Overview of renal function Overview of renal function and the clearance conceptand the clearance concept

Prof Harbindar Jeet SinghProf Harbindar Jeet SinghFaculty of MedicineFaculty of Medicine

Universiti Teknologi MARAUniversiti Teknologi MARAShah AlamShah Alam

Objectives

1. Describe the structure of kidney and nephron

2. List the functions of the kidney

3. Describe clearance concept

4. Define renal clearance

5. Discuss the uses of the clearance concept in the measurement of renal function, like g.f.r., renal blood flow etc.

6. Define glomerular filtration rate

7. Describe free water clearance

Major functions of the kidney

1. Water and electrolyte balance

2. Acid-base balance

3. Excretion of metabolic waste

4. Blood pressure regulation (RAAS)

5. Vitamin D metabolism

6. Secretion of erythropoietin

The human kidneys are bean-shaped organs lying behind the peritoneum oneach side of the vertebral column.

Each kidney weighs between 125 – 170 g in the male and between 115 – 155 g in the female. They make up less than 0.5% of the total body weight.

The single functional unit of a kidney is the nephron.

The number of nephrons found in mammals varies, the number increasing with increasing body mass

e.g. a mouse has 12,400 nephrons per kidneyhuman 1.2 million per kidneyan elephant has 7.5 million per kidneya fin whale has 192 million per kidney

Most mammalian kidneys generally contain two types of nephronsshort-looped nephronslong-looped nephrons

(except the dog, which has all long-looped nephrons and a mountain beaver that has all short looped nephrons)

Although the division of nephrons according to the position of the corpuscle does not coincide with the division based on the length of their loops, generally superficial glomeruli give rise to short-looped nephrons and mid-corticol and juxta-medullary glomeruli give rise to long-looped nephrons.

Short-looped and long-looped nephrons differ in two important aspects:

Short-looped nephrons turn near the inner-outer medullary border and lack a thin ascending limb.

Thin ascending limbs are found only in the inner medulla.

Generally there are more short-looped nephrons than long-looped nephrons.

The long-looped nephrons tend to exhibit substantial variation in the depth reached within the inner medulla.

Some mammalian kidneys, such as human kidneys, also have nephrons whose loops of Henle do not reach into the medulla; these nephrons are the cortical (superficial) nephrons.

Structure of a nephron

Tubule segments of the nephron

Tubule segment Abbreviation

Proximal convoluted tubule - PCTProximal straight tubule - PSTThin descending limb of loop of Henle - tDLHThin ascending limb of loop of Henle - tALHThick ascending limb of loop of Henle - TAL/TALHDistal convoluted tubule - DCTCortical collecting duct - CCTOuter medullary collecting duct - OMCDInner medullary collecting duct - IMCD

In 1842, William Bowman proposed that the glomerular capillaries secrete water, which flushes out solutes secreted by the renal tubules

In 1844 Carl Ludwig concluded that urine is formed by ultrafiltration of plasma at the glomerulus, and that it merely passes down the nephron without further alteration save for concentration of solutes by passive reabsorption of water

In 1917 Arthur Cushny modified Ludwig’s view by proposing that not only water but also solutes are reabsorbed by the tubules, in proportions, which are determined by their normal values in the plasma

In 1923 E.K. Marshall proved that tubular secretion also occurs

FilteredFiltered ExcretedExcreted ReabsorbedReabsorbed %%

HH22OO L/dL/d 180180 1.51.5 178.5178.5 99.299.2

NaNa++ mM/dmM/d 25,00025,000 150150 24,85024,850 99.499.4

ClCl-- mM/dmM/d 18,00018,000 150150 17,85017,850 99.299.2

HCOHCO33-- mM/dmM/d 4,5004,500 22 4,4984,498 99.9+99.9+

GlucoseGlucose

UreaUrea

mM/dmM/d

g/dayg/day

800800

46.846.8

0.50.5

23.423.4

799.5799.5

23.423.4

99.9+99.9+

5050

The process of urine formation involves a combination of ultrafiltration at the glomerulus, followed by selective tubular reabsorption of water and solutes, and selective tubular secretion of solutes.

The clearance concept is a central concept in renal physiology.

It is a concept or technique that allows us to assess renal function using only analysis of urine and plasma.

Instead of talking about the rate of excretion of a substance in urine, we talk about the rate of removal of this substance from the plasma.

Definition:

Renal clearance can therefore be defined as the volume of plasma in mls passing through the kidneys that has been completely cleared of a substance in a given period of time.

Thus, if the plasma contains 0.1 gm of a substance in 100 ml,

and 0.1 gm of the substance also passes into the urine each minute,

then, 100 mls of plasma is cleared of the substance per minute.

The equation for the calculation of plasma clearance is

Renal clearance (ml min-1) =

urine excretion rate of the substance (mg/min)plasma concentration of the substance mg/ml

Urine flow (ml/min) Conc. of substance in urine (μmol/ml)

Conc. of substance in plasma (μmol/ml)

Or Uc V

P

E.g.

Urine flow = 1 ml/minConc in plasma = 100 μmol/mlConc in urine = 12000 μmol/ml

12000 x 1= 120 ml/min

100

Uses of the clearance concept.

1. In the measurement of glomerular filtration rate (gfr)

For the measurement of g.f.r., we need a substance that is

a) freely filtered b) not secreted by the tubular cells, c) not reabsorbed by the tubular cells.d) should not be toxice) should not be metabolisedf) easily measurable.

An example of such a substance is inulin. It is a polysaccharide with a molecular weight of about 5200 and it fits all the above requirements.

So, if we say,

Plasma conc. of inulin = 100μg/100mlUrinary conc of Inulin = 12000 μg /100mlUrine flow (UV) = 1 ml /min

then, the clearance of inulin will be

Ci = U V = 12000 1= 120 ml/ min

P 100

As inulin is neither secreted nor reabsorbed the clearance of inulin can be taken as a measure of the glomerular filtration rate.

Other sugars that could be used include sorbitol, mannitol.

In clinical practice creatinine clearance is used as a measure of g.f.r.

2. For the measurement of Renal Blood Flow.

Inulin clearance or g.f.r. only reflects the volume of plasma that is filtered and not that remains unfiltered and yet passes through the kidney.

It is known that only 1/5 of the plasma that enters the kidneys gets filtered.

To measure renal blood flow we will have to measure renal plasma flow first and then from the hematocrit we calculate the actual blood flow.

For the measurement of renal plasma flow, we will again need a substance that is

a) freely filteredb) rapidly and completely secreted by the renal tubular cellsc) not reabsorbedd) not toxice) and easily measurable.

An example of the substance is Para-aminohippuric acid (PAH).

So if the concentration of PAH in the urine and plasma and the urine flow are as follows:

Conc. of PAH in urine = 25.2 mg/mlUrine flow = 1.1 ml/minConc of PAH in arterial blood = 0.05 mg/ml

Then CPAH or Renal Plasma Flow = 25.2 1.1

= 560ml 0.05

Lets say the hematocrit is 45%, then renal blood flow will be

560 100= 1018 ml/min

100 - 45

3. Clearance can also be used to determine renal handling of a substance.

Clearance values can also be used to determine how the nephron handles a substance filtered into it.

In this method the clearance for inulin or creatinine is calculated and then compared with the clearance of the substance being investigated.

Free water clearance

Free water refers to water that is free of solutes.

Free water clearance may be defined as the amount of distilled water that must be subtracted from or added to the urine in order to render that urine iso-osmotic with plasma.

CH20 = V - Cosm

where Cosm = the osmolal clearance

= U V P

e.g.

urine osmolality = 100 mOsm/kg H2O, urine flow rate = 10 ml per minute, plasma osmolality = 300 mOsm/kg H2O

Then,

100 10 CH2O = 10 - = 6.7 ml/min

300

When hyperosmotic urine of 1,000 mOsm/kg is formed, and the urine flow is 0.5 ml per minute,

Then,

1000 0.50.5 ml - = -1.2 ml/min

300

When hypoosmotic urine is formed, the free water clearance has a positive value,

and when hyperosmotic urine is formed this clearance has a negative value

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