Respiratory Physiology

Question 1

Where does tracheal collapse occur during the cycle of respiration?

Answer 1

Inspiratory collapse occurs in the cervical trachea

Expiratory collapse occurs in the intrathoracic trachea

Question 2

What is the pathophysiologic mechanism of laryngeal collapse in pugs?

Answer 2

Chondromalacia

Question 3

Explain anatomic dead space

Answer 3

Conducting airways incapable of gas exchange

Question 4

Describe airflow within the thorax

Answer 4

Air travels 16 generations to get to the alveoli
mainstem bronchi: semicircular cartilage rings similar to the trachea
bronchi in the lungs: irregular cartilate plates
bronchioles: muscular layer intermingled with elastic fibers. outer layer of the wall is made of connective tissue and elastic fibers

Question 5

Airflow through the large airways is via ______. The size and distribution of the airways occurs based off ________ + ________.
Beginning at the bronchioles, gas flow is determined by ______ instead of bulk flow.

Answer 5

bulk flow
resistance and dead space.
diffusion

Question 6

What cell types occur in the majority of the respiratory tract?
What replaces goblet cells in the bronchioles?

Answer 6

1. Ciliated pseudostratified columnar or cuboidal cells interspersed with mucus-secreting goblet cells
2. Club cells

Question 7

Describe cell types in the alveoli

Answer 7

cuboidal cells become squamous epithelial cells to maximize surface area and minimize the barrier to gas exchange.

Question 8

Name and describe the two principle cell types in the alveoli

Answer 8

type 1 modified squamous epithelial cells - line 95% of the alveolar surface
type 2 cuboidal alveolar cells - larger, 2x as many, produce surfactant

Question 9

Name the third alveolar cell type

Answer 9

phagocytic macrophages - patrol the alveolar surface engulfing bacteria and ingesting inhaled particulates

Question 10

Each alveolus is surrounded by how many pulmonary capillaries?

Answer 10

Approx. 500

Question 11

What are the pores of Kohn?

Answer 11

Pores that perforate the alveolar septae, which are supported by elastic and connective tissue fibers
They enable interalveolar communication and are a means of collateral ventilation
Collateral ventilation can occur to some extent through these pores enabling pressure equalization in adjvecent alveoli to prevent overdistension and/or collapse

Question 12

Explain the purpose and mechanisms of collateral ventilation

Answer 12

Without it, alveoli distal to obstructed airways would become atelectatic. 
interalveolar communications through the Pores of Kohn, between the bronchioles and
the alveoli (canals of Lambert), and the interbronchiolar communications of Martin. 
The relative contribution of these three different types of collateral ventilation is unknown. It is unlikely that any of these pathways are important in health. However, in the patients with lung diseases such as emphysema, there may be substantial collateral ventilation, which may be critical to maintaining oxygenation and ventilation

Question 13

Explain lung interdependence

Answer 13

The lungs are interdependent on each contiguous unit of connective tissue as well as on the adjacent chest wall.
If during inspiration there is a delay filling a part of the lung then the shape of the
chest wall in that area will be subtly distorted which then decreases pleural pressure in the area over the slowly filling lung. This decrease in pressure is transmitted to the alveoli thereby producing a greater pressure differential between the mouth and the alveoli which then augments or helps to correct flow to the area where it was lagging

Question 14

Name the components of the chest wall and additional muscles of respiration

Answer 14

Rib cage, external and internal intercostal muscles, parietal pleura, and the diaphragm

Accessory muscles including the rectus abdominis, laryngeal abductor, and sternohyoid and scalenus muscles are recruited to assist the ventilatory pump when the body needs to increase the rate and depth of breathing or when the ventilatory pump is impaired

Question 15

Name compensatory behaviors to increased work of breathing

Answer 15

patients will straighten the head and neck to reduce airway resistance resulting
from flexion of pharynx and trachea.
they will open their mouth to greatly increase oropharyngeal airway diameter (occurs early in dogs, and late in cats)
most patients will reduce their activity level to minimize oxygen consumption.

later postural adjustments may include a sitting position with forelimbs abducted. recruitment of the accessory muscles will then occur but can be difficult to detect, and hence abduction of the nasal cartilages may be the first indication beyond postural adjustments that respiratory workload has increased.
additional orofacial muscles may also contract with each inspiration, protruding the tongue, depressing the mandible and retracting the lip folds.
recruitment of the abdominal musculature may occur to provide active expiration and to shorten the expiratory time as part of increasing the frequency of breathing

Question 16

Describe the extrathoracic pressure changes during inspiration

Answer 16

During inspiration, the extrathoracic airways experience a transmural pressure gradient towards the lumen (pressing on the lumen) such that they collapse (low pressure within, high pressure outside). This is what leads to the characteristic inspiratory dyspnea seen in upper airway conditions such as upper airway obstructions, laryngeal paralysis, laryngeal collapse, tracheal collapse, tracheal stenosis and BAS. The reflexive behavior to increase inspiratory effort worsens the transmural pressure gradient

Question 17

What is transmural pressure

Answer 17

the pressure inside relative to outside of a compartment.
Under static conditions, the transmural pressure is equal to the elastic recoil pressure of the compartment.
The transmural pressure of the lungs is also called transpulmonary
pressure .

Question 18

What is paradoxical abdominal movement

Answer 18

When the abdominal wall moves in during inspiration.
It has just a few differentials – URT obstruction, pleural effusion, reduced pulmonary compliance (stiff lungs) and diaphragmatic rupture or paralysis

Question 19

Describe the intrathoracic pressure changes during inspiration

Answer 19

During inspiration, sub-atmospheric pressure is transferred from the intrathoracic space to the airways (-> air moves from the outside world (atmosphere) into the alveoli via pressure gradient). This pressure gradient is created through the concerted action of the respiratory muscles, most especially the diaphragm. The transmural pressure gradient created by the increase in intrathoracic volume expands the alveoli.

Question 20

Explain how the intrathoracic transmural/transpulmonary pressure is formed and the WOB

Answer 20

the respiratory muscles perform the work of breathing to generate the intrathoracic transmural pressure gradient by overcoming the elastic recoil of the lung tissue and chest wall. Some of this work is stored as elastic recoil energy in the lung tissues and
chest wall, enabling passive exhalation. The other major component of the work of breathing (WOB) is overcoming airway resistance, which is very small in normal lungs
- Since the WOB must overcome a combination of airway resistance and lung elastic recoil, any disease process which significantly increases either airway resistance or increases the tendency of the lungs to recoil / makes the lungs less compliant will increase the work of breathing. The phenomenon of an increased WOB is a very significant contributor to respiratory failure in patients with severe pulmonary disease. It is also the most difficult to objectively quantify. It has been estimated that the metabolic cost of breathing for patients with severe lung disease may increase from the normal 2% of total VO2 to between 25-50%. As a result, the increased oxygen content that can be obtained from this increased WOB is offset in large part by the increased oxygen consumption required to obtain it. The increased work of breathing also increases heat production and animals may become hyperthermic as a result, further increasing
respiratory drive.

Question 21

Describe what occurs at the end of inspiration

Answer 21

At the end of inspiration, there is potential energy stored in the elastic tissues of the lung, which during normal quiet expiration, returns the lung to its original volume.
The chest wall is elastic like the lung, but normally tends to expand
such that at the functional residual capacity, the inward recoil of the lung and the outward recoil of the chest wall are balanced.

Question 22

Describe functional residual capacity

Answer 22

the volume of gas in the lungs at the end of a passive exhalation. It is not the minimum volume of gas in the lungs (the residual volume), since an active expiratory effort can push additional air out of the lungs (the expiratory reserve volume).

Question 23

Describe what happens at FRC (the end of passive exhalation)

Answer 23

At the FRC, intrapleural pressure is sub-atmospheric (negative) because the pleural liquid is between the opposing forces of the inward recoil of the lungs and the outward
recoil of the chest wall. The lungs are above their resting volume while the thorax is below its resting volume. This potential space does not fill with gas from capillary blood because the partial pressure of the gasses dissolved in capillary blood is 706 (PH2O 47 + PCO2 46 + PN2 573 + PO2 40), which remains < atmospheric pressure of 760mmHg,
The natural sub-atmospheric intrapleural pressure is also the reason that air from a pneumothorax is absorbed if the communication between the alveoli and the pleural space closes.

Question 24

Describe pressure changes during forced expiration

Answer 24

During forced expiration, intrapleural pressure becomes positive causing a transmural closing pressure that drives gas out of the alveoli and collapses small airways. This dynamic compression of the airways during forced expiration results in flow that is effort-independent. = alveolar pressure minus intrapleural pressure
- this is why lower airway disease leads to expiratory dyspnea! In lower airway disease, inflamed and thickened intima and reduction of airway lumen by mucus leads to collapse of the lower airways as the transmural gradient is applied. In severe disease, air trapping occurs and requires some degree of active exhalation. A compensatory forced exhalation occurs which increases transmural pressure, exacerbates lower airway collapse and manifests as expiratory effort.

Question 25

What is Ohm's law

Answer 25

Pressure = V (flow) x R

Question 26

Define pulmonary resistance and Poiseuille’s law

Answer 26

resistance = Change in pressure/ the flow (V) resistance in the airways (Raw) = P alveolus - P mouth / V (flow) Raw = 8 η l / π r4 where n = viscosity, l = length and r = radius meaning the greater the radius of the tubular airway, the lower the resistance ( inversely proportional to the radius to the 4th power) medium sized bronchi have the greatest resistance in the lower airways (apart from the nose and upper airway)

Question 27

Describe the levels of resistance along the airway

Answer 27

the nose has the single largest fraction of resistance constituting 0.5-0.66 of the total resistance at low flow rates. Mouth, pharynx, larynx, and trachea provide 20-30% of the airway resistance. Most of the remainder of the airway resistance is associated with bronchi with diameters >2mm nasal resistance increases disproportionately with increasing flow rates such that during exercise, or when respiratory distress occurs, there is a switch from nasal breathing to mouth breathing.

Question 28

Describe nervous system control of the bronchioles

Answer 28

Bronchial smooth muscle is under control of the autonomic nervous system whereby parasympathetic stimulation causes bronchoconstriction and sympathetic stimulation causes bronchodilation. Contraction of the bronchial smooth muscles narrows their airway lumen and leads to increased resistance. - This is why Stimulation of the irritant receptors in the tracheobronchial tree induces bronchoconstriction reflexively via the vagal parasympathetic fibers. In patients with lung disease, mucosal edema, hypertrophy and hyperplasia of mucous glands, increased production of mucus, and hypertrophy of the bronchial smooth muscle all tend to decrease airway caliber and contribute to airway resistance. Administration of beta-adrenergic or methylxanthine bronchodilators or antiinflammatory corticosteroids to patients with airway diseases such as asthma leads to a significant decrease in the airway resistance and reductions in respiratory distress

Question 29

Describe compliance and elastic recoil

Answer 29

change in volume divided by the change in pressure C = ΔV/ΔP C = V2-V1 / P2-P1 Highly compliant lungs: small increase in pressure causes a large increase in volume. These lungs would have a steep slope on their pressure-volume curve. Less compliant lungs require a large distending pressure to effect a small change in volume. These lungs would have a shallow slope on their pressure-volume curve. - If a 20kg dog has normal lungs and a 2 cmH2O inspiratory pressure change causes a volume change of 200 mL then its pulmonary compliance is 100 mL/cmH2O. What if the same took in a normal 10mL/kg tidal volume and then decided to sniff in some more air? If the dog generated an additional 2 cmH2O of inspiratory pressure once the lungs were already distended by 200 mL of tidal volume there would not be a further increase in lung volume of 200 mL. The dog would need to generate a greater inspiratory pressure to achieve this. In other words, alveoli are more compliant (with less elastic recoil) at low volumes and are less compliant (with more elastic recoil) at high volumes

Question 30

Define hysteresis

Answer 30

At any given volume, the pressure is lower on expiration than inspiration meaning there is a loss of energy, which is mostly heat, and is described as hysteresis. For any given pressure, the volume of the lung is greater in expiration than inspiration. For the same change in pressure, the change in volume on inspiration is greater than the change in volume on expiration, meaning expiratory compliance is worse (less).

Question 31

List the five determinants of hysteresis

Answer 31

1. elastic recoil & surface tension - the pressure generated by elastic recoil on expiration is always less than the distending transmural pressure gradient required to inflate the lung - surfactant reduces surface tension and thus elastic recoil, which then improves compliance. When the lung is inflated, liquid molecules along the air-liquid interface are spread apart and the pull to one another, or the surface tension (the pressure causing elastic recoil, with the addition of lung connective tissue), is greater, thus the lung is less compliant than when it is deflated. Once the lung is fully inflated, the initial stages of the deflation curve have lower compliance or a lower change in volume over substantial changes in pressure 2. alveolar recruitment on inspiration and derecruitment on expiration - collapsed alveoli require greater pressure to open them up (change their volume) 3. gas absorption during measurement - not really a property of the lung parenchyma itself but rather an artifact of measurement. As mentioned above, measurement of static lung compliance has a certain built-in pause in every step, which allows some of the gas to become absorbed in living systems, leading to an apparent change in volume and pressure 4. relaxation of lung tissue: refers to the loss of energy in the lung parenchyma which occurs with stretch. This resembles the classical definition of hysteresis, as the quantity of unrecovered energy which results from something being imperfectly elastic. The imperfect lung stretches, consumes energy, and then wastes it on changing the shape of its collagen and elastin fibres instead of storing it for later release 5. differences in expiratory and inspiratory air flow (for dynamic compliance)

Question 32

What is surfactant

Answer 32

Complex liquid made by type II pneumocytes, consisting of 65% dipalmitoyl phosphatidylcholine, 20% phosphatidyl glycerol and 15% surfactant proteins SP-A, -B, -C and -D. Decreases surface tension and thus keeps alveoli from becoming atelectic A number of factors may modulate surfactant secretion, including glucocorticoids, epidermal growth factor, cyclic adenosine monophosphate (cAMP), and lung distention.

Question 33

Explain what happens in premature neonatal hypoxia

Answer 33

Premature neonates have inadequately developed surfactant production. As a result their lungs are unstable and have very low compliance. This instability leads to complete atelectasis of many alveoli, producing a right-to-left shunt and profound hypoxemia. The decreased compliance and atelectatic alveoli lead to alveolar hypoventilation. Surfactant production can be augmented by the administration of corticoids to the mother prenatally, which results in a decreased incidence of the syndrome.

Question 34

A lung unit with normal compliance and resistance will fill within a normal length of time and with a normal volume. Lung units with normal resistance but low compliance fill rapidly (fast lung unit / short time constant). Lung units with normal compliance but high resistance fill slowly (slow lung unit / long time constant). Flow applied to these two distinct lung units for the same amount of time will result in a difference in volume, or in other words to fill each unit equally requires different amounts of time. These units have different time constants. - Decisions regarding how best to ventilate an individual patient should be based on an assessment of the time constant of the majority of the lung and on how the patient responds to therapy. In patients with airway disease or abnormal compliance there can be transient gas movements out of some alveoli and into others as a result of lung units with different time constants, even when bulk flow has ceased at the mouth. Filling of a lung region with a partially obstructed airway will lag behind the rest of the lung such that it may continue to fill even when the rest of the lung has begun to empty with gas moving into from adjoining lung units. This phenomenon is called the pendelluft effect. As the breathing frequency is increased, the proportion of the tidal volume that goes to this partially obstructed region becomes smaller and smaller. Thus, less and less of the lung is participating in the tidal volume changes, and therefore the lung appears to become less compliant.

Answer 34

See above

Question 35

Differentiate static from dynamic compliance

Answer 35

Static compliance is a measure of the lung's true compliance after flow has stopped (after any inspiratory hold) Dynamic compliance occurs during a state of flow (measured using peak inspiratory pressure) and takes into account airway resistance Static compliance is always greater than dynamic compliance

Question 36

Define alveolar ventilation

Answer 36

Alveolar ventilation is the volume of fresh gas entering the respiratory zone per minute and is = to CO2 output divided by the concentration of CO2 in the expired gas. The concentration of CO2 in arterial blood and hence its partial pressure in alveolar gas is inversely related to alveolar ventilation. PaCO2 ∝ V̇ CO2 / V̇ A

Question 37

Differentiate physiologic from anatomic dead space

Answer 37

The physiologic dead space is the sum of the anatomic dead space and the alveolar dead space. Anatomic dead space is the amount of air remaining in the conducting airways at the end of inspiration that doesn't participate in gas exchange Alveolar dead space is the alveoli that are ventilated but not perfused and thus do not pick up CO2 Alveolar ventilation = minute ventilation – anatomic dead space ventilation V̇ A = V̇ T - V̇ D(Anatomic)

Question 38

Explain the PaCo2 vs EtCo2 gradient

Answer 38

If expired air originates from the alveolus, and all of the alveoli are able to participate in gas exchange, then the pCO2 of arterial blood and expired gas are the same. Because of alveolar dead space, however, the alveolar/EtCO2 are always lower than arterial as ventilated/inspired alveoli without perfusion (alveolar dead space) has the same pCO2 of inspired air, the EtCo2 is diluted and leads to a typical gradient of 2-5mmHg in healthy lungs.

Question 39

Give the equation for alveolar dead space

Answer 39

VD(Alveolar)/VT = (PaCO2 – PETCO2)/ PaCO2

Question 40

Things that increase anatomic dead space include

Answer 40

* Increasing body size * Increasing age * Increasing lung volume * Sat posture * Hypoxia (bronchoconstriction) * Lung disease (emphysema) * Endotracheal intubation

Question 41

Define physiologic dead space and list things that increase it

Answer 41

Physiologic dead space is the sum of the anatomic and alveolar dead spaces • Increasing age • Decreased pulmonary artery pressure (Increase in Zone 1) • IPPV (increased PVR, decreased pulmonary blood flow, short Ti causing maldistribution) • Increasing tidal volume • Hyperoxic vasodilatation • Anesthetic gases • Lung diseases (ALI/ARDS, PTE, atelectasis etc.)

Question 42

Define Fick's law of diffusion

Answer 42

Diffusion of the gas (V gas diffusing per minute) = (As x D × ∆P)/ T As = Membrane surface area D = Diffusion coefficient of the gas ΔP = Partial pressure gradient across the barrier between the pulmonary capillary and the alveoli T = Membrane thickness The volume of gas diffusion is inversely proportional to its thickness The surface area and membrane thickness can be altered by changes in the pulmonary capillary blood volume, the cardiac output or the pulmonary artery pressure, or by changes in lung volume.

Question 43

Describe diffusion-limited gases and give an example

Answer 43

In a diffusion-limited gas, the partial pressure of the gas in the pulmonary capillary equilibrates fully with the partial pressure of the gas in the alveoli during the time that the blood spends adjacent to the alveolus. The properties of the barrier and the diffusability of the gas limit its transfer across the barrier. Carbon monoxide Increasing the available surface area for diffusion can increase its uptake.

Question 44

Describe perfusion-limited gases and given an example

Answer 44

diffuse extremely rapidly and therefore the alveolar pressures of these gases equilibrates completely with the mixed venous blood before the blood has left the alveolar-capillary unit. nitrous oxide Additional diffusion of these gases is only possible once new mixed venous blood arrives at the alveolus. Under normal conditions the transfer of both oxygen and carbon dioxide are perfusion-limited but some diffusion limitation may occur under some conditions, including intense exercise, thickening of the blood-gas barrier, and alveolar hypoxia. Carbon dioxide is inherently much more diffusible than oxygen, however the alveolar-mixed venous gradient for CO2 is much lower than that for O2 (60 mmHg versus 5 mmHg), which offsets this inherent difference.

Question 45

List the 5 causes of hypoxemia

Answer 45

Hypoventilation Ventilation/Perfusion mismatch Shunting (right to left) whereby there is no ventilation at all Diffusion impairment - processes that thicken the barrier including interstitial or alveolar edema and interstitial or alveolar fibrosis - processes that decrease surface area including low cardiac output, tumors, or emphysema - processes that decrease erythrocyte uptake including anemia and low pulmonary capillary blood volumes Low inspired oxygen

Question 46

Explain how alveolar hypoventilation contributes to hypoxemia

Answer 46

Hypoventilation leads to an increase in the arterial partial pressure of Co2 which inevitably decreases the partial pressure of oxygen because it is more readily diffusible frm the arterial blood into the alveolus than oxygen is and because maximal alveolar pressure cannot exceed that of atmospheric pressure (760mmHg). This means the alveolar partial pressure of oxygen is capped by 760 - (partial pressure of nitrogen + water + CO2).

Question 47

Define the alveolar oxygen equation and its use with the alveolar-arterial oxygen gradient

Answer 47

The alveolar oxygen equation = [Fi02 x (Patm - PH20)] - pCO2/0.8 = 0.21 x (760-47) - pCO2/R (R=0.8) PH20 is the partial pressure of water or water vapor pressure 0.8 is the respiratory quotient which depends on the diet and metabolic state of the patient. the quotient = the amount of CO2 produced / the amount of O2 consumed. It is 0.8 for the standard human, carinovirous diet. As the proportion of carbohydrates increases, the quotient becomes closer to 1. The A-a gradient helps determine the effect of hypoventilation Increasing Fi02 has a substantial effect on increasing alveolar oxygen even in the face of increased pCO2 concentrations as it displaces nitrogen An elevated A-a gradient is caused by a normal shunt, VQ mismatch, and/or diffusion limitation

Question 48

Describe the P:F ratio and it's utility

Answer 48

This is the index of the partial pressure of paO2 and the Fi02. Normal P:F ratio is 500 whereby patient's breathing 21% oxygen have a paO2 of 100, 100% oxygen have a paO2 of 500 PAO2/FiO2 <300 = ALI and <200 = severe ARDS

Question 49

Describe pulmonary blood flow

Answer 49

Receives 2% of blood from the bronchial circulation and 98% of blood from the pulmonary circulation

Question 50

Describe pulmonary circulation to the lungs

Answer 50

The lungs receive the entire cardiac output of the right ventricle via pulmonary arteries. This blood is mixed venous blood and is pumped under low pressure to the lungs for gas exchange. Pulmonary vasculature is also a low pressure system with thin walls and minimal smooth muscle making it easily distensible and compressible. Gravity, body position, lung volume, alveolar and intrapleural pressures, intravascular pressures, and right ventricular cardiac output all impact pulmonary vascular resistance. In the pulmonary circulation, capillaries are the main capacitance vessels whereas the veins are in the systemic circulation.

Question 51

Describe bronchial circulation to the lungs

Answer 51

In addition to the mixed venous blood supplied to the lungs via the pulmonary circulation via the pulmonary "arteries," lungs also receive arterial blood from the left heart via the bronchial circulation.

Question 52

Describe vascular drainage of the lungs

Answer 52

The venous drainage of the bronchial circulation is split between the azygos vein (which drains the bronchoesophageal veins, intercostal muscles, and dorsal body wall before draining into the cranial vena cava as it enters the right atrium) and the pulmonary veins which enters the left atrium. As such the bronchial venous blood entering the pulmonary venous blood is part of the normal anatomic right-to-left shunt.

Question 53

Describe how PVR changes during phases of respiration

Answer 53

PVR is high at the residual volume (maximal forceful expiration), is low around FRC (end of passive expiration), and increases again during a normal inspiration towards TLC (total lung capacity) which is at the end of maximal inspiration

Question 54

Describe the changes in PVR during PPV (mechanical ventilation) and PEEP

Answer 54

During PPV, alveolar pressure and intrapleural pressure (ie intrathoracic pressures) are both positive during inspiration. As lung volume increases, alveolar and extra-alveolar vessels are compressed and PVR increases in addition to a decrease in venous return, right ventricular output, and pulmonary blood flow. This is even greater when there is PEEP as alveolar pressure and intrapleural pressure are positive during both inspiration and expiration. PEEP also increases RV workload and decreases venous return by increasing CVP and compressing intrathoracic vessels such as the vena cava. In healthy lungs with normal compliance, only 25% of PEEP is transmitted to the central veins. However this number increases with worsening compliance and is even more exacerbated with hypovolemia whereby substantial reductions in cardiac output, stroke volume, and pulse pressure variation can occur.

Question 55

Describe the effect of PEEP in the setting of CHF

Answer 55

In the setting of CHF where intravascular volume is increased by sodium and water retention, and preload is enhanced with sympathetically driven arteriolar constriction, PEEP decreases the work load of the heart by decreasing left and right ventricular preload. Additionally, PEEP decreases left ventricular afterload, which is = to the pressure generated by the myocardium + the pressure added to it by PEEP, or the transmural pressure (which is reduced in the setting of CHF and PEEP) and SVR. By reducing the transmural pressure, PEEP reduces afterload and assists left ventricular systolic function. The presence of PEEP in the setting of CHF pushes it into a more efficient part of the Frank-Starling curve.

Question 56

Explain the Frank Starling Curve

Answer 56

The heart is obligated to push out the volume which is delivered to it, or basically how ventricular preload affects stroke volume. The heart's ability to pump changes in response to the amount of venous return delivered to it. The x axis contains the left ventricular end diastolic pressure (LVEDP, mmHg) and the y axis represents stroke volume (SV, mLs)

Question 57

Explain how mean pulmonary artery pressure is maintained in the face of increasing cardiac output.

Answer 57

At normal cardiac output, not all pulmonary capillaries are perfused due to gavirty induced hydrostatic effects and variations in critical opening pressures. Thus once cardiac output is increased and pulmonary arterial blood flow increases, additional pulmonary capillaries are recruited and also distended. As the perfusion pressure increases, the transmural pressure gradient of the pulmonary blood vessels increases, distending vessels and reducing resistance. Derecruitment caused by low right ventricular output or high alveolar pressures decreases the surface area for gas exchange and may increase alveolar (physiologic) dead space.

Question 58

Explain hypoxic pulmonary vasoconstriction

Answer 58

When sections of alveoli are hypoxic or atelectic, there is compensatory vasoconstriction to shunt blood away from these poorly ventilated regions to well ventilated areas. The mechanism is poorly understood but is believed to be secondary to local vasoconstriction of hypoxia on pulmonary vascular smooth muscle at a pa02 of 20-100mmHg. Hypoxia inhibits an outward potassium current, depolarizing cells and allowing calcium entry, enabling contraction. Oxidation of the potassium channel opens it. High pulmonary artery pressures and alkalosis interfere with the response. When global hypoxia is encountered such as at altitude, whole-lung hypoxic pulmonary vasoconstriction can occur, greatly increasing pulmonary hydrostatic pressure and the workload on the right ventricle and may lead to high altitude pulmonary edema Hypercapnia also causes hpvc however the mechanism is in question.

Question 59

What determines V/Q and alveolar PO2 and PCO2. What is the normal VQ.

Answer 59

Well ventilated alveoli deliver O2 and take up CO2 as well perfused pulmonary capillaries deliver CO2 and take up O2. For a given lung unit, normal V/Q is 0.8-1.2 If V/Q increases for a given unit, alveolar pO2 will increase as pCO2 decreases If V/Q falls then oxygen delivery and carbon dioxide removal decreases, thus alveolar p02 will decrease and pCO2 will increase

Question 60

Explain normal inspired pO2 and pCO2 concentrations among alveoli and pulmonary capillaries under normal V/Q conditions

Answer 60

Inspired air entering the alveolus has a pO2 of 150mmHg and a pCO2 of 0 where as mixed venous blood entering pulmonary caillaries as a pO2 of 40 and pCO2 of 45mmHg. Following equilibration of both gases, the alveolar pAO2 is 100mHg and the pACO2 is 40. The gradient for the diffusion of oxygen into capillaries is thus 60 while the gradient for diffusion of CO2 into alveolar gas is about 5. These gradients are what make oxygen and carbon dioxide perfusion limited despite CO2 being 20x more soluble than oxygen.

Question 61

Describe what happens in a right to left shunt

Answer 61

No fresh inspired air enters that alveolus as it is not ventilated. The alveolar CO2 and O2 concentrations match new mixed venous blood that enters as a result of normal perfusion . Blood perfusing this alveolus leaves exactly as it entered and joins circulation in the pulmonary venous system going to the left side of the heart however the CO2 concentration is greater and the O2 concentration is lower.

Question 62

Describe what happens when V/Q and what this means

Answer 62

When V/Q is infinite, blood flow is completely occluded. None of the available oxygen can enter the pulmonary capillary blood and no carbon dioxide can enter the alveolus. Thus the gas composition of this unperfused alveolus is the same as inspired air. This is alveolar dead space. Each individual alveolar-capillary unit can exist anywhere on the continuum between infinite V̇ /Q̇ and zero V̇ /Q̇ . Units with low V̇ /Q̇ have low PO2 and high PCO2 while units with high V̇ /Q̇ have high PO2 and low PCO2.

Question 63

List ways V/Q mismatch can be calculated

Answer 63

calculating physiologic shunt, the physiologic dead space and differences between the alveolar and arterial PO2 and PCO2 values.

Question 64

Describe right-to-left shunts

Answer 64

when mixed venous blood enters the left side of the heart (arterial circulation) without having been (fully) oxygenated. shunts can be anatomic, absolute (true) intrapulmonary shunts, or shunt-like states (Small areas of low V/Q) Physiologic anatomic shunts include systemic venous blood entering the left ventricle without having traversed pulmonary vasculature and accounts for 1-2% of cardiac output - bronchial veins, thesbian veins, and pleural veins. Pathologic anatomic shunts include right to left intra-cardiac shunt such as tetralogy of fallot (pulmonic stenosis, VSD, right ventricular hypertrophy, and overriding aorta; clinical signs include cyanosis and polycythemia) True (intrapulmonary) shunts occur when mixed venous pulmonary blood that traverses along unventilated or collapsed alveoli. Shunt like states are just smaller regions of this

Question 65

Calculate shunt fraction

Answer 65

Shunt fraction = Qs/Qt = (CcO2 - CaO2)/ (CcO2 - CvO2) Qs = shunt flow Qt = total cardiac output CcO2 = calculated pulmonary capillary blood oxygen content = ([Hb (g/L)] x 1.34 (hufner's factor, a constant (1.34-1.39), maximum oxygen carrying capacity of blood in ml O2/g Hb) x SaO2 (% saturation of hemoglobin)) + (0.003 (oxygens water solubility at 37*C 0.03 ml/L/mmHg) x pa02 (partial pressure of arterial oxygen)) CaO2 = arterial oxygen content CvO2 = mixed venous oxygen content venous admixture is the resulting amount of shunt flow that enters arterial flow to assess for the presence of shunt, calculate this at room air or slightly increased Fi02 levels then again after a few minutes on 100% oxygen. After a few minutes, the areas of low V/Q should have enough pa02 values to fully saturate the hemoglobin perfusing them to leave you with a shunt fraction that is representative of true/absolute shunts.

Question 66

Describe oxygen solubility in plasma

Answer 66

At 37°C, 1 mL of normal arterial blood with a PaO2 of ~100 mmHg contains ~0.003 mL O2, thus most of the oxygen within the bloodstream is combined with hemoglobin

Question 67

Describe the structure of hemoglobin

Answer 67

Hemoglobin consists of four, linked polypeptide chains each attached to a ferrous iron containing heme group. Each molecule of hemoglobin can combine to 8 oxygen molecules. Each gram of Hb is capable of combining with about 1.34-1.39 mL O2 depending upon the level of other Hb forms such as methemoglobin (Fe3+ iron), or carboxyhemoglobin (CO + Fe2+).

Question 68

Describe the oxygen hemoglobine dissociation curve

Answer 68

Describes how the availability of oxygen (pa02) affects hemoglobin's affinity for it pa02 is on the x-axis Hemoglobin saturation (Sp02) is on the y-axis The curve is sigmoid as a result of positive cooperativity whereby the binding of one hb unit encourages binding of more until they are all bound ``` Oxygen loading (left-shift) is facilitated by conditions that occur in the lungs (low [H+], low PCO2, low temperature) and unloading (right-shift) is favored by conditions prevalent in the tissues (high [H+], high PCO2, high temperature, high 2,3DPG). The influence of pH (and PCO2) on the oxyhemoglobin dissociation curve is referred to as the Bohr effect. ```

Question 69

Explain why stored red blood cells do not increase oxygen

Answer 69

The increased oxygen affinity of stored packed red blood cells is due to the decrease in the concentration of 2,3-DPG that occurs during storage. These reductions can be mitigated through the use of red cell preserving additives such as (phosphate-adenine-glucose-guanosine-gluconate-mannitol [PAGGGM], saline-adenine-glucose-mannitol [SAGM], Adsol and Optisol). The increased oxygen affinity of hemoglobin in stored blood results in a reduction in the efficiency of oxygen unloading at the tissues and hence limitations in the effectiveness of stored blood transfusions to improve tissue oxygen delivery.

Question 70

Describe CO2 transport in the blood

Answer 70

Most of the CO2 in the blood is in the form of bicarbonate with smaller amounts as dissolved and carbamino compounds. 5-10% of total CO2 is in the form of actual CO2. CO2 is 20x more soluble in plasma than oxygen. Carbon dioxide can combine chemically with the terminal amine groups in blood proteins, particularly Hb, forming carbamino compounds, notably carbaminohemoglobin. This reaction occurs rapidly without the aid of catalysis and liberates protons. Deoxyhemoglobin can bind more carbon dioxide as carbamino groups than can oxyhemoglobin. Therefore, as the Hb in the venous blood enters the lung and combines with oxygen, it releases carbon dioxide from its terminal amine groups. About 5-10% of the total carbon dioxide content of blood is in the form of carbamino compounds. Most (80-90%) CO2 is carried as bicarbonate ions (HCO3-) generated by this carbonic anhydrase catalyzed reaction

Question 71

Define the Henderson Hasselbach Equation

Answer 71

In the presence of carbonic anhydrase, H20 readily combines with CO2 to from carbonic acid. CAH is present in concentrated forms in erythrocytes. In this way, Hb is present to accept the protons as a result of the dissociation of CA. H20 + CO2 -><- H2CO3 -><- H + HCO3 The carbon dioxide dissociation curve is much steeper than the oxygen hemoglobin curve. the cure is shifted to the right in the presence of oxyhemoglobin and shifted to the left in the presence of deoxyhemoglobin

Question 72

Describe the Haldane Effect

Answer 72

This is when the carbon dioxide dissociation curve is shifted to the right in the presence of oxyhemoglobin (for unloading of CO2 to the lungs) and shifted to the left in the presence of deoxyhemoglobin (for better binding at the tissue bed).

Question 73

Describe CO2 and O2 binding at the tissue bed

Answer 73

In tissues, PO2 is low and PCO2 is high CO2 dissolves in the plasma and diffuses into the erythrocyte. where it then dissolves into the cytosol and forms carbamino compounds and is then hydrated by CA to form carbonic acid. This carbonic acid then dissociates into protons and bicarbonate within erythrocytes. At low PO2 values, most Hb exists as deoxyhemoglobin. The deoxyhemoglobin is able to accept the protons liberated by the dissociation of carbonic acid and the formation of carbamino compounds (the Haldane effect). The hydrogen ions released by the dissociation of carbonic acid and formation of carbamino compounds bind to specific amino acid residues on the globin chains facilitating the release of oxygen from Hb (the Bohr effect). Both the Bohr and the Haldane effects occur because oxyhemoglobin is a stronger acid than deoxyhemoglobin, which more readily accepts protons in a transfer called the isohydric shift.

Question 74

Describe release of CO2 from erythrocytes

Answer 74

Bicarbonate ions diffuse out of the erythrocyte through the cell membrane much more readily than do hydrogen ions. Because more bicarbonate ions than hydrogen ions leave the erythrocyte, electrical neutrality is maintained by the exchange of chloride ions for bicarbonate ions by the bicarbonate-chloride carrier protein (band-3). This is called the chloride shift. Small amounts of water also move into the cell to maintain the osmotic equilibrium, increasing the erythrocyte MCV and hence increasing the venous PCV (typically ~2% above arterial PCV). At the lung, the PO2 is high and the PCO2 is low. As oxygen combines with Hb, the hydrogen ions that were taken up when it was in the deoxyhemoglobin state are released. They combine with bicarbonate ions, forming carbonic acid. This breaks down into carbon dioxide and water. At the same time, carbon dioxide is also released from the carbamino compounds. Carbon dioxide then diffuses out of the red blood cells and plasma and into the alveoli. A chloride shift occurs in the opposite direction to that in the tissues to maintain electrical neutrality. Water also moves out of cells to maintain osmotic equilibrium thereby reducing MCV and hence reducing the arterial PCV

Question 75

Describe fluid leakage in the lungs

Answer 75

Fluid mostly leaks from alveolar capillaries. At the capillary is the microvascular endothelium and the alveolar epithelium. Fluid accumulates in the interstitium before entering the alveoli.

Question 76

List and Explain the Modified Starling Equation

Answer 76

Jv = K[(Pcap – Pint) – sigma (πcap − πsg)] Jv = microvascular filtration rate K = hydraulic conductivity (a measure of water permeability) Pcap = hydrostatic pressure in the capillaries (~10mmHg) Pint = hydrostatic pressure in the interstitium sigma = Osmotic reflection coefficient (protein permeability of microvascular barrier) πcap = colloidal oncotic pressure in the capillaries πsg = colloid osmotic pressure subglycocalyx - increases in capillary hydrostatic pressure will lead to endothelial fluid flux into the interstitium -> if lymphatic drainage is overwhelmed, edema may results - the pressure within the interstitium is slightly positive. Any condition that rapidly/drastically decreases the pressure can also lead to edema formation such as chronic effusions or pneumothorax - interstitial hydrostatic pressure can also be reduced when pulmonary surfactant production is decreased such as in ARDS and PTE - decreases in the reflection coefficient will permit more solutes to leave the capillaries and promote edema formation. This is especially true when synthetic colloids leak into the interstitium and attract more fluid to it - damage to capillary endothelium from infection, toxins, oxygen toxicity, etc. - although decreases in the plasma colloid osmotic pressure may promote pulmonary edema, this is very rarely the cause of edema formation without additional pathology. - the lungs have a 10 fold ability to increase lymphatic drainage however if this is overwhelmed or compressed by tumors or parasitic disease, edema can occur

Question 77

Describe electrolyte transport in the lungs

Answer 77

The alveolar epithelial apical membrane contains sodium and chloride channels, while the basolateral membrane contains the Na+K+ATPase. active sodium transport in the distal lung epithelium is the primary mechanism that drives fluid clearance from the distal airspaces. The rate of fluid transport from the airspaces can be increased by cAMP stimulation. Catecholamine independent mechanisms including hormones, growth factors, and cytokines also contribute to epithelial fluid clearance in the lung. Fluid reabsorption is dependent on the active transport of sodium and other electrolytes, which drives water reabsorption. Manipulation of endogenous alveolar fluid clearance mechanisms in patients with ALI or ARDS has been the focus of much research interest and was the basis for the BALTI trials where beta-agonists were theorized to accelerate the rate of alveolar fluid clearance due to an increase in intracellular cAMP, resulting in increased sodium transport across alveolar type II cells through up-regulation of the apical sodium and chloride pathways and Na+K+ATPase. The first BALTI trial suggested that intravenous salbutamol in ALI/ARDS reduces extravascular lung water. Despite this promise however, the BALTI-2 trial concluded that treatment with intravenous salbutamol early in the course of ARDS was poorly tolerated, was unlikely to be beneficial and could worsen outcomes (increased 28-day mortality (34% versus 23%) at the interim analysis, mostly due to tachyarrhythmias)

Question 78

Describe control of breathing

Answer 78

The brain (PONS and MEDULLA of the BRAINSTEM) controls breathing as the respiratory muscles do not contract spontaneously. Spontaneous breathing is generated by neurons in the medulla. The respiratory control centers are organized into three groups of neurons located bilaterally in the medulla and pons. 1. The pontine group contains the pneumotaxic center in the upper pons- which transmits inhibitory signals to the inspiratory group. 2. The dorsal medullary group is located in the ventrolateral region of the nucleus of the solitary tract and in the reticular formation and are active during inspiration. They project information to the contralateral spinal cord drive by pattern generators in the rostral ventromedial medulla. 3. The ventral medullary group in the ambiguus and retroambiguus nuclei contains both inspiraton and expiratory neurons that project contralaterally to the spinal cord. The medullary respiratory center is also the final integration point for influences from higher brain centers; for information from central and peripheral chemoreceptors; and for afferent information from neural receptors in the airways, joints, muscles, skin, and elsewhere in the body. - Traumatic brain injuries are associated with alterations in the breathing pattern likely as a result of disruption of the neuronal pathways that transmit information between the various respiratory centers in the medulla and integrate information from the cerebrum, thalamus, pons and midbrain. Lesions in the diencephalon have been associated with Cheyne-Stokes respiration (irregular periods of fast and shallow, slow and heavy, and then apnea); midbrain lesions with hyperventilation; pontine lesions with apneustic breathing (regular deep inspirations with an inspiratory pause followed by inadequate expiration) and medullary lesions with ataxic breathing (abnormal pattern of breathing characterized by complete irregularity of breathing, with irregular pauses and increasing periods of apnea).

Question 79

What parameter controls ventilation?

Answer 79

PaCO2 most of the control is via the central chemoreceptors which are located near the ventral surface of the medulla and respond to changes in pH of the CSF, which in turn are caused by diffusion of CO2 from brain capillaries. The central chemoreceptors do not respond to changes in oxygen tension. PaO2 does not normally affect ventilation, but it may be more important at altitude and in lung disease. The response to hypoxia is enhanced by hypercarbia, acidosis and exercise.

Question 80

Explain how chemoreceptors react to rising paCO2 levels

Answer 80

As PaCO2 rises, PCSFCO2 increases, lowering the pH and stimulating respiration, thereby reducing PaCO2. With prolonged changes in PaCO2, the response of the central chemoreceptors is blunted by compensatory normalization of the CSF pH by production of HCO3- ions by glial cells.

Question 81

Explain peripheral chemoreceptors and list what they are stimulated by

Answer 81

peripheral chemoreceptors are rapid-response sensors of the arterial blood they are stimulated by hypoxemia, hypercapnia, academia, fever and hypoperfusion. The carotid bodies, located at the bifurcation of the carotid artery are largely responsible for the ventilatory response. These bodies contain large sinusoids with a very high rate of perfusion. Hypoxemia is sensed by oxygen-dependent potassium channels in the glomus cells that synapse with neurons supplying the carotid bodies whose cell bodies are located in the glossopharyngeal nerve (CN 9). Stimulation of peripheral chemoreceptors leads to an increase in ventilation and the carotid bodies undergo hypertrophy and hyperplasia under conditions of chronic hypoxia.

Question 82

Describe the lungs' response to increased lung volume

Answer 82

There are slowly adapting stretch receptors in bronchi and bronchioles with vagal myelinated afferents (towards CNS). Increased lung volume triggers the Hering–Breuer inflation reflex which terminates inspiration

Question 83

Explain the Hering–Breuer inflation reflex

Answer 83

reflex triggered to prevent the over-inflation of the lung. Pulmonary stretch receptors present on the wall of bronchi and bronchioles of the airways respond to excessive stretching of the lung during large inspirations

Question 84

Rapid lung expansion stimulates what receptors

Answer 84

Rapidly adapting pulmonary stretch receptors in the trachea and large bronchi responsible for cough, bronchospasm, and mucus production

Question 85

Describe J receptors

Answer 85

J-receptors are unmyelinated juxta-pulmonary capillary C-fibers that respond to increased pulmonary interstitial volume and mediate hyperventilation associated with increased left atrial pressure such as occurs in mitral valve disease

Question 86

Describe pulmonary muscle stretch stimulation

Answer 86

afferent fibers from muscle spindles and Golgi tendon organs within the inspiratory muscles that are stimulated by muscular stretch. These form part of segmental spinal cord reflexes and are involved in the sensation and regulation of breathing.

Question 87

Describe the function of pulmonary irritant receptors

Answer 87

they are vagal myelinated afferents in the airway epithelium. These sensory nerves respond to inhaled chemical irritants and to endogenous chemical stimuli, e.g. histamine released by mast cells and by reduced lung compliance and lung deflation.

Question 88

Describe the effects of furosemide on vagally mediated mechanoreceptors in the airways

Answer 88

Vagal afferent fibers are involved in modulation of the sensation of dyspnea and hence inhaled furosemide might modify the sensation of dyspnea. A double blind, randomized, crossover study in people suggested that inhaled furosemide might diminish and delay the onset of respiratory discomfort during loaded breathing.

Question 89

Get the table on page 14 of the physiology notes

Question 90

Explain the lung's effect on immunity

Answer 90

Pulmonary mast cells produce and release histamine, lysosomal enzymes, prostaglandins, leukotrienes, plateletactivating factor, neutrophil and eosinophil chemotactic factors, and serotonin in response to conditions such as pulmonary embolism and anaphylaxis. They may cause bronchoconstriction or immune or inflammatory responses, or they may initiate cardiopulmonary reflexes. The pulmonary circulation filters a variety of particles out of the circulation including microthrombi, gas bubbles, fat cells, cancer cells, platelet-leukocyte aggregates and particulates from intravenous drugs and fluid therapies. - The huge pulmonary reserve means that temporary occlusion of a few pulmonary capillaries by such particulates is largely clinically irrelevant – which is why we can get away with small gas bubbles in our fluid lines. Most of these particles are removed by the action of phagocytic cells, moved into lymphatics, broken down by the inherent fibrinolytic properties of the pulmonary vasculature or diffuse away into the adjacent alveoli. Cancer cells that metastasize and lodge in the pulmonary vasculature can begin to replicate however, which is ultimately why the lungs are a common site for metastatic disease.

Question 91

What is the equation of motion for inspiration

Answer 91

P = RxQ + V/C Pmus (pressure generated by inspiratory muscles) + Paw (airway pressure) = resistance x flow + elastance (inverse of compliance) x volume (TV) = resistive forces + elastic forces

Question 92

Describe transmural pressure

Answer 92

pressure inside relative to the outside. In the case of the lungs, this is called the transpulmonary pressure Under static conditions, transmural pressure is equal to the elastic recoil pressure of the lungs.

Question 93

Describe Boyle's law

Answer 93

Boyle's law explains that pressure and volume are always inversely proportional at a given temperature of a gas. It explains that when the volume of the lung increases during inspiration, the pressure in the lung will decrease. This causes air at atmospheric pressure to rush in and fill the lung

Question 94

Describe Ohm's law

Answer 94

``` Pressure = flow x resistance Resistance = pressure/flow Airway resistance = (Palv - Pmouth)/flow Ex: Volume controlled breath with constant flow • Raw = (PIP – PPlat) / Flow rate • Raw = (30 – 15) / 60 L/min • Raw = 15 **cmH2O/L/s** ```

Question 95

What is Pouiselle's law

Answer 95

Raw = 8 η l / π r4

Question 96

What is the formula for static compliance

Answer 96

Cs = TV/ (Pplat - Peep) = 300/(3 - 0) = 100

Question 97

When should supplemental oxygen be provided

Answer 97

at a pa02 < 70 or sp02 <93

Question 98

Describe the path of oxygen from room air to the alveoli and the resultant change in room air

Answer 98

When the air mixture reaches the lung, it has been humidified. The pressure of the water vapor in the lung does not change the pressure of the air, but it must be included in the partial pressure equation. For this calculation, the water pressure (47 mm Hg) is subtracted from the atmospheric pressure 760 mm Hg − 47 mm Hg = 713 mm Hg and the partial pressure of oxygen is (760 mm Hg − 47 mm Hg) × 0.21 = 150 mm Hg. These pressures determine the gas exchange, or the flow of gas, in the system. Oxygen and carbon dioxide will flow according to their pressure gradient from high to low.

Question 99

Explain how high altitude affects pa02 levels

Answer 99

At high altitudes, Patm decreases but the concentration (%) of gases does not change; the partial pressure decrease is due to the reduction in Patm

Question 100

Explain partial pressures of atmospheric gases

Answer 100

Air is a mixture of gases, primarily nitrogen (N2; 78.6 percent), oxygen (O2; 20.9 percent), water vapor (H2O; 0.5 percent), and carbon dioxide (CO2; 0.04 percent). Each gas component of that mixture exerts a pressure. The pressure for an individual gas in the mixture is the partial pressure of that gas. Approximately 21 percent of atmospheric gas is oxygen. Carbon dioxide, however, is found in relatively small amounts, 0.04 percent. The partial pressure for oxygen is much greater than that of carbon dioxide. The partial pressure of any gas can be calculated by: P = (Patm) × (percent content in mixture). Patm, the atmospheric pressure, is the sum of all of the partial pressures of the atmospheric gases added together, Patm= PN2+ PO2+ PH2O+ PCO2= 760 mm Hg × (percent content in mixture). The pressure of the atmosphere at sea level is 760 mm Hg. Therefore, the partial pressure of oxygen is: PO2= (760 mm Hg) (0.21) = 160 mm Hg

Question 101

If hypoxemia can lead to tissue hypoxia as DO = CaO2 x CO, what is the body's way of trying to prevent hypoxia?

Answer 101

Increasing CO

Question 102

What cause of hypoxemia will not improve with oxygen supplementation?

Answer 102

intrapulmonary shunt

Question 103

What is the primary and secondary benefit to supplementing oxygen?

Answer 103

Increase the arterial oxygen content (CaO2) in patients at risk of hypoxemia but secondarily is of venefit to all patients at risk of tissue hypoxia CaO2 = (Hb g/dL) 1.34 (mLs of oxygen/g Hb Huffners constant, theoretical oxygen carrying capacity of hemoglobin, maximum saturation) x SaO2 %) + (0.003 content, in mlO2/dL/mmHg, of dissolved oxygen in blood x PaO2mmHg)

Question 104

Reasons why supplemental oxygen should be humidified after a couple of hours and how to do so

Answer 104

Otherwise nasal mucosa becomes dry and dehydrated, respiratory epithelium become degenerated, mucosal secretion is impaired, and there is increased risk of infection Supplemental oxygen can be bubbled through a canister of sterile saline or water.

Question 105

How close should supplemental oxygen ideally be to the patient

Answer 105

2cm of the nostrils

Question 106

Oxygen flow rates of 2-4L/min generally provides what approx range of Fi02?

Answer 106

25-40%

Question 107

Give the delivery of oxygen formula

Answer 107

DO2 (rate of oxygen delivery in **mLs/min**) = CO x CaO2 CO = HR X SV; units are **L/min** MAP = CO x SVR CaO2 = (Hb (g/dL) x SaO2(%) x 1.34(mLsO2/gHb)) + (0.003 (**mLs**O2/dLhb/mmHg) x PaO2 (mmHg))

Question 108

What flow rates and Fi02s can be achieved with a tight fitting face mask

Answer 108

at 8-12L/min, the FiO2 can be up to 50-60% | Downside: rebreathing CO2, not tolerated in awake patient

Question 109

What flow rates and FiO2 can be achieved with a homemade oxygen hood

Answer 109

0.5-1L/min can achieve flow rates of 30-40% | Downsides: rebreathing Co2 and excessive humidification

Question 110

What is the ideal temperature of oxygen cages to prevent hyperthermia?

Answer 110

70*F

Question 111

List invasive methods of oxygenation

Answer 111

``` Nasal prongs Nasal cannulas (Nasopharyngeal oxygen) Endotracheal intubation Transtracheal oxygenation Mechanical ventilation High flow nasal oxygen Hyperbaric oxygen ```

Question 112

List the dosing range for nasopharyngeal/nasal cannula oxygen

Answer 112

50mLs/kg/min achieves avg **TRACHEAL** FiO2 of around 30% +/- 5 100mLs/kg/min achieves avg FiO2 of around 40% +/- 5 - usually the comfort level 200mLs/kg/min = ~60% +/- 10! 400 (if bilateral nasal cannulas) achieves avg FiO2 ~80%! +/- 10! Downside: nasal can lead to sneezing, nasopharyngeal bypasses the nasal turbinates which warms the air, both can cause epistaxis and irritation

Question 113

Describe the steps and locations of nasal vs nasopharyngeal oxygen cannulas

Answer 113

Numb the nares with proparacaine 5-10Fr red rubber catheter pre-measured to the lateral canthus of the eye for nasal and ramus of the mandible for nasopharyngeal Lube the tube and insert through the ventral medius. For nasopharyngeal, push the lateral aspect of the nostril medially and the nasal philtrum dorsally secure in place attach to oxygen tubing from humidified source with 1mL syringe or christmas tree adapter.

Question 114

List two methods of transtracheal oxygen supplementation

Answer 114

1. Through the needle large bore catheter percutaneously through the skin and underlying tissues directly into the trachea 2. incision into the trachea with placement of a large bore, multifenestrated catheter

Question 115

Describe how to place a through the needle large bore catheter percutaneously into the trachea

Answer 115

Shave just caudal to the larynx to the thoracic inlet aseptically scrub, block 3-5th tracheal rings with 2% lidocaine through the skina dn SQ scrub again knick the skin with #11 blade the needle of the catheter is then inserted through the skin, SQ, and sternohyoideus muscle into the trachea the catheter is then advanced into the trachea to its hub, which can be down to the carina in some cases it is sutured in place or connected to the patient with white tape and connected to humidified oxygen (tubing) with 1mL syringe or christmas tree adapter at a rate of 50-150mL/kg/min downsides/risks: kinking, occlusion, or dislodging of the catheter with cervical ventroflexion or excessive skin folds,

Question 116

Describe the more invasive method of intratracheal administration

Answer 116

The same set up as the percutaneous catheter except for the addition of a sterile field and a cut down is made similar to a tracheostomy. Heavy sedation or short acting anesthetic such as propofol or fentanyl/diazepam. Blocked with 1-2mg/kg of 2% lidocaine. Vertical skin incision made atop the 3rd through 5th rings, bluntly dissect SQ and sternohydoieus muscles down to tracheal rings using metzembaums or curved hemostats. A small incision is made between the 4th and 5th tracheal ring using a #11 blade and taking care not to cut more than 50% of the circumference of the trachea. A curved hemostat is used to poke through the hole followed by a groove director froma spay pack for insertion into the tracheal lumen. A large bore multifenestrated catheter with a stylette is then inserted. The stylette and grooved director can then be removed. the catheter is similarly secured with lengths of white tape and a 4x4 quaze with antimicrobial ointment placed over the site. Benefits: large bore catheters can be used, positive airway pressure can be delivered Oxygen flow rates of 50mLs/kg/min can get you 40-60% FiO2 Downsides: need for sedation, more invasive, introduction of bacteria into the tracheal lumen in an already respirator-ally compromised patient, and jet lesions and other damage to the trachea

Question 117

Describe hyperbaric oxygen therapy

Answer 117

Delivery of 100% oxygen at supra-atmospheric pressures (> 760mmHg) to increase the % of dissolved oxygen in the patient's blood stream by 10-20%!!! Allows for a greater concentration gradient (paO2) to promote diffusion of oxygen into tissues with otherwise inadequate perfusion and oxygen delivery. Indications include severe soft tissue lesions, burns, shearing injuries, infections, osteomyelitis. Downsides/risks: pneumothorax, ruptured tympanum, expense, need to keep them in a chamber, inability to rapidly access the patient should emergency occur

Question 118

Describe the potential complications of oxygen therapy in a chronic respiratory patient

Answer 118

Hypercapnia is normally the primary stimulus for respiration, however in patients with chronic respiratory disease and hypercapnia, their hypercapneic respiratory drive is diminished and the patient becomes largely dependent on hypoxia as a respiratory stimulant. If oxygen is administered to these patients aggressively, their hypoxic respiratory drive or drive to breathe is severely depressed and can lead to severe hypoventilation and respiratory failure. In severe cases, mechanical ventilation is required to treat the severe hypercapnia and hypoxia that develop.

Question 119

How is oxygen toxic

Answer 119

Oxygen is directly toxic to pulmonary epithelium and is characterized by 5 stages 1. Initiation phase: oxygen-derived free radical species such as superoxide anion, peroxide, and hydroxyl radicals cause direct damage to pulmonary epithelial cells as cellular antioxidant stores become depleted. Occurs withing 72 hours of having 100% FiO2 2 & 3. Inflammatory & Destruction phase: the pulmonary epithelial lining is destroyed leading to airway inflammation ad recruitment of activated inflammatory cells to the site. This massive influx and activation of inflammatory mediators leads to increased tissue permeability and pulmonary edema. 4. Proliferation phase: production of type 2 pneumocytes and monocytes increase 5. Collagen deposition and interstitial fibrosis occurs and is often permanent

Question 120

Describe the lesions caused by neonatal oxygen toxicity

Answer 120

Neonatal oxygen toxicity can lead to a retinal lesion called retrolental fibroplasia

Question 121

Define hypoxemia

Answer 121

pa02 < 80mmHg on room air | sp02 < 95% on room air

Question 122

List three cases of hypoxemia

Answer 122

``` Hypoventilation Low inspired Fi02 Venous admixture - anatomic right to left shunt - low V/Q regions of lung - small airway & alveolar collapse (atelectasis) - diffusion impairment - intrapulmonary (true) shunt - ```

Question 123

List three cases of hypoxemia

Answer 123

``` Hypoventilation Low inspired Fi02 Venous admixture - anatomic right to left shunt - low V/Q regions of lung - small airway & alveolar collapse (atelectasis) - diffusion impairment ```

Question 124

What level of hypoxemia is severe and life-threatening?

Answer 124

pa02 < 60mmHg on room air | sp02 < 90%

Question 125

Describe measurement of pa02

Answer 125

measured via a blood gas analyzer using a silver anode and platinum cathode system in an electrolyte solution (polarography) separated from the unknown solution (blood) by a semipermeable membrane (membrane is semipermeable to xoygen) don't want it exposed to room air otherwise changes partial pressures of O2 and CO2 analyzed quickly as in vitro metabolism and diffusion into and through the plastic also changes partial pressures avoid excessive anticoagulation as well pa02 stands for the partial pressure (vapor pressure) of oxygen dissolved in the plasma of artial blood is a measure of the lung's ability to oxygenate arterial blood normal at sea level is 80-110mmHg

Question 126

Describe the measurement of SaO2 and Sp02

Answer 126

SaO2 = hemoglobin saturation with oxygen Measurement of Sa02 occurs through an oximeter using many wavelengths of red to infrared light Measurement of Sp02 occurs via a pulseoximeter using only two wavelengths red 940 and ultra-violet 660 used to measure **only oxyhemoglobin**

Question 127

Describe the oxygen-hemoglobin dissociation curve

Answer 127

pa02 on the x axis sa02 on the y axis sigmoidal severe hypoxemia (pa02 < 60) still has a fairly robust Hb saturation % but it is the pa02 partial pressure concentration gradient that drives oxygen diffusion into the tissue mitochrondria

Question 128

What does cyanosis represent and what amount of deoxygenated hemoglobin is present?

Answer 128

It takes 5g/dL of deoxygenated hemoglobin to cause cyanosis grey to bluish mucous membrane color takes the local tissue, observer's vision, and incandescent light

Question 129

Explain why cyanosis becomes a very late but very severe marker of hypoxemia

Answer 129

If the patient only has 15gm/dL of hemoglobin (= HCT of 45%, cyanosis is present when the sa02 of 67%, which is equivalent to a pa02 of about 37mmHg This late indicator becomes even more important in an anemic patient that may only have 5g/dL of hemoglobin

Question 130

List examples of global hypoventilation

Answer 130

``` elevated paco2, etco2, pvCO2 neuromuscular dysfunction airway obstruction abdominal distension chest wall dysfunction pleural space filling defect ```

Question 131

List an additional cause of hypoxemia (Dr. Goggs' favorite one to list)

Answer 131

Decreased venous oxygen content - low cardiac output, sluggish return of peripheral blood/shock, high oxygen extraction by the tissues (ex: seizures) this is measured by central or mixed venous oxygen saturation

Question 132

List causes of low inspired oxygen

Answer 132

improperly connected anesthetic equipment or depleted oxygen tank high altitude - barometric pressure decreases but the % of oxygen in the air remains the same so the partial pressure value decreases leading to decreased alveolar and thus arterial oxygen. compensate to some extent by hyperventilating

Question 133

Define hypoventilation

Answer 133

hypoventilation is defined as decreased alveolar minute ventilation. a decrease in amv leads to decreased oxygen delivery to the alveolus which leads to decreased arterial oxygenation and hypoxemia **in patient's breathing room air** paCO2> 45 etCO2 > 40 pcentralvenousCO2 > 50 very effective treatment of hypoxemia caused by hypoventilation is administration of supplemental oxygen to increase the alveolar partial pressure of oxygen and thus arterial. caveat being the cases of chronic respiratory disease and chronic hypercarbia whereby hypoxemia is the stimulus for respiration

Question 134

Describe the alveolar air equation

Answer 134

The four gases that make up the alveolar air include O2, CO2, water vapor, and nitrogen water vapor pressure at sea level = 50 oxygen = 105 CO2 = 40mmHg - regulated by the brainstem respiratory control center nitrogen = 560 If an animal hypoventilates to a pCO2 of 80, the water vapor pressure and nitrogen levels remain unchanged but the oxygen partial pressure would fall to about 65 and the animal would become hypoxemic If you are breathing 100% oxygen, the nitrogen is displaced while the water vapor pressure and CO2 would remain the same giving oxygen the ability to increase to 665

Question 135

List the 4 causes of venous admixture

Answer 135

1. Regions of low V/Q (regional hypoventilation 2. Diffusion impairment 3. Small airway or alveolar collapse/atelectasis 4. Anatomic right to left shunt

Question 136

Describe venous admixture

Answer 136

all the ways in which venous blood goes from the right to the left side of the heart without becoming oxygenated normally < 5% in healthy lungs When there is increased venous admixture, in the absence of an intra-cardiac shunt, intrinsic pulmonary parenchymal disease exists

Question 137

Describe low V/Q regions

Answer 137

region of lung where there is reduced ventilation > increased perfusion examples include when lower airways are reduced in size such as brochospasm, fluid build-up along the walls, epithelial edema common mechanism in moderate pulmonary disease and is very responsive to oxygen therapy

Question 138

Describe how regions of zero V/Q occur

Answer 138

Regions where there is zero ventilation but adequate perfusion Also described as physiologic shunt as opposed to a true anatomic shunt where blood never passes by zero ventilated lung Most commonly occurs when alveoli and small airways are collapsed when there is accumulation of airway fluid (interstitial edema, hemorrhage, exudate) It is also common in dependent regions of lung in a recumbent animal hypoxemia due to this is nonresponsive to oxygen!! as the oxygen cannot make it down to the alveoli to arterialize the blood Collapsed small airways and alveoli can be reactivated by increasing airway or transpulmonary pressure, by taking a deep spontaneous breath, or augmenting airway pressure. Common cause of severe pulmonary disease - evidenced by improvement in oxygenation with PEEP and PPV

Question 139

Diffusion Impairment

Answer 139

results from a thickened respiratory membrane - **uncommon cause of hypoxemia in veterinary medicine** as the membrane is very thin fluid is forced by the low compliance of interstitial tissues and lymphatics upwards towards the loose interstitial tissues surrounding the medium-sized arterioles, venules, and bronchioles towards the hilus of the lung. With enough fluid accumulation, the interstitial fluid builds up enough pressure that they break into the airways and distribute along the airway surface first causing airway narrowing (low V/Q) and then small airway and alveolar collapse (0 V/Q) In order for true diffusion impairment, the flat type 1 alveolar pneumocytes must be damaged by inhalation or inflammatory injury and be replaced by thick, cuboidal type 2 pneumocytes at the site for alveolar/capillary gas exchange surface - which takes oxygen toxicity or severe ARDS. eventually the type 2s can mature to type 2 partially responsive to oxygen

Question 140

Describe anatomic shunts causing hypoxemia

Answer 140

Anatomic shunts are vascular abnormalities where the blood flows from the right side to the left side of the heart bypassing all alveoli in the process not a common mechanism most commonly in young animals with congenital defects not responsive to oxygen therapy or positive pressure ventilation and require surgery

Question 141

Categories of V/Q mismatch

Answer 141

1. No flow (0 V/Q) - physiologic shunt, decreases pa02, small airway and alveolar collapse, requires PPV or PEEP but is not responsive to oxygen in those units themselves 2. Low flow (low V/Q) - regional hypoventilation, decreases pa02, small airway narrowing secondary to lower airway disease, oxygen responsive 3. High flow (high V/Q) - regional hyperventilation, increases pa02 - hypovolemia, high tidal volumes 4. Alveolar dead space ventilation (infinite V/Q because 0 perfusion) - **no impact on pa02 as there is no blood flow to or from these regions** - pulmonary thromboembolism

Question 142

What do PaCO2 and PaO2 tell us

Answer 142

PaCO2 is representative of alveolar minute ventilation | PaO2 is representative of blood oxygenation (sum of the the efficiency of gas exchange and venous admixture)

Question 143

Describe the rule of 120

Answer 143

PaCO2 + PaO2 = 120 40 + 80 < 120 = venous admixture

Question 144

At sea level and breathing room air, what can the alveolar air equation be shortened to?

Answer 144

PAO2 = 150-PaCO2 If at a different altituted or FiO2 the entire calculation must be used normal A-a is less than 10 > 20 represents venous admixture Normal A-a gradients increase with increasing Fi02

Question 145

Describe the P:F ratio

Answer 145

used to compared oxygenating efficiency amongst differing levels of supplemental oxygen PaO2: FiO2 normal is 500 < 300 is ALI < 200 is ARDS Not good when used at room air as significant changes in paO2 can occur from abnormalities in pCO2 when breathing room air

Question 146

Give the venous admixture/shunt equation

Answer 146

mixed venous (pulmonary arterial) sample Qs/Qt = (CcO2 - CaO2)/ (CcO2 - CvO2) where Qs is the shunt fraction, Qt is the entire cardiac output, Qs/Qt is the venous admixture presented as a percentage of total cardiac output CcO2 = capillary oxygen content - assumed to be = to alveolar oxygen content (CAO2) and so it should be calculated CaO2 = arterial oxygen content, measured CvO2 = oxygen content of mixed venous blood, measured Normal venous admixture is < 5% >10% is indicative of abnormally elevated venous admixture severe lung disease can have an admixture as high as 50% units of oxygen content is mLs/dL The utility of this is that when taken breathing 100% oxygen, regions of low V/Q and diffusion defects are resolved/minimized such that you are truly assessing the degree of shunting- both physiologic (pulmonary) from atelectasis and true/anatomic shunt Affected by airway pressures (open/closed alveoli) and <100% oxygen (affect on low V/Q regions) and so best to do it with set settings across time points

Question 147

Give the anatomic landmarks of the upper respiratory tract

Answer 147

nasal passages and choanae, nasopharynx, oropharynx, larynx, and trachea

Question 148

Describe stertor

Answer 148

low pitched snoring like inspiratory and/or expiratory noise

Question 149

Describe stridor

Answer 149

high pitched noise associated with inspiration

Question 150

Describe anatomic landmarks for coughing

Answer 150

laryngeal, tracheal, mainstem bronchi - dry, nonproductive lower airway, parenchymal - wet, productive coughing upon exhalation is characteristic of intratracheal tracheal and mainstem bronchial collapse

Question 151

Describe how upper airway distress exacerbates disease

Answer 151

Increased work of breathing against an obstruction contributes to edema and inflammation, further airway narrowing, worsened obstruction, patient distress, and increased effort to breathe.

Question 152

Describe how upper airway disease leads to inspiratory difficulty

Answer 152

Often have loud noisy breathing and increased inspiratory time Inspiratory noise and distress result from collapse of the upper airway rostral to the thoracic inlet trachea because generation of negative intrathoracic pressure upon inspiration collapses the weakened airway structure into the lumen. This collapse prolongs the inspiratory phase and creates noise from air and tissue reverberation in the lumen. Increased intrapleural pressure upon expiration collapses the upper airway caudal to the thoracic inlet (intrathoraic trachea and mainstem bronchi) and results in prolonged expiration, expiratory dyspnea, and lower airway noises (wheezes) on auscultation

Question 153

List effective respiratory sedatives and doses

Answer 153

In the cardiovascularly stable patient, Ace is great. Since cardiovascular assessment is often limited/difficult in the true respiratory emergency, low doses of 0.005-0.02mg/kg IV or twice that if going IM), repeat as needed, takes up to 15 minutes for full effect after IV admin) Butorphanol 0.1-0.5mg/kg IM or IV If you have to induce, propofol 0.05-01mg/kg acknowledging cardiopulmonary depression. Can also add in diazepam to reduce the propofol dose or ketamine + diazepam or etomidate

Question 154

Describe the V acid base balance of upper airway disease

Answer 154

severe obstruction between the larynx and trachea often leads to increased pvCO2 with chronic obstructions and hypoventilation, have a compensatory retention of renal bicarb that improves acidosis and base excess but with continued pCO2 elevation

Question 155

List sedatives that do not affect laryngeal function during laryngeal exam

Answer 155

Propofol titrated to effect with or without doxapram (1.1 -2.2mg/kg IV) or premedication with acepromazine (0.2mg/kg IM - go lower in most patients) and butorphanol (0.4mg/kg IM) 20 minutes before mask induction with isoflurane

Question 156

Explain paradoxical laryngeal motion

Answer 156

inward movement of arytenoids during inspiration secondary to negative pressure generation

Question 157

What is assessed for during sedated laryngeal exam

Answer 157

laryngeal motion presence of laryngeal collapse inflammatory or proliferative abnormalities of the arytenoid cartilages eversion of laryngeal saccules changes in the diameter of the rima glottis cancer less common occurrence - epiglottic entrapment where there is caudal displacement of the epiglottis into the rima glottis length and appearance of the soft palate, epiglottis, and tonsillar crypts the caudal aspect of the soft palate should contact the epiglottis or extend no more than a few mm past it palpate the hard and soft palate for the presence of masses or defects

Question 158

List the characteristic findings in BAS

Answer 158

``` stenotic nares elongated soft palate tracheal hypoplasia As well as the following secondary to chronic increased resistance to airflow on inspiration - everted saccules - everted tonsills - laryngeal collapse - tracheal collapse - chronic GI signs (sliding hiatal hernia, regurgitation) - syncope ```

Question 159

Give the frequency of nasal turbinates extending into the nasopharynx in BAS dogs

Answer 159

20% of symptomatic cases- 80% of which were pugs

Question 160

Give the frequency of detectable bronchial collapse on fluroscopy of BAS dogs

Answer 160

87% had some degree of collapse or stenosis | more severe the bronchial collapse led to laryngeal collapse

Question 161

Give the frequency of endoscopic evidence of GI signs in BAS dogs

Answer 161

80% had some degree of reflux, esophagitis, hiatal hernia, ulceration, gastritis, pyloric stenosis there has also been increased cardiac troponin 1 levels suspect secondary to myocardial injury from chronic hypoxia normal c reactive protein and haptoglobin levels however, suggesting minimal systemic inflammation

Question 162

What is the most important component of surgical correction of BAS

Answer 162

widening the nares as this is what causes the most resistance to airflow on inspiration

Question 163

Describe tracheal hypoplasia and bronchial collapse in BAS

Answer 163

have not been found to be associated with outcome in dogs undergoing BAS surgery cricoarytenoid lateralization and arytenoid caudolateralization (arytenoid laryngoplasty) is a viable alternative to permanent tracheostomy in dogs with sever laryngeal collapse extraluminal tracheal rings or endotracheal stenting may be necessary

Question 164

Describe the different treatment approaches to nasopharyngeal polyps

Answer 164

can be up to 1/3 of cats with UR obstruction arise from mucosa of the auditory tube or middle ear and grow into the nasopharyngx or external ear canal etiology unknown, no viral causation found Treatment options include traction-avulsion which has a 40-50% rate or recurrence and 40% rate of horner's syndrome vs VBO which has a lower rate of recurrence, especially if in the auditory or middle ear disease, but a 60% chance of horner's as well as higher risk of vestibular dysfunction, and facial (7, facial assymetry and inability to blink) and hypoglyossal nerve paralysis (12, paralysis of the tongue) Most horner's resolves in about 4 wks but hearing deficits pre-op seem to be permanent use of steroids is hit or miss- one study said there was decreased rate of recurrence long term antibiotic therapy if evidence of otitis

Question 165

Describe nasopharyngeal stenosis

Answer 165

uncommonly reported partial or complete narrowing of the nasopharynx by a membrane caudal to the choanae and rostral to the caudal aspect of the soft palate can be congenital or secondary to chronic inflammation (infectious, aspiration/reflux, neoplasia, trauma) surgical correction through midline of the soft palate has been described in cats retroflex ballooning recurrence and need for repeated balloonings is common in failed cases, a stent can be placed in dogs - complications include dysphagia, hair entrapment, tissue in-growth, and prolonged nasal discharge- and in the most serious of cases, erosion of the stent through the soft palate

Question 166

Describe congenital choanal atresia

Answer 166

rare in cats, only reported once in dogs similar clinical signs to NPS results from abnormal bone or soft tissue obstructing the caudal nasal passage just rostral to the common nasopharynx. can be unilateral or bilateral treatment includes transnasal puncture with temporary stenting and surgical approaches to the nasopharynx