Lect 12: Regulation of Plasma Osmolarity Flashcards
(35 cards)
Body Balance
gain of water = loss of water
Urine excretion is Regulated
How does the kidney preserve normal ECF osmolarity?
it is achieved by the ability of the kidney to maximize (concentrate) or minimize (dilute) the osmolarity of the urine in response to increases and decreases in ECF osmolarity
When the ratio of urine to plasma osmolarity (U/P) = 1
the urine is isotonic and ECF osmolarity is isotonic?
When the ratio U/P osmo > 1
the urine is hyperosmotic and ECF osmolarity is also hypertonic.
==>The kidney compensates for this by excreting a urine that is much more hypertonic than the plasma
When the ratio of U/P osmo < 1
the urine is hypotonic and ECF osmolarity is hypotonic? But the urine is much more hypotonic (dilute) than the plasma
When you are dehydrated and are volume contracted with high ECF osmo, the way that the kidney corrects for this is
excreting a urine that is more conc than plasma (higher osmolarity) …..and vice versaa
The formation and excretion of a hypertonic urine indicates
the kidney is responding to minimize plasma hyperosmolarity by retaining water in excess of solutes or “free water” which is water not osmotically obligated to remain in the tubular fluid due to the presence of solutes.
The formation of a hypotonic urine indicates
the kidney is responding to minimize and correct plasma hypoosmolarity by eliminating water in excess of solutes
increased plasma osmolarity does what to ADH?
it will induce an increase in ADH , which results in retention of free water by the kidney (water pulled out of the fluid and rreturned to the circ) and a hypertonic/concentrate urine.
decreased plasma osmolarity does what to ADH
decreases ADH which results in the elimination of free water by the kidney and a hypotonic urine.
The challenge of the kidney is to maintain plasma and ECF osmolarity within the normal range (280-300) is met by regulating the excretion of
water
Where water intake is high the solutes were excreted in large volumes of dilute urine (600/30 = 20mOsm/L). Lots of water consumed, lots of water excreted.
Normally: Osmolar excretion (E) = Uosm x V
Uosm=600
V = 1.5L
So typically E= 600/1.5L/day ==400mOsm. So our typical urine osmolarity is 400mOsm.
Based on how much you consume this number can go up or down. Let’s say that you increase water intake and your Uosm reduces to 30mOsm. How much would your new osmolar volume (V) be? 600/30 = 20L/day…this is it max that your urine volume can become.
Where water intake is low the solutes were excreted in small volumes of concentrated urine. Little water taken in, little water excreted (600/1200)
Normally: Osmolar excretion (E) = Uosm x V
Uosm=600
V = 1.5L
So typically E= 600/1.5L/day ==400mOsm. So our typical urine osmolarity is 400mOsm.
Based on how much you consume this number can go up or down. Let’s say that you reduce your water intake, the kidney can increase Uosm to as high as 1200mOsm. So therefore your new volume would be 1200/600 = 1.5L/day (Notice that Uosm DOES NOT CHANGE. It was the volume of urine that changed based on your water intake)
Where water intake was high the solutes were excreted in large volume of dilute urine
and where water intake was low, the solute were excreted in a small volume of concentrated urine
A deficit of water consumption concentrates and increases plasma osmolarity
The most effective way to minimize this increase in plasma osmolarity is to minimize the excretion of the water without a change in the excretion of solute.
==>A decrease in solute excretion (increased solute reabsorption) would increase plasma osmolarity.
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 TF and returning it to plasma. This dilutes the plasma osmolarity and decreases hyperosmolarity, raising the volume.
So i response to a hyperosmotic plasma, the kidney has made a concentrated urine with a much higher osmolarity than plasma. It can take it to 1200mOm/L
The formation of a hypotonic urine indicates the kidney is responding to minimize and correct plasma and ECH hypoosmolarity by (resorbing or eliminating???)
eliminating water in excess of solutes
Changes in ADH affects the ECF and plasma osmolarity. An increase in osmolarity will induce an increase in ADH, which results in
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, gain or loss of free water serves to normalize ECF and plasma osmolarity
When the kidney excretes a dilute urine and the urine osmolarity is less than the plasma osmolarity, the kidney is
taking water in excess of solutes FROM the plasma and adding it to the tubular fluid. So the kidney has made a more dilute urine with a lower osmolarity than plasma
Free water clearance,
I understand this…it is a lot of equations so see notes p. 1319
Effects of solute consumption on free water clearance
positive and negative free water clearance is limited by the magnitude of solute consumed.
When you consume a diet low in solute, this will increase NEGATIVE free water clearance. Why?
low solute diet means U/P ratio is >1 because the denominator is smaller than the numerator, so overall the U/P ratio is larger.
plugging in this to Cwater = V- Cosm
makes V minus a very larger number, making the whole thing negative overall. So free water clearance is negative. Consuming a diet low in solute makes you increase negative free water clearance.
This means that the amt of water obligated to remain in the TF fluid is reduced and more water is avalaiblefor reabsorption.
When you consume a diet high in solute, this will increase POSITIVE free water clearance. Why?
high solute diet means U/P ratio is <1 because the denominator is larger than the numerator, so overall the U/P ratio is smaller.
plugging in this to Cwater = V - Cosm
makes V minus a very smaller number, making the whole thing positive overall. So free water clearance is positive. Consuming a diet high in solute makes you increase positive free water clearance.
So basically the amt of free water obligated to remain in the TF is increased and less water is available for reabsorption