He who has Medicosis Perficionalis can bear almost any topic. You're truly a genius Mashallah!! thank you so much for dedicating your time and effort to share these videos with medical students world wide.
OH MY GOD I JUST FINISHED THE VIDEO AND I’M IN LOVE WITH THE POETRY AT THE END 🤣🤣🤣🤣 your channel will be my all time favorite It's so so unique Thanks man
In a nut shell : PA CO2 = Pa CO2 . Because CO2 readily equilibrates in both alveoli & arterial blood as it has a higher diffusion coefficient than that of oxygen , so there is NO A-a gradient for CO2 ! The PA CO2 is the driving force for the diffusion of PA O2 to the pulmonary circulation across the (alveolar - capillary interface) ; where the metabolic activities of the alveolar pneumocytes & capillary endothelium determine the respiratory quotient (RQ) for O2 extraction from the PA O2 . Normal A-a gradient = 5 - 15 mm Hg .
But how can it equilibrate that much ? See pulmonary artery Pv CO2 is 46 mmHg. And pulmonary vein Pa CO2 is 40 mmHg so if its diffusing maximally due to high diffusion coefficient then it should have Pa CO2 as 6 mmHg but it's 40 mmHg this I'm not understanding. I know I'm making some silly mistake in it but I'm unable to find what error I'm making.
@@dr.allwyndsouza5246 Thank you doctor for the question & the comment ! As far as I understand it , the change in partial pressures of gases between arterial & venous bloods is NOT a simple algebraic addition or subtraction of values in blood & alveoli , even in a gas with the highest ever diffusion coefficient ! If you've better explanation , please let me know it !
@@WhyNot-si4pj sure I will let you know, Ive spent a significant time on this concept since 2 days, but I'm still unable to understand it. It doesn't seem like it's happening in the way we've been taught diffusion in physics, more deeper Im going into it, the more confusion Im getting from it.
@@dr.allwyndsouza5246 I agree with you doctor ! Some aspects of human physiology had been taught by the old schools of medicine as if they were ROCKET SCIENCE equations ; or ELECTRIC ENGINEERING graphs of deflections beyond the scope & understanding of medical scholars ! EKGs , acid - base balance , ABG interpretation , ..... etc are few of many examples to mention ! This is of course because of poor basis in natural science disciplines like chemistry , physics , mathematics ..... among the teaching staff ! I wonder myself how comes CO2 the end product of metabolism determines the O2 extraction from the PA O2 RETROGRADELY ?
bro, all of the questions i freaking had, you answered, every one of these steps where theres an underlying assumption, you answered, FINALLYYY, except i need some more physics with how CO2 and O2 affect each other in the alveolus, like i guess theyre enemies makes sense, but raelistically, how does the physics add up. do we just reject any extra O2 bc theres a certain amount already present. idk maan but thanky ou
@@MedicosisPerfectionalis hi doc 🤝i don't know who put that comment from my mobile may be my grandson,or his cousin🤷 would did that,if that action of previous comment hurted you much i apologize for that 🙏,🙏
I think Pco2 in atmosphere would be 0.2852 mm hg not 28.52 mm hg. Cause 0.04%=0.0004 and ×713 = 0.2852 mm hg Same with pAco2 5.6%=0.056 × 713 =39.98 mm hg
I think this video mis-represented the relationship between PACO² and PaCO² -- they are in fact the same because the diffusion rate of carbon dioxide through the blood-gas barrier is so high (Which means CO2 has not A-a gradient). However, the reason why you divide PaCO² by the respiratory quotient (0.8 under normal homeostasis) to calculate the A-a gradient for OXYGEN is because the PACO² value often underestimates PaCO² values WHEN you are calculating for the A-a gradient of OXYGEN. The respiratory quotient is defined as CO2 produced / O2 consumed. When you only metabolize carbohydrates, the RQ = 1 because 6 molecules of O2 is converted to 6 molecules of CO2. When you metabolize fats and proteins, the RQ drops below 0.8. When you divided PaCO² by 0.8, you over-estimate PACO². This overestimation is important because it allows you to account for the fact that oxygen consumption and carbon dioxide production is not always proportional.
"People do not like EQUATION, they would rather listen to JUSTIN BIEBER" 😅😅 This is hilarious, being here a big fan of JB n also a med student!😀 Thanks a trillion MEDICOSIS, medicine bcm so much fun now 🤗
Hello, at 10:24 when you are explaining the math behind PiCO2, the amount of CO2 in the air is in percentage, so i dont know if the math is actually right or maybe i misunderstood, anyways thanks for the help with the lecture, luv ur vids :)
Serious concept flaw that I have to point out (I was quite enjoy reading your series s sincerely) CO2 is freely diffuse into alveoli and assumbly no A-a gradient of CO2, and this concept get through alot resp physio, if u explain the derive of this equation for having A-a PCO2 difference it is completely wrong, pls consider make big ammendment.
The subtraction of PaCO2 divided by RQ is the estimation of used up of O2 by body, and the equation is derived from equilibrium at steady state of O2 gain (from ventilation) and O2 loss (from body consumption and ventilation): As at equilibrium: 1. ProducedCO2 = CO2 loss (MV * PACO2) 2. O2 gain ( MV * PinspirO2) = O2 loss (usedupO2 + MV * PAO2) with RQ of 0.8 for cell O2Consumption and CO2production ratio: UsedupO2 = 1.25 * ProducedCO2 = 1.25 * MV * PACO2 Thus combining equations we have PinspirO2 = 1.25 * PACO2 + PAO2; as we have PinspirO2 = FiO2 * ATMgasPa; PACO2 assumed equal PaCO2, the final equation PAO2 = FiO2(ATMgasPa) - 1.25(PaCO2). thanks
@@MeOKMeO Yes, I find this derivation of the equation more appealing. I have a question though. Say, for example, if one hyperventilates until PaCO2 drops to 20 mmHg, according to the equation, PAO2 will have to increase. But how can this then be explained? Is it okay to explain that PAO2 increases because O2 consumption decreases (since PaCO2/R decreases)? In this case, I think it would make more sense to look at PaCO2/R as PACO2, which can then be used as a ground to state that since PACO2 decreases, less alveolar O2 is "eaten up", so PAO2 increases. Do you get what I mean?? I'm feeling like I understand this equation but then not quite.
Hey , in your previous videos you have mentioned that we can use DLCO as a parameter to differentiate between extrinsic and intrinsic so why do we need this"¿
Isn't the PACO2 supposed to be less than the PaCO2? This is the only logical way for CO2 to diffuse from the pulmonary capillary into the alveoli since gases move along the pressure gradient
Don´t forget that the lungs are normally continuously ventilated. So during inspiration, you have definitely a low PACO2, so that CO2-excretion can take place, but it is not all the time like that.
honestly, i don't understand why PaCO2 have to minus 2 or 4. I know you said that alveoli also excrete CO2. But PaCO2 is PaCO2, de we have to minus it in reality? And you just said PACO2 is from "alveoli CO2 excretion" and "PvCO2". Because there is alveoli excretion, so we have to use PaCO2/0.8 to get the right PACO2, which is "alveoli CO2 excretion" and "PvCO2" . But later you said we have to substract some value from PaCO2 and divided by 0.8 to make it 45mmHg. I cannot understand from this point. Because we use 0.8 to make it closer to 50 which is from PvCO2 and alveoli CO2, and now you subsided 2 or 4 from it to make it closer to 45. Why? I mean alveolar CO2 is also part of PACO2, right? Why you ignore it?? I really don't understand and hope for an answer, please. thank you very much.
0.8 is respiratory quotient RQ= CO2 generated/O2 consumed (by carbohydrates and lipid (food) metabolic calculation) by use of 1 O2 , 1.25 CO2 generate. During inspiration O2 diffuse from alveoli but in exchange CO2 enter in alveoli for accurate pAo2 calculation have to extract O2 which diffused. Presence of CO2 @alveoli give idea of howmuch O2 diffused But by 1 O2 use 1.25 CO2 generate That's why (paco2*1.25)= CO2 in alveoli =CO2 generated by use of O2=that much O2 diffused from alveoli , *pAo2 = pio2 - pAco2(diffused O2 during inspiration)*
The amount of research you have done in each topic shows your extreme level of dedication towards Medicine.Thank you for everything you do.
You’re very welcome 😊
Thank you 🙏
its beautiful really
He who has Medicosis Perficionalis can bear almost any topic. You're truly a genius Mashallah!! thank you so much for dedicating your time and effort to share these videos with medical students world wide.
I am honored!
Thank you so much for your kind words!
OH MY GOD I JUST FINISHED THE VIDEO AND I’M IN LOVE WITH THE POETRY AT THE END 🤣🤣🤣🤣 your channel will be my all time favorite
It's so so unique
Thanks man
My pleasure 😇
Thank you so much for watching!
The poetry is absolutely spectacular! In love with all of your videos, thank youuu!
My pleasure!
Wow 😲😲.thank you..may God keep you safe and priliant
Thanks 🙏
You too!
You have an amazing sense of humour brother.
Your research and concepts are second to none.
Preparing for my exit exam.
Thank you 🙏
Good luck 🍀
i was tired from studying all day but your lecture helped me push forward to understand this topic as well! thankss
Glad it helped!
Thank you.
It was a real struggle to understand such equation.... Finally, it makes sense
Great 👍
In a nut shell :
PA CO2 = Pa CO2 . Because CO2 readily equilibrates in both alveoli & arterial blood as it has a higher diffusion coefficient than that of oxygen , so there is NO A-a gradient for CO2 !
The PA CO2 is the driving force for the diffusion of PA O2 to the pulmonary circulation across the (alveolar - capillary interface) ; where the metabolic activities of the alveolar pneumocytes & capillary endothelium determine the respiratory quotient (RQ) for O2 extraction from the PA O2 .
Normal A-a gradient = 5 - 15 mm Hg .
But how can it equilibrate that much ? See pulmonary artery Pv CO2 is 46 mmHg. And pulmonary vein Pa CO2 is 40 mmHg so if its diffusing maximally due to high diffusion coefficient then it should have Pa CO2 as 6 mmHg but it's 40 mmHg this I'm not understanding. I know I'm making some silly mistake in it but I'm unable to find what error I'm making.
@@dr.allwyndsouza5246 Thank you doctor for the question & the comment ! As far as I understand it , the change in partial pressures of gases between arterial & venous bloods is NOT a simple algebraic addition or subtraction of values in blood & alveoli , even in a gas with the highest ever diffusion coefficient ! If you've better explanation , please let me know it !
@@WhyNot-si4pj sure I will let you know, Ive spent a significant time on this concept since 2 days, but I'm still unable to understand it. It doesn't seem like it's happening in the way we've been taught diffusion in physics, more deeper Im going into it, the more confusion Im getting from it.
@@dr.allwyndsouza5246 I agree with you doctor ! Some aspects of human physiology had been taught by the old schools of medicine as if they were ROCKET SCIENCE equations ; or ELECTRIC ENGINEERING graphs of deflections beyond the scope & understanding of medical scholars ! EKGs , acid - base balance , ABG interpretation , ..... etc are few of many examples to mention ! This is of course because of poor basis in natural science disciplines like chemistry , physics , mathematics ..... among the teaching staff ! I wonder myself how comes CO2 the end product of metabolism determines the O2 extraction from the PA O2 RETROGRADELY ?
So 50 mmHg of O2 is consumed by the alveolar pneumocytes & capillary endothelium and 40 mmHg of CO2 is produced?
bro, all of the questions i freaking had, you answered, every one of these steps where theres an underlying assumption, you answered, FINALLYYY, except i need some more physics with how CO2 and O2 affect each other in the alveolus, like i guess theyre enemies makes sense, but raelistically, how does the physics add up. do we just reject any extra O2 bc theres a certain amount already present. idk maan but thanky ou
thanks!! now it's so easy, but didn't seem so at all
My pleasure 😇
Thank you so much for watching!
@@MedicosisPerfectionalis hi doc 🤝i don't know who put that comment from my mobile may be my grandson,or his cousin🤷 would did that,if that action of previous comment hurted you much i apologize for that 🙏,🙏
I think Pco2 in atmosphere would be 0.2852 mm hg not 28.52 mm hg.
Cause 0.04%=0.0004 and ×713 = 0.2852 mm hg
Same with pAco2 5.6%=0.056 × 713 =39.98 mm hg
Well explained, as always.
Thank you 😊
100% perfect lecture!
Interesting way of explaining.
Thank you 😊
at 4 atm pressure & 80% concentration. what is the amount of o2 dissolve ?
a-4ml
b-4.4ml
c-4.8ml
d-5.2ml
Sir earlier during A-a gradient lecture,you said there was no A-a gradient because PACO² and PaCO² are same , but here it's PACO² > PaCO² . How😵
No A-a gradient for which situation?
I think this video mis-represented the relationship between PACO² and PaCO² -- they are in fact the same because the diffusion rate of carbon dioxide through the blood-gas barrier is so high (Which means CO2 has not A-a gradient). However, the reason why you divide PaCO² by the respiratory quotient (0.8 under normal homeostasis) to calculate the A-a gradient for OXYGEN is because the PACO² value often underestimates PaCO² values WHEN you are calculating for the A-a gradient of OXYGEN.
The respiratory quotient is defined as CO2 produced / O2 consumed. When you only metabolize carbohydrates, the RQ = 1 because 6 molecules of O2 is converted to 6 molecules of CO2. When you metabolize fats and proteins, the RQ drops below 0.8. When you divided PaCO² by 0.8, you over-estimate PACO². This overestimation is important because it allows you to account for the fact that oxygen consumption and carbon dioxide production is not always proportional.
"People do not like EQUATION, they would rather listen to JUSTIN BIEBER" 😅😅
This is hilarious, being here a big fan of JB n also a med student!😀
Thanks a trillion MEDICOSIS, medicine bcm so much fun now 🤗
My pleasure 😇
Perfect video!!! And a great explanation. Thank you so much.
My pleasure 😇
17:43 Why did you decrease the PaCO2 from 40 to 36?I couldn't understand that point?
Hello, at 10:24 when you are explaining the math behind PiCO2, the amount of CO2 in the air is in percentage, so i dont know if the math is actually right or maybe i misunderstood, anyways thanks for the help with the lecture, luv ur vids :)
Thank you
You’re very welcome!
6:26 “ okay shut up” got me laugh so hard 🤣😭
Haha 😂
Serious concept flaw that I have to point out (I was quite enjoy reading your series s sincerely)
CO2 is freely diffuse into alveoli and assumbly no A-a gradient of CO2, and this concept get through alot resp physio, if u explain the derive of this equation for having A-a PCO2 difference it is completely wrong, pls consider make big ammendment.
The subtraction of PaCO2 divided by RQ is the estimation of used up of O2 by body, and the equation is derived from equilibrium at steady state of O2 gain (from ventilation) and O2 loss (from body consumption and ventilation):
As at equilibrium:
1. ProducedCO2 = CO2 loss (MV * PACO2)
2. O2 gain ( MV * PinspirO2) = O2 loss (usedupO2 + MV * PAO2)
with RQ of 0.8 for cell O2Consumption and CO2production ratio:
UsedupO2 = 1.25 * ProducedCO2 = 1.25 * MV * PACO2
Thus combining equations we have PinspirO2 = 1.25 * PACO2 + PAO2;
as we have PinspirO2 = FiO2 * ATMgasPa; PACO2 assumed equal PaCO2,
the final equation PAO2 = FiO2(ATMgasPa) - 1.25(PaCO2).
thanks
@@MeOKMeO MV=minute ventilation PinspirO2= PO2 in the inspired air ATMgasPa=atmopheric gas pressure
@@MeOKMeO Yes, I find this derivation of the equation more appealing. I have a question though. Say, for example, if one hyperventilates until PaCO2 drops to 20 mmHg, according to the equation, PAO2 will have to increase. But how can this then be explained? Is it okay to explain that PAO2 increases because O2 consumption decreases (since PaCO2/R decreases)? In this case, I think it would make more sense to look at PaCO2/R as PACO2, which can then be used as a ground to state that since PACO2 decreases, less alveolar O2 is "eaten up", so PAO2 increases. Do you get what I mean?? I'm feeling like I understand this equation but then not quite.
10:32 last equation, it's partial pressure in alveoli
Hey 👋 Gayathri,
Is this a question or a statement?
@@MedicosisPerfectionalis statement
Hey , in your previous videos you have mentioned that we can use DLCO as a parameter to differentiate between extrinsic and intrinsic so why do we need this"¿
Another way to do it!
Just like ESR and CRP.
why fractions change between atmospheric air and alveoli air ,,,, and these values are before or after exchanging gas
really appreciate your effort , keep it up xoxoxo
Thanks 🙏
Can you please help me by sharing?
Isn't the PACO2 supposed to be less than the PaCO2? This is the only logical way for CO2 to diffuse from the pulmonary capillary into the alveoli since gases move along the pressure gradient
Don´t forget that the lungs are normally continuously ventilated. So during inspiration, you have definitely a low PACO2, so that CO2-excretion can take place, but it is not all the time like that.
@@MohammadYounosI must have been tired when I wrote this. I mixed up the PaCO2 with the PvCO2.
Exactly my question... As per we calculated pAco2=~40 and pVco2=45
does negative A-a gradient exists?
Sir can you make something for ECG
That’s a great idea 💡
I have a question; why don't we use FiO2=15% as it is in the alveoli?
cause we are assuming that 15% is normal value of composition of O2 in aleveoli. In restrictive lung disease, the 15% is actually decreasing
Respect .
Thank you so much 😊
I love you. That is all.
honestly, i don't understand why PaCO2 have to minus 2 or 4. I know you said that alveoli also excrete CO2. But PaCO2 is PaCO2, de we have to minus it in reality?
And you just said PACO2 is from "alveoli CO2 excretion" and "PvCO2". Because there is alveoli excretion, so we have to use PaCO2/0.8 to get the right PACO2, which is "alveoli CO2 excretion" and "PvCO2" .
But later you said we have to substract some value from PaCO2 and divided by 0.8 to make it 45mmHg.
I cannot understand from this point. Because we use 0.8 to make it closer to 50 which is from PvCO2 and alveoli CO2, and now you subsided 2 or 4 from it to make it closer to 45. Why? I mean alveolar CO2 is also part of PACO2, right? Why you ignore it??
I really don't understand and hope for an answer, please. thank you very much.
11:04 pAco2 = 40
But with equation pAco2=45
🥵
0.8 is respiratory quotient RQ= CO2 generated/O2 consumed (by carbohydrates and lipid (food) metabolic calculation) by use of 1 O2 , 1.25 CO2 generate. During inspiration O2 diffuse from alveoli but in exchange CO2 enter in alveoli for accurate pAo2 calculation have to extract O2 which diffused.
Presence of CO2 @alveoli give idea of howmuch O2 diffused
But by 1 O2 use 1.25 CO2 generate
That's why (paco2*1.25)= CO2 in alveoli =CO2 generated by use of O2=that much O2 diffused from alveoli , *pAo2 = pio2 - pAco2(diffused O2 during inspiration)*
0.8 is respiratory quotient
RQ=CO2 generated/O2 consumed
That means by use of 1 O2 ,
0.8 CO2 generated not 1.25
@@darshandodia8500 O2 consumed = CO2 generated/RQ
O2 consumed=CO2 generated/0.8
O2 consumed=1.25co2 generated as 1/0.8=1.25
Ur calculation is right
O2 consumed= 1.25 CO2
It means for 1 oxygen consumed body produce 0.8 co2
Why is Korg teaching me physiology?? 🤔🤔
Haha 😂
Wow !!,☹️✨
What happened?
I apologize
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