Counter Current Mechanism
Learning Objectives:
1) Give approximate values for urine osmolality during water diuresis and during maximal anti-diuresis (dehydration)
a. Value of urine osmolality during water diuresis = <290 mOsm/L (which is plasma osmolality)
b. Value of urine osmolality during water anti-diuresis (dehydration) = 500-700 mOsm/L
i. Anti-diuresis – urine osmolality is usually greater than plasma osmolality reflecting the activity of
ADH, and free water clearance is negative
2) Explain the role of the THICK ascending limb and the collecting duct in the formation of DILUTE urine
a. The separation of solutes occurs primarily in the thick ascending limb of the loop of Henle, where NaCl is
reabsorbed by water is not, due to the low permeability of this segment to water
3) Compare the interstitial fluid osmolality of the renal cortex to that of the renal medulla during antidiuresis and
during water diuresis
a. During anti-diuresis: when ADH is present, urine in the collecting duct in the medulla equilibrates
osmotically with the hyperosmotic interstitium
i. Thus, cortex osmolality is more than medulla osmolality
b. During water diuresis: NaCl reabsorption in collecting duct continues in the renal medulla to further dilute
the urine
i. Thus, cortex osmolality is less than medulla osmolality
4) Explain how NaCl reabsorption by the thick ascending limb of the loop establishes the medullary interstitial hyperosmolality
a. As NaCl is reabsorbed from the thick ascending limb by the Na+, K+, 2Cl- cotransport it creates a gradient in
the interstitium
i. 200 mOsm/kg is the max at one time
b. Urine in the descending limb no equilibrates osmotically with the interstitium
i. Water leaves
c. Flow of urine now moves hyperosmotic urine into the ascending limb and the NaCl transport creates
another gradient
d. Urine in the descending limb equilibrates osmotically with interstitium
e. Again, flow of urine most hyperosmotic urine into the ascending limb
f. Again, NaCl transport creates a gradient
g. The loop configuration creates a counter-current multiplier for the effect of the Na+ pump to create the
cortico-medullary gradient (300-1200 mOsm/Kg)
i. So when ADH is present, urine in the collecting duct can equilibrate osmotically with the
hyperosmotic interstitium
5) Explain how the vasa recta act as a countercurrent exchanger in the renal medulla
a. Vasa recta are countercurrent exchangers
i. Water diffuses along the osmotic gradient; NaCl diffuses along its concentration gradient
1. Descending vasa recta = 300 mOsm/kg
2. Ascending vasa recta = 325 mOsm/kg
ii. Urine osmolality is inversely related to medullary (vasa recta) blood flow
1. The faster the blood flows, the less time for equilibration
iii. Vasa recta act to preserve the cortico-medullary osmotic gradient
6) Compare the volume of water reabsorbed in the cortical versus the medullary collecting duct, when ADH is present
a. Cortical water reabsorption with ADH =
b. Medullary water reabsorption with ADH =
7) Compare the osmolality of the final urine to the osmolality of the renal medullary interstitium when ADH is present
a. When ADH is present the osmolality of the final urine is similar to the medullary interstitium
8) Explain the difference between a water diuresis and an osmotic diuresis
a. Osmotic diuresis – an increase in urine flow accompanied by an increase in osmolar clearance
i. The osmolar clearance is defined by the plasma osmolality:
1. Cosm = (Uosm x V) / (Posm)
ii. Why would osmolar clearance change?
1. For example, in diabetes mellitus, an osmotic diuresis occurs because the capacity of the
kidney to reabsorb glucose is exceeded, glucose is excreted along with osmotically held
water  Cosm increases
b. Water diuresis 9) Give the formulas for calculating the osmolar and free water clearances and describe the mechanism by which a
positive or negative free water clearance is produced by the kidney
a. Cosm = (Uosm x V) / (Posm)
b. Free water clearance – Ch2o, ml/min, is calculated as the difference between V, the urine flow, and the
osmolar clearance
i. Ch2o = V – (Uosm x Vosm) / (Posm)
ii. For example, if V = 5 ml/min, Uosm = 145 mOsm/L, Posm = 290 mOsm/L (normal plasma
osmolality) then  Ch2o = 5ml/min – (145 x 5) / (290) = +2.5 ml/min
1. In this example, the urine osmolality is ½ the plasma osmolality, so dilute urine is produced
c. The free water clearance will always be positive when the urine is more dilute than the plasma, indicating
decreased activity of ADH
d. The negative free water clearance represents the amount of water reabsorbed from tubular fluid that has
an osmolality equal to plasma osmolality, in order to produce the final hyperosmotic urine
10) Describe the relationship between plasma [ADH] and free water clearance
a. ADH increases maximal urine osmolality by:
i. Increasing cortical and medullary collecting duct permeability to water
ii. Permitting increased reabsorption of water in the cortical collecting duct, which decreases urine
flow through the medullary collecting ducts allowing more time for water & urea to equilibrate with
the interstitium
iii. Decreasing blood flow in the vasa recta allowing more time for osmotic equilibration with the
interstitium
iv. Stimulating NaCl transport in the thick ascending limb
11) Describe the effect on urine osmolality and free water clearance of a drug that inhibits NaCl reabsorption in the
thick ascending limb of the loop of Henle
a. If NaCl reabsorption in the thick ascending limb is stopped by damage to the renal medulla or
administration of loop diuretics, the capacity to concentrate urine diminishes
b. Free water clearance moves to ZERO (urine osmolality approaches the value for plasma osmolality)
12) Describe the normal relationship between the state of water balance, the free water clearance, and the plasma
[ADH]
a. Positive free water clearance = diluted urine (lots of free water)
b. Negative free water clearance = ADH allows for concentrated urine (usually from 500-700 mOsm/L)