Physiology 🫁 Flashcards

Question

What is the mechanism of normal inspiration?

Answer 1

- Contraction of the diaphragm and external intercostal muscles increases all diameters of the chest and the lungs follow the chest passively "pressure is distributed". This causes increased volume of the lungs & decreased the intrapulmonary pressure by about -2mm Hg (below the atmospheric pressure) resulting in rush of 500 ml air into the lungs (Tidal volume).

Answer 2

- It is a passive process that occurs as a result of relaxation of the diaphragm and external intercostal muscles.

Answer 3

This relaxation decreases all diameters of the chest and lung recoils by its elasticity. This increases the intrapulmonary pressure to about + 2 mmHg (above the atmospheric pressure) resulting in rush out of 500 ml air (tidal volume).

Answer 4

It is an active process that occurs as a result of: a) Forceful contraction of the diaphragm and external intercostal muscles. b) Contraction of other thoracic muscles which are called (accessory muscles of respiration). They include: Sternocleiomastoid, serratus anterior, scaleni muscles, pectoralis minor and latismus dorssi muscles.

Answer 5

- Contraction of all these muscles causes marked increase of the dimensions of the chest & marked decrease in the intrapulmonary pressure to -20 mmHg causing rush of large volumes of air in.

Answer 6

It is an active process that occurs as a result of: a) Relaxation of the diaphragm and external intercostal muscles. b) Contraction of muscles of forced expiration (abdominal muscles and internal intercostal muscles)

Answer 7

- This causes marked reduction of the dimensions of the chest & marked increase in the intrapulmonary pressure to + 30 mmHg causing rush of large volumes of air out.

Answer 8

1) Resistance of the respiratory passage. 2) Pressure relationship in the thoracic cavity. 3) Surfactant. 4) Pulmonary compliance.

Answer 9

Poiseuille’s law

Answer 10

where: R = 8 η l / πr 4 ``` R = resistance η = viscosity of the inspired gas l = length of the airway r = radius of the airway ```

Answer 11

A. Contraction or relaxation of bronchial smooth muscle "Altering radius" B. Lung volume "Altering radius" C. Viscosity or density of inspired gas "Altering viscosity" D- Bronchial Mucosa

Answer 12

■ Changes airway resistance by altering the radius of the airways. (1) Parasympathetic stimulation, irritants, and the slow-reacting substance of anaphylaxis (asthma) constrict the airways, decrease the radius, and increase the resistance to airflow. (2) Sympathetic stimulation and sympathetic agonists (isoproterenol) dilate the airways via β2 receptors, increase the radius, and decrease the resistance to airflow.

Answer 13

- The autonomic nervous activity shows "diurnal variation" in which the sympathetic activity reaches its maximum in the afternoon period whereas the parasympathetic activity is maximum in the late evening & early morning. - This is why attacks of bronchial asthma are most common at late times of the day.

Answer 14

■ alters airway resistance because of the radial traction exerted on the airways by surrounding lung tissue. (1) High lung volumes are associated with greater traction and decreased airway resistance. Patients with increased airway resistance (e.g., asthma) “learn” to breathe at higher lung volumes to offset the high airway resistance associated with their disease. (2) Low lung volumes are associated with less traction and increased airway resistance, even to the point of airway collapse.

Answer 15

■ changes the resistance to airflow. ■ During a deep-sea dive, both air density and resistance to airflow are increased. ■ Breathing a low-density gas, such as helium, reduces the resistance to airflow.

Answer 16

- The presence of secretion (mucus) or an increase in the thickness of the mucosa (mucosal edema) increases the airway resistance as in bronchial asthma.

Answer 17

A. Intra pulmonary pressure. B. Intra pleural pressure. C. Intrathorathic pressure.

Answer 18

- It is the pressure inside the lung alveoli, with the atmosphere

Answer 19

- Before the start of inspiration, it equals zero (i.e., equal to atmospheric pressure). - During normal inspiration, It is equals -2 mmHg (i.e., 2 mmHg below the atmospheric pressure). This is due to contraction of the diaphragm and external intercostal muscles that increases all dimensions of the chest. - At the end of inspiration, pressure becomes atmospheric (zero) again. This is due to rush of about 500 ml air (tidal air) air that fills the lungs at the end of inspiration. - During normal expiration: the intrapulmonary pressure increases to about +2 mmHg (i.e., 2 mmHg above the atmospheric pressure). This is due to relaxation of the diaphragm and external intercostal muscles that decreases all dimensions of the chest cavity. This forces 500 ml of air out of the lungs helped by its elasticity. Thus, pressure returns zero again at the end of expiration. N.B.: in forced respiration, the alveolar pressure follows the same course of change BUT at different values during forced inspiration (where it reaches -20 mmHg) and during forced expiration (where it reaches +20 mmHg). "‏كلام فاضي كله"

Answer 20

- Each lung is enclosed in a double-walled sac called "pleura". Both layers of the pleura are formed of serous membrane. - The part of the pleura that lines the thoracic cavity is called "parietal pleura". It is reflected at the hilum of the lung to cover firmly the outer surface of the lung as "visceral pleura". - Between the two layers of pleura, there is an extremely narrow "pleural cavity" or "pleural sac".

Answer 21

- This cavity (sac) is filled with "pleural fluid", which is serous fluid secreted by pleura. This fluid acts as lubricant to reduce friction between the two layers during respiratory movements. So, it causes lung to slide on the chest wall & resists their separation. Thus, lungs follow passively movement of chest wall.

Answer 22

It is the pressure in the pleural cavity (sac).

Answer 23

Normal values of I.P.P. (usually a negative pressure): ▪ - 3 mmHg at the end of normal expiration. ▪ - 6 mmHg at the end of normal inspiration. ▪ -30 mmHg in forced inspiration with closed glottis (Muller's experiment). ▪ +40 mmHg in forced expiration with closed glottis (Valsalva's maneuver).

Answer 24

1. Lack of air in the pleural cavity. 2. Continuous tendency of the lung to recoil AGAINST tendency of chest wall to expand. Both lungs and chest wall have a "relaxation volume" at which they are neither stretched nor compressed. This relaxation volume is 0.6 L for the lungs and 3.5 L for the chest wall. However, at the end of normal expiration (when respiratory muscles are relaxed), the volume of both lung and chest wall equals 2.2 L (functional residual capacity). Thus, lungs continuously tend to recoil (from 2.2 to 0.6) and chest wall tends continuously to expand (from 2.2 to 3.5). "Easy"

Answer 25

- If air is introduced into the pleural sac (in pneumothorax), the negativity of I.P.P. is lost, the lung collapses and the chest is expanded.

Answer 26

- The elastic tissue in the bronchial wall and the interstitial tissues of the lungs. - The arrangement of the muscle fibers of the bronchi and bronchioles in a network. So that, on contraction, the bronchial tree not only becomes constricted but also shortened in length (recoils). - The surface tension of the fluid lining the alveolar walls.

Answer 27

Negative pressure is a suction force so: 1. It helps expansion of the lung during inspiration and prevents its collapse during expiration. 2. Responsible for the negativity of intrathorathic pressure which helps: a. Expansion of the lung. b. Venous return to the heart: This is called "respiratory pump". During normal and more during forced inspiration; the intrathorathic pressure becomes negative lead to suction of more blood from the extrathorathic vein increasing venous return and cardiac output. c. The negative intrathorathic pressure helps lymphatic flow in the thoracic duct through the thoracic cavity. d. Helps pulmonary blood flow.

Answer 28

■ lines the alveoli and reduces surface tension by disrupting the intermolecular forces between liquid molecules. This reduction in surface tension prevents small alveoli from collapsing and increases compliance.

Answer 29

■ is synthesized by type II alveolar cells and consists primarily of the phospholipid dipalmitoyl phosphatidylcholine (DPPC).

Answer 30

■ In the fetus, surfactant synthesis is variable. Surfactant may be present as early as gestational week 24 and is almost always present by gestational week 35.

Answer 31

■ Generally, a lecithin:sphingomyelin ratio greater than 2:1 in amniotic fluid reflects mature levels of surfactant.

Answer 32

■ Neonatal respiratory distress syndrome can occur in premature infants because of the lack of surfactant. The infant exhibits atelectasis (lungs collapse), difficulty reinflatingthe lungs (as a result of decreased compliance), and hypoxemia (as a result of decreased V/Q).

Answer 33

1. It decreases the surface tension of the fluid lining the alveoli through: a) It scatters among the fluid molecule decreasing the attraction between them. b) It also spreads over the fluid preventing air-fluid interface. This decreased surface tension: ▪ Helps lung expansion during inspiration esp. ▪ Protects against pulmonary edema as it decreases the filtration forces for the fluid from pulmonary capillaries into alveoli. ▪ Prevents lung collapse during expiration. ▪ Decreases the work of breathing needed. 2. Surfactant activates alveolar macrophages. 3. It has a bacteriolytic effect, making the bacteria susceptible to alveolar macrophages. 4. The phospholipids of surfactant may act directly on T-lymphocytes cell membrane to prevent excess immune response. 5. It assists ciliary movement in upper respiratory tract (so, helps to remove any particles and mucus away from air passage).

Answer 34

Infant RDS (hyaline membrane disease) Adult respiratory distress syndrome (ARDS)

Answer 35

is a deficiency of surfactant in infants

Answer 36

- In fetal life, surfactant is not produced in sufficient amounts until about the eighth month. Thus, premature babies lacking sufficient surfactant are born with collapse alveoli. - Because of the high surface tension, plasma fluid leaks into the alveoli producing pulmonary edema. - Due to “glistening membrane” appearance, the condition is also called "hyaline membrane disease".

Answer 37

- RDS is a fatal condition. It is more sever and more common in infants with low plasma level of thyroid hormone. Thyroid hormone increases the size and number of lamellar bodies in type II alveolar cells.

Answer 38

- RDS babies can be saved by mechanical ventilators that keep them alive until their lung can manufacture sufficient surfactant. Otherwise, bovine or synthetic surfactant can be administrated. - When premature delivery (before 8 months) is life-saving for the mother, elective caesarean section may result in baby with collapsed lungs. It is better in this condition to inject the mother with cortisol hormone which is essential for surfactant formation.

Answer 39

Acute lung injury.

Answer 40

A- Bloodstream (sepsis): develops from injury to the pulmonary capillary endothelium, leading to interstitial edema and increased lymph flow. B- Airway (gastric aspirations): direct acute injury to the lung epithelium increases permeability of the epithelium followed by edema. C- Cigarette smoking decreases surfactant. It leads to destruction of type II alveolar cells. D- After cardiac surgery during which the pulmonary circulation is interrupted, alveolar epithelium fails to secrete surfactant. E- Long-term inhalation of 100% O2.

Answer 41

In respiratory distress syndrome, the lung becomes stiffer and compliance decreases. This greatly increases the work of breathing. Also, a more negative pressure is required to maintain a given lung volume.

Answer 42

C = Δ V/P C = compliance (mL/mm Hg) ,V = volume (mL), P = pressure (mm Hg)

Answer 43

- the extent to which the lung expands for each unit increase in its transmural pressure.

Answer 44

- For the lungs, transmural pressure = intrapulmonary pressure – intrapleural pressure. It is called "transpulmonary pressure".

Answer 45

- approximately 200 ml/cm of water pressure. That is every time, the transpulmonary pressure increase by one centimeter of water, the lungs expand 200 milliliters.

Answer 46

Two types of pulmonary compliance can be measured: 1- Static pulmonary compliance. 2- Dynamic lung compliance.

Answer 47

- Can be estimated from static (relaxation) pressure – volume curve of the whole respiratory system (pressure is recorded after the muscles of respiration are relaxed).

Answer 48

- The relaxation pressure of the lung and chest wall equals zero when the lung volume is equal to functional residual capacity (FRC) which is the volume of the lung and chest wall at the end of normal expiration. Thus, FRC is the relaxation volume of the total respiratory system. At this volume, lung tendency to recoil and chest tendency to expand balance. - At the FRC volume, the curve of the lung alone shows +ve pressure (as the lung tends to collapse to reach its relaxation volume which is about 0.6 L) whereas the curve of chest wall alone shows –ve pressure (as the chest wall tends to expand to reach its relaxation volume which is about 3.5 L).

Answer 49

- At volumes higher than FRC, the relaxation pressure of total respiratory system is positive "instead of zero" mainly due to recoil tendency of the lung "strong" (as chest wall is approaching its relaxation volume). "Weaker"

Answer 50

- At volumes lower than FRC, the relaxation pressure of total respiratory system is negative mainly due to expansion tendency of the chest wall (as lungs are approaching their relaxation volume).

Answer 51

- Can be estimated from pressure – volume curve of the lungs recorded during the respiratory cycle (pressure is recorded in steps during inspiration and expiration).

Answer 52

- The compliance of the lung is more during expiration (BZA) than during inspiration (AXB). - This is known as "hysteresis" (i.e., forward path is different from the reverse path).

Answer 53

- The compliance of the lung is determined by the diagonal line (BYA) connecting the two ends of the curve (average lung compliance).

Answer 54

1. Factors in the lung: - It is affected by elastic forces of the lung (1/3 of compliance) and surface tension of the fluid lining the alveoli (2/3 of the compliance). - Lung compliance is decreased in pulmonary congestion, pulmonary fibrosis and pulmonary edema whereas it is increased in emphysema. (CFE - E) 2. Factors in the chest wall: - It is affected by the elastic properties of the thorax caused by elasticity of the muscles, tendons and connective tissues in the chest wall. - Chest wall compliance is decreased in case of: * Deformities of spine * Arthritis of vertebra. * Skeletal muscle disease e.g. poliomyelitis * Obesity.

Answer 55

Emphysema and fibrosis

Answer 56

- often caused by smoking, results in destruction of the alveolar septa and capillaries. Therefore, lung compliance is increased and the tendency of the lungs to collapse is decreased. Therefore, at the original FRC, the tendency of the lungs to collapse is less than the tendency of the chest wall to expand. The lung–chest wall system will seek a new, higher FRC so that the two opposing forces can be balanced; the patient’s chest becomes barrel-shaped, reflecting this higher volume. "Easy"

Answer 57

- Fibrosis has increased collagen fiber deposition, which increases the tissue component of elastic recoil. Therefore, lung compliance is decreased and the tendency of the lungs to collapse is increased. Therefore, at the original FRC, the tendency of the lungs to collapse is greater than the tendency of the chest wall to expand. The lung–chest wall system will seek a new, lower FRC so that the two opposing forces can be balance. "Easy"

Answer 58

- During normal quiet respiration, work is done only during inspiration whereas expiration is entirely a passive process. At rest, work of breathing is about 1-2% of the total body energy expenditure. Even during heavy exercise, work is only about 3-5% of the total energy expenditure of the body.

Answer 59

1- Compliance work (elastic work) (65% of total work) 2- Tissue resistance work (7% of total work) 3- Airway resistance work (28% of total work)

Answer 60

- It is the work required to expand the lung against its elastic force. - This can be calculated as follow: Compliance work = volume x pressure needed to inflate this volume

Answer 61

- It is the work required to overcome the resistance of non-elastic tissue of the lungs and thoracic cage.

Answer 62

- It is the work required to overcome airway resistance during the movement of air into the lung. - The major factor, affecting airway resistance is the diameter of the passage ways. The larger the diameter of the airway, the less work done.

Answer 63

1- Decreased elasticity (fibrosis) - Lung congestion (heart failure). 2- Increased tissue resistance work: - Diseases affecting thoracic cage e.g. kyphoscoliosis. - Muscle disease. 3- Increased airway resistance work: - Obstruction of air way. - Chronic bronchitis. - Asthma.

Answer 64

- A very thin respiratory membrane which consists of: 1) Layer of fluid lining the alveoli and containing surfactant that reduce the surface tension of the alveolar fluid. 2) Single layer of thin alveolar epithelial cells. 3) An epithelial basement membrane. 4) A very thin interstitial space between the alveolar epithelium and the capillary membrane. 5) Capillary basement membrane. 6) Capillary endothelial membrane.

Answer 65

Thickness of the respiratory membrane Surface area "inc in exercise"

Answer 66

- The rate of diffusion of gases through the membrane is inversely proportional to the thickness of the membrane.

Answer 67

- the overall thickness of the membrane is 0.6 μm.

Answer 68

- Any factor that increases the thickness to more than two to three times normal (such as pulmonary edema and pulmonary fibrosis can interfere with exchange of gases.

Answer 69

- The greater the surface area, the more will be the rate of gas diffusion.

Answer 70

❑ Removal of an entire lung will decrease surface area to 1⁄2 normal. ❑ Emphysema: a condition in which the walls of the alveoli become over distended causing many alveoli to coalesce together with dissolution of alveolar walls. Therefore, the total surface area by loss of the alveolar walls.

Answer 71

A. Pressure gradient | B. Lipid solubility and molecular weight (MW) of gases

Answer 72

100 100 40

Answer 73

- According to the pressure gradient, O2 diffuses from the alveoli into the capillaries and CO2 diffuses from the capillary to the alveoli.

Answer 74

- All respiratory gases are highly lipid soluble. The rate of gas diffusion is directly proportional to the solubility and inversely proportional to the MW of the gas.

Answer 75

- The solubility of CO2 is 24 times greater than that of O2. | - The MW of CO2 is 44 and 32 for O2. So, CO2 diffuses about 20% slower than O2 due to its high MW.

Answer 76

It is a constant which measures the amount of dissolved gas diffusing through a membrane when concentration gradient is one unit.

Answer 77

- The net result is that the diffusion rate of CO2 through the respiratory membrane is about 20 times that of O2. Oxygen diffuses twice as rapidly as nitrogen.

Answer 78

Diffusion is described by Fick’s law of diffusion as follow: V gas = A × D ( P1-P2) / T Where, ``` Vgas = rate of gas diffusion across a membrane. A = Surface area of the membrane. D= Diffusion coefficient. T= Thickness of the membrane. P1-P2 = Pressure gradient along which gas diffusion occurs. ```

Answer 79

- It is the volume of a gas that diffuses through the membrane each minute for a pressure difference of 1 mmHg.

Answer 80

- The diffusion capacity for O2 under resting condition is about 20ml./ min./ mmHg. The diffusion capacity for CO2 is 400 ml./min./mmHg (20 times that of O2).

Answer 81

- This capacity increases during muscular exercise to 80 ml./min./mmHg for O2 and 1200 ml./min./mmHg for CO2.

Answer 82

- This increase is due to increase surface area of blood for O2 to diffuse and better ventilation/perfusion ratio (all zone 3) (see below).

Answer 83

- situation in which the substance equilibrates between the capillary and interstitium.

Answer 84

- situation in which the substance does not equilibrate between the capillary and interstitium.

Answer 85

- Carbon monoxide is a unique gas in that it typically doesn’t equilibrate between the alveolar air and the capillary blood. Thus, it is a diffusion-limited gas. This is taken advantage of clinically, and the measurement of the uptake of CO in mL/min/mm Hg is referred to as the diffusing capacity of the lung (DLCO). DLCO is an index of the lung’s structural features.

Answer 86

- This is the ratio between alveolar ventilation “air that reaches the alveoli” and pulmonary blood flow per minute.

Answer 87

- The normal alveolar ventilation in an adult is about 4 L / min. and the total perfusion (the blood flow through the lungs) is equal to the cardiac output of the right ventricle (about 5 L/min). So, V/P ratio is about 0.8.

Answer 88

- This ratio differs from one part of the lung to another. This is due to effect of gravity on both pulmonary ventilation and pulmonary blood flow. Thus, these differences appear in erect posture and disappear in recumbent position.

Answer 89

- In relaxation (resting) position of the chest (i.e., FRC): ❑During standing, under the effect of gravity, the weight of lung in relaxation (resting) position keeps the upper parts of the lung more stretched than the lower parts. So, upper parts contain more air in resting position than the lower ones. ❑ Also, because of gravity, I.P.P. is more negative at the apex and less negative at the base. Since the negativity of I.P.P acts as expanding pressure, the upper lung zones have larger resting volumes than the lower ones.

Answer 90

- When ventilation starts in standing position, most of tidal volume is shifted to the lower lung zones which are of low volumes (i.e., more compliant). The upper lung zones receive only small ratios of tidal air.

Answer 91

- Gravity also affects the pressure in the pulmonary blood vessels. The lung is 30 cm high causing a difference in pressure of about 30 cm H2O or 23 mm Hg. The blood pressure is less at the top and more at the base. - Thus, normally in standing position the blood flow at the upper lung zones is considerably less than that of the lower lung zones.

Answer 92

- both ventilation and perfusion is less in the upper lung zones than in the lower. But, since the blood has a higher density than the lung tissues, changes in perfusion is much greater than changes in ventilation per unit distance down the lungs.

Answer 93

Zone I Zone II Zone III

Answer 94

- Present at the apex of the lung. - It is hypoperfused (blood flow is little). This is due to high alveolar pressure which compresses the pulmonary vessels. PA > Pa > Pv. - V/P ratio = 3.3.

Answer 95

- The blood flow starts to increase as: Pa > PA > Pv | - The pulmonary vessels are now partly open. The blood flow increases gradually from the top to the bottom of this zone.

Answer 96

- Present at the bottom of the lung. - It is hyperperfused. i.e., Pa > Pv > PA. So, the blood vessels are continuously opened. - V/P ratio = 0.6.

Answer 97

- Poor gas exchange occur when the alveolar air flow is either inadequate or excessive in relation to the blood flow i.e., in zone I and zone III.

Answer 98

- To compensate for this poor gas exchange, the following local autoregulatory mechanisms contribute to match alveolar air flow and blood flow

Answer 99

I. Pulmonary vessels II. Respiratory air way

Answer 100

❑ Low PO2 causes V.C of pulmonary blood vessels and decrease in pulmonary blood flow. ❑ While, high PO2 causes V.D. of pulmonary blood vessels and increase in pulmonary blood flow. ❑ This helps to redistribute blood flow from areas poor in O2 to areas rich in O2 producing normal V/P ratio

Answer 101

❑ High CO2 level dilates the airways and increases airflow in the alveoli. ❑ While low CO2 level constricts the airways and decrease airflow in the alveoli. ❑ This helps to match alveolar air with the pulmonary blood flow.

Answer 102

1. V/Q ratio in airway obstruction ■ If the airways are completely blocked, then ventilation is zero. If blood flow is normal, then V/Q is zero, which is called a shunt. ■ There is no gas exchange in a lung that is perfused but not ventilated. The PO2 and PCO2 of pulmonary capillary blood (and, therefore, of systemic arterial blood) will approach their values in mixed venous blood. ■ There is an increased A–a gradient. 2. V/Q ratio in pulmonary embolism ■ If blood flow to a lung is completely blocked (e.g., by an embolism occluding a pulmonary artery), then blood flow to that lung is zero. If ventilation is normal, then V/Q is infinite, which is called dead space. ■ There is no gas exchange in a lung that is ventilated but not perfused. The PO2 and PCO2 of alveolar gas will approach their values in inspired air.

Answer 103

Physical solution (Dissolved Oxygen) Chemical combination

Answer 104

- Oxygen dissolves in blood and this dissolved oxygen exerts a pressure. Thus, PO2 of the blood represents the pressure exerted by the dissolved gas, and this PO2 is directly related to the amount dissolved.

Answer 105

- Each 100 ml of arterial blood contains 0.3 ml of O2 dissolved physically in plasma. - Each 100 ml of venous blood contains only 0.13 ml of dissolved O2.

Answer 106

- Although it is small amount but it has the following significance: a) It reflects and determines the O2 tension in the blood. - At O2 tension 100 mmHg (arterial blood), the amount of O2 dissolved is 0.3 ml, while at O2 tension 40mm Hg (venous blood), the amount of O2 dissolved is 0.13ml. b) It determines the rate and direction of oxygen flow. - At the systemic capillary, it is the dissolved oxygen that diffuses to the tissues. when dissolved oxygen decreases, PO2 also decreases, and oxygen is released from Hb. Oxygen comes off Hb and dissolves in the plasma to maintain the flow of oxygen to the tissues. "As a compensatory mechanism"

Answer 107

- More than 98% of the O2 is bound to hemoglobin of the red blood cells.

Answer 108

- It is an oxygen carrying pigment. Each molecule consists of: Globin and heme

Answer 109

1) Globin: a protein composed of four polypeptide chains. There are many types of polypeptide chains according to their amino acid sequence. e.g., alpha-chain, β-chain, γ- chain, and δ-chain. 2) Heme: Four organic pigment molecules containing a single ferrous ion in the center of each molecule. Each ferrous ion can combine with one molecule of O2.

Answer 110

* The reaction is reversible and rapid. * The reaction is oxygenation not oxidation. i.e., the iron in the heme remains in the reduced form “Fe++ or ferrous iron”. * The combination of Hb and O2 occurs in steps and each step accelerates the next one.

Answer 111

- It is amount of O2 present in chemical combination with Hb in 100 ml of blood. - It depends on hemoglobin content and O2 tension.

Answer 112

- It is the amount of oxygen present in chemical combination with Hb in 100 ml of blood when it is fully saturated. - In normal subjects, Hb content is about 15 gm % and each gram of Hb can combine with 1.33 ml O2. O2 capacity = 15 x 1.33 = 19.95 ml (about 20 ml O2 / 100 ml.) - Oxygen capacity depends upon Hb content "Not oxygen tension as we measure the max". So, It is decreased in anaemia.

Answer 113

This equal = O2 content/O2 capacity * 100

Answer 114

- Under normal condition, the Hb of systemic arterial blood is only 97% saturated with O2. This is due to addition of venous blood from bronchial and coronary veins to the arterial blood which is called physiological shunt.

Answer 115

- It is the % of the blood that gives its oxygen as its passes through the tissue capillaries.

Answer 116

- It equals oxygen content in arterial blood minus oxygen content in Venus blood divided by oxygen content in arterial blood

Answer 117

- The total quantity of oxygen bound with Hb in normal arterial blood is approximately O2 content in arterial blood 19.5 ml/100 ml blood. - On passing through tissue capillaries, this amount is reduced to about 14.5 ml /100 ml blood. Thus, during rest about 5 ml of O2 is utilized by the tissues by each 100 ml of blood. - The normal value for utilization coefficient is approximately 25%. In muscular exercise, It is increased to 75%.

Answer 118

- The hemoglobin–O2 dissociation curve is a plot of percent saturation of hemoglobin as a function of PO2.a. At a PO2 of 100 mm Hg (e.g., arterial blood): hemoglobin is 100% saturated b. At a PO2 of 40 mm Hg (e.g., mixed venous blood): hemoglobin is 75% saturated.

Answer 119

sigmoid or S shape.

Answer 120

- This is because the combination of O2 with the haem groups of the Hb occurs in steps as the affinity of haem to O2 is increased after the haem group is oxygenated.

Answer 121

- The upper flat part plateau This enables persons living at high altitudes or with lung disease to get enough O2 from this blood. - Middle curved part slope This satisfies the tissues needs during rest. - lower vertical part steep as occurring in muscular exercise.

Answer 122

This part shows that: ❑ At O2 tension (100 mmHg.), % Hb saturation is 100%. ❑ At O2 tension (80 mmHg), % Hb saturation is 93 %. ❑ At O2 tension (60 mmHg), % Hb saturation is 90%. "Slight decrease" - So, decrease O2 tension from 100 to 60 mmHg causes decrease of % Hb saturation from 100% to 90% (i.e., only 10%). - The functional significance of the flat part of the curve is that the arterial O2 saturation does not change much until PO2 has decreased to about 60 mmHg.

Answer 123

- At O2 tension (40 mm Hg), the % Hb saturation is 70%. So, decrease O2 tension from 100 to 40 mmHg, causes decrease of % Hb saturation from 100% to 70% (i.e., 30% decrease of % O2 saturation which are given to the tissues during rest).

Answer 124

- With further decrease of O2 tension (below 40 mmHg), there is marked decrease in the % Hb saturation (i.e., more O2 supply to the tissues)

Answer 125

- It is the oxygen that remains in venous blood after supplying the tissues. - This reserve equals 14 ml O2 i.e., 70% during rest. - This reserve is utilized during emergency e.g., muscular exercise.

Answer 126

- Shift to right: Hb gives more O2 to tissues even under high O2 tension. - Shift to left: Hb gives less O2 to the tissue even under low O2 tension.

Answer 127

The following factors shift the curve to the right: 1) Increased CO2(Bohr effect). 2) Increased hydrogen ion (decrease pH). "Acidosis" 3) Increased temperature. 4) Increased 2,3-bisphosphoglycerate (2,3-BPG). "Produced by anaerobic glycolysis"

Answer 128

* Stored blood loses 2,3-bisphosphoglycerate, causing a left shift in the curve, while hypoxia stimulates the production of 2,3-bisphosphoglycerate, thereby causing a right shift. * Carbon monoxide (CO) has a greater affinity for Hb than does oxygen. CO shifts O2– Hb dissociation curve to the left. "CoHB & HBF"

Answer 129

- It is an iron containing pigment. | - It is present in slow twitch (red) skeletal and cardiac muscle fibers.

Answer 130

- It has higher affinity for O2 than Hb. So, Its O2 dissociation curve is shifted to the left of Hb dissociation curve.

Answer 131

It acts as O2 store * In skeletal muscle: During muscular exercise, the muscle blood supply is compressed by muscle contraction and then Myoglobin provide O2 to the muscles. * In cardiac muscle: During diastole, when the coronary blood flow is greatest, the Myoglobin can load up with O2. This stored O2 can be released during systole, when the coronary arteries are squeezed and closed by the contracting myocardium.

Answer 132

In arterial blood each 100 ml carries 48-52 ml CO2. This large amount is carried in two forms "While it has a maximum of 20 ml of oxygen"

Answer 133

❑ Physically dissolved (5%): in the plasma and RBCs. "2.5 ml" ❑ Chemically combined (95%): in two forms 1- Carbamino compounds (6%) 2- Bicarbonate ions (89%)

Answer 134

- It is the combination of CO2 with the terminal amino group of polypeptide chains of blood proteins as Hb "in RBCs" and plasma protein "in plasma" . R NH2 + CO2 ----------------> R NH COOH

Answer 135

- CO2 combines with water to form carbonic acid. Carbonic acid being a weak acid dissociates into bicarbonate ion (HCO3-) and H+ ion. CO2 + H2O ------------> H2CO3 --------> H+ + HCO3- - This reaction occurs more rapidly in the presence of carbonic anhydrase enzyme which that present in RBCs and absent from plasma.

Answer 136

- It represents a store for strong base (NaHCO3) it is called “alkali reserve”. - At the pulmonary capillaries, as O2 combines to Hb, its affinity to CO2 decreases and it releases the CO2 that diffuses into the alveoli.

Answer 137

- it is the amount of CO2 added by the tissues to the blood | - It equals 5ml. "55ml/100ml in venous blood"

Answer 138

same way as the CO2 in the arterial blood.

Answer 139

▪ Physically dissolved: in plasma 10%. ▪ Carbamino compounds: formed by combination of CO2 with terminal amino group of blood proteins 20%. ▪Bicarbonate: 70%.

Answer 140

* As a result of dissociation of H2CO3 the released H ions helps unloading of oxy- Hb. "Shift oxygen disassociation curve to the right" * H+HbO2 → H.Hb+O2. * due to released HCO3, HCO3 becomes much greater in RBCs than in plasma, thus many of bicarbonate ions diffuse to plasma. * The inside of RBCs gains more positive charge, this attracts chloride ions from plasma to enter the RBCs in exchange.

Answer 141

✓ Increased bicarbonate content of both plasma & RBCs. ✓ Increase Cl in RBCs and its decrease in plasma. ✓ water shift: shift of water from plasma to RBCs to maintain osmotic equilibrium. ✓ Hematocrit value of venous is about 3% greater than that of arterial blood due to increase RBCs volume by water shift.

Answer 142

- Respiration is generated by the activity of respiratory centers.

Answer 143

- They are poorly defined collection of neurons which are located in the brain stem.

Answer 144

- medullary and pontine centers.

Answer 145

Dorsal Respiratory Group (DRG) (Mainly inspiratory) Ventral Respiratory Group (VRG) (inspiratory & expiratory "mainly")

Answer 146

- The dorsal respiratory group of neurons extends in dorsal region of the medulla. Most of its neurons are located within the nucleus of the tractus solitarius.

Answer 147

❑ DRG has the property of intrinsic periodic firing "at regular intervals" . Thus, the basic rhythm of respiration is generated mainly in DRG. ❑ The neurons of the medullary inspiratory center are connected with the motor neurons of the diaphragm present in cervical segments (3,4,5) of the spinal cord and motor neurons of intercostal muscles present in the upper thoracic segments (1-10) by descending respiratory tract which run in the spinal cord. ❑ The nervous signals discharged by DRG begin weakly and increases steadily in a ramp manner (inspirtory ramp signal) for about 2 seconds then, it ceases abruptly for approximately the next 3 seconds causing relaxation of inspiratory muscles and produce expiration.

Answer 148

- Located in each side of the medulla, anterior and lateral to the dorsal respiratory group of neurons.

Answer 149

❑ The neurons of the ventral respiratory group remain almost totally inactive during normal quiet respiration, since expiration is a passive process. ❑ They become active only during forced expiration.

Answer 150

A) Pneumotaxic center B) Apneustic center

Answer 151

It is located in the upper pons, transmitting signals to the inspiratory area.

Answer 152

❑ The primary effect of this center is to control the “switch-off” point of the inspiratory ramp, i.e., limit respiration, thus controlling the duration of the filling phase of the lung cycle and regulates the respiratory rate.

Answer 153

- It is located in the lower 1/3 of pons.

Answer 154

- It continuously sends excitatory impulses to the medullary DRG center. Therefore, it tends to prolong the time span of the inspiration.

Answer 155

Mechanism of inspiration: - continuous excitatory impulses from the apneustic center to the DRG send excitatory impulses to motor neurons of inspiratory muscles. - Impulses in the phrenic & intercostal nerves cause contraction of the diaphragm & external intercostal muscles with subsequent increase in all diameters of the thorathic cavity and expansion of the lung. - The intrapulmonary pressure decreases to -2 mmHg and air rushes in.

Answer 156

- Inhibitory impulses from both the pneumotaxic center mainly and (Herring-Breuer inflation reflex) from lung suppress the spontaneous activity of DRG and apneustic center. Therefore, inspiration is terminated & expiration starts passively. - The intrapulmonary pressure increases to +2 mmHg and air rushes out.

Answer 157

- They are initiated by certain receptors in the lung parenchyma and the wall of the bronchioles.

Answer 158

There are two reflexes: 1) Inflation reflex 2) Deflation reflex

Answer 159

- Inflation of the lungs at the end of normal inspiration stimulates stretch receptors. - These receptors send afferent inhibitory impulses along the vagus to inhibit apneustic center and inspiratory center. So, inspiration stops & expiration occurs.

Answer 160

- Excessive deflation of the lung at the end of deep expiration stimulates deflation receptors that send inhibitory impulses to the expiratory center. So, expiration is inhibited and inspiration is produced.

Answer 161

Respiratory centers are under the control of: A) Nervous factors. “Kinda in specific cases” B) Chemical factors. “In rest” "the most imp"

Answer 162

i) Impulses from other nerve centers “rather than RESP. Centers” ii) Afferent impulses from different systems

Answer 163

1. Cortex | 2. Hypothalamus and Limbic system

Answer 164

1) Respiratory system: - From upper airway a) Nose (Sneezing reflex) “dep ins. Then forced exp.” b) Pharynx (Deglutition apnea) - From lower airway and lungs a) Stretch receptors: b) Irritant receptors c) C- receptors 2) Cardiovascular system: "in cases of increased VR and ABP" a. Harrison’s reflex: b. Effect of arterial blood pressure (ABP):

Answer 165

❑ Voluntary hyperventilation. ❑ Voluntary Apnea. ‘Both are limited”

Answer 166

❑ Pain. ❑ Temperature. ❑ Emotions. “Mild to moderate inc while severe dec”

Answer 167

- It is a protective reflex that expels out foreign material from nasal cavity. “By trigeminal nerve”

Answer 168

- Breathing is inhibited during 2nd phase of deglutition (pharyngeal phase). This prevents aspiration of food into the airways during swallowing. “By glossopharyngeal nerve”

Answer 169

- Receptors in the lungs and airway are of 3 types: 1. Stretch receptors: 2. Irritant receptors 3. C- receptors

Answer 170

- They are mechanoreceptors. - They are slowly adapting receptors. - These receptors are responsible for Herring-Breuer inflation and deflation reflexes.

Answer 171

- They respond to irritant stimuli which may be: chemical stimuli (as histamine & PGS) or mechanical stimuli as (aspiration of foreign body). - They are rapidly adapting receptors - which are responsible for several reflexes resulting in cough, bronchoconstriction, and mucous hypersecretion. “ Respiration is affected during eating, sneezing and the cough”

Answer 172

- they were previously termed J receptor signifying juxta pulmonary capillary receptors. - These receptors are stimulated by: chemical stimuli (as histamine, bradykinin, serotonin and some prostaglandins) and mechanical stimuli (as interstitial pulmonary edema). - Activation of these receptors causes laryngeal closure and apnea followed by rapid shallow breathing.

Answer 173

- Increased venous return to right side of the heart leads to an increase in the respiratory rate.

Answer 174

- This increase in ventilation helps oxygenation and removal of CO2 from the excess venous return. This helps to maintain V/P ratio constant. "VR & RR are friends"

Answer 175

❑ If ABP increases above normal: This will increase the rate of discharge of inhibitory impulses from arterial baroreceptors in the walls of aortic arch and carotid sinus. Impulses are transmitted along sinus nerve (branch from 9th cranial nerve) and aortic nerve (branch from 10th cranial nerve) to the medullary centers. This causes inhibition of respiration.

Answer 176

- It is temporary stoppage of respiration on injection of large dose of adrenaline. “As if it tightens the bronchioles” “Of course it doesn’t, this is just a mnemonic”

Answer 177

- V.C. which increase ABP. This stimulates arterial baroreceptors in the aortic arch and carotid sinus leading to inhibition of R.C.

Answer 178

- At rest, ventilation is mainly regulated according to the level of metabolism "the main controller of RR" which produces changes in partial pressures of O2, CO2 and H+ concentration in the arterial blood.

Answer 179

- CO2 “more specifically the main controller of RR” (the major metabolic end product) is the principal chemical factor that regulates ventilation.

Answer 180

- Chemoreceptors are receptors that respond to changes in the concentration of O2, CO2, and H+ in their environment.

Answer 181

- central and peripheral

Answer 182

- They are sensitive to the pH of the cerebrospinal fluid (CSF). Decreases in the pH of the CSF “indirect indication of CO2 levels” produce increases in breathing rate (hyperventilation). “To wash excess CO2”

Answer 183

▪ H+ does not cross the blood–brain barrier as well as CO2 does. a) CO2 diffuses from arterial blood into the CSF because CO2 is lipid-soluble and readily crosses the blood–brain barrier. b) In the CSF, CO2 combines with H2O to produce H+ and HCO3. The resulting H+ acts directly on the central chemoreceptors.

Answer 184

■ The carotid bodies are located at the bifurcation of the common carotid arteries. ■ The aortic bodies are located above and below the aortic arch.

Answer 185

❑ Decreases in arterial PO2: “Main stimulant” PO2 must decrease to low levels (<60 mm Hg) before breathing is stimulated. ❑ Increases in arterial PCO2: “mild stimulation” ■ stimulate peripheral chemoreceptors and increase breathing rate. ■ potentiate the stimulation of breathing caused by hypoxemia. “Potentiation each other so that they produce an effect” ■ The response of the peripheral chemoreceptors to CO2 is less important than the response of the central chemoreceptors to CO2 (or H+). ❑ Increases in arterial [H+] ■ stimulate the carotid body peripheral chemoreceptors directly, independent of changes in PCO2.

Answer 186

Site: - Present in medulla oblongata. - Present in Carotid bodies and Aortic bodies. "Connected to RC by buffer nerves" Connection: - Direct connection to the inspiratory centers. "Near it" - Connect the resp. centers in M.O. via sensory nerve fibers "buffer nerves" Stimulus: - Very sensitive to changes in CO2 level in the blood (↑ PCO2). - Sensitive to ↓in arterial PO2 and pH and ↑ in PCO2.