Renal Physiology April, 2011 J. Mohan, PhD. Lecturer, Physiology Unit, Faculty of Medical Sciences, U.W.I., St Augustine. Office : Room 105, Physiology Unit. References: Koeppen B.E. & Stanton B.A. (2010). Berne & Levy Physiology. 6th Edition. Mosby, Elsevier. Costanzo L.S. (2006) Physiology. 3rd Edition, Elsevier, Saunders. Marieb, E. & Hoehn, K. (2010). Human Anatomy & Physiology. 8th Edition, Pearson, Benjamin Cummings. Stanfield, C.L. & Germann W.J. (2008). Principles of Human Physiology. 3rd Edition, Pearson, Benjamin Cummings. Hall, J.E. (2011). Guyton and Hall Textbook of Medical Physiology. 12th Edition, Elsevier, Saunders. April 18, 2011 1
Physiology Objectives 1. Explain, with the use of equations, the principles of clearance, renal plasma flow, renal blood flow and urinary excretion. 2. Define renal plasma flow (RPF) and renal blood flow (RBF). 3. Justify the use of creatinine clearance over that of inulin in the estimation of GFR. 4. Explain the relationship between clearance and excretion for inulin, glucose, urea and penicillin. 5. Define tubular transport maximum (Tm), renal threshold and renal splay. 6. Explain, with the use of graphs, how renal glucose-handling varies with plasma glucose concentration. Today s Topics Sample calculation of Urinary Excretion. Sample calculations of. Some calculations of clearance of various substances. Measurement of Renal Plasma Flow and Renal Blood Flow. Sample calculation of Renal Plasma Flow and Renal Blood Flow. Renal handling of glucose. April 18, 2011 2
Excretion of a Substance Excretion by the kidneys = elimination of solute and water from the body in the form of urine For any substance, the amount of a substance that is excreted over a period of time is determined by this simple rule: the amount of substance that enters the renal tubule is excreted unless it is reabsorbed What determines entry into the renal tubule? Filtration or Secretion Therefore : amount of substance excreted (E) = amount filtered (F) + amount secreted (S) amount reabsorbed (R) i.e. E = F + S - R Excretion of a Substance Example: Amount of substance excreted (E) = amount filtered (F) + amount secreted (S) amount reabsorbed (R) i.e. E = F + S - R If the GFR= 150 ml/min, the plasma concentration of a solute is 0.08 mmol/ml. Assume that the solute is freely filterable is secreted at a rate of 3 mmol/min and reabsorbed at a rate of 6 mmol/min. What is the excretion rate of the solute? F = GFR x [Plasma] of solute = 150 ml/min x 0.08 mmol/ml = 12 mmol/min April 18, 2011 3
Excretion of a Substance Example: Amount of substance excreted (E) = amount filtered (F) + amount secreted (S) amount reabsorbed (R) i.e. E = F + S - R F = 12 mmol/min S = 3 mmol/min R = 6 mmol/min Therefore, solving for E = 12 + 3-6 = 9 mmol/min Recall : Principles of clearance from lecture 1 and how to calculate clearance of a substance x as given in the example That clearance of creatinine and inulin can be used to determine GFR Figure 32.12; Koeppen & Stanton, 2010 April 18, 2011 4
C x = U x X V P x ml/min represents a volume of plasma from which all of substance x has been removed and excreted into urine per unit time i.e. rate of removal of substance x from plasma by kidneys Note : ratio of the amount of x excreted in urine to the amount of x in plasma C x = U x X V Pa x ml/min E.g. if Ux = 100 mg/ml; V= 1 ml/min Then excretion rate of x = 100 mg/ml x 1 ml/min = 100 mg/min If Px = 1 mg.ml, then Cx = 100 mg/min 1 mg/ml = 100 ml/min April 18, 2011 5
Example 2 : A urine sample revealed that 450 ml of urine was collected in 1 hour and this had a concentration of Na+ of 15 mmol/l. The plasma concentration of Na+ was 145 mmol/l. What was the Na+ clearance? Express V in ml/min : 450 ml/60 mins = 7.5 ml/min U Na+ = 15 mmol/l; V= 7.5 ml/min; P Na+ = 145 mmol/l Clearance of Na+ = 15 mmol/l x 7.5 ml/min 145 mmol/l = 0.78 ml/min Example 2 : In a 24 hour period, 1.44.L of urine is collected from a man receiving an infusion of inulin (the fructose polymer used to estimate GFR). In his urine the [inulin] is 150 mg/ml and in his plasma the [inulin] in 1 mg/ml. What is his inulin clearance? Express V in ml/min : 1.44 L /24 hours = 1440 ml / 1440 min = 1.0 ml /min U inulin = 150 mg/ml; V= 1.0 ml/min; P inulin = 1 mg/ml Clearance of inulin = 150 mg/ml x 1.0 ml/min 1 mg/ml = 150 ml/min April 18, 2011 6
From previous examples, it can be seen that clearance can be calculated for any substance Clearance an vary from zero to ~ 600 ml/min depending on the renal handling of the substance A substance that is not normally filtered across the glomerular capillaries will have a clearance of approx. zero e.g. albumin Renal clearance of glucose = 0; glucose is filtered, unlike albumin, but is completely reabsorbed back into blood Substances that are filtered but partially reabsorbed e.g. Na+, urea, phosphate, Cl- have a clearance higher than zero Recall that the clearance of inulin, the fructose polymer, is used to estimate GFR Like creatinine, (the clearance of which is also used to estimate GFR), inulin is freely filtered across the gloerular capillaries, but is neither reabsorbed nor secreted, therefore its clearance measures GFR Organic acids e.g. para - aminohippuric acid (PAH) is both filtered & secreted, therefore has the highest clearance values Drugs such as penicillin is also secreted, therefore has a high clearance April 18, 2011 7
why is the use of creatinine preferred over that of inulin for the estimation of GFR? inulin is freely filtered across the glomerular capillaries, but is neither reabsorbed nor secreted, therefore its clearance is the most accurate measure of GFR available however, the use of inulin clearance is limited in the clinical situation because of several factors: 1. an intravenous injection followed by constant infusion is required 2. complete emptying of the bladder is necessary before the beginning of the clearance period, in order to remove all urine not containing inulin, and again at the end of the period in order to obtain all the urine produced during this period 3. the urine flow must be high so that enough urine may be obtained in a short period of time to permit analysis and to reduce possible errors introduced by urine remaining in the bladder at the beginning and end of the clearance period these requirements can be difficult to meet in a patient with compromised renal function April 18, 2011 8
the use of creatinine clearance overcomes some of these practical problems, but creatinine is secreted into the PT, and certain drugs e.g. cimetidine (organic cations) can inhibit tubular secretion of creatinine, thereby reducing creatinine clearance (but not actual GFR) Otherwise : 1. Creatinine is an end product of protein metabolism and is always present in the blood; its concentration remains fairly constant over a 24 hour period 2. therefore there is no need for an intravenous infusion, and a clearance period can extend for as long as 24 hours so that adequate amounts of urine can be collected and the frequency of bladder emptying is minimised 3. only 1 blood sample is needed and it can be taken during any point in the collection period 4. plasma creatinine alone can be used to follow changes in GFR in a patient with chronic renal disease April 18, 2011 9
Measurement of Renal Plasma Flow and Renal Blood Flow Recall from Lecture 1 the definition and regulation of Renal Blood Flow Recall that red blood cells are not filtered, so the flow of plasma that enters the kidney is Renal Plasma Flow Renal Plasma Flow (RPF) can be estimated from the clearance of para - aminohippuric acid (PAH) Renal Blood Flow can then be calculated from RPF and the haematocrit Measurement of Renal Plasma Flow and Renal Blood Flow Measuring True Renal Plasma Flow (RPF) by applying the Fick Principle to the kidney : i.e. the amount of substance entering the kidney via the renal artery = amount of substance leaving the kidney via the renal vein + the amount excreted in the urine For PAH : Amt. of PAH entering kidney= Amt. of PAH leaving kidney Amt. of PAH entering kidney = [RA] PAH x RPF Amt. of PAH leaving kidney = [RV] PAH x RPF + [U] PAH x V April 18, 2011 10
Measurement of Renal Plasma Flow and Renal Blood Flow Amt. of PAH entering kidney= Amt. of PAH leaving kidney [RA] PAH x RPF = [RV] PAH x RPF + [U] PAH x V RPF = [U] PAH x V [RA] PAH - [RV] PAH where RPF = Renal Plasma Flow [U] PAH = [PAH] in urine V = urine flow rate [RA] = [PAH] in renal artery [RV] = [PAH] in renal vein Measurement of Renal Plasma Flow and Renal Blood Flow PAH is ideal for measuring RPF because : 1. PAH is neither used nor produced by the kidney 2. PAH does not alter RPF 3. The kidneys remove most of PAH from renal arterial blood by filtration & secretion leaving very little in renal vein so [RV] is nearly zero 4. No organ apart from the kidney removes PAH from blood so [PAH] in the renal artery = [PAH] in any peripheral vein, which is easier to sample from humans than the renal artery April 18, 2011 11
Measurement of Renal Plasma Flow and Renal Blood Flow The measurement of True Renal Plasma Flow involves the infusion of PAH into the body, taking urine samples and blood samples Certain simplifications can be made which estimate Effective Renal Plasma Flow which approximates true RPF to 10% 1. [RV] PAH = assumed to be 0 2. [RA] PAH = [PAH] in any peripheral vein Measurement of Renal Plasma Flow and Renal Blood Flow Therefore: Effective RPF = [U] PAH x V = Clearance of PAH [P] PAH where [U] PAH V [P] = [PAH] in urine = urine flow rate = [PAH] in plasma April 18, 2011 12
Measurement of Renal Plasma Flow and Renal Blood Flow Renal Blood Flow can then be calculated from RPF and the haematocrit : RBF = RPF 1 Hct Where : RBF = Renal Blood Flow (ml/min) RPF = Renal Plasma Flow (ml/min) Hct = haematocrit = fraction of blood volume occupied by red blood cells 1-Hct = fraction of blood volume occupied by plasma Measurement of Renal Plasma Flow and Renal Blood Flow Example : A man with a urine flow rate of 1 ml/min has a plasma concentration of PAH of 1 mg/dl and urine concentration of PAH of 600 mg/dl, and a haematocrit of 0.45. What is his RBF? Effective RPF = 600 mg/dl x 1 ml/min 1 mg/dl = Clearance of PAH = 600 ml/min RBF = 600 ml/min --------------- 1-0.45 = 1091 ml/min April 18, 2011 13
Today s Topics Sample calculation of Urinary Excretion. Sample calculations of. Some calculations of clearance of various substances. Measurement of Renal Plasma Flow and Renal Blood Flow. Sample calculation of Renal Plasma Flow and Renal Blood Flow. Renal handling of glucose. Renal handling of glucose Recall from Lectures 1& 2 that glucose is freely filtered across glomerular capillaries and reabsorbed in the 1 st half of PT Recall the reabsorption of glucose involves : active transport of Na+ at basolateral membrane via Na+/K+- ATPase Na+ enters PT cell at apical membrane with glucose (Na+ - glucose symport or Na+ - glucose cotransport) on the Na+ - glucose transporter (SGLUT) glucose leaves the cell at basolateral membrane via passive mechanisms (facilitated diffusion) on the glucose transporters GLUT 1 & GLUT 2 Because of limited number of glucose transporters, the mechanism is saturable i.e. it has a transport maximum T m April 18, 2011 14
Na+ Reabsorption in 1 st half of PT Figure 33.1, Koeppen & Stanton, 2010 Filtration, Reabsorption & Excretion of glucose true theoretical renal threshold splay Figure 19.16 ; Germann & Stanfield. April 18, 2011 15
Renal handling of glucose Graph shows the relationship between plasma glucose concentration and glucose reabsorption. For comparison, the filtered load of glucose and the excretion rate of glucose are also shown Known as glucose titration curve which is obtained experimentally by infusing glucose and measuring its plasma concentration and the rate of glucose reabsorption Filtration : glucose is freely filtered across glomerular capillaries; filtered load is calculated as GFR x plasma concentration of glucose [P glucose] As plasma concentration of glucose increases, filtered load increases linearly (see graph) Reabsorption : Renal handling of glucose Tm for glucose reabsorption = 375 mg/min = maximum concentration of glucose that can be transported per unit time from the tubular fluid back into the blood the amount of glucose transported is proportional to the amount present in tubular fluid (amount filtered) up to the Tm for glucose a higher concentrations, the Tm is saturated and there is no increase in the amount transported Normal plasma glucose level = 80-100 mg/dl when plasma conc = 100 mg/dl, assuming GFR = 125 ml/min then filtered load = GFR x plasma concentration of glucose = 1.25 dl/min x 100 mg/dl = 125 mg/min 125 mg/min, which is < Tm, so all glucose is reabsorbed & none excreted in urine April 18, 2011 16
Renal handling of glucose glucose level eventually the amount of glucose in the filtrate > the capacity for reabsorption & some glucose will be excreted in the urine theoretical renal threshold = 300 mg/dl = plasma concentration at which the amount of glucose in the filtrate exceeds the transport maximum and at which glucose appears in the urine (see graph) can be calculated as follows : GFR x renal threshold = transport maximum Renal threshold = transport maximum GFR = 375 mg/min ------------------- 1.25 dl /min = 300 mg /dl Renal handling of glucose however, true renal threshold = 160-180 mg/dl ; at this plasma concentration of glucose, the filtered load of glucose = 225 mg/min (see graph) which is < Tm, but glucose starts to appear in the urine at this plasma concentration - this creates a rounded curve in the graph called splay Splay is the portion of the graph where reabsorption is approaching saturation, but is not fully saturated; because of splay, glucose appears in the urine before reabsorption levels off at the Tm value April 18, 2011 17
Renal handling of glucose Explanations for splay : Low affinity of Na+ glucose transporter for glucose as the glucose concentration in the tubular fluid rises, if glucose molecules detach from their transporter protein SGLT, those glucose molecules will be excreted because of the few remaining free transporters to which they may re-attach Heterogeneity of nephrons Tm represents the Tm for the whole kidney all the nephrons in the kidney do not have the same Tm; some will reach Tm at lower plasma concentrations than others and glucose will be excreted in the urine in those nephrons before others Renal handling of glucose at normal plasma glucose concentrations, 80 100 mg/dl, all of filtered glucose is reabsorbed and none is excreted (see graph) in uncontrolled diabetes mellitus, the plasma glucose concentration > Tm and glucose in excreted in the urine during pregnancy, GFR is, filtered load of glucose so that it may > reabsorption of glucose congenital abnormalities of the Na+- glucose cotransporter Tm glucose excreted in urine April 18, 2011 18