Lecture 25: Glomerular Filtration Rate & Renal Clearance - Fall 2024

1502 Glomerular Filtration Rate & Renal Clearance Fall 2024 Lecture 25 Monday, Oct 21...

1502 Glomerular Filtration Rate & Renal Clearance Fall 2024 Lecture 25 Monday, Oct 21: 9am Dr. Layla Al-Nakkash [email protected] 1 © L. Al-Nakkash 2020  Learning Objectives 1. Describe the filtration barrier in the glomerulus. 2. Identify the Starling forces that drive glomerular filtration rate (GFR). 3. Describe how the Starling forces influence GFR. 4. Assess how GFR is calculated using inulin or creatinine. 5. Examine the relationship between GFR and RPF. 6. Describe renal clearance. 7. Examine the relationship between a given substance and the clearance of inulin (clearance ratio). 8. Examine the differences between filtered load and excretion rate. © L. Al-Nakkash 2024 2  Glomerular Filtration This is the first step in the formation of urine. RBF enters the glomerular capillaries- a portion of that blood is filtered into the Bowman’s space (the first part of the nephron). This filtered fluid is like interstitial fluid : called the ULTRAFILTRATE. – Ultrafiltrate contains water & all the small solutes in blood. – Ultrafiltrate does not contain proteins and blood cells. The forces responsible for glomerular filtration are like those in the capillaries These are the Starling Forces The physical characteristics of the glomerular filtration wall control what is filtered into the Bowman’s space 3 © L. Al-Nakkash 2024 1. Describe the filtration barrier in the glomerulus.  Glomerular Filtration 30,000 x mag Endothelium: pores ~70-100 nm diameter.Quite large – Solutes plasma proteins are all filtered across it. Basement membrane Epithelium: contains specialized cells-podocytes. – Attached to the basement membrane by foot processes. A barrier to filtration. – Between the foot processes are filtration slits 4 © L. Al-Nakkash 2024 1. Describe the filtration barrier in the glomerulus.  Glomerular Filtration 30,000 x mag In addition to the pore sizes, the glomerular barrier has negative charge. – Due to the presence of negative charged glycoproteins. present throughout the glomerular-capillary barrier Negative charges add an electrostatic component to the filtration process. – Positive solutes are attracted to the negative charged barrier → more readily filtered. – Negative solutes are repelled by the negative charged barrier → less readily filtered. Small solutes are freely filtered across the barrier. i.e., ions: Na, K, Cl, HCO3 Plasma proteins have net negative charge and are less readily filtered (due to both size and charge) Proteinuria: can result from glomerular diseases where the negative charge on the barrier is lost, resulting in increase filtration of proteins. 5 © L. Al-Nakkash 2024 1. Describe the filtration barrier in the glomerulus. © L. Al-Nakkash 2024 1. Describe the filtration barrier in the glomerulus. 6  Starling Forces across Glomerular Capillaries Starling forces/pressures drive filtration across the glomerular capillary wall. Assume the oncotic pressure of the Bowman’s space = 0. This is analogous to the interstitial space but here filtration of proteins is negligible GFR: Glomerular filtration rate Kf: Hydraulic conductance or filtration coefficient GFR = Kf [(PGC - PBS) - πGC] PGC: Hydrostatic pressure in glomerular capillary PBS: Hydrostatic pressure in Bowman’s space πGC: Oncotic pressure in glomerular capillary Kf filtration coefficient: hydraulic conductance/water permeability of the glomerular capillary wall. What contributes towards it? Water permeability/total surface area – A much greater volume of fluid is filtered here versus systemic capillaries PGC hydrostatic pressure in glomerular capillaries: A force that favors filtration. PGC remains high (45 mmHg) along the entire length of glomerular capillaries. How is that different to the systemic side? PGC decreases along the capillary length PBS hydrostatic pressure in Bowman’s space: A force that opposes filtration. This pressure (10 mmHg) comes from the fluid in the nephron lumen. πGC oncotic pressure in the glomerular capillaries: A force that opposes filtration. Determined by the [protein] in the glomerular capillary blood. It increases as fluid is filtered out of the capillary. 7 © L. Al-Nakkash 2024 2. Identify the Starling forces that drive glomerular filtration rate (GFR).  Starling Forces & Net filtration Pressure GFR = Kf [(PGC - PBS) - πGC] GFR: Glomerular filtration rate Kf: hydraulic conductance, or, filtration coefficient Bowman’s Glomerular capsule Capillary PBS Note: πGC – πBS PGC πGC But πBS =0 No proteins in the Bowman’s space PGC: hydrostatic pressure in glomerular capillary (driving force for GFR) PBS: Hydrostatic pressure in Bowman’s space (slows GFR) π GC: Oncotic pressure in glomerular capillary (due to unfiltered proteins in the capillary – can not cross the barrier, slows GFR) 8 8 © L. Al-Nakkash 2024 2. Identify the Starling forces that drive glomerular filtration rate (GFR).  Starling Forces & Net Filtration Pressure GFR = Kf [(PGC - PBS) - πGC] =NFP GFR: Glomerular filtration rate Kf: hydraulic conductance, or, filtration coefficient Sample problem: What is the net filtration pressure, NFP, given the following pressures? PGC = 45 mmHg PBS = 10 mmHg PBS π GC = 26 mmHg Kf = 14 ml/min/mmHg PGC [(PGC - PBS) - πGC] πGC NFP = (45 - 10) - 26 = 9 mmHg What is the GFR? GFR = Kf [(PGC - PBS) - πGC] i.e. Kf x NFP PGC: hydrostatic pressure in glom. cap. (driving force for GFR) PBS: Hydrostatic pressure in Bowman’s space (slows GFR) GFR = 14 x 9 = 126 ml/min π GC: Oncotic pressure in glom. Cap. (slows GFR) 9 9 © L. Al-Nakkash 2024 2. Identify the Starling forces that drive glomerular filtration rate (GFR).  Starling Forces across Glomerular Capillaries GFR = Kf [(PGC - PBS) - πGC] A: The pressures at the beginning of the glomerular capillary. Here blood has just come from the afferent arteriole and no filtration has yet occurred. The sum of the three Starling pressures (the ultrafiltration pressure) is +16 mmHg, favoring filtration (45 -10) -19 A + value favors filtration B: The pressures at the end of the glomerular capillary. Here blood has been filtered and is about to leave the glomerular capillary to enter the efferent arteriole. The sum of the three Starling pressures (the ultrafiltration pressure) is 0, no filtration occurs, at this point which is the filtration equilibrium. Arrow direction indicates filtration out of, or, Which occurs at the end of the glomerular capillary absorption into the capillary. Arrow size indicates magnitude of the pressure. + indicates the pressure favors filtration. - Indicates the pressure favors absorption. 10 © L. Al-Nakkash 2024 2. Identify the Starling forces that drive glomerular filtration rate (GFR).  Starling Forces across Glomerular Capillaries How does filtration equilibrium occur? Comparing A (the start of the glomerular capillary) and B (towards the end of the glomerular capillary) the only pressure that changed was πGC. The oncotic pressure of the glomerular capillary blood. As fluid filters out of the glomerular capillary blood, what is left behind? Proteins Thus, the [protein] and πGC INCREASE. At the end of the glomerular capillary, πGC increases such that net ultrafiltration Arrow direction indicates filtration out of, or, pressure will become 0. absorption into the capillary. Arrow size indicates magnitude of the pressure. + indicates the pressure favors filtration. - Indicates the pressure favors absorption. 11 © L. Al-Nakkash 2024 2. Identify the Starling forces that drive glomerular filtration rate (GFR).  Changes in Starling Pressure and GFR Changes in PGC (hydrostatic pressure in the glomerular capillary) are a result of changes in the ___________ resistance of the afferent and efferent arterioles Constriction of the afferent arteriole will  _____ afferent arteriole resistance. – This will result in decreased RPF, and decreased GFR. – As less blood flows into the glomerular capillary → PGC will also decrease → reducing the net ultrafiltration pressure. When might this occur? Activation of the SNS or high levels of PGC: Hydrostatic pressure in glomerular capillary Angiotensin II (seen during hemorrhage) 12 © L. Al-Nakkash 2024 3. Describe how the Starling forces influence GFR.  Changes in Starling Pressure and GFR Changes in PGC (hydrostatic pressure in the glomerular capillary) are a result of changes in the ___________ resistance of the afferent and efferent arterioles Constriction of the efferent arteriole will _____  efferent arteriole resistance. – This will result in decreased RPF, and increased GFR. – As less blood flows out of the glomerular capillary → PGC will increase → increasing the net ultrafiltration pressure. When might this occur? Low levels of angiotensin II 13 © L. Al-Nakkash 2024 3. Describe how the Starling forces influence GFR.  Measurement of GFR A marker for nephron function Measured using the clearance of a glomerular marker. Something that: – is freely filtered across the glomerular capillaries. – is not reabsorbed or secreted by the renal tubule. – does not alter the GFR Inulin fits these criteria. It is very large, not charged, doesn’t bind to plasma proteins, It is freely filtered across the glomerular capillary wall, It is neither reabsorbed nor secreted by renal tubular cells. The amount of inulin filtered across the glomerular capillaries = the amount excreted in urine. Inulin must be infused (IV). It is not found endogenously Clearance of inulin = GFR. = excretion rate of inulin or creatinine RPF: Renal plasma flow GFR = _[U]inulin /Cr x V_ = Cinulin/Cr [U]inulin/Cr: [inulin or creatinine] in urine V = Urine flow rate [P]inulin/Cr: [inulin or creatinine] in plasma [P]inulin/Cr Cinulin/Cr: clearance of inulin or creatinine CREATININE is used clinically- even though it is a product of muscle metabolism, it is filtered, not reabsorbed/secreted. 14 © L. Al-Nakkash 2024 4. Assess how GRF is calculated using inulin or creatinine.  Measurement of GFR using inulin Sample problem: A patient is infused with inulin to measure GFR. The patient’s urine flow is varied by drinking large volumes of water. [P]inulin is maintained constant at 1 mg/ml via the infusion. Before drinking water After drinking water [U] inulin = 100 mg/ml [U] inulin = 20 mg/ml V = 1 ml/min V = 5 ml/min What is the effect of the increase in urine flow on the patient’s GFR? _[U]inulin x V_ = Cinulin GFR = [P]inulin GFR before _100 mg/ml x 1 ml/min_ GFR after _20 mg/ml x 5 ml/min_ drinking = 1 mg/ml drinking = 1 mg/ml = 100 ml/min = 100 ml/min ANSWER: GFR remained constant, even though urine flow changed. While V increased 5-fold, [U]inulin decreased the same amount. 15 © L. Al-Nakkash 2024 4. Assess how GRF is calculated using inulin or creatinine.  Measurement of Filtration Fraction This is the relationship between GFR and RPF. Filtration fraction is the fraction of RPF filtered across the glomerular capillaries. Filtration fraction = _GFR_ RBF: Renal plasma flow GRF: glomerular filtration rate RPF The normal filtration fraction is ~ 20% (0.2). 20% of the RPF is filtered and 80% is not filtered. 80% that is not filtered leaves the glomerular capillaries via the efferent arterioles and becomes the peritubular capillary blood. 16 © L. Al-Nakkash 2024 5. Examine the relationship between GFR and RPF.  Renal Clearance The rate at which a substance is removed/cleared from plasma. Renal clearance is the volume of plasma completely cleared of a substance by the kidneys per unit time. = urinary excretion or excretion rate _[U]x_x V C = clearance C= [U]x: urine concentration of substance x [P]x V= urine flow [P]x: plasma concentration of substance x Renal clearance is the ratio of urinary excretion/ plasma concentration. The higher the renal clearance, the more plasma that is completely cleared of that substance. A substance with a high renal clearance could be removed with a single pass through the kidneys (organic acids and bases: PAH, morphine), whereas a substance with a lower renal clearance may not even be removed at all (albumin). 17 © L. Al-Nakkash 2024 6. Describe renal clearance.  Measurement of Clearance Ratio for Na+ Sample problem: A patient is infused with inulin. In a 24-hour period 1.44 L of urine is collected. The patient’s urine [inulin] is 150 mg/ml and urine [Na+] is 200 mEq/L. The patient's plasma [inulin] is 1 mg/ml and plasma [Na+] is 140 mEq/L. What is the clearance ratio for Na+ and how much was reabsorbed? Urine volume, V = 1.44 L / 24 hr. = 1440 ml/1440 min = 1.0 ml/min Needs to be in ml/min CNa = [U]Na x V = 200 mEq/L x 1 ml/min =1.43 ml/min [P]Na 140 mEq/L Cinulin = [U]inulin x V = 150 mEq/L x 1 ml/min = 150 ml/min [P]inulin 1 mEq/L C Na = _1.43 ml/min_ = 0.01 or 1% C inulin 150 ml/min ANSWER: Sodium is freely filtered across glomerular capillaries. A clearance ratio of 0.01 means that 1% of sodium filtered, is excreted. 99% Therefore, how much was reabsorbed ? _____. 18 © L. Al-Nakkash 2024 7. Examine the relationship between a given substance and the clearance of inulin (clearance ratio).  Filtered Load & Excretion Rate Filtered Load = GFR x [P]x Excretion Rate = [U]x x V Filtered load: is the amount of substance filtered into the Bowman’s space per unit time. This fluid in the Bowman’s space and in the nephron lumen is ‘tubular fluid' Luminal fluid Excretion rate: is the amount of substance excreted per unit time. This is the sum of filtration, reabsorption and secretion. Reabsorption: many substances are reabsorbed ( ions, glucose) into the peritubular capillary – required, otherwise they are lost in urine Secretion: less substances are secreted from the peritubular capillary (K, organic acids/bases). Allows excretion of these substances 19 © L. Al-Nakkash 2024 8. Examine the differences between filtered load and excretion rate.  Filtered Load & Excretion Rate = Transport Rate for a substance Reabsorption or secretion rate = Filtered Load – Excretion rate Comparing FL with ER provides information as to whether a substance has been reabsorbed or secreted If Filtered load > Excretion rate : – what do you expect has happened to the substance? Net reabsorption If Filtered load < Excretion rate : – what do you expect has happened to the substance? Net secretion 20 © L. Al-Nakkash 2024 8. Examine the differences between filtered load and excretion rate.  Filtered Load & Excretion Rate Reabsorption or secretion rate = Filtered Load – Excretion rate Sample problem: A patient has a GFR of 180 L/day and the plasma [Na+] is 140 mEq/L. The patients voids 1L of urine in 24 hours and the urine [Na+] is 100 mEq/L. What is the filtered load and excretion rate for sodium? Is this patient undergoing net sodium secretion or reabsorption? Filtered Load = GFR x [P]x = 180 L/day x 140 mEq/L =25,200 mEq/day Filtered Load – Excretion rate Excretion Rate = V x [U]x = 25,200 - 100 mEq/day = 1 L/day x 100 mEq/L =25,100 mEq/day =100 mEq/day FL > ER = Reabsorption 21 © L. Al-Nakkash 2024 8. Examine the differences between filtered load and excretion rate.  Glomerular filtration is the last step in urine formation True / False Starling Forces modify GFR True / False GFR is indirectly proportional to NFP True / False Unfiltered proteins will ↑ the oncotic P of glomerular capillary blood True / False Filtration fraction = RPF / GFR True / False Inulin is a glomerular marker True / False If Filtered load < excretion rate, we expect a substance is reabsorbed True / False © L. Al-Nakkash 2024 22  Glomerular filtration is the last step in urine formation True / False First step Starling Forces modify GFR True / False GFR = Kf [(PGC - PBS) - πGC] GFR is indirectly proportional to NFP PBS) - πGC] True / False GFR = f GC K [(P - NFP= [(PGC - PBS) - πGC] Unfiltered proteins will ↑ the oncotic P of glomerular capillary blood True / False ↑ πGC Filtration fraction = RPF / GFR True / False =GFR/RPF Inulin is a glomerular marker True / False As is creatinine If Filtered load < excretion rate, we expect a substance is reabsorbed True / False If FL