Many modern methods are available to assess kidney function.
However, clearance of substances from plasma is still a useful way to measure kidney function.
Clearance helps quantify how effectively the kidneys remove substances from the body.
Clearance can be used to measure:
- Glomerular filtration rate (GFR)
- Renal blood flow
- Tubular reabsorption
- Tubular secretion
Renal clearance = the volume of plasma completely cleared of a substance by the kidneys per unit time.

Clearance Equation

$C_s \times P_s = U_s \times \dot V$ Cs×Ps=Us×V˙

Where:

Cs = Clearance of substance
Ps = Plasma concentration of the substance
Us = Urine concentration of the substance
V̇ = Urine flow rate

Solving for Clearance

Original equation: $C_s \times P_s = U_s \times \dot V$ Cs×Ps=Us×V˙

Divide both sides by Ps: $C_s=\frac{U_s \times \dot V}{P_s}$ Cs=PsUs×V˙

Easy Meaning

Amount entering kidneys through plasma:

$C_s \times P_s$ Cs×Ps

Amount leaving in urine:

$U_s \times \dot V$ Us×V˙

These two amounts are equal.

Example of Clearance Principle

Suppose plasma contains 1 mg of a substance in each mL.
Suppose kidneys excrete 1 mg of that substance each minute in urine.
Then:

$Clearance = 1\; mL/min$ Clearance=1mL/min

This means 1 mL of plasma is completely cleared of the substance every minute.

Clearance Formula Used in Practice

$C_s=\frac{U_s \times \dot V}{P_s}$ Cs=PsUs×V˙

Renal clearance is calculated by dividing the urinary excretion rate by the plasma concentration of the substance.

KEY CONCEPT

Clearance measures how efficiently kidneys remove a substance from plasma.
Clearance = Volume of plasma cleared of a substance per unit time.
Urinary excretion rate:

$U_s \times \dot V$ Us×V˙

Clearance equation:

$C_s=\frac{U_s \times \dot V}{P_s}$ Cs=PsUs×V˙

Clearance is useful for measuring GFR, renal blood flow, tubular reabsorption, and tubular secretion.

USE OF CLEARANCE TO QUANTIFY KIDNEY FUNCTION — EASY FORMULA GUIDE

Think of Clearance = Kidney Cleaning Power

👉 Question every clearance formula asks:
“How much plasma is completely cleaned of this substance each minute?”

1. CLEARANCE RATE (Cs)

Formula

$C_s=\frac{U_s \times \dot V}{P_s}$ Cs=PsUs×V˙

Easy Meaning

$\text{Clearance}=\frac{\text{Amount leaving in urine}}{\text{Amount present in plasma}}$ Clearance=Amount present in plasmaAmount leaving in urine

Example

Given: $U_s=100\;mg/mL$ Us=100mg/mL $\dot V=2\;mL/min$ V˙=2mL/min $P_s=4\;mg/mL$ Ps=4mg/mL

Step 1: $U_s\times \dot V$ Us×V˙ $100\times2=200$ 100×2=200

Step 2: $C_s=\frac{200}{4}$ Cs=4200 $C_s=50\;mL/min$ Cs=50mL/min

Interpretation

Kidneys completely clear 50 mL plasma/min of this substance.

2. GLOMERULAR FILTRATION RATE (GFR)

Formula

$GFR=\frac{U_{Inulin}\times \dot V}{P_{Inulin}}$ GFR=PInulinUInulin×V˙

Easy Meaning

Inulin is:

Freely filtered
Not reabsorbed
Not secreted

Therefore: $\text{Inulin Clearance = GFR}$ Inulin Clearance = GFR

Example

Given: $U_{Inulin}=125\;mg/mL$ UInulin=125mg/mL $\dot V=1\;mL/min$ V˙=1mL/min $P_{Inulin}=1\;mg/mL$ PInulin=1mg/mL $GFR=\frac{125\times1}{1}$ GFR=1125×1 $GFR=125\;mL/min$ GFR=125mL/min

Interpretation

Kidneys filter 125 mL plasma/min

(Normal GFR ≈ 125 mL/min)

3. CLEARANCE RATIO

Formula

$\text{Clearance Ratio} = \frac{C_s}{C_{Inulin}}$ Clearance Ratio=CInulinCs

Example

Substance clearance: $C_s=250$ Cs=250

nulin clearance: $C_{Inulin}=125$ CInulin=125 $\text{Ratio}=\frac{250}{125}=2$ Ratio=125250=2

Interpretation

Ratio > 1

➡ Substance is secreted.

Another Example $\frac{125}{125}=1$ 125125=1

Ratio = 1

➡ Filtered only.

Another Example $\frac{60}{125}=0.48$ 12560=0.48

Ratio < 1

➡ Reabsorbed.

4. EFFECTIVE RENAL PLASMA FLOW (ERPF)

Formula

$ERPF= \frac{U_{PAH}\times \dot V} {P_{PAH}}$ ERPF=PPAHUPAH×V˙

Easy Meaning

PAH is almost completely removed from plasma in one kidney passage.

Therefore: $PAH\ Clearance \approx ERPF$ PAH Clearance≈ERPF

Example

$U_{PAH}=600$ UPAH=600 $\dot V=1$ V˙=1 $P_{PAH}=1$ PPAH=1 $ERPF=\frac{600\times1}{1}$ ERPF=1600×1 $ERPF=600\;mL/min$ ERPF=600mL/min

Interpretation

About 600 mL plasma reaches kidneys each minute

5. RENAL PLASMA FLOW (RPF)

Formula

$RPF= \frac{U_{PAH}\times \dot V} {P_{PAH}-V_{PAH}}$ RPF=PPAH−VPAHUPAH×V˙

Easy Meaning

PAH is not removed 100%.

So correct ERPF using venous PAH.

Example

$U_{PAH}=600$ UPAH=600 $\dot V=1$ V˙=1 $P_{PAH}=1$ PPAH=1 $V_{PAH}=0.1$ VPAH=0.1

Step 1: $1-0.1=0.9$ 1−0.1=0.9

Step 2: $RPF=\frac{600}{0.9}$ RPF=0.9600 $RPF=667\;mL/min$ RPF=667mL/min

Interpretation

Actual plasma flow ≈ 667 mL/min

6. RENAL BLOOD FLOW (RBF)

Formula

$RBF= \frac{RPF} {1-Hematocrit}$ RBF=1−HematocritRPF

Easy Meaning

Plasma is only part of blood.

Need to convert plasma flow into total blood flow.

Example

$RPF=600$ RPF=600 $Hematocrit=0.45$ Hematocrit=0.45

Step 1: $1-0.45=0.55$ 1−0.45=0.55

Step 2: $RBF=\frac{600}{0.55}$ RBF=0.55600 $RBF=1090\;mL/min$ RBF=1090mL/min

Interpretation

Kidneys receive about 1.1 L blood/min

7. EXCRETION RATE

Formula

$\text{Excretion Rate} = U_s\times \dot V$ Excretion Rate=Us×V˙

Example

$U_s=50$ Us=50 $\dot V=2$ V˙=2 $50\times2=100$ 50×2=100 $Excretion=100\;mg/min$ Excretion=100mg/min

Interpretation

100 mg leaves body each minute.

8. REABSORPTION RATE

Formula

$\text{Reabsorption} = Filtered\ Load – Excretion$ Reabsorption=Filtered Load−Excretion

or $= (GFR\times P_s) – (U_s\times \dot V)$ =(GFR×Ps)−(Us×V˙)

Example

$GFR=125$ GFR=125 $P_s=2$ Ps=2

Filtered Load: $125\times2 = 250$ 125×2=250

Excretion: $50$ 50

Reabsorption: $250-50 = 200\;mg/min$ 250−50=200mg/min

Interpretation

200 mg/min returned to blood.

9. SECRETION RATE

Formula

$Secretion = Excretion – Filtered\ Load$ Secretion=Excretion−Filtered Load

Example

Excretion: $300$ 300

Filtered Load: $250$ 250 $300-250 = 50\;mg/min$ 300−250=50mg/min

Interpretation

Kidney tubules added 50 mg/min into urine.

KEY CONCEPT (EXAM MEMORY TRICK)

Compare Clearance with Inulin

Clearance Value	Meaning
= Inulin Clearance	Only filtered
< Inulin Clearance	Reabsorbed
> Inulin Clearance	Secreted
= 0	Completely reabsorbed
= RPF (~600 mL/min)	PAH
= GFR (~125 mL/min)	Inulin

Super Shortcut

Excretion $U\times V$ U×V

Clearance $\frac{U\times V}{P}$ PU×V

Filtered Load $GFR\times P$ GFR×P

Reabsorption $Filtered\ Load-Excretion$ Filtered Load−Excretion

Secretion $Excretion-Filtered\ Load$ Excretion−Filtered Load

Inulin → GFR

PAH → RPF

RPF → RBF using Hematocrit

INULIN CLEARANCE CAN BE USED TO ESTIMATE GLOMERULAR FILTRATION RATE (GFR)

If a substance is filtered freely through the glomerulus like water, it enters the filtrate easily.
If that substance is not reabsorbed by the tubules, none of it returns to the blood.
If that substance is not secreted by the tubules, no extra amount is added to the urine.
Therefore, the amount filtered by the kidneys equals the amount excreted in the urine.

Relationship Between Filtration and Excretion

$GFR \times P_s = U_s \times \dot V$ GFR×Ps=Us×V˙

Where:

GFR = Glomerular filtration rate
Ps = Plasma concentration
Us = Urine concentration
V̇ = Urine flow rate

Solving for GFR

Start with: $GFR \times P_s = U_s \times \dot V$ GFR×Ps=Us×V˙

Divide both sides by Ps $GFR=\frac{U_s \times \dot V}{P_s}$ GFR=PsUs×V˙

Easy Concept

$GFR = Clearance$ GFR=Clearance

for any substance that is:

Why Inulin Is Used

Inulin is a polysaccharide molecule.
Molecular weight ≈ 5200.
It is not produced naturally in the body.
It must be given intravenously.
Inulin is freely filtered by the glomerulus.
Inulin is not reabsorbed.
Inulin is not secreted.

Therefore

$Inulin\ Clearance = GFR$ Inulin Clearance=GFR

Inulin Example

Given

Plasma concentration: $P_{Inulin}=1\;mg/mL$ PInulin=1mg/mL

Urine concentration: $U_{Inulin}=125\;mg/mL$ UInulin=125mg/mL

Urine flow rate: $\dot V=1\;mL/min$ V˙=1mL/min

Step 1: Calculate Excretion Rate

$U_{Inulin}\times \dot V$ UInulin×V˙ $125\times1$ 125×1 $=125\;mg/min$ =125mg/min

Meaning

125 mg of inulin appears in urine every minute.

Step 2: Calculate Inulin Clearance

$C_{Inulin} = \frac{U_{Inulin}\times \dot V} {P_{Inulin}}$ CInulin=PInulinUInulin×V˙

Substitute values: $= \frac{125\times1}{1}$ =1125×1 $=125\;mL/min$ =125mL/min

Interpretation

125 mL of plasma is completely cleared of inulin each minute.
Because inulin is neither reabsorbed nor secreted:

$GFR = 125\;mL/min$ GFR=125mL/min

Visual Memory Trick

Inulin Journey

Blood → Glomerulus → Filtrate → Urine

✅ Freely filtered

❌ No reabsorption

❌ No secretion

Therefore: $Filtered\ Amount = Excreted\ Amount$ Filtered Amount=Excreted Amount

Thus: $Inulin\ Clearance = GFR$ Inulin Clearance=GFR

Other Substances Used to Estimate GFR

Iothalamate
Chromium-EDTA
Cystatin C
Creatinine

KEY CONCEPT

An ideal GFR marker must be:
- Freely filtered
- Not reabsorbed
- Not secreted
Inulin fulfills all these conditions.
Filtration rate equals excretion rate:

$GFR \times P_s = U_s \times \dot V$ GFR×Ps=Us×V˙

GFR equation:

$GFR=\frac{U_s \times \dot V}{P_s}$ GFR=PsUs×V˙

Therefore:

$\boxed{Inulin\ Clearance = GFR}$ Inulin Clearance=GFR

Normal inulin clearance (GFR) ≈

$125\;mL/min$ 125mL/min

This figure explains:

How Inulin Clearance is used to measure Glomerular Filtration Rate (GFR).

Because:

✅ Inulin is freely filtered

❌ Not reabsorbed

❌ Not secreted

Therefore:

Amount Filtered = Amount Excreted

This makes Inulin the perfect substance for measuring GFR.

STEP 1: WHAT IS GFR?

GFR (Glomerular Filtration Rate)

= Amount of filtrate formed by all glomeruli per minute.

Normal:

GFR = 125 mL/min

Meaning:

The kidneys filter about 125 mL of plasma every minute.

STEP 2: WHY IS INULIN SPECIAL?

Most substances are:

Reabsorbed
OR
Secreted

Therefore they cannot accurately measure filtration.

Inulin

After filtration:

❌ Not reabsorbed

❌ Not secreted

❌ Not metabolized

Only filtered.

Therefore

Everything filtered appears in urine.

Filtered = Excreted

This is the key concept.

STEP 3: UNDERSTAND THE EQUATION

Since:

Amount Filtered = Amount Excreted

We can write: $GFR \times P_{inulin} = U_{inulin} \times \dot V$ GFR×Pinulin=Uinulin×V˙

What Does Each Term Mean?

GFR

Amount filtered per minute

(mL/min)

P₍inulin₎

Plasma inulin concentration

(mg/mL)

U₍inulin₎

Urine inulin concentration

(mg/mL)

V̇

Urine flow rate

(mL/min)

EASY MEMORY

Filtered Side

GFR × Plasma Concentration

↓

Amount entering nephron

Excreted Side

Urine Concentration × Urine Flow

↓

Amount leaving body

Since filtered = excreted:

Both are equal.

STEP 4: SOLVING THE EQUATION

Start with: $GFR \times P_{inulin} = U_{inulin}\times \dot V$ GFR×Pinulin=Uinulin×V˙

Move Plasma Inulin to the other side: $GFR= \frac{U_{inulin}\times \dot V} {P_{inulin}}$ GFR=PinulinUinulin×V˙

This is the famous GFR equation.

SOLVING THE FIGURE EXAMPLE

Given

Plasma Inulin: $P_{inulin}=1 \; mg/mL$ Pinulin=1mg/mL

Urine Inulin: $U_{inulin}=125 \; mg/mL$ Uinulin=125mg/mL

Urine Flow: $\dot V=1 \; mL/min$ V˙=1mL/min

Step 1

Write formula $GFR= \frac{U\times \dot V} {P}$ GFR=PU×V˙

Step 2

Insert values $GFR= \frac{125\times1} {1}$ GFR=1125×1

Step 3

Calculate $GFR=125$ GFR=125

ANSWER

GFR = 125 mL/min

✅ Normal GFR

EXAMPLE 2

Suppose:

Plasma Inulin

$P=2\;mg/mL$ P=2mg/mL

Urine Inulin

$U=200\;mg/mL$ U=200mg/mL

Urine Flow

$V=2\;mL/min$ V=2mL/min

Formula

$GFR= \frac{U\times V} {P}$ GFR=PU×V

Put values

$GFR= \frac{200\times2} {2}$ GFR=2200×2 $GFR= \frac{400}{2}$ GFR=2400 $GFR=200$ GFR=200

ANSWER

GFR = 200 mL/min

EXAMPLE 3

Suppose:

Plasma Inulin

$P=4$ P=4

Urine Inulin

$U=100$ U=100

Urine Flow

$V=5$ V=5

Formula

$GFR= \frac{100\times5} {4}$ GFR=4100×5 $GFR= \frac{500}{4}$ GFR=4500 $GFR=125$ GFR=125

ANSWER

GFR = 125 mL/min

SUPER EASY CONCEPTUAL STORY

Imagine:

🩸 Blood contains 1 mg of inulin in each mL

The kidneys filter:

125 mL/min

Therefore:

Amount filtered each minute: $125 \times 1 =125\;mg/min$ 125×1=125mg/min

Because inulin is NOT reabsorbed and NOT secreted:

The same amount must appear in urine.

So: $125\;mg/min$ 125mg/min

is excreted.

This is exactly what the urine data shows: $125\;mg/mL \times 1\;mL/min = 125\;mg/min$ 125mg/mL×1mL/min=125mg/min

Filtered = Excreted

Therefore:

GFR = 125 mL/min

EXAM SHORTCUT

For any substance: $Clearance= \frac{U\times V}{P}$ Clearance=PU×V

For Inulin: $Inulin\ Clearance = GFR$ Inulin Clearance=GFR

Because:

✅ Freely filtered

❌ Not reabsorbed

❌ Not secreted

KEY CONCEPT

Inulin is the ideal marker of GFR because everything that is filtered is excreted unchanged. Therefore, the amount of inulin filtered by the glomerulus equals the amount excreted in urine, and GFR can be calculated using: $\boxed{ GFR= \frac{U_{inulin}\times V} {P_{inulin}} }$ GFR=PinulinUinulin×V

Easy Memory:

“Inulin goes in, gets filtered, and comes out exactly the same—so Inulin Clearance = GFR.” 🎯

CREATININE CLEARANCE AND PLASMA CREATININE CONCENTRATION CAN BE USED TO ESTIMATE GLOMERULAR FILTRATION RATE

Creatinine is a by-product of muscle metabolism.
Creatinine is removed from body fluids mainly by glomerular filtration.
Therefore, creatinine clearance can be used to estimate GFR.
Measuring creatinine clearance does not require intravenous infusion.
For this reason, creatinine clearance is used more commonly than inulin clearance in clinical practice.
Creatinine clearance is not a perfect measure of GFR.
A small amount of creatinine is secreted by the renal tubules.
Therefore, the amount of creatinine excreted is slightly greater than the amount filtered.
There is also a small error that tends to overestimate plasma creatinine concentration.
These two errors tend to offset each other.
Therefore, creatinine clearance provides a reasonably accurate estimate of GFR.
In some patients, urine collection for creatinine clearance measurement may not be practical.
In such cases, changes in GFR can be estimated by measuring plasma creatinine concentration.
Plasma creatinine concentration is inversely related to GFR.

EASY EQUATION

GFR ≈ Creatinine Clearance

$GFR \approx \frac{U_{Cr} \times V}{P_{Cr}}$ GFR≈PCrUCr×V

Where:

UCr = Urine creatinine concentration
V = Urine flow rate
PCr = Plasma creatinine concentration
GFR = Glomerular filtration rate

EASY CONCEPT

↑ Plasma Creatinine = ↓ GFR
↓ Plasma Creatinine = ↑ GFR
If GFR suddenly decreases by 50%, the kidneys filter less creatinine.
The kidneys also excrete less creatinine.
Creatinine begins to accumulate in the body fluids.
Plasma creatinine concentration rises.
Plasma creatinine continues to rise until a new balance is reached.
At the new balance, filtered creatinine equals excreted creatinine.
At the new balance, creatinine production equals creatinine excretion.

EASY MATHEMATICAL RELATIONSHIP

Normal Condition

GFR = 100%
Plasma Creatinine = 1×

If GFR Falls to 50%

GFR = 1/2 of normal
Plasma Creatinine = 2× normal

Easy Concept

$GFR \downarrow 50\% \Rightarrow P_{Cr} \uparrow 2\times$ GFR↓50%⇒PCr↑2×

If GFR Falls to 25%

GFR = 1/4 of normal
Plasma Creatinine = 4× normal

Easy Concept

$GFR \downarrow 75\% \Rightarrow P_{Cr} \uparrow 4\times$ GFR↓75%⇒PCr↑4×

If GFR Falls to 12.5%

GFR = 1/8 of normal
Plasma Creatinine = 8× normal

Easy Concept

$GFR \downarrow 87.5\% \Rightarrow P_{Cr} \uparrow 8\times$ GFR↓87.5%⇒PCr↑8×

MEMORY RULE

GFR	Plasma Creatinine
Normal	1×
1/2 Normal	2×
1/4 Normal	4×
1/8 Normal	8×

Under steady-state conditions, creatinine excretion equals creatinine production.
This balance is maintained even when GFR decreases.
However, maintaining this balance requires a higher plasma creatinine concentration.
Therefore, plasma creatinine rises as GFR falls.

KEY CONCEPT

Creatinine is filtered mainly by the glomerulus.
Creatinine clearance is commonly used to estimate GFR.
GFR ≈ (Urine Creatinine × Urine Flow Rate) ÷ Plasma Creatinine.
Plasma creatinine is inversely related to GFR.
↓ GFR → ↑ Plasma Creatinine.
50% fall in GFR → 2× rise in plasma creatinine.
75% fall in GFR → 4× rise in plasma creatinine.
87.5% fall in GFR → 8× rise in plasma creatinine.
High plasma creatinine usually indicates reduced kidney filtration function.

Figure 28.21

This graph shows what happens when GFR suddenly falls by 50% while the body continues to produce creatinine at the same rate.

1️⃣ TOP GRAPH: GFR (Glomerular Filtration Rate)

Red Line (Top)

100 mL/min  ──────────┐
                      │
                      │ Sudden drop
                      ▼
45-50 mL/min ─────────────────────

What does this mean?

Before Day 1, kidneys filter about 100 mL/min.
On Day 1, GFR suddenly drops to about 50 mL/min.
After that, GFR stays low.

Simple Concept

Think of kidneys as a water filter.

Before:

Filter works at 100% speed.

After:

Filter suddenly works at only 50% speed.

Therefore:

✅ Less creatinine can be removed from blood.

2️⃣ MIDDLE GRAPH: Serum Creatinine Concentration

Red Curved Line

1 mg/dL ────────
                \
                 \
                  \
                   \
                    2 mg/dL

What happens?

Before Day 1:

Serum creatinine = 1 mg/dL

Immediately after GFR falls:

Creatinine removal decreases.
But creatinine production remains normal.

So:

More creatinine enters blood than leaves blood.

Therefore:

Blood creatinine starts accumulating.

Why does the curve rise gradually?

Because creatinine does not instantly double.

Day by day:

Extra creatinine accumulates.
Blood level slowly rises.

So the curve climbs gradually.

Why does it stop near 2 mg/dL?

Because eventually a new equilibrium is reached.

At that point:

Creatinine Excretion = Creatinine Production

Again.

Easy Memory Trick

GFR	Serum Creatinine
100 mL/min	1 mg/dL
50 mL/min	2 mg/dL

Rule

GFR ↓ by 50% → Serum Creatinine doubles

This is a very important clinical concept.

3️⃣ BOTTOM GRAPH: Creatinine Production and Excretion

This graph is the most important.

Solid Horizontal Red Line

“Production”

This upper solid line represents:

Creatinine Production

≈ 1.8 g/day

Muscles keep producing creatinine continuously.

The body does NOT know GFR has fallen.

So production remains constant.

Dotted Red Line

“Excretion”

This dotted line shows:

Creatinine Excretion by kidneys

Remember:

Excretion ≈ GFR × Plasma Creatinine

Before Day 1

Production = Excretion

Produced = 1.8 g/day
Excreted = 1.8 g/day

Balance exists.

No accumulation.

At Day 1

GFR suddenly falls by 50%.

The kidneys can now remove only about half as much creatinine.

Therefore:

Dotted line suddenly drops downward

Excretion
1.8 g/day
      ↓
~1.0 g/day

This sharp vertical fall in the dotted line shows:

👉 Kidney filtering ability suddenly decreased.

Why Does the Dotted Line Rise Again?

After Day 1:

Blood creatinine starts increasing.

Remember:

Excretion = GFR × Plasma Creatinine

Although GFR is low,

Plasma creatinine becomes higher.

So:

Low GFR × Higher Creatinine

gradually increases excretion again.

Therefore the dotted curve rises upward.

Eventually

Serum creatinine becomes about 2 mg/dL.

Now:

50% GFR × 2× Creatinine

gives almost the same excretion as before.

So:

Excretion ≈ 1.8 g/day

again.

The dotted line reaches the production line.

A new steady state is achieved.

🟨 YELLOW Shaded Area (Most Important Part)

The yellow area is labeled:

“Positive Balance”

This area represents:

Creatinine retained inside the body

What is happening in this yellow area?

For a short period:

Production > Excretion

Example:

Produced = 1.8 g/day

Excreted = 1.0 g/day

Difference:

0.8 g/day

stays inside the body.

That retained creatinine accumulates in blood.

This accumulation is shown by the yellow shaded region.

So:

🟨 Yellow Area = Total creatinine retained in the body during adaptation.

Easy Example

Imagine:

A water tank receives:

Inflow = 10 liters/day

But outlet suddenly becomes:

5 liters/day

Then:

5 liters/day

accumulates inside the tank.

The accumulated water is equivalent to the:

🟨 Yellow shaded area.

Meaning of the Dotted Line

Dotted Line = Creatinine Excretion

Starts equal to production.
Suddenly drops when GFR falls.
Gradually rises as serum creatinine increases.
Finally reaches production again.

Complete Story of the Figure

Step 1

Normal state

GFR = 100
Creatinine = 1 mg/dL
Production = Excretion

↓

Step 2

GFR falls by 50%

100 → 50 mL/min

↓

Step 3

Excretion suddenly falls

(Dotted line drops)

↓

Step 4

Production remains unchanged

(Solid line stays flat)

↓

Step 5

Production > Excretion

🟨 Positive balance develops

↓

Step 6

Creatinine accumulates in blood

Serum creatinine rises

↓

Step 7

Higher serum creatinine increases excretion

(Dotted line rises)

↓

Step 8

New steady state

GFR = 50 mL/min
Serum Creatinine ≈ 2 mg/dL
Production = Excretion

KEY CONCEPT

✅ GFR suddenly decreases by 50%.

✅ Creatinine production remains constant.

✅ Creatinine excretion initially falls (dotted line drops).

✅ Production becomes greater than excretion, creating the 🟨 positive balance area.

✅ Retained creatinine accumulates in blood, causing serum creatinine to rise.

✅ As serum creatinine rises, excretion gradually increases again.

✅ A new equilibrium is reached when:

Creatinine Production = Creatinine Excretion

and serum creatinine stabilizes at approximately double the original value (2 mg/dL).

Figure 28.22

This figure shows the relationship between:

X-axis

GFR (Glomerular Filtration Rate)

and

Y-axis

Plasma Creatinine Concentration

The Main Message of This Graph

GFR and Plasma Creatinine have an INVERSE relationship

Meaning:

GFR ↓  → Creatinine ↑

GFR ↑  → Creatinine ↓

Think of Creatinine Like Garbage

Imagine:

Creatinine = Garbage
Kidneys = Garbage removal system
GFR = Speed of garbage collection

Normal Situation

Garbage Produced = Garbage Removed

No accumulation occurs.

Therefore:

Plasma Creatinine ≈ 1 mg/dL
GFR ≈ 125 mL/min

This is the point marked:

“Normal”

on the graph.

Understanding Every Point on the Curve

Point 1 (Normal)

GFR = 125 mL/min
Creatinine = 1 mg/dL

Kidneys work normally.

Creatinine is removed efficiently.

Point 2

GFR ≈ 60 mL/min
Creatinine ≈ 2 mg/dL

Notice:

GFR becomes HALF

125 → 60

But creatinine becomes DOUBLE

1 → 2

IMPORTANT RULE

GFR ↓ by 50%
↓
Plasma Creatinine ↑ 2 times

This is the most tested concept.

Point 3

GFR ≈ 30 mL/min
Creatinine ≈ 4 mg/dL

Again:

GFR halves

60 → 30

Creatinine doubles

2 → 4

Point 4

GFR ≈ 15 mL/min
Creatinine ≈ 8 mg/dL

Again:

GFR halves

30 → 15

Creatinine doubles

4 → 8

Point 5

GFR ≈ 7.5 mL/min
Creatinine ≈ 16 mg/dL

Very severe kidney failure.

Very little filtration occurs.

Creatinine accumulates massively.

Why Is the Curve Not a Straight Line?

Notice:

The curve bends.

It is not:

Instead it is:

shaped.

Reason

Creatinine and GFR are related by: $\text{Plasma Creatinine} \propto \frac{1}{\text{GFR}}$ Plasma Creatinine∝GFR1

or $P_{Cr} = \frac{K}{GFR}$ PCr=GFRK

where K is constant creatinine production.

Easy Formula

If creatinine production remains constant: $GFR \times Plasma\ Creatinine = Constant$ GFR×Plasma Creatinine=Constant

Example

Normal:

125 × 1
=
125

After GFR falls to 60:

60 × 2
=
120

Approximately same.

After GFR falls to 30:

30 × 4
=
120

Still same.

Why Does Creatinine Rise So Much at Low GFR?

Look at the left side of the graph.

A small decrease in GFR causes a huge rise in creatinine.

Example

125 → 100 mL/min

Creatinine:

1 → 1.2 mg/dL

Only small change.

But:

15 → 7.5 mL/min

Creatinine:

8 → 16 mg/dL

Huge change.

Clinical Meaning

In early kidney disease:

Creatinine may look almost normal.

Even though GFR has already fallen significantly.

Why Creatinine Is a Good Marker of Kidney Function

Because:

Low GFR
↓
Less filtration
↓
Less creatinine excretion
↓
Creatinine accumulates
↓
Plasma creatinine rises

Thus:

High Creatinine = Low GFR

Visual Story of the Graph

Imagine a sink.

Normal Kidney

Tap = Creatinine production

Drain = GFR

Drain is wide open.

Water doesn’t accumulate.

Creatinine = 1 mg/dL

GFR Falls

Drain becomes narrow.

Water starts collecting.

Creatinine rises

GFR Falls Further

Drain becomes almost blocked.

Water accumulates rapidly.

Creatinine rises dramatically

Quick Memory Table

GFR (mL/min)	Plasma Creatinine (mg/dL)
125	1
60	2
30	4
15	8
7.5	16

Exam Shortcut

Whenever GFR is halved:

GFR Change	Creatinine Change
125 → 60	1 → 2
60 → 30	2 → 4
30 → 15	4 → 8

Rule:

✅ GFR ↓ 50% → Plasma Creatinine ↑ 100% (doubles)

✅ Plasma Creatinine and GFR are inversely proportional.

KEY CONCEPT

The graph shows an inverse relationship between GFR and plasma creatinine.
Normal GFR ≈ 125 mL/min corresponds to plasma creatinine ≈ 1 mg/dL.
When GFR decreases, creatinine accumulates in blood.
If GFR falls by 50%, plasma creatinine approximately doubles.
Therefore, high plasma creatinine indicates reduced kidney filtration and impaired renal function.

PARA-AMINOHIPPURIC ACID CLEARANCE CAN BE USED TO ESTIMATE RENAL PLASMA FLOW

If a substance is completely removed from the plasma by the kidneys, its clearance equals the renal plasma flow (RPF).
In this situation, the amount of substance delivered to the kidneys equals the amount excreted in the urine.
Therefore, renal plasma flow can be calculated from the clearance of that substance.

EASY EQUATION

$RPF = \frac{U_s \times V}{P_s}$ RPF=PsUs×V

Where:

Us = Urine concentration of the substance
V = Urine flow rate
Ps = Plasma concentration of the substance
RPF = Renal Plasma Flow
GFR is only about 20% of total renal plasma flow.
Therefore, a substance that is completely cleared from plasma must be both filtered and secreted.
No known substance is completely cleared by the kidneys.
Para-aminohippuric acid (PAH) is about 90% cleared from the plasma.
Therefore, PAH clearance can be used to estimate renal plasma flow.
A correction is needed because about 10% of PAH remains in the blood leaving the kidneys.
The percentage of PAH removed from blood is called the extraction ratio of PAH.
The normal extraction ratio of PAH is about 90%.
In kidney disease, the extraction ratio may decrease.
This decrease occurs because damaged tubules cannot secrete PAH efficiently.

EASY CALCULATION OF PAH CLEARANCE

Given:

Plasma PAH concentration (PPAH) = 0.01 mg/mL
Urine PAH concentration (UPAH) = 5.85 mg/mL
Urine Flow Rate (V) = 1 mL/min

Step 1: Calculate PAH Clearance

$PAH\ Clearance = \frac{U_{PAH} \times V}{P_{PAH}}$ PAH Clearance=PPAHUPAH×V

Substitute values: $PAH\ Clearance = \frac{5.85 \times 1}{0.01}$ PAH Clearance=0.015.85×1 $PAH\ Clearance = \frac{5.85}{0.01}$ PAH Clearance=0.015.85 $PAH\ Clearance = 585\ mL/min$ PAH Clearance=585 mL/min

Answer

$PAH\ Clearance = 585\ mL/min$ PAH Clearance=585 mL/min

This value is called Effective Renal Plasma Flow (ERPF).

EASY CALCULATION OF TRUE RPF

Extraction Ratio (EPAH) = 90% = 0.9

Formula

$RPF = \frac{PAH\ Clearance}{Extraction\ Ratio}$ RPF=Extraction RatioPAH Clearance

Substitute values: $RPF = \frac{585}{0.9}$ RPF=0.9585 $RPF = 650\ mL/min$ RPF=650 mL/min

Answer

$RPF = 650\ mL/min$ RPF=650 mL/min

EXTRACTION RATIO FORMULA

$E_{PAH} = \frac{P_{PAH} – V_{PAH}}{P_{PAH}}$ EPAH=PPAHPPAH−VPAH

Where:

PPAH = Renal arterial PAH concentration
VPAH = Renal venous PAH concentration
EPAH = PAH extraction ratio

EASY CONCEPT

If 90% of PAH is removed:

$E_{PAH}=0.9$ EPAH=0.9

If 100 PAH molecules enter the kidney:
- 90 are removed
- 10 remain in renal venous blood

CALCULATION OF RENAL BLOOD FLOW (RBF)

Given:

RPF = 650 mL/min
Hematocrit = 0.45

Formula

$RBF = \frac{RPF}{1-Hematocrit}$ RBF=1−HematocritRPF

Substitute values: $RBF = \frac{650}{1-0.45}$ RBF=1−0.45650 $RBF = \frac{650}{0.55}$ RBF=0.55650 $RBF = 1182\ mL/min$ RBF=1182 mL/min

Answer

$RBF = 1182\ mL/min$ RBF=1182 mL/min

KEY CONCEPT

PAH is filtered and actively secreted by the renal tubules.
About 90% of PAH is removed from plasma in one pass through the kidneys.
PAH clearance is used to estimate renal plasma flow (RPF).
PAH Clearance = (UPAH × V) ÷ PPAH
PAH Clearance = 585 mL/min
True RPF = PAH Clearance ÷ Extraction Ratio
RPF = 585 ÷ 0.9 = 650 mL/min
RBF = RPF ÷ (1 − Hematocrit)
RBF = 650 ÷ 0.55 = 1182 mL/min
PAH clearance provides a useful estimate of renal plasma flow and renal blood flow.

Measurement of Renal Plasma Flow (RPF) Using PAH — Simplified Concept

STEP 1: What is PAH?

PAH (Para-Aminohippuric Acid) is a special substance because:

✅ Freely filtered by the glomerulus
✅ Secreted by renal tubules
✅ Almost all PAH entering the kidney is removed in one pass

So very little PAH remains in renal venous blood.

BIG IDEA

Think of PAH as a colored dye entering the kidney.

Before Kidney

Blood enters kidney carrying PAH.

Arterial PAH concentration $P_{PAH}=0.01 \text{ mg/mL}$ PPAH=0.01 mg/mL

Inside Kidney

Kidney removes PAH by:

Filtration
Tubular secretion

Therefore almost all PAH is transferred into urine.

After Kidney

Very little PAH remains in blood.

Renal venous PAH $0.001 \text{ mg/mL}$ 0.001 mg/mL

Only a tiny amount escapes removal.

WHY PAH IS USED?

Because the kidney removes almost all PAH from plasma.

Therefore:

Amount of PAH entering kidney

≈

Amount of PAH appearing in urine

This allows us to calculate how much plasma flowed through the kidneys.

THE FORMULA

Guyton gives: $RPF=\frac{U_{PAH}\times V}{P_{PAH}}$ RPF=PPAHUPAH×V

Where:

$U_{PAH}$ UPAH

PAH concentration in urine $=5.85\; mg/mL$ =5.85mg/mL

$V$ V

Urine flow rate $=1\; mL/min$ =1mL/min

$P_{PAH}$ PPAH

PAH concentration in arterial plasma $=0.01\; mg/mL$ =0.01mg/mL

EASY CALCULATION

Step 1

Write formula $RPF=\frac{U_{PAH}\times V}{P_{PAH}}$ RPF=PPAHUPAH×V

Step 2

Insert values $RPF=\frac{5.85\times1}{0.01}$ RPF=0.015.85×1

Step 3

Multiply numerator $RPF=\frac{5.85}{0.01}$ RPF=0.015.85

Step 4

Divide $RPF=585$ RPF=585

ANSWER

$\boxed{RPF=585\;mL/min}$ RPF=585mL/min

WHAT DOES 585 mL/min MEAN?

It means:

Every minute

About $\boxed{585\;mL}$ 585mL

of plasma passes through both kidneys.

VISUAL STORY

🩸 Blood enters kidney

⬇

PAH concentration = 0.01 mg/mL

⬇

Kidney filters + secretes PAH

⬇

Almost all PAH moves into urine

⬇

Urine PAH becomes very high

(5.85 mg/mL)

⬇

Very little PAH remains in venous blood

(0.001 mg/mL)

⬇

From this removal of PAH

⬇

Calculate Renal Plasma Flow $\boxed{585\;mL/min}$ 585mL/min

WHY IS RENAL VENOUS PAH SHOWN?

The figure shows: $0.001\;mg/mL$ 0.001mg/mL

remaining in renal venous blood.

This tells us:

❌ PAH removal is not exactly 100%

✅ About 90% is removed

So for a more accurate calculation, Guyton sometimes applies a correction called the PAH extraction ratio.

KEY CONCEPT

PAH is filtered and secreted by the kidney.
Almost all PAH entering the kidney appears in urine.
Therefore PAH clearance approximates Renal Plasma Flow (RPF).
Formula:

$RPF=\frac{U_{PAH}\times V}{P_{PAH}}$ RPF=PPAHUPAH×V

Using the figure values:

$\boxed{RPF=585\;mL/min}$ RPF=585mL/min

Meaning: about 585 mL of plasma flows through the kidneys each minute. 🚑🩸🩺

FILTRATION FRACTION IS CALCULATED FROM GFR DIVIDED BY RPF

Filtration fraction (FF) is the fraction of plasma that is filtered through the glomerular membrane.
To calculate filtration fraction, both GFR and RPF must be known.
GFR can be measured using inulin clearance.
RPF can be estimated using PAH clearance.
If RPF is 650 mL/min and GFR is 125 mL/min, filtration fraction can be calculated.

EASY EQUATION

$FF = \frac{GFR}{RPF}$ FF=RPFGFR

GIVEN VALUES

GFR = 125 mL/min
RPF = 650 mL/min

STEP 1

$FF = \frac{125}{650}$ FF=650125

STEP 2

Divide numerator and denominator by 25: $FF = \frac{5}{26}$ FF=265

STEP 3

$FF = 0.192$ FF=0.192

STEP 4

Rounded value: $FF \approx 0.19$ FF≈0.19

STEP 5 (Convert to Percentage)

$FF = 0.19 \times 100$ FF=0.19×100 $FF = 19\%$ FF=19%

FINAL ANSWER

$FF = 0.19$ FF=0.19

or $FF = 19\%$ FF=19%

KEY CONCEPT

Filtration Fraction (FF) = GFR ÷ RPF.
FF represents the percentage of plasma filtered by the glomeruli.
Normal GFR ≈ 125 mL/min.
Normal RPF ≈ 650 mL/min.
FF = 125 ÷ 650 = 0.19.
FF ≈ 19%.
This means about 19% of the plasma entering the kidneys is filtered into Bowman’s capsule, while about 81% continues into the peritubular capillaries.

CALCULATION OF TUBULAR REABSORPTION OR SECRETION

If the glomerular filtration rate and urinary excretion rate of a substance are known, tubular reabsorption or secretion can be calculated.
If urinary excretion is less than the filtered load, the substance has been reabsorbed by the renal tubules.
If urinary excretion is greater than the filtered load, the substance has been secreted by the renal tubules.
The amount appearing in the urine may represent filtration alone or filtration plus secretion.

EASY RULE

Filtered Load > Excretion Rate = Net Reabsorption
Filtered Load < Excretion Rate = Net Secretion

GIVEN VALUES

Urine Flow Rate (V) = 1 mL/min
Urine Sodium Concentration (UNa) = 70 μEq/mL
Plasma Sodium Concentration (PNa) = 140 μEq/mL
GFR = 100 mL/min

STEP 1: CALCULATE FILTERED LOAD

Formula

$Filtered\ Load = GFR \times P_{Na}$ Filtered Load=GFR×PNa

Substitute Values

$Filtered\ Load = 100 \times 140$ Filtered Load=100×140 $Filtered\ Load = 14,000\ \mu Eq/min$ Filtered Load=14,000 μEq/min

Answer

$Filtered\ Load = 14,000\ \mu Eq/min$ Filtered Load=14,000 μEq/min

STEP 2: CALCULATE URINARY EXCRETION RATE

Formula

$Excretion\ Rate = U_{Na} \times V$ Excretion Rate=UNa×V

Substitute Values

$Excretion\ Rate = 70 \times 1$ Excretion Rate=70×1 $Excretion\ Rate = 70\ \mu Eq/min$ Excretion Rate=70 μEq/min

Answer

$Excretion\ Rate = 70\ \mu Eq/min$ Excretion Rate=70 μEq/min

STEP 3: CALCULATE TUBULAR REABSORPTION

Formula

$Tubular\ Reabsorption = Filtered\ Load – Excretion\ Rate$ Tubular Reabsorption=Filtered Load−Excretion Rate

Substitute Values

$Tubular\ Reabsorption = 14,000 – 70$ Tubular Reabsorption=14,000−70 $Tubular\ Reabsorption = 13,930\ \mu Eq/min$ Tubular Reabsorption=13,930 μEq/min

FINAL ANSWER

$Tubular\ Reabsorption = 13,930\ \mu Eq/min$ Tubular Reabsorption=13,930 μEq/min

Therefore, most of the filtered sodium was reabsorbed.
Only a small amount of sodium was excreted in the urine.
Comparison of a substance’s clearance with inulin clearance helps determine how the kidney handles that substance.
If the clearance of a substance equals inulin clearance, the substance is filtered but neither reabsorbed nor secreted.
If the clearance of a substance is less than inulin clearance, the substance is reabsorbed.
If the clearance of a substance is greater than inulin clearance, the substance is secreted.