Pulm Physio

Question 1

tidal volume

Answer 1

what enters our lungs during quiet resting breathing

Question 2

inspiratory capacity

Answer 2

a forced inspiration, and this inspiration is performed to total lung capacity—maximum inflation maneuver.

volume of air that can be inspired from the resting position (end-expiratory) - decreased by diseases that increase lung stiffness and/or decrease muscle strength

Question 3

expiratory reserve volume

Answer 3

Expiratory reserve volume is when we exhaled below our tidal volume all the way out.

volume of air that can be exhaled from the resting position (end-expiratory) - decreased by diseases that reduce expiratory airflow

Question 4

vital capacity

Answer 4

The air we can move in and out of our lungs during a maximum maneuver like this one is called vital capacity (abbreviated VC).

inspiratory capacity together with expiratory reserve volume form our vital capacity

Maximal volume of air that can be exhaled when the lung is fully inflated.

Question 5

functional residual capacity

Answer 5

. As you can see the functional residual capacity is the baseline resting volume. We are now, right now breathing above FRC. We take a breath and relax back down to FRC. We are not thinking about going back to FRC, it just happens. This is a point of equilibrium in our respiratory system.

Question 6

lung volumes assesd my spirometry

Answer 6

TV

IC

ERV

VC

Question 7

determinants of FRC

Answer 7

balance between chest and lung tensions

chest wants to go out

lungs want to go in

the point at which the inward recoil of the lung is balanced out by the tendency of the chest wall to expand outwards and there the respiratory system is at equilibrium

balace (with no muscle effort!!!!!!!) determines FRC

Question 8

intrapleural space

Answer 8

assessed as intrapleural pressure (Pip)

equivalent to P around the heart, in the great vessels, around and inside the esophagus

vacuum - lungs and chest pulling in opposite directions - at rest, slight negative P (with respect to atm)

Question 9

transpulmonary pressure

Answer 9

equal to elastic recoil!!!!!!!

alveolar pressure - interpleural pressure

alveolar pressure - measure at mouth

ip - measure in esophagus

tp is the difference!

under static conditions (zero airflow, glottis open) = P(alv) = 0 so P(tp) = - P(ip)

If we are in a situation of rest (no flow) and are communicating the atmospheric pressure with the alveolus, at FRC, the alveolar pressure is 0 cm of H2O. Atmospheric pressure is 0 cm of H2O and 760mm of mercury. In a static situation there is no gradient of pressure between alveolus and the atmosphere and therefore we have no flow of air.

Question 10

inspiration muscles

Answer 10

diaphragm = push abd down - increases most planes of the chest cavity when contracts (descends down and gets smaller)

external intercostal - expand all planes of the chest

Question 11

pressures on inspiration

•

Answer 11

•It is a process that requires utilization of energy. And if we start off at FRC which I have represented with the green dot in the lower left hand image, and we have transpulmonary pressure = 5 and intrapleural pressure = -5, we should now think about taking a breath in so the diaphragm will descend and generate negative intrapleural pressure.

If I were to to contract our diaphragm with a muscle strength of 3 by how much would our intrapulmonary pressure descend? By the same amount so our intrapleural pressure will go from -5 to -8. So that generation of a negative intrapleural pressure by muscle activity will be transmitted to the alveoli and drop our intraalveolar pressure in a proportional manner. What has happened is we have created a pressure gradient of 3cm of H2O between the atmosphere and the intralveolar space which is now sub atmospheric. This generates influx of air and a change in volume is generated inside the alveoli and they inflate

Question 12

transpulmonary pressure at TLC

Answer 12

In order to keep your lungs inflated you have to continually pull down on your diaphragms. You have now equilibrated pressures between your alveolus and the atmospheric pressure, gradients disappear and no longer air will enter the lungs

At the point of maximal inflation the transpulmonary pressures are dictated by the negative intrapleural pressure that has been generated, and this negative intrapleural pressure tells us a lot about the properties of the lung.

P(alv) is zero (equlibrated, glottis is open, no air coming in) but P(tp) is high

Question 13

pressure of elastic recoil

Answer 13

cork system but relax diaphragm - P transvered from ip to alv (negative interthoracic P is transfered to alveoli

Before we had no pressure gradients, all the energy to inflate the lungs was exerted by the diaphragm and now the energy used to deform the lungs is transferred onto the elastic tissue of the lungs and there is a pressure exerted onto the alveoli. There now is a gradient between the alveolus and the air of about 35cm of H2O. That gradient is big enough to deflate the lungs without needing to activate the expiratory muscles. That pressure is called pressure of elastic recoil.

. We have transferred the energy from the respiratory muscles to the parenchyma, and the elastic tissue in the lungs, created an alveolar to atmospheric pressure gradient and we are ready to exhale in a passive manner—just utilizing the pressure of elastic recoil that the lungs naturally have and tend to deform down to a low lung volume.

Question 14

elastic recoil

Answer 14

inherent resistance of a tissue to changes in shape, tendency oto revert to original shape when deformed

Question 15

recoil pressure

Answer 15

pressure attributable to the elastic properties that acts to return the lungs to their resting or unstressed position

****provides driving force for expiration!

Question 16

lung volumes that requrie measurement in addition to spirometry

Answer 16

TLC, RV, FRC

Question 17

active v passive breathing

Answer 17

We have gone from FRC, we used our inspiratory muscles to generate pressure gradients and drop alveolar pressure until lungs inflate to TLC. And we released that energy and transferred that energy to the elastic tissue in the form of recoil and elastic recoil pressure by generating a gradient and allowing a passive exhalation back to FRC. Exhalation back to FRC is a passive process utilizing the energy that we had to contract our diaphragm.

Question 18

residual volume

Answer 18

RV is a volume that we cannot expel from our lungs, we cannot deflate our lungs any further past RV. This requires energy expenditure

If we were to try to exhale from FRC to ERV we would need to use energy because we are moving our lungs from resting position down to residual volume.

Question 19

lung compliance

Answer 19

Lung compliance can be defined as a pressure gradient needed to inflate lungs to a given volume. At low lung volumes such as FRC the changes in pressure needed to inflate the lung is small—lungs are compliant at low volumes. As we stretch the lungs we require larger pressures to inflate the lungs the same volume. Since the elastic fibers have been stretched almost to their maximum it is much harder to continue inflating them.

Question 20

maximal recoil pressure

Answer 20

determined at TLC

pressure at TLC

Question 21

lung compliance in fibrosis

Answer 21

For any given change in pressure in a normal patient we have a large change in volume, but in patients with fibrosis we have small changes in pressure.

Lung compliance has decreased. At total lung capacity we haven’t reach a predicted lung volume (equivalent to a population of this person’s height and age). The transpulmonary pressures needed to inflate that lung are much higher. The diaphragm of this patient with fibrosis needs a generation of a much higher intrapleural pressure in order to inflate the lungs and they can’t even do it maximally because the recoil is very elevated–the lung have fibrosed and elastic tissue is replaced.

Question 22

lung compliance in emphysema

Answer 22

For the same change in pressure we have a much higher change in volume. It is very easy to distend the lungs of an emphysematous person. We measure transpulmonary pressures at maximal inflation. The patient can super inflate the lungs but it takes very little pressure. The recoil pressure of the lungs is very low at the that maximal inflation.

Question 23

static balance in TLC

Answer 23

between max inspiratory force by res[iratory muscles and elastic recoil force inward

Question 24

static balance in RV

Answer 24

between max epiratory force by respiratory muscles and F gen by outward elastic F of chest

Question 25

FRC at base vs apex

Answer 25

at apex - high gravitational force - more negative IPP IPP is less negative at the base higher FRC at the apex! Higher--- the pleural pressure is more negative and there is a greater distending force at the apex because of gravity than at the base where the transpulmonary pressure is much smaller.

Question 26

differences in pressure and ventilation

Answer 26

alveoli are more inflated at the apex (bigger transpulmonary pressure) BUT have small changes in vol for changes in P - less ventilation ## Footnote If now I use our change in volume for change in pressure curve, you can see that despite being more inflated at the apex, since the alveoli are at a different resting lung volume they ventilate less. The change in volume for any given change in pressure is smaller up here whereas down here [pointing to green arrow] the change in volume for the same change in pressure is higher. So smaller lung volumes at the bottom, greater change in volume for any given change in pressure at the base of the lungs.

Question 27

measurement of FRC

Answer 27

use Boyle's Law - V1P1 = V2P2 •We can measure changes in pressures/ volumes that happen in the box and apply law of proportion to the lung and estimate functional residual capacity.

Question 28

P(chest wall)

Answer 28

Pcw = Pip - Patm

Question 29

base chest wall resting potential

Answer 29

happiest at halfway to TLC ## Footnote At FRC it wants to go to resting position outwards and at TLC it wants to move back inwards.

Question 30

combined chest wall and lung complaince

Answer 30

When we combine the chest wall and lung compliance we can see that the -5 is perfectly well balanced with the +5 pressure gradient of the lung. The white circled point is the functional residual capacity. The purpose of this diagram is to see the balance of forces. chest wall pressures are negative

Question 31

fibrosis and lung compliance

Answer 31

FRC decreases!

Question 32

effect of scoliosis and lung compliance

Answer 32

Scoliosis is a disease of the chest wall and the spine. If the chest wall is compromised it will be stiffer and harder for the chest wall to pull outwards and FRC will decrease and the lung volumes will be small. It’s almost as if patients with scoliosis have a cage put across the lungs and they can’t inflate properly.

Question 33

alveolar stability

Answer 33

LaPlace's law - hard to open alveoli w small radius ## Footnote Therefore if we were just to purely obey LaPlace’s law those smaller alveoli would empty out into ones that are larger just because the surface tension would be so high in those alveoli we would require very high pressures to open them up. decrease ST with surfactant allows us to maintain the discrepancy in alveolar size in base and apex without actually popping everything open over to the apices.

Question 34

effect of surface tension on lung compliance

Answer 34

Surface tension as an impediment to inflation is strongest at low lung volumes were the radius is low. When radius is high and we have inflated our lungs to total lung capacity the greatest impediment to inflation are the tensile forces.

Question 35

what is the maximal airflow?

Answer 35

500 L/min

Question 36

what is airflow at rest?

Answer 36

5 L/min

Question 37

intrapleural pressure during tidal inhale

Answer 37

, the intrapleural pressure if you look is always more negative than what you would have predicted the path the line should have taken. Your respiratory muscles are generating more pressure than you would have predicted just from the static line. You have to overcome resistance in the airways. So this is the overcoming of the friction if you will of the airflow moving down the airway passage. So the intrapleural pressure you generate is always going to be more negative than you would have predicted from just the static. In fact, this area therefore, if I color it in, is a perfectly valid measurement of airway resistance. It is how much more negative the intrapleural pressure is compared to thestatic conditions. That is overcoming the airway resistance.

Question 38

interpleural pressure during exhale

Answer 38

the intrapleural pressure during exhalation is actually less negative than the static line. What did Beno tell you yesterday that we do to exhale?We are very passive. There is normal energy stored in the lung that pushes the flow of gas out. In fact that is what you are seeing here, the pressure that is stored in the lung is greater than what is required to overcome the resistance, so the airflow proceeds totally passively because the recoil of that pressure stored in the lung, the normal elasticity, has enough force to overcome the airway resistance

Question 39

airflow rate in tracheobronchial tree

Answer 39

each branch - slows to 1/2 of the speed

Question 40

Reynolds Number

Answer 40

if greater than 2000 favors turbulent flow increases if: bigger radius, faster, bigger density

Question 41

large airway and flow

Answer 41

fast, turbulent flow small cross sectional area

Question 42

small airway and flow

Answer 42

slow, laminar flow large cross sectional area

Question 43

transitional flow

Answer 43

between small and large vessels intermediate flow and cross sectional area mostly laminar but turbulent at bifurcations

Question 44

elevated airway resistance

Answer 44

when constricted - R prop to 1/r^4 50% decrease in radius = 16x resistance! paitents will increase power output to overcome decrease due to resistance - airflow has to be FASTER to get through laminar flow to turbulent flow - wheeze!

Question 45

treatment of elevated airway resistance

Answer 45

turbulent airflow! increase airway lumen OR decrease density of inspired gas (give He)

Question 46

flow equation

Answer 46

flow = alveolar driving pressure/resistance (Ohm's law) driving P: muscle strength (Pmus = Pip), lung recoil (Pel) IN EXHALE Resistance: size of the airpways, number of parallel aiways, collapsibility of airwy walls

Question 47

What will be compromised first, resting or forced airflow?

Answer 47

forced airflow

Question 48

where in the tracheobronchial tree is airway resistance the highest?

Answer 48

trachea!

Question 49

airway resistance and lung volume

Answer 49

as lung volume increases, resistance decreases! the luminal diameter is very dependent on those tensile forces within that fibroelastic system in the lung. What happens during inspiration? I stretch those fibers. So if I stretch the fibers [video], as the tension increases during inspiration, the airway dilates. As you are inhaling, the airways are dilating. As you are exhaling, the airways are constricting.

Question 50

collapsibility of airway walls

Answer 50

in inspiration - negative prssure in the thorax is distending the airways in expiration - positive pressure surrounding the negative pressure in the airways - compress and collapse (increase resistance) flow is determined by alveolar pressure minuepleural pressure - independent of effort

Question 51

FVC

Answer 51

forced vital capacity The distance from full inflation and full deflation is the vital capacity. We put the letter F here, which denotes the fact that it is a forceful maneuver, a maximal forceful exhalation,

Question 52

FEV1

Answer 52

this is the forced expiratory volume in the first second.

Question 53

airflow in obstructive disease

Answer 53

i.e. asthma lower percent is exhaled takes longer to fully exhale

Question 54

restrictive expiratory airflow

Answer 54

airflow is fast but VC is LOW The lung is stiffer than normal, the patient can’t distend it as much, so their vital capacity falls. The FEV1 can’t be 4 liters if they only exhale 3 liters, so it’s got to fall also. But if you look at that FEV1, compared to the FVC, you see that these airways are really quite efficient. He exhaled 90 percent of his vital capacity in the first second, so the airway resistance is not high in this patient, even though the FEV1 is low.

Question 55

spirometry curve

Answer 55

Question 56

maximal expiratory curve

Answer 56

at total lung capacity, you have the capability of generating the highest pressure from muscle forces, your lung recoil is at its highest, and your airways are maximally dilated so you get very high flows. As you exhale, your airways are getting smaller and smaller. Your muscles are becoming more disadvantaged in terms of their length/tension relationship. The diameter is shrinking, and so the flow progressively falls.

Question 57

effort dependent exhalation

Answer 57

first half of the curve The initial peak flow that you can generate at the early part of the maneuver is very much dependent on that effort

Question 58

effort independent exhalation

Answer 58

as the lung deflates, all these curves converge, so this latter half of the flow-volume curve has become actually not dependent on your effort

Question 59

why is there effort independent exhalation

Answer 59

in exiration - airway collapse becomes pinched harder does not lead to more flow during an inspiration, you generate a negative intrapleural pressure. As I showed you, the airway walls dilate. During exhalation, you generate a positive intrapulmonary pressure and the airways tend to constrict and even though you may push harder, that also tends to constrict the airway more. You don’t necessarily get more flow in the latter half of the curve.

Question 60

minimum work concept

Answer 60

as increase respiratory rate: decrease elastic work (the work load to distend the lung is highest at large volumes and low frequencies and is minimal at small volumes but high respiratory rates) increase non elastic work (Smaller volumes at a very high volume generates high flow rates, and so there is a much higher work load required to overcome the resistance.) Now, you can sum these 2 curves, and get the total work that the respiratory muscles have to do, and you get a U-shaped curve. And in fact, we tend to breathe at the nadir of that curve, which is shown by the crossing of the 2 lines, individual components.

Question 61

convection

Answer 61

airways active - requires E long distances

Question 62

diffusion

Answer 62

alveoli passive - no E required acts only in short distances

Question 63

At 21% concentration at ambient air, what is the partial pressure of oxygen in an alveolus?

Answer 63

160, give or take (~20% of 760mm Hg)

Question 64

D(L)

Answer 64

Area, thickness and diffusion constant can be lumped into a term we called D(L), lung diffusion. ## Footnote he flux of gas is equal to lung diffusion times the partial pressure differences.

Question 65

components of D(L)

Answer 65

C1: membrane diffusion C2: binding to Hgb - capillary blood vol

Question 66

components of membrane diffusion

Answer 66

RBC membrane blood plasma capillary endothelial cell interstitium alveolar epithelial cell

Question 67

D(m)

Answer 67

everything up to alveolus (all diffusion)

Question 68

V(c)

Answer 68

oxygen binding to hemoglobin

Question 69

theta

Answer 69

rate that O2 binds to hemogloben

Question 70

inhale CO

Answer 70

diffusion limited!! Binds 100x more avidly to hemoglobin than oxygen does. Carbon monoxide is readily taken up by hemoglobin. Hemoglobin acts a carbon monoxide sink and soaks it all up. Therefore when we inhale a mixture with a known mixture of carbon monoxide, and it encounters blood that has no carbon monoxide, hemoglobin will take up all the carbon monoxide it has been presented with. Therefore the partial pressure of carbon monoxide in solution, does not rise very much, because all the carbon monoxide is being taken up by hemoglobin. What we can see over here is by the end of the transitioning of the capillary, exit of the capillaries the concentration of carbon monoxide rises slowly but doesn’t achieve very high levels. Therefore the only way, usually, carbon monoxide flux from alveolus into the blood stream is impaired is if there alterations in the membrane per se. That’s why we call carbon monoxide a gas that is diffusion limited.

Question 71

Diffusioncharacteristics of N2O

Answer 71

perfusion limited!! initial PP in blood = 0 binds very little to Hgb PP increases very fast and equilibrates immediatly with PP in alveolus Since hemoglobin doesn’t not capture NO, the partial pressures in the capillaries rise very quickly. In happens in .25 seconds. The partial pressure of NO in solution in the capillary is almost the same as the one present in the alveolus. So, there is no longer a pressure gradient of NO in the capillary and alveolus, therefore diffusion occurs no longer. Memebrane has no role is limiting the diffusion of NO because the partial pressures rapidly equilibrate so what is really needed for diffusion to occur again is for a fresh aliquot of blood to come into the alveolar capillary membrane area void of NO so that the partial pressure gradient is reestablished and diffusion can occur again. The concept of needing NO to be clear from that area in order to generate a diffusion gradient again is called a Perfusion limitation. This is a gas that is perfusion limited, it needs to be cleared to restore the partial pressure gradients

Question 72

Diffusion characteristics of O2

Answer 72

in normal conditions, oxygen acts like a gas that is perfusion limited partial pressure is NOT zero partial pressures have to rise in the capillaries in order for this to occur. More or less at around a quarter of the way into the capillary, partial pressures are almost equilibrated and hemoglobin as a consequence has been saturated and most of the gas has been picked up. We say it has been a 100% saturated when it has picked up all the oxygen it can carry.

Question 73

diffusion characteristics of O2 - pathological

Answer 73

Remember, the goal of rising those partial pressures is to be able to saturate the hemoglobin so the oxygen is carried in appropriate quantities to the body. This now will not happen a third of the way into the capillary but as the hemoglobin is leaving the capillary, this is a reserve that we all have built in. As disease progresses, we don’t alter the capability of oxygenating our blood and we do not compromise the oxygen carrying capacity of the blood because essentially there is a good 2/3 of time left for oxygen to saturate hemoglobin in a normal scenario. In disease states as they progress, partial pressures will equilibrate with the alveolus later on and hopefully by that time, the disease doesn’t progress, hemoglobin will still exit fully saturated. , the person with interstitial edema now has to walk briskly. Their diffusion is already impaired but compensated at rest. The addition of a situation that would produce tachycardia and speed up circulation time will actually not allow enough time for hemoglobin to pick up the oxygen that is needed.

Question 74

measuring diffusion capacity

Answer 74

inhale CO D(L) = Vco/P1-P2 Because it is diffusion limited, its membrane limited. So any changes in gas transfer are because the membrane is diseased- I can’t say whether its an alteration of thickness or area but the membrane is responsible and the lungs consequently are always know P1, P2 = 0 We know the partial pressure inhaled, we assume the partial pressure in the capillary is 0 and therefore by rearranging and solving this equation we can calculate the actual Diffusion for the lungs.

Question 75

Dissolved O2

Answer 75

insufficient to eet metabolic needs proportional to partial pressure (henry's law)

Question 76

Hgb molecule

Answer 76

4 chains - 2 alpha, 2 beta heme group has iron molecule at the center It is usually stored in the ferrous state which is readily releasable from iron When the hemes don’t contain oxygen, the molecule is in a deoxygenated state or called a tight binding structure

Question 77

Hg-O2 dissocation curve

Answer 77

due to cooperatviity steep = easy unloading of )2 flat = stable O2 sat

Question 78

O2 content

Answer 78

O2 bound to Hg + O2 dissolved about 20 ml/100 ml

Question 79

Effect of Hgb concentration on O2 sat and content

Answer 79

hgb doesn't change much in disease but if shot or something, saturdation doesn't cahnge (no prob w diffusion) but O2 content does The amount of hemoglobin plays a very important role in the oxygen carrying capacity of the blood in comparison to the built-in fail-safe systems that we have with oxygen partial pressures

Question 80

P50 point

Answer 80

partial pressure at which Hgb is 50% saturated

Question 81

Temp and Hgb-O2 dissocation curve

Answer 81

as febrile - shift right less saturated for each PP

Question 82

acidosis and Hg-O2 dissocation curve

Answer 82

as more acidic - shfit right less O2 bound

Question 83

3 ways of CO2 in the blood

Answer 83

1. dissolved 2. carbamino Hgb (bound) 3. bicarbonate

Question 84

carbonic anhydrase

Answer 84

catalyzes combining of CO2 and H20 - FAST in RBC

Question 85

importance of proton on RBC

Answer 85

That proton will reduce hemoglobin and we saw what happens to the hemoglobin disassociation curve. It facilitates hemoglobin reduction and the reduction of hemoglobin actually favors release of oxygen and binding of carbon dioxide in order to generate carbaminohemoglobin. Now oxygen is ready to be dissolved and be brought back into the tissues for utilization. That’s how oxygen gets released.

Question 86

Haldane effect

Answer 86

If we have bicarbonate floating around the redox state will be altered. Therefore what happens to the bicarbonate is that it exits the cell by a process called chloride shift (via a pump) that, in order to maintain electro neutrality of the cell, favors exit of bicarbonate into the plasma and an influx of chloride ions to keep electroneutrality. Chloride sometimes binds to sodium/potassium to release salt.

Question 87

CO2 dissascociation curve

Answer 87

Scenarios where oxygen saturation is high, the concentration of CO2 is lower – where would this happen? Maybe In the lung tissues. This is the reverse of what happens in the body because now what is needed is for CO2 to be released from the hemoglobin in order for the hemoglobin molecule to be able to pick up an oxygen that is coming in from the alveoli and become fully oxygenated. At the level of the alveolar capillary membrane, the inverse effect occurs between oxygen and carbon dioxide release.

Question 89

hypoxic vasoconstriction

Answer 89

alveolar hypoxia constricts small pulmonary arteries probably direct effect of O2 on vascular smooth muscle critical at birth to transition for placental to air breathing directs blood from poorly ventilated and diseased lung

Question 90

column of blood

Answer 90

how blood moves through pulmonary system - no muscular arteries!

Question 91

caliber of alveolar vessels

Answer 91

determined by alveolar pressure

Question 92

caliber of extra alveolar vessels

Answer 92

caliber is determined by lung volume

Question 93

transmural pressure

Answer 93

difference between inside and outside of the capillary

Question 94

Pulmonary Vascular Resistance

Answer 94

measure of esistance to flow in the pulmonary vasculature system = input P - output P/blood flow low! and unique - becomes even lower as pressure increases in pulm SVR is 10x higher

Question 95

pressure-flow relationship in pulmonary vasculature

Answer 95

non linear! depends on flow rate higher flows - lower resistances SO as CO increases, pulmonary vascular resistance decreases

Question 96

why does PVR fall as flow increases?

Answer 96

1. capillaries that were already open distend further (compliant) 2. capillary that were once closed now open (recruitment)

Question 97

blood flow distribution and the lung

Answer 97

greatest in the bottom of the lung due to gravity

Question 98

transmural pressure

Answer 98

difference in P between the inside and outside of the capillaries collapse or distend

Question 99

transmural pressure of alveolar vessel

Answer 99

Pt = Pc - Pa ## Footnote •The capillary vessels are very prone to changes in the arterial venous pressure and also changes in the alveolar pressure since they are juxtaposed to the alveolus. exposed to alveolar pressure and are compressed if this increases

Question 100

transmural pressure of extraalveolar vessels

Answer 100

PT= Pv - PpI ## Footnote •The extra-alveolar vessels are less prone to this individual alveolar pressure and how it’s distend and more prone to changes in pleural pressure. So the extra-alveolar vessels are subject to pressures transmitted through the pulmonary artery and the pressures transmitted by the pleural pressures. are exposed to a pressure less than alveolar and are pulled open by the radial traction of the surrounding parenchyma

Question 101

total pulmonary vascular resistance

Answer 101

highest at RV (extraalveolar is super high) or at TLC (alveolar is super high) * Alveolar vessel * At low lung volumes = when the alveolus is not distended: blood flow in the alveolar capillary is increased. * At high lung volume = alveolus is hyperextended (green arrow): blood flow may be impaired and therefore resistance increases. * At the same time, the opposite is occurring in the extra-alveolar vessels. * So at low lung volumes= when lung is not distended: the resistance in the extra-alveolar vessels is very high (red arrow). * As you inflate to TLC, the resistance in the extra-alveolar vessels becomes much lower. **The combination of these two factors (alveolar and extra-alveolar) working together in the pulmonary circulation allows for the total pulmonary vascular resistance to be lowest at tidal breathing at FRC [she repeated this twice]**

Question 102

Zone I of the lung

Answer 102

PA\>Pa\>Pv almost no flow! P in alveoli is such that there is little flow and it isn't used much alveolar pressure is greater than BOTH local pulmonary arterial and venous vessels are compressed annd there is no flow, high pleural P * Further from the heart so that the effects of gravity are less; therefore, the driving pressure, the arterial pressure is less than it is at the base of the lung * the pressure in the alveoli is higher * Therefore, forward flow is determined most by the pressure in the alveolus and less by the difference between the alveolar and venous pressure. * This means that those alveolar-capillary units are closed = they are not actively being used at any given moment in time.

Question 103

Zone II of the lung

Answer 103

pulmonary arterial pressure is greater than alveolar pressure alveolar pressure is greater than venous pressure (so downstream pressure is alveolar) flow is independent of venous pressure -only ep on arterial and alveolar Pa \>PA \> Pv * Characteristics: * Closer to the heart, closer to the ground so gravity plays bigger role. Arterial pressure is higher than alveolar pressure in the same area of lung

Question 104

Zone III

Answer 104

pulmonary artery pressure is greater than venous and alveolar pressure venous pressure is greater than alveolar pressure (depends on arterio-venous difference) Pa\>Pv\>PA

Question 105

Summary of Zones and what determines flow

Answer 105

Recap Zone 1: Alveolar pressure determines flow. It is highest in this region and higher than the driving pulmonary artery pressure. There is little or no flow in this region. Zone 2: Flow is determined by the difference between arterial and alveolar pressure. Venous pressure is lower than alveolar pressure. Zone 3: Flow is determined by difference between arterial and venous pressure just like it would in the other arterial pressures.

Question 106

norepinephrine

Answer 106

will constrict pulmonary vessels but not very effective

Question 107

histamine

Answer 107

causes vasoconstriction in the pulmonary circuit - vasodilation in system

Question 108

hydrostatic pressure

Answer 108

pressure in vessels is higher than in interstitium favors movement of fluid into interstitium

Question 109

oncotic pressure

Answer 109

protein content higher in the vessels favors movement of fluid inward toward the vessels

Question 110

low pulmonary capillary pressure

Answer 110

favors reabsorption

Question 111

Starling equation

Answer 111

describes capillary filtration

Question 112

why is driving force in pulmonary so low?

Answer 112

Whereas in the pulmonary circulation, the driving pressure is low because it is a low pressure system, This favors reabsorption and allows the alveolus to remain dry so that O2 can go across the membrane.

Question 113

normal filtration

Answer 113

Figure A: Normal Pc-Pi: forward pressure that drives fluid into interstitium Πc-Πi: backward pressure that drives fluid back into capillary.

Question 114

interstitial edema

Answer 114

Figure B: If there is excess fluid due to perturbation of one of the pressures. Increased fluid being filtered à lymphatics will become more active in trying to filter fluid thereby keeping the alveolus dry.

Question 115

alveolar edema

Answer 115

Figure C: Something changes and there is a diseased state. If lymphatics are overwhelmed, fluid eventually goes into the alveolus.

Question 116

stages of pulmonary edema

Answer 116

1. There is increase in Pc that drives fluid into interstitium. + No change between the oncotic pressure in the interstitium and capillaries à overall increase in filtration 2. Initially lymphatics take up excess fluid 3. At some point lymphatics become overwhelmed à fluid goes into the alveolus and causes pulmonary edema.

Question 117

what causes increased capillary hydrostatic pressure

Answer 117

MI mitral stenosis fluid overload

Question 118

what causes increased capillary permeability

Answer 118

toxins sepsis ARDS

Question 119

what causes reduced lymph drainage

Answer 119

increased CVP

Question 121

pH

Answer 121

-log[H+]

Question 122

pH of blood in healthy

Answer 122

7.4

Question 123

Henry's Law

Answer 123

I have a beaker of water sitting in a chamber with CO₂ in the air. The partial pressure of CO₂ is [initially] in the air but not in water, so immediately there will be some CO₂ entering the beaker into solution. How much CO₂ will be in that beaker if you wanted to calculate it? It would eventually come into equilibrium with the atmosphere until partial pressure in beaker equaled the partial pressure in the atmosphere. I want to calculate the amount of CO₂ in the beaker. You guys were given this in Munger’s lecture: **Henry’s law**. Multiply pCO₂ by s, which is the solubility coefficient for CO₂. s=.03 for CO₂. The CO₂ content [in the beaker] is .03 x pCO2.

Question 124

Henderson Hasselbach for acid-base status in patients

Answer 124

pH = pKA + (log [HCO3-])/(.03 x PCO2) the pH in the blood is going to be dictated by the ratio of the bicarbonate to the pCO2

Question 125

2 routes of bicarbonate change

Answer 125

1. adding H+ or HCO3- 2. changes in CO2 (mass action)

Question 126

Davenport diagram axes

Answer 126

pH, [HCO3-]

Question 127

respiratory acidosis

Answer 127

i.e. opoid overdose decrease ventilation (hypoventilating) --\> increased Pco2 buffer effect = why it's not horizontal to the new PCO2 (generate more H+ and HCO3- but less acidic because of buffer!) to PCO2= 80 line

Question 128

compensation for respiratory acidosis

Answer 128

renal compensation retain some HCO3- stay at high PC)2 but incerase HCO3 so become less acidic) not full compensation

Question 129

respiratory alkalosis

Answer 129

i.e. panic attack and hyperventilation go to higher RR and lower PCO2 - more basic

Question 130

compensation for respiratory alkalosis

Answer 130

renal compensation kidney excrete HCO3- to become more acidic stay at same PCO2

Question 131

metabolic acidosis

Answer 131

i.e. diabetic ketoacidosis incrase H+ in the body - decrease HCO3 (because it is buffering the H) - pCO2 is constant

Question 132

compensation for metabolic acidosis

Answer 132

respiratory compensation hyperventilation to decrease CO2 buffering is the same and the lines are horizontal

Question 133

metabolic alkalosis

Answer 133

i.e. vomiting lose a lot of H no change in Pco2 but mecome more basic

Question 134

compensation for metabolic alkalosis

Answer 134

respiratory compensation hypoventilate to retain CO2 and make blood more acidic

Question 135

how is bicarbonate generated?

Answer 135

incrase in CO2 metabolic processes

Question 136

compensation in metabolic disorders

Answer 136

respiratory! begins immediately

Question 137

compensation in respiratory disorders

Answer 137

metabolic takes a few days to have effects

Question 138

isohydric principle

Answer 138

all buffers are in equilibrium - only have to analyze one system!