Respiratory Physiology II Past Papers 2022 - eejan

1 Rawan Asrawi + Ola Qutaiba Rawan Asrawi + Ola Qutaiba Yanal Shafagoj Respiratory system Respiratory system deals with the HOMEOSTASIS of its gases, which are O2, CO2, and H2, in order to maintain the Arterial Blood Gases (ABG’s). Note: you should memorize numbers of PO2, PCO2 In the system of ABG’...

1 Rawan Asrawi + Ola Qutaiba Rawan Asrawi + Ola Qutaiba Yanal Shafagoj Respiratory system Respiratory system deals with the HOMEOSTASIS of its gases, which are O2, CO2, and H2, in order to maintain the Arterial Blood Gases (ABG’s). Note: you should memorize numbers of PO2, PCO2 In the system of ABG’s we rename the gases as PaO2, PaCO2, PH, with their normal approximated values 100mmHg, 40mmHg, and 7.4 respectively. - If these potential gases are within the normal limits, we know that the lung is functioning properly, otherwise there’ll be some abnormalities. - We are focusing mainly on O2, because it’s the essential molecule for cells in order to live. We will talk about CO2 later Now moving into a normal living cell, in order to give energy in the form of ATP, we need to provide it with a GLUCOSE during the stages of Cellular Respiration: 1. Glycolysis, which occurs outside the mitochondria (1 glucose is converted to 2 pyruvates) and that gives 2 ATP. Occurs inside the cytosol 2. The entry of the pyruvate to mitochondria (in the presence of O2), that will end with 34 ATP. So, in total 1 glucose gives 36 ATP (Max). ❖ Hypoxia: Is the decrease in O2 utilisation by the cells, and that can happen due to many reasons like decrease in O2 availability, or maybe the cell itself suffers from toxins like Cyanide poisoning, or Gram- negative bacteria. The main question here, what are the POTENTIAL causes of HYPOXIA? To answer the previous question, we have to clear some points, Firstly, the O2 in atmospheric pressure is actually taken from the SEA LEVEL, so the atmosphere makes a column of air that has weight (force), and force on area gives Pressure (Atmospheric pressure). Equals 760 mmHg (at sea level) - Column A is longer, which gives us an indication that the force is bigger and so the pressure. Patm= 760 mmHg Sea level - The atmospheric pressure is composed of 2 gases mainly, which are N2 (79%), and O2 (21%), we can find CO2, but in a very small amounts (0.3%) nothing to be compared with the previous two, so we consider it as 0%. The atmospheric pressure =760mmHg, and so the PO2=160mmHg, and the PN2=600mmHg, (we get these values by timing the atmospheric pressure with the gas percentage). 760X21%=159 or just make it 160 mmHg ▪ Examples: 1- We have a mountain, which its height is 5.5 Km (5500m) away from the sea level, the atmospheric pressure here drops to ½ the original amount (760mmHg), because it is inversely proportional to altitude, so the Patm=380mmHg, and the PO2=80mmHg, also the PN2 will drop to 300mmHg. Commercial jetliners travel at about 33,000 ft (10,000 m) and of course pressurized cabins. Notice that 11Km is actually double the height in the previous example, so you can solve it immediately either by dividing the original pressure by 4 or dividing the values of the previous example with 2. The answers will be Patm=190mmHg, PO2=40mmHg, and PN2=150mmHg. A question to ask!!! What will be the atmospheric pressure for a person standing on top of mountain EVEREST, which is approximately 8888m above sea level (easy number to memorize) (Exactly:8848 m, all those numbers are just examples… )‫)لست مضطرا لحفظ أي منها‬ - Final answer: Patm=234mmHg, PO2~45mmHg (no one can breathe nor live there, due to high altitude.) Every 5500 m, atmospheric pressure decreases by 50% In conclusion, High altitude is considered one of the reasons in Hypoxia. Respiratory system is composed (in an Anatomical point of view) of 2 zones: A- The conductive zone, in that zone the main function is only conducting air in and out, so it must be opened, and there’s no gas exchange, due to absence of alveoli lack of capillaries, that zone is called (Anatomical Dead Space). B- The respiratory zone, there’s a gas exchange, where O2 diffuses to blood as if the biological membrane doesn’t exist, and the CO2 diffuses even more easily (in fact 20 times easier than O2, because it’s 20 times more soluble). Previously, we mentioned that the airways (conducting zone) have to be opened, if any obstruction occur, that will increase the Resistance (R). - According to a law that (R) is directly proportional to the length (L) of the vessel and the viscosity (η) of the blood, and inversely proportional to the radius to the fourth power (r4), so any small change in the radius will give a HUGE change in Resistance. - Any obstruction in the conductive zone leads to an obstructive disease, and if it lasts for a period of time, we call it a Chronic Obstructive Pulmonary Disease (COPD), further more the resistance here will increase and the patient will suffer from a difficulty in breathing either in inhaling or exhaling (mainly exhaling as it will be explained later), and so we prescribe a Branchiodilation drugs for them, in order to decrease the resistance. - Some vocabs you need to know: Inflatable=compliant, which means if we apply a little force, we can deform it by changing the shape and structure, and that is a MUST for the respiratory zone if not it’ll be rigid and spiffed. By comparison, 100 times more distending pressure is required to inflate a child toy balloon than to inflate the lung. Obstruction always deals with the conducting zone. Restriction always deals with respiratory zone. ❖ Lung diseases are categorised into 3 families: 1- The conducting zone diseases (COPD) around 70%, the most common ones. 2- The respiratory zone diseases (Restriction pattern) around 20%, ex: pulmonary fibrosis. 3- The vascular zone diseases (vascular pattern) around 10%, the least common, ex: pulmonary hypertension. As we mentioned previously the conductive zone is called anatomical dead space, but is it dead? Obviously NO, it has many functions including filtration, moving mucus, brings the temperature of air to 37° & humidifying the dry air by adding water vapor (H2Og) → a function of goblet cells (if dry air reaches the alveoli, it causes physical injury). By adding water vapor a new gas contributes to the pressure in the dead anatomical space→ PH2O, PH2O at 37° Celsius equals 47 mmHg. So, in the conductive zone there are 3 gases contributing to the pressure: PO2, PN2, PH2O and the total pressure is 760 mmHg → 47 mmHg of it is PH2O, then → 760 – 47= 713mmHg → 21% of it is PO2 & 79% is PN2. Final value PO2=150mmHg, which represents the PO2 in an Anatomical space, notice it is kind of close to the atmospheric one (160mmHg).150 mmHg is humidified atmospheric air. Then what are the values inside alveoli? PO2→ 100 mmHg PCO2→ 40 mmHg PH2O→ 47 mmHg PN2→760-(100+40+47)= 573 mmHg IMPORTANT NOTES: - We can’t take N2 from air and add to it H2 or O2 to get biological compounds - PN2 doesn’t have any importance; N2 is a spectator molecule and has no role in the RS. You’re not required to memorize it but it’s very easy to calculate (760-the rest of the partial pressure values of other gases). - We need just to know PO2 and PCO2 because they always affected - Our physiological temperature is 37° Celsius thus PH2O in RS is always = 47mmHg. - Some abbreviations that will be used later on: PvO2 → venous O2 pressure, Pv-O2 (v bar) → mixed venous pressure, PAO2 → alveolar pressure, PaO2 → arterial pressure, PeO2 → expiratory pressure, PiO2 → interstitial pressure. →In vascular: A → Alveolar a → Arterial V → Venous → Mixed venous i → Interstitial E: expired Ebar: mixed expired Pv-O2= 40mmHg, Mixed venous blood is the combination of venous blood from both the superior and inferior vena cava systems. (Rt atrium… Rt ventricle… Pulmonary artery) - Oxygen diffuses from high partial pressure to low partial pressure. - O2 attaches to hemoglobin in RBCs (we get HbO2) and the time that RBCs spend in capillaries equals one cardiac cycle duration which can be 0.8 second in case the heart rate is 75 (at rest), or 0.4 second in case of exercising (heart rate=150). 0.2 when HR=300 bpm - When the RBC reaches a point where the time=0.3sec the PO2 is at equilibrium with PAO2 (PO2=100mmHg) & till 0.8 sec the PO2 is 100 mmHg So only one third of the Respiratory Membrane is being used. The other two thirds are kept for reserve (for example exercise) - The Respiratory Membrane is composed of 6 layers (The doctor mentioned 3: Alveolar epithelium, interstitium, and capillary endothelium), O2 must cross these 6 layers. - The oxygen has to diffuse through the 6 layers of the respiratory membrane: 1. Surfactant (surface acting agent) 2. Alveolar epithelium. 3. Basement membrane of alveoli. 4. Interstitium. 5. Basement membrane of capillary. 6. Endothelium Respiratory membrane is also known as alveolar capillary membrane or air-blood barrier - Thickness of the respiratory membrane is only 0.25 – 0.6 . Imp: Oxygen can cross any biological membrane as if the membrane doesn’t exist so as we know O2 doesn't need any channels to transport it. Thus, hypoxia is unlikely due to diffusion problems. (Oxygen is not diffusion limited) - As we’ve mentioned before the diffusibility of CO2 is 20 times more than O2, then if O2 can cross any biological membrane as it doesn’t exist, CO2 will diffuse 20 times easier than O2. *In case there’s a problem in the lungs which gas is affected first? O2, because its diffusibility is less. In respiratory failure type 1 → O2 is affected, & in type 2 both O2 & CO2 are affected. ❖ Systemic Circulation: Remember Interstitium: is the space between cells and capillaries. o In respiratory system: PO2 in pulmonary capillaries = 40mmHg and PAO2= 100mmHg - As we said O2 diffuses from high PO2 (alveolar) to low one (capillaries). o In circulation: PaO2= 100mmHg and PiO2= 40mmHg, O2 diffuses from arterioles to interstitial. Pcell O2 < 40mmHg, O2 diffuses from interstitial to the cell. *Why is the PO2 in venous = PO2 in interstitial? Because of the difference in volume. - Quick revision: ▪ Blood = 7% of our weight = 5L = 5000ml 3000 ml in systemic veins (60%) 750 ml in sys arteries (15%) 350 ml in sys capillaries (7%) 450 in lungs (9%) 350 in the heart (7%) ▪ Water in our body: 60% of our weight, for ex: 70kg x 60% = 42L 2/3 inside cells = 28L 1/3 outside cells= 14L (divided into: 11L in interstitial and 3L in plasma) The End of sheet #1 2 Ola Qutaiba Rawan Asrawi Yanal shafagoj Respiratory system To make diagnosis, prognosis, or even seeing the development of any medication, physicians always attend to make some tests by using different devices, one of them is the Spirometer. Spirometer: A device that is used to measure volume of the air coming in and out (inspired or expired). Ventilation: We can find in a normal lung (before inspiration) about 2.2L (FRC) of air, and by inhaling (quite breathing) it’ll reach 2.7L, about 0.5L (500ml), we call that amount as the Tidal Volume (VT) in case of a forceful inspiration, the inspiration muscles are going to inhale an additional 3L (5.7L), and that one is called the Inspiratory Reserved Volume (IRV), which is reached only in doing exercises. Same thing if we make a forceful expiration, the expiration muscles of the abdomen and internal intercostal muscles are going to exhale half of the resting volume (2.2L), which gives a (1.1L), we call that volume as the Expiratory Reserved Volume (ERV). In young adult the ERV is more than RV and in older age is the opposite. The reason behind that is because lung looses elastic recoil and the stiffening of the chest wall with age. The volume of air remaining after the forceful expiration (we can’t expire it) is called the Residual Volume (RV). Notice that we ended up with 4 volumes, VT, IRV, ERV, RV. Knowing that the last one (RV) can not be measured by the Spirometer, because it stays in the lungs without moving. If we add more than one lung volume together, we get a lung capacity, which could be containing 2,3, or4 volumes, (Maximum 4 volumes). We use these volumes and capacities to It’s known to have 4 lung capacities in that system. know which lung disease we are dealing with. 1-Inspiration Capacity (IC): the maximum amount of air that can be inspired following a normal expiration. IC= VT+IRV 2-Functional Residual Capacity (FRC): the amount of air remaining in the lungs following a normal expiration (before inhaling or after exhaling). FRC=ERV+RV 3-Vital capacity (VC): the maximum amount of air that can be expired following a maximal inspiration. VC=IRV+VT+ERV 4-Total Lung Capacity (TLC): the amount of air in the lungs at the end of a maximal inspiration =6L. TLC=IRV+VT+ERV+RV A question to ask!!! Which of the following capacities, does the Spirometer measure??? Answer: the ones that do not deal with the Residual Volume (RV), which are the IC & VC. The values giving above could vary between different countries and different regions, you’re going to find that these values could be higher (like in German), or lower (like in Japan), as their biological composition vary but in a few amounts. ————————————————————— The type of tests that shows how good the lungs are, are called the Pulmonary Function Test (PFT), or Kidney Function Test (KFT), as Liver Function Test (LFT). Each organ has its own tests Mechanics of Breathing (How do we breathe in): Of course, containing the inspiration (), and the expiration (), and these 2 terms are indicating the Flow, which happens per unit of time like (L/min) … -Flow is directly proportional to the Driving Force (DF), and inversely proportional to the Resistance. F=DF/R this is Ohm’s law Take a note the driving force could be expressed as the difference in concentrations (if we are talking about ions or molecules such as glucose), or as the difference in the pressure (if we are referring to the blood movement) & that term is also used in the air movement (DF=ΔP). If the pressure in A = the pressure in B, then ΔP (Driving Force) Is ZERO, and that means there’s No Flow. ————————————————————————————Ok, now pay attention, you know that the Patm=760mmHg, Right? That’s true, but here to avoid big numbers you have to consider It as Zero, Ok? 760 is our reference which is equal to zero mmHg So, whenever we say the Patm is zero, deep down you know it’s 760mmHg more or less than the actual number, we’ll use signs. Patm =760 0, Patm=759 -1, Patm=761 +1 ———————————————————————————————————— If we want to make a driving force between A and B, we could Do one of two options, either by increasing the PA or decreasing the PB, in order to make a Flow. 1. By positive Pressure Breathing: resuscitator: P at the nose or mouth is made higher than the alveolar pressure (Palv). This is artificial pattern of breathing 2. By negative Pressure Breathing: Palv is made less than Patm. This is normal physiological pattern of breathing Option 1 Option 2 PA =+1 PA =0 Flow Flow PB=0 PB =-1 However, we can’t do such a thing in normal breathing it’s impossible, but we can insert an endotracheal tube (intubation) and connect it with a machine to control the pressure (by increasing (we call it positive pressure breathing) such a mechanism is called the Artificial breathing. Btw, this machine is called a Respirator/Ventilator/Resuscitator. You can find it inside the hospital’s ICU, or if you love watching Some medical Dramas, you’ll notice this machine a lot in Grey’s Anatomy ;) Now, in order to understand how we breath in or out, we use Boyle’s law, which describes the relationship between pressure and volume as if we multiplied them in a closed chamber, it’ll give a constant, so any increase in one the other will decrease. P x V = Constant The lungs are surrounded by 2 membranes (Visceral & Parietal), and between them we have thepleural cavity, which pintrapleural = -4 mmHg (756 mmHg or 4 mmHg below the Patm). Pleural cavity A comparison between Inspiration and quite Expiration: Inspiration …Active process (ATP) Quite Expiration ….Passive process The diaphragm contracts (it is a The diaphragm relaxes. skeletal muscle) and that contraction needs ATP. There’ll be increasing in the volume of There’ll be decreasing in the thoracic cavity. volume of thoracic cavity. The intrapleural/ pressure becomes The intraplural pressure becomes more negative. less negative Lung inflates (its volume increases) Lung volume decreases Intrathoracic pressure is more negative Intrathoracic pressure =-6mmHg negative =-4mmHg Air enters the lungs. is less Air is pushed out of lungs. First lung inflate….pressure inside the lung becomes sub atmospheric…and air enters. Notes: 1- The muscles of a respiratory system consume 2% of total Oxygen consumption (very efficient), and that’s because we only need them in contraction, so it’s a one way ticket, but if the lungs suffer from certain disease, it’ll affect the whole mechanism, in which we have to make a lot of effort in Oxygen consumption and ATP usage, and that could use about 80% of Oxygen, and so the body could die from a Fatigue. Such as in IRDS 2- The air isn’t inflating the lungs, it is the lungs that inflate to make the air enters them. Schematic View of Respiration Po2 in ADS =150mmHg, Pco2 in ADS =0mmHg Po2 in Respiratory zone = 100, Pco2 =40mmHg Pco2 in venous side =45, Po2 in arterial side =40mmHg Pco2 in interstitial fluid/tissue =45mmHg Intracellular Pco2 must be >45mmHg, in order to make The diffusion out the cell. ‫ إلى‬، ‫تم تغيير ارقام الجدول الموجود في ساليدات الجزء األول‬.‫األرقام المعتمدة من قبل د ينال‬ A Revision from the Cardiac System CO = SV x HR = 70 x 70 = 4900~5000 ml/min And that is applying on the Respiratory system as well. Respiratory Minute Ventilation (RMV), which is the volume of air inhaled or exhaled per minute. RMV=RR * VT Every time a person breath, he inhales and exhale 0.5 L, which is equal to the Tidal Volume (VT). When multiplying it to 12 (RR), it gives 6L/min, which is the total amount of air moved into and out of respiratory system per minute, ( close to the number above). Respiratory rate or frequency RR: Number of breaths taken per minute. …‫ هههههههه‬، ‫ركزوا اللي جاي يعتبر أهم المهم‬ Ok, when we exhale the air (breathing out), the air composition in the ADS will be the same as the alveolar air, because the air is coming directly from that zone, End of Exhaling so the values of pressure in the ADS are Po2=100mmHg, Pco2=40mmHg. Same thing with the inhalation (breathing in), the air Composition in the alveolar zone will be the same as The ADS zone, so the values of pressure are Po2 =150mmHg, Pco2 =0mmHg Po2=100, Pco2=40 End of Inhaling Po2=150, Pco2=0 * Tidal volume is used to describe the volume of air that is exhaled or inhaled at rest which = 500 ml. During inspiration: If an adult male is 75 kg, Anatomic dead space volume is 150 ml (2 ml/kg body weight). As this 500 ml enters, the first 150 ml will push the air that is present in Anatomic dead space (the remaining expired air) Into the alveoli as if nothing happened (PO2 100, PCO2 40). No fresh air has entered yet/ After that, the fresh air which is inhaled will displace the expired air that has been pushed (in the ADS). This fresh air has a PO2 = 150, PCO2 = 0 So, the first 150 LITERALLY did nothing. The second 150 ml will push the first 150 into the alveoli and the other 150 will displace the 2nd one. Therefore, the fresh air that entered is 350 ml. -Alveolar ventilation: the amount of volume of fresh air that enters the lungs per minute. Alv. Ventilation = 350 ml * 12 = 4.2 L ADS ventilation = 150 ml *12 = 1.8 L Inside the thorax, we find that the heart is neighboring the lungs, and that site is actually helping the lungs in receiving air and blood (ventilation and perfusion), and these two are a Must for the lungs to work properly, as we have a ratio between them (V/Q), any volume that reaches the alveoli without being per fused is a Wasting Volume (we can’t benefit from it as it does not have any blood in it). Physiological dead space is equal to anatomic plus wasted volume, which is the volume of air in the respiratory zone that does not take part in gas exchange, knowing that alveolar wasted volume is equal to zero ml, leaving the physiological dead space equal to the Anatomical dead space. A suggested question from Dr. Yanal!!!! Can the PDS be smaller than the ADS????? (Think about it) Some information from the slides (Read them): End of sheet #2 3 Mera Masalmeh Rawan Asrawi & Asma’a Abu-Qtaish Yanal Shafagoj Respiratory system Before we start our lecture today, we have to answer the doctor’s question at the end of the previous sheet: “Can the PDS be lesser than the ADS?” The answer: PDS can’t be less than ADS. PDS = ADS + WV ADS (constant =150mm), WV can be: 1) WV=0  PDS= ADS 2) WV>0  PDS will be higher than ADS In previous sheets, we focused on the pressure of air compositions in each part of the respiratory system. Now, let's consider mixed air. During expiration, the air you exhale first comes from the anatomical dead space (150 ml), followed by air from the alveoli (350 ml). Let's calculate the partial pressures of O2 and CO2 in mixed air situation. = = 15𝑂 𝑚𝑙 (𝑖𝑛 𝐴𝐷𝑆)×150 𝑚𝑚𝐻𝑔 (𝑃𝑂2)+350 (𝑖𝑛 𝐴𝑙𝑣𝑒𝑜𝑙𝑖) ×100 𝑚𝑚𝐻𝑔 (𝑃𝑂2) = 116 500 𝑚𝑙 150 𝑚𝑙 (𝑒𝑥ℎ𝑎𝑙𝑒)×𝑍𝑒𝑟𝑜 𝑚𝑚𝐻𝑔 (𝑃𝐶𝑂2 )+350 𝑚𝑙 (𝑖𝑛ℎ𝑎𝑙𝑒)×40 (𝑃𝐶𝑂2) 500 𝑚𝑙 -Past paper question! = 28 𝑚𝑚𝐻𝑔  mixed air expiration PO2 is highest in: A. Arterial blood B. Alveolar air C. Intersttal fuid D. Mixed expiraton air  Basic of the respiratory system - The respiratory tree is composed of 23 generations/divisions/branches. - Trachea is generation #0, Primary bronchi is generation #1, Secondary bronchi is generation #2 and so on. - You have to know the first two generations (trachea and Primary bronchi), generation #16 which is the terminal bronchiole, generation #17 which is the respiratory bronchiole and generation #23 which is the alveolus (closed bulb). 𝑚𝑚𝐻𝑔 - Generations #0  #16 are located in the conductive zone (no gas exchange occurs here). - Generations #17  #23 are located in the respiratory zone (the main sites for gas exchange) - Generations #0  #10 have cartilage which give them support and therefore they are not a collapsible structure unlike the rest generations (#11 #23) which are supported by smooth muscles, if the generation collapse the R will increase so we should increase ∆𝑃 to maintain the same flow.  The components of alveolus There are three types of cells: 1. Type1 alveolar: thin squamous because of its function (gas exchange), it occupies 95% of the alveolus. 2. Type2 alveolar: rounded and cuboidal epithelial cells containing microvilli, they secrete alveolar fluid which include surfactant (90% of its component is phospholipids), surfactant reduce surface tension, this type of cell occupies 5% of the alveolus. Immune cell: Alveolar macrophages Alveolar macrophage: Is the garbage man of the alveoli and thus clean the alveoli. - Alveolus diameter is 300 micrometer. - If the amount of the fluid in the interstitial space increase this will cause edema, therefore the thickness of the respiratory membrane will increase NOTE: the thickness is inversely related to the diffusion  𝐽𝑂2 ∝ 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠  Resistance - Flow (RMV: respiratory minute volume )= 6L/min The flow needs driving force (∆𝑃) to overcome R (if we have too much R, we will need high ∆𝑃) Flow = ∆𝑃 𝑅 But how we measure resistance? 𝑹 = 𝟖 𝜼𝑳 𝝅𝒓𝟒 →𝑹 ∝ 𝟏 𝒓𝟒 (We can’t calculate r4 to all respiratory structures, we express it by cross sectional area) 𝑨𝟐 = 𝝅𝒓𝟒 → 𝑹 ∝ 𝟏 𝒓𝟒 →𝑹 ∝ 𝟏 𝑨𝟐 Airway resistance is a vague expression (it’s difficult to be measured in direct ways), it’s ∆𝑷 ∆𝑷 measured indirectly. 𝑭 = 𝑹 → 𝑹 = 𝑭 - The opposite of resistance is permeability (K) and it’s inversely related to R  𝐾 ∝ 1 𝑅 - permeability is also a vague concept - Physiologically we have significant airway resistance which = 1 ,so we need pressure difference = 1 to keep the flow constant (6L/min). Flow = ∆𝑃⁄𝑅  R =1 ∆𝑃 = 1  R= 10 ∆𝑃 = 10 Now let’s compare the resistance between cardiovascular system and Respiratory one: Cardiovascular Respiratory CO=5L/min Flow =6L/min P1= 0 & P2= 100 → ∆𝑃 = 100 ∆𝑃 = 1 TPR=100 R=1 ✓ Airway resistance is small and negligible because we need very little ∆𝑃 to overcome R. ✓ Vascular resistance is 100 times more than airway resistance although their flow is almost the same. - Most of the airway resistance normally realized in large divisions, but why?! Because of the NET cross sectional area. - NET Cross sectional area in small generations is greater than the large ones ,so the R will be higher in large airways. We have bronchioles 12% or 200ml or more then the Asthma is reversable. Spirometry measurements are usually done before and after administration of a 2 agonists (salbutamol, dobutamine, albuterol, fenoterol, terbutaline…YOU don’t need to memorize those drugs). Reversibility with the use of a bronchodilator is defined as an increase in FEV 1 of 12% or 200 ml. Patients with severe asthma may need a short course of oral steroid therapy before they demonstrate reversibility THE END OF SHEET #3 4 Hiba misleh Adel Abu-Awad & Rawan Asrawi Yanal shafagoj Compliance of lungs Compliance of lung means how easy to inflate the lung (It about stretch ability) - If you inflate lungs with little force → Compliant lung - If you inflate lungs with too much force → uncompliant lung But, How we can inflate lungs? Lung is like a balloon, we put it in a closed box, its floor is mobile, moves upward or downward. The pressure in the box is 0 (760mmHg), if we increase the volume the pressure will decrease depending on boyle's law of gases V×P =constant which will result in decreasing the pressure. Imp: We change the pressure that surrounds the lung and we measure changing in volume of the lung (how much pressure we need to inflate the lung) →ΔV/ ΔP Airway Increase V We always put the independent variable on x axis (P) and the dependent variable (V) on y axis. The Slope= ΔV/ ΔP which is the Compliance (C) As we said, lung is an elastic balloon which has resting state (when we don't have any force applied on it) *resting volume of lung= 150 ml we also call it minimal volume (MV) or unstressed volume At resting volume →Lung loses its tendency to collapse because it’s at resting state. → What is the benefit of MV? When a baby was born and took only one breath ,then died he will have minimal volume ,then Medical examiner Resting volume for a hollow organ ‫ الطبيب ر‬forensic medicine) will take some pieces of ( ‫الشعي‬ is the volume that is not tending to collapse nor to expand the dead baby’s Lung and put it in water, if these pieces float → they have minimal volume. But if the baby was born dead (stillbirth), he didn’t take any breath, his lung pieces will sink down → don’t have minimal volume. → Can normal people like us reach minimal volume? No we can’t , we only can reach residual volume 1.1L, in RDS baby can reach MV Resting volume and residual volume of lung are not the same To inflate the lung there are 3 stages: ▪ Stage 1: the lung is Uncompliant Hysteresis Note: it is not wise to inflate a totally deflated lung because it will make too much effort (too much work means too much ATP) and ATP should not overcome 2% in this case it will reach 20%. ▪ Stage 2: Compliant lung Note: it is wise to inflate a partially inflated lung → little force and too much changing in V (high compliance) ▪ Stage 3: Uncompliant lung → it is not wise to inflate a totally inflated lung - Deflation curve is not the same as inflation curve - We call the path during deflation it hysteresis (when backward process is different than forward process) ** By convention we take compliance we take the Deflation curve and not the inflation - Approximately the lung's compliance is 200ml/1cmH2O - Respiratory system is composed of two balloons, internal balloon is the lung and the external balloon is the thorax. ** It's too difficult to inflate 2 balloons one inside the other rather than inflating one balloon only so the compliance in vitro (outside the body) Total Compliance(CT) =100 CT= Total Compliance CL= compliance of lung CT= compliance of thorax - The lung at FRC (functional residual capacity 2.2L) has this much tendency to collapse (number 1) - Resting volume of thorax is 4.5 L (75% of the TLC) as if the thorax is compressed this is how much is the tendency of thorax to expand (number 2) 2 1 So, we have Lung-Thorax System For lung-thorax system, the tendency of thorax to expand equals the tendency of lung collapse and opposites in direction so the system is at rest ** Resting volume of the thorax when it is no more tending to expand ▪ Resting volume of the lung 150 ml ▪ Resting volume of the thorax 4.5 L ▪ Resting volume of the system 2.2 L Characteristics (features) of elastic structure: - To move it from resting state you must apply force → Active - But to bring it back to its resting state you just take the force away → Passive (recoil tendency of any elastic structure) Past question: If a person took tidal volume (500ml) o What will be the tendency of the lung to collapse comparing to FRC (2.2L )? - Answer: Increase o What will be the tendency of the thorax to expand comparing to FRC - Answer: decrease o What will be the tendency of the lung-thorax system ? - Answer: It has a tendency to collapse compared to FRC because it’s its resting volume → FRC is very important - According to what we are talking about FRC is resting state of the system at which the tendency of lung to collapse equals and opposite to the tendency of thorax to expand - It’s value 2.2 L - It’s the volume of air present in the lungs before taking tidal volume. ❖ Emphysema The airways are open due to elastic fibers made from protein elastin. ** Smoking inhibits anti proteases (for example anti trypsin) → proteases are now free to act and will start destroying elastic fibers Emphysema has high compliance and this is a problem which result: Collapsing tendency decreased to half while thorax tendency does not affected in this case → system is not at rest , what is the solution ?? The solution is to increase FRC for two purposes: 1. To increases the tendency to collapse of lung 2. To decrease the tendency to expand of thorax To make them equal and have new equilibrium - FRC’ in Emphysema patient is more than in a healthy person. - Emphysema patient will have Active Expiration, he will need compression and muscle contraction for expiration → energy expenditure is more, while in a healthy person expiration is Passive - Another patient has collapsing tendency more than normal opposite to Emphysema like in pulmonary Fibrosis , or IRDS (Infant Respiratory Distress Syndrome) will have two arrows in the tendency to collapse→ FRC will decrease This will lead to: 1. Decrease in tendency to collapse of lung 2. Increase in tendency to expand of thorax So the tendency of lung collapsing will equal the tendency of thorax expansion. - At residual volume what is the tendency to collapse of lung compared to FRC (always we compare to FRC)?? ▪ Tendency of lung to collapse →decreases but still present ▪ Tendency of thorax to expand→ increases ▪ The tendency for the system to expand → increases, here Inspiration is Passive - Work in the respiratory system is of 2 major types: 1. Work to overcome elastic forces Static (70%): that required to expand the lungs against the lung and chest elastic forces. Two thirds are due to surface tension and one third is duo to elastic fibers. 2. Work to overcome non-elastic forces dynamic (30%) that required to overcome: a. The viscosity of the lung and chest wall structures (20%). b. Airway resistance work (80%): that required to overcome airway resistance to movement of air into the lungs. →From slides: Past questions: →The maximum expiratory flow- volume curves in the diagram above were obtained from a healthy individual (curve A) and a 57 year old man who complains of shortness of breath (curve B). - Which of the following disorders does the man most likely have? A. Asbestosis B. Emphysema C. Fibrosis D. Acute asthmatic attack E. ARDS →In normal male adult person Which of the following Is it true at functional residual capacity? A- Lung compliance is low B- It is about 4.5 C- The elastic recoil of the long thorax system is inward D- The elastic recoil of the chest wall is outward. →B →D End of sheet #4 5 Ola Qutaiba & Naseem Al-Ta'mari Dental student Dr. Yanal Shafagoj Before talking about our main topic, we have to Make a quick revision between the pulmonary and the systemic circulations: Comparison (arterial pressure if it was) Systolic Diastolic Mean (1/3 systolic + 2/3 diastolic) capillary ‫المحاضرة شوي صعبة‬ ‫ياريت تكونوا مركزين منيح بدراستها‬ Good Luck!!! Pulmonary circulation Systemic circulation 25mmHg 8mmHg 14-15mmHg 100mmHg 80mmHg 95mmHg 7-10mmHg 30mmHg (if it was surrounding a muscle knowing that the starting point is 40 and the ending point is 20, we measure the pressure in the center to get 30mmHg). Notice, there’s a tiny amount of blood pressure going to the pulmonary capillary, that’s because we do not want to make Filtration, it’s not the main function here! Lungs can get some nutrients and filtration from another sites, and just by imagining if the amount of blood going to the pulmonary capillary was huge, we would suffer from a Pulmonary Edema, and the problem never stops here, the case will get worse to have an interstitial edema, ending with an alveolar edema, why so? Well, as we know the diffusion is inversely related to the thickness, in Edema we have a great increase in the thickness, resulting in a very poor diffusion, and that could rupture the alveoli, so the gas exchange won’t happen properly (Hypoxia later), that’s why we have the Lymphatic System to correct such actions (we will talk about it later). Ok, going inside a pulmonary capillary, we will see 2 main forces, the capillary hydrostatic pressure (Pc) (10mmHg), and the πcap (from proteins) (28mmHg, dividing into 20 from the Albumin, and 8 from the Globulin), while outside the capillary there’ll be the interstitial pressure (Pi) (-5mmHg) and πinterstitial (14mmHg) the last one is kind of big due to the presence of proteins in interstitial. The previous forces are involved in one law, that helps to know Whether there’s an absorption or filtration, which is the Starling Forces Law (Forces going outside – the ones going inside) (πinterstitial + interstitial pressure + oncotic pressure =29mmHg) – (πcap =28mmHg) = +1 mmHg {FILTRATION, but so little and we discussed why above}. - Qleft heart area = Pa /TPR = 100/100 = 1 and that means the blood output is 5L per min. - QRight heart area = Pa / PVR = 14/14 = 1 and that means the blood output is 5L per min. From the two statements above, you clearly can see that the Total Peripheral Resistance (TPR) is higher than Pulmonary Vascular Resistance (PVR), nearly by 7 times more and that’s why the thickness of the left area is bigger than the right one (High R = High P = more the thickness). The lung has an apex and a base, by looking at the figure You can sense where the blood prefers to stay, which is at BASE , due to Gravity. Apex So, Q base > Q apex (By 20 times). In a standing person, the plural pressure is differing in different. sites (Regional Difference), in apex it might reach -8mmHg, while in base it reaches -2 mmHg, with that amount of pressure. The apical alveoli are already inflated, and the ones in the base are partially inflated (very compliance), so in any action of taking a breath, most air will just go directly to the base, making the ventilation higher than the one at the apex. Base Remember! It’s not wise to inflate a totally inflated lung. So, V base > V apex. Now, if you remembered the Alveolar Ventilation was equal to 4.2 L per min ( we reached it by multiplying 350 with 12), dividing the result with 5 (cardiac output), we will get 0.84 (we also refer to it as 1) and that’s the Average V/Q. Here, the V/Q apex > V/Q base. V/Q in the apex = 3.4 which means we have air more than blood so this region helps to have alveolar wasted volume (so PDS>ADS) Don’t memorize the numbers. But you have to know that V and Q are more in the base but V/Q RATIO more in apex A question to ask!!! If we have an alveolus surrounded with a Closed blood vessel (No blood perfusion), that case is called a Pulmonary Embolism, can you tell the V/Q value and the value of the O2 pressure inside the respiratory zone ? Answer: We have an air Ventilation (V) but no Perfusion (Q), so the Ratio will become ∞ (Any number / 0 = infinity), in addition to that the O2 pressure will be equal to that in the ADS (150mmHg), due to the poor diffusion, same thing applies to the Pco2 (0mmHg). Same case but replacing the closed BV with a poorly ventilated lung (due to a closure in the airways), the ratio will be equal to ZERO & the Po2 will equal the one in the capillary venous side (40mmHg) (O2 will diffuse from the capillary to the alveoli until PO2 in alveoli become as much as PO2 in capillary ). The V/Q in the Apex >1, according to that the PAO2 in an apical alveolus will be MORE than a 100, could be 130mmHg, while in a basal alveolus the PAO2 will be Less than a 100 (90mmHg). The blood drained from the Apical region as PAO2= 130mmHg (in a standing person), same thing applies to the basal region, the drainage of PAO2=90mmHg, but if we are asking about the Quantity, the base is higher by 20 times than the Apex (Q of the base > Q of the apex) PAO2 =130 PAO2 =90 So when we calculate PO2 for blood that goes from lungs to the heart it will be 100 mmHg (actually it`s 95 and we will know why in the next lec). Talking about the aerobic bacteria, which prefers to live in the apical alveolar regions, where the higher oxygen pressure Do you still remember the Respirator??? Yes, that one machine we’ve talked about in sheet2! Well, we said that using it with an inserted tracheal tube (intubation) helps in the breathing mechanism (artificially), if that machine was calibrated to give pressure (positive pressure breathing), the alveoli might inflate way too much (because of high pressure) and it could affect the surrounding capillaries, and that might lead to a capillary closure, resulting in V/Q = infinity. CARDIAC SYSTEM ☺ As you can see, the blood flow is fluctuating between The Systolic and the Diastolic (Systolic is higher by 35%). In that case, we call the blood flow Pulsatile (Fluctuate more in systole, less in diastole), with the only exception is the Coronaries (in the heart), they receive more blood during Diastole than the systole because in contraction the arteries close, so no blood entry. →Zones of pulmonary blood flow: ▪ Zone 1 Blood flow in pulmonary capillaries is pulsatile but in systemic capillaries is not pulsatile, while the blood flow is no more continuous with respect to time but it's intermittent situation, because 1/3 capillaries are opened at any time while the other 2/3 are closed, (don’t forget that they have pre capillary sphincter). → No flow at cardiac cycle because alveolar pressure> systolic pressure> diastolic pressure (PA >Pa>Pv ) - This zone doesn’t exist in normal human lungs. - Occur in some cases like bleeding or when we set the ventilator at very high pressure. ▪ Zone 2 Systolic pressure> alveolar pressure>diastolic pressure (Pa>PA>Pv) → Flow during systole only - Intermittent or pulsatile flow - Occur in small area in apex of the lung. ▪ Zone 3 Systolic pressure > diastolic pressure> alveolar pressure (Pa>PV>PA) - It flows through the entire cardiac cycle. - Occur in our normal lungs. NOTE: During exercise there is only zone 3 :(lymphatic system) ‫اوك هسا الزم تركيز حاد ألنه تكملة للجزء اللي بأول صفحة‬ Normally, we have filtration and reabsorption happened at the wall of capillaries, and these events are controlled by what we called starling forces. - Starling forces are 4 forces, and they are: 1- Pc; Capillary Hydrostatic pressure - a filtration force (pushes fluid outside the capillary) it equals (10 mmHg) in pulmonary capillaries. 2-πc; osmotic (oncotic) force, due to plasma protein concentration – a reabsorption force, *It equals (28mmHg) in all capillaries of the body 3-Pi; interstitial fluid Hydrostatic pressure – it’s variable but ranges from zero to slightly negative. *It equals (-5mmHg) in the lungs >>represent the intra-pleural pressure 4-πi; osmotic force due to interstitial fluid protein concentration - a filtration force. *It equals (14 mmHg). - The summation of these forces: 10+14+5= 29 (filtration forces outward) “due to proteins”. -28 inside (reabsorption force). So, there is a difference of +1 between filtration and reabsorption, there is a little filtration which is taken up by lymphatic system (scavengers of our bodies) it removes everything (proteins, dead cells, access fluids….etc). →The lung should always be dry, we cannot tolerate pulmonary Edema, top medical emergency (threatening situation). - If the capillary Hydrostatic pressure becomes 28 mmHg instead of 10 mmHg the extra filtration will become +19 so more fluid filtration which causes edema. → In acute MI (Myocardial infarction), patient will have pump failure (left ventricle can't pump the blood) so he will have accumulation of blood in left ventricle and pulmonary veins and capillaries so Pc will increase to 28 mmHg, even if Pc increases and reaches up to 25 mmHg, (This is called pulmonary edema safety factor), lymphatics can still take care of the extra filtered fluid →In chronic conditions -like chronic left heart failure: even if Pc reaches 45 mmHg, lymphatics can still take care of the extra filtered fluid. This means that if a person developed pulmonary edema, his situation must have been severe (and his body has undergone severe damage). - However, the lung is full of lymphatics and can take care of the excess filtrated fluids and edema does not occur. Bronchial Circulation: That circulation arises from the Aorta, it also considers as a part of the systemic circulation (oxygenated), it receives about 1-2% of the left ventricular output (polluted and mixed circulation). Look at the figure, there are 2 alveoli, and between them there’s an alveolar capillary, on the outside we can see an Extra capillary, when we inhale these 2 alveolus will get bigger compressing on the capillary between, while the intraplural pressure becomes -6 mmHg, which leads the extra capillary to dilate, now there are 2 different R, the first one is high and the other is low, in that case we take the Total R by saying: TR= R1 + R2 Notice→the graph shows the least point of TR, which is at the FRC. In emphysema (FRC is higher than normal) and fibrosis (FRC is lower than normal), the pulmonary vascular resistance (PVR) will be high enough to reach the AFTER LOAD which lead to (Right Heart Failure). After load: the amount of blood that remains in heart in right ventricle after contraction. - The load imposed on the ventricle after contraction: it is the systolic ABP End of sheet #5 6 Majed zeyadeh Tareq zain alabdeen Yanal shafagoj Gas transport In this lecture we are going to discuss gasses transportation through the blood, specifically about oxygen. Oxygen is transported in blood in 2 ways: 1-dissolved in plasma 2-bound to Hb  dissolved in plasma 1.5% To know oxygen concentration that is dissolved in plasma we apply (HENRY’S LAW) [O2] = PaO2 x SO2 Pa: arterial pressure, 100 x 0.003 = 0.3 ml O2/dl blood S: solubility dL=100ml  Bound to hemoglobin (oxyhemoglobin HbO2) 98.5% Hemoglobin concentration in the blood  for males (14-16) g/dl with average=15  for females (12-14) g/dl (avg=13) *each 1 gram of hemoglobin can maximally bind reversibly to 1.34 ml O2 , we will consider the concentration of hemoglobin is (15 g/dl) (15*1.34=20 ml O2, which means 15 g of Hb contains 20 ml O2 for each dl ) if we have 1 dl of blood (0.3 ml is dissolved in blood and 20 ml are bound = 20.3 ml) As a result 1.5% dissolved, 98.5 % bound. Blood volume = 7% of body weight, considering human weight is 70. Thus we have 5 liter blood =5000ml=5x106 µL. Each 1 µL blood has 5 million RBCs, in each RBC we have 280 million Hb molecule Each Hb contains 4 peptide chains (2α 2β), Total MW = 64,500 (Numbers are not required). All you have to know that MW of Hb < 70,000 (70k) α = 141 x 2 = 282 amino acid “we will use this number for the coming lectures” β = 146 x 2 = 292 aa total aa (574) x avg MW (110) = 6 One RBC can transport (280*10 * 4)O2 molecules 63,000 63,000 + heme MW = 64,500 (Fe+2) Iron molecule in heme must be in ferrous form in order to achieve transporting O2, if its ferric (Fe+3) it can't be used for transporting, luckily in RBC we have reductase (meth-hemoglobin reductase) which convert ferric to ferrous form. Fe+2 binds and releases When hemoglobin binds to 4 (O2) molecules its fully saturated 100% Fe+3 binds and doesn't When hemoglobin binds to 3 (O2) molecules its 75% saturated When hemoglobin binds to 2 (O2) molecules its 50% saturated release When hemoglobin binds to 1 (O2) molecule its 25% saturated When hemoglobin binds to 0 (O2) molecule its 0% saturated. Binding of oxygen to Hb is achieved in 3 zones, where the first zone is difficult, second one is easier and third one (to be fully saturated) it gets difficult again, this phenomena creates a curve called sigmoid. It means sigma-like or S-like if a person breathes pure O2 (760 mmHg is only oxygen) PO2 in blood becomes 650 which result  O2 will become free radicals (oxygen species) which destroy DNA, proteins, cell membrane…. In the previous sheet we said that high apical PO2 unable to correct low PO2 (base) when they get mixed in left ventricle or left atrium, but why? ANSWER: The answer is: HbO2 dissociation curve is sigmoidal (not linear) At Po2 = 130 (from apex), curve is in plateau and Hb here is fully saturated which means 30 is not an excess so it doesn't correct Po 2 that comes from base (equal 90) (sigmoidal shape). When Po2 = 100 (PaO2) Hb are almost 100% saturated as we mentioned before 15 g of Hb has 20 ml O2/dl (20 ml O2/dl is achieved when Po2 = 100) When Po2 = 40 (as it is in venous blood) the content of 15 g Hb = 15 ml O2/dl , 75% saturated When Po2 = 60 represents the transition point from zone 3 to zone 2 , the content of 1 g Hb = 18 ml O2/dl (steep decline in o2) 90% saturated In general medulla oblongata has 3 centers (cardiac, vascular and respiratory), respiratory center sends impulses through neurons lead to either more contraction as result more ventilation and vice versa. At rest Extraction ratio = 𝑎−𝑣 𝑎 = 20−15 20 = 0.25 = 25%  Homeostasis : is to keep environment around the cell (interstitium) constant With regard to the dangerous effect in changing concentrations for (H 2 or Na+), its different for O2 , if Po2 is decreased up to 40 % it will have slight effect on the amount of oxygen due to the sigmoidal shape. As long as we are in the range (60-100) which is zone 3, notice that when Po2 = 60 or more the content of O2 = 18 or higher and all what we need for our cells ( in relaxation ) is 5 ml, When Po2 is doubled (Po2 = 200) respiratory centers will not send impulses and saturation stays 100% because all Hbs are fully saturated In respiratory system the feedback, isn’t two tails as it is for (H2 or Na+, K+, Ca++ etc) , its called half tail. two tails : if concentration is increased the feedback will decrease it and vice versa, for half tail feedback only works when its below certain point which is 60. If we go for high altitudes as long as we are ascending the Po 2 will be decreasing gradually and ventilation will be normal, the moment we reach Po 2 = 60 Hyperventilation occurs. So far we’ve discussed 3 numbers in the previous figure ( Po2 = 100, 60, 40). Now we will talk about PO2 when it equals 26 and in this case it represents P50 Po2 = 26 = P50 , at this pressure Hbs are 50% saturated PaO2 = 100  arterial pressure Po2 = 60  is the point that respiratory center depends on PvO2 = 40  venous pressure Po2 = 26  P50 ‫الدكتور ركز على هاي االرقام مهم‬ ‫تعرفو كل واحد عن شو بعبِر‬ When we do exercise cells in our body release Co 2 which binds to hemoglobin molecule in a different site as noncompetitive inhibitor cause a change in Hb structure from R state to T state which will result in releasing O2,at the same time the cell release H+ which will bind to another site on the Hb cause more releasing of O2 by lowering the affinity and that’s called Bohr’s effect. Increase in temperature and binding to 2-3 DPG (diphosphoglycerate) will also cause releasing of oxygen. At exercise Exam Question: when Po2 = 50, how much is the saturation percentage of Hb? Ans: more than 75%, why? When Po2 = 40 saturation will be 75% , logically it must be higher IN THE FIRST FIGURE, The curve represents cells in exercising condition (shifted to the right). At any Po2 amount of O2 bound is less and amount of O2 released is more (lower affinity) IN SECOND FIGURE, the Shifted to the left curve occurs when we have low (concentration Co2, temperature, H+ , 2-3 DPG)  Embryo hemoglobin fetal Hb or HBF Hb structure for embryo is different than adult’s Hb ( 2 α ,2 γ ), this structure prevents binding of 2-3 DPG to HB which causes shift to the left (higher affinity). When baby gets his first breath the bone marrow starts to release (2α 2β) End of sheet # 6 7V2 Abdullah Rawashdeh Sarah Abu hammad Dr. Yanal Shafagoj Transport of gases Resistance is a vague expression, and permeability is inversely proportional to the resistance. K=1/R Diffusion of O2 is directly proportional tothe difference in pressure, which makes it the driving force that causes the flow. Flow =Driving force/resistance or driving force*K, where K is the permeability K can be calculated by using this equation: K=(A/dx) *(Sol/√MW) where A is the surface area (knowing that the surface area of the respiratory membrane is 50100 m2) and dx is the thickness (the thickness of alveolar membrane is between 0.2-0.6 Micrometer), sol is the solubility of the gas. The Molecular weight is the least important factor in gas diffusion, because in the equation above it’s under a square root, so big changes in it will be small under the square root. Instead of sol/√MW we’re going to use Diffusion Coefficient, we’re going to consider O2 the reference so it’s equal to 1, for CO2 it’s equal 20 (because its solubility is 20 times that of oxygen), lastly for CO it’s 0.8, remember those numbers. The DLCO (diffusion capacity of the lung for CO) equals 17ml/min/mmHg then we can calculate that the DLO2 equals 17/0.8=21ml/min/mmHg. DLCO2 equals DLO2 times 20 which equals around 400, note that we don’t measure DLO2 directly, but indirectly through measuring from DLCO (not required of us to know how DLCO is measured and why CO is used instead of O2) During exercising, the entire lung becomes zone 3, which means it works about 3 times more efficiently, due to the distension of the capillaries and the improvement in the V/Q. This makes DLO2 around 63 (21*3). We know from previous knowledge that CO is a competitive inhibitor to O2, it binds 250 more tightly than O2. To understand this let’s imagine a cup with hemoglobin protein and PO 2 =100 mmHg and PCO=0.4 mmHg, in this cup half the hemoglobin bind to O2 and the other half to CO. 100/0.4=250 A patient with CO poisoning will have normal ABG for oxygen (Po2=100 mmHg) but the concentration of it will be very small as well as the O 2 saturation. The main problem with this is that the neurons in the brain (Respiratory center in the brain stem) won’t see any issue, they are reading normal ABG for O 2 and won’t alert the patient that his body isn’t getting enough oxygen. But in HBO2 sat curve, the curve will be shifted to the left. A patient that has 7.5g/dl of hemoglobin in blood (anemic) will have normal ABG for O2 and 100% O2 saturation. But the oxygen concentration in blood will be half of normal. Remember that normal hemoglobin equals 15g/dl and half of it will cause the normal concentration of O2 (20 ml/dl) decrease in half (10 ml/dl). Normally 5 ml of oxygen goes through the capillaries (Extraction Ratio) while the rest will remain in blood and go with the veins. And in this patient still the same 5 ml of oxygen go through the capillaries but only 5 ml go with the veins, which decreases PVO2 (in this situation PVO2=P50=26mmHg …Venous PO2 drops from 40 mmHg to 26 mmHg). Also, the extraction ratio will increase to 50% instead of the normal 25%. (5 ml out of 10 mml instead of out of 20 ml) Let’s say a muscle was paralyzed, the PO2 in the interstitial will increase to 100mmhg because the muscle isn’t taking any O2 anymore. If we give him a vasodilator the flow will increase but the PO2 will not change in the interstitial and will remain 100mmhg. If the muscle was normal but with higher metabolism the PO2 in the interstitial will decrease. Lastly, in exercise both the metabolism and flow are increasing but the PO 2 would not be changed that much or maybe not at all. During exercise ABGs stay the same. Uptake of oxygen in the lung (comparison between a normal and damaged lungs): In the first case (the red line) the PO2 reaches 100mmgh in the first 1/3 of the capillary —> o2 is not diffusion limited, normal case. In the 2nd case (orange line), there’s a damage in the lungs, so the DLO2 is lower than normal (1/4 the original value) but we reach 100mmgh PO2, (X3 the time in comparison with the first case). In 3rd case (green line) here O2 becomes (diffusion is limited) because of the thickness of the membrane of the lung (A lot of damage), So we didn't reach normal PO2, and the diffusing capacity is 1/8 the original value. We know that normal ABG for O2 is 100, but actually it’s 95, the 5 difference here occurs because of the bronchial circulation and cardiac veins empty in both left and right portions of the heart, so some venous blood mixes with the blood in aorta (venous admixture). You don’t have to know the percentage of the passing venous blood in the pulmonary circulation. -The heart pumps 50 dl of blood per minute (each 1dl contain 20ml O2) and the cells take 5ml/dl which makes the oxygen consumption 250ml/min (5*50=250), we need oxygen mainly for energy. But food molecules consume O2 and produce CO2 in different ratios, for glucose 6 molecules of O2 are consumed and 6 molecules of CO2 are produced, which makes the ratio VCO2/VO2 equal to one, proteins have a ratio of 0.8 and fats 0.7, mixed food has a ratio of 0.8 (0.82 tbh), This is called respiratory exchange ratio or respiratory quotient R.Q. Questions: 1-Systemic arterial PO2 is 100mmHg and hematocrit is 40%, what is systemic arterial PO2 if blood is added to increase hematocrit to 50? A. PO2=50 mmHG B. PO2=70 mmHG C. PO2=100 mmHG D. PO2=120mmHG E. PO2=149 mmHG 2-Arterial PO2 is 100 mmHg and content is 20ml O2/dl, what is arterial PO2 if half of all the red cells is removed? A. PO2=0 mmHG B. PO2=30 mmHG C. PO2=50 mmHG D. PO2=60 mmHG E. PO2=100 mmHG Answers 1: C 2: E End of sheet 7 8 Sondos Al-Amarat Khaled Alsadeq Dr. Yanal Shafagoj Transport of CO2 in blood CO2 is transported in three forms: The majority is transported as bicarbonate, then as carbaminoHb (bound to Hemoglobin), and the lowest percentage as dissolved in plasma. ❖ Dissolved in plasma: Numbers are not required. To know CO2 concentration that is dissolved in plasma, we apply (HENRY’S LAW) [CO2]a = Pa CO2 * S CO2. [CO2]v = PvCO2 * S CO2 Pa = arterial pressure =40 * 0.06. = 46*0.06 S = solubility = 2.4 ml co2/dl blood. =2.7 ml co2/dl blood SO2= 0.003 SCO2= 0.003 * 20 = 0.06 So, CO2 dissolved in plasma much more than O2 dissolved in plasma, which is 0.3. Cardiac output = 5L = 50dl 50 * 4 = 200ml/ min => Co2 production per minute Respiratory exchange ratio (RQ) = CO2 production per minute/ O2 consumption per minute=> 200/250 = 0.8. Question: ‫ كيف أحسب كمية األكسجين اللي باخذها بس من البالزما؟‬,5 ‫كمية األكسجين كاملة اللي بتوخذها الخاليا‬ PaO2 * SO2 => 100 * 0.003 = 0.3 PvO2 * SO2 => 40 * 0.003 = 0.12 0.3 – 0.12 = 0.18 ml of O2 which is dissolved in plasma. The rest of 5ml (4.82) is transported via hemoglobin. For CO2: in bicarbonate (4 * 60% = 2.4) In carbaminoHb (4 * 30% = 1.2) In plasma (4 * 10% = 0.4). Why does CO2 transport as bicarbonate in a high percentage? Because of the presence of carbonic anhydrase, which is enzyme catalyzed the reaction between Co2 & H2O nearly six thousand times. CO2 + H2O H2CO3 H+ + HCO3-. This enzyme present inside RBCs converts the CO2 immediately to H2CO3 then to HCO3This HCO3- must get out from RBCs to plasma to prevent it from stopping the reaction then chloride ion will enter to compensate the negative charge (Electroneutrality). We have three compartments in our bodies (intracellular, interstitiam, intravascular) in each compartment we must conserve the equilibrium cation =anion. Question: In which site the Cl- is higher? o In arteries plasma o In Veins plasma Ans: in arteries plasma The same question for bicarbonate? Ans: in veins plasma Why the Hb present in RBCs not in plasma? ✓ Plasma has an enzyme to destroy the protein (protease). ✓ MW for Hb < 70,000 so, it will be filtered in kidney and we will lose it in the urine (Hemoglobinuria). ✓ The presence of reductase inside the RBCs which reduces ferric to be ferrous (NADPHMethemoglobin reductase). ✓ If Hb presents on plasma the viscosity will increase and therefore the resistance will increase. ✓ The presence of CA inside RBCs (the most imp reason). In addition:. 2,3 BPG inside RBC. If no BPG the HbO2 curve is no more sigmoidal, it becomes like that for myoglobin. Bohr's effect: when CO2 or H+ binds to HB, O2 is released (In tissues). (In lungs) when O2 binds to Hb the CO2 is released and H+ is released, this is what called (reverse Bohr effect / Haldane effect). When CO2 is released, the reaction will reverse. CarbaminoHb Co2 + Hb ———————————————————————— CO2 in the capillary is 45, but before cross first 1/3 of the capillary it becomes 40, then continues with no more exchange (normal person). Two means by which oxygen consumption by tissue can be increased. Name them… 1. Increase blood flow: CO (cardiac output) = 5L The distribution of blood from the heart (according to the liter base) as follows: ✓ Skeletal muscles(40%*70kg=28kg their weight): one liter ✓ GIT: one liter ✓ Brain: one liter ✓ 2 Kidneys(250g…their weight): one liter ✓ Others (such as coronary blood flow 250ml and etc... 2. Increase extraction ratio which is discussed in sheet 6 ‫حاشى الحياة بأنها تشقيه‬ ‫وعدوه يمسي ويضحى فيه‬ ‫والمرء ال تشقيه إال نفسه‬ ‫ويظن أن عدوه في غيره‬ END OF SHEET 8 9 Dental student Rawan Asrawi + Dental student Yanal Shafagoj Regulation of Respiration We are going to talk about the controller system (control of breathing ) →The purpose of control breathing is to maintain normal ABGs (Homeostasis of O2 , CO2 , H+) (PaO2=100 , PaCO2= 40 , PH= 7.4) → What are the tools? ↑ventilation or ↓ventilation *What is the feedback system…(nature of the receptor)? ↓ PaCO2 , ↑PaCO2 , ↓ PaO2 (below 60 mmHg, if it was more than 60 there would be no effect because respiratory centers in medulla oblongata will not work), ↓ H+, and finally ↑H+ Note: increasing in We mean by feedback system: if there PaO2 has no effect on is a problem how it will be corrected controller or feedback →This is the brain stem → consists of mid-brain + pons + medulla oblongata (connected with spinal cord). -In Medulla oblongata we have respiratory center -In medulla 2 collections of neurons: 1 is located Dorsally (DRS) and 1 is located ventrally (VRN) DRN → inspiratory:I neurons VRN → inspiratory and expiratory.: I +E neurons systems Center: collection of neurons having related functions, ex: cardiac center and the vascular center DRN= dorsal respiratory neurons VRN: ventral respiratory neurons →These neurons will send signals to phrenic neurons located At (C3-C5) SPINALNEVES (the origin of phrenic nerve) then the signal will be transmitted to the diaphragm through phrenic n ( also called diaphragmatic n ) and this will cause contraction of diaphragm. The upper 1/3 of the pons (called Pneumotaxic center) switches off DRN ,and the Lower 1/3 (Apneustic center) switches on DRN →There are signals from periphery (as another feedback) from lung and arterial system acts on DRN -Diaphragm is skeletal muscle it is voluntary (needs motor neuron..no automaticity as in the heart),so can you stop breathing?? Yes, but up to certain limit. -In medulla there is other center which sensitive to H+ called chemo-sensitive area it is sensitive to change in [H+], ex: when there is acidosis (↑H+) that activates through chemosensitive area which stimulate DRN → PHRENIC N → contraction of diaphragm → more ventilation →In aortic arch and in the carotid artery there is aortic bodies and carotid bodies, respectively (carotid bodies are more important, they are very small and they send impulses to DRN about PaO2 Aortic body → through vagus n, carotid body → through glossopharyngeal n *let's revise things from previous sheets: O2 uptake from alveoli = 350 ml (1/7 from FRC) and PAO2 = 100 (should O2 (intake)= O2 (uptake)) but if ventilation increases (O2 intake > uptake) So PAO2> 100, specifically it will be 150 (not 160 due to H2O vapor) PACO2 becomes 0, Here we have Hyper ventilation. Hyper ventilation is when you try to make the composition of air inside alveoli closer to Note: PaO2 is as composition of outside air by definition when same as PAO2 PaCO2 decreases below 40 mmHg How can we make PAO2 more than 150? PaO2 or We make the percentage of O2 more than PAO2 The dot means liter/min V 21% When we reverse axis (PO2 on the x axis → independent): →If we increased PaO2 above 60 the ventilation will not be affected →if we decreased PaO2 below 60 there will be Hyperventilation 1= 4.2 L → For CO2 the opposite will occur since when we increase the alveolar ventilation Pco2 will decrease until 0 When we reverse axis: If PaCO2 becomes more than 40 there will be hyperventilation to wash it out (linear relation ) PaCO2 If someone decided to hold his breath, (the signal come out from the cortex of the brain)→ this (negative or decreasing) signal goes to the phrenic neurons → to the diaphragm →the ventilation stops, but you can’t hold your breath for ever, So: If ventilation decreases→ PaCO2 will increase→ then CO2 will cross the arterial blood to the BBB (blood brain barrier/CSF barrier) →CO2 will bind with H2O to produce H2CO3 that will dissociate into H+ and HCO3- →THEN H+ will go to chemo-sensitive area in medulla that will activate DRN → PHRENIC N → contraction of diaphragm → more ventilation. H+ can't cross BBB but CO2 can cross any membrane as if CO2 works indirectly here it is not exist H+ can cross but slowly. (through the H+) → This negative feedback happen by the H+ increment (acidosis), as example: after taking one pack of aspirin, the H+ takes time to enter the CSF (slow, because it is a charged molecule). We mentioned before carotid and aortic bodies, but how are they going to be able to tell the DRN about the levels of arterial PO2? Arterial PO2 is 100, interstitial PO2 is 40 and inside cells