Respiratory Physiology II PDF

Summary

This document details respiratory physiology, focusing on the transport of carbon dioxide in blood. It outlines the different forms in which CO2 is transported (bicarbonate, carbaminoHb, dissolved). It also discusses the role of carbonic anhydrase and the Bohr effect in this process.

Full Transcript

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 requir...

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). 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): one liter  GIT: one liter  Brain: one liter  2 Kidneys(250g): 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

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