.. ‫مرحبا‬ Last Lecture we where talking about How to measure Renal Plasma Flow (aka RPF) , Recall that we used a substrance that is 100 % CLEARED from the plasma ONCE it enter the kidney . Today’s lecture will talk about CLEARANCE and its Very important Topic and its slightly complicated So Brace your self and Lets Start ,, BUT FIRST LET ME TAKE A SELFIE :P IF you know that 650 ml of plasma is entering the kidney in 1 minute then you know RPF = 650 ml/min . And your given a lab result that says the concentration of substance X in plasma is 1 mg/ml . THEN you can conclude that there is 650 mg of (X ) entering the kidney each minute ! Now if (X) is removed COMPLETELY via the kidney to the URINE and NOTHING is reabsorbed Back to the renal vein , I wont surprise you if I told you that you have cleared 650 ml of plasma . In other words : CLEARANCE = volume of plasma which provide (X) for excretion per minute . So the Unit of clearance is volume / time = (ml/min) The assumption that we did is Substance X is removed Totally ONCE it Enter the kidney . Okay ‫??? شى استفدت‬ We can use the clearance to correlate it with the Renal Plasma Flow which will correlate with Glomerular Filtrate Rate which give us an image about the Kidney Health Status . ( The Concept of Glomerular marker ) we usually use PAH - a substance called para-amino-hippouric acid – to measure Clearance because it has the following characteristics :  Low Molecular Wieght  can be easily FILTERED .  Not Reabsorbed BUT ITS SECRETED ! Since PAH is 100 % cleared from The Blood then CL of PAH = RPF So we are expecting Clearance of PAH to be 650 ml/min If I told you that CL of PAH in reality is 585 ml/min ! NOT 650 , you will ask Okay why ?? Because 10 % of The blood going to kidney will NOT reach the glomerulus , it NOURCHES the kidney Parenchyma . SO in real life There is NO way to remove a substance 100 % from the renal artery . So what do we call the 585 ?? we call it EFFECTIVE RPF (eRPF = 90 % * True RPF) ** So the CL of PAH is equal to eRPF ** Example : lets assume that a blood sample contain concentration of 1 mg/ml of PAH , how much of PAH is filtered in one minute ?? Answer : From last lecture you MUST be knowing that 20 % of RPF is Filtered through the Glomerulus= GFR So GFR = 20 % * 650 = 125 ml/min Time = 1 minure , So PAH filtered = GFR * Time = 125 mg . Tayeb we said that 100 % cleared ! the rest - the 525 mg - are SECRETED . What is the Filter Load ? it’s the amount of X filtered per minute ( in mg ) * DO you know that FILTRATION is a passive process  so No Energy  and No carrier Okay then we can conclude that filtration has a linear relation as the graph below tells ,, Wait a moment and the picture will complete in your mind !  DO you know that SECRETION is an active process  so it consume Energy  and their is a carrier . And as any other active process it can be saturated (when all carriers are fully working) this maximum working load is called T.max , Take PAH as an example , If the concentration lies below T.max it will be 100 cleared as usual , But If the concentration of PAH is greater than the T.max , then some PAH will leave the kidney via the renal vein . so its NOT completely Cleared at high concentrations . Given that the T.max of PAH = 80 mg/min Ex. If the blood reaching the kidney in 1 minute contain 100 mg PAH , calculate: i) the Filtered load . ii) the secreted amount answer : i) 20 % of 100 mg is 20 mg /min ii) Tmax = 80 mg/min , and the residual quantity in the peritubular vein is within the Tmax So the Secreted amount is 80 mg And Nothing goes to RENAL VEIN . But if the blood reaching the kidney in 1 minute contain 1000 mg PAH , calculate: i) the Filtered load . ii) the secreted amount answer : i) 20 % of 1000 mg is 200 mg /min see the graph ii) Tmax = 80 mg/min , the residual quantity in the peritubular vein is much greater than Tmax . so also the secreted amount will be 80 mg/min And 720 mg will go to the RENAL VEIN . (1000-200-80 ) . So if we Repeat Ex. For 10 000 mg/min the secreted 80 mg/min will be negligible ! So we will conclude that at very high concentration PAH clearance will Decrease !! ‫اتعقدت ؟؟؟ انا اكتر وهللا الدكتىر عجق الدنيا ببعضها‬ Dr says that last lecture he gave us this formula Clearance = ( conc. In urine * volume of urine ) / conc. In plasma Lets repeat But with inulin , inulin has MW=5000 ( small ) , its Not reabsorbed nor secreted . -Any thing below 70 000 ( 70 thousand ) is considered small and can be filtered easily – Tayeb lets Start : As you know RPF = 650 mg/min , 20 % filtered = GFR = 125 mg/min Since inulin is only filterable Not reabsorbed or secreted ONLY glomerular filtration provide excretion of inulin to urine Example : if Concentration of inulin in Plasma is 1 mg/ml how much is the Filter Load ? 125 mg of inulin is filtered per minute If we Double The Concentration of inulin in plasma what happens to the filter load ?? DOUBLES Tayeb what happens to the clearance ?? Nothing . Prove : In the last Example conc. = 1 mg/ml , Filter load = 125 mg in 1 minute Double conc.  2mg/ml , the amount filtered is 250 mg Recall Filter load = GFR * Conc. And clearance = (conc. In urine * volume of urine ) / conc. In plasma Measure CL for both to make our work easier take urine Volume as a constant ( same quantity ) in both so it DOES NOT have an effect when we compare them . Clearance 1 = 125/1  125 Clearance 2 = 250/2  125 So clearance is constant for inulin , because of the relation between filter load and plasma concentration is linear ( the graph last page ) Inulin is considered a good Exogenous glomerular marker . What is a glomerular marker : it’s a substance used to measure GFR , has low MW , Not reabsorped and Not Secreted So it is only filtered , ( filtered  G filtrate R ) . But the problem with inulin is that it may cause headache when given to patients so we are trying to Find an endogenous Glomerular Marker !! Creatine is produced Daily From break down of proteins in Muscles in quantities between 1.5-2.0 grams and its absorbed by the systematic circulation and blood conduct it to the kidneys where creatine is execreted at 1.5-2.0 grams Daily , so the Concentration of creatine in the blood is Constant . Creatine has a low MW (114 ) ,, Not reabsorebed ,, slightly Secreted (10 % ) So ***Creatine Can be USED as an ENDOGENOUS Glomerular Marker *** Now as we said it slightly secreted  10 % of Creatine is Secreted Actively So Does this mean that we are overestimating clearance ( which will be used to estimate GFR )?? NO ! Why ?? Because when we measure concentration in plasma we are measuring 100 % of the Creatine which has 2 forms ,, 90 % free  filtralable . 10 % Bound to proteins  Not filtered So total effect Both 10 % cancel out and The clearance in NOT overestimation ;) So Creatine IS A GOOD ESTIMATE For GFR . So to measure clearance of creatine : According to the Equation : i) take blood sample and measure conc. Of creatine in plasma ( btw its conc. Is constant and its NOT related to food intake or exercise ) ii) take the urine sample and measure conc. Of creatine in Urine . taking a 24 hr urine sample give a better value than 1 urine sample ( more accurate ) iii) urine volume ( Usually taken over a 24 hr period ) Disadvantage of taking a urine sample over 24 hr is we cannot easily take it from a child or an old patient especially with dementia ! IF we measure the above 3 variables we can calculate True GFR But to escape the 24 hr urine sample collection we calculate something called Estimated GFR (eGFR) , using equations , their are many equations one for adults , other for children , one for end stage renal failure and so many , we are concerned only with 1 equation in the Exam : Where age in years , ideal body mass = patient hight in cm – 100 ( your hight 170 cm  you IBM = 70 kg ) . Why ideal body mass Not patient Body mass ??? because ideal body mass is more accurate indicator for muscle mass Not Fat ! because creatine is directly proportional to creatine . Normally creatine plasma concentration : In males range : 0.7 – 1.2 mg/dl (NOTE : decilitre ) In Female range : 0.62 – 1.1 mg/dl . American Unit is mg/dl Canada and Europe use micromole unit The conversion factor between the two is 88.4 So a patient of Creatine conc. = 1 mg/dl = 88.4 micromole You Must KNOW BOTH units  NOTICE that a body builder with plasma conc of 1.4 mg/dl is Normal . But an old lady with plasma conc of 1.4 mg/dl is Not Normal . The eGFR equation For Females is the same for males but multiplied by .85 ( because Females have lower muscle mass than males ) eGFR is more than 95 % close to the True GFR , but sometimes it Does NOT ! for Example in end stage renal failure patient T GFR maight me 3 ml/min , but e GFR calculated to be 15 ml/min due to secretion Not Filtration , so here e GFR is an over estimation of T GFR . ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Lets assume there is a patient having an auto immune disease and you gave him voltarin , - and as a medical doctor you know that voltarin is Nephrotoxic , it inhibit Prostaglandins formation which act as a local vasodilators on afferent arteriole , so blockage of PGs synthesis means afferent vasoconstriction which will results in diminishing GFR , and (affect the kidney FUNCTION ) . So You are dealing with 2 problems 1st the auto immune disease . 2nd the risk of nephrotoxcity So You have to screen that GFR every time the patient comes to the follow ups , you DO SO BY taking a blood sample and measure plasma conc. Of creatine and calculate eGFR at the time you started the treatment . Then reDO the test after six weeks , 3 months , 6 months and then yearly . For Example : the Plasma conc. Of creatine at begging of treatment was 0.5 mg/dl Then after six weeks when yo did the second test it was 1.0 mg/dl What Does this means ?? notice that 1.0 mg/dl is within the normal range ! In this patient the raise in creatine conc. ( Double ) means that the GFR has been reduced to (HALF) , so half of the nephrons in this patient had Stopped Functioning ! although its within the Normal Range , So we conclude That absolute reading is NOT enough to GRADE ranal function . Grading Renal Functions : Renal Function decrese normally with age so that 1 % of GFR is decresed per year after the age of 30 , so a patient of 80 years his normal GFR is about 60 ml/min (50% of Normal GFR (125)) . Kidney Dysfunction is Divided into 4 Stages (Normal ) IF GRF of patient is 100% of Normal GFR then you are dealing with a normal Kidney . ( Stage 1 ) IF GRF of patient is between 50 - 100% of Normal GFR then you are dealing with a decresed renal reserve . in this stage Urea and creatine are usually within normal range and patient can maintain homeostasis , usually patient is asymptomatic . (stage 2 ) IF GRF of patient is between 20 - 50 % of Normal GFR then you are dealing with a renal insufficiency . (stage 3 ) IF GRF of patient is between 5 - 20 % of Normal GFR then you are dealing with a renal Failure . (stage 4 ) IF GRF of patient is between 0 - 5 % of Normal GFR then you are dealing with an end stage renal failure and you have to put patient in dialysis and start searching for a kidney Donor . within 1 week the patient may die if left untreated . Stage Name Normal Stage 1 % of Normal GFR 100 50-100 Clinically Normal Kidney Decresed renal reserve Stage 2 Stage 3 Stage 4 20-50 05-20 00-05 Renal insufficiency Renal failure End stage Renal failure Notes: Urea and creatine Normal . maintain homeostasis . Asymptomatic . Dialysis / Kidney transplant Home massage of this lecture is that we have 2 important uses for GFR i) to screen renal functions in patients taking NSAIDS ( Voltarin ) ii) classify ( GRADE ) kidney Dysfunction . Done ………………………………………….. The Exam By : In Bakir Luck Jaber Good 