The kidney concentrates urine by maintaining a gradient of increasing osmolality from the cortex to the medulla.
This gradient is produced by the loops of Henle acting as countercurrent multipliers.
The gradient is maintained by the vasa recta acting as countercurrent exchangers.
A countercurrent system is one in which fluid flows in opposite directions through two closely placed parallel pathways.
Both the loops of Henle and the vasa recta form countercurrent systems in the renal medulla.
The loop of Henle acts as a countercurrent multiplier because:
- The thin descending limb is highly permeable to water through Aquaporin-1.
- The thick ascending limb actively transports Na⁺ and Cl⁻ into the interstitium.
- Tubular fluid continuously enters from the proximal tubule and leaves through the distal tubule.
Initially, the descending limb, ascending limb, and medullary interstitium all have an osmolality of about 300 mOsm/kg H₂O.
The thick ascending limb actively pumps about 100 mOsm/kg of Na⁺ and Cl⁻ into the medullary interstitium.
This increases the interstitial osmolality to about 400 mOsm/kg H₂O.
Water then leaves the thin descending limb until its fluid reaches the same osmolality as the surrounding interstitium.
New tubular fluid with an osmolality of 300 mOsm/kg H₂O continuously enters from the proximal tubule.
This allows more Na⁺ and Cl⁻ to be pumped into the interstitium.
Hypotonic fluid moves into the distal tubule.
Isotonic and then hypertonic fluid enters the thick ascending limb.
This process repeats continuously.
As a result, a gradual increase in osmolality develops from the top to the bottom of the loop of Henle.
Juxtamedullary nephrons have longer loops of Henle than cortical nephrons.
Their thin ascending limbs are relatively impermeable to water but permeable to Na⁺ and Cl⁻.
Na⁺ and Cl⁻ passively diffuse into the medullary interstitium.
This produces additional passive countercurrent multiplication.
The longer the loop of Henle, the greater the osmolality that can be produced at the tip of the medulla.
The high medullary osmolality is preserved by the vasa recta.
The vasa recta act as countercurrent exchangers.
Na⁺ and urea diffuse from blood leaving the medulla into blood entering the medulla.
Water diffuses from the descending vessels into the ascending vessels.
As a result, solutes remain within the medulla while water is carried away into the circulation.
The vasa recta also remove the water reabsorbed from the collecting ducts.
Countercurrent exchange is a passive process.
It cannot maintain the medullary osmotic gradient if countercurrent multiplication by the loop of Henle stops.
The countercurrent system spreads a large osmotic gradient over a long length of tubules (about 1 cm or more), rather than across a very thin cell layer.
Countercurrent exchange also occurs in other parts of the body.
For example, arteries and accompanying veins in the limbs exchange heat, helping conserve body heat, especially in mammals living in cold environments.

Figure: Figure 37–3, Figure 37–15, Figure 37–16

KEY CONCEPT

The countercurrent mechanism allows the kidney to concentrate urine. The loop of Henle acts as a countercurrent multiplier by actively transporting Na⁺ and Cl⁻ while allowing water to leave the descending limb, creating a medullary osmotic gradient. The vasa recta act as countercurrent exchangers, preserving this gradient by retaining Na⁺ and urea in the medulla while removing reabsorbed water. Longer loops of Henle produce a greater medullary osmolality and increase the kidney’s ability to concentrate urine.

Operation of the Loop of Henle as a Countercurrent Multiplier (Figure 37-15)

Easiest & Most Conceptual Summary for self learners

This is one of the most important kidney physiology figures.

It explains how the Loop of Henle creates the high osmolarity (hyperosmotic medulla) that allows the kidney to produce concentrated urine.

⭐ One-Line Concept

The Thick Ascending Limb pumps salt out, the Descending Limb loses water, and continuous fluid flow multiplies a small difference into a large medullary osmotic gradient.

Before Understanding the Figure

Imagine a U-shaped tube buried inside a sponge.

Left side = Thin Descending Limb (TDL)
Middle = Medullary Interstitium (MI)
Right side = Thick Ascending Limb (TAL)

Tubular Fluid
      ↓
 Thin Descending Limb
        │
        │
      Bottom
        │
        │
 Thick Ascending Limb
      ↑
Tubular Fluid Leaves

Very Important Properties

Descending Limb (TDL)

✅ Water can leave easily

❌ Salt cannot leave easily

Remember:

Descending = Water Leaves

Thick Ascending Limb (TAL)

❌ Water cannot leave

✅ Salt is actively pumped out

Remember:

Ascending = Salt Leaves

Main Characters

Part	Main Job
Descending limb	Loses water
Ascending limb	Pumps NaCl
Interstitium	Becomes salty

Now Let’s Understand Every Panel

A. Starting Situation

The figure begins with:

300
300
300

Everywhere is 300 mOsm/kg.

Descending limb = 300
Interstitium = 300
Ascending limb = 300

Everything is equal.

Easy Concept

Imagine three glasses of water.

All contain exactly the same amount of salt.

No movement occurs.

B. Salt Pump Starts Working

Now look at Panel B.

The Thick Ascending Limb pumps NaCl into the interstitium.

The figure shows:

Interstitium

Ascending limb

What happened?

Salt moved:

Ascending Limb
      ↓
Interstitium

Result

Interstitium becomes:

400

Ascending limb becomes:

200

Why?

Because:

Salt leaves

Water cannot follow.

So the ascending limb becomes dilute.

Easy Concept

Imagine removing all the salt from soup,

but leaving the water.

The soup becomes dilute.

C. Descending Limb Responds

Now look at Panel C.

Descending limb becomes:

Why?

The surrounding interstitium is now very salty.

Water leaves the descending limb.

Salt stays inside.

Therefore,

the descending limb becomes concentrated.

Easy Concept

Imagine grapes drying in the sun.

Water leaves.

Sugar stays.

The grapes become sweeter.

D. New Fluid Keeps Flowing

Fresh filtrate (300 mOsm) enters from above.

Older concentrated fluid moves downward.

Result

The bottom becomes:

while the top remains closer to 350.

Why?

Continuous flow pushes concentrated fluid deeper into the medulla.

Easy Concept

Imagine adding fresh water to the top of a river.

Older water is pushed further downstream.

E. Ascending Limb Pumps Again

Now the TAL again pumps salt.

Notice:

Upper ascending limb becomes:

Lower part remains more concentrated.

Result

Another 200 mOsm difference develops.

Easy Concept

The salt pump keeps working every moment.

F. Descending Limb Again Loses Water

Now the descending limb again equilibrates.

Numbers become:

325
425
600

Why?

Water continues leaving wherever the surrounding interstitium is saltier.

Easy Concept

Each time salt is pumped,

water follows in the descending limb.

G. Continuous Flow Again

Fresh fluid enters.

Old fluid moves down.

Gradient becomes larger.

Now bottom reaches:

Easy Concept

The concentrated fluid is pushed deeper into the loop.

H. Final Gradient

Now the figure reaches:

Top

Bottom

Notice:

The medulla has become much more concentrated.

This is the Countercurrent Multiplier

A small difference

↓

Repeated again and again

↓

Becomes a huge gradient.

Why Is It Called “Countercurrent”?

Because fluid moves in opposite directions.

Descending Limb
        ↓↓↓↓

Ascending Limb
        ↑↑↑↑

Two opposite streams.

Why Is It Called “Multiplier”?

Initially,

only about 200 mOsm difference exists between:

Interstitium

and

Ascending limb.

But repeated cycles produce:

Top
300

↓

Bottom
700

In humans,

this eventually becomes:

≈ 1200–1400 mOsm/kg in the inner medulla.

Easy Concept

Think of climbing stairs.

Each step is only 20 cm.

But after many steps,

you reach the roof.

Small differences become a large overall gradient.

What Creates the Gradient?

Two things working together:

1. Salt Pump

Ascending Limb
      ↓
Pumps NaCl Out

2. Water Movement

Descending Limb
      ↓
Water Leaves

Together

Salt Out
+
Water Out
+
Continuous Flow
=
Large Osmotic Gradient

What Happens After This Gradient Is Created?

Later,

the Collecting Duct passes through this salty medulla.

When ADH is present,

water leaves the collecting duct.

The urine becomes concentrated.

Without this gradient,

ADH cannot concentrate urine effectively.

Complete Story of the Figure

Everything Starts at 300
        ↓
Ascending Limb Pumps NaCl
        ↓
Interstitium Becomes 400
Ascending Limb Becomes 200
        ↓
Descending Limb Loses Water
        ↓
Descending Limb Becomes 400
        ↓
Fresh Fluid Enters
Old Fluid Moves Down
        ↓
Salt Pump Works Again
        ↓
Water Leaves Again
        ↓
Repeated Hundreds of Times
        ↓
Top ≈300
Bottom ≈700
(≈1200 in Human Kidney)
        ↓
Hyperosmotic Medulla Created
        ↓
ADH Can Produce Concentrated Urine

Easy Waterfall Analogy

Imagine a mountain waterfall.

Every minute:

New water enters from the top.
Water keeps flowing downward.
Workers keep throwing salt onto the ground beside the stream.
The ground becomes saltier and saltier.

Eventually,

the bottom of the mountain becomes extremely salty.

The Loop of Henle works in the same way.

Easy Memory Trick

Remember: “SWF”

S = Salt pumped out (Ascending limb)
W = Water leaves (Descending limb)
F = Flow multiplies the gradient

↓

Countercurrent Multiplier

Important Points from Figure 37-15

The Loop of Henle functions as a countercurrent multiplier to generate a progressively hyperosmotic renal medulla.
Initially, the fluid in the descending limb, interstitium, and ascending limb is approximately 300 mOsm/kg.
The Thick Ascending Limb (TAL) actively pumps NaCl into the medullary interstitium but is impermeable to water, making the tubular fluid progressively dilute.
The increased interstitial osmolarity causes water to leave the Thin Descending Limb (TDL), concentrating the tubular fluid in that segment.
Continuous movement of fresh filtrate into the descending limb and concentrated fluid around the loop repeatedly re-establishes the transverse osmotic difference.
Repeated cycles multiply a small horizontal osmotic difference (about 200 mOsm/kg) into a large vertical corticomedullary osmotic gradient.
In the human kidney, this mechanism ultimately creates an interstitial osmolarity that may reach approximately 1200–1400 mOsm/kg in the deepest medulla.
The hyperosmotic medulla is essential for water reabsorption from the collecting ducts in the presence of ADH, allowing the kidney to produce concentrated urine.

KEY CONCEPT (Figure 37-15)

Figure 37-15 demonstrates how the Loop of Henle acts as a countercurrent multiplier. The Thick Ascending Limb continuously pumps NaCl into the medullary interstitium without permitting water to follow, while the Thin Descending Limb, which is permeable to water but relatively impermeable to salt, loses water until it equilibrates with the increasingly hyperosmotic interstitium. Continuous tubular flow repeatedly re-establishes these small osmotic differences, progressively multiplying them into a large corticomedullary osmotic gradient. This hyperosmotic medulla provides the driving force for ADH-mediated water reabsorption in the collecting ducts, enabling the kidneys to produce concentrated urine.

Figure 37-16: Operation of the Vasa Recta as Countercurrent Exchangers

Easiest & Most Conceptual Explanation for self learners

This figure explains how the vasa recta preserves (does not create) the high osmolarity of the renal medulla.

Loop of Henle = Creates the medullary osmotic gradient (Countercurrent Multiplier).
Vasa Recta = Preserves the gradient (Countercurrent Exchanger).

Golden Rule

The Loop of Henle makes the salty medulla.
The Vasa Recta protects the salty medulla.

First, What is the Vasa Recta?

The vasa recta are long, U-shaped capillaries that run parallel to the Loop of Henle.

Their job is not to create concentration.

Their job is to:

supply oxygen and nutrients to the medulla,
remove excess water,
prevent the medullary salt from being washed away.

Why is the Vasa Recta Needed?

Imagine the medulla is like a deep salt lake.

The Loop of Henle spends a lot of energy making this lake salty.

If ordinary blood vessels flowed straight through:

fresh blood would carry away all the salt,
the kidney would lose its osmotic gradient,
concentrated urine could never be formed.

So the kidney uses a special U-shaped blood vessel called the vasa recta.

Understand the Figure First

The figure shows one U-shaped blood vessel.

          Cortex
            │

Descending Vasa Recta
        ↓↓↓↓↓↓

Outer Medulla

Inner Medulla

Bottom (1200)

Ascending Vasa Recta
        ↑↑↑↑↑

Back to Cortex

Osmolarity in the Figure

Notice the numbers.

At the top

Blood enters:

Normal plasma osmolarity.

Going Down

Blood becomes:

Coming Up

Blood becomes

Eventually leaving the kidney at about:

Almost the same as normal blood.

Why Does Blood Become More Concentrated While Descending?

Look at the arrows.

The figure shows:

Water
      →

NaCl
←

Urea
←

Meaning:

Water leaves the blood.

Salt enters the blood.

Urea enters the blood.

Step 1

Blood enters at

The surrounding medulla is

The medulla is much saltier.

What happens?

Water leaves blood.

Salt enters blood.

Urea enters blood.

Therefore,

blood osmolarity rises.

Easy Concept

Imagine walking into a desert.

You lose water.

Everything left behind becomes concentrated.

Exactly the same thing happens in descending vasa recta.

At the Bottom of the Loop

Now blood reaches

It is now almost equal to the surrounding medulla.

So there is almost no net movement.

Blood Turns Upward

Now blood starts ascending.

The surrounding medulla becomes progressively less concentrated.

What happens now?

Exactly the opposite.

Water enters blood.

Salt leaves blood.

Urea leaves blood.

The figure shows

H2O
←

NaCl
→

Urea
→

Easy Concept

The blood is now “too salty.”

It gives salt back to the medulla.

It receives water.

Blood Leaves the Kidney

Finally blood leaves as

Almost normal plasma.

Why Doesn’t the Vasa Recta Wash Away the Salt?

This is the most important point.

When blood goes down,

it picks up salt.

When blood comes up,

it gives almost all that salt back.

So,

very little salt actually leaves the medulla.

Easy Analogy

Imagine carrying sand in a bucket.

You pick up sand going downhill.

You drop almost the same sand while coming uphill.

When you return,

almost no sand has been removed.

Exactly what the vasa recta does.

Why is it Called a Countercurrent Exchanger?

Because blood flows in opposite directions.

Descending
↓↓↓↓↓

Ascending
↑↑↑↑↑

Substances exchange between the two limbs.

Nothing is wasted.

Compare Loop of Henle and Vasa Recta

Loop of Henle	Vasa Recta
Tubular fluid	Blood
Creates gradient	Preserves gradient
Countercurrent multiplier	Countercurrent exchanger
Pumps NaCl	Exchanges NaCl and water
Forms hyperosmotic medulla	Prevents washout of medulla

Understanding Every Number in the Figure

Top

Blood entering

Normal plasma.

Descending Limb

Reason:

Water leaves
Salt enters
Urea enters

Blood becomes concentrated.

Bottom

Blood equals surrounding interstitium.

No major diffusion.

Ascending Limb

Reason:

Water enters
Salt leaves
Urea leaves

Blood becomes dilute again.

What Happens to Water?

Descending

Water leaves blood.

Blood
↓

Water Out

Ascending

Water enters blood.

Blood
↑

Water In

What Happens to Salt?

Descending

Salt enters blood.

Interstitium

↓

Blood

Ascending

Salt leaves blood.

Blood

↓

Interstitium

What Happens to Urea?

Exactly like salt.

Descending

Urea enters blood.

Ascending

Urea returns to medulla.

One Complete Cycle

Blood Enters Cortex
300
      ↓
Water Leaves
NaCl Enters
Urea Enters
      ↓
Blood Becomes 425
      ↓
725
      ↓
1200
      ↓
Turns Upward
      ↓
Water Enters
NaCl Leaves
Urea Leaves
      ↓
775
      ↓
475
      ↓
325
      ↓
Returns to Circulation

Everyday Analogy

Imagine a sponge soaked with salty water.

A sponge must stay salty.

Now imagine a pipe passing through it.

If water flows quickly,

the sponge loses all its salt.

Instead,

the pipe is folded into a U.

As water goes down,

it absorbs salt.

As it comes back,

it releases the same salt.

The sponge remains salty.

That sponge is the renal medulla.

That pipe is the vasa recta.

High-Yield Exam Points

Vasa recta act as countercurrent exchangers, not multipliers.
They preserve the corticomedullary osmotic gradient created by the Loop of Henle.
In the descending vasa recta, water diffuses out, while NaCl and urea diffuse into the blood, increasing blood osmolarity.
At the hairpin turn, blood osmolarity approaches that of the surrounding deepest medulla (about 1200 mOsm/kg).
In the ascending vasa recta, water diffuses into the blood, while NaCl and urea diffuse back into the medullary interstitium, decreasing blood osmolarity.
Blood leaves the medulla only slightly more concentrated than when it entered (about 325 mOsm/kg), carrying away excess water but retaining most of the medullary solute.
The slow blood flow through the vasa recta minimizes solute washout while still supplying oxygen and nutrients to the renal medulla.

KEY CONCEPT (Figure 37-16)

Figure 37-16 illustrates the vasa recta functioning as countercurrent exchangers. As blood descends into the increasingly hyperosmotic medulla, water leaves the blood while NaCl and urea enter, causing blood osmolarity to rise progressively. At the hairpin bend, the blood nearly equilibrates with the surrounding medulla. As blood ascends toward the cortex, the opposite exchange occurs: water enters the blood while NaCl and urea diffuse back into the medullary interstitium. Because nearly all of the solute gained during descent is returned during ascent, the vasa recta prevent washout of the medullary osmotic gradient, while simultaneously removing excess water reabsorbed from the nephron. This preservation of the hyperosmotic medulla is essential for ADH-dependent urine concentration.

ROLE OF UREA

Urea helps create the osmotic gradient in the medullary pyramids.
This osmotic gradient helps the kidneys produce concentrated urine.
Urea moves through cell membranes by urea transporters.
This transport occurs mainly by facilitated diffusion.
The kidneys contain at least four UT-A urea transporters:
- UT-A1
- UT-A2
- UT-A3
- UT-A4
UT-B is present in:
- Red blood cells (erythrocytes)
- Descending limbs of the vasa recta
Urea transport in the collecting ducts is mainly carried out by UT-A1 and UT-A3.
Both UT-A1 and UT-A3 are regulated by vasopressin (ADH).
During antidiuresis (high vasopressin levels), more urea is deposited in the medullary interstitium.
This increases the osmotic gradient in the medulla.
As a result, the kidney can produce more concentrated urine.
The amount of urea in the medullary interstitium depends on the amount of urea filtered by the kidneys.
The amount of filtered urea depends on dietary protein intake.
A high-protein diet increases urea production and improves the kidney’s ability to concentrate urine.
A low-protein diet decreases urea production and reduces the kidney’s ability to concentrate urine.

KEY CONCEPT

Urea is an important contributor to the medullary osmotic gradient that allows the kidneys to produce concentrated urine. Urea transport is mainly mediated by UT-A1 and UT-A3 in the collecting ducts under the influence of vasopressin. High vasopressin and a high-protein diet increase urea accumulation in the medulla, enhancing urine-concentrating ability, whereas a low-protein diet reduces this ability.

OSMOTIC DIURESIS

Osmotic diuresis is an increase in urine volume caused by large amounts of unreabsorbed solutes in the renal tubules.
Unreabsorbed solutes remain inside the tubules and attract water.
As a result, water stays in the tubular fluid instead of being reabsorbed.
These solutes also reduce the ability of the proximal tubule to reabsorb Na⁺.
Normally, water reabsorption prevents a large Na⁺ concentration gradient from developing.
When unreabsorbed solutes are present, less water is reabsorbed.
The Na⁺ concentration in the tubular fluid falls.
The maximum concentration gradient for Na⁺ reabsorption is reached.
Further Na⁺ reabsorption from the proximal tubule is reduced.
More Na⁺ remains inside the tubular fluid.
Water stays with the Na⁺.
As a result, a much larger volume of isotonic fluid enters the loop of Henle.
Although the Na⁺ concentration is lower, the total amount of Na⁺ reaching the loop of Henle increases.
In the loop of Henle, reabsorption of both water and Na⁺ decreases.
This occurs because medullary hypertonicity decreases.
The reduced medullary hypertonicity is mainly due to decreased reabsorption of Na⁺, K⁺, and Cl⁻ in the ascending limb.
More fluid then reaches the distal tubule.
Because the medullary osmotic gradient is reduced, less water is reabsorbed in the collecting ducts.
The final result is:
- Marked increase in urine volume
- Increased excretion of Na⁺
- Increased excretion of other electrolytes
Osmotic diuresis can be produced by substances that are filtered but not reabsorbed, such as mannitol and related polysaccharides.
It can also occur when naturally occurring substances exceed the kidney’s reabsorptive capacity.
In diabetes mellitus, high blood glucose increases the filtered glucose load.
When the filtered glucose exceeds the transport maximum (TmG), glucose remains in the renal tubules.
The retained glucose causes osmotic diuresis and produces polyuria.
Large infusions of sodium chloride or urea can also cause osmotic diuresis.
Osmotic diuresis is different from water diuresis.
In water diuresis, water reabsorption in the proximal nephron remains normal.
The maximum urine flow during water diuresis is about 16 mL/min.
In osmotic diuresis, urine flow increases because water reabsorption decreases in the proximal tubule and loop of Henle.
Therefore, much larger urine volumes can be produced.
As solute excretion increases, urine osmolality gradually approaches that of plasma, even when vasopressin secretion is maximal.
This happens because an increasing proportion of the urine consists of isotonic fluid from the proximal tubule.
If osmotic diuresis occurs in diabetes insipidus, urine concentration also increases for the same reason.

Figure: Figure 37–17

KEY CONCEPT

Osmotic diuresis occurs when unreabsorbed solutes remain in the renal tubules and hold water inside the tubular fluid. This decreases water and Na⁺ reabsorption, increases the amount of fluid reaching the distal nephron, and produces a large increase in urine volume and electrolyte excretion. Common causes include mannitol, severe hyperglycaemia in diabetes mellitus, and large amounts of sodium chloride or urea. Unlike water diuresis, osmotic diuresis begins with reduced water reabsorption in the proximal nephron and can produce much larger urine volumes.

Figure 37-17 – Relationship Between Urine Flow and Urine Concentration During Osmotic Diuresis (Easy Conceptual Summary for self learners)

This figure explains how urine volume (urine flow) and urine concentration (urine osmolality) change under three different conditions:

Maximum Vasopressin (ADH)
Isosmotic urine
Diabetes insipidus (No ADH action)

The figure has two graphs, and both explain the same physiology from different viewpoints.

Main Concept of the Figure

The kidney always tries to balance water and solute (osmoles).

Think of urine as:

Urine = Water + Dissolved Solutes (Na⁺, Urea, etc.)

The amount of urine produced depends mainly on:

How much solute must be excreted
How much water ADH allows the kidney to reabsorb

So,

More ADH → More water reabsorbed → Less urine → More concentrated urine
Less ADH → Less water reabsorbed → More urine → Dilute urine

TOP GRAPH

Axes

X-axis

Solute load (mOsm/min)

This means:

How many osmoles the kidney must excrete every minute.

More solute load means:

➡️ More waste must leave the body.

Y-axis

Urine flow (mL/min)

This means:

How much urine is produced every minute.

Three different lines are shown.

1. Green Line — Maximum Vasopressin (Maximum ADH)

This is the lowest curve.

What happens?

Even if solute load increases,

urine flow increases only a little.

Why?

ADH makes the collecting duct highly permeable to water.

Most water is reabsorbed back into the blood.

Only a small amount of water leaves as urine.

Example

Suppose the kidney must remove 100 osmoles.

With ADH,

the kidney packs all those osmoles into very little water.

Result:

✔ Small urine volume

✔ Highly concentrated urine

Easy Concept

Imagine putting 10 spoons of sugar into one small glass of water.

Very concentrated.

Very little water.

Exactly what ADH does.

Key Point

Maximum ADH

↓

Maximum water reabsorption

↓

Smallest urine volume

↓

Highest urine concentration

2. Purple Dashed Line — Isosmotic Urine

This is the middle straight line.

What does Isosmotic mean?

Urine has the same osmolality as plasma.

About

300 mOsm/L

What happens?

As solute load increases,

urine flow increases proportionally.

Why?

Every litre of urine always contains about the same concentration of solute.

Therefore,

if twice as much solute must be excreted,

twice as much urine must be produced.

Easy Concept

Imagine every bottle always contains:

300 grams of salt per litre.

If you must remove more salt,

you simply fill more bottles.

Key Point

Isosmotic urine has

constant concentration,

so urine volume rises directly with solute load.

3. Red Line — Diabetes Insipidus

This is the highest line.

What happens?

Even with small solute loads,

urine flow is already very high.

As solute load increases,

urine volume becomes even larger.

Why?

There is no effective ADH.

Collecting ducts cannot reabsorb water.

Large amounts of water are lost.

Example

Suppose the kidney must remove 100 osmoles.

Without ADH,

it cannot concentrate urine.

So,

it must use a huge amount of water.

Result:

Large volume of dilute urine.

Easy Concept

Imagine trying to dissolve one spoon of sugar in a whole bucket of water.

Very dilute.

Very large volume.

Key Point

Diabetes insipidus

↓

No ADH effect

↓

Water cannot be reabsorbed

↓

Very large urine volume

↓

Very dilute urine

Comparison of the Top Graph

Condition	Urine Volume
Maximum ADH	Lowest
Isosmotic	Intermediate
Diabetes insipidus	Highest

BOTTOM GRAPH

This graph looks at the relationship differently.

Instead of plotting solute load,

it compares:

Urine Flow

with

Urine Osmolality

Axes

X-axis

Urine flow

(mL/min)

Y-axis

Urine osmolality

(mOsm/L)

The dashed horizontal line represents:

Isosmotic urine

Approximately

300 mOsm/L

Green Curve — Maximum Vasopressin

What happens?

When urine flow is very low,

urine osmolality is extremely high.

As urine flow increases,

urine becomes less concentrated.

Why?

ADH allows maximum water reabsorption.

Small urine volume.

Same solute in less water.

Very concentrated urine.

As urine volume gradually increases,

more water stays in urine.

Therefore,

urine concentration falls.

Easy Concept

Imagine mixing one spoon of salt with:

100 mL water

Very concentrated.

Now mix the same salt with:

500 mL water

Less concentrated.

Key Point

Low urine flow

↓

High urine osmolality

Purple Dashed Line — Isosmotic Urine

This horizontal line stays constant.

Why?

Because urine concentration remains equal to plasma.

Around

300 mOsm/L

regardless of urine volume.

Key Point

Isosmotic urine always has the same concentration.

Red Line — Diabetes Insipidus

What happens?

Urine flow is always high,

but urine osmolality stays very low.

Why?

Without ADH,

water cannot be reabsorbed.

Urine contains lots of water.

Therefore,

it is always dilute.

Easy Concept

Think of adding one spoon of salt to a swimming pool.

Extremely dilute.

Key Point

Diabetes insipidus produces:

Large urine volume

Very low osmolality

How ADH Changes Urine

Maximum ADH

Water returns to blood.

↓

Urine becomes concentrated.

↓

Urine volume decreases.

No ADH

Water stays inside tubules.

↓

Urine becomes dilute.

↓

Urine volume increases.

Clinical Importance

Diabetes Insipidus

Problem:

ADH is absent or ineffective.

Patient develops:

Massive polyuria
Dilute urine
Polydipsia
Risk of dehydration

SIADH

Opposite condition.

Too much ADH.

Results:

Very little urine
Highly concentrated urine

Osmotic Diuresis

Seen in:

Diabetes mellitus
Mannitol therapy

Extra osmoles remain in tubules.

Water follows these osmoles.

Urine volume increases even if ADH is present.

Quick Comparison Table

Condition	ADH	Urine Volume	Urine Osmolality
Maximum vasopressin	High	Very low	Very high
Isosmotic urine	Moderate	Moderate	~300 mOsm/L
Diabetes insipidus	None	Very high	Very low

Easy Memory Trick

Maximum ADH = “Save Water” 💧

Small urine
Concentrated urine

Isosmotic = “Same as Plasma” ⚖️

About 300 mOsm/L

Diabetes Insipidus = “Dump Water” 🚿

Huge urine volume
Very dilute urine

Flow Diagram

More ADH
     ↓
More water reabsorbed
     ↓
Less urine produced
     ↓
Higher urine osmolality

No ADH
     ↓
Water stays in tubules
     ↓
More urine produced
     ↓
Lower urine osmolality

Higher Solute Load
        ↓
Kidney must excrete more osmoles
        ↓
Urine volume increases
        ↓
Increase depends on ADH level

Key Concept

This figure demonstrates how urine flow and urine concentration are determined by both the solute load and the action of vasopressin (ADH). In the presence of maximum ADH, the collecting ducts become highly permeable to water, allowing extensive water reabsorption. As a result, even when the solute load increases, urine volume remains low and urine becomes highly concentrated. During isosmotic urine formation, urine has approximately the same osmolality as plasma (about 300 mOsm/L), so urine volume increases directly with the amount of solute that must be excreted. In diabetes insipidus, where ADH is absent or ineffective, the collecting ducts cannot reabsorb water efficiently. Consequently, large volumes of dilute urine are produced, even with relatively small solute loads. Thus, ADH is the major hormone that determines whether the kidney conserves water by producing a small volume of concentrated urine or loses water by producing a large volume of dilute urine.

RELATION OF URINE CONCENTRATION TO GFR

The concentration of urine depends partly on the glomerular filtration rate (GFR).
When the flow of fluid through the loops of Henle decreases, the osmotic gradient in the medullary pyramids becomes greater.
A stronger osmotic gradient allows the kidneys to produce more concentrated urine.
During dehydration, GFR decreases.
A lower GFR delivers less fluid to the countercurrent mechanism.
As a result, the flow of fluid through the loops of Henle decreases.
This increases the kidney’s ability to concentrate urine.
Therefore, urine becomes more concentrated when GFR is low.
When GFR is greatly reduced, urine can become highly concentrated even in the absence of vasopressin (ADH).
If one renal artery is narrowed (constricted), the GFR decreases in that kidney.
The kidney with the narrowed renal artery produces hypertonic (concentrated) urine.
The opposite kidney, with a normal GFR, continues to produce hypotonic (dilute) urine.

KEY CONCEPT

A decrease in GFR reduces the flow of fluid through the loops of Henle, which strengthens the medullary osmotic gradient and increases the kidney’s ability to concentrate urine. During dehydration or renal artery constriction, urine becomes more concentrated, and this can occur even without vasopressin.

“FREE WATER CLEARANCE”

Free water clearance (CH₂O) is used to measure whether the kidneys are losing or conserving water.
It is calculated by finding the difference between:
- Urine flow rate (V̇)
- Osmolar clearance (COsm)
Formula: CH₂O = V̇ − COsm
Since osmolar clearance (COsm) is calculated as: COsm = (UOsm × V̇) / POsm
The complete formula becomes: CH₂O = V̇ − (UOsm × V̇) / POsm

Easy Concept of the Formula

V̇ (Urine flow rate) = Total amount of urine produced each minute.
UOsm (Urine osmolality) = Concentration of dissolved particles in urine.
POsm (Plasma osmolality) = Concentration of dissolved particles in plasma.
COsm (Osmolar clearance) = The amount of water needed to remove dissolved particles if the urine had the same concentration as plasma.
CH₂O (Free water clearance) = Extra water that is either removed from the body or conserved by the kidneys.
If CH₂O is positive (+):
- The urine is dilute (hypotonic).
- The kidneys are excreting extra free water.
If CH₂O is negative (−):
- The urine is concentrated (hypertonic).
- The kidneys are conserving water.
Using the values in Table 37–6:
- During maximal antidiuresis:
  - CH₂O = −1.3 mL/min (−1.9 L/day)
  - This means the kidneys are conserving water by producing concentrated urine.
- In the absence of vasopressin (ADH):
  - CH₂O = +14.5 mL/min (20.9 L/day)
  - This means the kidneys are excreting large amounts of free water by producing dilute urine.

Table: Table 37–6

KEY CONCEPT

Free water clearance (CH₂O) measures whether the kidneys are conserving or excreting water. A negative CH₂O means concentrated urine and water conservation, while a positive CH₂O means dilute urine and increased water excretion.

further simplified Formula

$\boxed{C_{H_2O}=\dot V-\frac{U_{Osm}\times \dot V}{P_{Osm}}}$ CH2O=V˙−POsmUOsm×V˙

Easiest Concept

Think of the formula in 2 simple steps.

Step 1: Start with all the urine produced

$\dot V$ V˙

V̇ = Total urine flow per minute
This is the total amount of urine made by the kidneys.

Step 2: Remove the water that is needed to carry the dissolved particles

$\frac{U_{Osm}\times \dot V}{P_{Osm}}$ POsmUOsm×V˙

This part is called Osmolar Clearance (COsm).

It tells us:

“How much water is actually needed to remove all the dissolved solutes (osmoles) from the body?”

Step 3: What remains is “Free Water”

$\boxed{C_{H_2O}= \text{Total Urine} – \text{Water needed for solutes}}$ CH2O=Total Urine−Water needed for solutes

So,

Free Water Clearance = Total urine − Water required to carry dissolved particles

Meaning of Each Symbol

Symbol	Easy Meaning
CH₂O	Free water clearance
V̇	Total urine produced per minute
UOsm	Urine osmolality (concentration of urine)
POsm	Plasma osmolality (concentration of blood plasma)
COsm	Water needed to excrete the dissolved solutes

How to Remember

If CH₂O is Positive (+)

More water is excreted than needed.
Urine is dilute (hypotonic).
Kidney is losing free water.

Example:

Urine = 10 mL/min
Water needed for solutes = 6 mL/min

$CH_2O=10-6=+4$ CH2O=10−6=+4

✅ 4 mL/min of free water is lost.

If CH₂O is Negative (−)

Less water is excreted than needed.
Urine is concentrated (hypertonic).
Kidney is conserving water.

Example:

Urine = 2 mL/min
Water needed for solutes = 4 mL/min

$CH_2O=2-4=-2$ CH2O=2−4=−2

✅ The kidney is retaining 2 mL/min of free water.

One-Line Memory Trick

$\boxed{\textbf{Free Water Clearance = Total Urine − Water Needed for Solutes}}$ Free Water Clearance = Total Urine − Water Needed for Solutes

Positive CH₂O → Water is being excreted → Dilute urine
Negative CH₂O → Water is being conserved → Concentrated urine

Made by Self learning CEO AND FOUNDER Dr sheen