Pulm Physio Flashcards

1
Q

tidal volume

A

what enters our lungs during quiet resting breathing

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2
Q

inspiratory capacity

A

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

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3
Q

expiratory reserve volume

A

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

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4
Q

vital capacity

A

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.

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5
Q

functional residual capacity

A

. 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.

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6
Q

lung volumes assesd my spirometry

A

TV

IC

ERV

VC

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7
Q

determinants of FRC

A

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

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8
Q

intrapleural space

A

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)

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9
Q

transpulmonary pressure

A

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.

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10
Q

inspiration muscles

A

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

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11
Q

pressures on inspiration

A

•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

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12
Q

transpulmonary pressure at TLC

A

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

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13
Q

pressure of elastic recoil

A

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.

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14
Q

elastic recoil

A

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

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15
Q

recoil pressure

A

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

****provides driving force for expiration!

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16
Q

lung volumes that requrie measurement in addition to spirometry

A

TLC, RV, FRC

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17
Q

active v passive breathing

A

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.

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18
Q

residual volume

A

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.

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19
Q

lung compliance

A

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.

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20
Q

maximal recoil pressure

A

determined at TLC

pressure at TLC

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21
Q

lung compliance in fibrosis

A

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.

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22
Q

lung compliance in emphysema

A

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.

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23
Q

static balance in TLC

A

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

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24
Q

static balance in RV

A

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

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25
Q

FRC at base vs apex

A

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.

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26
Q

differences in pressure and ventilation

A

alveoli are more inflated at the apex (bigger transpulmonary pressure) BUT have small changes in vol for changes in P - less ventilation

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.

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27
Q

measurement of FRC

A

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.

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28
Q

P(chest wall)

A

Pcw = Pip - Patm

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29
Q

base chest wall resting potential

A

happiest at halfway to TLC

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

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30
Q

combined chest wall and lung complaince

A

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

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31
Q

fibrosis and lung compliance

A

FRC decreases!

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32
Q

effect of scoliosis and lung compliance

A

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.

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33
Q

alveolar stability

A

LaPlace’s law - hard to open alveoli w small radius

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.

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34
Q

effect of surface tension on lung compliance

A

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.

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35
Q

what is the maximal airflow?

A

500 L/min

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36
Q

what is airflow at rest?

A

5 L/min

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37
Q

intrapleural pressure during tidal inhale

A

, 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.

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38
Q

interpleural pressure during exhale

A

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

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39
Q

airflow rate in tracheobronchial tree

A

each branch - slows to 1/2 of the speed

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40
Q

Reynolds Number

A

if greater than 2000 favors turbulent flow

increases if: bigger radius, faster, bigger density

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41
Q

large airway and flow

A

fast, turbulent flow

small cross sectional area

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42
Q

small airway and flow

A

slow, laminar flow

large cross sectional area

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43
Q

transitional flow

A

between small and large vessels

intermediate flow and cross sectional area

mostly laminar but turbulent at bifurcations

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44
Q

elevated airway resistance

A

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!

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45
Q

treatment of elevated airway resistance

A

turbulent airflow!

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

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46
Q

flow equation

A

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

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47
Q

What will be compromised first, resting or forced airflow?

A

forced airflow

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48
Q

where in the tracheobronchial tree is airway resistance the highest?

A

trachea!

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49
Q

airway resistance and lung volume

A

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.

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50
Q

collapsibility of airway walls

A

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

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51
Q

FVC

A

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,

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52
Q

FEV1

A

this is the forced expiratory volume in the first second.

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53
Q

airflow in obstructive disease

A

i.e. asthma

lower percent is exhaled

takes longer to fully exhale

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54
Q

restrictive expiratory airflow

A

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.

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55
Q

spirometry curve

A
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56
Q

maximal expiratory curve

A

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.

57
Q

effort dependent exhalation

A

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

58
Q

effort independent exhalation

A

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

59
Q

why is there effort independent exhalation

A

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.

60
Q

minimum work concept

A

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.

61
Q

convection

A

airways

active - requires E

long distances

62
Q

diffusion

A

alveoli

passive - no E required

acts only in short distances

63
Q

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

A

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

64
Q

D(L)

A

Area, thickness and diffusion constant can be lumped into a term we called D(L), lung diffusion.

he flux of gas is equal to lung diffusion times the partial pressure differences.

65
Q

components of D(L)

A

C1: membrane diffusion

C2: binding to Hgb - capillary blood vol

66
Q

components of membrane diffusion

A

RBC membrane

blood plasma

capillary endothelial cell

interstitium alveolar epithelial cell

67
Q

D(m)

A

everything up to alveolus (all diffusion)

68
Q

V(c)

A

oxygen binding to hemoglobin

69
Q

theta

A

rate that O2 binds to hemogloben

70
Q

inhale CO

A

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.

71
Q

Diffusioncharacteristics of N2O

A

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

72
Q

Diffusion characteristics of O2

A

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.

73
Q

diffusion characteristics of O2 - pathological

A

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.

74
Q

measuring diffusion capacity

A

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.

75
Q

Dissolved O2

A

insufficient to eet metabolic needs

proportional to partial pressure (henry’s law)

76
Q

Hgb molecule

A

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

77
Q

Hg-O2 dissocation curve

A

due to cooperatviity

steep = easy unloading of )2

flat = stable O2 sat

78
Q

O2 content

A

O2 bound to Hg + O2 dissolved

about 20 ml/100 ml

79
Q

Effect of Hgb concentration on O2 sat and content

A

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

80
Q

P50 point

A

partial pressure at which Hgb is 50% saturated

81
Q

Temp and Hgb-O2 dissocation curve

A

as febrile - shift right

less saturated for each PP

82
Q

acidosis and Hg-O2 dissocation curve

A

as more acidic - shfit right

less O2 bound

83
Q

3 ways of CO2 in the blood

A
  1. dissolved
  2. carbamino Hgb (bound)
  3. bicarbonate
84
Q

carbonic anhydrase

A

catalyzes combining of CO2 and H20 - FAST in RBC

85
Q

importance of proton on RBC

A

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.

86
Q

Haldane effect

A

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.

87
Q

CO2 dissascociation curve

A

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.

88
Q
A
89
Q

hypoxic vasoconstriction

A

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

90
Q

column of blood

A

how blood moves through pulmonary system - no muscular arteries!

91
Q

caliber of alveolar vessels

A

determined by alveolar pressure

92
Q

caliber of extra alveolar vessels

A

caliber is determined by lung volume

93
Q

transmural pressure

A

difference between inside and outside of the capillary

94
Q

Pulmonary Vascular Resistance

A

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

95
Q

pressure-flow relationship in pulmonary vasculature

A

non linear!

depends on flow rate

higher flows - lower resistances SO as CO increases, pulmonary vascular resistance decreases

96
Q

why does PVR fall as flow increases?

A
  1. capillaries that were already open distend further (compliant)
  2. capillary that were once closed now open (recruitment)
97
Q

blood flow distribution and the lung

A

greatest in the bottom of the lung due to gravity

98
Q

transmural pressure

A

difference in P between the inside and outside of the capillaries

collapse or distend

99
Q

transmural pressure of alveolar vessel

A

Pt = Pc - Pa

•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

100
Q

transmural pressure of extraalveolar vessels

A

PT= Pv - PpI

•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

101
Q

total pulmonary vascular resistance

A

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]

102
Q

Zone I of the lung

A

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.
103
Q

Zone II of the lung

A

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

104
Q

Zone III

A

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

105
Q

Summary of Zones and what determines flow

A

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.

106
Q

norepinephrine

A

will constrict pulmonary vessels but not very effective

107
Q

histamine

A

causes vasoconstriction in the pulmonary circuit - vasodilation in system

108
Q

hydrostatic pressure

A

pressure in vessels is higher than in interstitium

favors movement of fluid into interstitium

109
Q

oncotic pressure

A

protein content higher in the vessels

favors movement of fluid inward toward the vessels

110
Q

low pulmonary capillary pressure

A

favors reabsorption

111
Q

Starling equation

A

describes capillary filtration

112
Q

why is driving force in pulmonary so low?

A

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.

113
Q

normal filtration

A

Figure A: Normal

Pc-Pi: forward pressure that drives fluid into interstitium

Πc-Πi: backward pressure that drives fluid back into capillary.

114
Q

interstitial edema

A

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.

115
Q

alveolar edema

A

Figure C: Something changes and there is a diseased state.

If lymphatics are overwhelmed, fluid eventually goes into the alveolus.

116
Q

stages of pulmonary edema

A
  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.
117
Q

what causes increased capillary hydrostatic pressure

A

MI

mitral stenosis

fluid overload

118
Q

what causes increased capillary permeability

A

toxins

sepsis

ARDS

119
Q

what causes reduced lymph drainage

A

increased CVP

120
Q
A
121
Q

pH

A

-log[H+]

122
Q

pH of blood in healthy

A

7.4

123
Q

Henry’s Law

A

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.

124
Q

Henderson Hasselbach for acid-base status in patients

A

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

125
Q

2 routes of bicarbonate change

A
  1. adding H+ or HCO3-
  2. changes in CO2 (mass action)
126
Q

Davenport diagram axes

A

pH, [HCO3-]

127
Q

respiratory acidosis

A

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

128
Q

compensation for respiratory acidosis

A

renal compensation

retain some HCO3-

stay at high PC)2 but incerase HCO3 so become less acidic)

not full compensation

129
Q

respiratory alkalosis

A

i.e. panic attack and hyperventilation

go to higher RR and lower PCO2 - more basic

130
Q

compensation for respiratory alkalosis

A

renal compensation

kidney excrete HCO3- to become more acidic

stay at same PCO2

131
Q

metabolic acidosis

A

i.e. diabetic ketoacidosis

incrase H+ in the body - decrease HCO3 (because it is buffering the H) - pCO2 is constant

132
Q

compensation for metabolic acidosis

A

respiratory compensation

hyperventilation to decrease CO2

buffering is the same and the lines are horizontal

133
Q

metabolic alkalosis

A

i.e. vomiting

lose a lot of H

no change in Pco2 but mecome more basic

134
Q

compensation for metabolic alkalosis

A

respiratory compensation

hypoventilate to retain CO2 and make blood more acidic

135
Q

how is bicarbonate generated?

A

incrase in CO2

metabolic processes

136
Q

compensation in metabolic disorders

A

respiratory! begins immediately

137
Q

compensation in respiratory disorders

A

metabolic

takes a few days to have effects

138
Q

isohydric principle

A

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

139
Q
A