Respiratory Physiology II PDF

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.  Diffusion of O2 equals the difference in pressure, which makes it the driving force that causes the flow.  Flow =Driving...

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.  Diffusion of O2 equals the 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 membrane is 50-100 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 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 D LO2 directly, but indirectly through measuring from DLCO (not required of us to know how DLCO is measured)  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.  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 O2 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 O2 and won’t alert the patient that his body isn’t getting enough oxygen.  But in PO2-O2 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). Also, the extraction ratio will increase to 50% instead of the normal 25%.  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. 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 the (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, we need oxygen mainly for energy. But food molecules consume O 2 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 PO 2 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 O 2/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