6. renal plasma clearance.ppt

physiology

1. RENAL PLASMA CLEARANCE
Dr. Ramadan Saad

2. Learning Objectives
•Describe the concept of renal plasma clearance.
•Use the formula for measuring renal clearance.
•Use clearance principles for inulin, creatinine etc.
for determination of GFR.
•Explain why it is easier for a physician to use
creatinine clearance Instead of Inulin for the
estimation of GFR.
•Describe glucose and urea clearance.
•Explain why we use of PAH clearance for
measuring renal blood flow.
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3. Basic Renal Processes
1.Filtration (F)
2.Reabsorption (R)
3.Secretion (S)
Excretion = F + S -R
Urine Formation
Afferent Arteriole Efferent Arteriole
Glomerulus
Bowman’s
Capsule
Renal
Tubule
Peritubular
Capillary
Conceptual Point:
Filtration is the most basic “mode” of renal substance handling.
-solutes need pass the filter barrier
-no specific transport processes
-if there is no reabsorption or secretion, then the substance will be excreted.
There are very few “filtered-only” solutes.
Most are also reabsorbed and/or secreted.

4. Basic Renal Processes
1.Filtration (F)
2.Reabsorption (R)
3.Secretion (S)
Excretion = F + S -R
Urine Formation
Afferent Arteriole Efferent Arteriole
Glomerulus
Bowman’s
Capsule
Renal
Tubule
Peritubular
Capillary
Some Examples: Substances that are Filtered then Reabsorbed
>99.9%799.5∼0.5800Glucose mM/day
>99.9%4,498∼24,500HCO
3
mM/day
99.3%19,85015020,000Cl mM/day
99.4%24,85015025,000Na mM/day
99.2%178.5∼1.5180H
2
O L/day
% of filtered load
reabsorbed
Amount
Reabsorbed
Amount
Excreted
Amount
Filtered
Substance, units
►This “filtered then almost completelyreabsorbed” scenario is certainly not the case for all solutes.

5. Some Important Renal Physiology Numbers
►Renal Blood Flow RBF 1.2 L/min (1200ml)
►Renal Plasma Flow RPF 625 ml/min
RPF = RBF x (1 –hematocrit)
typical hematocrit is ~0.43, so RPF is 1.1x0.57.
►Glomerular Filtration Rate GFR 125 ml/min
►Urine Flow Rate 1 ml/min
►Filtration Fraction GFR/RPF 20%
20% of plasma entering a glomerulus is filtered.
Thus, 20% of any freely-filtered solute present enters Bowman’s space.
Note that values given above can vary in different circumstances.
Also remember that RBF far exceeds what kidney cells need to stay
alive so RBF can vary dramatically without affecting kidney cell vitality.

6. Concept of Clearance
Clearance is just a way to quantify renal handling of a substance.
Clearanceis defined as the volume of plasma“cleared” of a substance
by the kidneys per minute. ( ml/min )
Clearance of a substance is often used to evaluate renal function.
First…. we will define clearance in words.
Volume of Plasma Cleared
Now….Let’s see how we
can calculate clearance.
Note: Clearance units are volume per time. Clearance is notthe amount of the
substance removedbut instead the volume of plasmafrom which it was removed.
▼
Every minute-625 ml plasma goes to the kidney (RPF)
125ml/min are filtered forming filtrate (GFR)
Remaining 500ml/min remain in the blood and enter into PC

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9. Concept of Clearance
To calculate clearance of substanceX-(C
X),
first need to calculateamount of X excretedin urine per unit time.
Urine volume per min (ml/min)
Urine X concentration
Another hand theamount of substance X in “cleared plasma”can be expressed
This formula is convenient because U
X, P
Xand V are easily measured.
you will need to remember this formula
Product of plasma volume per unit time
Plasma X concentration
amount of X excreted in urine= UX· V
amount of X in cleared plasma= PX·
CX
(amount X in
cleared plasma)
PX· CX = UX· V
(amount X excreted
in urine)
PX
CX=
UXX V

10. Inulin Clearance
Inulin:polysaccharide, not a naturally occurring substance in body
freely filtered but not reabsorbed or secreted….so all inulin
that is filtered willend up in the urine
C
INULINis the “gold standard” for measuring GFR
Volume filtered
is volume cleared. (125ml/min)
GFR = C
INULIN =
U
INULIN· V
P
INULIN
V = urine produced in ml/min
U
INULIN= urine inulin concentration
P
INULIN= plasma inulin concentration
The main clinical drawback here is that inulin must be continuously
infused while urine is collected (this is usually a day or so).

11. P
A(inulin)x RPF P
V(inulin)x RPF
Filtered = P
in
X GFR
Excreted =
U
in X V
Filtered = Excreted
P
in X GFR = U
in X V
No reabs; No
secretion.
Inulin clearance

12. PAH Clearance
PAH: para-aminohippurateis also not naturally in the body
it is freely filtered and completely secreted….so both filtered and
secreted PAH willend up in the urine
C
PAHis clinically used to estimate RPF
RPF = C
PAH =
U
PAH· V
P
PAH
V = urine produced in ml/min
U
PAH= urine PAH concentration
P
PAH= plasma PAH concentration
Cleared volume
much larger than
filtered volume.
So large in fact that it
“effectively” approaches
RPF
Recall that…. RPF = RBF x (1-0.45) so…. RBF =
C
PAH
0.55
hematocrit

13. P
A
PAH x RPF P
V
PAHx RPF = 0
Near-complete
extraction in one
passage through
kidney.
U
PAHV

14. Glucose Clearance
Glucose:
is freely filtered like inulin. However that no glucose appears
in the urine because glucose is completely reabsorbed as it passes
through tubules
This means all of the glucose that comes to the kidney is
saved and leaves the kidney in the plasma membrane
Glucose 0 Completely reabsorbed
Inulin 125 Not reabsorbed and not secreted
PAH 625 Completely secreted
Normal Clearance Value (ml/min)

15. Creatinine Clearance
Creatinine: produced from creatine metabolism in muscle
production rate usually very constant (if muscle mass constant)
freely filtered and not reabsorbed …little bit is secreted
(this makes it a good but imperfect substitute for inulin)
C
CRis clinically used to routinely access GFR
“GFR” = C
CR =
U
CR· V
P
CR
V = urine produced in ml/min
U
CR= urine CR concentration
P
CR= plasma CR concentration
There is a nice inverse relationship between
P
CRand “GFR”
Normal P
CR= 1 mg/dl
P
CR

16. If GFR drops by 50%, then P
CRdoubles ( 2 mg/dl ).
Creatinine Clearance
Creatinine: produced from creatine metabolism in muscle
production rate usually very constant (if muscle mass constant)
freely filtered and not reabsorbed …little bit is secreted
(this makes it a good but imperfect substitute for inulin)
C
CRis clinically used to routinely access GFR
“GFR” = C
CR =
U
CR· V
P
CR
V = urine produced in ml/min
U
CR= urine CR concentration
P
CR= plasma CR concentration
There is a nice inverse relationship between
P
CRand “GFR”
Normal P
CR= 1 mg/dl
50% GFR
●Thus, a single P
CRvalue can be
used to roughly estimate GFR.
(level of creatinine excreation = level creatinine
production in the body -balance concept)

17. P
A
cr x RPF P
V
crx RPF
Filtered =
P
crX GFR
Excreted =
U
cr X V
Filtered = Excreted
P
cr X GFR = U
cr X V
No reabs; No
secretion.
Small amount is
secreted

18. Urea Clearance Test:
•The patient evacuates his bladder, then drinks a glass of water.
•After 1 hr blood & urine samples are taken & he drinks another glass of
water.
•After 2 hrs another urine sample is taken.
•The urine vol. /min. is calculated.
–If it is above 2 ml /min we get the maximal urea clearance.
MC = (U) x (V) / (P) = 75 ml /min. (normally).
–If it is below 2 ml /min we get the standard urea clerance.
SC = (U) x (V) / (P) = 54 ml /min. (normally)
(U) = Conc. of urea in 1 ml urine.
(V) = Vol. of urine /min.
(P) = Conc. of urea in 1 ml plasma.
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19. Renal Clearance
Summary
•Plasma clearance tests can be used for:
–Measurement of the glomerular filtration rate.
–Measurement of the renal plasma flow rate (&
from there we can calculate the renal blood
flow rate).
–Determining the renal handling of the different
substances; whether or not the substance is
reabsorbed or secreted by the renal tubules.
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20. Application
The following information was obtained from a human
subject:
•Plasma inulin= 1 mg / ml.
•Urine inulin =150mg/ml.
•Urine flow rate = 1 ml / min.
•Plasma level of substance X = 2 mg / ml.
•Urine substance X = 100 mg / ml.
Assuming that X is a freely filtered substance, which of the
following statements is most correct?
•a) There is net secretion of substance X.
•b) There is net reabsorption of substance X.
•c) There is both reabsorption and secretion of substance X.
•d) The clearance of substance X could be used to measure the GFR.
•e) The clearance of substance X is greater than the clearance of inulin.
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21. Application
If plasma inulin concentration is 0.2 μg/mL,
urine flow is 2 mL/min, and urine inulin
concentration is 9 μg/mL, what is GFR?
•A. 90 mL/min
•B. 45 mL/min
•C. 100 mL/min
•D. None of the above are correct.
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22. Application
If plasma inulin concentration is 0.2 μg/mL,
urine flow is 2 mL/min, and urine inulin
concentration is 9 μg/mL, what is GFR?
•A. 90 mL/min
•B. 45 mL/min
•C. 100 mL/min
•D. None of the above are correct.
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23. Application
If plasma inulin concentration is 0.2 μg/mL,
urine flow is 2 mL/min, and urine inulin
concentration is 9 μg/mL, what is GFR?
•A. 90 mL/min
•B. 45 mL/min
•C. 100 mL/min
•D. None of the above are correct.
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24. Application
Renal clearance. If Us = 500 mg/L, V = 0.1 L (in
an hour), Ps = 5 mg/L, then Cs equals:
•a. 10 L/hr
•b. 25000 L/hr
•c. 250 L/hr
•d. 50 L/hr
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25. Application
A person is infused with PAH and, after
equilibration, PPAH is 0.02 mg/ml. At this
time, urine flow is 2ml/min and UPAH is 4
mg/ml. What is renal plasma flow?
•a. 200 ml/min
•b. 300 ml/min
•c. 500 ml/min
•d. 1000 ml/min
•e. none of the above are correct.
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