Respiratory Flashcards

Question

maximal/forced inspiration

Answer 1

stronger contraction of diaphragm recruitment of accessory muscles → further expansion of thoracic cavity - sternocleidomastoid + scalenes move sternum up and out - pectoralis minor elevates ribs

Answer 2

no active contraction of resp muscles relaxation of inspiratory muscles → recoil of lungs = air moves out

Answer 3

abdominal muscles contract → compress organs + force diaphragm higher = push air out internal intercostal muscles contract → pull ribs down and inward

Answer 4

pulmonary function test to determine amount + rate of inspired + expired air test lung volume + capacity use spirometer

Answer 5

Vt volume of air moved in/out of lungs during normal breath ~500 mL

Answer 6

IRV amount of air forcefully inhaled after normal inspiration additional capacity ~3000 mL

Answer 7

ERV amount of air forcefully exhaled after normal exhalation ~1100 mL (smaller than IRV)

Answer 8

RV volume of air remaining in lungs after maximal forceful expiration ~1200 mL prevents lungs from collapsing cannot be measured with spirometry

Answer 9

volume of air in lungs at end of maximal inspiration Vt + IRV + ERV + RV 5000-6000 mL

Answer 10

Vt + IRV total capacity for inspiration

Answer 11

Vt + IRV + ERV max vol of air that can be forcibly exhaled after maximal inspiration

Answer 12

ERV + RV vol remaining in lungs at end of normal expiration

Answer 13

total amount of air moved into respiratory system per minute = Vt x resp frequency = 0.5L x 12 brpm = ~6L/min

Answer 14

amount of air moved into alveoli per minute less than minute ventilation depends on anatomical dead space = 0.15 L =(Vt - dead space) x resp frequency = (0.5 - 0.15 L) x 12 brpm = ~4.2 L/min dead space is constant regardless of breath size increased depth of breathing is more effective in increasing alveolar ventilation than increasing breathing rate

Answer 15

spirometry test: forced expiratory volume in one second in healthy person: should approximate vital capacity

Answer 16

spirometry test: forced vital capacity total amount of air that is blown out in one breath after max inspiration as fast as possible ~ vital capacity

Answer 17

proportion of the amount of air that is blown out in one second normal = ~80%

Answer 18

ratio = 83% high FEV1 + high FVC

Answer 19

FEV1 is significantly reduced, FVC is normal/reduced = ratio <70% → reduced difficulty exhaling all air from lungs → shortness of breath slower exhalation of air due to damage to lungs or narrowing of airways high air vol in lungs after full exhalation ex. bronchial asthma, COPD, cystic fibrosis

Answer 20

reduced vital capacity FEV1 + FVC are reduced ratio = 90% (almost normal) cannot fill lungs completely (restricted from full expansion) due to stiffness in lungs ex. pulmonary fibrosis, neuromuscular disease, scarring of lung tissue

Answer 21

F = ΔP/R proportional to pressure difference between two points inversely proportional to resistance

Answer 22

P1V1 = P2V2 for a fixed amount of an ideal gas at a fixed temp, P and V are inversely proportional

Answer 23

↓ V = ↑ P expiration

Answer 24

↑ V = ↓ P inspiration

Answer 25

- lungs have tendency to collapse - chest wall pulls thoracic cage outward at equilibrium: inward elastic recoil of lungs balances outward elastic recoil of chest wall

Answer 26

reduces friction of lungs against thoracic wall during breathing

Answer 27

pressure within pleural cavity fluctuates with breathing always sub-atmospheric (-) → opposing directions of elastic recoil of lungs + thoracic cage if P(ip) = P(alv) → lungs would collapse

Answer 28

pressure of air inside alveoli changes during ventilation + determines direction of air flow dynamic - directly involved in producing air flow when P alv = P atm, F=0

Answer 29

governs gas exchange between lungs and atmosphere = flow

Answer 30

force responsible for keeping alveoli open pressure gradient across alveolar wall: P(tp) = P(alv) - P(ip) static - does not cause airflow determines lung volume P(tp) > 0 to maintain lungs expanded in thorax (P alv > P ip)

Answer 31

1. diaphragm + inspiratory intercostals contract 2. thorax expands 3. P(ip) becomes more subatmospheric = ↑ P(tp) 4. lungs expand 5. P(alv) becomes subatmospheric 6. air flows into alveoli

Answer 32

1. diaphragm + inspiratory intercostals relax 2. chest wall recoils inward 3. ↑ P(ip) = ↓ P(tp) - return to preinspiration value 4. lungs recoil 5. air in alveoli becomes compressed = P(alv) > P(atm) 6. air flows out of lungs

Answer 33

small → small ΔP is enough to cause airflow

Answer 34

1. inertia 2. friction

Answer 35

negligible

Answer 36

1. lung tissue past itself during expansion (negligible) 2. lung and chest wall tissue surfaces past each other (reduced by IPF) (negligible) 3. frictional resistance to flow of air = 80% of total airway resistance

Answer 37

gas particles move in linear fashion low resistance found in small airways (distal to terminal bronchioles)

Answer 38

extra energy required → produce vortices ↑ resistance found in most of bronchial tree

Answer 39

highest resistance large airways (trachea, larynx, pharynx)

Answer 40

for laminar flow resistance is proportional to viscosity of gas + length of tube inversely proportional to airway radius (^4)

Answer 41

each small airway has high individual resistance terminal bronchioles aligned in parallel = lower aggregated resistance combined → lowest R

Answer 42

R is determined primarily by small airways → easily occluded by: - smooth muscle contraction around airway - edema in walls of alveoli + bronchioles - mucous collecting in lumen of bronchioles

Answer 43

measure of elastic properties of lungs how easily lungs can expand determined by - elastic components of lungs + airway tissue (elastin + collagen) - surface tension

Answer 44

= (ΔV)/(ΔP(tp)) change in lung volume produced by given change in P(tp) slope of pressure-volume curve ↑ C(L) = easier to expand (at given change in P(tp))

Answer 45

↑ compliance floppy lungs due to destruction of elastin in walls of alveoli = collapse of smaller airways less elastic recoil large changes in lung vol due to little changes in P(tp) low O2 due to less SA for gas exchange

Answer 46

↓ compliance collagen deposition in alveolar walls stiff lungs higher P(tp) changes are necessary to generate changes in lung volume poor gas exchange causes ↓ O2

Answer 47

property of surface of liquid that allows it to resist external force → cohesion of molecules gives lungs elastic recoil (makes lungs collapse) decreases lung compliance

Answer 48

air entering lungs is humidified → saturated with water water molecules cover alveolar surface cohesion of water → create surface tension = creates inward recoil → alveolar collapse decreases gas volume inside alveolar = increases pressure to prevent collapse

Answer 49

P = (2T)/r [T = surface tension] at equilibrium, tendency of increased pressure to expand the alveolus balances the tendency of surface tension to collapse the smaller the radius, the greater the pressure needed to keep alveolus inflated

Answer 50

smaller alveoli = greater pressure → gas movement from high to low pressure = (POTENTIALLY) small alveoli collapse into larger ones do not collapse because of surfactant → lowers surface tension = similar pressure across alveoli

Answer 51

produced by type II alveolar cells made of lipoproteins = hydrophobic + hydrophilic properties → decreases density of water molecules covering alveolar surface reduces surface tension of alveolar fluid = ↑ lung compliance stabilize alveoli against collapse

Answer 52

infant respiratory distress syndrome born premature = under developed lungs → immature type II alveolar cells can't produce enough surfactant = ↓ O2 have abnormally low lung compliance

Answer 53

P increases with increased motion → due to change in temperature, concentration of gas molecules

Answer 54

in a mixture of gases, each gas acts independently total pressure = sum of individual pressures

Answer 55

= P (N2) + P (O2) + P (H2O) + P (CO2) = 760 mmHg P (p) = % of air x P total 78% N2 → P (N2) = 593 mmHg 21% O2 1% H2O 0.04% CO2

Answer 56

V(gas) ∝ [A/T] x D x (P1 - P2) rate of transfer of a gas through membrane (Vgas) - is proportional to: A = area of diffusion ΔP = difference in gas partial pressure between two sides D = diffusion constant - is inversely proportional to: T = thickness of membrane

Answer 57

D ∝ (sol)/√ (MW) sol = solubility MW = molecular weight

Answer 58

solubility: CO2 > O2 similar molecular weights CO2 diffuses 20x more rapidly than O2 P (CO2) is higher in alveoli than atm P (O2) is higher in atm than alveoli

Answer 59

1. loss of O2 to blood by diffusion = ↓ P (O2) 2. mixing of inspired air with functional residual vol = ↓ P (O2) (3.) warming + humidification of air in resp tract = ↓ P (O2)

Answer 60

determined by - P (O2) + P (CO2) in atm - alveolar ventilation [V(A) = (V(T) - V(D)) x resp frequency] - metabolic rate - perfusion

Answer 61

= ↑ O2 intake + ↑ CO2 production = ↑ P (O2) + ↓ P (CO2)

Answer 62

= ↑ O2 intake + ↑ CO2 production = ↓ P (O2) + ↑ P (CO2)

Answer 63

amount of blood that passes through pulmonary capillary system in the lung using O2 from alveoli + adding CO2

Answer 64

high pressure system overcome high resistance to deliver blood to peripheral tissue

Answer 65

low pressure system only deliver blood to lungs

Answer 66

V/Q ratio balance between ventilation + perfusion one of major factors affecting alveolar (+ arterial) levels of O2 + CO2

Answer 67

local perfusion decreases to match a local decrease in ventilation (+ vice versa) ↓ airflow / blood flow to region of lung ↓ P (O2) in p. blood / ↓ P (CO2) in alveoli vasoconstriction of p. vessels / bronchoconstriction ↓ blood flow / airflow diversion of blood flow and airflow away from local area of disease to healthy areas of the lung

Answer 68

blockage of airway leads to poor gas exchange in alveoli → blood flow is diverted to better ventilated alveoli to compensate

Answer 69

blocked airway → ↓P(O2) + ↑P(CO2) decreases V/Q ratio

Answer 70

blocked blood flow → ↑P(O2) + ↓P(CO2) increases V/Q ratio

Answer 71

arterial blood = highly oxygenated similar partial pressures of O2 and CO2 as seen in alveoli majority of O2 is bound to Hb → in RBC small amount is dissolved in plasma

Answer 72

venous blood = poorly oxygenated ↓ P(O2) + ↑ P(CO2) compared to alveoli majority of CO2 is transported as bicarbonate some is bound to Hb = carbaminohemoglobin small amount is dissolved in plasma

Answer 73

- diffuses from alveoli (↑[]) to p. capillary (↓[]) - dissolved in plasma - moves to RBC - binds with Hb

Answer 74

- diffuses from RBC to plasma - diffuses into tissue = dissolved O2 in ISF - moves into cells - consumed in cell mitochondria

Answer 75

x = P (O2) - systemic venous: ↓ P(O2) - systemic arterial: ↑ P(O2) y = Hb saturation (%) - ↓ as O2 dissociates from Hb difference between Hb sat of arterial and venous = amount of O2 unloaded in tissue capillaries

Answer 76

DPG = metabolic waste product no DPG = high sat (L) added DPG = low sat (shift curve to R) same effect seen with CO2

Answer 77

high metabolism increases temperature ↑t = ↓ sat shifts curve to R

Answer 78

high metabolic activity = high acidity (low pH = high [H+]) shifts curve to R

Answer 79

high saturation during fetal development decreases with aging adult Hb = shift curve to R

Answer 80

CO2 exits cells and is dissolved in ISF → diffuses to blood in blood = 4 pathways 1. remains in plasma as dissolved CO2 2. enters RBC and remains as dissolved CO2 3. enters RBC and binds to deoxyHb = HbCO2 4. enters RBC and reacts with H2O → carbonic acid (CA catalyst) → dissociates to HCO3- + H+ [HCO3- exits RBC; H+ increases in venous blood = ↓ pH]

Answer 81

HbCO2 dissociates → dissolved CO2 HCO3- moves back into RBC → binds with H+ → hydrolyzed into H20 + dissolved CO2 ↓[CO2] in plasma drives diffusion of dissolved CO2 from RBC into plasma → diffuses into alveoli

Answer 82

Hb from HbO2 + H+ from H2CO3- bind together H+ is not dissolved in RBC or plasma to maintain stable pH Hb buffers production of H+ in peripheral tissues + capillaries

Answer 83

high [H+] as a result of ↑ P(CO2) hypoventilation causes CO2 production > CO2 elimination = ↑ P(CO2) + ↑ [H+]

Answer 84

low [H+] as a result of ↓ P(CO2) hyperventilation causes CO2 production < CO2 elimination = ↓ P(CO2) + ↓ [H+]

Answer 85

high blood [H+] as a result of metabolism change is independent of changes in P(CO2)

Answer 86

low blood [H+] as a result of metabolism change is independent of changes in P(CO2)

Answer 87

automatic process regulated by neuronal network in pons + medulla (brainstem)

Answer 88

in medulla by specialized neurons (PreBotC)

Answer 89

dorsal respiratory group ventral respiratory group Pre-Botzinger Complex: sets basal resp rate

Answer 90

fires during inspiration stimulates inspiratory muscles = diaphragm + external intercostals phrenic nerve → diaphragm intercostal nerves → external intercostals

Answer 91

expiration contains Pre-Botzinger Complex in upper portion = respiratory rhythm generator intercostal nerves → internal intercostals

Answer 92

- pneumotaxic centre - apneustic centre = pontine respiratory group

Answer 93

centres help modulate activity of medullary inspiratory neurons - termination of inspiration at appropriate time - smooth transition between inspiration + expiration

Answer 94

descending inhibition to apneustic centre + DRG

Answer 95

DRG + VRG send signals to one another = local regulation of each other

Answer 96

by: higher structures of CNS inputs from central + peripheral chemoreceptors inputs from mechanoreceptors in lung + chest wall inputs from other peripheral receptors

Answer 97

voluntary control of breathing Hering-Breuer reflex (stretch receptors in lungs)

Answer 98

voluntary control of breathing medullary chemoreceptors carotid + aortic body chemoreceptors proprioceptors in muscles + joints receptors for touch, temp, + pain stimuli

Answer 99

carotid bodies + aortic bodies respond to changes in arterial blood: - hypoxia = ↓ P(O2) - metabolic acidosis = ↑ [H+] - respiratory acidosis = ↑ P(CO2)

Answer 100

located in medulla respond to changes in brain extracellular fluid: - ↑ P(CO2) via associated changes in [H+]

Answer 101

stable ventilation between 60-120 mmHg arterial P(O2) arterial P(O2) drops below 60 → stimulation of peripheral chemoreceptors ↓ inspired P(O2) = ↓ alveolar P(O2) = ↓ arterial P(O2) ↑ firing of peripheral chemoreceptors → reflex via medullary respiratory neurons (activation of DRG + VRG to control activity of resp muscles centrally = ↑ resp rate + tidal vol) ↑ contraction of resp muscles = ↑ ventilation = return of P(O2) to normal

Answer 102

stable ventilation at 40 mmHg arterial P(CO2) arterial P(CO2) increases above 40 → stimulation of central chemoreceptors (also peripheral) ↑ P(CO2) = ↑ alveolar P(CO2) = ↑ arterial P(CO2) ↑ brain extracellular fluid P(CO2) + ↑ brain extracellular fluid [H+] = respiratory acidosis ↑ firing of central chemoreceptors → reflex via medullary respiratory neurons (activation of DRG + VRG to control activity of resp muscles centrally = ↑ resp rate + tidal vol) ↑ contraction of resp muscles = ↑ ventilation = return of P(CO2) to normal

Answer 103

H+ stimulates peripheral chemoreceptors (does not cross blood brain barrier)