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Could somebody please explain why in respiratory alkalosis, there is a left shift in the oxygen binding curve??
Whereas, at high altitude,where there occurs resp alkalosis, there is right shift?
In Goljan book it has been explained on the page of 5 (incresed 2,3 BPG production because of increased glycolysis)
Hi! As you mentioned, high altitude means resp alk, so the hb-oxy dissociation curve will shift to the LEFT, meaning increased affinity for oxygen, making oxygen less available for tissues. Alkalosis = low P a CO2 = lower levels of carbon dioxide in arterial blood i.e. blood that contains oxyhemoglobin. The co2 produced in the cellular respiration cycle will diffuse inside the red blood cells, where it will combine with water ad with the help of carbonic anhydrase will turn into carbonic acid, which will spontaneously dissociate into H and HCO3-. The bicarbonate ion will exit the cell throufh a special antiporter (HCO3/Chloride) but the hydrogen ion has no channel to exit through. Instead it will bind to the hemoglobin. This binding will alter the shape of the Hb and thus Hb will release the oxygen to the tissues. So, when there is alkalosis, there is little co2 and in the end little H bound to Hb and Hb will hold on to the oxygen, when there is resp acidosis, there is a lot of co2 and lots of protons altering the shape of Hb, making it easier to release oxygen, hence the left shift in alkalosis and right shift in acidosis.
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