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 Physiology, Sheet 6, Dr.yanal 13\2\2012

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Majed Sharayha



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PostSubject: Physiology, Sheet 6, Dr.yanal 1322012   Wed Feb 15, 2012 4:47 am

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PHYSIOLOGY SHEET “6”

We don’t put a lot of effort to expand a lung , so inflating a lung is more easier than inflating a balloon , but that’s not always guaranteed ; because sometimes the lung is not compliable , or the collapse forces of the lung are more than the normal .

The lung is tending to collapse , to overcome this collapsing force we need an expanding force , or pressure and that needs contraction of muscles to make intraplural pressure negative .

The collapsing forces of the lung are manifested mainly ( 70 % ) during static ( when there is no air flow ) the entire collapsing forces will be elastic forces .

These elastic forces are due to two mainly forces :-
Surface tension means that the intermolecular attaching of water molecule is trying to shrink the lung , to bring it to the center which means collapsing the lung .
Surface tension is found only when there is air-water interface.
Elastic fibers .

2/3 of the lung forces ( work of breathing ) is to overcome surface tension forces , and 1/3 to overcome elastic fibers .

We can separate these two forces ( surface tension force , elastic fibers force ) by :
-if you know how much you need for the total pressure , and how much you need for the elastic fibers , then you can calculate the surface tension needed : Total = surface tension + elastic fibers
Surface tension is cavern by laplac’s law :

P=2T / r P: the inflation pressure .
T: the tension .
r:radius .
Because( PRESSURE = force / area ) and (TENTION = force / length ):


Now we fill the lungs with air to see how much force we need to inflate the lung , we can take the lung outside the body in vitro NOT in vivo , where the lung in vitro is not surrounded by negative intraplural pressure , its surrounded by atmospheric pressure .
We put the lung in a closed box and the floor has a hand which is able to move downward and increasing the volume of the box .

each time we make the volume in the box ( intraplural ) instead of zero -1 or -2 , and then we measure how much inflation we get by changing the volume of the box .The balloon will expand more .

now we will see how pressure and volume changes during inflation the lung:-

first we make the inflation pressure more negative , more negative means more force ,( that why the negative sign in the chart) .

phase 1 :
we start from minimal volume ( empty lung ) , because in vitro there is no intraplural pressure to hold the lung , which means that we don’t have an volume in the lung before making inspiration like in vivo .
now we start changing the intraplural pressure , we find that we need to make too much pressure change ( too much force ) , to get very little change in volume .

To explain how we get the compliance :
Tan of the = y2-y1
X2 – x1
= compliance
Of the lung .

6 L

(V) phase 1 phase 2
mv
-
(P)
( Inflation – compliance curve )
compliance means how much change in the volume per unit change in the pressure :
c= v / p
if we change the pressure too much , and get little change in the volume the compliance will be too small.
If we change the pressure slightly , and get too much change in volume its called the compliance of the lung .

So we can compliance also means that we apply force and we deform the structure ( deformity = elongation ) , if we change the deformity too much and little applying force we will get high compliancy and vise versa.

Collapsing forces is the opposite of expanding force , and it’s the opposite of the compliance.,( elasticity it’s the recoiling tendency ).


Phase 2 :
At critical point ( critical opening pressure ) means that any little pressure increasing , we will get too much change in the volume ( remember the popcorn ! ).
So in phase 2 the lung become compliance , we apply too little force and get too much change .
The slop in the figure presenting the compliance, so whenever we increase the angle the slop is going to increase so the compliance will increase .
First in phase 1 the slop was small so the compliance decreases , then in phase 2 the slop was big so the compliance increases , finally at the end the slop return small and the compliance decreases .

6 L is the total lung capacity .


If you deflate the lung during normal breathing to zero volume , then you start to inhale again , it will coast you too much because you are inhaling from zero volume ( minimal volume in vitro).
Its not wise to inflate your lung from empty lung , its much better to inflate your lung from partial inflated lung .
Its very hard to expand the walls of lungs when they are adjacent closely to each other , because we will need too much force , too much ATP expenditure therefore work of breathing increases .
_ in normal breathing , the work of breathing doesn’t increase too much , because when we take 0.5 L ( to increase from 2.2 -2.7 ) we just need a force equal ( 1 to 2 ) mmHg , but her the pressure increase too much so the work of breathing increased too .


Important messages :-

In each time we exhaled , we empty the lung totally , and to reinflate the lung will be very difficult , very hard , it will cost too much .
Inflating the lung from partially inflating lung is an easy process , the compliance of the lung at this range is high .
To inflate already inflated lung is hard , so we don’t want to inflate already inflated lung.

If we look to these alveoli , which one is very difficult to inflate and which one is very easy to inflate :
the first one and the last one are difficult to inflate
The middle one is easy to inflate .


1 2 3
We say that to inflate the lung we change the pressure surrounding the lung , make it more negative , and then we see how much change in volume.

But if we want to deflate the lung it will follows a different path .


Hysteresis : means that the forward process is not the same as the backward process .

Hysteresis because in logical thinking the width must be thin not wide .
So what is happening ?
The doctor explain it as an example :
Take for example 4 L which we can reach it by two ways :
We can go from 3.5 to 4 by accelerating .
We can go from 4.5 to 4 by decelerating .
When you hold an object 40 KG , however you are accelerating or
Decelerating , we measure how muscle tension we need to
Overcome the weight of this object you hold .
To measure that we need to stop moving , which means no muscle length change , isometric contraction ( in static position no movement ) .
There is a tone of muscle , tone here in static position reflect the weight of the object .
( Tone = 40 )
But you realize that when you are accelerating the tone was 50 , and when you are decelerating the tone was 40 , while we are having the same object weight !! that’s why we call it hysteresis .

the force that we need To hold the lung inflated ( which means how much negative intraplural pressure you need to over come the collapsing forces ) during inflation , is more than during deflation .



Why there is hysteresis ?
We have surfactant ( glucolipoprotein mainly phospholipids ) .
Phospholipids has a unique structure , it has polar head and non- polar tail .


Non –polar
here we have two tails facing each other there is no intermolecular attraction between them .



So if we have the alveolus and we lined it from the inside with phospholipids how phospholipids will be orientated ?

the hydrophilic portion faces the epithelium and the hydrophobic portion faces the air way , there will be no intermolecular attraction .
so its not only how much surfactant do you have its also how does it orientated .


lets say that we have 4 surfactant , if the alveolus very big the concentration of the surfactant become very small , which is not efficient , because the number of molecules are fixed , whatever the size of the alveolus .
But if the size of the alveolus become small . these 4 surfactant closes the position of water , then the surfactant will be very efficient , and the surface tension very low .



To summarize :
The amount of surfactant is fixed but the size of alveolus is not.
The more the alveolus size , the less the concentration , the less the effectiveness of the surfactant , the more of the surface tension .
But that is not applied to the 4 L example because however accelerating or decelerating we have the same size , so the term of concentration is the same .
So its not only the concentration , which means its not the quantity it’s the quality .



Now how the surfactant is orientated :-
During inflation , the surfactant is spread in a non- desirable way ( the surfactant is exist but its not effective ) because its not effective the surface tension is high
During deflation it seems that we are compact the surfactant in a desirable way , so it becomes orientated in way which is effective ,making surface tension minimum and thus need less intraplural pressure .


Q : do you think that when we inflate the lung we follow the same path as deflating the lung , or there is major difference between the two pathways ?
There is a major difference between the two pathways , this is called hysteresis , the surfactant is responsible for this phenomena , because of its shape and its structure .


How we can remove the surface tension ?
By removing the water- air interface , by starting to fill the lung with normal saline .09% NACL ( isotonic solution ) .
When we fill the lung with normal saline we don’t need to overcome any surface tension , because there is no surface tension , because we filled the lung with fluid not with air .
So in normal saline we need to overcome only one force , which is elastic fibers .

In saline filled lung there is no surface tension just elastic force .
In air filled lung there are two forces ( surface tension and elastic ).

With normal saline
If we took a volume , we need -6 to inflation curve ,
and -2 to deflation curve .
-2
-6
If -2 reflects the elastic fibers and -6 reflects the total forces ,
so the surface tension is 4 .
From here we can conclude that surface tension effect by 2/3 , and elastic fiber by 1/3 .


Now we return to laplac’s law :
P = 2t / r
You don’t have surfactant , the surface tension is high then the inflation pressure must be high too , so too much inflation pressure , too much work of breathing , it cost much .
Surfactant make decreasing in the work of breathing , decreasing t in surface tension , and increasing in the compliance of the lung .



When baby born premature ( before term ) , like for example the baby was born between week ( 26-28 ) , the gestational period is 4o weeks , there is 50 % possibility of death for this premature baby , we called that immature lung .

The explanation
In premature baby there is no surfactant the radius decreases and the surface tension increases , the inflating pressure is too much , which mean too much muscle contraction .
Mortality rate is one baby out of two .
The baby is going to face bronchial problems because the collapsing forces are high , then each time he exhaled he goes to minimal volume ( the inflating from zero is very difficult ) , so he is going to use most of his ATP and he might die from lack of ATP , muscle fatigue .

The premature baby has :
His expiratory rate is 60 .
Air sever dyspnea .
He is using all his respiratory and accessory muscles ( neck , intercostals ) .
In cardiovascular :
We took that capillaries have starling forces which determine whether filtration or reabsorption will occur .
We have 4 starling forces :
2 forces inside ( Pc [ capillary hydrostatic pressure ] , πc [colloid osmotic pressure ])
2 forces outside ( pi , πi )
Because the baby has to make too much negative pressure to draw air to the lung ( to overcome his collapsing forces ) , the pressure in the interstitial is very negative ( -20 , -30 ) which means filtration force .



So this baby is having problems in inflating the air and pulmonary edema , then there will be outward movement of fluids because the capillaries are surrounded by huge negative pressure .

There will be interstitial edema , oxygen diffuses with no difficulties assuming that the respiratory membrane is 0.2 μm , but now the interstitial is filled with water , the respiratory membrane become thicker and might interfere with oxygen diffusion , then their will be oxygen diffusion limited ( we cant deliver oxygen ) .

These patients will have alveolar edema , and they might die from pulmonary edema , sever hypoxia .

This baby is facing a major problem called infant respiratory distress syndrome IRDS , means prematurity causes lack of surfactant .

Now if the baby was born between week ( 30-32 ) mortality rate goes down to 25% , one day makes difference .



If there is a thread of early delivery what should we do :

The pregnant must have a medical care for prematurity .
We can stimulate the production of surfactant in the embryo by giving the mother glococorticoid ( hydro cortizon ) injection .
The surfactant is produced with the help of glucocorticoid , thyroid hormone , estrogen , so the female baby has chances to live better than the male .

How do you know if its allowed to deliver the baby or not ?
You have to get information about the maturity of the lung , by taking surfactant markers from the amniotic fluid ( fluid surrounded the baby ) by amenotheosis .
We measure how much surfactant (mg ) to albumin ratio ( g ) if its :
55 and above the baby is mature .
35-55 in between .
Less than 35 the baby is immature .





Done by : ALA’A MOHAMMAD YOUSEF
Date : 13/ 2 / 2012 .
Lec. Num . 6
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Physiology, Sheet 6, Dr.yanal 13\2\2012
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