GFR and renal clearance Flashcards

1
Q

why do we measure GFR (glomerular filtration rate)

A

assesses renal function

the sum of filtration rate of each of the functioning nephrons

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2
Q

what does a reduction in the number of functioning nephrons lead to

A

a fall in total GFR

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3
Q

where are most drugs and certain antibiotics excreted from the body primarily from

A

glomerular filtration in the kidneys

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4
Q

what happens to drugs if GFR is reduced

A

the excretion of the drug will fall

it will accumulate in the body to possible toxic levels - to prevent this - dosage must be reduced appropriately

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5
Q

what are normal values for GFR in young adults (range included)

A

120 mL/min per 1.73m2 (typical value for body surface area of an adult)
range = 90-150 mL/min per 1.73m2

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6
Q

how much does GFR fall by after the age of 40

A

falls by about 10 mL/min per decade

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7
Q

what is the definition for glomerular filtration rate

A

GFR
the amount of fluid filtered from the glomeruli into the bowmans capsule per unit time (mL/min) - sum of filtration rate of all functioning nephrons

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8
Q

what is the definition for a freely filtered substance

A

conc of x in plasma = conc of x in glomerular filtrate

a freely filtered substance may be absorbed or secreted later on, freely filtered refers to the glomerulus

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9
Q

how can we measure GFR using substance Y as an eg and EQUATION

A

consider a substance Y present in the plasma (small enough to be freely filtered into Bowman’s capsules but neither reabsorbed nor secreted into the tubules) therefore the amount of Y excreted per unit time will be the same as the amount filtered per unit time

GFR x plasma conc of Y (Py) = urine flow rate (V) x urine conc of Y (Uy)
GFR (mL/min) = V x Uy all/Py

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10
Q

what is an example substance we can use to measure GFR and some characteristics that allow this to occur

A
inulin
a plant polysaccharide
freely filtered 
not toxic
measurable in urine and plasma
not found in mammals - needs to be transfused
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11
Q

what is the definition for renal clearance

A

the renal clearance of a substance is equal to the volume of plasma which would be required to supply that amount of the substance excreted by the kidneys per unit time - always expressed as a volume of plasma per unit time

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12
Q

what is the equation to calculate renal clearance

A

for any substance Z

the renal clearance Cz is calculated as Cz = V x Uz/Pz

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13
Q

when can a substances GFR = renal clearance

A

if the substance is freely filtered and neither reabsorbed nor secreted

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14
Q

which one has a lower clearance - sodium or inulin

A

sodium has a lower clearance than inulin so more sodium is reabsorbed

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15
Q

what is indicated by a substance whose renal clearance is higher than that of inulin

A

when a substance has a renal clearance higher than inulin - it is being secreted from the blood into tubular fluid

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16
Q

what is indicated by a substance whose renal clearance is lower than that of inulin

A

if the renal clearance of a substance is lower than that of inulin - it means more of the substance is being reabsorbed than secreted

17
Q

how can we measure renal plasma flow (RPF)

A

using PAH - para amino hippuric acid

18
Q

where is PAH secreted into

A

proximal tubules

19
Q

what happens to PAH

A

if the plasma conc is not too high - the combination of filtration and secretion ensures that practically all of the PAH arriving at the kidneys in the plasma appears in the urine - and almost none leaves in the renal vein (needs to be infused)

20
Q

what can PAH clearance be used to give an estimate for

A

the volume of plasma perfusing the kidneys per unit time (renal plasma flow rate) because it is both filtered and secreted

21
Q

what is the equation for the rate at which PAH enters the kidneys per minute in the renal arteries

A

rate at which PAH enters the kidneys per minute in the renal arteries =
RPF x Ppah

determined by conc x flow

22
Q

example
if PAH clearance = 625 mL/min
inulin clearance = 125 mL/min
what proportion of the plasma entering the glomeruli is filtered

A

125/625 = 0.2 = 20%

23
Q
example 
if PAH clearance = 625 mL/min 
inulin clearance = 125 mL/min
if the arterial plasma inulin conc was 1mmol/L - what would be the plasma inulin conc in
1) the efferent arteriole
2) the renal vein
A

1) 1mmol/L as when something is freely filtered the conc does not change
2) 0.8 mmol/L as it is diluted by 20%

24
Q

why is PAH clearance rarely performed clinically when renal disease is suspected

A
  • the value of using PAH relies upon it being completely cleared by filtration and secretion - these processes are not guaranteed to be working properly in renal diseases
  • the kidney needs to actively secrete all PAH back into urine - the secretory mechanisms are impaired during renal disease - not diagnostic
25
Q

what are the problems when practically determining GFR using inulin clearance

A
  • does not occur naturally in the body - to measure its renal clearance it has to be infused intravenously
  • this causes several problems - rarely used clinically
  • time consuming and tedious
  • requires several blood samples and due to short duration over which measurements are made - bladder catheterisation (since voluntary bladder emptying may be incomplete)
    not given orally as it may be ingested
26
Q

how can we use creatinine clearance (Ccreatinine)

A
  • it is an endogenous substance produced from the metabolism of muscle creatine
  • released into the blood at a relatively constant rate - plasma conc of creatinine remains fairly stable in a given individual
  • requirements : 24hr urine collection, a single plasma sample taken at some time during the clearance period
  • it is freely filtered and neither reabsorbed nor metabolised by the kidneys
  • however a small amount of creatinine enters the urine by secretion into the proximal tubule - the amount excreted slightly exceeds the amount filtered
27
Q

what are the problems when using inulin and creatinine to measure clearance

A

clearance of inulin is accurate but inconvenient

clearance of creatinine is convenient but not so accurate

28
Q

how can the clearance of new substances be determined

A

monitoring their disappearance from the plasma after administering a given dose (as they emit radiation) - thus eliminating the need to collect urine

29
Q

what is the substance used to measure clearance

A

EDTA
51CrEDTA - a gamma emitter
administer a single injection and then measure the plasma activity of 51Cr at subsequent intervals

30
Q

how can we measure the clearance of 51CrEDTA from the slope of the straight line

A

the greater the slope the greater the clearance

31
Q

what are the advantages of using 51CrEDTA in practice

A

requires only a single dose of 51CrEDTA
collection of 2-3 plasma samples taken at appropriate times after the injection
free to permeate the whole of the extracellular fluid
can be IV
bolus injection
high conc starts to fall as starts to be removed in kidney
if someone is suffering from severerenal failure - reduced GFR - less steep gradient - takes longer to be removed due to impaired GFR