14.8 Exercise Problems

Q1. Define vital capacity. What is its significance?

Answer: Vital capacity is the maximum volume of air that can be exhaled after a maximum inspiration. It is about 3.5 – 4.5 liters in the human body. It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissues and the environment.

Q2. State the volume of air remaining in the lungs after a normal breathing.

Answer: The volume of air remaining in the lungs after a normal expiration is known as functional residual capacity (FRC). It includes expiratory reserve volume (ERV) and residual volume (RV). ERV is the maximum volume of air that can be exhaled after a normal expiration.
It is about \(1000 \mathrm{~mL}\) to \(1500 \mathrm{~mL}\). RV is the volume of air remaining in the lungs after maximum expiration. It is about \(1100 \mathrm{~mL}\) to \(1500 \mathrm{~mL}\).
\(
\begin{aligned}
& \therefore \mathrm{FRC}=\mathrm{ERV}+\mathrm{RV} \\
& \cong 1500+1500 \\
& \cong 3000 \mathrm{~mL}
\end{aligned}
\)
The functional residual capacity of the human lungs is about \(2500-3000 \mathrm{~mL}\).

Q3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?

Answer: Each alveolus is made up of highly-permeable and thin layers of squamous epithelial cells. Similarly, blood capillaries have layers of squamous epithelial cells. Oxygen-rich air enters the body through the nose and reaches the alveoli. The deoxygenated (carbon dioxide-rich) blood from the body is brought to the heart by the veins. The heart pumps it to the lungs for oxygenation. The exchange of \(\mathrm{O}_2\) and \(\mathrm{CO}_2\) takes place between the blood capillaries surrounding the alveoli and the gases present in the alveoli. Thus, the alveoli are the sites for gaseous exchange. The exchange of gases takes place by simple diffusion because of pressure or concentration differences. The barrier between the alveoli and the capillaries is thin and the diffusion of gases takes place from higher partial pressure to lower partial pressure. The venous blood that reaches the alveoli has a lower partial pressure of \(\mathrm{O}_2\) and higher partial pressure of \(\mathrm{CO}_2\) as compared to alveolar air. Hence, oxygen diffuses into the blood. Simultaneously, carbon dioxide diffuses out of the blood and into the alveoli.

Q4. What are the major transport mechanisms for \(\mathrm{CO}_2\)? Explain.

Answer: Plasma and red blood cells transport carbon dioxide. This is because they are readily soluble in water.
(1) Through plasma:
About \(7 \%\) of \(\mathrm{CO} 2\) is carried in a dissolved state through plasma. Carbon dioxide combines with water and forms carbonic acid.
\(
\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{CO}_3
\)
                                  (Carbonic acid)
(2) Through RBCs:
About \(20-25 \%\) of \(\mathrm{CO}_2\) is transported by the red blood cells as carbaminohaemoglobin. Carbon dioxide binds to the amino groups on the polypeptide chains of hemoglobin and forms a compound known as carbaminohaemoglobin.
(3) Through sodium bicarbonate:
About \(70 \%\) of carbon dioxide is transported as sodium bicarbonate. As \(\mathrm{CO} 2\) diffuses into the blood plasma, a large part of it combines with water to form carbonic acid in the presence of the enzyme carbonic anhydrase. Carbonic anhydrase is a zinc enzyme that speeds up the formation of carbonic acid. This carbonic acid dissociates into bicarbonate \(\left(\mathrm{HCO}_3{ }^{-}\right)\)and hydrogen ions \(\left(\mathrm{H}^{+}\right)\).
\(
\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \stackrel{\text { Carbonic anhydrase }}{\longrightarrow} \mathrm{H}_2 \mathrm{CO}_3
\)
\(
\mathrm{H}_2 \mathrm{CO}_3 \underset{\text { anhydrase }}{\stackrel{\text { Carbonic }}{\longrightarrow}} \mathrm{HCO}_3^{-}+\mathrm{H}^{+}
\)

Q5. What will be the \(\mathrm{pO}_2\) and \(\mathrm{pCO}_2\) in the atmospheric air compared to those in the alveolar air?
(i) \(\mathrm{pO}_2\) lesser, \(\mathrm{pCO}_2\) higher
(ii) \(\mathrm{pO}_2\) higher, \(\mathrm{pCO}_2\) lesser
(iii) \(\mathrm{pO}_2\) higher, \(\mathrm{pCO}_2\) higher
(iv) \(\mathrm{pO}_2\) lesser, \(\mathrm{pCO}_2\) lesser

Answer:

\(
\text { (ii) } \mathrm{pO}_2 \text { higher, } \mathrm{pCO}_2 \text { lesser }
\)
The partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. In atmospheric air, \(\mathrm{pO}_2\) is about \(159 \mathrm{~mm}\) Hg. In alveolar air, it is about \(104 \mathrm{~mm} \mathrm{Hg}\). The partial pressure of carbon dioxide in atmospheric air is lesser than that of carbon dioxide in alveolar air. In atmospheric air, \(\mathrm{pCO}_2\) is about \(0.3 \mathrm{mmHg}\). In alveolar air, it is about \(40 \mathrm{~mm} \mathrm{Hg}\).

Q6. Explain the process of inspiration under normal conditions.

Answer: 

Inspiration or inhalation is the process of bringing air from outside the body into the lungs. It is carried out by creating a pressure gradient between the lungs and the atmosphere. When air enters the lungs, the diaphragm expands toward the abdominal cavity, thereby increasing the space in the thoracic cavity for accommodating the inhaled air.
The volume of the thoracic chamber in the anteroposterior axis increases with the simultaneous contraction of the external intercostal muscles. This causes the ribs and the sternum to move out, thereby increasing the volume of the thoracic chamber in the dorsoventral axis. The overall increase in the thoracic volume leads to a similar increase in the pulmonary volume. Now, as a result of this increase, the intra-pulmonary pressure becomes lesser than the atmospheric pressure. This causes the air from outside the body to move into the lungs.

Q7. How is respiration regulated?

Answer: The respiratory rhythm center present in the medulla region of the brain is primarily responsible for the regulation of respiration. The pneumotaxic center can alter the function performed by the respiratory rhythm center by signaling to reduce the inspiration rate.
The chemosensitive region present near the respiratory center is sensitive to carbon dioxide and hydrogen ions. This region then signals to change the rate of expiration for eliminating the compounds.
The receptors present in the carotid artery and aorta detect the levels of carbon dioxide and hydrogen ions in the blood. As the level of carbon dioxide increases, the respiratory center sends nerve impulses for the necessary changes.

Q8. What is the effect of \(\mathrm{pCO}_2\) on oxygen transport?

Answer: \(\mathrm{pCO}_2\) plays an important role in the transportation of oxygen. At the alveolus, the low \(\mathrm{pCO}_2\) and high \(\mathrm{pO}_2\) favors the formation of hemoglobin. At the tissues, the high \(\mathrm{pCO}_2\) and low \(\mathrm{pO}_2\) favor the dissociation of oxygen from oxyhemoglobin. Hence, the affinity of hemoglobin for oxygen is enhanced by the decrease of \(\mathrm{pCO}_2\) in blood. Therefore, oxygen is transported in the blood as oxyhemoglobin and oxygen dissociates from it at the tissues.

Q9. What happens to the respiratory process in a man going up a hill?

Answer: As the altitude increases, the oxygen level in the atmosphere decreases. Therefore, as a man goes uphill, he gets less oxygen with each breath. This causes the amount of oxygen in the blood to decline. The respiratory rate increases in response to the decrease in the oxygen content of the blood. Simultaneously, the rate of heartbeat increases to increase the supply of oxygen to the blood.

Q10. What is the site of gaseous exchange in an insect?

Answer: In insects, gaseous exchange occurs through a network of tubes collectively known as the tracheal system. The small openings on the sides of an insect’s body are known as spiracles. Oxygen-rich air enters through the spiracles. The spiracles are connected to the network of tubes. From the spiracles, oxygen enters the tracheae. From here, oxygen diffuses into the cells of the body. The movement of carbon dioxide follows the reverse path. The \(\mathrm{CO}_2\) from the cells of the body first enters the tracheae and then leaves the body through the spiracles.

Q11. Define the oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Answer: The oxygen dissociation curve is a graph showing the percentage saturation of oxyhemoglobin at various partial pressures of oxygen. The curve shows the equilibrium of oxyhemoglobin and hemoglobin at various partial pressures. In the lungs, the partial pressure of oxygen is high. Hence, hemoglobin binds to oxygen and forms oxyhemoglobin. Tissues have a low oxygen concentration. Therefore, at the tissues, oxyhemoglobin releases oxygen to form hemoglobin. The sigmoid shape of the dissociation curve is because of the binding of oxygen to hemoglobin. As the first oxygen molecule binds to hemoglobin, it increases the affinity for the second molecule of oxygen to bind. Subsequently, hemoglobin attracts more oxygen.

Q12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Answer: Hypoxia is a condition characterized by an inadequate or decreased supply of oxygen to the lungs. It is caused by several extrinsic factors such as the reduction in \(\mathrm{pO}_2\), inadequate oxygen, etc. 

Q13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity.

Answer: (a) IRV vs ERV

Inspiratory reserve volume (IRV)	Expiratory reserve volume (ERV)
It is the maximum volume of air that can be inhaled after a normal inspiration.	It is the maximum volume of air that can be exhaled after a normal expiration.
It is about 2500 – 3500 mL in the human lungs.	It is about 1000 – 1100 mL in the human lungs.

(b) Inspiratory capacity (IC) Vs Expiratory capacity (EC)

Inspiratory capacity (IC)	Expiratory capacity (EC)
It is the volume of air that can be inhaled after a normal expiration	It is the volume of air that can be exhaled after a normal inspiration.
It includes tidal volume and inspiratory reserve volume. IC = TV + IRV	It includes tidal volume and expiratory reserve volume. EC = TV + ERV

(c) Vital capacity (VC) Vs Total lung capacity (TLC) 

Vital capacity (VC)	Total lung capacity (TLC)
It is the maximum volume of air that can be exhaled after a maximum inspiration. It includes IC and ERV.	It is the volume of air in the lungs after a maximum inspiration. It includes IC, ERV, and residual volume.
It is about 4000 mL in the human lungs.	It is about 5000 – 6000 mL in the human lungs.

Q14. What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.

Answer: Tidal volume is the volume of air inspired or expired during normal respiration. It is about 6000 to \(8000 \mathrm{~mL}\) of air per minute. The hourly tidal volume for a healthy human can be calculated as: Tidal volume \(=6000\) to \(8000 \mathrm{~mL} / \mathrm{minute}\) Tidal volume in an hour \(=6000\) to \(8000 \mathrm{~mL} \times(60 \mathrm{~min})\) \(=3.6 \times 105 \mathrm{~mL}\) to \(4.8 \times 10^5 \mathrm{~mL}\)
Therefore, the hourly tidal volume for a healthy human is approximately \(3.6 \times 10^5 \mathrm{~mL}\) to \(4.8 \times 10^5 \mathrm{~mL}\).

Exemplar Section

VERY SHORT ANSWER TYPE QUESTIONS

Q1. Define the following terms?
a. Tidal volume
b. Residual volume
c. Asthma

Answer: a. Tidal volume: Volume of air inspired or expired during a normal respiration. It is approx. 500 mL, i.e., a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.
b. Residual volume: Volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL. Residual air mainly occurs in alveoli.
c. Asthma: Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. In asthma, due to flattening of tracheal vessels, alveoli are deprived of oxygen. Asthma is characterised by spasm in bronchial muscle.

Q2. A fluid filled double membranous layer surrounds the lungs. Name it and mention its important function.

Answer: Pleural fluid is found in between the two membranes of lung and it reduces the friction on the lung surface.

Q3. Name the primary site of exchange of gases in our body?

Answer: Alveoli

Q4. Cigarette smoking causes emphysema. Give reason.

Answer: Cigarette smoking causes damage of the alveolar walls leading to decreased respiratory surfaces for exchange of gases.

Q5. What is the amount of \(\mathrm{O}_2\) supplied to tissues through every 100 ml . of oxygenated blood under normal physiological conditions?

Answer: 5 mL of oxygen/100 mL of oxygenated blood.

Q6. A major percentage ( \(97 \%\) ) of \(\mathrm{O}_2\) is transported by RBCs in the blood. How does the remaining percentage ( \(3 \%\) ) of \(\mathrm{O}_2\) transported?

Answer: Through Plasma

Q7. Arrange the following terms based on their volumes in an ascending order
a. Tidal Volume (TV)
b. Residual Volume (RV)
c. Inspiratory Reserve Volume (IRV)
d. Expiratory Capacity (EC)

Answer: a. Tidal Volume (TV): 500 mL
b. Residual Volume (RV): 1100 mL-200 mL
c. Inspiratory Reserve Volume (IRV): 2500 mL-3000 mL
d. Expiratory Capacity (EC): 1500 mL-1600 mL

Hence, the arrangement of these terms according to their volume in ascending order would be :
Tidal Volume (TV) <Residual Volume (RV) <Expiratory Capacity (EC) <Inspiratory Reserve Volume (IRV).

Q8. Complete the missing terms
a. Inspiratory Capacity (IC) = _____ +IRV
b. ___________________ = TV + ERV
c. Functional Residual Capacity (FRC) = ERV + ____ 

Answer: a. Inspiratory Capacity (IC) = TV + IRV
b. EC = TV + ERV
c. Functional Residual Capacity (FRC) = ERV + RV

Q9. Name the organs of respiration in the following organisms:
a. Flatworm – _________________________________________
b. Birds – _____________________________________________
c. Frog- ______________________________________________
d. Cockroach – ________________________________________

Answer: a. Flatworm—Entire body surface
b. Birds—Lung
c. Frog—Lung and moist skin
d. Cockroach—Tracheal tubes

Q10. Name the important parts involved in creating a pressure gradient between lungs and the atmosphere during normal respiration.

Answer: The diaphragm and a specialised set of external and intercostals muscles between the ribs, help in the generation of pressure gradient during normal respiration.

SHORT ANSWER TYPE QUESTIONS

Q1. State the different modes of \(\mathrm{CO}_2\) transport in blood.

Answer: Nearly \(20-25 \%\) of \(\mathrm{CO}_2\) by RBCs.
Nearly \(70 \%\) of \(\mathrm{CO}_2\) as bicarbonates Nearly \(7 \%\) of \(\mathrm{CO}_2\) as dissolved state in plasma.

Q2. Compared to \(\mathrm{O}_2\), diffusion rate of \(\mathrm{CO}_2\) through the diffusion membrane per unit difference in partial pressure is much higher. Explain.

Answer: Solubility is an important factor deciding diffusion rate. As the solubility of \(\mathrm{CO}_2\) is \(20-25\) times higher than \(\mathrm{O}_2\), diffusion of \(\mathrm{CO}_2\) through the diffusion membrane per unit difference in partial pressure is much higher.

Q3. For completion of respiration process, write the given steps in sequential manner
a. Diffusion of gases \(\left(\mathrm{O}_2\right.\) and \(\left.\mathrm{CO}_2\right)\) across alveolar membrane.
b. Transport of gases by blood.
c. Utilisation of \(\mathrm{O}_2\) by the cells for catabolic reactions and resultant release of \(\mathrm{CO}_2\).
d. Pulmonary ventilation by which atmospheric air is drawn in and \(\mathrm{CO}_2\) rich alveolar air is released out.
e. Diffusion of \(\mathrm{O}_2\) and \(\mathrm{CO}_2\) between blood and tissues.

Answer: Respiration involves the following steps:
1. Breathing or pulmonary ventilation by which atmospheric air is drawn in and \(\mathrm{CO}_2\) rich alveolar air is released out.
2. Diffusion of gases \(\left(\mathrm{O}_2\right.\) and \(\left.\mathrm{CO}_2\right)\) across alveolar membrane.
3. Transport of gases by the blood.
4. Diffusion of \(\mathrm{O}_2\) and \(\mathrm{CO}_2\) between blood and tissues.
5. Utilisation of \(\mathrm{O}_2\) by the cells for catabolic reactions and resultant release of \(\mathrm{CO}_2\).

Q4. Differentiate between
a. Inspiratory and expiratory reserve volume
b. Vital capacity and total lung capacity
c. Emphysema and occupational respiratory disorder

Answer: (a)
\(
\begin{array}{|l|l|}
\hline \text { Inspiratory Reserve volume } & \text { Expiratory Reserve Volume } \\
\hline \begin{array}{l}
\text { Additional volume of air a person can inspire by a } \\
\text { forcible inspiration is called inspiratory reserve } \\
\text { volume (IRV). }
\end{array} & \begin{array}{l}
\text { Additional volume of air a person can expire by a } \\
\text { forced expiration is called expiratory reserve } \\
\text { volume (ERV). }
\end{array} \\
\hline \text { In a normal individual, IRV is about } 2500 \text { to } 3000 \mathrm{~mL} & \text { In a normal human, ERV is about } 1000 \text { to } 1100 \mathrm{~mL} . \\
\hline
\end{array}
\)
(b)
\(
\begin{array}{|l|l|}
\hline \text { Vital capacity } & \text { Total Lung Capacity } \\
\hline \begin{array}{l}
\text { The maximum volume of the air one can breathe } \\
\text { after a forceful expiration, is called Vital Capacity. }
\end{array} & \begin{array}{l}
\text { The total volume of air in lungs at the end of a } \\
\text { forced inspiration is called Total Lung Capacity. }
\end{array} \\
\hline
\end{array}
\)
(c)
\(
\begin{array}{|l|l|}
\hline \text { Emphysema } & \text { Occupational Respiratory disorder } \\
\hline \begin{array}{l}
\text { Alveoli walls are damaged in } \\
\text { emphysema }
\end{array} & \begin{array}{l}
\text { any other part of the respiratory system is damaged in occupational } \\
\text { respiratory disorder }
\end{array} \\
\hline \text { Caused due to smoking } & \text { Caused due to fine particles generated } \\
\hline
\end{array}
\)

LONG ANSWER TYPE QUESTIONS

Q1. Explain the transport of \(\mathrm{O}_2\) and \(\mathrm{CO}_2\) between alveoli and tissue with diagram.

Answer:

• Transport of gases: Blood is the medium of transport for \(\mathrm{O}_2\) and \(\mathrm{CO}_2\). About \(97 \%\) of \(\mathrm{O}_2\) is transported by RBCs in the blood. The remaining \(3 \%\) of \(\mathrm{O}_2\) is carried in a dissolved state through the plasma. Nearly \(20-25 \%\) of \(\mathrm{CO}_2\) is transported by RBCs whereas \(70 \%\) of it is carried as bicarbonate. About \(7 \%\) of \(\mathrm{CO}_2\) is carried in a dissolved state through plasma.
• Transport of oxygen: Haemoglobin is a red coloured iron containing pigment present in the RBCs. \(\mathrm{O}_2\) can bind with haemoglobin in a reversible manner to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of \(\mathrm{O}_2\). Binding of oxygen with haemoglobin is primarily related to partial pressure of \(\mathrm{O}_2\). Partial pressure of \(\mathrm{CO}_2\), hydrogen ion concentration and temperature are the other factors which can interfere with this binding. A sigmoid curve is obtained when percentage saturation of haemoglobin with \(\mathrm{O}_2\) is plotted against the \(\mathrm{pO}_2\). This curve is called the Oxygen dissociation curve and is highly useful in studying the effect of factors like \(\mathrm{pCO}_2, \mathrm{H}^{+}\) concentration, etc., on binding of \(\mathrm{O}_2\) with haemoglobin. In the alveoli, where there is high \(\mathrm{pO}_2\), low \(\mathrm{pCO}_2\), lesser \(\mathrm{H}^{+}\)concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low \(\mathrm{pO}_2\), high \(\mathrm{pCO}_2\), high \(\mathrm{H}^{+}\)concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. This clearly indicates that \(\mathrm{O}_2\) gets bound to haemoglobin in the lung surface and gets dissociated at the tissues. Every 100 mL of oxygenated blood can deliver around 5 mL of \(\mathrm{O}_2\) to the tissues under normal physiological conditions.
• Transport of carbon dioxide: \(\mathrm{CO}_2\) is carried by haemoglobin as carbamino-haemoglobin (about \(20-25 \%\) ). This binding is related to the partial pressure of \(\mathrm{CO}_2 \cdot \mathrm{pO}_2\) is a major factor which could affect this binding. When \(\mathrm{pCO}_2\) is high and \(\mathrm{pO}_2\) is low as in the tissues, more binding of carbon dioxide occurs whereas, when the \(\mathrm{pCO}_2\) is low and \(\mathrm{pO}_2\) is high as in the alveoli, dissociation of \(\mathrm{CO}_2\) from carbamino-haemoglobin takes place, i.e., \(\mathrm{CO}_2\) which is bound to haemoglobin from the tissues is delivered at the alveoli. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and minute quantities of the same is present in the plasma too. This enzyme facilitates the following reaction in both directions
  \(
  \mathrm{CO}_2 \stackrel{\substack{\text { Carbonic } \\ \text { anhydrase }}}{\rightleftarrows} \mathrm{H}_2 \mathrm{CO}_3 \stackrel{\substack{\text { Carbonic } \\ \text { anhydrase }}}{\rightleftarrows} \mathrm{HCO}_3^{-}+\mathrm{H}^{+}
  \)

At the tissue site where partial pressure of \(\mathrm{CO}_2\) is high due to catabolism, \(\mathrm{CO}_2\) diffuses into blood ( RBCs and plasma) and forms \(\mathrm{HCO}_2\) and \(\mathrm{H}^{+}\). At the alveolar site where \(\mathrm{pCO}_2\) is low, the reaction proceeds in the opposite direction leading to the formation of \(\mathrm{CO}_2\) and \(\mathrm{H}_2 \mathrm{O}\). Thus, \(\mathrm{CO}_2\) trapped as bicarbonate at the tissue level and transported to the alveoli is released out as \(\mathrm{CO}_2\). Every 100 mL of deoxygenated blood delivers approximately 4 mL of \(\mathrm{CO}_2\) to the alveoli.

Q2. Explain the mechanism of breathing with neat labelled sketches.

Answer: Breathing involves two stages:
a. Inspiration: Inspiration is initiated by the contraction of diaphragm, which increases the volume of thoracic chamber in the anteroposterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of thoracic chamber in the dorso-ventral axis also. Such an increase in thoracic volume leads to a similar increase in pulmonary volume resulting in decreased intra- pulmonary pressure to less than atmospheric pressure. This causes the movement of external air into the lungs, i.e., inspiration.

b. Expiration: The inter-costal muscles return the diaphragm and sternum to their normal positions with relaxation of the diaphragm. This reduces the thoracic volume and thereby the pulmonary volume. As a result an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causes the expulsion of air from the lungs i.e., expiration.

Q3. Explain the role of neural system in regulation of respiration.

Answer: Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues. This is done by the neural system. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre. Neural signal from this centre “can reduce the duration of inspiration and thereby alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to \(\mathrm{CO2}\) and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated. Receptors associated with aortic arch and carotid artery also can recognise changes in \(\mathrm{CO2}\) and \(\mathrm{H}+\) concentration and send necessary’ signals to the rhythm centre for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.