Clearance: FeNa

Question 1

Equation to calculate renal elimination

Answer 1

Renal elimination = Filtered Load - Reabsorbed Load + Secreted Load

Question 2

What is renal clearance?

Answer 2

The volume of plasma from which a specific compound is completely eliminated (cleared) per unit time

Thus, clearance is a flow rate

Question 3

Input occurs via the renal artery. How is input calculated?

Answer 3

Input = (Pax) x (RPFa)

Where:
Pax = arterial concentration of X (mg/ml)

RPFa = renal arterial plasma flow (ml/min) aka rate of entry into kidney

Question 4

Output occurs via the renal vein. How is the venous effluent calculated?

Answer 4

Venous effluent = (Pvx) x (RPFv)

Where:
Pvx = renal v. plasma concentration of x (mg/min)

RPFv = renal plasma flow

Question 5

Output is via urine; how is urinary excretion rate calculated?

Answer 5

Urinary excretion rate = (Ux) x (V)

Where:
Ux = urine concentration of x (mg/ml)

V = urine flow rate (ml/min)

Question 6

A clearance calculation provides the answer to, “From how many mLs of plasma entering the kidney per min was substance X removed?”. How is clearance calculated?

Answer 6

Cx = [(Ux) x (V)]/Px

Or:
Cx = (urinary excretion rate)/Px

Question 7

When does the renal elimination rate (i.e., clearance) equal the GFR?

Answer 7

When a substance is freely filtered but neither reabsorbed nor secreted

Question 8

Explicit requirements of a substance suitable for measurement of GFR

Answer 8

Freely filtered across glomerulus (i.e., not bound to plasma proteins)

Undergoes no tubular reabsorption or secretion

Is neither metabolized nor synthesized in the kidney

Does not alter GFR

Question 9

What water-soluble substance is freely filtered but neither reabsorbed nor secreted, and thus its clearance can be used to measure GFR?

Answer 9

Inulin

Question 10

What endogenous water-soluble substance is freely filtered but neither reabsorbed nor secreted and can be used to approximate the GFR based on its clearance?

Answer 10

Creatinine

Question 11

What substance is completely eliminated (i.e., filtered and secreted but not reabsorbed) from the plasma during passage through the kidney, and what can it be used to measure?

Answer 11

Para-amino hippurate (PAH)

Its clearance = effective RPF (because ~10% of RPF does not enter peritubular capillaries), note that it works for low plasma concentrations only

Question 12

How is RBF calculated?

Answer 12

RBF = RPF/(1-Hct)

Question 13

Equation for filtered load

Answer 13

FL = GFR x Px

Question 14

What is the meaning of a filtered load > excreted load and vice versa?

Answer 14

FL > EL means there is net reabsorption of the filtered solute

EL > FL means there is net secretion of the filtered solute

Question 15

What is the range of values for clearance?

Answer 15

0-RPF

[you get 0 if substance is filtered but completely reabsorbed, like glucose, amino acids, often HCO3, etc]

Question 16

______ is widely available since it is used in radiological procedures; it is also the current default means of measuring GFR in pts since it is freely filtered and not reabsorbed. There is error in this method because, unlike inulin, it undergoes some tubular secretion, but this can be blocked if desired with ______

Answer 16

125-I-iothalamate (GloFil); probenecid

Question 17

___ is a 120 aa peptide that is freely filtered, then digested/reabsorbed by proximal tubules. With constant rate of synthesis, the plasma concentration should be a reflection of the GFR

Answer 17

Cystatin C

Question 18

With normal sodium intake, what is considered a normal fractional excretion of sodium (FeNa)?

Answer 18

Typically near 1%

Question 19

If GFR is low (i.e., serum creatinine is high), and FeNa is very low (e.g., 0.1%), what does that mean for the pts renal function?

Answer 19

Tubules are alive with massive reabsorption of a small amount of tubular fluid (acute kidney injury due to pre-renal ischemia, kidney can recover relatively quickly)

Question 20

If GFR is low (i.e., serum creatinine is high), and FeNa high, what does that mean for the pts renal function?

Answer 20

Significant nephron death is occurring, causing disrupted reabsorption of filtrate (acute kidney injury due to acute tubular necrosis, slow to recover)

Question 21

Equation used to calculate fractional excretion of sodium

Answer 21

FeNa = Na excreted/Na filtered = [U(Na) x V] / [P(Na) x GFR]

Note GFR = [U(Cr) x V]/P(Cr), so can be simplified to FeNa = C(Na)/C(Cr)

Question 22

Spot urine protein allows a rough assessment of daily protein loss when there is protein in the urine. What does it mean if > 3-3.5 g is lost per day?

Answer 22

It is being lost faster than the liver can replace it

Question 23

Spot urine protein equation

Answer 23

Protein excreted (g/day) = U(pro)/U(Cr)

Question 24

Only a fraction of the plasma that enters the glomerulus is filtered. What must happen for the clearance of a solute to exceed the GFR?

Answer 24

Tubular secretion

Question 25

Clearance of PAH is considered a measurement of effective Renal Plasma Flow (or eRPF), what is meant by “effective”?

Answer 25

Blood perfusing the renal pelvis and some juxtamedullary nephrons bypasses the proximal tubule basolateral membrane (roughly ~10% is not included in calculation)

Question 26

PCr will increase until UCrV equals ____

Answer 26

mg creatinine produced per minute

Question 27

As GFR falls, the fraction of the total mass of creatinine ending up in the urine that was contributed by the proximal tubule ______, also helped in part by the ______ PCr

[increase/decrease?]

Answer 27

Increases; increased

Question 28

Cockcroft-Gault GFR calculation

Answer 28

[(140-age) x kg] / [ serum creatinine x 72]

Multiply by 0.85 for females

Question 29

Assume that a pt was supposed to have a timed collection of urine and plasma to assess renal function. An IV infusion was set up for 0.5 ml/min infusion of normal saline containing inulin and PAH. The anticipated steady-state plasma levels for inulin and PAH were 1 mg/ml and 0.1 mg/ml, respectively. However, due to a math error during infusion solution preparation, the inulin concentration in the infusion solution was doubled.

As a result of this error, what changes take place in C(inulin), C(PAH), U(In)V, FE(inulin)

Answer 29

C(inulin) = unchanged

C(PAH) = unchanged

U(In)V = doubled

FE(inulin) = unchanged

Question 30

Assume that a pt was supposed to have a timed collection of urine and plasma to assess renal function. An IV infusion was set up for 0.5 ml/min infusion of normal saline containing inulin and PAH. The anticipated steady-state plasma levels for inulin and PAH were 1 mg/ml and 0.1 mg/ml, respectively. However, due to a math error during infusion solution preparation, the inulin concentration in the infusion solution was doubled.

Had there not been this math error, but the steady-state plasma level for inulin for this pt was nevertheless 2 mg/ml, what change would you expect in the C(inulin)?

Answer 30

C(inulin) was halved

Question 31

Assume a pt was supposed to have a timed collection of urine and plasma to assess renal function using a 0.5 ml/min infusion of inulin and PAH intended to generate plasma concentrations of 1 mg/ml and 0.1 mg/ml, respectively. The nurse accidentally dilutes the solutes to twice their intended volume of normal saline (i.e., concentrations are halved). In an attempt to hide this error, the nurse doubles the infusion rate to 1.0 ml/min.

What changes would you expect in C(inulin), C(PAH), U(In)V, and FE(inulin)?

Answer 31

All unchanged

Question 32

Creatine phosphate serves as the storage form for ATP in muscle to be used during contraction. Without this ATP storage, muscles would consume all their ATP in a few seconds. In an attempt to increase the size of this rapidly-available energy resource, athletes began taking creatine as a nutritional supplement. This caused considerable alarm/confusion amongst healthcare providers because blood tests of the young athletes showed serum creatinine values indicative of severe renal failure.

If a 24 hr urine collection was examined, what changes would be expected in the C(Cr) and U(Cr)V?

Answer 32

C(Cr) would likely be unchanged because the U(Cr)V was increased

Question 33

What must happen to obtain a value other than 0 for renal glucose clearance? what are the consequences of this?

Answer 33

Glucose transporters must be saturated

Consequences include osmotic retention of tubular fluid manifesting as polyuria offset by polydipsia

Question 34

Why is the microalbuminuria test used in diabetic patients?

Answer 34

Provides a sensitive test for kidney damage; measures trace amounts of albumin below the detection limit of urine dipstick