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Superficial explanation
I understand that something must match, but why should one liter of air match one liter of blood??
I think the explanation with busses and passengers is particularly poor (but very commonly accepted). In expired alveolar air there is about 14% O2 (140 ml pr L), while in blood that has just passed the alveolar wall there is 100% saturation (no seats left). This means that there is nowhere close to a match between passengers (molecules of O2) and busses (Hb molecules), there is a lot more O2 molecules. So, can anyone explain the stoichiometry behind a V/Q of one. |
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Respectfully,
v/q is explained as such. 14% expired o2 refers to the partial pressure of 0.14 in the expired air. Oxygen partial pressure in fresh air is 0.21. so for simplicity, the exchange descreases fresh air by approximately 0.7 of oxygen. Oxygen saturation in blood refers to places on hemoglobin occupied by oxygen. The saturation is determined by the partial arterial oxygen (dissolved oxygen). The higher the pressure, the closer to 100% hemoglobin saturation can be. this pressure rises when it is juxtaposed with "fresh air" on alveolar surface to create equilibrium. so the closer the exposures of both sides of the alveolar surface, the better the equilibrium can be. This helps to increase oxygen arterial partial pressure of the blood and subsequently oxygen saturation. v/q does not mean that for every liter of air in the lungs, there is one liter of blood. The total lung volume is about 5 liters, and blood is about 5 liters as well. But, not all of the blood is in the lungs at one time. if I misread your question, then ignore my reply. Last edited by Water : 02-21-2008 at 05:33 PM. |
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V/Q = 1 is an approximately value.. even the MORE correct version of V/Q = 0.8 itself is approximate provided that certain conditions are met such as cardiac output, breathing frequency etc.
To specifically answer your question: Ventilation of the lung is 4L per minute and Perfusion of the lung is 5L per minute - both being approximate because obviously each breath you are taking as you are reading this sentence is not gonna be the same rite. So, the approximate number comes from 4/5 = 0.8 To go beyond your question: The significance of the ratio is what we should be most concerned about and especially for step1 purposes. Even inside our lung, the ratio is different and im going to assume that you have grasped the concept why apex of the lung would have higher V/Q value and a lower value at the base. Then you have to apply this to pathological states and also be able to connect this to Pa O2 and Pa CO2 values. And of course have some understaning of A-a values connected to this. Keep working hard
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Discharge status: Alive but w/out permissionSJSM'09 |
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And dont be surprised if anyone stands on one feet and says.. "i read somewhere ventilation is 5L/min and the average cardiac output is 6L/min .. etc, etc." You tell em' "more power to ya" and ask if they really understood what they read.
As for the bus situation.. what if we have a smoker in the bus.. and if all passengers are oxygen.. oh NO...
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Discharge status: Alive but w/out permissionSJSM'09 |
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OK, I will try myself if nobody else dare.
Humans produce a certain amount of CO2 and uses a certain amount of O2 pr minute, and we need to fill our lungs with the volume of air that contains at least the amount of O2 consumed, and removes the exact amount of CO2 produced. The volumes are therefore more critical for CO2, and I will do the calculation for CO2. The difference between the amount of CO2 dissolved at venous blood pCO2 (46 mm Hg) and the amount dissolved at arterial pCO2 (40 mm Hg) is approximately 40 ml CO2 pr liter blood. Therefore, we need an amount of air (pr liter of blood) that can take away the excess 40 ml CO2 pr liter blood, but not take away too much, because then the pCO2 in arterial blood would fall below 40 mm Hg (which our biochemistry doesn’t like). In other words, we would like the amount of air that dilutes 40 ml 100% CO2 down to 5.3% CO2 (which corresponds to 40 mm Hg). 40 ml CO2 in 800 ml air gives 5% CO2 (which is close enough for this demonstration). Therefore, for each liter of blood through the lungs (giving off 40 ml CO2) we need 800 ml of air to dilute the CO2 down to ~5% (or 40 mm Hg). Last edited by mortenra : 02-26-2008 at 04:41 PM. |
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something to think about
im sorry bud, i didnt think you wanted to break it down this much out of your original question but the explanation you gave only allows to ponder even more.
Quote:
standard solubility of O2 in blood = 0.03 mL O2/L/mm Hg (BRS pg.128) standard solubility of CO2 in blood = 0.6 ml CO2/L/mm Hg (20X greater than O2) so with 46 mmHg (venous) = 27.6 mL CO2/ L of blood with 40 mmHg (arteries) = 16.6 mL CO2/ L of blood difference would be 11 mL (not 40 mL) While for partial pressure of oxygen its safer, we have to understand this would explain only the DISSOLVED CO2 which is about 5-10%. Pay much more attention to the MAJOR FORM of CO2 transport to lungs which is HCO3- (about 80%) due to "Cholride shift" in RBC's with the help of carbonic anhyrdase. Other would be carbonic compounds about 15%. Quote:
Atmosphere breakdown: 78% nitrogen, 21% oxygen, 0.03% CO2. They do teach kids its 1% because decimals havent been introduced to them yet. .03% of 800mL (the air that we are supposed to intake) does not give 40 mL of CO2. The percent of CO2 EXHALED is 5% and INHALED is 0.03 - 0.06%. Your calculations mixed it up maybe? Quote:
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Discharge status: Alive but w/out permissionSJSM'09 Last edited by Dr. X : 02-27-2008 at 12:13 AM. Reason: carbonic dehydrase --> carbonic anhydrase.. ;) |
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