Lect 10: Proximal Tubule

Question 1

The PT receives the ultrafiltrate from the BS. How does it maintain ECF solute and fluid volume?

Answer 1

it maintains homeostasis by a process of separating solute and water from the plasma (filtration) and returning solute and water to the plasma (reabsorption).

Question 2

Each segment of the nephron may be considered for its constitutive and regulatory function with regard to the renal handling of solutes and water.

Answer 2

CONSTITUITVE function (not regulatory-regardless of changes in ion homeostasis) occurs with little regulation and mediates a lesser renal response to changes in solute or water balance. The regulatory function of the different segments of the nephron mediates the renal response to changes in solute or water balance.

Question 3

The PT is the nephron segment mediating

Answer 3

REABSORPTION of 67% both filtered water and salt and put back into circulation.

Question 4

The PT fluid reabsorption occurs isosmotically, without a change in NaCl conc in the 33% of the tubular fluid remaining in the PT

Answer 4

The 33% had the same conc as the plasma conc. So there is equal proportion of water and solute in the PT.

Question 5

The PT is a “leaky” epithelium meaning

Answer 5

that its resistance to water reabsorption is very low. A small difference in peritubular osmolarity relative to tubular osmolarity is enough to cause a drive this isosmotic fluid reabsorption from the tubular into the peritubular (BLOOD SIDE) out to renal vein

Question 6

After the solutes are filtered in the glomerulus, the job of the PT is to reabsorp and return all of this back into

Answer 6

the blood via transcellular and pericelluar transport.

Question 7

If the filtered load is more than the saturable load in the PT

Answer 7

it will appear in the urine….because there is no other way to reabsorb any of the solutes/glucose. This also means that the presence of glucose in the tubular lumen will raise the osmolarity pf the tubular fluid. By doing this you oppose the ability of the downstream segments to reabsorp fluids. This is why the pt. becomes dehydrated. Bc you can’t reabsorb water in excess of solutes as well as you could if the solute wasn’t there.

Question 8

The PT mediates the SECRETION of organic anions (drugs, metabolites, diuretics) from the peri- capillaries to the lumenal fluid

Answer 8

where they remain until they are excreted into the urine.

Question 9

The PT is also the nephron segment mediating the secretion of organic cations (drugs, metabolites) from

Answer 9

the peri- to the lumenal fluid where they remain and are excreted in the urine.

Question 10

Solute conc from beginning to end

Answer 10

at the beginning the TF/P ratio is 1

Question 11

Changes for Inulin: an increase in TF/P

Answer 11

Inulin is freely filtered at the glomerulus and is neither reabsorbed from nor secreted into the tubular fluid. It is pretty much trapped in the tubular fluid. An increase in TF/P for inulin indicates an increased conc of inulin in the TF resulting from the reabsorption of water from the tubular fluid. WATER IS LEAVING but inulin cannot leave

Question 12

Changes for Na

Answer 12

A TF/P close to 1 for Na indicates the equivalent reabsorption of Na and water from the proximal tubular fluid resulting in a constant Na conc in the tubular fluid. 67% of the filtered Na and water is reabsorbed in the PT, the Na conc does not change and remains constant. Na conc in glomerular filtrate is 145

Question 13

Changes for Cl-

Answer 13

The incerase in TF/P reflects the preferential reabsorption of HCO3 (transcellular) rather than Cl– in the early PT.

Question 14

Changes for HCO3, a.a and glucose–

Answer 14

The decrease in TF/P for HC03, amino acids and glucose indicates reabsorption from the tubular fluid, which contributes, as osmotic equivalents to driving the reabsorption of water

Question 15

The change in transepithelial voltage from –3mV at the beginning of the PT to +3mV “downstream” results from the exit of more positive cationic charged solutes in first 25% of the PT

Answer 15

So there is more negative anionic charge left back. The slight excess of positive charge remaining in the TF in the form of cations creates the lumen positive transepithelial voltage difference measured across the tubule cell layer.

Question 16

The change in the transepithelial voltage from –3mV to +3mV results from the net efflux and reabsorption of more positively charged, cationic solutes in the first 25% of the PT

Answer 16

and the net efflux anf reabsorption of more negatively charged, anionic solutes downstream in the remaining 75% of the PT. The slight excess of negative or positive charge in the tubular fluid in the form of anions or cations creates the lumen positive transepithelial voltage difference in the first 25% of the PT and creates the lumen net negative transepithelial voltage difference downstream in the remaining 75%

Question 17

The conc HCO3, a.a. and glucose falls because

Answer 17

the PT is extracting them out of the tubular fluid and putting them back into the circulation

Question 18

Paracellular and Trancellular

Answer 18

Paracellular: Na and Cl travels between cells through tight junctions

Transcellular: uptake across the lumenal membrane and efflux across the basal membrane or the lateral membrane

Question 19

Na and Cl Reabsorption

Answer 19

The major driving force for Na uptake is the difference in electrochem potential. More Na is outside the cell than inside (15mM inside compared to 145mM outside) (due to Na/K pump)
Bc Na is positively charged all that negative charge inside is another driving force
3. There is a backleak: –3mV is the driving force that pulls Na that has been reabsorbed. 33% of the Na absorbed transcellularly back leaks into the TF paracellular

Question 20

Transporters in the early PT

Answer 20

the Na/Glucose co-transporter transfers these two into the cell. Resulting in a a lumen negative charge (which is what drives the paracellular backleak of Na into the lumen of early PT. Co-transport creates the lumen neg voltage difference

Question 21

Cl- transport occurs both betw cells (paracellular and transcelular) in the

Answer 21

EARLY and LATE PT. It is mediated by uptake of apical membrane and efflux at the basal & lateral membrane in the LATE PT.

Question 22

What is responsible for paracellular Cl– transport?

Answer 22

paracellular transport in the early convoluted PT (and late prox straight tubule) is driven by a lumen negative transepithelial voltage difference. it drives by repulsion (negitve lumen repels negative ions). Negative due to preferential HCO3- reabsorption in the early PT, lumenal Cl- conc is elevated above plasma Cl- conc in the late PT and an outward transepithelial Cl- conc gradient exists driving passive paracellular efflux of lumenal Cl-. The paracellular efflux of Cl- creates a lumen potential diffusion potential, which in turn drives paracellular efflux of Na in the late PT

Question 23

Transcellular transport of Cl-

Answer 23

happens mainly in the late PT.
The active uptake of Cl- across the PT apical membrane occurs by Cl solute antiport driven by an outwardly directed anion conc gradient. The passive efflux of intracellular C–across the basolateral membrane is mediated by a Cl- channel and by a KCl symporter

Question 24

Water Reabsorption

Answer 24

passive!
Most of the glomerular filtrate is returned to the circulation at the PT where reabsorption of tubular fluid occurs w/o a change in osmolarity.

Question 25

The primary driving force for paracellar and transcellular water reabsorption is

Answer 25

the small osmotic gradient resulting from active solute reabsorption across the PT. This PT is very leaky and permits lots of water movement in response to small changes in the gradient. - aquaporins allow transcellular - ABSORPTION > filtration

Question 26

How is paracellular movement of water happening?

Answer 26

it is driven by active Na-dependent increase in osmolarity in the lateral space entrains or sweeps Na and Cl within the flow in the direction of the peritubular capillary by a process of “solvent drag”

Question 27

Role of PT in acid base balance (1)

Answer 27

Reabsorb and return most of the filtered HC03 to the circulation maintaining ECF HCO3 conc constant (24mL)

Question 28

Role of PT in acid base balance (2)

Answer 28

Secretes H+ protons generated from acid metabolism, production of organic acids such as lactic acid and from intestinal bicarb loss, which may decrease THE BICRB conc in the plasma. Losing bicarb upsets the balance of acid-base so that extra proton that remain can also be secreted by the PT

Question 29

Role of PT in acid base balance (3)

Answer 29

The process of secreting H generates “new” bicarb, which replaces HCO3 lost in buffering organic and inorganic acid. (When HCO3 buffers the H+ generated it becomes H2CO3 (it gets “lost”) and needs to be replaced. If we lost HCO3 in the urine we would become acidotic and die. HCO3 conc has to be around 24mM

Question 30

HCO3 reabsorption

Answer 30

NOT A DIRECT WAY
On the lumenal side, there is a carbonic anhydrase enzyme which dehydrates HCO3 into OH and CO2. CO2 is extremely diffusible. OH combines with H to form HOH (water). The CO2 and H2O recombine intracellular to form HCO3 which moves out of the cell via a Na/HCO3 transporter. The Na/bicarb transporter (3HCO3: 2Na) has a –2 charge which serves as a driving force to push Na and HCO3 out of the cell against their conc difference

Question 31

PT Secretion participates in the renal excretion of H+ ions as

Answer 31

1) titratable acid and as 2) NH4+

Question 32

Renal excretion of H as titratable acid arises from the

Answer 32

titration of dibasic phoshoric acid (HPO42-) to monobasic phosphoric acid (HPO4-). A phosphate buffer in the urine. The pK of phosphoric acid makes it an excellent buffer of H at the pH of the PT fluid

Question 33

Renal excretion of H as NH4+ arises from

Answer 33

the titration of NH3 to NH4. The PT. thick ascending loop of Henle and the collecting ducts participate in the excretion of H as NH4. In the PT, the origin of NH3 is intracellular glutamine metabolism secondary to active glutamine uptake across the luminal and basolateral membrane

Question 34

For each H secreted as a titratable acid, a new HCO3- ion is made intracellularly and

Answer 34

is returned to the circulation to replace the HCO3– ion lost in the buffering of metabolic acid. New HCO3- replenishes the continuous depletion of HCO3 in the ECF as acid is generated from metabolism and buffered by HCO30

Question 35

PT excretion of H and NH4+

Answer 35

protons secreted by the PT are also excreted as ammonium (and HPO4-) which is trapped in the tubular fluid and is excreted.

Question 36

Diffusion Trapping

Answer 36

when a small amount of NH3 freely diffuses across the lumen into the tubular fluid where it is trapped by titration to NH4+ (impermeable to cell membrane). It is then excreted in the urineThe protons titrating NH3 to NH4 in the PT fluid arise from the Na/H exchanger or the H-ATPase transporting H across the lumenal membrane

Question 37

Renal compensatory response to RESPIRATORY acidosis (or hypoventilation - a increase in ECF pCO2)

Answer 37

includes an 1) increase in PT H+ secretion as NH4
2) an increase in HCO3 synthesis, which increases ECF HCO3 conc proportional to the increase in Pco2. More HCO3 around can generate CO2 for the blood.

The PT is to raises the HCO3– in order TO MAINTAIN THE RATION OF HCO3/CO2 conc close to 20 and to maintain ECF pH close to 7.4.

Question 38

Renal compensatory response to RESPIRATORY alkalosis (or hyperventilation - decrease ECF pCO2) includes

Answer 38

1) decrease PT in H secretion as NH4+
2) decrease in the HCO3 synthesis, which decreases the exctracellular fluid HCO3 conc proportional to the decrease in Pco2.

CO2 went down so you decrease HCO3 amount

Question 39

Renal compensatory response to METABOLIC acidosis (a primary decrease in HCO3- conc), includes

Answer 39

1. increase in H+ secretion as NH4+
2. increase in HCO3- synthesis.
3. increased glutamine metabolism generating additional NH3 for H+ secretion.

They serve to maintain the HCO3/CO2 ratio close to 20 and maintain the ECF pH close to 7.4

Question 40

Renal compensatory response to METABOLIC alkalosis (a primary increase in ECF HCO3 conc) includes

Answer 40

1. decrease in H+ secretion as NH4
2. decrease in HCO3 synthesis
3. decreased glutamine metabolism which decreases intracellular NH3 available for H secretion.