69: Control of Osmolarity Flashcards
Osmolarity is the _______ of all solute concentrations in a given compartment and, for the purpose of quantification, is assigned the unit osmoles of solute per liter of water (mOsm/L). For example, the normal range of plasma osmolarity is approximately ___ to ____ mOsm/L and represents the sum total of all the organic and inorganic, anionic and cationic solute concentrations in the plasma. __ (145 mMole/L) and ___ (116 mMole/L) solute concentrations are the major determinants of plasma osmolarity.
In contrast to the well-known active transport of solutes, active transport of water from a compartment of low “water concentration” to a compartment of high “water concentration” does ____ occur.
Osmolarity is the sum total of all solute concentrations in a given compartment and, for the purpose of quantification, is assigned the unit osmoles of solute per liter of water (mOsm/L). For example, the normal range of plasma osmolarity is approximately 280 to 300 mOsm/L and represents the sum total of all the organic and inorganic, anionic and cationic solute concentrations in the plasma. Na (145 mMole/L) and Cl (116 mMole/L) solute concentrations are the major determinants of plasma osmolarity.
In contrast to the well-known active transport of solutes, active transport of water from a compartment of low “water concentration” to a compartment of high “water concentration” does not occur.
In the steady state, daily (24 hr) input (gain) and output (loss) of water are essentially equal. The major sources of water are from ingested fluids and foods, with a lesser amount from metabolism. The major route of water loss is the renal excretion of urine, with lesser amounts lost in the feces, from the skin (perspiration), and in exhaled air (ventilation).
The production of sweat during exercise is primarily for body _____ regulation rather than for body fluid regulation.
Maintenance of plasma osmolarity in the normal range (280 - 300 mOsm/L) occurs by regulating the loss of water in the urine. The kidney excretes more or less water depending on whether the plasma is ____-osmotic or _____-osmotic, respectively. The maintenance of plasma osmolarity in the normal range (280 - 300 mOsm/L) is very important and deviations of only 15% (322 mOsm/L to 238 mOsm/L) can severely compromise ____ function, due primarily to changes in the transmembrane Na and K concentration gradients, which, in turn, disrupt normal neuronal impulse propagation.
In the steady state, daily (24 hr) input (gain) and output (loss) of water are essentially equal. The major sources of water are from ingested fluids and foods, with a lesser amount from metabolism. The major route of water loss is the renal excretion of urine, with lesser amounts lost in the feces, from the skin (perspiration), and in exhaled air (ventilation).
The production of sweat during exercise is primarily for body temperature regulation rather than for body fluid regulation.
Maintenance of plasma osmolarity in the normal range (280 - 300 mOsm/L) occurs by regulating the loss of water in the urine. The kidney excretes more or less water depending on whether the plasma is hypo-osmotic or hyper-osmotic, respectively. The maintenance of plasma osmolarity in the normal range (280 - 300 mOsm/L) is very important and deviations of only 15% (322 mOsm/L to 238 mOsm/L) can severely compromise CNS function, due primarily to changes in the transmembrane Na and K concentration gradients, which, in turn, disrupt normal neuronal impulse propagation.
When plasma is hyperosmotic, we should excrete a urine that has a _____ solute concentration than the plasma.
The ratio of urine to plasma osmolarity (U/P)osm normalizes the concentrating/diluting ability of the kidney to changes in plasma osmolarity produced by an excess or deficit in water consumption.
- When (U/P)osm = 1, the urine is _____ and ECF osmolarity is Isotonic.
- When (U/P)osm > 1, the urine is _____ and ECF osmolarity is hypertonic.
- When (U/P)osm
When plasma is hyperosmotic, we should excrete a urine that has a higher solute concentration than the plasma.
The ratio of urine to plasma osmolarity (U/P)osm normalizes the concentrating/diluting ability of the kidney to changes in plasma osmolarity produced by an excess or deficit in water consumption.
- When (U/P)osm = 1, the urine is isotonic and ECF osmolarity is Isotonic.
- When (U/P)osm > 1, the urine is hypertonic and ECF osmolarity is hypertonic.
- When (U/P)osm
An increase in osmolarity will induce an _____ in ADH, which results in a retention of free water by the kidney and a hypertonic urine. A decrease in osmolarity will induce a ____ in ADH, which results in an elimination of free water by the kidney and a hypotonic urine. In both instances, the gain or loss of free water serves to normalize plasma osmolarity.
An increase in osmolarity will induce an increase in ADH, which results in a retention of free water by the kidney and a hypertonic urine. A decrease in osmolarity will induce a decrease in ADH, which results in an elimination of free water by the kidney and a hypotonic urine. In both instances, the gain or loss of free water serves to normalize plasma osmolarity.
Osmolar Excretion = _ x _
Osmolar Excretion = U x V
The challenge of maintaining plasma and extracellular fluid osmolarity within the normal range (280 – 300 mOsm/L) is met by regulating the excretion of ____, not the excretion of solute.
Osmolar excretion is typically ___ mosmoles/day in a urine
volume of ___ L/day. Thus, typical Uosm = 600/1.5 = 400 mOsm/L.
High water intake: the kidney can reduce Uosm to as low as ___ mOsm/L, approx. 10% of plasma osmolarity. Thus, at a typical osmolar excretion, a maximum urine volume of ___ = 20 L/day or ~ 1 L/hr may be achieved.
Restricted water intake (or excessive water loss): the kidney can increase Uosm to as high as ___ mOsm/L, approx. 400% of plasma osmolarity. Thus, at a typical osmolar excretion, a minimum urine volume ___ = 0.5 L/day or ~ .02 L/hr may be achieved.
The challenge of maintaining plasma and extracellular fluid osmolarity within the normal range (280 – 300 mOsm/L) is met by regulating the excretion of water, not the excretion of solute.
Osmolar excretion is typically 600 mosmoles/day in a urine
volume of 1.5 L/day. Thus, typical Uosm = 600/1.5 = 400 mOsm/L.
High water intake: the kidney can reduce Uosm to as low as 30 mOsm/L, approx. 10% of plasma osmolarity. Thus, at a typical osmolar excretion, a maximum urine volume of 600/30 = 20 L/day or ~ 1 L/hr may be achieved.
Restricted water intake (or excessive water loss): the kidney can increase Uosm to as high as 1200 mOsm/L, approx. 400% of plasma osmolarity. Thus, at a typical osmolar excretion, a minimum urine volume 600/1200 = 0.5 L/day or ~ .02 L/hr may be achieved.
The kidney functions to maintain plasma osmolarity in the normal range by retaining or excreting water in excess of solute, “_____”.
When the kidney excretes a concentrated urine and the urine osmolarity is greater than the plasma osmolarity, the kidney is taking water in _____ of solutes (“free water”) from the tubular fluid and returning it to the plasma. This dilutes the plasma osmolarity and decreases hyperosmolarity and raises plasma volume. This is free water because water does not follow the solute out of the urine & it is reabsorbed.
The ability of the kidney to defend against an increase in plasma osmolarity is dependent upon making a concentrated urine with a _____ osmolarity than plasma. The kidney can increase the osmolarity of the urine up to _____ mOsm/L, an osmolarity much greater than can occur in plasma.
The ability of the kidney to defend against a decrease in plasma osmolarity is dependent upon making a ____ urine with a _____ osmolarity than plasma. The kidney can decrease the osmolarity of the urine to approximately ___ mOsm/L, an osmolarity much less than can occur in
plasma.
The kidney functions to maintain plasma osmolarity in the normal range by retaining or excreting water in excess of solute, “free water”.
When the kidney excretes a concentrated urine and the urine osmolarity is greater than the plasma osmolarity, the kidney is taking water in excess of solutes (“free water”) from the tubular fluid and returning it to the plasma. This dilutes the plasma osmolarity and decreases hyperosmolarity and raises plasma volume. This is free water because water does not follow the solute out of the urine & it is reabsorbed.
The ability of the kidney to defend against an increase in plasma osmolarity is dependent upon making a concentrated urine with a higher osmolarity than plasma. The kidney can increase the osmolarity of the urine up to 1200 mOsm/L, an osmolarity much greater than can occur in plasma.
The ability of the kidney to defend against a decrease in plasma osmolarity is dependent upon making a dilute urine with a lower osmolarity than plasma. The kidney can decrease the osmolarity of the urine to approximately 30 mOsm/L, an osmolarity much less than can occur in
plasma.
Free water clearance quantifies a time-dependent rate function of the kidney to eliminate or retain water, free of solutes, in preserving plasma osmolarity. Urine flow = V, Osmolar Clearance = Cosm.
CH2O = __ - ___, where Cosm = ____
When (U/P)osm = 1, Cosm = V and CH2O = _; the kidney neither adds nor removes free water from the plasma.
When (U/P)osm > 1, Cosm _ V and CH2O _ 0; the kidney is retaining free water to add to the plasma. _____ free water clearance.
When (U/P)osm
Free water clearance quantifies a time-dependent rate function of the kidney to eliminate or retain water, free of solutes, in preserving plasma osmolarity. Urine flow = V, Osmolar Clearance = Cosm.
CH2O = V – Cosm, where Cosm = UV/P
When (U/P)osm = 1, Cosm = V and CH2O = 0; the kidney neither adds nor removes free water from the plasma.
When (U/P)osm > 1, Cosm > V and CH2O 0; the kidney is eliminating free water taken from the plasma. Positive free water clearance.
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet low in solute will _____ negative free water clearance and ____ positive free water clearance. The amount of water “osmotically obligated” to remain in the tubular fluid is _____ and ____ “free water” is available for reabsorption.
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet high in solute will _____ negative free water clearance and _____ positive free water clearance. The amount of water “osmotically obligated” to remain in the tubular fluid is _____ and _____ “free water” is available for reabsorption.
From the range of free water clearances indicated above, we are better adapted to withstand an _____ of water than a deficit (dehydration).
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet low in solute will increase negative free water clearance and decrease positive free water clearance. The amount of water “osmotically obligated” to remain in the tubular fluid is reduced and more “free water” is available for reabsorption.
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet high in solute will decrease negative free water clearance and increase positive free water clearance. The amount of water “osmotically obligated” to remain in the tubular fluid is increased and less “free water” is available for reabsorption.
From the range of free water clearances indicated above, we are better adapted to withstand an excess of water than a deficit (dehydration).
In solute balance, where solute consumption and excretion is 600 mOsmoles/day, the largest volume of urine excreted will have the lowest osmolarity of ___ mOsm/L.
600 mOsmoles/day divided by 30 mOsm/L = 20L/day (____ free water clearance).
In solute balance, where solute consumption and excretion is 600 mOsmoles/day, the smallest volume of urine excreted will have the highest osmolarity of _____ mOsm/L:
600 mOsmoles/day divided by 1200 mOsm/L = 0.5L/day (____ free water clearance)
In solute balance, where solute consumption and excretion is 1200 mOsmoles/day, the largest volume of urine excreted will have the _____ osmolarity of 30 mOsm/L:
1200 mOsmoles/day divided by 30 mOsm/L = 40L/day (_____ free water clearance)
In solute balance, where solute consumption and excretion is 1200 mOsmoles/day, the smallest volume of urine excreted will have the _____ osmolarity of 1200mOsm/L:
1200 mOsmoles/day divided by 1200 mOsm/L = 1.0L/day (_____ free water clearance)
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet low in solute (600 mOsm/day) will _____ negative free water clearance and _____ positive free water clearance.
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet high in solute (1200 mOsm/day) will _____ negative free water clearance and _____ positive free water clearance. Increased solute excretion forces the kidney to excrete more water to eliminate the extra solute in the urine.
In solute balance, where solute consumption and excretion is 600 mOsmoles/day, the largest volume of urine excreted will have the lowest osmolarity of 30 mOsm/L.
600 mOsmoles/day divided by 30 mOsm/L = 20L/day (positive free water clearance).
In solute balance, where solute consumption and excretion is 600 mOsmoles/day, the smallest volume of urine excreted will have the highest osmolarity of 1200 mOsm/L:
600 mOsmoles/day divided by 1200 mOsm/L = 0.5L/day (negative free water clearance)
In solute balance, where solute consumption and excretion is 1200 mOsmoles/day, the largest volume of urine excreted will have the lowest osmolarity of 30 mOsm/L:
1200 mOsmoles/day divided by 30 mOsm/L = 40L/day (positive free water clearance)
In solute balance, where solute consumption and excretion is 1200 mOsmoles/day, the smallest volume of urine excreted will have the highest osmolarity of 1200mOsm/L:
1200 mOsmoles/day divided by 1200 mOsm/L = 1.0L/day (negative free water clearance)
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet low in solute (600 mOsm/day) will increase negative free water clearance and decrease positive free water clearance.
When in solute balance, where urinary solute excretion equals solute consumption, consuming a diet high in solute (1200 mOsm/day) will decrease negative free water clearance and increase positive free water clearance. Increased solute excretion forces the kidney to excrete more water to eliminate the extra solute in the urine.
DILUTE URINE: Solute (NaCl) reabsorption, in the _____ of water reabsorption, in tubule segments with low water permeability (___ and ___ ascending limb of the loop of Henle and ____ tubule), effectively dilutes the tubular fluid to approx. ____ mOsm/L permitting ____ free water clearance and excretion of a dilute, _____-osmotic urine.
CONCENTRATED URINE: Solute (NaCl) reabsorption in the thick ascending limb of the loop of Henle generates a hypertonic interstitial fluid surrounding the collecting tubules in the renal medulla due to countercurrent amplification. Osmotic equilibration of the tubular fluid in the collecting duct with the hypertonic interstitium results in water _____, which concentrates the tubular fluid to maximum of approx. 1200 mOsm/L, permitting _____ free water clearance and excretion of a concentrated, hyper-osmotic urine.
DILUTE URINE: Solute (NaCl) reabsorption, in the absence of water reabsorption, in tubule segments with low water permeability (thin and thick ascending limb of the loop of Henle and distal tubule), effectively dilutes the tubular fluid to approx. 120 mOsm/L permitting positive free water clearance and excretion of a dilute, hypo-osmotic urine.
CONCENTRATED URINE: Solute (NaCl) reabsorption in the thick ascending limb of the loop of Henle generates a hypertonic interstitial fluid surrounding the collecting tubules in the renal medulla due to countercurrent amplification. Osmotic equilibration of the tubular fluid in the collecting duct with the hypertonic interstitium results in water reabsorption, which concentrates the tubular fluid to maximum of approx. 1200 mOsm/L, permitting negative free water clearance and excretion of a concentrated, hyper-osmotic urine.
The kidney lags in time, in its response to excess consumption of water or other dilute beverages, which, upon absorption from the GI tract, dilute the plasma and extravascular fluid osmolarity. The ability of the kidney to compensate and correct for decreases in plasma osmolarity depends upon it making a urine with an osmolarity _____ than the osmolarity of the plasma. The greater the decrease in plasma osmolarity, the greater will be the ____ in urine osmolarity, always less than the plasma osmolarity, to a lower limit of ___ mOsm/L. This plasma to urine difference in osmolarity reflects the ability of the kidney to remove water in excess of solutes (free water) from plasma and excrete it in the urine.
When rapidly consuming a liter of bottled water and adding it to the plasma, the most effective way to correct for the resulting decrease in plasma osmolarity is to take the volume of water consumed from the plasma and ____ in the urine. This is how the kidney corrects a decrease in plasma osmolarity, by effectively removing the volume of water consumed from the plasma, which has the effect of ____ the plasma osmolarity toward the normal range. The evidence for this process is the appearance and measurement of a larger volume of a ____ urine.
The kidney does ___ correct a decrease in plasma osmolarity by increasing the reabsorption of solute in excess of water from the tubular fluid. While this would correct the decrease in plasma osmolarity, it would have the effect of increasing plasma volume and cause isotonic fluid retention (hypervolemia), essentially substituting hypervolemia for hypoosmolarity. By excreting water in excess of solute, the kidney corrects hypoosmolarity without a significant change in plasma ____.
The kidney lags in time, in its response to a deficit in consumption of water or consumption of other dilute beverages, which has the effect of concentrating the plasma and increasing extravascular fluid osmolarity. The ability of the kidney to compensate and correct for the increase in plasma osmolarity depends upon it making a urine with an osmolarity ____ than the osmolarity of the plasma. The greater the increase in plasma osmolarity, the ____ will be the increase in urine osmolarity, always greater than the plasma osmolarity, to an upper limit of _____ mOsm/L. This urine to plasma difference in osmolarity reflects the ability of the kidney to return water in excess of solutes (free water) from the tubular fluid to the plasma and ____ the excretion of free water in the urine. This has the effect of correcting, in the short term, the increase in plasma osmolarity. In the long term, dehydration will ultimately be corrected by satisfying _____, the drive to find and consume water.
Dehydration and an increase in plasma osmolarity results from ___ of water from the plasma in the feces, by ___ in the lungs, and by ____ from the skin, without sufficient replacement by water consumed. The most effective way for the kidney to correct for the increase in plasma osmolarity is to replace the water lost by dehydration with water taken from the tubular fluid. This is how the kidney corrects an increase in plasma osmolarity, by effectively increasing the ____ of water in excess of solutes from the tubular fluid to the plasma, which _____ the plasma osmolarity toward the normal range. The evidence for this process is the appearance and measurement of a ____ volume of a concentrated urine. The kidney does not correct an increase in plasma osmolarity by secreting solute in excess of water from the plasma to the tubular fluid or by decreasing solute reabsorption. While these would correct the increase in plasma osmolarity, it would have the effect of decreasing plasma volume and cause isotonic fluid loss (hypovolemia), essentially substituting hypovolemia for hyperosmolarity. By absorbing water in excess of solute, the kidney corrects hyperosmolarity without a significant change in plasma _____.
The kidney lags in time, in its response to excess consumption of water or other dilute beverages, which, upon absorption from the GI tract, dilute the plasma and extravascular fluid osmolarity. The ability of the kidney to compensate and correct for decreases in plasma osmolarity depends upon it making a urine with an osmolarity less than the osmolarity of the plasma. The greater the decrease in plasma osmolarity, the greater will be the decrease in urine osmolarity, always less than the plasma osmolarity, to a lower limit of 30 mOsm/L. This plasma to urine difference in osmolarity reflects the ability of the kidney to remove water in excess of solutes (free water) from plasma and excrete it in the urine.
When rapidly consuming a liter of bottled water and adding it to the plasma, the most effective way to correct for the resulting decrease in plasma osmolarity is to take the volume of water consumed from the plasma and excrete in the urine. This is how the kidney corrects a decrease in plasma osmolarity, by effectively removing the volume of water consumed from the plasma, which has the effect of increasing the plasma osmolarity toward the normal range. The evidence for this process is the appearance and measurement of a larger volume of a dilute urine.
The kidney does not correct a decrease in plasma osmolarity by increasing the reabsorption of solute in excess of water from the tubular fluid. While this would correct the decrease in plasma osmolarity, it would have the effect of increasing plasma volume and cause isotonic fluid retention (hypervolemia), essentially substituting hypervolemia for hypoosmolarity. By excreting water in excess of solute, the kidney corrects hypoosmolarity without a significant change in plasma volume.
The kidney lags in time, in its response to a deficit in consumption of water or consumption of other dilute beverages, which has the effect of concentrating the plasma and increasing extravascular fluid osmolarity. The ability of the kidney to compensate and correct for the increase in plasma osmolarity depends upon it making a urine with an osmolarity greater than the osmolarity of the plasma. The greater the increase in plasma osmolarity, the greater will be the increase in urine osmolarity, always greater than the plasma osmolarity, to an upper limit of 1200 mOsm/L. This urine to plasma difference in osmolarity reflects the ability of the kidney to return water in excess of solutes (free water) from the tubular fluid to the plasma and minimize the excretion of free water in the urine. This has the effect of correcting, in the short term, the increase in plasma osmolarity. In the long term, dehydration will ultimately be corrected by satisfying thirst, the drive to find and consume water.
Dehydration and an increase in plasma osmolarity results from loss of water from the plasma in the feces, by evaporation in the lungs, and by evaporation from the skin, without sufficient replacement by water consumed. The most effective way for the kidney to correct for the increase in plasma osmolarity is to replace the water lost by dehydration with water taken from the tubular fluid. This is how the kidney corrects an increase in plasma osmolarity, by effectively increasing the return of water in excess of solutes from the tubular fluid to the plasma, which decreases the plasma osmolarity toward the normal range. The evidence for this process is the appearance and measurement of a smaller volume of a concentrated urine. The kidney does not correct an increase in plasma osmolarity by secreting solute in excess of water from the plasma to the tubular fluid or by decreasing solute reabsorption. While these would correct the increase in plasma osmolarity, it would have the effect of decreasing plasma volume and cause isotonic fluid loss (hypovolemia), essentially substituting hypovolemia for hyperosmolarity. By absorbing water in excess of solute, the kidney corrects hyperosmolarity without a significant change in plasma volume.
Tubule fluid is ____ in the proximal tubule, ____osmotic during the loop of henle, ____-osmotic at the end of the Loop of Henle, and is more dilute (hypo-osmotic) or more concentrated (hyper-osmotic) at the end of the collecting duct, depending on circulating levels of ____.
Tubule fluid is isosmotic in the proximal tubule, hyperosmotic during the loop of henle, hypo-osmotic at the end of the Loop of Henle, and is more dilute (hypo-osmotic) or more concentrated (hyper-osmotic) at the end of the collecting duct, depending on circulating levels of ADH=AVP.
Antidiuresis refers to _____ water intake (hydropenia) when the kidneys maximally concentrate the urine and excrete a minimal volume of urine because of increased absorption of water in excess of solutes.
Water diuresis refers to ingestion of _____ water when the kidneys maximally dilute the urine and excrete a large volume of urine because absorption of water in excess of solutes does not occur (lower, blue line above).
Although net absorption of water can occur in all segments of the nephron, as the tubular fluid advances from the proximal tubule to the collecting duct, the osmolarity of the tubular fluid does ____ necessarily change, relative to plasma, in all segments of the nephron.
For example, in the proximal tubule, approximately _____% of the glomerular filtrate is reabsorbed and fluid reabsorption occurs isosmotically, _____ a change in the osmolarity of the remaining 33% of tubular fluid, relative to plasma.
In the thin descending limb of the Loop of Henle, ____ water than solute is reabsorbed causing a relative ____ in the tubular fluid osmolarity. In the thin and thick ascending limb of the Loop of Henle _____ solute than water is reabsorbed due to the relative water impermeability of the thin and thick ascending limb of the Loop of Henle. Accordingly, the tubular fluid entering the distal tubule is relatively ______osmotic, with an osmolarity of approximately 100 mOsm/L.
Antidiuresis refers to restricted water intake (hydropenia) when the kidneys maximally concentrate the urine and excrete a minimal volume of urine because of increased absorption of water in excess of solutes.
Water diuresis refers to ingestion of excess water when the kidneys maximally dilute the urine and excrete a large volume of urine because absorption of water in excess of solutes does not occur (lower, blue line above).
Although net absorption of water can occur in all segments of the nephron, as the tubular fluid advances from the proximal tubule to the collecting duct, the osmolarity of the tubular fluid does not necessarily change, relative to plasma, in all segments of the nephron.
For example, in the proximal tubule, approximately 67% of the glomerular filtrate is reabsorbed and fluid reabsorption occurs isosmotically, without a change in the osmolarity of the remaining 33% of tubular fluid, relative to plasma.
In the thin descending limb of the Loop of Henle, more water than solute is reabsorbed causing a relative increase in the tubular fluid osmolarity. In the thin and thick ascending limb of the Loop of Henle more solute than water is reabsorbed due to the relative water impermeability of the thin and thick ascending limb of the Loop of Henle. Accordingly, the tubular fluid entering the distal tubule is relatively hypoosmotic, with an osmolarity of approximately 100 mOsm/L.
Importantly, the tubular fluid entering the distal tubule is relatively ____osmotic, regardless of whether the plasma osmolarity is above or below the normal range due to dehydration or excess consumption of water (overhydration) and also regardless of the osmolarity of the urine.
The urine osmolarity, whether dilute or concentrated, depends on whether or not water is reabsorbed in the initial and ____ collecting tubules (ICT and CCT) and the outer and inner medullary ____ ducts (OMCD and IMCD). Importantly, Arginine VasoPressin (AVP), otherwise known as AntiDiuretic Hormone (ADH), regulates the variable water permeability in these four distal nephron segments and, therefore, regulates the variable reabsorption of water in excess of solutes in these nephron segments resulting in a smaller volume of a more _____ urine with higher osmolarity or a larger volume of more dilute urine with lower osmolarity.
The important difference in making a more or less concentrated urine is the presence or absence of circulating levels of ADH/AVP, which increases the water permeability of the distal nephron and permits negative free water clearance, _____ of water in excess of solutes. In the absence of ADH/AVP, the distal nephron has a very ____ water permeability, which prevents negative free water clearance and allows ____ free water clearance, the excretion of water in excess of solutes.
Importantly, the tubular fluid entering the distal tubule is relatively hypoosmotic, regardless of whether the plasma osmolarity is above or below the normal range due to dehydration or excess consumption of water (overhydration) and also regardless of the osmolarity of the urine.
The urine osmolarity, whether dilute or concentrated, depends on whether or not water is reabsorbed in the initial and cortical collecting tubules (ICT and CCT) and the outer and inner medullary collecting ducts (OMCD and IMCD). Importantly, Arginine VasoPressin (AVP), otherwise known as AntiDiuretic Hormone (ADH), regulates the variable water permeability in these four distal nephron segments and, therefore, regulates the variable reabsorption of water in excess of solutes in these nephron segments resulting in a smaller volume of a more concentrated urine with higher osmolarity or a larger volume of more dilute urine with lower osmolarity.
The important difference in making a more or less concentrated urine is the presence or absence of circulating levels of ADH/AVP, which increases the water permeability of the distal nephron and permits negative free water clearance, reabsorption of water in excess of solutes. In the absence of ADH/AVP, the distal nephron has a very low water permeability, which prevents negative free water clearance and allows positive free water clearance, the excretion of water in excess of solutes.
In summary, a hypoosmotic tubular fluid is generated in the distal nephron, regardless of whether the plasma is hyperosmotic or hypoosmotic. A hypertonic/hyperosmotic medullary interstitium is generated surrounding the medullary nephron segments, regardless of whether the plasma is hyperosmotic or hypoosmotic. In effect, the kidney is ready and waiting for the “command” to excrete more or less water in excess of solute (free water) to reverse plasma hypo- and hyperosmarity. The “command” occurs in the form of more or less ADH/AVP, which opens or closes the “door” of collecting duct water permeability, allowing return, or not allowing return, of water in excess of solutes from the hypoosmotic tubular fluid to the plasma. Allowing return of water in excess of solutes (free water) to the plasma mitigates plasma hyperosmolarity. Not allowing return of water in excess of solutes (free water) to the plasma mitigates plasma hypoosmolarity.
