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Genitourinary 1 > Topic 4 - Quantification of Renal Function > Flashcards

Topic 4 - Quantification of Renal Function Flashcards

1
Q

How do you quantify the kidney’s conservation of mass?

A

input = output

input (via renal artery) = output (via renal veins + urine)

ignored in this equation: relatively small amount of lymph drainage from kidneys

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2
Q

How do you quantify the kidney’s fluid volume balance?

A

renal arterial blood flow = renal venous blood flow + urine

renal arterial blood flow typically ~20% resting cardiac output = 1L/min (1000mL/min or 1400L/d)

urine flow = 1-3mL/min (1.5-4L/d)

renal arterial blood flow and renal venous blood flow are nearly equal, just call it RBF

Renal plasma flow = RBF * (1-Hct)

e.g. if RBF = 1000mL/min and Hct = 40%, RPF = (1000mL/min)* 0.6 = 600mL/min

RPF (like RBF) is much larger than urine flow, so renal arterial plasma flow and renal venous plasma flow are nearly equal, just call it RPF

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3
Q

How do you quantify the kidney’s mass balance of solutes?

A

for solutes dissolved in plasma: input = output

amount of solute entering in arterial plasma = amount of solute leaving in venous plasma + amount of solute leaving in urine

C=A/V, so A = C*V

For solute x:

  • amount of solute entering in arterial plasma = Cx.art*RPF
  • amount of solute leaving in venous plasma = Cx.vein*RPF
  • amount of solute leaving in urine = Cx.urine*urine flow
  • Cx.art*RPF = Cx.veins*RPF + Cx.urine*urine flow rate
  • P = concentration of solute in plasma
  • U = concentration of solute in urine
  • V = urine flow rate
  • RPF*Px.arterial = RPF*Px.veins + Ux*V
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4
Q

How do you calculate the mass balance of solutes for glucose?

How much of the incoming renal arterial plasma flow was ‘cleared’ of glucose?

What was the renal clearance of glucose?

A

RPF*Pglucose.art = RPF*Pglucose.vein + Ug*V

600mL/min*80mg/dL = 600mL/min*80mg/dL + 0mg/dL*1mL/min = 480mg/min

none of it

0mL/min

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5
Q

How do you calculate the mass balance of solutes for para-aminohippurate (PAH)?

How much of the incoming renal arterial plasma flow was ‘cleared’ of PAH?

We say ‘renal clearance’ of PAH was:

A

RPF*Ppah.art = RPF*Ppah.vein + Upah*V

600mL/min*1mg/dL = 600mL/min*0mg/dL + 600mg/dL*1mL/min = 6mg/min

all of it

600mL/min

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6
Q

How do you calculate the mass balance of creatinine (byproduct of muscle metabolism)?

Another substance that is treated like creatinine in kidneys?

What is the renal clearance of creatinine (or inulin)

A

RPF*Pcr.art = RPF*Pcr.vein + Ucr*V

600mL/min*1mg/dL = 600mL/min*0.8mg/dL + 120mg/dL*1mL/min

6mg/min = 4.8mg/min + 1.2mg/min

100% = 80% + 20%

inulin

20%*600mL/min = 120mL/min

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7
Q

What is inulin?

A

a fructose polymer that does not naturally occur in the body, is nontoxic, is not metabolized by the kidneys, and is stable and easy to measure in plasma and urine

inulin is used in classic investigations of renal functions

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8
Q

If no trees have been cut in 100acre woods, what is the clearance of trees?

what is this like the renal clearance of?

A

0 acres

glucose

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9
Q

If all the trees have been cut in 100acre woods, what is the clearance of trees?

what is this like the renal clearance of?

A

100 acres

PAH

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10
Q

If 20% of trees have been cut in 100acre woods, what is the clearance of trees?

if instead, scattered or selective logging cuts 20% of the trees, what is the clearacne of trees?

what is this like the renal clearance of?

A

20 acres

20 acres

inulin or creatinine

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11
Q

What is the definition of renal clearance?

A

the volume of renal arterial plasma that would have to be completely cleared of a substance (each minute) to account for the rate of appearance of the substance in the urine

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12
Q

Using creatinine as an example, how would you calculate renal clearance?

A

rate of appearance of Cr in urine = Ucr*V = 1.2mg/mL*1mL/min = Ecr

Clearance.cr = volume of arterial plasma (per minute) needed to acount for Ecr

what volume of plasma would have 1.2mg of Cr in it?

volume = amount/concentration

Clearance.cr = 1.2mg/min / 1mg/dL = 120mL/min

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13
Q

How is renal clearance calculated?

A

Clearance.x = Ux*V/Px = Ex/Px

Cx = Cin is filtered

Cx < Cin is reabsorbed

Cx > Cin is secreted

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14
Q

Why do we want to measure GFR?

A

clearance of creatinine or inulin

GFR is the volume of plasma water that is filtered at the glomeruli per unit of time, usually reported in mL/min

GFR of the kidney = sum of filtration rates of all functioning nephrons

  • index of kidney function

a decrease in GFR generally means that kidney disease is progressing, whereas movement toward a normal GFR generally suggests recuperation

  • knowledge of the patient’s GFR is essential in evaluating the severity and course of kidney disease
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15
Q

How do we measure GFR?

A

by determining the clearance of a substance that is only excreted by the process of glomerular filtration

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16
Q

What substances are used for determining GFR?

A

Inulin - plant polysaccharide

  • only filtered; is NOT reabsorbed or secreted
  • biologically inert and non-toxic
  • not bound to plasma proteins or metabolized and is considered gold-standard for measuring GFR

Creatinine

  • used to estimate GFR in the muscle mass
  • not perfect substance for measuring GFR because it is secreted to a small extent in the proximal tubule
17
Q

What is the equation for measuring GFR?

A

amount filtered (F) = amount excreted (E)

Pcr*GFR = Ucr*V

Ccr = GFR = Ucr*V/Pcr

18
Q

How do we measure renal plasma flow?

A

clearance of PAH

  • requires low PAH plasma concentrations
  • PAH clearance and renal plasma flow (RPF)
    • renal plasma flow (RPF)
    • renal blood flow (RBF)
19
Q

What is the equation for measuring renal plasma flow?

A

measuring renal plasma flow = clearance of PAH

Ppah*RPF = Upah*V

Cpah = RPF = Upah*V/Ppah

20
Q

What is PAH?

A

para-aminohippurate

  • weak organic acid
  • actively secreted into the proximal tubule
  • by filtration and secretion, PAH is almost entirely cleared from the plasma in one pass through the kidney
21
Q

How do you determine filtration fraction (FF)?

A

not all creatinine (or other substances used to measure GFR) that enters the kidney in the renal arterial plasma is filtered at the glomerulus

  • virtually, all the plasma that enters the kidneys pass through the glomerulus
  • FF = GFR/RPF = Ccr/Ppah

a small fraction of RPF goes directly to some capsular and supporting tissue without going through the glomerulus

22
Q

How do you calculate creatinine clearance and how is GFR normalized?

A

Ccr = Ucr*V/Pcr

GFR normalization:

Ccr, normalized = (Ccr, measuredx1.73m2)/SA, measured

23
Q

How does creatinine excretion compare between one with a normal GFR of 120mL/min and stable chronic kidney disease with GFR or only 10mL/min?

A

both patients have the same excretion of creatinine

  • the production of creatinine by muscle is relatively constant from day to day, and the kidneys are solely responsible for creatinine excretion
  • daily creatinine excretion equals daily creatinine production by the muslce. because muscle mass is a function of body size, daily creatinine excretion is a function of body size
  • creatinine excretion does not change when GFR decreases, unless GFR falls so low that uremia develops
24
Q

How do changes in serum creatinine result from an acute fall in GFR and how is attaining a new steady state?

A

between days 0 and 1, the patient is excreting all the creatinine that is produced, and serum creatinine is stable at 1.0mg/100mL

A 50% reduction in GFR on day 1 results in an abrupt fall in filtered (and therefore excreted) creatinine. Release from muscle, however, remains constant, and as a result creatinine is reatined and the serum concentration increases

As creatinine concentration progressively rises, filtered (and excreted) creatinine also increases until excreted creatinine returns to control levels and again matches creatinine production

this new steady state (days 3-4) is achieved by a doubling of serum creatinine concentration, which maintains the filtered creatinine load at control levels in the face of a halving of the GFR

25
Q

How can the clearance of creatinine (Ccr) be estimated from plasma creatinine [cr]p?

A

because creatinine is excreted by glomerular filtration, the filtered load of creatinine must stay constant for excretion to be constant. for filtered load to remain constant, Pcr must increase as Ccr decreases

  • except for very low GFRs, Pcr is inversely related to Ccr
    • exponential relationship
    • Pcr doubles when Ccr decreases by 0.5
  • early in progressive renal disease, small changes in Pcr reflect large changes in Ccr
  • later changes in Pcr reflect small changes in Ccr
26
Q

How do you interpret Ccr during progressive nephron injury?

A

a decrease in GFR may be the first and only clinical sign of kidney disease

  • measuring the GFR is important when kidney disease is suspected
  • whole kidney GFR is the sum of all GFRs for single nephrons
  • when nephrons are lost gradually, the tendency for wkGFR to decline is buffered by the residual nephrons
  • the residual nephrons increase snGFR which compensates for the lost nephrons
27
Q

How do you interpret Ccr during acute kidney injury?

A

Ccr cannot be determined when [cr]p is changing

  • [cr]p will increase minimally immediately following traumatic reduction of GFR to zero
  • even after GFR has been at 0 for 24hrs, [cr]p may only have increased to 2 suggesting GFR is 50% of normal, not zero
28
Q

A selective decrease in the efferent arteriolar resistance results in…

a. an increase in GFR and an increase in RBF
b. a fall in GFR and an increase in RBF
c. an increase in GFR and a decrease in RBF
d. a fall in GFR and a fall in RBF
e. an increase in filtration fraction (FF)

A

b.

29
Q

The dysproteinemia (abnormality of the plasma proteins) of multiple myeloma is frequently associated with a marked elevation of plasma-globulin. Which of the following would you expect…

a. potassium loss
b. extracellular fluid volume concentration
c. renal ischemia
d. a decrease in GFR
e. respiratory alkalosis

A

d.

30
Q

What is the clearance of a substance P given that Up=360mg/L, V=0.1L/hr, Pp=4mg/L?

a. 220L/day
b. 2.2L/day
c. 9mg/day
d. 18mg/day
e. none of the above

A

e.

Cp = (Up*V)/Pp

Cp = 360*0.1/4 = 9L/hr = 216L/24hr

31
Q

Substance x is freely filterable and is neither metabolized nor stored in the kidneys. The plasma concentration of x was 500mg/dL and 250mg/min appeared in the urine. The inulin clearance was 100mL/min. The tubular reabsorption of x was:

a. 25mg/min
b. 250mg/min
c. 5mg/min
d. 50mg/min
e. 500mg/min

A

Rx = Fx - Ex

Rx = (Cin*Px) - Ex

Rx = 100mL/min*500mg/dL - 250mg/min

Rx = 100mL/min*5mg/mL - 250mg/min

Rx = 500mg/min - 250mg/min

Rx = 250mg/min

32
Q

Urea excretion rate was 10mg/min. Urea reabsorption rate was 10mg/min. Plasma urea concentration was 10mg/dL. GFR was:

a. 100mL/min
b. 200mL/min
c. 300mL/min
d. 350mL/min
e. 400mL/min

A

Ru = Fu - Eu

Ru = GFR*Pu - Eu

10mg/min = GFR*10mg/dL - 10mg/min

10mg/min = GFR*0.1mg/mL - 10mg/min

GFR*0.1mg/mL = 10mg/min + 10mg/min

GFR = 20mg/min / 0.1mg/ml = 200mL/min

33
Q

Which of the following is true?

a. a substance filtered at the glomerulus cannot also be secreted
b. if the tubular reabsorption of a substance is inhibited, its clearance will decrease
c. if the secretion of PAH is inhibited, it excretion will increase
d. clearance of a substance that is excreted exclusively by filtration (neither reabsorption nor secretion) varies in direct proportion to the plasma concentration
e. none of the above

A

e. none are true

34
Q

The following data were obtained on a 60Kg patient during a renal function test:

Plasma PAH = 2.0mg/dL

Plasma inulin = 94.7mg/dL

Urine PAH = 226mg/dL

Urine Inulin = 3500mg/dL

Based on the data and given a urine flow rate of 2.3mL/min, calculate the glomerular filtration rate.

a. 2.3mL/min
b. 85mL/min
c. 260mL/min
d. 82.7mL/min
e. none of the above

A

b. 85mL/min

GFR = Cin = 35mg/mL*2.3mL/min / 0.947mg/mL

GFR = Cin = 85mL/min

35
Q

The following data were obtained on a 60Kg patient during a renal function test:

Plasma PAH = 2.0mg/dL

Plasma inulin = 94.7mg/dL

Urine PAH = 226mg/dL

Urine Inulin = 3500mg/dL

Based on the data and a urine flow rate of 2.3mL/min, what is the rate of water reabsorption by the tubule?

a. 2.3mL/min
b. 85mL/min
c. 260mL/min
d. 82.7mL/min
e. none of the above

A

d. 82.7mL/min

the rate of water reabsorption is the difference between GFR (cin, 85mL/min) and urine flow rate (2.3mL/min)

Genitourinary 1 (17 decks)

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