Case Study Questions Class 11 Biology Chapter 17 Breathing and Exchange of Gases
CBSE Class 11 Case Study Questions Biology Breathing and Exchange of Gases. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Breathing and Exchange of Gases.
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CBSE Case Study Questions Class 11 Biology Breathing and Exchange of Gases
CASE 1
Blood is the medium of transport for O2 and CO2. About 97 per cent of O2 is transported by RBCs in the blood. The remaining 3 per cent of O2 is carried in a dissolved state through the plasma. Nearly 20-25 per cent of CO2 is transported by RBCs whereas 70 per cent of it is carried as bicarbonate. About 7 per cent of CO2 is carried in a dissolved state through plasma.
Haemoglobin is a red coloured iron containing pigment present in the RBCs. O2 can bind with haemoglobin in a reversible manner to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of O2. Binding of oxygen with haemoglobin is primarily related to partial pressure of O2. Partial pressure of CO2, 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 O2 is plotted against the pO2. This curve is called the Oxygen dissociation curve and is highly useful in studying the effect of factors like pCO2, H+ concentration, etc., on binding of O2 with haemoglobin. In the alveoli, where there is high pO2 , low pCO2 , lesser H+ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low pO2 , high pCO2 , high H+ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. This clearly indicates that O2 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 O2 to the tissues under normal physiological conditions.
CO2 is carried by haemoglobin as carbamino-haemoglobin (about 20-25 per cent). This binding is related to the partial pressure of CO2. pO2 is a major factor which could affect this binding. When pCO2 is high and pO2 is low as in the tissues, more binding of carbon dioxide occurs whereas, when the pCO2 is low and pO2 is high as in the alveoli, dissociation of CO2 from carbamino-haemoglobin takes place, i.e., CO2 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.
1.) ________________ of O2 and CO2 is carried in a dissolved state through the blood plasma.
a) 3% and 8%
b) 70% and 20%
3% and 9%
3% and 7%
2) Identify the correct statement
Statement 1 – 7 per cent of O2 is carried in a dissolved state through the plasma
Statement 2 – 3 per cent of CO2 is carried in a dissolved state through plasma.
Statement 3 – 70 per cent of CO2 is carried as bicarbonate.
Statement 4 – 97 per cent of CO2 is transported by RBCs.
a) Both 1 & 2 are correct
b) Both 3 & 4 are correct
c) Only 1 is correct
d) None of the above
3.) Name the factors which play key role in binding of oxygen and haemoglobin.
4.) How Oxygen dissociation curve are obtained.
5.) How much oxygen can deliver to tissue per 100 ml of oxygenated blood in normal conditions?
Answer key
1) d
2) c
3.) Following factors play key role in binding of oxygen and haemoglobin.
- Partial pressure of CO2
- Hydrogen ion concentration
- Temperature
4.) Oxygen dissociation curve is obtained when percentage saturation of haemoglobin with O2 is plotted against the pO2. This curve also known as sigmoid curve. It is highly useful in studying the effect of factors like pCO2, H+ concentration.
5.) Every 100 ml of oxygenated blood can deliver around 5 ml of O2 to the tissues under normal physiological conditions.
CASE 2
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. Inspiration can occur if the pressure within the lungs (intra-pulmonary pressure) is less than the atmospheric pressure, i.e., there is a negative pressure in the lungs with respect to atmospheric pressure. Similarly, expiration takes place when the intra-pulmonary pressure is higher than the atmospheric pressure. The diaphragm and a specialised set of muscles – external and internal intercostals between the ribs, help in generation of such gradients. Inspiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber in the antero-posterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs, i.e., inspiration. Relaxation of the diaphragm and the inter-costal muscles returns the diaphragm and sternum to their normal positions and reduce the thoracic volume and thereby the pulmonary volume. This leads to an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causing the expulsion of air from the lungs, i.e., expiration. We have the ability to increase the strength of inspiration and expiration with the help of additional muscles in the abdomen. On an average, a healthy human breathes 12-16 times/minute. The volume of air involved in breathing movements can be estimated by using a spirometer which helps in clinical assessment of pulmonary functions.
1.) Identify the correct statement
Statement 1 – In Inspiration atmospheric air is drawn in
Statement 2 – In Expiration the alveolar air is released out
Statement 3 – In Inspiration atmospheric air is drawn out
Statement 4 – In Expiration the alveolar air is released in
a) Both statement 1 and 2 are correct
b) Both statement 2 and 3 are correct
c) Both statement 3 and 4 are correct
d) Both statement 1 and 2 are incorrect
2.) Inspiration can occur if the pressure within the lungs is __________________
a) Positive with respect to atmospheric pressure
b) Negative with respect to atmospheric pressure
c) Positive with respect to intra-pulmonary pressure
d) Negative with respect to intra-pulmonary pressure
3.) Give pre-condition for expiration can take place?
4.) What is mechanism behind inspiration?
5.) Give the use of spirometer.
Answer key
1.) a
2.) b
3.) Expiration takes place when the intra-pulmonary pressure is higher than the atmospheric pressure.
4.) Increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs, this is how inspiration take place.
5.) The volume of air involved in breathing movements can be estimated by using a spirometer which helps in clinical assessment of pulmonary functions.
CASE 3
Tidal Volume (TV) – 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.
Inspiratory Reserve Volume (IRV) – Additional volume of air, a person can inspire by a forcible inspiration. This averages 2500 mL to 3000 mL.
Expiratory Reserve Volume (ERV) – Additional volume of air, a person can expire by a forcible expiration. This averages 1000 mL to 1100 mL.
Residual Volume (RV) – Volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL. By adding up a few respiratory volumes described above, one can derive various pulmonary capacities, which can be used in clinical diagnosis.
Inspiratory Capacity (IC) – Total volume of air a person can inspire after a normal expiration. This includes tidal volume and inspiratory reserve volume (TV+IRV).
Expiratory Capacity (EC) – Total volume of air a person can expire after a normal inspiration. This includes tidal volume and expiratory reserve volume (TV+ERV).
Functional Residual Capacity (FRC) – Volume of air that will remain in the lungs after a normal expiration. This includes ERV+RV.
Vital Capacity (VC) – The maximum volume of air a person can breathe in after a forced expiration. This includes ERV, TV and IRV or the maximum volume of air a person can breathe out after a forced inspiration.
Total Lung Capacity (TLC) – Total volume of air accommodated in the lungs at the end of a forced inspiration. This includes RV, ERV, TV and IRV or vital capacity + residual volume.
1.) Total Lung Capacity is equal ________
a) Vital capacity + Residual volume
b) Vital capacity + Functional Residual Capacity
c) Functional Residual Capacity + Tidal Volume
d) Inspiratory Capacity + Vital capacity
2.) Identify the correct formula
a) Total Lung Capacity = Vital capacity + Inspiratory Capacity
b) Expiratory Capacity = Tidal Volume + Tidal Volume
c) Total Lung Capacity = Vital capacity + Inspiratory Capacity
d) Expiratory Capacity = Residual Volume + Expiratory Reserve Volume
3.) Give the formula to calculate inspiratory capacity.
4.) Define vital capacity.
5.) Give the formula to calculate volume of air remains in the lungs under normal physiological conditions after a normal breathing.
Answer key
1) a
2) d
3) Tidal Volume + Inspiratory Reserve Volume = Inspiratory Capacity
4) Vital capacity is defined as the maximum volume of air a person can breathe in after a forced expiration it is about 4000mL in a normal adult person. Vital capacity is higher in athletes and singers. Cigarette smokers have a lower vital capacity of the lungs. This includes ERC, TV, and IRV, or the maximum volume of air a person can breathe out after a forced inspiration.
5) The volume of air remaining in the lungs after normal respiration under normal physiological condition is called as functional residual capacity.
FRC = ERV + RV
Functional residual capacity = Expiratory reserve volume + residual volume.
CASE 4
We have a pair of external nostrils opening out above the upper lips. It leads to a nasal chamber through the nasal passage. The nasal chamber opens into the pharynx, a portion of which is the common passage for food and air. The pharynx opens through the larynx region into the trachea. Larynx is a cartilaginous box which helps in sound production and hence called the sound box. During swallowing glottis can be covered by a thin elastic cartilaginous flap called epiglottis to prevent the entry of food into the larynx. Trachea is a straight tube extending up to the mid-thoracic cavity, which divides at the level of 5th thoracic vertebra into a right and left primary bronchi. Each bronchi undergoes repeated divisions to form the secondary and tertiary bronchi and bronchioles ending up in very thin terminal bronchioles. The tracheae, primary, secondary and tertiary bronchi, and initial bronchioles are supported by incomplete cartilaginous rings. Each terminal bronchiole gives rise to a number of very thin, irregular-walled and vascularised bag-like structures called alveoli. The branching network of bronchi, bronchioles and alveoli comprise the lungs. We have two lungs which are covered by a double layered pleura, with pleural fluid between them. It reduces friction on the lung-surface. The outer pleural membrane is in close contact with the thoracic lining whereas the inner pleural membrane is in contact with the lung surface. The part starting with the external nostrils up to the terminal bronchioles constitute the conducting part whereas the alveoli and their ducts form the respiratory or exchange part of the respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also brings the air to body temperature. Exchange part is the site of actual diffusion of O2 and CO2 between blood and atmospheric air.
The lungs are situated in the thoracic chamber which is anatomically an air-tight chamber. The thoracic chamber is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and on the lower side by the dome-shaped diaphragm. The anatomical setup of lungs in thorax is such that any change in the volume of the thoracic cavity will be reflected in the lung (pulmonary) cavity. Such an arrangement is essential for breathing, as we cannot directly alter the pulmonary volume.
1.) ___________________ is undergoes repeated divisions to form bronchi and bronchioles
a) Alveoli
b) Bronchi
c) Pleura
d) Trachea
2.) Bronchiole gives rise to very thin, irregular-walled and vascularised bag-like structures known as
a) Bronchi
b) Pleura
c) Larynx
d) Alveoli
3.) What is pleura? Give its function.
4.) Give reason – Why Larynx is known as sound box?
5.) What is epiglottis? Give its function.
Answer key
1.) b
2.) d
3.) Pleura – Both lungs are covered by a double layered pleura, with pleural fluid between them. It reduces friction on the lung-surface.
4.) Larynx is a cartilaginous box which helps in sound production and hence called the sound box.
Epiglottis is a thin elastic cartilaginous flap which covers glottis.
5.) Function – prevent the entry of food into the larynx.
CASE 5
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 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 CO2 and 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.
Respiratory disorders or diseases are diseases of lungs and human airways that affect human respiration.” A disorder is defined as a state of irregular functioning of the body.
Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles.
Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. One of the major causes of this is cigarette smoking.
Occupational Respiratory Disorders: In certain industries, especially those involving grinding or stone-breaking, so much dust is produced that the defence mechanism of the body cannot fully cope with the situation. Long exposure can give rise to inflammation leading to fibrosis (proliferation of fibrous tissues) and thus causing serious lung damage. Workers in such industries should wear protective masks.
1.) ___________________ from this centre neural signal can reduce the duration of inspiration and alter the respiratory rate.
a) Pneunotaxic centre
b) Pneumotaxic centre
c) Chemosensitive centre
d) Rhythm centre
2.) Increase in CO2 and hydrogen ions substances activates,
a) Respiratory rhythm
b) Pneumotaxic centre
c) Rhythm centre
d) Chemosensitive centre
3.) What is Pneumotaxic centre? Give its function.
4.) What is emphysema? What is its major cause?
5.) What is fibrosis?
Answer key
1.) a
2.) b
3.) Pneumotaxic centre – Is the centre present in the pons region of the brain. It 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.
4.) Emphysema – Emphysema is a chronic condition in which alveolar walls are damaged due to this damage the surface area for exchange of gases is reduced. It is caused mainly by cigarette smoking.
5.) Fibrosis is condition in which proliferation of fibrous tissues causes serious lung damage.It this condition surface area for exchange of gases is reduced.