Respiratory Physiology II 2022 PDF

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 To know oxygen concentratio...

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 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) 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 +2 Iron molecule in heme must be in ferrous form (Fe ) 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. 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: This is because the relative blood flow is low and gaseous exchange can take place at a more leisurely pace. In addition, the basilar alveoli are lessstretched than the apical ones and can “give more” with inflation (i.e., they are more compliant), that’s why 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 Po2 that comes from base (equal 90) (sigmoidal shape). You should memorize these numbers. When Po2 = 100 (PaO2) Hb are 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 15 g Hb = 18 ml O2/dl (steep decline in o2) 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 (H2 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+) , 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 Po2 will be decreasing gradually and ventilation will be normal, the moment we reach Po2 = 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 Co2 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 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