Renal Clearance and Its Clinical Applications | DR RAI M. AMMAR | ALL MEDICAL DATA by DR RAI M. AMMAR www.facebook.com/drraiammar www.twitter.com/drraiammar www.instagram.com/drraiammar www.linkedin.com/in/drraiammar www.medicall.com.pk/blog/auther/drraiammar/ For Any Book or Notes Visit Our Website: www.allmedicaldata.wordpress.com www.drraiammar.blogspot.com YOUTUBE CHANNEL : https://www.youtube.com/channel/UCu-oR9V3OdFNTJW5yqXWXxA ANY QUESTION ?? Get in touch with us at Any of the Above Social Media or Email at drraiammar@gmail.com allmedicaldata@gmail.com

1. B Y
D R R A I M . A M M A R M A D N I

2. Renal Handling of
Different Substances

3. Plasma Clearance
 The amount of plasma “cleared” of a substance
 Plasma clearance for a substance that is FILTERED,
but not REABSORBED or SECRETED == GFR ==
125ml/min
 EXAMPLE = INULIN

4. Renal Handling of Water
and Solutes
Filtration Reabsorption Excretion
Water (liters/day) 180 179
Sodium (mmol/day) 25,560 25,410
Glucose (gm/day) 180 180
Creatinine (gm/day) 1.8 1.8
1
0
0
150

5. Clearance Technique
Renal clearance of a substance is the volume of
plasma completely cleared of a substance per min.
Cs = Us x V
Ps
Cs x Ps = Us x V
Where: Cs = clearance of substance S
Ps = plasma conc. of substance S
Us = urine conc. of substance S
V = urine flow rate

6. Plasma Clearance
 Plasma clearance for a substance that is FILTERED
AND REABSORBED, but NOT SECRETED < GFR
 EXAMPLE = GLUCOSE, UREA
 Clearance rate can be anywhere from 0 up to normal
clearance (125 ml/min) depending on amount
reabsorbed

7. Plasma Clearance
 Plasma clearance for a substance that is FILTERED
AND SECRETED > GFR
 EXAMPLE = PAH (Para-aminohippuric acid)
 Not only is the plasma that is filtered cleared of that
substance, but additional amt cleared from plasma
which was not filtered
 Plasma clearance rate for PAH=RENAL PLASMA
FLOW

11. For a substance that is freely filtered, but not reabsorbed or
secreted (inulin, 125 I-iothalamate, ~creatinine), renal clearance
is equal to GFR
Use of Clearance to Measure GFR
amount filtered = amount excreted
GFR x Pin = Uin x V
GFR =
Pin
Uin x V
Copyright © 2006 by Elsevier, Inc.

12. Calculate the GFR from the following data:
Pinulin = 1.0 mg / 100ml
Uinulin = 125 mg/100 ml
Urine flow rate = 1.0 ml/min
GFR =
125 x 1.0
1.0
= 125 ml/min
GFR = Cinulin =
Pin
Uin x V

13. Steady-state
relationship
between
GFR and serum
creatinine
concentration

14. Clearances of Different Substances
Clearance of inulin (Cin) = GFR
if Cx < Cin: indicates reabsorption of x
Substance Clearance (ml/min)
inulin 125
glucose 0
sodium 0.9
urea 70
Clearance creatinine(Ccreat) ~ 140 (used to estimate GFR)
if Cx > Cin: indicates secretion of x

15. COMPLETE CLEARANCE OF A SUB FROM PLASMA
GIVES THE RENAL PLASMA FLOW
 RENAL PLASMA FLOW= PLASMA FITERED 20%
PERITUBULAR CAPILLARYPLASMA =80%
PAH IS ABOUT 90% CLEARED, 20% FILTERED
AND 70% SECRETED SO ITS EXTRACTION RATIO
IS 90%,MEAANING THAAT 10% REMAINS IN THE
RENAL VENOUS BLOOD.
PAH CLEARANCE IS A GOOD INDEX OF RENAL
PLASMA FLOW.

17. ASSUME
PPAH=0.01mg/ml
 UPAH=5.85mg/ml
 V=1ml/min
CPAH=585ml/min
 Actual RPF= 585/0.9(extraction ratio of
PAH)=650ml/min
 Total blood flow=CPAH/1-Hematocrit
 650/0.55=1182 ml/min

18. ONCE AGAIN
 EXCRETION RATE OF A SUBSTANCE IS=US*V
 FILTERED LOAD OF SUBSTANCE=PS*GFR
 IF EX RATE IS MORE THAN FILT LOAD=the sub
must be filt and secreted
 IF EX RATE IS LESS THAN FILTERED LOAD=the
sub must have been reabsorbed

19. assume
 V=1ml/min
 PNa=140meq/L
 UNa=70meq/L
 GFR=100ml/min
 What is tubular reabsorption of Na
 Calculate filtered load of
sodium=100*140=14000meq/min
 Calculate excretion rate of
sodium=1*70=70meq/min
 14000-70=13930meq/L is tubular reabsorption
of Na.

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