Majed Sharayha
Posts : 130 Reputation : 5 Join date : 2011-08-23 Age : 31
| Subject: Physiology, Sheet 6, Dr.yanal 1322012 Wed Feb 15, 2012 4:47 am | |
| Download link ---------> Click here-------------------------------------------------------------------- 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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