AN EXERCISE IN SOLVING RENAL CLEARANCE
PROBLEMS
INTRODUCTION
This laboratory exercise is a series of renal clearance problems. The method of calculating the
answers will be discussed in lecture. Solve each of the problems as presented on the worksheets
here on these pages.
Renal Physiology Problems
1. A person is suspected of renal failure and it is decided to run an inulin renal clearance test to
determine the Glomerular Filtration Rate (GFR.) The person is asked to empty their bladder, then
an IV is started with inulin. A catheter is used to collect the urine sample during the test and
when completed both the urine sample and a blood sample are sent to the laboratory. Using an
ACCU-STAT instrument the following laboratory results were obtained:
Plasma inulin = 0.14 mg/ml urine
Urine inulin = 20 mg/ ml
urine volume = 9 ml / 10 minutes
Calculate the GFR for this person. Is the GFR normal? Does the patient’s GFR indicate any
possible Goldblatt Renal Hypertension? (Ans. 128.5 ml/min; yes; no)
2. Given that the Tm (reabsorption) for glucose is 320 mg/min, calculate how much glucose will spill
over in the urine of the same person described in question 1, given that an ACCU-STAT reading of
the plasma glucose is: Plasma glucose = 450 mg %
(Ans. 258.3
mg/min)
3. A person with a BP of 163/105 mm Hg was tested for suspected renal (Goldblatt) hypertension.
PAH and inulin tests were run on the person with the following laboratory results:
Plasma inulin = 1.0 mg/ml plasma
PAH plasma = 0.1 mg/ml
Urine inulin = 41 mg /ml urine
PAH urine = 20 mg /ml
Urine flow rate = 2.0 ml/min
hematocrit (HCT) = 40%
Tm (secretion) for PAH = 80 mg/min
MCV = 93 cubic microns
Use 90% bladder clearance for PAH
a) Calculate the renal clearance value for PAH. (Ans a) 400 ml/ min;
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b) Calculate the secretion rate for PAH. (ans: b) 31.8 mg /min)
c) Calculate the total whole blood flow through the renal system per minute. ( ans: c) 741 ml/min)
What role did the Tm for PAH play in the calculation? Explain.
Does the data indicate a possible renal hypertension?
If this person does have renal hypertension, how would you explain the 160/105 torr BP?
Do these additional renal problems
One of the roles of the kidney is to remove substances from the blood. Clearance is defined as the
volume of plasma per minute (ml plasma / min) from which a particular substance is totally removed
or “cleared.”
For any substance, A, renal clearance value (RCV) is determined as follows:
RCV A = UA x V
PA
Where,
U = urine concentration of substance A
V = urine flow rate (ml/min)
P = plasma concentration of substance A
Marker substances:
Inulin is an exogenous substance that is freely filtered but is neither reabsorbed nor secreted by renal
tubules. Therefore, the clearance of inulin represents the volume of plasma filtered each minute (ml
plasma/min), or GFR.
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Any substance that is freely filtered, but that also undergoes net reabsorption will have a clearance
value less than inulin. Any substance that is freely filtered but undergoes net secretion will have a
clearance value higher than inulin.
Creatinine is a byproduct of muscle metabolism. It is endogenous to the body. It is freely filtered
and is only slightly secreted. Concentrations of this substance are fairly level in people with normal
renal function. Creatinine clearance can be used to estimate GFR, but since it undergoes some net
secretion, its value will be a slight overestimate of GFR.
Ex. = CCR = 140 ml/min when GFR is 125 ml/min
PAH (para amino hippurate) is an exogenous substance and is a weak acid. It is freely filtered and is
secreted by a Tm limited process along the renal tubules. Since it enters the nephron by two routes
(filtration and secretion), it is a marker for any blood reaching the nephron (at the glomerulus and
peritubular capillaries) but not for blood that reaches other non-nephron regions of the kidney. The
clearance value of PAH represents effective renal plasma flow (ERPF.)
Total RBF is the value of interest. To arrive at RBF, ERPF must be corrected by taking into account:
a) the ratio of whole blood to plasma (use the HCT)
b) 90% of the total blood flow to the kidney reaches nephrons. 10% does not.
Use this information to solve the following problem set:
4. What is the tubular load of glucose (TL glu) when plasma glucose concentration is 450 mg% and
GFR is normal (125 ml/min)?
5. Knowing the Tm for glucose (320 mg/min) what is the rate of glucose excretion in this person?
6. Substance “A” is freely filtered and undergoes net secretion, but no reabsorption. The following
data were obtained from the patient:
Plasma concentration A = 200 mg %
Urine concentration A 190 mg/ ml
Clearance inulin (RCV inulin) = 115 ml/min
Urine flow rate = 20 ml urine collected over 10 min
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Determine
a) This person’s GFR
b) The tubular load of substance A
c) The rate of excretion of substance A
d) The rate of secretion of substance A
7. Determine the clearance of the following substance:
Plasma concentration = 0.4
mg/ml
urine concentration = 40 mg
/ml
Urine flow
1ml/min
rate
=
Does this substance undergo any other processes besides filtration? Explain.
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8. A person presents with the following lab results.
Urine flow rate = 15 ml / 10 min
Plasma PAH = 0.2 mg /ml
Urine PAH = 110 mg / ml
HCT = 42%
Determine
a) RPF
b) RBF
c) Is this RBF normal?
9. A patient received a piggyback IV infusion of inulin and PAH over a 20 min interval. 15 ml of
urine was collected. Lab values returned as follows:
Urine inulin = 160 mg/ml
Plasma inulin = 2 mg / ml
HCT = 43%
Urine PAH = 78 mg/ml
Plasma PAH = 0.2 mg /ml
Determine
a) GFR
b) RBF
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c) Discuss these values relative to normal values.
Study Questions:
 Make sure you know how to define and perform all calculations for: CO, RBF, ERBF, RPF,
ERPF, GFR, RCV, and excretion.
 What is the importance of PAH? Inulin? Creatinine?
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