Answer

(a) Insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body. These openings lead to trachae. The cells exchange $O_{2} / CO_{2}$ directly with the air in the spiracles present on insects body.

• (b) The tissues do not exchange $O_{2} / CO_{2}$ directly with coelomic fluid in insects. Insects have a tracheal system that directly delivers air to the tissues, bypassing the need for a circulatory fluid like coelomic fluid for gas exchange.

• (c) The tissues do not exchange $O_{2} / CO_{2}$ directly with the air outside through the body surface. Insects have a specialized tracheal system that allows air to enter through spiracles and travel through tracheal tubes directly to the tissues.

• (d) The tracheal tubes do not exchange $O_{2} / CO_{2}$ directly with the haemocoel. Instead, the tracheal system delivers air directly to the tissues, and gas exchange occurs at the cellular level without involving the haemocoel.

Answer

(c) Diffusion of gases is a physical phenomenon that takes place between the tissue and blood vessels, and does not occur during breathing whereas bringing air to the body temperature, its cleaning and warming occurs during the process of breathing.

• Bringing air to body temperature: This is a correct statement because during breathing, the air is brought to body temperature as it passes through the respiratory tract.
• Warms up the air: This is a correct statement because the air is warmed up as it travels through the nasal passages and respiratory tract.
• Cleans up the air: This is a correct statement because the respiratory system has mechanisms like mucus and cilia that help in cleaning the air by trapping dust and other particles.

Answer

(d) The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere. The pressure within the lungs is less than the atmospheric pressure so there is a negative pressure in the lungs with respect to atmospheric pressure.

The puncture in the chest affects this pressure gradient maintained by the lungs and thus may cause cessation of breathing.

• (a) reduced breathing rate: A puncture in the chest cavity would likely disrupt the pressure gradient necessary for normal breathing, leading to more severe issues than just a reduced breathing rate. The body would struggle to maintain adequate ventilation, making this option less likely.

• (b) rapid increase in breathing rate: While a rapid increase in breathing rate can occur in response to various conditions, a puncture in the chest cavity would more likely cause severe disruption to the mechanics of breathing, potentially leading to cessation rather than just an increased rate.

• (c) no change in respiration: A puncture in the chest cavity would almost certainly affect the pressure gradient required for normal lung function, making it highly unlikely that there would be no change in respiration.

Thinking Process

The reaction between haemoglobin and $\mathrm{CO}_{2}$ is reversible, whereas it is irreversible in case of $\mathrm{CO}$.

Answer

(b) Haemoglobin consist of a protein globin and pigment here. The four portion of iron in name combine with molecule of oxygen. It is an easy reversible reaction to form oxyhaemoglobin

$$ \mathrm{Hb}+O_{2} \rightleftharpoons HbO_{2} $$

Whereas, the complex formed by the reaction of carbon monooxide and haemoglobin is incredibly strong

$\underset{\text { (Haemoglobin) }}{\mathrm{Hb}+\mathrm{CO}} \rightarrow \underset{(\substack{\text { Carboxy } \\ \text { haemoglobin }})}{\mathrm{HbCO}}$

As a result of this strong between the haemoglobin and carbon monooxide the haemoglobin looses its affinity to oxygen thus may lead to choking or even death.

• (a) It reduces $CO_{2}$ transport: This option is incorrect because carbon monoxide primarily affects the transport of oxygen, not carbon dioxide. The binding of carbon monoxide to haemoglobin forms carboxyhaemoglobin, which prevents haemoglobin from carrying oxygen effectively, but it does not significantly impact the transport of carbon dioxide.

• (c) It increases $CO_{2}$ transport: This option is incorrect because carbon monoxide does not increase the transport of carbon dioxide. Instead, it binds to haemoglobin, reducing its ability to carry oxygen. The transport of carbon dioxide is not directly affected by carbon monoxide exposure.

• (d) It increases $O_{2}$ transport: This option is incorrect because carbon monoxide exposure actually decreases the transport of oxygen. When carbon monoxide binds to haemoglobin, it forms a stable complex called carboxyhaemoglobin, which reduces the haemoglobin’s ability to bind and transport oxygen, leading to reduced oxygen delivery to tissues.

Answer

(b) Inspiration is a active process whereas expiration is a passive process. Inspiration occur when the muscles of diaphragm contrac to increase the overall volume of thoracic cavity.

Thus the pressure within the lungs (intra-pulmonary pressure) is less in comparison to the atmospheric pressure, i.e., there is a negative pressure in the lungs with respect to atmospheric pressure. As the muscles use energy for contraction inspiration is called active process.

Whereas, during the expiration diaphragm muscles relax without the use of energy intra-pulmonary pressure becomes higher than the atmospheric pressure and air noshes out. Thus, it is a passive process.

B. Expiration (chest cavity reduced)

• (a) inspiration is a passive process whereas expiration is active

  • This statement is incorrect because inspiration is an active process that requires the contraction of the diaphragm and intercostal muscles, which increases the volume of the thoracic cavity and decreases the intra-pulmonary pressure. Expiration, on the other hand, is typically a passive process that occurs when these muscles relax, allowing the thoracic cavity to decrease in volume and increase intra-pulmonary pressure, pushing air out of the lungs.

• (c) inspiration and expiration are active processes

  • This statement is incorrect because while inspiration is an active process requiring muscle contraction and energy expenditure, expiration is generally a passive process that does not require energy. Expiration occurs due to the natural recoil of the lungs and relaxation of the diaphragm and intercostal muscles.

• (d) inspiration and expiration are passive processes

  • This statement is incorrect because inspiration is an active process that involves the contraction of the diaphragm and intercostal muscles, which requires energy. Expiration, however, is usually a passive process that occurs when these muscles relax and the elastic recoil of the lungs pushes air out without the need for energy.

Answer

(c) The maximum volume of air that a person can breathe in after forced expiration or the maximum volume of air that a person can breathe out after forced inspiration is called vital capacity

$ \underset{\text { (Inspiratory reserve volume) }}{\mathrm{V} \mathrm{C}=\mathrm{I} \mathrm{R} \mathrm{V}} \underset{\text { (Expiratory reserve volume) }}{ + \mathrm{E} \mathrm{R} \mathrm{V}} + \underset{\text { (Tidal volume) }}{\mathrm{T} \mathrm{V}}$

The value of vital capacity varies from $3400 \mathrm{~mL}$ to $4800 \mathrm{~mL}$.

On the other hand, tidal volume is the air inspired or expired during normal breathing. Total lung capacity is the volume of air present in lungs and respiratory passage after maximum inspiration. Whereas, inspiratory capacity is total volume of air that a person can inspire after normal inspiration.

• (a) Total lung capacity: This is the volume of air present in the lungs and respiratory passages after maximum inspiration, not the volume of air taken in after forced expiration.

• (b) Tidal volume: This is the volume of air inspired or expired during normal breathing, not the volume of air taken in after forced expiration.

• (d) Inspiratory capacity: This is the total volume of air that a person can inspire after normal inspiration, not the volume of air taken in after forced expiration.

Answer

(d) There are various factors which affect the binding of $\mathrm{O}_{2}$ with $\mathrm{Hb}$.

These factors are

(i) Low temperature

(ii) Low $\mathrm{H}^{+}$concentration (low pH).

(iii) Low diphosphoglyceraldehy

So, higher $\mathrm{pO}_{2}$ is the in correct statement.

• Higher pH: Higher pH (lower $\mathrm{H}^{+}$ concentration) increases the affinity of hemoglobin for oxygen, facilitating $\mathrm{O}_{2}$ binding.

• Lower temperature: Lower temperature enhances the binding of oxygen to hemoglobin, as it stabilizes the oxygen-hemoglobin complex.

• Lower $p CO_{2}$: Lower partial pressure of carbon dioxide ($p CO_{2}$) reduces the concentration of $\mathrm{H}^{+}$ ions, leading to an increased affinity of hemoglobin for oxygen.

Answer

(d) The diaphragm and a specialised set of muscles, called external muscles present between the ribs are involved in the normal breathing in humans. They are involved in generating pressure gradient of air between the lungs and the atmosphere, so as to faeilitate the intake of air.

• (a) External and internal intercostal muscles: This option is incorrect because while the external intercostal muscles are involved in normal breathing, the internal intercostal muscles are primarily involved in forced expiration, not in normal breathing.

• (b) Diaphragm and abdominal muscles: This option is incorrect because the abdominal muscles are not typically involved in normal breathing. They are more involved in forced expiration and other activities like coughing or heavy breathing.

• (c) Diaphragm and external intercostal muscles: This option is partially correct but not fully accurate. While the diaphragm and external intercostal muscles are indeed involved in normal breathing, the answer provided in the question specifies “a specialised set of muscles, called external muscles present between the ribs,” which is a more precise description.

Answer

(b) Emphysema is a chronic disorder in which alveolar walls are damaged due to the infacation or obsomal distersion. It is a respiratory disorder caused by ciggrette smoking and inhalation of other smoke or toixic substences over a period of time.

• (a) The bronchioles are not primarily damaged in emphysema; the main issue lies with the alveolar walls. While bronchioles can be affected in other respiratory conditions like chronic bronchitis, emphysema specifically targets the alveoli.

• (c) The plasma membrane is not the primary structure affected in emphysema. The disorder specifically involves the destruction of alveolar walls, not the cellular plasma membranes.

• (d) The respiratory muscles are not directly damaged in emphysema. While the disease can lead to increased effort in breathing and potential muscle fatigue, the primary damage occurs in the alveolar walls.

Answer

(b) Pneumotaxic Centre Located in the dorsal part of pons varoli of the brain can reduce the duration of inspiration and thus alter the respiratory rate.

Apneustic Centre Whereas is located in the lower part of pons varoli is responsible for promoting inspiration process.

Chemosensitive Centre is situated adjacent to the rhythm centre which is highly sensitive to $CO_{2}$ and hydrogen ions. Increase in $CO_{2}$ and $\mathrm{H}^{+}$in body and activates this centre for the elimiration of $\mathrm{CO}_{2}$ and $\mathrm{H}$

Medullary Inspiratory Centre is a specialised region present in medulla of the brain. and is primarily responsible for regulating the respiratory rhythm.

• Medullary Inspiratory Centre: This centre is primarily responsible for regulating the respiratory rhythm, not for reducing the inspiratory duration.

• Apneustic Centre: This centre is responsible for promoting the inspiration process, not for reducing the inspiratory duration.

• Chemosensitive Centre: This centre is highly sensitive to $CO_{2}$ and hydrogen ions and activates to eliminate $CO_{2}$ and $\mathrm{H}^{+}$, but it does not reduce the inspiratory duration.

Thinking Process

$CO_{2}$ is carried by haemoglobin as carbamino haemoglobin. This binding is related to the partial pressure of $CO_{2}$.

Answer

(b) When, the $p CO_{2}$ is low and $pO_{2}$ is high as in the lung alveoli, dissociation of $CO_{2}$ from carbamino-haemoglobin takes place, $CO_{2}$ which is bound to haemoglobin from the tissue is delivered at the alveoli, to maintain the concentration of $CO_{2}$ thus increasing $pCO_{2}$.

Exchange of gases takes place between tissue capillary and tissue cells. Capillary cells with high $pO_{2}$ causes diffusion of $O_{2}$ into tissue cells via tissue fluid on the other hand high $p CO_{2}$ in the tissue cells causes diffusion of $CO_{2}$ into tissue capillary via tissue fluid.

• (a) $p CO_{2}$ is high and $p O_{2}$ is low: This condition is typically found in the tissues where CO₂ is being produced as a waste product of cellular respiration. Under these conditions, CO₂ binds to hemoglobin to form carbamino-hemoglobin rather than dissociating from it.

• (c) $p CO_{2}$ and $p O_{2}$ are equal: This scenario is not physiologically relevant because the partial pressures of CO₂ and O₂ are never equal in the body. The exchange of gases relies on a gradient, and equal partial pressures would not facilitate the necessary diffusion processes.

• (d) None of the above: This option is incorrect because there is a specific condition (high $p O_{2}$ and low $p CO_{2}$) under which CO₂ dissociates from carbamino-hemoglobin, as described in option (b).

Answer

(d) Spirometer is the device used to measure the volume of air involved in breathing movements and it also helps in clinical assessment of pulmonary functions.

Stethoscope is a medical device used for listening the internal sounds of an animal or human body.

Hygrometer is a device used for measuring the moisture content in the atmosphere, i.e., humidity.

Sphygmomanometer is a device that is used to measure blood pressure.

• Stethoscope: It is a medical device used for listening to the internal sounds of an animal or human body, not for measuring air volume in breathing movements.

• Hygrometer: It is a device used for measuring the moisture content in the atmosphere, i.e., humidity, not for estimating air volume in breathing movements.

• Sphygmomanometer: It is a device that is used to measure blood pressure, not for estimating air volume in breathing movements.

Answer

(b) (i) Inspiratory Capacity (IC) =Tidal Volume + Inspiratory Reserve Volume (TV + IRV).

(ii) Vital Capacity (VC) Tidal Volume + Inspiratory Reserve Volume + Expiratory Reserve Volume. (TV + ERV + IRV)

(iii) Residual Volume (RV) Volume of air remaining in the lungs after a forcible expiration.

(iv) Tidal Volume (TV) Volume of air inspired or expired during a normal respiration.

• Option (a):

  • (ii) Incorrect: The given statement is actually correct. Vital Capacity (VC) is indeed the sum of Tidal Volume (TV), Inspiratory Reserve Volume (IRV), and Expiratory Reserve Volume (ERV).
  • (iii) Incorrect: The given statement is incorrect. Residual Volume (RV) is not equal to Vital Capacity (VC) minus Inspiratory Reserve Volume (IRV). RV is the volume of air remaining in the lungs after a forcible expiration.

• Option (c):

  • (i) Correct: The given statement is incorrect. Inspiratory Capacity (IC) is the sum of Tidal Volume (TV) and Inspiratory Reserve Volume (IRV), not Residual Volume (RV).
  • (iii) Incorrect: The given statement is incorrect. Residual Volume (RV) is not equal to Vital Capacity (VC) minus Inspiratory Reserve Volume (IRV). RV is the volume of air remaining in the lungs after a forcible expiration.

• Option (d):

  • (i) Correct: The given statement is incorrect. Inspiratory Capacity (IC) is the sum of Tidal Volume (TV) and Inspiratory Reserve Volume (IRV), not Residual Volume (RV).
  • (ii) Incorrect: The given statement is actually correct. Vital Capacity (VC) is indeed the sum of Tidal Volume (TV), Inspiratory Reserve Volume (IRV), and Expiratory Reserve Volume (ERV).
  • (iv) Incorrect: The given statement is actually correct. Tidal Volume (TV) is indeed equal to Inspiratory Capacity (IC) minus Inspiratory Reserve Volume (IRV).

Answer

(a) A sigmoid curve obtained when percentage saturation of haemogblobin with $O_{2}$ is plotted against the $pO_{2}$.

The oxygen haemoglobin dissociation curve is shifted to right under following condition.

(i) Decrease in partial pressure of oxygen.

(ii) Increase in partial pressure of carbonoxide.

(iii) Increase in hydrogen concentration.

(iv) Decrease in $\mathrm{pH}$ activity.

(v) Increased body temperature.

• (b) high $\mathrm{pO}_{2}$: High partial pressure of oxygen ($\mathrm{pO}_{2}$) would not cause a right shift in the oxygen-haemoglobin dissociation curve. Instead, it would increase the affinity of haemoglobin for oxygen, leading to a left shift.

• (c) low $p \mathrm{CO}_{2}$: Low partial pressure of carbon dioxide ($p \mathrm{CO}_{2}$) would not cause a right shift. It would decrease the concentration of hydrogen ions, leading to a left shift in the curve.

• (d) less $\mathrm{H}^{+}$ concentration: A lower concentration of hydrogen ions (higher pH) would not cause a right shift. Instead, it would increase the affinity of haemoglobin for oxygen, resulting in a left shift in the curve.

Answer

(b) Earthworm respire through their moist cuticle and aquatic arthropods, respire through trachea.

Fishes respire through gills, and birds/reptiles respire through lungs.

• Option (a):

  • Earthworm: Incorrect because earthworms respire through their moist cuticle, not gills.
  • Aquatic arthropods: Incorrect because aquatic arthropods respire through gills or trachea, not a moist cuticle.
  • Fishes: Incorrect because fishes respire through gills, not trachea.
  • Birds/Reptiles: Correct because birds and reptiles respire through lungs.

• Option (c):

  • Earthworm: Correct because earthworms respire through their moist cuticle.
  • Aquatic arthropods: Incorrect because aquatic arthropods respire through gills or trachea, not lungs.
  • Fishes: Correct because fishes respire through gills.
  • Birds/Reptiles: Incorrect because birds and reptiles respire through lungs, not trachea.

• Option (d):

  • Earthworm: Correct because earthworms respire through their moist cuticle.
  • Aquatic arthropods: Incorrect because aquatic arthropods respire through gills or trachea, not gills.
  • Fishes: Incorrect because fishes respire through gills, not trachea.
  • Birds/Reptiles: Correct because birds and reptiles respire through lungs.

Thinking Process

The quantity of air that lung can receive, hold or expel under different condition is called pulmonary volume. Combination of two or more pulmonary volume is called pulmonary capacities.

Answer

(a) Tidal Volume (TV) is the volume of air inspired or expired during normal breath. This is about $500 \mathrm{~mL}$, i.e., a healthy man inspire or expire about 6000 to $8000 \mathrm{~mL}$ of air per minute.

(b) Residual Volume (RV) is the volume of air remaining in the lungs even after a forcible expiration. It is about $1100 \mathrm{~mL}$ to $1200 \mathrm{~mL}$.

(c) Asthma It is a disease caused due to an allergic reaction to foreign substances. The major symptoms are difficulty in breathing causing wheezing and coughing. Due to the inflammation of bronchi.

Answer

A fluid filled double membranous layer that surrounds the lungs is called pleura, and fluid is pleural fluid in between them. The outer pleural membrane is in close contact with the thoracic lining whereas, the inner pleural membrane is in contact with the lung surface. These collectively reduce friction on lung’s surface.

Answer

The primary site for the exchange of gases in our body is alveoli. There are 300 millions of alveoli collectively in both the lungs. These alveoli have very thin wall consisting of squamous epithelium. With extensive network of blood capillaries.

The presence of blood capillaries in the alveoli, result in easy exchange of gases. Each alveolus is also called as miniature lung.

Answer

Emphysema is a chronic disorder of respiratory system, where inflation or abnormal distension of alveolar wall occurs. Cigarette smoking and the inhalation of other smoke or toxic substances over a period of time causes the damaging of septa between the alveoli, and of its elastic tissue is replaced by the connective tissue in lungs.

Hence, the respiratory surface decreases, thus causing the emphysema. It causes shortness of breath, production of sputum, chronic bronchitis, etc.

Answer

Every $100 \mathrm{~mL}$ of oxygenated blood can deliver around $5 \mathrm{~mL}$ of $\mathrm{O}_{2}$ to the tissue under normal physiological conditions.

Answer

About $97 \%$ of $O_{2}$ is transported by RBCs in the blood. The remaining $3 \%$ of $O_{2}$ is carried in a dissolved state through the plasma.

Answer

(d) Expiratory Capacity (EC) Approximate volume is $1000 \mathrm{~mL}$.

(b) Residual Volume (RV) Approximate volume is $1200 \mathrm{~mL}$.

(c) Inspiratory Reserve Volume (IRV) Approximate volume is 2500 to $3000 \mathrm{~mL}$.

(a) Tidal Volume(TV) Approximate volume is 6000 to $8000 \mathrm{~mL}$.

Answer

(a) Inspiratory Capacity (IC) $=(\mathrm{TV})+(\mathrm{IRV})$ Tidal Volume. Inspiratory Reserve Volume

(b) Expiratory Capacity $(E C)=(T V+E R V)$ Tidal Volume. Expiratory Reserve Volume

(c) Functional Residual Capacity (FRC) $=($ ERV ) Expiratory $+($ RV $)$ Reserve Volume. Residual Volume

Thinking Process

Mechanism of breathing vary among different groups of animals depending on their habitats and levels of organisation.

Answer

(a) Flatworm General body surface

(b) Birds Lungs

(c) Frog Lungs and moist skin

(d) Cockroach Tracheal tubes

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.

Answer

Carbon dioxide is carried by the blood in three forms

(i) In Dissolved State Under normal temperature and pressure, about $7 \%$ of $\mathrm{CO}_{2}$ is carried by physical solution.

(ii) As Carbamino Compounds

$CO_{2}$ binds directly with $Hb$ to form an unstable compound carbaminocompounds $(CO_{2} )$ About $23 \%$ $CO_{2}$ is transported in this form. When $p CO_{2}$ is high and $pO_{2}$ is low as in the tissues, more binding of carbon-dioxide occurs whereas, when $p CO_{2}$ is low and $pO_{2}$ is high as in alveol as tissue dissociation of $CO_{2}$ from carbamino-haemoglobin takes place.

$$ HbO_{2}+CO_{2} \rightleftharpoons HbCO_{2}+\mathrm{H}^{+}+O_{2} $$

(iii) As Bicarbonate lons $CO_{2}$ reacts with water to form carbonic acid $\left(H_{2} CO_{3}\right)$ in the presence of carbonic anhydrase in RBC. $H_{2} CO_{3}$ dissociates into hydrogen and bicarbonate ions $\left(HCO_{3}{ }^{-}\right)$.

The whole reaction proceeds as follows

$\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \underset{\text { Anhydrase }}{\stackrel{\text { Carbonic }}{\rightleftarrows}} \underset{\text { Carbonic acid }}{\mathrm{H}_2 \mathrm{CO}_3}$

$\underset{\text { Carbonic acid }}{\mathrm{H}_2 \mathrm{CO}_3} \rightleftharpoons \underset{\substack{\text { Hydrogen } \\ \text { ion }}}{\mathrm{H}^{+}}+\underset{\substack{\text { Bicarbonate } \\ \text { ion }}}{\mathrm{HCO}_3^{-}}$

The carbonic anhydrase reaction mainly occur in RBC as it contain high concentration of enzyme carbonic anhydrase and minute quantity of it is present in plasma too.

Thinking Process

Diffusing capacity can be defined as the volume of gas, that diffuses through the membrane per minute for a pressure difference of $1 \mathrm{~mm} \mathrm{Hg}$. It is further dependent on the solubility of the diffusing gases.

Answer

As, the solubility rate of $CO_{2}$ is $20-25$ times higher than that of the $O_{2}$, the amount of $CO_{2}$ that can diffuse through the diffusion membrane per unit difference in partial pressure is much higher compared to that of $O_{2}$.

Answer

(d) Pulmonary ventilation by which atmospheric air is drawn in and $CO_{2}$ rich alveolar air is released out.

(a) Diffusion of gases $\left(O_{2}\right.$ and $\left.CO_{2}\right)$ across alveolar membrane.

(b) Transport of gases by blood.

(c) Diffusion of $O_{2}$ and $CO_{2}$ between blood and tissues.

(e) Utilisation of $O_{2}$ by the cells for catabolic reactions and resultant release of $CO_{2}$.

Answer

Difference between these are as follows

(a)

(b)

(c)

Answer

Representing the transport of $O_{2}$ and $CO_{2}$ between alveoli and tisue with diagram

Answer

Mechanism of Breathing Breathing involves two stages, inspiration during which atmospheric air is drawn in and expiration by which the alveolar air is released out.

The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere, the help of diaphragm and inter costal muscles.

Answer

Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissue. This is done by the neural system.

Respiration regulated by neural system in following ways/ manress

(i) A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible in regulating respiration process. 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.

(ii) A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to $\mathrm{CO}_{2}$ and hydrogen ions. Increase in these substances activates this centre, which in turn signals the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated.

(iii) Receptors associated with aortic arch and carotid artery also recognise changes in $\mathrm{CO}_{2}$ and $\mathrm{H}^{+}$concentration and send necessary signals to the rhythm centre for remedial action because the role of oxygen in the regulation of respiratory rhythm is quite insignificant.