The Kidneys
Practical questions
Functions of the kidneys
• Regulation of water and inorganic ion
balance.
• Removal of metabolic waste products
from the blood and their excretion in the
urine.
• Removal of foreign chemicals from the
blood and their excretion in the urine.
• Gluconeogenesis.
• Production of hormones/enzymes (Renin,
Erythropoietin…)
3 main renal processes
1. Glomerular filtration.
(‫)סינון‬
2. Tubular secretion.
(‫)הפרשה‬
3. Tubular reabsorption.
(‫)ספיגה מחדש‬
excretion =
filtration - reabsorbtion + secretion
Glomerular Filtration Rate (GFR)
• The volume of fluid filtered from the
glomerulus into Bowman’s space per unit time.
• Affected by:
–
–
Net filtration pressure.
Permeability of the
corpuscular membrane.
– Surface area available
for filtration.
• Regulated by nerve and hormones.
• In a 70Kg person: GFR of ~180L/day.
Filtered Load
• The total amount of substance (S) filtered
into Bowman’s space per time unit.
• Filtered Load of S = GFR x [S]P
Example:
GFR = 180L/day
[Na+]P = 3.5g/L
What is the Na+ filtered load?
GFR x [Na+]P = 180L/day x 3.5g/L = 630g/day
Filtered Load
(continue)
• Once the filtered load is known, it can
be compared to the amount excreted.
• Indication of net tubular reabsorption
or net secretion.
Example:
Na+ filtered load = 630g/day
Na+ excreted = 3.2g/day
% reabsorbed = 99.5%
Renal clearance
• The volume of plasma completely cleared of a
specific substance (S) per unit time.
Mass of S excreted / time
[S]P
(example: if the mass of urea excreted in the urine is
5 g/hour, and the plasma concentration of urea is 5
g/L, then the urea clearance is 1 L/h)
• A useful way of quantifying
renal function.
Renal clearance
Renal clearance = GFR
(Inulin, Creatinine)
Renal clearance > GFR
No net reabsorption,
secretion, metabolism
Net secretion
(K+, H+…)
Renal clearance < GFR
(Glucose, Amino-acids…)
Net reabsorption
Kidney function assessment
Urine and blood test of a 65 years old male
gives the following results:
– Amount of urine collected over 24 hours:
4.32 L
– Plasma creatinine: 0.03 mg/mL
– Urine creatinine:
0.3 mg/mL
– Plasma K+:
0.005 mmol/mL
– Urine K+:
0.120 mmol/mL
What is this man’s GFR?
Assessing GFR
Urine volume per day: 4.32L
Plasma creatinine:
0.03 mg/mL
Urine creatinine:
0.3 mg/mL
+
Plasma K :
0.005 mmol/mL
+
Urine K :
0.120 mmol/mL
Mass of Cr excreted / time [Cr ]U × V / t
=
CCr =
[Cr ]P
[Cr ]P
CCr = renal clearance of Creatinine (L/day)
[Cr]U = Creatinine concentration in the urine
V/t = the volume of urine collected per time unit
[Cr]P = Creatinine concentration in the plasma
0.3mg / mL × 4.32 L / day
CCr =
= 43.2 L / day
0.03mg / mL
So, his GFR is 4 times lower than normal
(180 L/ day)…
What is the renal
clearance of K+?
CK +
Urine volume per day: 4.32L
Plasma creatinine:
0.03 mg/mL
Urine creatinine:
0.3 mg/mL
+
Plasma K :
0.005 mmol/mL
+
Urine K :
0.120 mmol/mL
0.120mmol / mL × 4.32 L / day
=
= 103.7 L / day
0.005mmol / mL
CK+ > GFR
K+ is net secreted
Urine volume per day: 4.32L
Plasma creatinine:
0.03 mg/mL
Urine creatinine:
0.3 mg/mL
+
Plasma K :
0.005 mmol/mL
+
Urine K :
0.120 mmol/mL
What is the renal
secretion rate of K+ in
this patient?
K+ excretion = K+ filtered - K+ reabsorbed + K+ secreted
Computing the difference between how much
is excreted and how much is filtered gives net
secretion rate
How much K+ is filtered?
Filtered Load = GFR × [K ]P =
+
43200mL/day × 0.005mmol / mL = 216mmol / day
What is the renal
reabsorption rate of K+ in
this patient? (continue)
Urine volume per day: 4.32L
Plasma creatinine:
0.03 mg/mL
Urine creatinine:
0.3 mg/mL
+
Plasma K :
0.005 mmol/mL
+
Urine K :
0.120 mmol/mL
K+ excretion = K+ filtered - K+ reabsorbed + K+ secreted
K+ secreted =
K+ excreted
0.120 mmol/mL x 4320 mL/day
= 518 mmol/day
K+ filtered
216 mmol/day
= 302 mmol/day
58% of the total K+ excretion is by secretion
K+ in the Kidney
Reabsorption:
• Not regulated
• In the proximal tube
and thick ascending tube
of the loop of Henle
• Absorption of 90% of K+.
Secretion:
• Coupled to Na+ reabsorption.
• Via K+ channels in the apical
membrane of the cortical
collecting duct.
• Regulated by the activity of
the of the Na/K ATPase and
by aldosterone activity.
Another example
A patient complains about a general
feeling of sickness (nausa, fatigue,
frequent urination…).
You decide to check his kidneys
function.
You are about to do blood and urine test
and you decide to use Inulin as a way to
assess GFR.
Test results
[In]P = 0.5 mg/mL
V (per min) = 1.1 mL
[In]U = 60 mg/mL
[Glu]P = 0.4 mg/mL
[Glu]U = 1.5 mg/mL
Assessing GFR
[In]P = 0.5 mg/mL
V (per min) = 1.1 mL
[In]U = 60 mg/mL
[Glu]P = 0.4 mg/mL
[Glu]U = 1.5 mg/mL
Mass of In excreted / time [ In]U × V / t
CIn =
=
[ In]P
[ In]P
60mg / mL × 1.1mL / min
CIn =
= 132mL / min = 190 L / day
0.5mg / mL
So his GFR is 190 L/day – normal,
more or less
What about his
glucose level?
What is the glucose
renal clearance?
[In]P = 0.5 mg/mL
V (per min) = 1.1 mL
[In]U = 60 mg/mL
[Glu]P = 0.4 mg/mL
[Glu]U = 1.5 mg/mL
CGlu
Mass of Glu excreted / time [Glu ]U × V / t
=
=
[Glu ]P
[Glu ]P
CGlu
1.5mg / mL × 1.1mL / min
=
= 4.125mL / min = 5.9 L / day
0.4mg / mL
CGlu << GFR
Glucose is net reabsorbed
What is the glucose
reabsorption rate?
[In]P = 0.5 mg/mL
V (per min) = 1.1 mL
[In]U = 60 mg/mL
[Glu]P = 0.4 mg/mL
[Glu]U = 1.5 mg/mL
Glu excretion = Glu filtered - Glu reabsorbed + Glu secreted
What is the glucose filtration load?
Filtered Load = GFR × [Glu]P =
132mL/min × 0.4mg / mL = 53mg / min
1.5 mg/mL x 1.1 mL/min = 53 mg/min - Glu reabsorbed
1.7 mg/min = 53 mg/min - Glu reabsorbed
Glu reabsorbed = 51.3 mg/min
(97% of the glucose that was filtered)
Glucose reabsorption in the kidney
Proximal
convoluted
tubule
Proximal
straight
tubule