Renal function tests Urine analysis Combined urine blood
Renal function tests Urine analysis
Combined urine & blood tests �Include: renal plasma clearance & TM (tubular maximin) �Importance: become abnormal in early phases of renal desease , before blood composition is changed.
Clearance �Definition: clearance of any substance is the volume of plasma cleared from this substance in one min. �Renal tubular handling of different substances: � 1 - reabsorbtion : partial or complete. 2 - secretion: partial or complete 3 - niether secreted or reabsorbed. 4 - formed in the renal tubules.
Calculaion of clearance: If substance X neither secreted or reabsorbed: 1. The amount of X cleared from plasma/min = the amount of X excreted in urine /min. (1) 2. The amount of X cleared from plasma/min = the volume of plasma cleared from X in min (C) X the conc. Of X in plasma (p) 3. = the amount of X excreted in urine /min = the volume of urine /min (V) X the concentration of X in urine (U).
as (1) so CXP=UXV so C= U X V/ P
Uses of clearance 1 - measurement of GFR �This is done by use of inulin clearance. �Inuline is a polymer of fructose (non toxic , not metabolized in liver , carried on PP or bound to RBCs) �Inulin is freely filtered neither secreted nor reapsorbed. �the amount of inulin filterd in 1 min = the amount present in urine in 1 min. �Inulin clearance = GFR = 125 ml/min
2 -measurement of RBF Fick’s principle states that tissue uptake or release of substance equals the blood flow F(volume/time) multiplied by the difference between arterial (CA) and venous (CV) concentrations of the substance: =F×(CA−CV)
� Substace used PAHA or diodrast? ? ? � Why? When those substances present at low concentration(< 2 mg%) comletely secreted by single circulatin through the kidney � Clearance of PAHA = RPF � RPF= U X V/ P = 630 ml/min � 10% of renal blood flow go to renal pelvis &renal capsule (non functional stracture)= extraction ratio. Total plasma flow = RPF X 100/90= 630 X 10090=700 mlmin.
�Renal blood flow RBF = RPF X 1/ 1 - HV RBF= 1200 ml/min
1 - creatinine clearance � The most common type of clearance to assec renal function. Why? endogenous, constant, not affected by daily protein intake. � Partially secreted � Method? ? 1 - collect 24 hr urine & calculate V (volume of urinrmin) 2 -serum P & urinary U conc of creatinine is measured. 3 - creatinine clearance = U X V P =140 ml/min
2 - urea clearance �Reabsorbed by collecting duct. �Can detect early stages of renal failure at 60% loss of renal function when clinical symtoms remain normal( until 95% lost). �Method? 1 -bladder first evacuated. 2 -give subject a glass of water at start. 3 -another glass given 1 hr later. 4 - blood sample is withdrown for urea
4 - if urine vol. >2 ml/min , C= U X VP = 75 mlmin or urea max. 5 - if urine vol. < 2 mlmin. use urea standard C=U x√V/P =54 ml/min.
3 - glucose �Freely filterd in glomeruli & copletely reabsorbed in PCTs. , not secreted. �Norlam glucose clearance = 0 ml/ min �Importance? Used to estimate the abbsrpative power of the kidney
Load of substance �Defenition : the amount of substance delivered to renal tubules min. Load of substance = conc. Of sabstance in plasma XGFR
TM OR tubular maximum �The maximum amount of substance in mg that can be absorbed or secreted by kidney tubules.
TMG or tubular max for glucose �The max amuont of glucose in mg that can be absorbed by kidney tubules. �In males 375 mg/min, in females 305 mg/min. �Tubula load for glucose at blood glucose level 100 mgdl =100 100 x 125= 125 mgmin �Tubular laod at renal theroshold 180 mg% = 180100 X 125=225 mgmin. �This is less than TMG? ? Why? ? ? Coz not all 2 million nephron has the same TMG.
Renal theroshold for glucose �The plasma glucose level at which glucose first appear in urine. �Normal 180 mg dl �May br higher in DM.
ANSWER THE FOLLOWING the following data was collected from an adult male �Plasma glucose level 135 mg/dl. �Concen. Of inulin in plasma= 50 mg dl. �Concentration of inulin in urine is 6000 mgdl. �Vol. of urine in 40 min is 60 ml. �Calculate the GFR �Calculate tubular load for glucose.
�If plasma inulin was =2 mg%, the urine inulin was 2 mg%, vol of urine =2 mgmin �Calculate GFR.
�Use the data below to calculate the net amount of substance X secreted by the kidney -clearance of inulin 120 mlmin. -plasma conc of X =10 mgdl. -urinr conc of X =10 mgdl. -urine flow rate 1. 5 mlmin
�Inulin is used in an experiment to measure the glomerular filtration rate (GFR). Inulin is continuously infused to achieve a steady-state concentration in the plasma of 1. 0 mg/d. L. Urine is collected over a 10 hour period. The total volume of urine is 1. 5 L, and the urinary concentration of inulin is 440 mg/L. What is the GFR, as determined from the inulin clearance?
�A 25 year old man weighing 60 kg has a plasma [creatinine] of 1. 4 mg/d. L (nl = 0. 8 - 1. 4 mg/d. L). A 24 hour urine collection is done to determine his creatinine clearance, and thereby estimate his GFR. The following data are obtained: Urine [creatinine] = 833 mg/L Urine volume = 1, 080 ml/24 hour What is his creatinine clearance?
answer �Creatinine clearance = _____44. 6____ ml/min Calculation: Urine flow rate = 1, 080 ml/1440 min = 0. 75 ml/min Urine [creatinine] = 833 mg/L = 0. 833 mg/ml Plasma [creatinine] = 1. 4 mg/d. L = 0. 014 mg/ml
answer �Inulin Clearance = GFR = _____110____ ml/min Calculations: Urine flow rate = 1, 500 ml/600 min = 2. 5 ml/min � Urine [inulin] = 440 mg/L = 0. 44 mg/ml �Plasma [inulin] = 1. 0 mg/d. L = 0. 01 mg/ml �Inulin clearance = GFR = (2. 5 ml/min x 0. 44 mg/ml)/0. 01 mg/ml = 110 ml/min
3 - handling of renal tubules with different substances � Clearance = 0 as glucose completely reabsorbed. � Inulin(neither reabsorbed nor secreted) so inulin clearance is used as a refrence value 125 ml/ min. � substance with clearance less than inulin are partially reapsorbed as urea 54 &75 ml/min. � substance with clearance more than inulin are partially secreted as creatinine =140 ml/min. � Substance with clearance=650 -700 ml/min (PAHA) completely secreted. � Substance with clearance higher than 700 ml/ min are completely secreted & synthezied by renal tubules as amonia
Urine examination �One of the most important & useful urologic test available. �Patient with urinary tract symptoms or signs shoukd undergo urine analysis. �Important for asseament of renal function.
Complete urine analysis include: 1. Physical or macroscopic examination. 2. Chemical examination. 3. Microscopic examination.
Urine collection: � Freshly voided few hours after the last meal � In the laboratory. � Examined withen 1 hour. Indication for morning specimen? sp. Gravity in minimal renal desease and diabetes inspidus. Protien in orthostatic protienurea, TB bacilli in urine
Method of collection 1. Midstream clean catch method. 2. Catherization. 3. Suprapupic needle aspiration under local anaethesia.
Physical examination 1 - Urine colour: �normal? Amber yellow. �Red? ? Blood, HB, myoglobin or drugs (l-dopa & metronidazole. �Dark yellow? Dehydration(conc. ) or heamolytic joundice. �Black in obstractive joundice �Colourless in diabetis inspidous.
2 - aspect �Normal: clear. �Cloudy or turbid urine(1) pyurea (urinary tract infection) or (2)more often amorphous phosphate (crystals prespitated in alkaline urine diappear if acid is added) or(3) excess protein( tested by heating). �Odor : uriniferous odor, amoniacal odor means that specimen stay too long to be reliable.
3 -urine volume �Normal 1 -1. 5 L per day. �Polyurea >2. 5 L per day un DM or D. insipidous or early renal failure. �Oligourea < 400 cc per day in hypotention & late phases of renal failure. �Anurea <100 cc in renal failure & urinary tract obstraction.
4 - specefic gravity �Normal: 1003 - 1025 according to urine( diluted {D. insipidoud} or cocentrated {dehydration or DM}). �Fixed SG at 1010 in renal failure. �Renal tubular function assested by 1 - dilution test. 2 - concentration test.
5 - reaction of urine �Normal PH: 4. 5 - 8. �Failure of kidney to secrete H+ renal tubular acidosis (suspected when early urine PH of acidotic patient is > 5. 4. �Urine acidification test: 100 mg/kg BW ammonium chloridein the morinig if PH in next 8 hrs is >5. 4 this is a case of renal tubular acidosis.
Chemical tests 1 - protein in urine �Signif. proteinurea more than 1 gm 24 hrs. �Cause: glomrulonephritis, cancer. �prolonged fever & exercise transient proteinurea. �Tested by heating coagulatin or by electrophoresis. �False +ve: concentrated urine, presence of WBCS or vaginal secretions or epithelial cells.
2 - glucose �Normal urine: no glocose �Presence of glucose: DM. �False +ve: large dode of asperin, vit. C, cephalosporins.
3 - HB, myoglobin �In heamolytic anemea & excessive RBCs destruction. �Heamoglobinurea.
4 - bilirubin & urobilinogen �Normal urine contain no bilirubin & little urobilinogen �Direct bilirubin appear in obstractive & hepatocellular joundice.
Microscopic examination �Early morning urine is the best specimen. �Presence of : RBCs, pus cells, bacteria , leucocytes, bilharizial ova, yeasts& crystals.
1 - RBCs �heamaturea: presence of blood in urine. �Causes: urinary tract stones, infections, malignancies , bilhariziasis & sternous execise.
2 - bacteria, leucocytes & pus cells �Indicate presence of infection. �Pyurea: more than 5 -8 WBCs.
3 - casts �Casts are formed in the DCTs & CDs. �Mixure of coagulated protein(gloulin) , mucos &shedded cells �In renal deseases where protein is concentrated & tubules acidified. renal tubular desease as in glomerulonephritis. �Hyaline , granular & albumin casts
4 -crystals �Presence of crystals does not indicate renal desease. �Types: cystine, uric acid, ca oxalate & triple phosphate (struvite).
Tests depends on blood analysis �Creatinine serum level : 0. 7 -1. 4 mg%. �Not influenced by dietry intake (endogenous substance). �Remain withen normal range till 50% of renal function is lost. �Blood urea: 10 - 40 mg%
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