Renal - Filtration and reabsorption

Question 1

Describe the filtration barrier in the kidney

Answer 1

1. Glomerular capillary endothelium (fenestrated)
2. Glomerular basement membrane (supportive)
3. Bowman’s capsule epithelium = Podocytes

Podocytes are foot-like projections that wrap around the glomerular capillaries leaving slit-like openings between them

Question 2

What is the basic functional unit of the kidney

Answer 2

The renal corpuscle

• The glomerulus
• Bowman’s Capsule

Question 3

What determines the size of molecules that can be filtered within the renal corpuscle. What size molecules can be filtered

Answer 3

Pore size / fenestration size in the glomerular capillaries

Particles < 7 kDa –> filtered

Particles 7 to 70 kDa –> partially filtered

Particles > 70 kDa –> not filtered

Question 4

Give examples of molecules below 7 kDa, 7 - 70 kDa and more than 70 kDA

Answer 4

< 7 kDa: glucose, Na, Cl, K, Cl-

7 - 70 kDa

1. Albumin 67 kDa (small enough to fit through pores but filtration prevented due to negative charge
2. Hb 69.8 kDa can fit but sequestered within RBCs

> 70 kDa
Immunoglobulins > 150 kDa

Question 5

Why are negatively charged particles less likely to be filtered in the glomerulus

Answer 5

The negative charge of the podocytes and basement membrane repel negatively charged proteins preventing filtration

Small anions are filtered because the do not come close enough in contact with the glomerular membrane proteins to be repelled

Question 6

How does protein binding affect filtration of drugs. What type of molecules bind to albumin? What type of molecules bind to alpha - 1 - glycoprotein

Answer 6

When protein bound, substances are not filtered

Albumin binds acids

alpha-1-glycoprotein binds bases

Question 7

What is reabsorbed in the PCT. State proportions reabsorbed

Answer 7

67% of Na, K, Cl-, H2O

85% of HCO3-

100% glucose and amino acids

Question 8

Describe the active reabsorption processes that occur in the PCT

Answer 8

Basolateral membrane Na/K ATPases: Extrude Na out of tubular cells into the capillaries in exchange for potassium. Low intracellular sodium drives

1. Co-transporters
   - -> SGLT-2 (sodium glucose linked transporter)
2. Counter transporters
   - -> Na/H counter transporter

Question 9

What is Tmax.

Answer 9

= transport maximum for glucose in the kidney
= 12 mmol/L (filtrate concentration)
= 300 mg of glucose per minute

Any glucose filtered above this is excreted in urine

Question 10

Why does glycosuria occur in pregnancy

Answer 10

GFR increases –> increased filtration of glucose into the renal tubules: Tmax is exceeded at a lower plasma glucose concentration (along with hormonal changes)

Question 11

What are the two different mechanism for tubular secretion in the kidney

Answer 11

ORGANIC ANION TRANSPORTER
- Uric acid
- Penicillin
- Probenecid
- Aspirin
WW2 --> co-administration of probenecid with penicillin maintained higher plasma penicillin levels (limited capacity for secretion as one transporter secretes all these substances)

ORGANIC CATION TRANSPORTER

• Creatinine
• Catecholamines
• Morphine

Question 12

Draw the graph that represents Glucose reabsorption and excretion in the PCT

Answer 12

X axis = plasma [glucose] in mmol/l
Y axis = Glucose transport in mg/min

Straight line increase until 12 mmol/l then plateau at 300 mg/min glucose transport.

Glucose excretion rises dramatically at glucose concentrations higher than 12 mmol/L

Question 13

What is meant by the term clearance

Answer 13

The clearance of a given solute quantifies the capacity of the kidney to eliminate that solute from blood, whether through filtration, reabsorption or secretion.

Clearance (mL/minute) is the volume of plasma completely cleared of a substance by the kidneys per unit time.

Question 14

How is the clearance equation derived

Answer 14

Amount (moles) of solute A cleared from plasma = Amount (moles) of solute A in urine

As moles (n) = volume x concentration

Volume plasma filtered x Plasma [A] = Volume urine excreted x Urine [A]

Volume plasma filtered = V urine x [A] urine
______________

                                              [A] plasma

Clearance = V urine x [A] urine
______________

                                               [A] plasma

Question 15

What is the glomerular filtration rate and how is it related to the Starling forces?

Answer 15

GFR is the volume of fluid passing into the tubules per unit time. Typically 125 ml/min.

GFR is determined by the balance of Starling forces

GFR = Kf (Pi - Po) - (πi - πo)

Kf is the glomerular filtration coefficient (reflects the permeability of the barrier)

Po and Pi are hydrostatic pressures inside and outside capillary respectively

πi and πo are oncotic pressures inside and outside the capillary respectively

Question 16

What are the typical values for the variables within the Starling equation as it relates to GFR

Answer 16

GFR = Kf (Pi - Po) - (πi - πo)

GFR = Kf (48 - 10) - (25 - 0)

GFR = Kf x 13 mmHg

There is a net driving pressure (filtration pressure) of 13 mmHg pushing fluid from the glomerular capillaries into the bowman’s space and then on to the renal tubules.

Question 17

What are the limitations of using creatinine for GFR estimation.

Answer 17

1. VARIABLE CREATININE PRODUCTION
   The rate of creatinine production is variable between patients and is dependent on muscle mass. Muscle mass is in turn influenced by age, sex, race etc. Numerous algorithms try to compensate for these factors e.g. Cockcroft Gault formula compensates for age, sex and weight.
2. CREATININE IS ACTIVELY SECRETED INTO PCT
   - -> accounting for 10 - 20% of excreted creatinine. This results in a slight overestimation of creatinine clearance and hence an overestimation of GFR. –> this error is tolerable in normal patients, those with advanced renal failure the error becomes proportionally much larger: filtration reduces with disease progression but secretory mechanisms are left intact.

Question 18

What is meant by the term filtration fraction?

Answer 18

The fraction of the plasma that is filtered by the glomerulus.

Kidney receives blood at 1000 ml/min
Hct = 40 %
Therefore 600ml/min is renal plasma flow.
GFR is 125 ml/min in health

Filtration fraction is then 600 ml/min / 125 ml/min
= 20%