
\[C{O_2}\] dissociates from carbamino haemoglobin when
a. $PC{O_2}$ is high and $P{O_2}$ is low
b. $P{O_2}$ is high and $PC{O_2}$ is low
c. $PC{O_2}$ and $P{O_2}$ are equal
d. None of the above
Answer
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Hint: Carbamino haemoglobin is known as a compound of haemoglobin and carbon dioxide. Single haemoglobin molecule can transport four carbon dioxide molecules back to the lungs, where they are released when the molecule changes back to the oxyhaemoglobin form.
Complete answer:
Carbon dioxide molecules are transported in the blood from body tissues to the lungs as dissolved carbon dioxide, carbonic acid, carbamino haemoglobin, and bicarbonate ions for elimination by pulmonary gas exchange. Among all carbon dioxide, about 5 to 7 percent is dissolved in the plasma. Secondly, carbon dioxide can bind to plasma proteins or can enter RBCs and bind to haemoglobin to form carbaminohemoglobin. This mode transports about 10 percent of the carbon dioxide.
Carbaminohemoglobin unloads the carbon dioxide in the lung, where the $PC{O_2}$ is lower. The process of loading and unloading of carbon dioxide is facilitated by the Haldane effect; the binding of oxygen with haemoglobin displaces carbon dioxide and hydrogen ions from the haemoglobin. The concept of the Haldane effect, works like that of the Bohr Effect in oxygen transport. Here, when haemoglobin is oxygenated in the lung to release hydrogen ions, carbonic acid and ultimately carbon dioxide are produced, with the effect being a reduced affinity to carbon dioxide in the lung resulting from oxygenation. The opposite occurs in the tissue, where haemoglobin releases oxygen and takes up or buffers hydrogen, leading to increased affinity for carbon dioxide. Thus, high $P{O_2}$ and low $PC{O_2}$ in the blood capillaries of lung favours separation of \[C{O_2}\] from carbamino haemoglobin.
Hence, the correct answer is option (B).
Note: When $P{O_2}$ shows an increase, as it does when the blood enters the alveoli and exchanges O$_2$ with alveolar air, carbaminohemoglobin dissociates to \[C{O_2}\] and $Hb - N{H_2}$. The reduction in \[C{O_2}\] content of the haemoglobin by increasing $P{O_2}$ is called the Haldane effect.
Complete answer:
Carbon dioxide molecules are transported in the blood from body tissues to the lungs as dissolved carbon dioxide, carbonic acid, carbamino haemoglobin, and bicarbonate ions for elimination by pulmonary gas exchange. Among all carbon dioxide, about 5 to 7 percent is dissolved in the plasma. Secondly, carbon dioxide can bind to plasma proteins or can enter RBCs and bind to haemoglobin to form carbaminohemoglobin. This mode transports about 10 percent of the carbon dioxide.
Carbaminohemoglobin unloads the carbon dioxide in the lung, where the $PC{O_2}$ is lower. The process of loading and unloading of carbon dioxide is facilitated by the Haldane effect; the binding of oxygen with haemoglobin displaces carbon dioxide and hydrogen ions from the haemoglobin. The concept of the Haldane effect, works like that of the Bohr Effect in oxygen transport. Here, when haemoglobin is oxygenated in the lung to release hydrogen ions, carbonic acid and ultimately carbon dioxide are produced, with the effect being a reduced affinity to carbon dioxide in the lung resulting from oxygenation. The opposite occurs in the tissue, where haemoglobin releases oxygen and takes up or buffers hydrogen, leading to increased affinity for carbon dioxide. Thus, high $P{O_2}$ and low $PC{O_2}$ in the blood capillaries of lung favours separation of \[C{O_2}\] from carbamino haemoglobin.
Hence, the correct answer is option (B).
Note: When $P{O_2}$ shows an increase, as it does when the blood enters the alveoli and exchanges O$_2$ with alveolar air, carbaminohemoglobin dissociates to \[C{O_2}\] and $Hb - N{H_2}$. The reduction in \[C{O_2}\] content of the haemoglobin by increasing $P{O_2}$ is called the Haldane effect.
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