Renal clearance

Question 1

Explain how inulin clearance is indicative of the glomerular filtration rate (GFR)

Answer 1

Properties of inulin:
o Freely filtered
o Not reabsorbed
o Not secreted
 Amount of inulin excreted per minute = Amount of inulin filtered per minute
 GFR =
Urine [inulin or creatinine]×Flow rate
Plasma [inulin or creatinine]
 Renal clearance: the volume of plasma that would be required to supply that amount of the substance excreted per unit tie.
 Renal clearance can be measured by using any compound. The value obtained will depend on how the substance is handled
by the kidneys.

Question 2

How is renal clearance determined

Answer 2

If the substance is freely filtered, not reabsorbed or secreted, then its clearance is the same as GFR.

 Assuming the solute is freely filtered, if it undergoes net reabsorption, its clearance will be less than that of inulin.
 Or, if it is not freely filtered (e.g. bound to plasma proteins) then its clearance will also be less than that of inulin.
 If the solute undergoes net secretion, its clearance will be greater than that of inulin.
 If the solute is not excreted in the urine, its clearance is 0. However, this can be because it is filtered and completely
reabsorbed or it is not filtered at all.

Question 3

What is the renal clearance of urea

Answer 3

Renal plasma flow: 650 ml/min (maximum possible clearance).
 [Urea] is 5 mmol/L.
 Urea excretion is 300 μmol/ml.
 1 ml urine produced per minute.

Question 4

What are the drawbacks of these clearing methods

Answer 4

Measuring clearance assesses overall nephron function (i.e. all 2-3 million nephrons in both kidneys).

 This gives the sum of all transport processes occurring along nephrons.
 Therefore, it provides no information about whether a substance is filtered or the precise sites and mechanisms of
transport.
 Nevertheless, clearance methods are widely used for clinical assessment of renal function.

Question 5

Suggest what the clearance of inulin, creatinine and PAH mean respectively

Answer 5

Clearance of inulin (Cin) = GFR
 Clearance of creatinine (Ccr) = GFR
 Clearance of para-amino hippurate (CPAH) = Renal plasma flow

Question 6

Explain how the clinical measurement of GFR is obtained through inulin and creatinine

Answer 6

1. Inulin clearance
    Polysaccharide of fructose which must be infused intravenously (45 minutes to equilibrate).
    Freely filtered, not reabsorbed, not secreted.
    Continual blood and urine sampling required.
    Chemical analysis cumbersome.
2. Creatinine clearance (is a useful alternative (not creatine))
    Endogenous (by-product of muscle metabolism).
    Released into blood at relatively constant rate.
    Plasma concentration fairly stable; therefore only need one blood sample.
    Freely filtered, not reabsorbed, some secretion.
    Therefore Ccr normally overestimates GFR by 10-20%.

Question 7

What is the link between plasma creatinine concentration and GFR

Answer 7

If GFR decreases, then urinary excretion of creatinine
will also decrease.

 Decreasing urinary excretion of creatinine increases
plasma creatinine concentration as it is not cleared.
 This increases urinary creatinine excretion due to
higher plasma creatinine concentration.
 Decrease in GFR leads to a proportional increase in
plasma creatinine concentration: plasma creatinine
concentration can be used to estimate GFR.

Within the normal range of plasma creatinine (50-120
μmol/L), there are relatively small changes of plasma
creatinine in normal renal functions. Small changes in
plasma creatinine concentration has large effects on
GFR.
 Not very sensitive at normal GFRs.
 Useful at low GFR

Question 8

What factors influence the normal GFR rate?

Answer 8

Depending on the source, the range given for GFR is
90-140 ml/min.
 This demonstrates that there can be considerable
variation between individuals with normal renal
function.

Question 9

Describe the measurement of renal plasma flow using Para amino hippurate (PAH)

Answer 9

 Organic anion not usually present in the body, so has
to be infused.
 Freely filtered
 Avidly secreted from peritubular capillaries by a Tm dependent mechanism into proximal tubule lumen. At low plasma
concentrations, the combination of filtration and secretion of PAH means that almost all PAH arriving in the renal plasma is
excreted in the urine.

Question 10

What is the transport maximum

Answer 10

Tm refers to the point at which an increase in concentration does not
result in an increase in movement of a substance across a m

Question 11

Describe how to the kidneys handle PAH

Answer 11

Plasma creatinine concentration and GFR
 If GFR decreases, then urinary excretion of creatinine
will also decrease.

 Decreasing urinary excretion of creatinine increases
plasma creatinine concentration as it is not cleared.
 This increases urinary creatinine excretion due to
higher plasma creatinine concentration.
 Decrease in GFR leads to a proportional increase in
plasma creatinine concentration: plasma creatinine
concentration can be used to estimate GFR.
GFR can be estimated from 𝐏𝐜𝐫

 Within the normal range of plasma creatinine (50-120
μmol/L), there are relatively small changes of plasma
creatinine in normal renal functions. Small changes in
plasma creatinine concentration has large effects on
GFR.
 Not very sensitive at normal GFRs.
 Useful at low GFRs.
What parameters influence the normal range of GFR?
 Depending on the source, the range given for GFR is
90-140 ml/min.
 This demonstrates that there can be considerable
variation between individuals with normal renal
function.
Measurement of renal plasma flow: Para-amino
hippurate (PAH)
 Organic anion not usually present in the body, so has
to be infused.
 Freely filtered
 Avidly secreted from peritubular capillaries by a Tm dependent mechanism into proximal tubule lumen. At low plasma
concentrations, the combination of filtration and secretion of PAH means that almost all PAH arriving in the renal plasma is
excreted in the urine.

Question 12

Explain the application of the Fick principle

Answer 12

Mass balance equation: amount of PAH entering the kidneys per minute is equal to the amount of PAH leaving the kidneys
per minute.
 PAH enters via the renal artery.

 PAH leaves via urine or the renal vein (5-10%).
 Renal plasma flow =
Excreted load of PAH
[PAH]arterial−[PAH]venous

 During low plasma concentration of PAH, the amount leaving the renal vein is 0. This means we can only estimate the renal
plasma flow:

Estimated renal plasma flow =
UPAH×Flow rate
PPAH

Question 13

What is the filitration fraction

Answer 13

The filtration fraction (FF) is defined as the ratio of the GFR to the RPF. The fraction of plasma filtered across the glomerular
capillaries.
 Represents the fraction of plasma which is filtered across the glomerular capillaries.
 Increasing FF (e.g. severe haemorrhage): ↓ hydrostatic pressure and ↑ oncotic pressure in peritubular capillaries  ↑
reabsorption.
 Decreasing FF: ↑ hydroastatic pressure and ↓ oncotic pressure in peritubular capillaries  ↓ reabsorption.

Question 14

Describe glucose handling by the kidneys

Answer 14

 Under normal plasma glucose concentrations:
o Glucose enters by the renal artery.
o 20% is freely filtered due to its size.
o All glucose is reabsorbed, so clearance is 0.
 In diabetes mellitus, with high plasma glucose concentrations, 20% is filtered and some of this is reabsorbed, but some
glucose is not reabsorbed, leading to glucose excretion in urine (glycosuria), because the glucose transporters are saturated.

Question 15

What is the relationship between solute clearance and plasma concentration

Answer 15

Increased glucose clearance at very high plasma levels of glucose reflects
saturation of the re-absorption process.

Question 16

Describe the SGLT2 transporter

Answer 16

 In the proximal convoluted tubule.
 Low affinity, high capacity
 Responsible for 97% of renal glucose transport.
 SGLT1 (in the proximal straight tubule) is responsible for the remaining
3% glucose transport.
 SGLT2 is a new target for the treatment of type II diabetes.
 SGLT2 becomes saturated during diabetes.

Question 17

Explain how glucose reabsorption has a tubular maximum

Answer 17

 At low glucose concentrations, glucose is freely filtered and
completely reabsorbed at higher glucose concentrations.
 As plasma concentration increases above 200 mg/dL, glucose
reabsorption plateaus off due to saturation of SGLT2, and more
glucose is excreted into the urine.
 Splay: as plasma glucose increases and as glucose reabsorption
reaches transport maximum, it does so gradually due to different
levels of SGLT2 present and so are saturated at different times to
others.

Question 18

How do you calculate Tm for glucose

Answer 18

 Filtered rate = [Plasma glucose] × GFR
 Excretion rate = Amount of glucose in urine × Rate at which urine is produced
 Tm = Filtered load− Excretion rate
 380 mg/min of glucose can be reabsorbed before glucose is excreted in urine.

Question 19

How do you calculate tubular transport rate

Answer 19

Transport rates can be calculated for any substance and gives an indication of how the nephron handles that substance.
 Transport rate = Filtered load− Excretion rate
 Filtered load = Px − GFR
 Excretion rate = Ux × Flow rate
 Transport rate of zero or close to zero: no reabsorption or secretion.
 Negative transport rate: secretion
 Positive transport rate: reabsorption