handouts pulmonary physiology

7/31/2019 Handouts Pulmonary Physiology

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PHYSIOLOGY HANDOUT RESPIRATORY

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Objectives:

Structure and non-respiratory functions of the lung

01. Describe the structure and general functions of the respiratory system

02. Identify the mechanism by which particles are cleared from the airways.

03. Describe the non-respiratory functions of the lung. Describe mechanisms for clearance ofvasoactive substances from the blood during passage through the lung. Identify a substance thatis almost completely cleared and one that is not cleared to any significant extent.

04. Be able to estimate the alveolar oxygen partial , and PAO2, pressure, PO2, PIO2using thesimplified form of the alveolar gas equation.

Alveolar ventilation

05. Define partial pressure and fractional concentration as they apply to gases in air, list normalvalues for each for O2, CO2, and N2.

06. List the normal alveolar, arterial, and mixed venous blood gas values for PO2, SatO2, PCO2, HCO3,and pH.

07. Draw a normal spirogram, labeling the four lung volumes and four capacities. List the volumes that

comprise each of the four capacities. Identify which volume and capacities cannot be measuredby spirometry.

08. Define the mechanisms that determine the clinically important boundaries of lung volume (i.e., TLC,FRC, and RV).

09. Contrast the causes and characteristics of restrictive and obstructive lung disease, including theabnormalities in lung volumes are associated with each.

10. Define and contrast the following terms: anatomic dead space, physiologic dead space, wastedventilation, and total ventilation per minute and alveolar minute ventilation. Calculate thesevolumes by using the proper formulas.

11. Contrast the proportional relationship between alveolar ventilation and the arterial blood gases PCO2and PO2.

12. Define the following terms: hypoventilation, hyperventilation, hypercapnea, eupnea, hypopnea, andhyperpnea. Identify hypo and hyperventilation based on concentrations of PCO2

Mechanics of respiration

13. On one page, diagram the lung volume, tracheal pressure, alveolar pressure, and pleural pressure

during a normal quiet breathing cycle. Identify on the figure the onset of inspiration, cessationof inspiration, and cessation of expiration. Relate the pleural and airway pressure values to themovement of air.


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14. Identify the forces that generate the negative intra-pleural pressure, and predict the direction thatthe lung and chest wall will move if air is introduced into the pleural cavity (pneumothorax) whenthe lung is at functional residual volume. Discuss the effect of the elastic properties of the chestwall and the lung on functional residual capacity.

15. Define compliance (lung and chest wall) and identify two common clinical conditions in whichcompliance is higher or lower than normal. Calculate compliance by using the formula.

16. Draw a normal pulmonary pressure volume curve (starting from residual volume to total lungcapacity and back to residual volume), labeling the inflation and deflation limbs. Explain the causeand significance of the hysteresis in the curves.

17. Define surface tension as it applies to the lungs, including the effects of alveolar size and the roleof surfactants. Define athelectasis and the role of surfactants in preventing it.

18. Based on changes in FEV, FEV1, FVC, TLC, and flow volume curves, characterize the pathologyas a restrictive and/or obstructive lung disease. Describe how FRC and residual volumes are

altered in each case.

19. Describe how airway resistance alters dynamic lung compliance.

20. Describe the roles of airway diameter and turbulent flow on airway resistance.

21. Define dynamic airway compression, and use this principle to explain the shift in the shape of flowvolume curves that occur with COPD (chronic obstructive pulmonary disease).

Diffusion of gases

22. State the Ficks law for diffusion and determine the limitations of gas transfer

23. Define oxygen diffusing capacity, and describe the use of carbon monoxide to determine oxygendiffusion capacity.

24. Name the factors that affect diffusive transfer of gas.

Pulmonary circulation

25. Contrast the systemic and pulmonary circulations with regards to pressures, resistance to bloodflow, and response to hypoxia.

26. Define zones I, II, and III in the lung, with respect to pulmonary vascular pressure and alveolar

pressure. Explain how the alveolar pressure , the blood flow and the gravity interfere in the functionof each zone.

27. Describe the roles of distention and recruitment of pulmonary vessels in changing pulmonary bloodflow and pulmonary vascular resistance. Identify the zones in which these two mechanisms apply.

28. Describe the consequence of hypoxic pulmonary vasoconstriction on the distribution of pulmonaryblood flow. Describe the effects of inspired nitric oxide on pulmonary vascular resistance andhypoxic vasoconstriction.

29. Contrast the airway and vascular control mechanisms that help maintain a normalventilation/perfusion ratio.


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30. Explain the development of pulmonary edema by: a) increased hydrostatic pressure, b) increasedpermeability, c) impaired lymphatic outflow or increased central venous pressure, and d)hemodilution (e.g., with saline volume resuscitation).

Ventilation perfusion

31. Name five causes of hypoxemia: hypoventilation, diffusion impairment, shunt, VA/Qmismatching, high altitude

32. Identify hypoventilation based on concentrations of PCO2 and differentiate from other pathologiesby using PCO2 and lung volumes

33. Describe the consequences of diffusion impairment and the effect of an exercise test.

34. Define right to left shunts and physiologic dead space (wasted ventilation). Describe theconsequences of each for pulmonary gas exchange.

35. Predict how abnormal ventilation/perfusion ratios will affect local alveolar oxygen and carbondioxide pressures and exchange.

36. Identify the average V/Q ratio in a normal lung. Explain how V/Q is affected by the verticaldistribution of ventilation and perfusion in the lung.

37. List the normal relative differences from the top to the base of the lung in alveolar and arterial PO2,PCO2, pH, and oxygen and CO2 exchange.

38. Define two causes of abnormal V/Q distribution.

39. Define the measurement of the alveolar to arterial PO2 difference, (A-a)DO2, the normal value for(A-a)DO2, and the significance of an elevated (A-a)DO2.

40. Name three compensatory reflexes for V/Q inequality.

Oxygen and carbon dioxide transport

41. Define percent hemoglobin saturation, oxygen tension, oxygen content as they pertain to blood.

42. Draw and label an oxyhemoglobin dissociation curve (hemoglobin oxygen equilibrium curve),showing the amount of dissolved oxygen and the relationships between oxygen partial pressure,hemoglobin saturation, and blood oxygen content.

43. How does the shape of the oxyhemoglobin dissociation curve influence the uptake and delivery ofoxygen?Define P50. and its relationship with the O2 Hb affinity

44. Show how the oxyhemoglobin dissociation curve is affected by changes in blood temperature, pH,PCO2, and 2,3-DPG.

45. Describe how anemia and carbon monoxide poisoning affect the shape of the oxyhemoglobindissociation curve, PaO2, PCO2, and SaO2.

46. List the forms in which carbon dioxide is carried in the blood. Identify the percentage of total CO2

transported as each form.


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47. Describe the importance of the chloride shift in the transport of CO2 by the blood.Identify the enzyme that is essential to normal carbon dioxide transport by the blood and itslocation.

48. Draw the carbon dioxide dissociation curves for oxy and deoxyhemoglobin. Define the interplaybetween CO2 and O2 binding on hemoglobin that causes the Haldane effect.

49. Mention and explain causes of hypoxia: hypoxic hypoxia, anemic hypoxia, stagnat hypoxia,histotoxic hypoxia

Regulation of acid base equilibrium

49. Define respiratory acidosis and alkalosis and give clinical examples of each.

50. Describe the mechanism and function of respiratory acid base compensations.

51. Describe in quantitative terms the effect of ventilation on PCO2 according to the alveolar ventilation

equation.

Control of respiration

52. List the anatomical locations of chemoreceptors sensitive to changes in arterial PO2, PCO2, and pHthat participate in the control of ventilation. Identify which chemoreceptor population is mostimportant in sensing short-term (acute) and long-term (chronic) alterations in blood gases.

53. Describe the respiratory drive in a COPD patient, and predict the change in respiratory drive whenoxygen is given to a COPD patient.

54. Describe the mechanisms for the shift in alveolar ventilation that occur immediately upon ascent to

high altitude, after remaining at altitude for two weeks, and immediately upon return to sea level.

55. Describe the physiological basis of shallow water blackout during a breath-hold dive.

56. Describe the interaction of hypoxia and hypercapnia in the control of alveolar ventilation.

57. Describe the significance of the feed forward control of ventilation (central command) duringexercise, and the effects of exercise on arterial and mixed venous PCO2, PO2, and pH.

58. Describe the effect of old age on lung volumes, lung and chest wall compliance, and blood gases.

Respiratory system under stress

59. Describe the effect of exercise on respiratory system

60. Describe the respiratory response to high altitude: acclimatization process


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I. Functions of Respiratory System

A. The primary function of the lugs is to allow oxygen to move from the atmosphere into the blood andcarbon dioxide to move out.

B. Other functions of the lungs:

1. Synthesis of lipids, proteins and carbohydrates e.g. lipids pulmonary surfactant DPPC,proteins:collagen,elastine,enzymes,etc.

2. Defense mechanisms:Alveolar macrophages engulfs inhaled particles

Alveolar macrophages secrete:enzymes, immune response componentsSurface enzymes:lisozymes found in leukocytes (bactericidal properties) ;Lymphocytes

3. Renoval of vasoactive substances by the endothelium of the vessels of Pulmonary circulationas:Prostaglandin E1,E2,F2:are completely removed in a single pass through the lungs:Norepinephrine: 30% in mixed blood is removed by lungs (epinephrine is unaffected)

4. Formation and release of chemical substances:Pulmonary surfactantHistamine,Prostaglandins, leukotrienes, Platelet activating factor, Serotonin,bradykinin, heparin,etc.

5. Coversion of hormones (Al All)6. Endocrine function7. Blood reservoir8. Role in coagulation: produce heparin, tromboplastic substances, plasminogen activator.

Over all function: destroy fibrin9. Acid-base balance

II. Structure of the Respiratory System

A. The airways are divided in:

1. Conducting zone:trachea,bronchi and its subdivisions and bronchioles, (last structureis the terminal bronchiole)

2. Transition and Respiratory Zone:Begins with respiratory bronchiole (bronchiole with alveoli, alveolar ducts and alveoli)


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CONDUCTING ZONE RESPIRATORY ZONE

Z1 to z16 Z 17-Z19 transitional zoneAnatomical dead space Z 20-Z 23 respiratory zoneVD= 150 ml normallySupplied by the bronchial artery Supplied by the pulmonary arteryCilia present No cilia presentMost airway resistance Least airway resistance

Bronchi or central airways BronchiolesFrom Z1 to Z10 From Z11 to Z16Contains cartilage No cartilageConstrict or dilate independent Lung perenchyma dilatesOf lung volume dependent of lung volume

B. Symbols used in respiratory physiology table 19-2

Primary symbolsC Concentration or complianceD DiffusionF Fractional concentration of a gasf Respiratory frequencyP Pressure or partial pressureQ Volume of bloodQ Volume of blood per unit time (blood flow or perfusion)R Gas exchange ratio or resistanceS SaturationV Gas volumeV Volume of gas per unit time (gas flow or ventilation)


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Secondary symbolsA Alveolara Arterialaw AirwayB Barometric

c CapillaryDS Dead spaceE ExpiredI Inspiredel ElasticL Lungp Physiologicpa Pulmonary arterypc Pulmonary capillarypc Pulmonary end capillarypl Pleuralpv Pulmonary venouspw Pulmonary wedges Shuntst Static

t Timetm Transmuralv Venousv Mixed venous

Examples of combinationsCL Lung complianceDLCO Lung diffusing capacity for carbon monoxideCvO2 Concentration of oxygen in mixed venous bloodPB Barometric pressurePCO2 Partial pressure of carbon dioxidePA Alveolar pressurePO2 Partial pressure of oxygenPaCO2 Partial pressure of oxygen in arterial bloodPACO2 Partial pressure of carbon dioxide in alveoli

PAO2 Partial pressure of oxygen in alveoliPIO2 Partial pressure of oxygen in inspired gasPvO2 Partial pressure of oxygen in mixed venous bloodPECO2 Partial pressure of carbon dioxide in expired gasPpl Pleural pressureSaO2 Saturation of hemoglobin with oxygen in arterial bloodRaw Airway resistanceFIO2 Fractional concentration of inspired oxygenVDS Volume of dead space airVA / Q Alveolar ventilation - perfusion ratioVA Alveolar ventilationVE Expired minute ventilationVO2 Oxygen consumption per minute .

Note: A dot above a primary symbol denotes flow per unit time

C. Laws of gases

1. Boyles lawAt a constant temperature, the volume of a given quantity of any gas varies inversely as the pressureto which that gas is subjected.

P1V1= P2V2

This is the basis for Plethysmography

Pbox ix Vbox i=Pbox f x (Vbox i- V)

Where: (Vbox i-V)= Vbox f


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2. CharlesLaw (Gay-Lussacs Law)Volume of a given mass of gas, at constant pressure, is directly proportional to theabsolute temperature.

3. Daltons LawTotal pressure exerted by a mixture of gases is equal to the sum of the separate or partial pressuresthat each gas would exert if it alone occupied the entire volume.

PB=PN2 + PO2 + PH2O + PCO2

a. What is the PO2 in room air, at sea level:

PO2=PB x FO2 Note:Atmospheric pressure (PH) at sea level=760 mmhg02 concentration in normal room air=20.93%F02=fractional concentration of 02=.209

P02=760x.209P02=158.8=160 mm hg

b. What is the partial pressure of 02 when inspiring room air at sea level?

Note:water vapor pressure must be subtracted from PB

PI02=

c. Calculation of PI02 at high altitude.Increased elevation does not affect 02 concentration but decreases atmospheric pressure.Subtract 120 mm Hg/mile of increased altitude.

Example of P02 at one mile above sea level:PI02=[760- (120+47)]x .209=124.1 mm Hg

4. Henrys LawThe total volume of that physically dissolved in a liquid depends on the partial pressure of the gas andin its solubility in the liquid.

Volume of dissolved gas = Px Cs where: P=partial pressureVolume of liquid Cs=coefficient of solubility

Examples

a. If the solubility coefficient of 02 is 0.003 ml and the partial pressure (Pa02) is 100 mmhg, how

much 02 is physically dissolved?

Dissolved 02=(0.003)(100)=0.3 ml dissolved

b. if 100% 02 is breathed at a Pa02 of 650, how much of gas is physically dissolved?

Dissolved 02=(0.003)(650)=1.95 ml dissolved


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Ventilation is the process of moving air into and of the alveoli.Is usually defined too as the volume of fresh air entering the alveoli per minute, and is implicit that a similarvolume of alveolar air is leaving the body per minute.

I. SpirometryDevice for measuring gas volume. Can meassure only the lugs volumes that the subject canexchange with it.

fig 3-4 A (Levitzky)A. Spirometer types

VVoolluummee--ddiissppllaacceemmeenntt ssppiirroommeetteerrssFFllooww--sseennssiinngg ssppiirroommeetteerrss

B. Lung Volumes and CapacitesTidal volume (VT): The volume of air inhaled or exhaled with each breath. In a 70 Kg adult is about500 mI at rest

Inspiratory reserve volume (IRV): Maximum volume of air that can be inhaled during a maximalforced inspiration starting at the end of a normal VT. About 3 L

Expiratory reserve volume (ERV): Volume of air that can be expelled during a maximal forcedexpiration that starts at the end of a normal VT. About 1.2 L

Residual volume (RV): Volume of air remaining in the lungs after a maximal expiration, about 1.2 L

Total lung capacity(TLC): Volume of air in the lung after a maximal inspiratory effort. Isapproximately of 6 L. TLC=VT+IRV+ERV+RV

Vital capacity (VC): The maximum volume of air that can be expelled after a maximum Inspiration.

VC= TLCRV

Inspiratory capacity (IC): Maximum volume of air that can be inhaled to total lung capacity Over andabove the tidal volume. IC=TV+IRV

Functional residual capacity (FRC): Volume of air remaining in the lungs at the end of a normal tidalexpiration. With respiratory muscles relaxed at this time, it represents the balance point between theinward elastic recoil of the lung and the outward elastic recoil of the chest wall. FRC=RV+ERV


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II. Pulmonary Lung diseaseDetermination of lung volumes is helpful in the diagnosis and differentiation of pulmonary disease:Obstructive and restrictive.

A. RestrictiveGroup of disorders that result in an increase in the connective tissue of the lung, increased elastic recoiland reduce compliance of the lung . e.g. Fibrosis, alveolar wall thickening.Decreased compliance, increasing difficulty to expand the lung. Increased elastic recoil.It is characterized by lower: FRC, TLC, VC, IVR, ERV, and even reduced RV

B. Obstructive

Resistance to airflow, airway may be obstructed (mocus) or destruction of connective and elastic tissue ofthe lung (emphysema): e.g. emphysema, chronic bronchitis, asthmaConditions that cause shortness of breath and obstruction of airflowLung compliance is increased, elastic recoil is decreased.It is characterized by Increased: RV, FRC, TLC. Decreased: VC and ERV.

III. Anatomical and Physiological dead space

A. Anatomical dead space1. An inspired breath entering the lungs is distributed, in part, to the alveoli where gas exchangeoccurs, but the rest of the breath remains in the conducting airways where there is no gas exchange.The conducting airways are referred as the anatomic dead space.


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Can be estimated by considering 1 mI of dead space per pound of body weight or more precisely byusing the Fowlers method.

2. Fowlers method:Single breath of 100% 02When the gas is exhaled through one-way valve, Nitrogen concentration at the mouth and the volumeexpired is monitored simultaneously.First part of the expired gas registers 0% nitrogen because is undiluted 100% 02 from the Anatomicdead space.

Transition period is when the expired gas registers a slowly rising nitrogen concentration, this indicatesthat the gas is a mixture of dead space gas and alveolar gas.Final portion of expired gas comes solely from the the alveoli forming the alveolar plateau, where thenitrogen concentration is constant and less than 80%The volume of anatomical dead space is the volume expired between the beginning of the expirationand the midpoint of the transitional phase.

fig 2-6 (West)

B. Physiological dead space

Is equal to the anatomical dead space plus the alveolar dead spaceAlveolar dead space is the volume of gas that enters unperfused alveoli per breath.

1. Calculated by using Bohr equation. C02 found in the mixed expired gas must come from alveoli that are bothventilated and perfused. Air coming from unperfused alveoli will not contribute to the C02 at the mixed expired air.

VD = PaCO2 - PECO2PaCO2


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IV. Alveolar ventilation

A. Minute volume volume of air entering and leaving through the nose or mouth per minuteAlveolar ventilation volume of air entering and leaving per minute

VT = VD+ VA VT=tidal volumeVD=dead space volumeVA=alveolar volume

Multiplying these volumes by number of breath/min (n):

Minute volume VE= VT x n

Alveolar ventilation VA= VA x n

VA= VE VD dots over Vs indicate per minute

Dead space ventilation VD= VD x n

B. Alveolar ventilation and Carbon Dioxide

1. The best way to evaluate edequate breathing in a patient is the partial pressure of Carbon dioxide.Normal PaC02= 40 mm Hg

2. The concentration of C02 in the alveolar gas depends on the alveolar ventilation and on the rate of C02production by the body.

3. The volume of C02 expired per unit of time (VEC02) = VA x FAC02

PAC02 VC02VA

In healthy person PAC02= Pa02 If PaC02 > 40 the patient is hypoventilating.If PaC02 < 40 the patient is hyperventilating.

4. Example:

a. Calculate the normal minute ventilationMinute vantilation = VT x f VT= tidal volume 500 mI

f =respiratory frequency (14-18 min)

Minute ventilation = 500mI x 14/min=7000 mI/min= 7 L/min

Minute ventilation should be NO less 7.0 L/min

b. Calculate the normal alveolar ventilation (VA)

VA= (VT VD) n VT = 500mIVD = 150 mI

n = 14-18 min

VA =(500 mI 150 mI) 14/min= 350 x 14= 4 900 mI/min= 4.9 l/min (normal is 4.9 to 5 L/min)


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In normal individuals the Anatomical dead space is equeal to the Physiological dead space.The ratio of dead space to tidal volume (VD/VT) is approximately 0.2-0.35

VD = 150 = 1 = PaC02 = PAC02VT 500 3

C. PO2 and PCO2 during normal Ventilation

The below schema of the pulmonary and vascular arterial and venous system, shows the normal PO2 andPCO2 at different levels during normal ventilation.

D. PO2 and PCO2 during hypoventilation.


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E. Example of patients :Determine if the patient is hyperventilating, hypoventilating or with normal alveolar ventilation

Patient A: VT = 500 mI n = 15 PaC02 = 35

Dx: __________________________________

Patient B : VT = 250 mI n = 30 PaC02 = 55

Dx: __________________________________

Patient C: VT = 700 mI n = 20 PaC02 = 55

Dx: ___________________________ , how is dead space? ____________

How is alveolar ventilation ? ___________________________________

Patient D: VT = 450 mI n = 12 PaC02 = 35

Dx: ___________________________ , how is the dead space? ___________


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I. Respiratory Muscles

A. Inspiratory muscles

1. DiaphragmInnervated by phrenic nervesAt rest, the muscle descend 1-2 cmDeep inspiration, may descend 10 cm

2. External intercostalsLift ribs up and out (increase anteroposterior dimensions )Innervated by intercostal nerves from T 1 to T 11

3. Accessory muscles

Scalene, sternocleidomastoidWork at high levels of ventilation or with obstruction of the airway

B. Expiratory muscles1. Abdominal flat muscles

Depress the lower ribs, pull down the anterior part of the lower chest, and compress the abdominalcontents

2. Internal intercostalDepress the ribs and move them downward and inward

II. Elastic properties of the lungs

A. Like an elastic structure such as a balloon, an applied force is required for inflation, but the deflation to

a low resting volume is a passive process and can ocour using energy stored by the elastic elements ofthe lung.

1. To produce the inflation

Alveoli expand passively in response to anincrease in distending pressure across thealveolar wall force per unit area (pressure) onthe inside surface of the lungs must be greaterthan pressure on the outside surface of thelung

2. To produce the deflation

Once the lungs are inflated, the pressuredifferences between the inside and outside ofthe lungs (often called elastic recoil pressure, abbr. Pel., or transpulmonary pressure)reflects the energy available to deflate thelungs.

B. Recoil tendencies


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The chest wall and lung have opposite recoil tendencies at rest.At the end of expiration the lung is tending to collapse and the chest wall is tending to spring out. Thiscreates a negative intrapleural pressure (-5 cm water ) and holds the lung expanded

C. Volume- pressure relationships

1. Transpulmonary pressure

Absolute pressure difference between the inside and outside of the lung. Pel is always positive.

2. Compliance

a. Change in lung volume per change in transpulmonary pressure (Pel)

b. Is the inverse of elastic recoil. Denotes the easy with which something can be stretched or distorted.

Human lung compliance = 200mI/cm H2 0

Lung and chest wall complianceUseful data for clinical evaluations

1 1 1--------------------------- = ----------------------------- + ----------------------

Total compliance Compliance of Complicance ofthe lung the chest wall

3. Specific compliance

Compliance is volume dependent. It is greater at low lung volumes and lower at high lung volumes.

4. Factors that affect lung compliance

a. Reduced by:Fibrosis-proliferation of connective tissue.

Pulmonary edema


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Musculoskeletal disordersIncreased pulmonary venous pressure obesity.

b. Increased by:EmphysemaAge

5. Measurement of the Pressure Volume curve


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6. Problems

a. Problem:Tidal volume =600 mIIntrapleural pressure before inspiration = 5 cm H2 0Intrapleural pressure after inspiration = 8 cm H2 0

Calculate lung compliance

b. Problem: given that the compliance of an individuals lung 0.5 L/cm H20 and meanintrapleural pressure is 7 cm H2 0 , what is the volume of exhaled gas if intrapleuralpressure rose to 5 cm H2 0?

c. Problem: Given that the compliance of an individuals lung is 0.5 L/cm H2 0 andintrapleural pressure was- 10 cm H2 0. What is the new intrapleural pressure if thisperson exhaled 1.0 L?

III. Surface tension

Is produced by the cohesive forces between adjacent liquid molecules at the liquid-air interface. In analveolus (or in bubble soap) the cohesive forces will produce a net force, which will tend to collapse thealveolus (or the buble). Within the bubble this force will be expressed as a pressure.Surface tension follows LaPlaces Law

A. Laplaces law

Give the relationship between the surface tension and the pressure inside the alveolus.

P = 2T P = Pressurer T = Tension

r = radius

What would happen if two alveoli of different size were connected to each other by common airway?.

Assume that the surface tension of the two alveoli is equal.How will be the direction of the airflow ?Which alveolus has the largest pressure ?


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B. Role of surface tension and proper functioning of the lung

1. Pressure within the airway during expiration is less than that which exists at the same lungvolume during the process of inflation.Hysteresis is an elastic property of the lung. Is represented, in the pressure volume curve by thedifference in volume during inflation and deflation at a given pressure.

2. Lungs inflated with saline have a much larger compliance their air-filled lungs

Air filled: formation of an air-fluid interfaceSaline-filled no generation of interface

3. Surface area vs. Surface tension


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When adding detergent, water or surfactant.

C. Surfactant

Is a Dipalmitoilphosphatidyl choline (DPPC) produced by type II epithelial cells.The molecules of DPPC are hydrophilic at one end hydrophobic at the other.

1. Functions

Decreases surface tensionStabilizes the alveolusIncreases lung complianceReduces the energy expenditure of breathing (from 2% with to 60% without surfactant)Helps to keep alveoli dry

2. Hyaline membrane disease (HMD) or Infant Respiratory Distress Syndrome (IRDS)Seen in premature infants due to lack of surfactant

a. Predisposing factors:

More common in males than in femalesMore common in children of diabetic women (Insulin destroys enzymes that produce surfactant)Decreased thyroid functionDecreased ACTH productionPrematurity

b. Prenatal diagnosis:

Amniocentesis (lecithin /sphyngomielin ratio)Lecitin sphingomyelin ratio of 2.0 may indicate HMD

c. Treatment

Prenatal treatment underlying cause (in mother):Give cortisol to mother prior to birth to simulate type ll cells

After baby is born:Positive end expiratory pressure ventilator (PEEP)lntroduce artificial surfactant

NOTE: PEEP helps to keep alveolar pressure above atmospheric during expiration.


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3. Adult respiratory distress Syndrome (ARDS)

a. May be produced by:

Hypoxia or hypoxemia (seen after trauma or surgery)AcidosisAbnormalities in pulmonary circulationAtelectasisPulmonary edemaHyperoxia

D. Interdependence:

All alveoli are physically connected to surrounding alveoli. For one alveolus to collapse anotheralveolus must be stretched. The resistance to stretch will reduce the tendency of small alveoli tocollapse.

IV. Regional differences in ventilationBase of the lung is better ventilated

A. Due to:

Gravity effectsWeight of the lung.

1. At the end of normal expiration the mean value for intrapleural pressure is-5 cm H2 0 in the uprightindividual as we move towards the apex (against gravity) intrapleural pressure decreases (more negative) andas we go more toward the lung base pressure increases (more positive).

2. At the apex intrapleural pressure is- 2.5 cm H2 0, that represents a low pressure but a large force

expanding the alveoli. Therefore, at the begining of inspiration alveoli at the apex are large and contain alarge volume or air.

3. At the base intrapleural pressure is- 2.5 cm H2 0, that represents a higher pressure but asmaller force expanding the alveoli. Therefore, at the begining of inspiration alveoli at the base aresmall and contain a small volume or air.

Fig 7-8 & 7-9 (West)

V. Airway resistance


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A. Resistance = P/Q

1. Factors that determine airway resistance

a. r4, viscosity and density of the gas (increasing density and viscosity, increased resistance)b Lung volume (Increased volume, decreased resistance )c. State of contraction of airway smooth muscle. Sympathetic and parasympathetic, PACO2 and histamined. Major site of airway resistance: medium size bronchi 80%

B. Turbulent VS laminar flowLarge airways Z1 Z 16 = turbulent flowSmall airways Z 17 Z 23= laminar flow

C. Dynamic compression of the airways

A flowvolume curve can be recorded by making a subject inspire to total lung capacity and then exhale ashard as he can to residual volume;if this is performed repeatedly, with progressively more effort, airflow is

increased at high lung volumes (effort dependent) but not at low lung volumes (effort independent ).At any given lung volume a certain flow rate can not be exceeded

1. A rise in intrapleural pressure at a given lung volume causes a rise in alveolar pressure, and arise in alveolar pressure will cause an increase in airflow, if airway resistance is constant (V = P)During forced expiration; intrapleural pressure can rise above atmospheric pressure.When airway resistance dissipates pressure in the airways as gas flows toward the mouth, pressure on theoutside of the airways can become greater than pressure on the inside of the airways.This transmural pressure difference, in turn, can compress the airway and increase airway resistance.Increases in Pel (e.g.with increase in lung volume) reduce the tendency for airway compression, forexample,schema C in the lower figure.

Forced vital capacity

(West pag. 153)


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Increases in the airway resistance will increase the tendency for airway compression.

2. Increasing the stiffness of the airways will reduce the tendency for airway compression.Completely rigid tubes would not be susceptible to airway compression. In contrast airways with weak wallswill easily collapse when pressure outside is increased above pressure inside:

3. At high lung volumes airway compression does not normally occur: Pel is very high, and airway resistanceis minimal.

D. Forced Expiratory Volume -1/ Forced Vital Capacity

1. A forced vital capacity determination is accomplished under maximum patient muscle effortto insure maximum flow rates at all lung volume

2. FEV1 is the forced expiratory volume at one second. It is normally about 80 % of the FVC

3. FEV1 is a funtion of: lung size, elastic ability to expand the lungs, and conducting airway

diameter (airway resistance).

4. The FVE 1/FVC ratio is good index of expiratory airway resistanceNormal subject FVE 1/FVC is greater than 0.80 (at least 80% of FVC is expired in the first second.

a. Obstructive disease FVE 1/FVC is below 0.80

b. Restrictive disease (such as fibrosis) both FEV and FVE1 are reduced but FEV 1/ FVC isnormal or increased.

5. Maximum mid-expiratory flow rate (MMER25-75%) is the average rate of air flow measured between 25and 75% of the total FVC. This measurement is often more sensitive than FVE 1 in detecting chronic diffuseobstrution.

E. How to determine if a problem is Restrictive or Obstrutive

1. Restrictive disease:

a. To determine a Restrictive problem, the FVC is obtained from the machine at ambient temperature(ATPS,AT=ambient temperature, PS=pressure standard) and has to be converted into BTPS (bodytemperature pressure standard) by multiplying by a conversion factor

Vbtps= 1.073. Vatps


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Then divide the measured BTPS by the predicted volume (can be obtained from heigt/weigh charts) andmultiply by 100 to get a percent.

Example: if the predicted volume is 5 liter and the one measured in the patient in BTPS is 4 liters then:

Measured FVC in BTPS x 100 4 liters x 100 = 80 % VCPredicted VC 5 liters

If the % vital capacity < 80 = restrictive problem, restriction of expansion of chest.Is characterized by increased respiratory frequency decreased, tidal volume, FRC, TLC.There is difficulty in breathing in all the air.

b. Examples of restrictive diseasesdiffuse interstitial pulmonary fibrosisatelectasisobesity

broken ribsdiseases of the pleuradiseases of the chest wallneuromuscular disorders

2. Obstructive:

a. Obtaining the FEV1 by spirometry and dividing by VC and multiplying by 100

Example:FVE1 x 100 65 x 100 = 76%VC 85

If the FEV1/VC % calculated is < 80% the problem is obstructive.Is characterized by decreased respiratory frequency, increased tidal volume, RV, FRC, TLC.

b. Examples of obstructive diseases:

COPD (emphysema, chronic bronchitis)AsthmaLocalized airway obstruction

Tracheal obstructionBronchial obstruction

F. Flow-volume curves (West pag. 110-113 and 153-154)

1. At high lung volumes the airflow rate is effort-dependent. At low lung volumes the expiratory flow is effort-independet. This is because resistance increases with increasing expiratory effort due to greater dynamiccompression of the airways with more positive intrapleural pressure

2. Restrictive disease: both maximum flow rate and total volume exhaled are reduced.Flow rate is often abnormally hign during the latter part of expiration because of the increased lung recoil

3. Obstructive disease: are associated with high lung volume, and high residual volume because the airwayclosure occurs at high lung volumes. Flow rate is very low in relation to lung volume. During the effort


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independent part the curve is depressed inward (flow rates are low for any relative volume). Examples:asthma, bronchitis, emphysema.

G. Normal flow curves under different physiological conditions


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H. Flow-volume loops help to distinguish between fixed and variable obstructions:

1. Fixed (intra or extrathoracic) obstruction: both the inspiratory and expiratory flow volume curves aretruncated, indicating a decreased peak during expiratory and inspiratory flows.

2. Variable obstruction: when the changes in the transmural pressure gradient caused by the inspiratoryor expiratory effort result in changes in the diameter of the obstruction. The loop can demonstrate when the

obstruction is intra or extrathoracic.

a. Variable extrathoracic obstruction: during forced expiration the cross-sectional area increases asthe pressure inside the airway increases. Then the expiratory curve is not affected. During forcedinspiration the pressure inside the upper airway decreases and the cross-sectional area willdecrease, and the flow volume curve is truncated.

b. Variable intrathoracic obstruction: during forced expiration positive intrapleural pressure is producedand the cross-sectional area decreases reducing then the forced expiratory flow, during forcedinspiration, the intrapleural pressure becomes more negative and the cross- sectional areaincreases, then flow-volume curves are not affected.


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The pulmonary circulation is characterized as a low pressure, low resistance and high compliancesystem.

Comparison of the systemic and pulmonary circulations:

Characteristics Systemic circulation Pulmonary circulation

1. Output 5-6 L/min 5-6 L/min2. Blood volume 4.5 L 0.5 L3. Pressure high low4. Resistance high low5. Place of resistance arterioles evenly divided

cap=vein=artery

6. Normal state constricted dilated7. Gravity little effect marked effect8. State of vessels maintains tone constrict capillaries9. Wall of the pump thick (LV) thin (RV)10.Effect of hypoxia dilates capillaries constrict capillaries11.Autonomic control marked minimal

(PS dilates, S constricts)12.A-V anastomosis none present (abundant at

capillary level)

A. Pulmonary vascular resistance

PVR = MPAP MLAP PVR = pulmonary vascular resistancePBF MPAP = mean pulmonary arterial pressure

MLAP = mean letf atrial pressure

PVR = 15-5 mm Hg / 6 L min = 1.6

SVR = 100 2 mm Hg / 6 L min = 16.3


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The low pulmonary vascular resistance is due to the fact that the pulmonary arterioles haverelatively low smooth muscle arterial contractile tone and the pulmonary capillaries have arelatively large cross sectional area.

1. Lung volume and PVR:

Vessels are divided in alveolar and extraalveolar.At high lung volumes the resistance of the alveolar vessels increases.At low lung volumes the resistance of alveolar vessels decreases.

At high lung volume the extraalveolar vesssels (larger arteries and veins) decrease theresistance, this is due to:a. The increase in the negativity of the intrapleural pressure that increases the transmural

pressure distending the vessel.b. Radial traction.

During forced expiration, at RV, intrapleural pressure becomes very positive and theresistance of extraalveolar vessels increases.

When the resistances of the alveolar and extraalveolar vessels is added, resistance islowest at functional residual capacity and increases at both high and low lung volumes.


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2. Recruitment and distensibilityAs the cardiac output increases, the pulmonary vascular resistance decreases.The mechanisms that explain this are:a. Recruitment: opening of unperfused vessels, it is the main mechanism to decrease

the PVR when the artery pressure is raised from low levelsb. Distention: increase in the radio of the vessels.

It is the mechanism to decrease the PVR at relatively high vascular pressures.

B. Regional distribution of blood flow

Determined by gravity effects and extravascular pressures.In the upright individual, blood flow decreases from the bottom to the top of the lung.

1. Zones of the lung.


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Zone 1 PA > Pa > Pv Pulmonary arterial pressure falls below alveolar pressure,Capillaries would be completely collapsed Theoretically thiszone is ventilated but not perfused (alveolar dead space)This zone does not exists under normal conditions.

Zone 2 Pa > PA > Pv Pulmonary arterial pressure increases becauses thehydrostatic effect and is greater than the PA

Zone 3 Pa > Pv > PA Venous pressure exceeds alveolar pressure, the flow isdetermined by the arterial-venous pressure difference

C. Hypoxic vasoconstriction

Probable mechanism:direct action of hypoxia on vascular smooth muscle, or release ofvassoactive subtances as histamine, angiotensisn, catecholamines, and protaglandins, inhibitorsof NO.The effect of hypoxic vasoconstriction is to move the blood flow away from hypoxic zones of thelung.At high altitude, generalized vasoconstriction may occur.

D. Pulmonary hypertension

May be due to:

1. Increase in left atrial pressure, e.g. mitral stenosis and left ventricular failure.2. Increase pulmonary blood flow,e,g.congenital left to right shunt, atrial septal defect,

patent ductus arteriosus.3. Increase pulmonary vascular resistance (most common cause)

a. Vasoconstrictive,e.g.decreased PA02 (high altitude)b. Obstructive,e.g.embolusc. Obliterative,e.g.emphysema


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E. Pulmonary edema

1. Starling equation

Qf = Kf {(Pc Pis ) o (pl is)}

2. Types of pulmonary edemaa. Interstitial: fluid in the interstitial spaceb. Alveolar: fluid comes into the alveoli

3. Mechanisms of pulmonary edema

a. Increase in pulmonary capillary hydrostatic pressure (most common cause),e.gmyocardial infarction,mitral stenosis,left ventricular failure, veno-occlusive disease,

and fluid overload.b. Increased capillary permeability,e.g.sepsis,02 toxicity,ARDS (adult respiratory

distress syndrome), endotoxins,inhaled toxins (S02,N02,Chloride).c. Lymph obstruction,e.g. silicosis,lymphatic CA, filariasis (parasite infection)d. Decreased interstitial pressure,e.g. rapid removal of pleural effusion of pneumothorax

or hemothorax.e. Decreased colloid osmotic pressure e.g. overdilution by intravenous solutions,

hypoproteinemia by renal failure.f. Idiopathic, e.g. HAPE (high altitude pulmonary edema), heroin addicts, pulmonary

edema efter head injuries.

1. Ficks law for diffusion

The Ficks law describes the factors that describe diffusion through the alveolar capillarymembrane:

A = area

Vgas A D(P1-P2 ) D = diffusion constantT T = thickness of the membrane

P1-P2= partial pressure difference

Then the rate of diffusion is directly proportional to the area of the barrier, the diffusivity, and tothe difference in concentration between the two sides. It is inversely proportional to thethickness of the membrane.

The diffusion constant is directly proportional to the solubility of the gas and inverselyproportional to the square root of the molecular weight of the gas.Oxygen is less dense than carbon dioxide but the solubility of CO2 in the liquid phase is about24 times that of oxygen, for that reason CO2 diffuses about 20 times more rapidly than does O2

2. Diffusion and Perfusion limitation

A. At resting conditions, red blood cells spend about 0.75 sec inside the pulmonary capillary.B. What happened when foreign gases such CO and N2 O are added to the alveolar gas ?

a. CO moves very rapidly thought the alveolar-capillary membrane and combineschemically with hemoglobin (its affinity is 210 times that of O2 for hemoglobin), andbecause CO is not chemically dissolved in plasma its partial pressure doest not risemaintainig the partial pressure gradient through the alveolar capillary membrane.Because CO diffusion is limited only by its diffusivity and by the surface area andthickness of the barrier it is considered diffusion limited.

b. N20 does not combine chemically with the Hb, then its partial pressure in the alveolar


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gas and in the blood equilibrates very rapidly (in about 0.1 sec)and no more N2Otransfer occurs until new blood enters the pulmonary capillaries whith no N20. For thatreason it is considered perfusion limited.

3. Diffusion of oxygen

A. Blood enters the pulmonary capil laries with a PO2 of 40 mm Hg and in about 0.25 sec, PO2has risen to a value equal to that in the alveolar gas.O2 moves easily through the alveolarcapillary membrane and combines chemically whith Hb until saturates and then the PaO2increases until equilibrate with the alveolar pressure. Under this normal condition, oxygentransfer is perfusion limited.

B. During exercise the red blood cell spends much less time in the capillary, but the equilibrium of PO2between gas and blood occurs (perfusion limited).

C. When alveolar-capillary membrane is thicker (due to fibrosis or edema) O2 transport may bediffusion limited.

D. During hypoxic hypoxia (arterial PO2 is depressed) the alveolar-capillary partial pressuregradient is decreased and PO2 takes longer to equilibrate. Then O2 becomes diffusion limited.

4. Measurement of Diffusing Capacity

Diffusing capacity is the rate at which oxygen or carbon monoxide is absorbed from the alveolargas into the pulmonary capillaries (ml/min/mm Hg).

Diffusion capacity of the lung

DI = VcoP1-P2

The best form to measure diffusion capacity of the lung is to measure Carbon monoxide afteradministration by the following formula:

Dlco2 = Vco This is because the PCO in pulmonary capillaries is neglectedPaco

Dl02 = 1.23 DICO This is useful to calculate the diffusion of the 02.


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The diffusion capacity of the lung CL for 02 or CO occurs in two stages:

1. Diffusion through the blood gas barrier (including plasma and red cell interior)2. Reaction with hemoglobin

The summation of the two resistances produces a diffusion resistance

VENTILATION PERFUSION RELATIONSHIPS

The amount of gas exchange in the lungs depends on:

1. The amount of ventilation reaching the alveoli per unit time (VA)2. The amount of blood circulating through the lung capillaries per unit time (Q)

B. Oxygen transport from air to tissues

Pl02 = 150

PA02 = 100-105

Pa02 = 90-95

Pcell 02 = 20-60

Pmitoch. 02 = 5-10

If a lab report gives the following values:

Pa02 = 92PA02 = 80

What is your dx ? ______________________________________________________________


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Hypoxemia

Decreased Pa02 (40)

PC02 = VC02VA

Increasing the P02 of the inspired gas can easily abolish the hypoxemia.

Causes:Depresion of respiratory center by drugs (barbiturates and morphine derivatives) oranesthesia.Diseases of the medulla or the spinal condusting pathways (poliomyelitis )

Diseases of the nerves as Guillain Barre syndrome, or of the neuromuscular junctionAs Myasthenia gravis.Disease of the respiratory muscles, thoracic cage or obstruction of the upperAiways.Extreme obesity (pickwickian syndrome).

2. Diffusion ImpairmentEquilibration does not occur between the P02 in the pulmonary capillary blood andalveolar gas.

Normal or low Pa02 at rest, but Pa02 descreaseswith exercise.Can be corrected by administering 100% 02.least common cause of hypoxemia

Causes:FibrosisInterstitial pneumoniaConnective tissue diseases as scleroderma, lupus erythematosus.

3. Increased venous admixture or Shunt

Blood that enters the arterial system without receiving oxygen when passing throughVentilated areas of the lung.

Pa02 is < 600 after 100% 02

a. Anatomical or true shunt

b. Shunt-like effect


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Causes:Congenital heart disease (atrial or ventricular septal defects,patent ductus arteriosus)Intrapulmonary shunts (arterial-venous fistulas).

4. Uneven V/Q

Commonest cause difference

Increased A a D02 (diffusion 02)

a. Uneven Vb. Uneven Q (increases VD alveolar )

In order to evaluate the ventilation/ perfusion inequality you simply look at the partialPressure of 02 difference of alveolar and arterial pressure:

PA02 _ pa02

Pa02 is obtainned from the arterial blood gas laboratory exam, but to know PA02 calculateOt by using the following equaton (review):

PA02 = PI02 PAC02 + FR

PA02 = PI02 1.2 paC02 + F R = VC02 = 200 = 0.8 1 = 1.2V02 250 0.8

PA02 = PI02 1.2 PaC02 + F where F = PaC02 (1-R)FI02= 2 to 5 mm Hg breathing room air

A. Consequences of high and VA/Q

In order to understand the effect of V/Q inequality on blood gases, consider the

following hypothetical alveolar- capillary units:

V/Q P02 PC02

# 1. 1 100 40

# 2. zero 40 46

# 3. infinity 149 zero


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Unit with V/Q = 0 the alveolus has blood flow but no ventilation, the P02 and PC02will equilibrate with systemic venous (pulmonary arterial) blood.

Unit with V/Q = the alveolus has ventilation but no blood flow, does not exchange02 or C02 and the final PA02 will be the same as dead space air,that is 150 mm Hg and C02 of zero.

Most alveoli have VA/Q values whose composition is between VA/Q = andVA/Q = 0

The VA/Q ratio is a measure of mismatched ventilation and perfusion. An increase inventilation (V) or a decrease in perfusion (Q), in any portion of the lung will result in anelevated V/Q ratio for that portion. The result will be an elevated alveolar P02 and a decreaseedPC02.The opposite will occur with a low V/Q ratio.


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PA02 =PAC02 =PAN2 =

___________713

V/Q = 1

PA02 =PAC02 =PAN2 =

Note: there is hing N2

_________ when V/Q is713

V/Q = 0 = 0

PA02 =PAC02 =PAN2 =

____________713

V/Q = 1 =


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B. 02-C02 diagramAll possible P02 and PC02 values resulting from differences in VA/Q are characterizedusing VA/Q line. To construct this curve, a specific inspired and venous 02 and C02composition is needed

C. Regional distribution of gas and blood flow along the lung:In normal lungs, gas and blood are not distributed uniformly throughout the lung and thisleads to differing ventilation perfusion ration (VA/Q) in diferrent regions of the lung.Both V and Q exhibit an increase from the apex of the lung to the base. However, theGradient for perfusion is much greater.

This means that VA/Q must increase from the base to the apexVA/Q at the apex = 3.3VA/Q at the base = 0.63

As consequence of this low VA/Q at the base, the PA02 is lowest at the base.


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D. Effects of Q on V02; effects of VA on VC02

a. For a given composition of blood entering the alveolus:- uptake of 02 (V02) in any portion of the lungs will tend to increase as Q

increases,but will be little influenced by VA.- VC02 will tend to increase as VA increases but will be little influenced by Q

b. Because V02 depends largely of Q, and VC02 depends largely on VA, it can beinferred that where ventilation is high and blood flow is low (high VA/Q),theelimination of C02 will be favored over the uptake of 02 (R is high). Conversely,when blood flow is high relative to ventilation (low VA/Q), the uptake of 02 will begreater than the elimination of C02(R is low).


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E. Distribution of ventilation perfusion ratios


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Clinical examples

Patient A:Comes to ER with a PA02 = 70, PaC02 = 58Patient has____________________________due to ________________________

Patient B:Pa02 = 70, PaC02 = 3970< 90 = _____________________________

70 + 39 = _________________ then it is consider a ____________________________

How is the paC02___________________, then is not a __________________________

You suspect a diffusion disorder and order an ______________________________ test.

Results: Pa02 = 60, PaC02 = 39

Which observation is determinat to confirm you diagnosis of diffusion Impairment?

___________________________________________________

Patient C:Pa02 = 70, PC02 = 40 - at rest

Pa02 = 74, PC02 = 39 -during exercise

Pa02 = 340, PC02 = 41 - when breathing 100% 02

a. Patient C has _____________________, 70 < 90

b. Adding Pa02 + PaC02 you have 70 + 40 = 110, this is less than ______, and

you suspect __________________________

c. The PaC02 of the patient is 40 (normal) then this is not a case of _____________

d. You order an exercise test because you suspect a _________________________

e. You find that Pa02 has increased from 70 to 74 eliminating the diagnosis ofdiffusion disorder.

At this point you have eliminated:Hypoventilation (patient has normal paC02)Diffusion impairment (Pa02 increased during exercise)How can you differentiate between the other disease problems:

1. Venous shunt2. Uneven ventilation


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Diagnostic test

Diagnosis of shunt problem in a patient that presents low Pa02

Patient breaths 100% oxygen and if P02 < 600If a patient whith any of the other three problems receives 100% oxygen the partial P02Increases to 650 mmHg.

When the obove patient (patient C) received 100% oxygen the Pa02 increased onlyto 340 mmHg, then the patients problem is v-a mixture (shunt).

How to determine the direction of the shuntRight to left shunt: low Pa02 in the artery.Left to right shunt:normal Pa02, high Pv02,high pressure in right Chamberts.

Diagnosis of a patient with uneven ventilation/perfusion

Increased A-a D02 (increased difference in the alveolar and arterial P02Could be uneven V or uneven Q ( VD alveolar).

Is most commonly due to alveolar dead space volume.

Normal A-a 02 difference is 5 to 10 mm Hg.

The A-a difference should be of 30 When breathing 100% 02 and 10 when breathingRoom air.

When breathing 100% 02, for every 20 mm Hg of A a difference, there is 1% shunt.


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The Transport of Oxygen and Carbon Dioxide by the Blood

Copyright 2000 M. G. LEVITZKY

I. Transport of O2 by the bloodA. Amount of O2 consumed at rest 250-300 ml/min.

B. Physical dissolved O2 in the blood1. 100 ml of plasma at 37C contains 0.003 ml O2/mmHg PO2.

Normal arterial blood with a PO2 of 100 mmHg contains only 0.3 ml O2/100 ml blood.

C. Chemical combination of O2 with the Hemoglobin (Hb).Approximately 15 g of Hb is present in each 100 ml of blood (15 g/dl) and each gram of Hb has thecapacity to carry 1.39 ml of O2.

1. O2 carrying capacity of Hb = 1.39 ml O2/gm Hb.That is, if the Hb were maximally combined with O2 (fully saturated , the blood couldmaximally carry 20.4 ml/dl or 20.4 ml of O2 per 100 ml of blood(15 g Hb/dl x 1.39 ml/g of Hb = 20.85 ml/dl) (100% O2 saturation at 250 mmHg PO2).

2. Therefore a person with the "normal" 15 gm Hb per 100 ml blood has a Hb O2 carryingcapacity of 20.1 ml O2/100 ml blood

a. % Hb saturation = % O2 saturation of Hb

b. PO2 vs % Hb saturation. Under normal circumstances PO2 of the plasma determines the

amount of O2 that combines with Hb.

Special properties of the reaction Hb +O2 HbO2.Rapid and reversible: Half time is 0.01 sec or less.

3. The oxyhemoglobin dissociation curve:a. The curve is sigmoid-shaped because hemoglobin molecule has four subunits. Each Hbmolecule can carry four O2 molecules. The interaction of each one of the subunits with the O2facilitates the successive combination of O2 with the next subunits. Similarly each dissociationfacilitates the next. This S-shape is extremely important physiologically.

b. At high PO2 values, large changes can occur in the PO2 value without significantly altering

the % HbO2 saturation, while at low PO2 values, small changes in PO2 cause large changes inthe % HbO2 saturation.

c. In the lungs, the flat portion of the curve assures that the blood will be nearly fully oxygenatedeven in the situation in which the alveolar PO2 decreased to a value as low as 60 mmHg. Atthe tissue level, on the other hand, large amounts of O2 can be released by the blood into thetissue with small changes in the PO2 value (increased slope zone).


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The reaction rate is influenced by pH, PCO2 and temperature.

D. Physiologic consequences of the oxyhemoglobin dissociation curve.

1. O2 loading - pulmonary capillaries

a. Relatively flat. This means that Hb will remain highly saturated with O2 (and therefore have ahigh O content) at fairly low PO2's. (e.g. at a PO2 of 100 mmHg Hb is 97.4% saturated; at aPO of 70 mmHg Hb is still 94.1% saturated). This is of great value to a person at altitude or apatient with respiratory disease. It also means that PO2 reflects the respiratory state betterthan Hb saturation.

b. pH and PCO2: decreased pH and increased PCO2 can decrease the loading of O2 (the Bohreffect). Also increased temperature can decrease O2 loading.

c. Other factors interfering with O2 loading:Carbon monoxide - displaces oxygen from hemoglobinMethemoglobinemia Fe

++Fe

+++(doesn't combine with O2)

Anemia - decreased Hb

2. O2 unloading - tissue capillariesa. Curve is very steep between 40 and 10 mmHg, which is in the range of metabolically active

tissues. (at a PO2 of 40 mmHg, Hb is 75% saturated; At a PO2 of 20 mmHg, Hb is 32%saturated).

At metabolically active tissues the pH is lower, the PCO2 is higher, the temperature is higher,

and the 2,3 DPG levels (increased in anaerobic metabolism) are higher. These shift the HbO2dissociation curve to the right and assist in unloading O2. They cause the mixed venousHbO2 dissociation curve to shift to the right.

E. Myoglobin: High affinity for O2 - can store O2.

F. Cyanosis: More than 5 gms deoxy Hb/100 ml arterial blood.

II. Transport of CO2 by the blood:

A. Amount produced at rest: 200-250 ml/min

B. Physical solution:1. CO2 is 20 X more soluble in plasma than is O 2. Therefore about 5-10% of total CO2

transported is physically dissolved.


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C. Carbamino CO2: About 5-10% of CO2 is transported as carbamino compounds. Deoxyhemoglobinforms more carbamino compounds than oxyhemoglobin.

D. Conversion into bicarbonate (HCO3-): About 80-90% of CO2 is transported as bicarbonate.

E. Role of hemoglobin in CO2 transport:

1. Deoxy Hb is a better H+

acceptor than Oxy Hb. That is, deoxy Hb is a weaker acid.2. Therefore:

H+Hb + O2 H

++ HbO2

3. Therefore deoxy Hb can accept the H+

produced by CO2 + H2O H2CO3 H+

+ HCO3-,

allowing CO2 to be transported as HCO.4. Oxyhemoglobin is a stronger acid than deoxyhemoglobin, so it allows less CO2 to be

transported as bicarbonate. Oxyhemoglobin also forms less carbamino compounds thandeoxyhemoglobin.

F. CO2 dissociation curve of whole blood:1. Relatively straight.2. Oxyhemoglobin shifts curve to the right (Haldane effect).

From: West. Respiratory Physiology


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III. Hypoxia Can Have Several Causes

A. Hypoxia is tissue oxygen deficiency

Brain is the most sensitive tissue to hypoxia: complete lack of oxygen can causeunconsciousness in 15 sec and irreversible damage within 2 min.

Type ofHypoxia

O2 Uptakein Lungs Hemoglobin Circulation

Tissue O2Utilization

Hypoxic Low Normal Normal Normal

Anemic Normal Low Normal Normal

Ischemic Normal Normal Low Normal

Histotoxic Normal Normal Normal Low

B. Causes:

Hypoxic: high altitude, pulmonary edema, hypoventilation, emphysema, collapsed lungAnemic: iron deficiency, hemoglobin mutations, carbon monoxide poisoningIschemic: shock, heart failure, embolismHistotoxic: cyanide poisoning (inhibits mitochondria)

1. Carbon monoxide (CO) poisoning:

CO binds to the same heme Fe atoms that O2 binds toCO displaces oxygen from hemoglobin because it has a 200X greater affinity forhemoglobin.Treatment for CO poisoning: move victim to fresh air. Breathing pure O2 can givefaster removal of CO


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2. Cyanide poisoning:o Cyanide inhibits the cytochrome oxidase enzyme of mitochondriaoo Two step treatment for cyanide poisoning: 1) Give nitrites

Nitrites convert some hemoglobin to methemoglobin. Methemoglobin pulls cyanideaway from mitochondria.

2) Give thiosulfate.

Thiosulfate converts the cyanide to less poisonous thiocyanate.

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