part 1 If substance clearance is = inulin clearance; only filtered not reabsorbed nor secreted < inulin clearance; reabsorbed by nephron tubules > inulin clearance; secreted by nephron tubules Examples of some of the clearances of known substances handled by the kidneys 1 Glucose clearance is zero (0) and is less than the clearance of Inulin. Therefore, Glucose is reabsorbed by the tubules of the nephron. 2 Same with Urea, it is also less than the clearance of Inulin. Therefore, Urea is also reabsorbed mostly by the tubules of the nephron. 3 Para-aminohippuric Acid (PAH) has a clearance that exceeds the clearance of Inulin by 4-5 times. Therefore, PAH is secreted mostly by the tubules. We can therefore make conclusions as to how the kidney handle different substances based on their: ● Filtration rate ● Excretion ● Clearance Table 1. Renal Handling of Solutes For any molecule X that is freely filtered at the glomerulus: Renal handling of X is: Filtration > excretion Net reabsorption of X Excretion > filtration Net secretion of X Filtration = excretion No net reabsorption or secretion Clearance of X < inulin clearance Net reabsorption of X Clearance of X = inulin clearance X is neither reabsorbed nor secreted Clearance of X > inulin clearance Net secretion of X 5. Inulin Clearance Substance is freely filtered and not reabsorbed or secreted, the rate of secretion in urine is equal to rate of substance filtered by kidney. Thus, ● GFR is nothing more than the product of the urinary secretion of that substance multiplied by the volume of urine divided by the plasma concentration of the substance. This is actually equal to the clearance of that particular substance. ● Inulin is a polysaccharide molecule with a molecular weight (MW) of 5200, fits these criteria. Inulin clearance (Cin ) is equal to the GFR which is around 125mL/min. INULIN CLEARANCE GFR = Cin = 125 ml/min 6. Creatinine Clearance CREATININE CLEARANCE CCr = UCr x V PCr ● CCr - Creatinine Clearance ● UCr - Urine Concentration of Creatinine ● PCr - Plasma Concentration of Creatinine ● V - Urine flow rate (ml/min) ● Creatinine clearance overestimates GFR by 10-20% due to creatinine secretion in very small amounts. ○ Because there is some amount of creatinine secretion that occurs in the tubules of the nephron ● Blood urea and creatinine will not be raised above normal range until 60% of total kidney function is lost. ● Hence, the more accurate GFR or creatinine clearance is measured whenever renal disease is suspected or careful dosing of nephrotic drug is required. Application Of Renal Clearance In Kidney Function ● Inulin clearance can be used to estimate GFR (by calculation) ● PAH clearance can be used to estimate renal plasma flow ● Measurement of filtration fraction = GFR/RPF ● Measurement of creatinine clearance Group 1A, 2A, & 3A | Tubular Function 1 (Renal 3) 4
II. RENAL TRANSPORT PROCESSES IN THE TUBULES ● Mainly: ○ Reabsorption ○ Secretion A. MECHANISMS ● Both mechanisms are identical ● Can be active or passive 1. DIFFUSION SIMPLE DIFFUSION FACILITATED DIFFUSION Movement of a substance from a higher to a lower concentration gradient. DOES NOT REQUIRE ENERGY. Similar to simple diffusion that moves a substance down its concentration gradient. SPECIFIC CHANNEL PROTEINS are required for movement Figure 9. Renal Transport Processes: Mechanisms Victor Mendoza Lecture Video Part 1 2. ACTIVE TRANSPORT ● Utilizes energy to move a substance across a membrane from a low to high concentration Figure 10. Renal Transport Processes: Mechanisms Victor Mendoza Lecture Video Part 1 3. SYMPORT ● Move two or more substances in the same direction at the same time Figure 11. Renal Transport Processes: Mechanisms Victor Mendoza Lecture Video Part 1 4. ANTIPORT ● Move two or more substances in opposite directions access the cell membrane Figure 12. Renal Transport Processes: Mechanisms Victor Mendoza Lecture Video Part 1 NOTE: Both symport and antiport mechanisms may utilize concentration gradients maintained by ATP pump – SECONDARY ACTIVE TRANSPORT. 5. UNIPORTER ● With a carrier protein that moves the substances in only one direction Figure 13. Renal Transport Processes: Mechanisms Victor Mendoza Lecture Video Part 1 Group 1A, 2A, & 3A | Tubular Function 1 (Renal 3) 5