Respiratory Physiology MCQs (for FCPS Part 1)

Respiratory Physiology MCQs (for FCPS Part 1)

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These MCQs are taken from the book “FCPS Pretest Series – Physiology”. To purchase the book, click here.

 

 

fcps pretest series physiology
IMPORTANT NOTE: You will get at least one MCQ from “oxygen – hemoglobin dissociation curve” in every FCPS Part 1 exam. Other important topics from respiratory Physiology are:

          1. Peripheral and central control of breathing.

          2. Ventilation-perfusion defects

          3. Lung volumes and capacities + Spirometry in asthma and COPD

1. In arterial blood, dissolved co2 (1.4 ml/100 ml) is more than the Dissolved oxygen (0.3 ml/100 ml) because

A. co2 is in equilibrium with hco3

B. In the lungs diffusing capacity for co2 is greater than that of 02

C. Most of the oxygen that enters the blood is combined with Hb

D. The alveolar concentration of co2 is higher than that of 02

E. The plasma solubility of co2 is 20 times greater than that of 02

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Ans. E. The plasma solubility of: CO > Co2 > Oxygen

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HB affinity of: CO > Oxygen > Co2

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2. At high altitude, increasing alveolar ventilation raises ph of Plasma because

A. Increased muscle work of increased breathing generates more co2

B. It activates neurons which remove acid from the blood

C. It decreases pco2 in the alveoli

D. It increases the p02 of the plasma

E. It makes hemoglobin a stronger acid

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Ans. C. HYPERVENTILATION: ↑Co2 removed out of alveoli → ↓Alveolar Pco2 → ↓Blood Pco2 → Respiratory alkalosis.

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HYPOVENTILATION: ↓Co2 removed from the alveoli → ↑Alveolar Pco2 → ↑Blood Pco2 → Respiratory acidosis.

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3. Following cardiac changes occur during inspiration

A. A decrease in pressure gradient between extrathoracic veins and right atrium

B. A decrease in systemic arterial pressure

C. A decrease in the heart rate

D. Decreased right ventricular filling

E. Decreased right ventricular output

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Ans. B

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INSPIRATION

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During inspiration, the intrapleural pressure decreases (more negative). This produces a suction force, thus pulling blood from systemic veins and left ventricle towards the heart. This increases cardiac output of the right heart and decreases cardiac output of the left heart. Due to ↓output of the left heart, systemic B.P decreases by 10 mmHg.

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Due to ↓B.P, the barorecptor reflex is activated which activates sympathetic system, resulting in ↑heart rate

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EXPIRATION

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During expiration, the thoracic pressure increases → ↑Venous return to the left heart → ↑Left heart output → ↑Systemic pressure → Baroreceptor reflex activates parasympathetic system → ↓Heart rate

4. A patient suffering from chronic respiratory failure

A. Does not increase his ventilation in response to oxygen lack

B. Is likely to have a raised calcium level

C. Is likely to have alkaline urine

D. Should be given 100% oxygen on admission to hospital

E. Shows an increased respiratory sensitivity to carbon dioxide

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Ans. B. The positively charged Ca+ and H+ compete for binding with negatively charged proteins. In chronic respiratory failure, respiratory acidosis occurs due to accumulation of Co2 in the body, resulting in increased H+ ions in the blood → More H+ bind with proteins → So, less Ca+ bind with proteins → ↑Plasma levels of free ca+

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5. The most common physiologic cause of hypoxemia is

A. Elevated erythrocyte 2,3 diphosphoglycerate level (2,3 DPG)

B. Hypoventilation

C. Incomplete alveolar oxygen diffusion

D. Pulmonary shunt flow

E. Ventilation perfusion inequality

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Ans. E. Although hypoventilation, incomplete oxygen diffusion and pulmonary shunts all are causes of hypoxemia, the most common cause is ventilation-perfusion inequality. The mismatch of ventilation and blood flow occurs to some degree in the normal upright lung but may become extreme in the diseased lung. The three indices used to measure ventilation-perfusion inequality are alveolar arterial Po2 difference, physiologic shunt (venous admixture), and alveolar dead space. Elevated 2, 3-diphoglycerate (2, 3-DPG) levels shift the oxygen dissociation curve to the right and thereby augment tissue oxygenation. This elevation does not result in hypoxemia.

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6. Which of the following act as oxygen binding proteins in Skeletal muscle

A. Actin

B. Myoglobin

C. Myosin

D. Tropomyosin

E. Troponin

.Ans. B. Myoglobin is an iron- and oxygen-binding protein found in the skeletal and cardiac muscles.

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7. In case of Inspiration, chest expansion occurs due to:

A. Increased 02 pressure in blood

B. CNS stimulation

C. Decreased 02 in the blood

D. Volunteer action

E. Decreased intrathoracic pressure

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Ans. E. During inspiration, diaphragm moves downward → Chest volume increases → Intrathoracic pressure (intrapleural pressure) decreases (pressure is inversely proportional to volume) → Air flows from outside atmosphere to alveoli (atmospheric pressure is zero and intrapleural pressure is less than zero; hence, air flows from higher to lower pressure) → chest expansion occurs

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8. Exchange of gases through lungs depends upon

a. Partial pressure difference of the gases

b. Surface area available for gas exchange

c. Diffusion distance

d. Solubility and molecular weight of the gases

e. All of the above

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Ans. E

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Diffusion= SA X ΔP X SolubilityTmw..

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SA = Surface area, ΔP = Difference in partial pressures, T = membrane thickness, mw = molecular weight of the gas

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This equation shows that diffusion of gases across a membrane is directly proportional to:

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  1. The partial pressure difference of the gas
  2. Surface area available for gas exchange
  3. Solubility of the gas

    .

    Diffusion of gases across a membrane is inversely proportional to:

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  4. Diffusion distance
  5. Molecular weight of the gas.

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9. Regarding the rate, at which a gas diffuses across pulmonary Alveolar level, following statements are correct except:

A. According to fick’s law

B. Inversely proportional to square root of molecular weight of the gas

C. Inversely proportional to the thickness of the membrane

D. Proportional to molecular weight of the gas

E. Proportional to the solubility of the gas in body fluids

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Ans. D. Diffusion of gas across a membrane is inversely proportional to molecular weight of the gas.

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10. Factors that affect rate of diffusion through the Respiratory membrane are:

A. Thickness of the membrane only

B. The pressure difference across the respiratory membrane

C. When the physiological dead space is less, then much of the work of Ventilation is wasted effort.

D. Smokers develop physiological shunt when ventilation perfusion ratio is Above normal.

E. When the normal person in the upright position both blood flow and alveolar Ventilation are great in the upper part of lung than lower pail.

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Ans. B

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11. The most efficient gasseous exchange takes place in the

A. Alveolar ducts

B. Alveolar sacs

C. Alveoli

D. Respiratory bronchioles

E. Terminal bronchiole

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Ans. C

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12. During the transport of oxygen in the arterial blood the Highest tension of oxygen (p02) is seen in:

A. Arterial blood

B. Left ventricle

C. Pulmonary capillaries

D. Right atrium

E. Venous blood

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Ans. C. Oxygen diffuses from the alveoli into the pulmonary capillaries surrounding the alveoli; so, pulmonary capillaries have the highest Po2.

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13. A 57-year-old clerk who is a chronic smoker presented with Cough and was cyanosed. The most likely findings will be:

A. Increased HBa

B. Increased HBs

C. Increased concentration of deoxygenated Hb

D. Increased percentage of oxygenated Hb

E. Normal Hb

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Ans. C. Smoking causes emphysema, which results in destruction of alveolar walls → ↓Surface area is available for exchange of gases → ↓Exchange of gases (oxygen and Co2) → ↓Po2 → ↓Hb saturation (i.e., ↑concentration of deoxygenated Hb) → Cyanosis

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14. Hypoxemia does not depend on:

a. PaCO2

b. Altitude

c. Hb

d. Fi02

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Ans. C. Oxygen in the blood exists in two forms:

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  1. Dissolved in blood
  2. Attached to HB

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    Hypoxemia means decreased PaO2 (partial pressure of oxygen in arterial blood). Pao2 depends upon:

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  3. Atmospheric pressure
  4. Fraction of air in inspired air

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    So, Pao2 does not depend upon HB concentration in the body (e.g, anemia, in which HB concentration is decreased, or polycythemia, in which HB concentration is increased,does not affect Pao2 and hence, does not cause hypoxemia).

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    15. Which of the following decreases arterial O2 saturation without decreasing arterial 02tension

    a. Anemia

    b. Low V/Q

    c. Carbon monoxide poisoning

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    Ans. C. Oxygen in the blood exists in two forms: (1) Dissolved in blood (on which Po2/oxygen tension depends) (2) Attached to HB (on which HB saturation/O2 saturation depends).

    .

    [Option A] – Anemia only decreases HB concentration, it does not have any effect on PO2 or HB saturation

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    [Option B] – Any ventilation perfusion mismatch decreases Po2 which in turn decreases HB saturation (HB saturation is directly proportional to Po2)

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    [Option C] – HB affinity of CO > Oxygen. So, CO binds with HB, displacing oxygen from it. So, CO decreases HB saturation but, it does not affect Po2.

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    16. Partial pressure of oxygen in arterialblood depends on

    a. Metabolic rate

    b. Hb concentration

    p. Oxygen bound to Hb

    d. Oxygen dissolved in plasma

    e. Amount of oxygen in air

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    Ans. D. As mentioned earlier, it’s the dissolved oxygen only (and not the oxygen attached to HB) which exerts pressure.

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    17. At the end of expiration:

    a. Intra-alveolar pressure is subatmospheric

    b. Intra-pleural pressure is subatmospheric

    c. No change in intra pleural pressure

    d. Intrapleural pressure is more thanatmospheric pressure

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    Ans. B. Atmospheric pressure is 760 mmHg, but while dealing with lung pressures, the atmospheric pressure is considered to be zero.

    .

    At the end of inspiration, the intra-pleural pressure is – 8, and at the end of expiration, the intra-pleural pressure is – 5. So, the intra-pleural pressure is always negative and less than atmospheric pressure.

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    18. Atmospheric pressure can be quoted as approximately

    A. 1 bar

    B. 1.5 lb/in2

    C. 100 nm

    D. 7.6m h

    E. 760cm hg

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Ans. A. Atmospheric pressure is equal to 760 mmHg = 1.01 bar

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19. Normal quiet expiration is brought about by contraction/recoil of

a. Diaphragm

b. Elastic tissue in thoracic and lung wall

c. Abdominal muscles

d. Sternocleidomastoid

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Ans. B. Inspiration is an active process while normal expiration is a passive process which occurs due to elastic recoil of the alveoli and the thoracic wall.

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Quiet inspiration

Diaphragm

Forced inspiration

External intercostals muscles

Quiet Expiration

Recoil of chest wall and lungs (passive process)

Forced expiration

Abdominal muscles (Rectus abdominis is the main muscle), internal intercostal muscles

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20. Forceful expiration does not involve contraction of

A. External I/C muscle

B. External oblique

C. Internal oblique

D. Rectus abdominus

E. Transverse abdominus

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Ans. A

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21. MOST POWERFUL EXPIRATORY MUSCLE IS

A. Diaphragm

B. Trapezius

C. Rectus abdominus

D. Serratus posterior

C. Serratus anterior

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Ans. C

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22. Which of the following muscles is critical for Expiration?

A. External intercostals

B. Innermost intercostals

C. Diaphragm

D. Abdominal muscles

E. Transverse thoracis

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Ans. D

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23. Muscle of quite inspiration

a. Diaphragm

b. Intercostal muscle

c. Sternocladomastoid

d. Scalenius posterior

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Ans. A

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24. Muscle of forceful inspiration

a. External intercostal muscle

b. Internal intercostal muscle

c. Nasaris

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Ans. A

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25. During exercise pt uses accessory muscle in deep breathing. Which muscle is involved

A. Stenocleidomastoid

B. Serratus anterior

C. Interval intercostais

D. Diaphragm

E. External intercostal

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Ans. E

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26. The percentage change during quiet respiration (inspiration) Movement of the diaphragm is

A. 45%

B. 50%

C. 60%

D. 75%

E. 85%

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Ans. D

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27. Major part of energy utilized during breathing is to overcome

a. Elastic recoil of lungs

b. Resistance of chest wall

c. Large airway resistance

d. Small airway resistance

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Ans. A. Most of the energy used during breathing is to overcome the elastic recoil of the lungs which opposes expansion of the lungs and inspiration.

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28. Almost 70 % of work of breathing

a. Is to overcome elastic recoil of lungs

b.. Counteract chest wall compliance

c. Resistance offered by small size bronchi

d. Resistance by large bronchi

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Ans. A

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29. Negative intrapleural pressure is due to:

a. Lymphatic drainage of pleura

b. Uniform distribution of surfactant over alveoli

c. Negative intraalveolar pressure

d. Presence of cartilage in upper airway

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Ans. A. The intra-pleural pressure is always negative, and it is negative mainly due to lymphatic drainage of the chest which produces a suction force.

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30. Which of the following is the best-known metabolic function of the lung

a. Inactivation of serotonin

b. Conversion of angiotensin I into angiotensin II

c. Inactivation of bradykinin

d. Metabolism of basic drugs bycytochrome P450 system

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Ans. B. The lungs produces ACE (angiotensin converting enzyme) which converts angiotensin I to angiotensin II.

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31. If the diameter of alveoli is reduced to half, resistance becomes

a. 1/4th

b. 1/16

c. 4th

d. 16th

e. 1/8th

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Ans. D

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Resistance= 1(Radius)4..

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It means that resistance is inversely proportional to radius (diameter).

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If the diameter (radius) is reduced to half, resistance increases 16 times, i.e, (radius)4 = (2)4 = 2 X 2 X 2 X 2 = 16

32. Compliance of the lungs:

A. Is affected in various lung diseases

B. Changes when there is surfactant deficiency

C. Depends upon the size of lungs

D. Is a measure of stretchability of the lungs

E. Is the change in the lung volume per unit change of transpulmonary Pressure

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Ans. E

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Compliance= ΔVΔP..

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Where ΔV = Change in volume, ΔP = Change in pressure

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33. Lung compliance is increased in

A. Emphysema

B. Bronchial fibrosis

C. Bronchiectasis

D. Pnumonectomy

E. Surfactant

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Ans.A. Two forces act on the lungs: (A) Intrapleuralpressure (B) Elastic recoil

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         (A) INTRAPLEURAL PRESSURE

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  • The intrapleural pressure (IP) is always negative.
  • IP expands the lungs, e.g., during inspiration, IP > elastic recoil. So, the lungs expand. It means that the lower is IP (i.e., the more negative is IP), the easier it will be to expand the lungs, and hence, the higher will be the compliance.


    (B) ELASTIC RECOIL

     

  • The elastic recoil is always positive.
  • It tends to collapse the lungs, e.g., during expiration, elastic recoil > IP. That’s why the lungs collapse. It means that the greater is leastic recoil, the more difficult it will be to expand the lungs, and hence, the lower will be the compliance.

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    EXPLANATION OF MCQ

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  • The elastic tissue in the walls of alveoli is responsible for elastic recoil of the lungs.
  • In emphysema, the walls of alveoli are destroyed with loss of elastic tissues (which are replaced by fibrous tissues), so the elastic recoil decreases and hence, compliance increases.

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34. surfactants are secreted by:

A. Alveolar macrophages

B. Capillary endothelial cells

C. Kupffer cells

D. Type I alveolar epithelial cells

E. Type II alveolar epithelial cells

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Ans. E. Surfactant is secreted by lamellar bodies of type II pneumocytes.

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1.jpg

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35. Lamellar bodies are found in

a. Clara cells

b. Type I pneumocytes

c. Type II pneumocytes

d. Macrophages

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Ans. C. Lamellar bodies/lamellar granules are present in type II pneumocytes; lamellar bodies secrete surfactant.

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http://synapses.clm.utexas.edu/atlas/7_1_4.gif

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36. Pulmonary alveoli has following characteristics except

A. Are lined by type I epithelial cells

B. Are not fully functional until birth

C. Begin to develop during the 6th month of fetal life

D. Continue to develop during early postnatal period

E. Lack type II epithelial cells (pneumocytes) at birth

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Ans. E. Type II epithelial cells secrete surfactant and are formed before birth.

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37. Regarding surfactant following statements are correct Except:

A. Deficiency is also associated with patchy atelactesis

B. Is decreased in the lungs of cigarette smokers

C. Is important at the time of birth

D. Is produced by type I alveolar epithelial cells

E. Reduces surface tension in alveolar lumen

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Ans. D. surfactant is produced by type II alveolar cells.

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38. Surfactant:

a. Decreases compliance work

b. Decreases when alveolar size decreased

c. Has a low turnover rate

d. Decrease compliance

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Ans. A.

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  • Surface tension is the force present between fluid and air interface. It is a collapsing force, inhibits lungs expansion, and hence, decreases lungs compliance.
  • Surfactant opposes surface tension and hence, increases lungs compliance.

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    39. Which statement is correct regarding surfactant

    a. Slow turn over

    b. Secreted by type Ipnuemocytes

    c. Forms macromolecular layer on fluid

    d. Decreases lung compliance

    e. Production decreases as the size of alveoli decrease

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Ans. C. Surfactant forms macromolecular assemblies of less than 100 nm.

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40. Surfactant secretion starts at which week of gestation:

a. 28 weeks

b. 32 weeks

c. 34 weeks

d. 36 weeks

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Ans. A. Surfactant secretion starts as early as 24 weeks of gestation.

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41. Surfactant is composed mainly of

a. Lecithin

b. Cephalin

c. Phosphatidylgiycerol

d. Protein

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Ans. A. Dipalmitoylphosphtidycholine (DPPD), also known as lecithin, is the main component of surfactant.

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42. Which one of the following will happen in surfactant deficiency

a. Increases compliance

b. Decreases compliance

c. No effect on compliance

d. Surface tension is decreased

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Ans. B. Surfactant decreases surface tension, and hence, increases compliance. In surfactant deficiency, surface tension is higher, and hence, compliance is decreased (this is the reason why surfactant deficiency in respiratory distress syndrome causes lungs collapse).

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43. Which of the following variables in most likely be lower than normal in a patient with infant respiratory distress syndrome?

A. Alveolar P02 difference

B. Compliance of lungs

C. Oncotic pressure of alveolar fluid

D. Surface tension of alveolar fluid

E. Work of breathing

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Ans. B

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44. Vital capacity:

A. Can be measured by a spirometer

B. Gives information about the respiratory muscles

C. Is less in females than in males

D. Is reduced in some lung diseases

E. Is the sum of tidal volume, inspiratory reserve volume and expiratory

Reserve volume

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Ans. E

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http://figures.boundless.com/19654/full/figure-39-02-01.jpe

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45. After full inspiration person forcefully expires, the volume and capacity contained in the lungs is

A. TLC

B. FRC

C. IC

D. VC

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Ans. D

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46. Vital capacity of the lung is –

A. Expiratory reserve volume plus residual volume

B. inspiratory reserve volume plus expiratory reserve volume

C. Tidal volume plus expiratory reserve volume

D. Tidal volume plus inspiratory reserve volume

E. Tidal volume plus inspiratory reserve volume plus expiratory

Reserve volume

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Ans. E

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47. FRC is

a. TV + ERV

b. IRV + ERV

c. ERV AND RV

cl. IRV AND RV

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Ans. C

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48. Lung function that cannot be measured by spirometry:

a. Residual volume

b. IRV

c. ERV

d. All can be measured

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Ans. A. Residual volume can not be measured by spirometry. So, any lung capacity which includes residual volume (e.g., total lung capacity) can not be measured by spirometry.

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49. Residual volume

a. Amount of air left in the lung after amaximal exhalation

b. Amount of air expired before tidalvolume

c. It is equal to tidal volume

d. It is 1000 ml

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Ans.A

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50. Pulmonary volumes and capacities:

A. The residual volume is the air that can be removed from the lungs.

B. Vital capacity is the measure of residual volume

C. The inspiratory reserve volume is the extra volume of air that can be inspired over and beyond the normal tidal volume.

D. The total lung capacity is equal to inspiratory reserve volume plus tidal volume.

E. As the lung compliance decreases the vital capacity increases

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Ans. C

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51. Significance of pulmonary volumes are:

A. It is increased when the person lies down and deceased in standing position

B. The residual volume represents the air that can be removed from the lungs by Forceful expiration.

C. Vital capacity is increased if the pulmonary compliance is increased

D. The inspiratory reserve volume of air inspired with each normal breath

E. All pulmonary volumes are about 40.45% percent less in female

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Ans. C

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52. Vital capacity

a. After maximum inspiration,maximum expired volume is VC

b. IRV AND ERV

c. Tidal volume and IRV

d. After maximal expiration maximumvolume inspired

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Ans.A

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53. Which of the following can be measured by spirometry

a. VC

b. RV

c. TLC

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Ans. A. All lung volumes can be measured by spirometry except residual volume. So, all capacities can be measured by spirometry except those which include residual volume (e.g, total lung capacity). As vital capacity = IC + IRV + ERV (does not include residual volume), so it can be measured by spirometry.

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54. Normal FEV1/FVC ratio is

a. 0.6

b. 0.8

c. 1

d. 5

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Ans. B. FVC (Forced vital capacity): FVC is the maximum amount of air that can be expelled after maximum inhalation, e.g, blowing a birthday candle. Normally, FVC = 5 liters

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FEV1(Forced expiratory volume – 1): The amount of air which can be forcibly exhaled from the lungs in the first second. Normally, FEV1 = 4 liters

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So, normal FEV1/FVC = 4/5 = 0.80 (or 80%)

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55. In bronchial asthma with clinical findings of bronchial spasm, the most important pulmonary function test which will be abnormal:

a. FEV1

b. Expiratory reserve volume

c. Inspiratory reserve volume

d. Residual volume

e. Tidal volume

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Ans. A. Normally FEV1/FVC = 80%. FEV1/FVC < 80% is diagnostic of COPD (asthma), and FEV1/FVC > 80% is diagnostic of restrictive lung diseases.

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Table Summary of obstructive Vs restrictive patttern

Variable

Obstructive pattern
(e.g., asthma, emphysema)

Restrictive pattern

(e.g., fibrosis)

TLC

↓↓

FEV1

↓↓

FVC

↓↓

FEV1/FVC

↑ or normal

Peak flow

FRC

RV

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56. The most important pulmonary function test in bronchial asthma

a. FEV1

b. Expiratory reserve volume

c. Inspiratory reserve volume

d. Residual volume

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Ans. A

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57. In asthma,which is the most characteristic finding:

a. Tidal volume = 350ml

b. FEV1/FVC < 65%

c. Increased FVC

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Ans. B

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58. COPD and restrictive lung disease has following features to differentiate

A. FEV1/FVC ratio

B. FRC

C. RV

D. Tidal volume

E. Vital capacity

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Ans. A

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59. On a young man of 20, suffering from bronchial asthma, lung function Tests were performed. Only one of the following relates to correct diagnosis:

A. Fev1/fvc< 75

B. P02 = 40 mmhg arterial

C. Respiratory rate = 12/ min

D. Tidal volume = 500 ml

E. Vital capacity = 3o ml

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Ans. A

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60. In pulmonary fibrosis all values are low except:

a. Residual volume

b. Total lung capacity

c. FEV1 to FVC

d. Vital capacity

.

Ans. C. Pulmonary fibrosis is a restrictive lung disease and in all restrictive lung diseases, FEV1/FVC > 80 (Vs. COPD in which FEV1/FVC < 80).

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61. In asthma, FEV1/FVC is

a. 90%

b. 80%

c. 70%

d. 60%

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Ans. B. In COPD (asthma), which causes airflow obstruction, FEV1/FVC < 80%

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62. In asthma

a. FEVI/FEV less than 65%

b. FVC is reduced

c. FEV1/FEV more than 0.8

d. No effect on FEV1

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Ans. A

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63. The functional residual capacity (FRC):

A can be measured by a spirometer

B. Is the sum of residual volume and inspiratory reserve volume

C. Is the sum of residual volume and vital capacity

D. Is the sum of tidal volume and residual volume

E. Is the volume of air in the lungs at the end of tidal expiration

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Ans. E

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64. Tidal volume is calculated by:

a. Inspiratory capacity minus the inspiratory reserve volume

b. Total lung capacity minus the residual volume

c. Functional residual capacity minus residual volume

d. Vital capacity minus expiratory reserve volumes

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Ans. A. Tidal volume = IC – IRV

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65. If a person inhales 500 ml of air during each breath with respiratory rate of 10 breaths/min. Calculate the alveolar ventilation

a. 5000 ml/min

b. 3000 ml/min

c. 3500 ml/min

d. 4000 ml/min

e. 4500 ml/min

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Ans. C

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Alveolar ventilation=(Tidal volumeDead space) X Respiratory rate

.

.

Normally, dead space = 150 ml, so:

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Alveolar ventilation = (500 – 150) X 10 = 350 X 10 = 3500 ml/min

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66. After forced expiration, residual volume will be

a. 500ml

b. 1000ml

c. 1200ml

d. 2400ml

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Ans. C

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Lungs Volume

Normal Values (ml)

Inspiratory reserve volume

3000

Tidal volume

500

Expiratory reserve volume

1100

Residual volume

1200

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67. Amount of air remaining in lung after maximum expiration

A. 1200ml

B. 2000ml

C. 1100ml

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Ans. A

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68. Anatomical dead space:

a. Increases with tracheostomy

b. Remove particulate matter from air before entering alveoli

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Ans. B.

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PHYSIOLOGICAL (RESPIRATORY) DEAD SPACE

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DEFINITION: The part of respiratory system which has ventilation but no perfusion (which contains air but is not exchanging oxygen and Co2 with blood) is called physiological (or respiratory) dead space.

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TYPES OF DEAD SPACE: There are two types of dead space:

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         (A) ANATOMICAL DEAD SPACE.

  • Airway regions that, because of inherent structure, are not capable of O2 and Co2 exchange with the blood are called anatomical dead space.
  • Anatomical dead space includes the conducting zone, which ends at the level of terminal bronchioles. Significant gas exchange occurs only in the alveoli.
  • The function of anatomical dead space is to warm, humidify, and filter the air.
  • The size of anatomical dead space in ml is equal to a person’s weight in pounds. For example, average weight of an adult is 150 pounds, so normally, anatomical dead space is 150 ml.

    .

    (B) ALVEOLAR DEAD SPCAE

    .

  • Alveoli which contain air but do not take part in gas exchange comprise alveolar dead space.
  • .Physiological (respiratory) dead space = Total dead space = Anatomical dead space + Alveolar dead space

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69. Dead space volume is decreased in

A. Deep inspiration

B. Old age

C. Shallow breathing

D. Standing

E. Tracheostomy

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Ans. E. Tracheostomy decreases anatomical dead space.

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70. The dead space:

A. Changes with increased depth of inspiration

B. Includes nasal cavities

C. Is the portion of inspired air which does not take part in Gaseous exchange

D. Varies with age

E. Varies with changes in posture

.

Ans. C

.

71. In a patient having high ventilation/perfusion ratio, there is:

A. A change in the composition of alveolar air

B. Adequate perfusion of alveoli

C. An increase in the dead space

D. Inadequate ventilation

E. Shunted blood

.

Ans. C

.

72. When a branch of Pulmonary Artery is blocked by embolus, following lung function increases

a. Alveolar Co2

b. Alveolar 02

c. Pulmonary artery 02

d. Pulmonary artery Co2

e. Ventilation Perfusion ratio

.

Ans. E. When a branch of pulmonary artery is blocked, ventilation does not change but perfusion decreases, so ventilation perfusion ratio increases.

.

73. Ventilation perfusion mismatch in which Ventilation > Perfusion is known as:

a. Shunt

b. Dead space

c. Normal response

d. Anatomic dead space

.

Ans. B

.

SHUNT

.

  • Loss of ventilation (perfusion > ventilation) is called shunt.
  • In shunt: V/Q = Zero/Q =0(zero divided by any thing equals zero).

    .

    1.jpg

    .

 

 

 

 

 

 

 

 

 

 

DEAD SPACE

.

  • Loss of perfusion (ventilation > perfusion) is called dead space.
  • In dead space: V/Q = V/zero =∞ (any thing divided by zero equals infinity).

    .

    47. Physiological shunt occurs in which case

    a. Perfusion is zero

    b. Ventilation is infinity

    c. Ventilation is zero

    .

    Ans. C. Zero ventilation is called shunt, while zero perfusion is called dead space.

    .

    74. Which one of the following is diffusion limited

    a. 02

    b. Co2

    c. 02 during exercise

    d. N2o

    .

    Ans. C

    .

    DIFFUSION LIMITED: A gas is diffusion limited if it does not equilibrate by the time the blood reaches the end of capillary. For example: (1) Oxygen in exercise, emphysema, fibrosis (2) CO (That’s why CO is used to measure diffusion capacity of the lung)

    .

    PERFUSION LIMITED: A gas is said to be perfusion limited when it equilibrates early along the length of capillary (the diffusion is limited by perfusion, and can only be increased by increasing perfusion). For example: (1) Oxygen (in normal health) (2) CO2

    .

    .

    http://pathwaymedicine.org/images/respiratory/Diffusion-Perfusion-limited-Gas-Exchange.png

    .

    .

 

 

 

 

 

75. Gas used to measure diffusion in lungs

a. CO

b. NO

c. CO2

d. Nitrogen

.

Ans. A. Carbon monoxide is diffusion limited and it never reaches equilibrium, because it has very high affinity for HB, and hence, all CO which comes to the blood binds to HB. That’s why, CO is used to measure diffusion capacity of the lungs.

.

76. Ventilation is very high at

a. Apex of lung

b. Base of lung

c. Middle zone

d. Lateral zone

.

Ans. A. Both ventilation and perfusion are higher at the base due to the effect of gravity. But in the apex, ventilation > perfusion, while at the base, perfusion > apex. So, the apex has extra ventilation (that’s why mycobacterium loves the apex), and the base has extra perfusion.

.

C:\Documents and Settings\Administrator\Desktop\1.jpg

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77. Which of the following is higher at the apex of the lung than at the base when a person is standing

a. V/Q ratio

b. Blood flow

c. Ventilation

d. PaCO2

.

Ans. A. Both ventilation and perfusion are greater at the base of the lungs than at the apex of the lungs due to the effect of gravity. But, in the apex: ventilation > perfusion, while at the base: perfusion > apex. So, ventilation perfusion ratio is higher at the apex (i.e, 3) than at the base (i.e, 0.6).

.

78. Ventilation-perfusion ratio is maximum at apex of lung due to:

a. Blood flow is increased considerablymore than ventilation

b. Blood flow is decreased considerablymore than ventilation

c. Base of lung has high ventilation inrelation to blood flow

d. Direct connection with trachea

.

Ans. B. V/Q ratio is higher at the apex because at the apex: blood flow < ventilation.

.

.

79. 20 Yrs old smoker while ascending stairs had sudden right sided sharp chest painand breathlessness, most likely cause ofbreathlessness:

a. Low o2 tension/pressure

b. Ventilation-perfusion mismatch

c. Low blood count

d. Depression of respiratory centre

e. Alveolar capillary block

.

Ans. B. Smoking causes emphysema, and emphysema destroys the walls of alveoli along with destruction of capillaries. This decreases perfusion, leading to mismatching of ventilation and perfusion. And any mismatching of ventilation and perfusion causes the partial pressure of oxygen in blood leaving the lungs to fall below the partial pressure that would occur with exact matching of ventilation and perfusion. The resulting decrease in partial pressure of oxygen causes breathlessness (dyspnea).

.

80. Under physiological conditions, highestPartial pressure of oxygen is observed in

a. Pulmonary Arteries

b. Pulmonary Capillaries

c. Pulmonary Veins

d. Aorta

.

Ans. C. Pulmonary artery contains deoxygenated blood, returning from the tissues, so it has low Po2. Pulmonary artery contains oxygenated blood which has high Po2.

.

81. In body, venous 02 saturation is highest in:

a. Hepatic vein

b. Renal vein

c. Coronary sinus

d. Jugular vein

.

Ans. B. In contrast to heart, which is a very active organ and receives only 5 – 10% of cardiac output, the kidneys receive 20 – 25% of cardiac output. So, the oxygen extraction of heart is very high while that of kidneys is very low. So, the venous oxygen saturation of the heart is the lowest, and that of kidneys is the highest.

.

82. Ventilation perfusion imaging is valuable in:

a. Pneumonia

b. Interstitial lung disease

c. Pulmonary embolism

d. Emphysema

.

Ans. C. In pulmonary embolism, perfusion is decreased while ventilation is normal. So, V/Q increases which helps in diagnosing pulmonary embolism.

.

83. Cyanosis is due to

a. Increased deoxyhemoglobin in blood

b. Decreased PO2 in blood.

c. Increased Co2 in blood

d. Increased myoglobin in blood

.

Ans. A. Cyanosis (bluish discoloration of the skin) occurs when the deoxyhemoglobin (hemoglobin not combined with oxygen) is greater than 5mg/dl.

.

84. Cyanosis

A. Can occur in anemic hypoxia

B. Does not depend on the total amount of hb in the blood

C. Is bluish discoloration of the tissues when concentration of reduced Hb Of the blood in the capillaries is more than 5g/dl

D. Its peripheral type is characterized by low arterial p02

E. Occurs in carbonmonoxide poisoning

.

Ans. C

.

85. Boy presented with central cyanosis,most likely cause is:

a. Cyanide poisoning

b. Histotoxic hypoxia

c. Deoxy-hemoglobin more than 5gm/dl

d. COPD

.

Ans. C (Refer to the above MCQ)

.

86. Respiratory acidosis occurs in:

a. Emphysema

b. Hyperventilation

c. Ingestion of excess amount of sodiumbicarbonate

d. Starvation

e. Vomiting

.

Ans. A. In any COPD disease (asthma, chronic bronchitis, emphysema), there is obstruction to expiration (Vs. restrictive lung diseases, where there is obstruction to inspiration), so the air high in CO2 remains in the lungs, and CO2 can not be expired properly. Accumulation of CO2 in the blood results in respiratory acidosis (Co2 is acidic because it combines with water to form carbonic acid).

.

87. In a patient under examination has leftshifted trachea and resonant note on right side. Diagnosis is

a. Right pneumothorax

b. Left pneumothorax

c. Pleural effusion

d. Consolidation

.

Ans. A. In pneumothorax, the pleural cavity freely communicates with alveoli, which freely communicate with outside atmosphere. So, air moves into the pleural cavity, increasing the intra-pleural pressure in pleural cavity. The increased intra-pleural pressure has two effects:

.

  • The increased intra-pleural pressure exerts pressure on mediastinum (trachea) and pushes it to the opposite side.
  • The alveoli collapse.

As the pleural cavity contains air, the percussion becomes resonant (air makes percussion resonant. Solids, e.g., tumors, make percussion dull).

.

88. A young man had a stab injury in the chest wall leading to Pneumothorax. What would be the most likely response?

A. Ipsilateral lung will collapse and chest wall will spring in

B. Ipsilateral lung will collapse and chest wall will spring out

C. Ipsilateral lung will collapse and opposite chest wall will spring out

O. Ipsilateral lung will expand and opposite lung will collapse

E. No change in lung volume and ipsilateral wall will spring out

.

Ans. B. If pleura is ruptured, air is introduced into the intrapleural space (pneumothorax), so the intrapleural pressure increases and becomes equal to atmospheric pressure. The lungs will collapse (their natural tendency) and the chest wall spring outward (its’s natural tendency).

.

89. The partial pressure of oxygen (p02) in the arterial blood is decreased in

A. Anemic hypoxia

B. Carbon monoxide poisoning

C. Cyanide poisoning

D. Hypovolemic shock

E. Hypoxic hypoxia

.

Ans. E

.

HYPOXIA

 

DEFINITION:Hypoxia means decreased oxygen delivery to the tissues..

TYPES OF HYPOXIA:There are four types of hypoxia:

.

         (1) HYPOXIC HYPOXIA

.

  • In hypoxic hypoxia, the Po2 in the blood is too low to saturate the HB (HB saturation is directly proportional to Po2).
  • Hypoxic hypoxia occurs by two mechanisms:

    .

  • A decrease in the amount of breathable oxygen. For example: at high altitudes where the atmospheric pressure is low.
  • Cardiopulmonary failure in which the lungs are unable to efficiently transfer oxygen from the alveoli to the blood. For example: COPD, interstitial lung diseases.

    .

    (2) ANEMIC HYPOXIA

    .

  • In anemic hypoxia, the amount of functional hemoglobin is too small, and hence the capacity of the blood to carry oxygen is too low.
  • Anemic hypoxia may occur by two mechanisms:

    .

  • The total amount of hemoglobin is reduced. For example, anemia, severe bleeding
  • Hemoglobin that is present is rendered nonfunctional. For example, carbon monoxide poisoning, in which CO combines with HB at the same site as oxygen, thus displacing oxygen from HB, and making HB unavailable for oxygen transport.

    .

    (3) STAGNANT HYPOXIA

    .

    In stagnant hypoxia, the blood is normal but the flow of blood to the tissues is reduced. For example: Congestive heart failure

    .

         (4) Histotoxic Hypoxia

.

  • In histotoxic hypoxia, the oxygen content in the blood is normal but, the tissue cells are poisoned and are therefore unable to make proper use of oxygen.
  • Although characteristically produced by cyanide, any agent that decreases cellular respiration may cause it. Some of these agents are narcotics, alcohol, formaldehyde, acetone, and certain anesthetic agents.

    .

    90. Normal oxygen tension with decreased oxygen carrying capacity isseen in:

    a. Histotoxic hypoxia

    b. Anemic hypoxia

    c. Stagnant hypoxia

    d. Hypoxic hypoxia

    .

    Ans. B

    .

    Total oxygen content of the blood = Oxygen carried by HB + Oxygen dissolved in blood

    .

    OXYGEN CARRYING CAPACITY

    .

  • Oxygen carrying capacity means “the maximum quantity of oxygen that can be carried by HB in a unit volume of blood”.
  • Each gram of Hb can combine with 1.34 ml of oxygen. Normally, blood contains 15g Hb/100ml of blood. Hence the oxygen carrying capacity of Hb is 15 X 1.34 = 20 ml oxygen/100 ml blood.
  • Oxygen carrying capacityis directly proportional to HB concentration. In anemia, HB concentration decreases, so the oxygen carrying capacity decreases.

    .

    OXYGEN TENSION (Po2)

    .

  • Oxygen tension (Po2) is directly proportional to the amount of dissolved oxygen (which does not depend upon HB concentration). In anemia, the amount of dissolved oxygen remains normal. So, the oxygen tension remains normal.

    .

    91. Half life of carboxyhemoglobin

    a. 5 min

    b. 2hr

    c. 6 hr

    d. 20 hr

    .

    Ans. C. Carboxy hemoglobin (COHb) is the hemoglobin that has carbon monoxide instead of the normal oxygen bound to it. The half life of COHb is 4 – 6 hours.

    .

92. Pulmonary 02 toxicity resultsfrom

a. Prolonged 02 therapy

b. Increase PCO2

c. At high altitude

d. In under water

.

Ans. A. Oxygen toxicity is a condition resulting from the harmful effects of breathing molecular oxygen at elevated partial pressures for prolonged periods. The lungs, as well as the remainder of the respiratory tract, are exposed to the highest concentration of oxygen in the human body and are therefore the first organs to show toxicity.

.

93. Which one of the following is caused by prolonged oxygen therapy:

a. Pulmonary embolism
b. Pulmonary edema
c. Tension penumothorax

.

Ans. B. Prolonged oxygen therapy may cause pulmonary oxygen toxicity which may result in inflammation in the lungs and pulmonary edema, which is evident on chest X- ray.

.

94. In chronic respiratory acidosis, how much Hco3 rise occurs for 10 mmHg of Pco2

a. 1 mEq

b. 3 mEq

c. 7 mEq

d. 10 mEq

e. 15 mEq

.

Ans. B. When Co2 combines with water, it forms carbonic acid (H2Co3) which then splits into hydrogen ions and bicarbonate ions. It means that when Co2 increases in respiratory acidosis, more Co2 binds with more water to form more carbonic acid, which then splits to form more hydrogen and bicarbonate ions.

.

Co2+H2O H2Co3 H++ HCo3+

.

ACUTE RESPIRATORY ACIDOSIS: In acute respiratory acidosis, for each 10 mmHg rise in Pco2, HCo3+ increases by 1 mEq.

.

CHRONIC RESPIRATORY ACIDOSIS: In chronic respiratory acidosis, for each 10 mmHg rise in Pco2, HCo3+ increases by 4, i.e, HCo3+ increases by 0.40 times the rise in Pco2 (0.4 X 10 = 4). As compared to acute resp. acidosis, the increase in HCo3+ is higher in chronic resp. acidosis because of renal compensation which attempts to reabsorb more bicarbonate.

.

So, here the best option is “3”, but “4” will be more appropriate if present in options.

.

95. PaCO2 increases during first minute of apnoea by:

a. 10 mmHg/min

b. 2 mmHg/min

c. 4 mrnHg/min

d. 6 mmHg/min

e. 8 mmHg/min

.

Ans. A. Apnoea means “suspension of external breathing”; it occurs when the breaths are shallow or there are pauses in breathing.

.

Apnoea increases PaCO2. In the first minute, PaCO2 increases by 6 mmHg. After the first minute, PaCO2 increases slowly at a rate of 3 mmHg/min.

.

96. Which of the following conditions causes a decrease in arterial 02 saturation without a decrease in 02 tension

a. Anemia

b. Carbon monoxide poisoning

c. low V/Q ratio

d. Hypoventilation

.

Ans.B. Oxygen in the blood exists in two forms: (1) Dissolved in blood (on which Po2 depends) (2) Attached to HB (on which HB saturation/O2 saturation depends).

.

Anemia decreases HB concentration only (Po2 and HB saturation is normal).

.

CO binds with HB, displacing oxygen from HB, and hence, decreases HB saturation/oxygen saturation (Po2 and HB concentration remain normal).

.

97. In which of the following most likelyreduction in arterial oxygen tension occurs

a. Anemia

b. CO poisoning

c. Moderate exercise

d. Hypoventilation

.

Ans.D. In hypoventilation, less oxygen comes to alveoli from the atmosphere, so less oxygen goes to the blood, so the dissolved oxygen decreases. As oxygen tension (also called partial pressure of oxygen, Po2) depends upon dissolved oxygen, oxygen tension will decrease. (Recall that anemia/polycythemiaaffect HB concentration only, and CO affects HB saturation only).

.

98. Blood in which of the following vessels normally has lowest Po2

a. Umbilical artery

b. Umbilical vein

c. Maternal artery

d. Maternal femoral vein.

Ans.A. Umbilical artery which returns blood from fetus to placenta has the lowest Po2.

.

99. In pulmonary artery embolism, what will happen:

a. Alveolar po2 increased

b. Alveolar Pco2 increased

c. Pulmonary artery pco2 increased

d. Pulmonary artery pco2 decreased

e. Ventilation perfusion ratio increases

.Ans. E. In pulmonary artery embolism, perfusion decreases but ventilation remains normal. So, ventilation/perfusion increases.

.

100. Hyperbaric oxygen is most likely used in

a. Tetanus

b. Gas gangrene

c. Botulism

.

Ans.B. Hyperbaric oxygen treatment is “giving oxygen at high pressure”.

.

Gas gangrene is caused by Clostridium perfringens. Clostridia lack superoxide dismutase, making them incapable of surviving in the oxygen-rich environment created within a hyperbaric chamber. This inhibits clostridial growth, exotoxin production, and exotoxin binding to host tissues. Hyperbaric oxygen therapy may also promote host polymorphonuclear cell function.

.

101. Hyperbaric oxygen is best explained as treatment of
A. CO poisoning
B.Gas gangrene

C. Decompression sickness

D. asthma

E. Type 2 respiratory failure

.

Ans. C. Option A, B, C all are correct but the most appropriate is option “C”.

.

  • CO poisoning: For CO poisoning, first 100% oxygen is given. If the patient does not improve, only then hyperbaric oxygen (HBO)is given.
  • Gas gangrene: The best treatment of gas gangrene is debridement, and not HBO.
  • Decompression sickness:This condition responds best to HBO.

    .

    So, the best option here is “decompression sickness” (also called caison’s disease).

    .

102. Patient having hypoxia with normal P02:

a. Atelectasis

b. Increase thickness of respiratory membrane.

c. AV shunt

d. Pulmonary fibrosis

.

Ans. C. An arteriovenous fistula (AV shunt) is an abnormal connection or passageway between an artery and a vein. It may be congenital, surgically created for hemodialysis treatments, or acquired due to pathologic process, such as trauma or erosion of an arterial aneurysm.

.

As the pressure in arteries is higher than veins, and blood flows from higher pressure to lower pressure, oxygenated blood from arteries flows to veins in AV shunts. The oxygenated blood bypasses the capillaries, resulting in hypoxia (decreased oxygen supply to the tissues), but Pao2 (partial pressure of oxygen in arterial blood) remains normal because Po2 depends upon dissolved oxygen which is not affected by AV shunt(In all left to right shunts, e.g, ASD, VSD, PDA, hypoxia occurs but Pao2 remains normal).

.

103. Which of the following forms of hypoxia is not corrected by 02 administration:

a. Poorly ventilated lung

b. Alveolar hypoventilation

c. Right to left shunts

d. Impairment of diffusion

.

Ans. C. In right to left shunts (tetralogy of fallot, transposition of great vessels), hypoxia is caused by shunting of deoxygenated blood from right side to the left side. Giving oxygen to the patient does not correct hypoxia inright to left shunt because hypoxia will exist as long as right to left shunt is present.

.

104. Damage to pneumotaxic center leads to:

a. Apneusis

b. Deep inspiration

c. Short expiratory phase

d. Irregular breathing

e. Apnea

.

Ans.B. Respiration is regulated by: (1) Respiratory center (2) Chemoreceptors

.

RESPIRATORY CENTER

.

DEFINITION:Groups of neurons located bilaterally in medulla and pons, which control all respects of respiration are called respiratory center”.

.

DIVISIONS: There are four divisions of respiratory center, as shown in the table below:.

.

Respiratory center

Location

Function

Dorsal respiratory group

Dorsal medulla

Upon stimulation, it initiatesinspiration

Ventral respiratory group

Ventral medulla

Takes part in forced respiration (not normal respiration

Pnemotaxic center

Upper pons

Decreases duration of inspiration, hence increases respiratory rate

Apneustic center

Lower pons

Prolongs duration of inspiration, hence decreases respiratory rate

.

As the pneumotaxic center decreases the duration of inspiration, so its damage will result in increased duration of respiration (deep inspiration).

.

CHEMORECEPTORS

.

DEFINITION:The chemoreceptors send signals to “dorsal respiratory group” of respiratory center to switch off inspiratory ramp signals in it, thus decreasing the durationof inspiration and increasing respiratory rate.

.

TYPES:There are two types of chemoreceptors: (1) Peripheral (2) Central

.

(A) PERIPHERAL RECEPTORS: The peripheral receptors (in carotid sinus + aortic arch) respond to:

.

  1. Po2 (<60 mmHg)
  2. Pco2
  3. Ph

(B) CENTRAL CHEMORECEPTORS: The central chemoreceptors (in medulla) respond to:

  1. Pco2
  2. Ph

.

105. Duration and rate of the inspiratory ramp signals are Controlled by impulses from the

A. Apneustic centre

B. Cerebral cortex

C hypothalamus

D. Peripheral chemoreceptors

E. Pneumotaxic centre

.

Ans. E

.

106. Regarding the respiratory center:

A. Ceases rhythmical activity if both vagi are cut

B. Is regulated by afferent vagal impulses

C. Is sensitive to ph alteration in the blood

D. Is situated in the medulla oblongata

E. Responds to impulses from the cerebral cortex

.

Ans. D

.

107. Vagus forms the efferent pathway of:

a. Hering – Breur reflex

b. J- receptors of pulmonary capillaries

c. Bain Bridge reflex

d. All of above

.

Ans.A. HERING – BREUER REFLEX:When lungs become over inflatedStretch receptors in bronchi and bronchioles are stimulated → Signals through vagiare sent to dorsal respiratory area → Inspiratory ramp signals are switched off → Duration of inspiration is decreased → Respiration rate is increased

.

108. Regarding Hearing breure reflex which statement is correct:

A. Afferents are carried by the vagi

B. It is stimulated when lungs are expanded

C. It causes decrease in blood pressure

D. It causes increase in respiratory rate

.

Ans. B

.

109. In hering breuer reflex, impulses from which receptors go Along the vagal afferents to the respiratory centers to limit The inspiration

A. Chemoreceptors in carotid and aortic bodies

B. Irritant receptors among airway epithelial cells

C. J-receptors close to pulmonary vessels

D. Proprioceptors around joints

E. Stretch receptors in the wail of bronchi and bronchioles

.

Ans. E. (See the above MCQ)

.

110. The central chemoreceptors in the medulla oblongata directly respond to changes in:

A. Arterial pco2

B. Arterial p02

C. Bicarbonate concentration in the csf

D. H+ concentration in the csf and brain interstitial fluid

E. Hydrogen ion concentration in the arterial blood

.

Ans. D. H+ is the major stimulus for central chemoreceptors; they directly respond to H+ concentration in the CSF.

.

111. Regarding the nervous control of respiration:

A. Dorsal respiratory group of neurons when stimulated always cause expiration

B. The basic rhythm of respiration is caused by ventral respiratory group of neurones.

C. The puemuotaxic center is not responsible for limiting the duration of inspiration and Decreasing respiratory rate.

D. Apneustic center in the lower part of pons provides extra drive for inspiration.

F. Herring- bruer reflex is controlled by the pneumotaxic center of respiration.

.

Ans. D

.

112. All of the following affect resting ventilation except:

a. Stretch receptors

b. J receptor

c. Oxygen

d. PCO2

.

Ans. A

.

[Option A] – The stretch receptors are present in smooth muscles of airways. They are stimulated by lungs over-distension, and decreases duration of inspiration and increases respiratory rate via Hering-Breuer reflex. As lungs over-distension does not occur at rest, stretch receptors do not affect resting ventilation.

.

[Option B] – J (juxta = close to) receptors are present in alveolar walls, close to capillaries. They are stimulated by engorgement of pulmonary capillaries (e.g, in left heart failure), and causes the sensation of dyspnea and increase respiratory rate. They can be active at rest and hence, can affect resting ventilation.

.

[option C + D] – Even if a person is at rest, changes in Po2 and Pco2 can activate chemoreceptors and affect resting ventilation.

.

113. Damage to which part of brain will result in central cease of respiration

a. Medulla

b. Midbrain

c. Pons

d. Cerebellum

.

Ans.A. Inspiration is initiated by “dorsal respiratory group” of respiratory center which is located in medulla. So, if medulla is damaged, inspiration will not be initiated and hence, respiration will stop.

.

114. A person having anemia, the systemic arterial blood oxygen content is decreased. What will be its effect on alveolar ventilation

a. Increases

b. Decreases

c. Remains the same

d. Increases than Decreases

e. Decreases than Increases

.

Ans. C. Anemia decreases the total oxygen content by decreasing hemoglobin concentration; it does not have any effect on Po2.

.

Alveolar ventilation is controlled by chemoreceptors which respond to Po2. As anemia does not affect Po2, so alveolar ventilation remains normal (unaffected) in anemia.

.

115. Peripheral and central chemoreceptorsrespond to

a. Increased PCO2

b. P02

c. Oxygen concentration

d. H+

.

Ans.A

.

[Option A] – Both central and peripheral chemoreceptors respond to increased Pco2.

.

[Option B + C] – Peripheral chemoreceptors respond to Po2 but central chemoreceptors do not.

.

[Option D] – Though both peripheral and central chemoreceptors respond to hydrogen ions (Ph), hydrogen ions in systemic blood can not cross the blood brain barrier and hence, changes in hydrogen ions in systemic blood have effect only on peripheral receptors but do not have effect on central chemoreceptors. This fact makes this option less suitable.

.

116. Low P02 Stimulates respiratory center through

a. Directly acting on medulla

b. Stimulating peripheral chemoreceptor

c. Acting on central chemoreceptor

.

Ans.B

.

  • Po2 is detected by peripheral chemoreceptors only.
  • Pco2 is mainly detected by central chemoreceptors, but it can also be detected by peripheral chemoreceptors.

    .

117. Peripheral chemoreceptors

a. Have increased 02 consumption

b. Detect increased 002/ PH

c. Respond to P02 less than 60 mmhg

d. Respond to po2 less than 40 mmhg

.

Ans.C. Normally, respiration is controlled by central chemoreceptors. Peripheral chemoreceptors take over control only when Pao2 < 60 mmHg, or Fio2 (fraction of oxygen in inspired air) < 17% (Normally: Pao2 = 60 – 80, Fio2= 21%)

.

118. Respiratory center in brain is sensitive to bloodchanges in

a. PCO2

b. P02

C. pH

d. H ion concentration

.

Ans.A. Central chemoreceptors respond to Pco2 and hydrogen ions (Ph). But hydrogen ions in systemic blood can not cross the blood brain barrier, while Co2 in systemic blood can cross the blood brain barrier. So, central chemoreceptors respond only to blood changes in Pco2

.

119. Rate of respiration is controlled by

a. PCO2

b. PH

c. PO2

d.Hco3

.

Ans. A. Normally, respiratory rate is controlled by central chemoreceptors which respond to both Pco2 and Ph (hydrogen ions). As hydrogen ions can not pass through blood brain barrier, PCo2 mainly controls respiratory rate under normal conditions.

.

110. Irritant receptors in airways

a. Present near epithelium

b. Stimulated by stretching

c. Causes coughing andbroncoconstriction

d. Rapidly adapting

.

Ans.C

.

  • Pulmonary Irritant Receptors are sensors present within the respiratory epithelium which can sense and respond to a variety of chemical irritants.
  • Afferent signals from these sensory cells may initiate coughing in response to a variety of inhaled irritants and might induce bronchoconstriction in those with asthma.

    .

111. Sympathetic stimulation causes

a. Bronchodilation

b. Vasodilation in skin

c. Bronchoconstriction

d. Increased GIT motility

.

Ans.A

.

  • Sympathetic stimulation causes bronchodilatation (that’s why beta agonists are used to treat bronchoconstriction in patients of asthma).
  • Parasympathetic system causes bronchoconstriction.

    .

    112. In which of the following vein, venous Pa02 and PaCO2 is appropriate witharterial blood

    a. Femoral vein

    b. Anticubetal vein

    c. Intenal jugular vein

    d. Subclavian vein

    e. Veins on dorsum of a warm hand

    .

    Ans.E. Hand heating (above 37%) is effective in arterializing venous blood.

    .

    113. Hypoxic pulmonary vasoconstriction accentuate by :

    a. Increase P02

    b. Decrease P02

    c. Increase PCO2

    d. Decrease PCO2

    .

    Ans.C. Hypoxia causes pulmonary vasoconstriction. This response is accentuated (increased) by ↑ Pco2 (hypercarbia).

    .

    114. Regarding respiratory center

    a. Basic rhythm lies in dorsal medulla

    b. Apneustic center shortens inspiration

    c. Pneumotaxic center lengthens inspiration

    d. Medullary chemoreceptors directly respond to hypoxia

    .

    Ans.A

    .

  • The dorsal respiratory group (DRG) is located in the dorsomedial region of the medulla.
  • The DRG is involved in the generation of respiratory rhythm, and is primarily responsible for the generation of inspiration.
  • It is stimulated by apneustic center and inhibited by Pneumotaxic center.

    .

115. Peripheral chemoreceptor do not respond to:

a. Increase plasma Na+

b. H+

c. 02

d. CO2

.

Ans. A. Peripheral chemoreceptors respond to ↓Po2, ↑Pco2, and ↓Ph (↑H+).

.

116. The stimulatory effect of decreased arterial P02 on the Respiratory center is through

A. Central chemoreceptors in the medulla oblongata

B. Direct action on the respiratory center

C. peripheral chemoreceptors in carotid and aortic bodies

D. The carotid sinus receptors

E. The cerebral cortex

.

Ans. C

.

117. 02-Hb dissociation curve delivering 02 in normal limits will have:

a. Pa02 40mm Hg SO2 60%

b. Pa02 68mmHg SO2 78%

c. Pa02 96mmHg SO2 90%

d. Pa02 123mmHg SO2 98%

e. Pa02 256mmHg SO2 99%

.

Ans.C. PaO2 means “partial pressure of oxygen in arterial blood”.

.

So2 means “oxygen saturation (or HB saturation), i.e, the percentage of HB combined with oxygen”. So2 increases with the increase in Po2.

.

PO2 (mmHg)

SO2 (%)

100

100

40

75

26 (T50)

50

.

Normally, PaO2 = 80 – 100 mmHg, and SO2 = 90 – 100%. So, option “C” is correct.

.

118. In right shift of oxygen dissociation curve, the most likely value of T50 would be

a. 20

b. 18

c. 26

d. 35

.

Ans. C. HB saturation varies with PO2. T50 (or P50) is PO2 at which 50% of HB is saturated. Normally, HB is 50% saturated when PO2 is equal to 26 mmHg. So T50 is equal to 26 mmHg.

.

119. The P02 (partial pressure of 02) in the blood at which Hemoglobin is 50% saturated with oxygen is about

A. 10 mmhg

B. 26 mmhg

C. 35 mmhg

D. 50 mmhg

E. 6 mmhg

.

Ans. B

.

120. Which of following causes shift of 02-Hb dissociation curve to left

a. Decrease in temperature

b. Increase in temp.

c. Increase 2-3DPG

d. Increase in H ions

.

Ans.A

.

Factors Causing Right Shift Of O2 – HB Curve

Factors Causing Left Shift Of O2 – HB Curve

↓Temperature

↑Temperature

↓H+ (↑Ph)

↑H+ (↓Ph)

↓Co2

↑Co2

↓DPG

↑DPG

.

Stored outdated blood (which loses DPG)

.

Fetal HB (has very high affinity for O2)

.

CO poisoning (Decreases HB saturation, but increases affinity of remaining unbound sites for oxygen)

.

121. An increase in which of the following parameters will shift the 02 dissociation curve to the left:

a. Temperature

b. Partial pressure of CO2

c. 2,3 DPG concentration

d. Oxygen affinity of hemoglobin

.

Ans.D

.

LEFT SHIFT

.

(1) ↑HB AFFINITY

.

If the affinity of HB increases for oxygen (e.g, fetal HB, CO poisoning), oxygen will not be dissociated from HB, more oxygen will be combined to HB, HB saturation will increase, and hence, the curve will be shifted to the left.

.

(2) ↓Po2

.

Decreases in Po2 decreases HB saturation, thus shifting the curve to the left.

.

RIGHT SHIFT

.

(1) ↓HB AFFINITY

.

If the oxygen affinity of HB decreases, less oxygen is combined with HB, so HB saturation decreases, shifting the curve to the right.

.

(2) ↑Po2

.

Increase in Po2 increases HB saturation, thus shifting the curve to the right.

.

122. Right ward shift of Oxygen Hb dissociation curve is due to decrease in

A. High attitude

B. Temperature

C. PH

D. H ion

.

Ans. C

.

123. One of the following led to decrease release of oxygen from hemoglobin

A. decrease in arterial pressure of oxygen

B. decrease in temperature

C. increase in atrterlal pco2

D. increase in atrterial H ions

E. increase in temperature

.

Ans. B

.

124. Which of the following shifts the 02-Hb dissociation curve to right:

a. Acidosis

b. Alkalosis

c. increase PH

d. Nitrous oxide

e. Fetal,hernoglobin

.

Ans.A. Acidosis (↑H+, ↓Ph) shifts the curve to the right. Alkalosis (↓H+, ↑Ph) shifts the curve to the left.

.

125. Right shift of Hb dissociation curve is caused by decrease in

a. Temperature

b. 2,3-DPG

c. Partial pressure of CO2

d. PH

.

Ans. D (Refer to the above MCQ)

.

126. Oxyhemoglobin dissociation curve is shifted to the left By

A. Fetal hemoglobin

B. Increased 2 3 diphosphoglycerate

C. Increased pco2

D. Increased temperature

E. Low ph

.

Ans. A. Fetal HB has high affinity for HB (so that it can effectively bind to placental oxygen); this increases HB saturation and shifts the curve to the left.

.

127. CO2 is transported from the alveoli of lungs into blood in lungs via:

a. Diffusion

b. Active transport

c. Facilitated diffusion

d. Secondary active transport

e. Osmosis

.

Ans.A. Gases are always transported by simple diffusion.

.

.

http://images.slideplayer.com/1/253726/slides/slide_10.jpg

.

.

 

 

 

 

 

 

 

 

 

 

 

 

 

128. CO2 diffusion in lungs is greater than O2 due to
A.Insolubility coefficient
B.IncCO2 partial pressur in lungs
C.Incdensity of CO2

.

Ans.A. Co2 is 20 times more soluble in blood than oxygen. Therefore, Co2 diffuses out first before oxygen can diffuse in.

.

129. Which one of the following describes the form in which Most of the carbon dioxide formed in the tissues of a Human is returned to the lungs

A. Bicarbonate ions in WBCs

B. Bicarbonate ions in plasma

C. Carbonic acid dissolved in lymph

D. Carbonic anhydrase in R.B.C

E. Carboxyhaemoglobin in R.B.C

.

Ans. B. 90% of Co2 is transported as “bicarbonate in the blood” (and not as bicarbonate in RBCs)

.

130. Following are the chemical forms in which carbon dioxide is Transported blood except:

A. Co2 (carbon dioxide)

B. Carbamino proteins

C. Carboxyhemoglobin

D. H2co3 (carbonic acid)

E. Hco3 (bicarbonate)

.

Ans. C

.

131. Solubility co-efficient of CO2 is

a. 0.018

b. 0.012

c. 0.003

d. 0.57

.

Ans.D. The solubility coefficient for CO2 is 0.57 and for oxygen is 0.024

.

132. Gas having maximum diffusion capacity in body fluids
a. CO2
b. CO
c. He
d. O2

.

Ans. A. The diffusion coefficient (capacity) of: CO2 > O2 > CO > Nitrogen> He

.

134. Carotid bodies respond to

a. Increased blood H+

b. Increased blood Co2

c. Alkalosis

d. Sympathetic stimulation

.

Ans. B. The peripheral chemoreceptors, located in carotid bodies, at the bifurcation of carotid artery, respond to:

.

  1. ↓Po2
  2. ↑Pco2
  3. ↓Ph (↑H+)

.

135. Deep inspiration leads to

a. Increased cardiac output

b. Increased pulmonary compliance

c. Decrease venous return

d. S1 splitting

.

Ans. D. Inspiration decreases the intrapleural pressure (IP), i.e, the IP becomes more negative. This produces a suction force which pulls blood from all over the body (including systemic veins, and the left heart) towards the lungs. This causes increased venous return, increased cardiac output from the right heart, and decreased cardiac output from the left heart.

As the cardiac output of the right heart is greater than the left heart during inspiration, the pulmonary valve takes longer than aortic valve to close, so the aortic valve closes earlier than pulmonary valve, resulting in splitting of S2 (second heart sound).

.

134. During muscular exercise, all of the following are seen except:

a. Increase in blood flow to muscles

b. Stroke volume increases

c. 02 dissociation curve shifts to left

d. 02 consumption increases

.

Ans. C

.

[Option A] – Increased lactic acid causes vasodilatation, which increases blood flow to the muscles.

.

[Option B] – Muscle contractions push blood towards the heart, so venous return increases.

.

[Option C] – ↑Temperature, ↑Co2 production, and ↓Ph shifts the curve to the right.

.

[Option D]Exercising muscles require more energy and more ATP, so oxygen consumption increases.

.

135. Early asthma attack is associated with:

a. Increased oxygen tension

b. Increased lung compliance

c. Decreased CO2 tension

d. Decreased lung compliance

.

Ans. B. COPD:In COPD (e.g., asthma), air trapping occurs which increases compliance; it means that increase in residual volume increases compliance.

.

RESTRICTIVE LUNG DISEASES:In restrictive lung diseases, lungs become stiff and hence, lungs compliance decreases.

.

136. Increased airway resistance causes:

a. Decrease in residual volume

b. Increase in residual volume

c. Increase in tidal volume

.

Ans. B. In COPD, the airway resistance increases due to constriction of the airways. It decreases the volume of air that can be expired, thus increasing residual volume.

.

137. Pulmonary vascular resistance increases in:

a. High altitude

b. Increase in O2

c. Increase in Co2

.

Ans. A

Resistance= 1(Radius)4..

.

At high altitude, atmospheric pressure is low which causes hypoxia. In contrast to other parts of the body, where hypoxia causes vasodilatation, it causes vasoconstriction in lungs → Vasoconstriction causes decrease inradius of blood vessels → Decrease in radius increases resistance

.

138. Erythropoetin production will increase

A. At high altitude

B. During exercise

C. During sleep

D. Flying in aeroplane

E. While performing valsalva manoeuvre

.

Ans. A. At high altitude, hypoxia occurs; hypoxia is a stimulus ofr erythropoietin production.

.

139. Measurement of intravascular pressure by a pulmonary catheter should be done:
a. At the end of expiration

b. At the peak of inspiration

c. During mid expiration

d. During inspiration

.

Ans. A. Intrapulmonary pressure can be recordedcorrectly only when extravascular pressure is zero, and it is zero (i.e., equal to atmospheric pressure) at the end of expiration (and at the end of inspiration); pressure in lungs is always equal to zero whenever there is no flow of air.

.

140. Cyanosis in traumais interpreted as:

a. Early sign of hypoxia

b. Late sign of hypoxia

.

Ans. B. Early sign of hypoxia is tachycardia, and late sign of hypoxia is cyanosis.

141. On the summit of Mt. Everest, where the barometric pressure is about 250 mm Hg, the partial pressure of O2 in mm Hg is about
A. 0.1
B. 0.5
C. 5
D. 50
E. 100

.

Ans. D. In ambient gas, the partial pressure of a gas can be calculated by the flowing formula:

Pgas= Fgas × Patm

Where Pgas = Partial pressure of gas in ambient air, Fgas = Concentration (fraction) of gas, Patm = atmospheric pressure

.

Here, Patm = 250, and fraction of oxygen in atmospheric pressure is always equal to 21% = 0.21, so:

.

PO2=0.21 ×250=52.5 50 mmHg

.

142. The forced vital capacity is
A. The amount of air that normally moves into (or out of) the lung with each respiration
B. The amount of air that enters the lung but does not participate in gas exchange
C. The amount of air expired after maximal expiratory effort
D. The largest amount of gas that can be moved into and out of the lungs in 1 min
.

Ans. C

143. The tidal volume is
A. The amount of air that normally moves into (or out of) the lung with each respiration.
B. The amount of air that enters the lung but does not participate in gas exchang

C. The amount of air expired afer maximal expiratory ef ort.
D. The amount of gas that can be moved into and out of the lungs in 1 min.
.

Ans. A

144. Which of the following is responsible for the movement of O2 from the alveoli into the blood in the pulmonary capillaries?
A. Active transport
B. Filtration
C. Secondary active transport
D. Facilitated diffusion
E. Passive diffusion
.

Ans. E. All gases in the body cross the membranes by passive (simple) diffusion.

145. Airway resistance
A. Is increased if the lungs are removed and inflated with saline
B.Does not affect the work of breathing
C.Is increased in paraplegic patients
D.Is increased in following bronchial smooth muscle contraction.
E. Makes up 80% of the work of breathing
.

Ans. D.

Resistance=1(Radius)4..

.

  • Bronchial smooth muscles contraction (e.g., by sympathetic stimulation) causes bronchoconstriction, thus decreasing radius and increasing resistance.
  • Bronchial smooth muscles relaxation (e.g, by parasympathetic system) has the opposite effects.

 

146. Surfactant lining the alveoli
A. Helps prevent alveolar collapse
B. Is produced in alveolar type I cells and secreted into the alveolus
C. Is increased in the lungs of heavy smokers
D. Is a glycolipid complex

.

Ans. A. Surfactant decreases surface tension (a collapsing force), thus increasing compliance, and prevent alveolar collapse. In RDS, decreased surfactant results in lungs collapse (atelectasis).

 

147. A deficiency of pulmonary surfactant would result in:

A. Decreased change in intrapleural pressure to achieve the tidal volume

B. Decreased surface tension in the alveoli

C. Decreased work of breathing

D. Increased functional residual capacity (frc)

E. Patchy atelectasis

.

Ans. E. Surfactant deficiency (e.g., in RDS) causes atelectasis (lungs collapse).

.

148. Surfactant:

A. It increases the surface tension of the small alveoli.

B. It stabilizes the size of alveoli during expiration and prevents its collapse.

C. Presence of surfactant can cause pulmonary edema.

D. It is a phospholipid and protein mixture secreted by the cells of pleura.

E. Respiratory distress syndrome of the newborn babies is due to its Excessive production at the time of birth

.

Ans. B

.

149. Which of the following has the greatest effect on the ability of blood to transport oxygen?
A. Capacity of the blood to dissolve oxygen
B. Amount of hemoglobin in the blood
C. Ph of plasma
D. CO2 content of red blood cells
E. Temperature of the blood
.

Ans. B. Oxygen is not very soluble in blood, so 99% of oxygen is transported in blood attached to HB. Therefore, decreases in HB (e.g., in anemia, hemorrhage) significantly affect the transport of oxygen. (Recall that most of the Co2 is transported as bicarbonate ions in the blood).

 

150. In comparing uncompensated respiratory acidosis and uncompensated metabolic acidosis which one of the following is true?
A. Plasma pH change is always greater in uncompensated respiratory acidosis compared to uncompensated metabolic acidosis.
B. There are no compensation mechanisms for respiratory acidosis, whereas there is respiratory compensation for metabolic acidosis.
C. Uncompensated respiratory acidosis involves changes in plasma [HCO3–], whereas plasma [HCO3–] is unchanged in uncompensated metabolic acidosis.
D. Uncompensated respiratory acidosis is associated with a change in Pco2, whereas in uncompensated metabolic acidosis Pco2 is constant

.

Ans. D. Respiratory acidosis occurs due to increase in Pco2.

.

Metabolic acidosis occurs due to increase in fixed acid (i.e, any acid except Co2). So, in uncompensated metabolic acidosis, Pco2 will be constant. However, when respiratory compensation occurs in metabolic acidosis, Pco2 will be decreased.

.

151. The main respiratory control neurons
A. Send out regular bursts of impulses to expiratory muscles during quiet respiration
B. Are unaffected by stimulation of pain receptors
C. Are located in the pons
D. Send out regular bursts of impulses to inspiratory muscles during quiet respiration
E.Are unaffected by impulses from the cerebral cortex
.

Ans. D. The main respiratory neurons are “dorsal respiratory group” of respiratory center. They are located in dorsal medulla, and upon stimulation, they send impulses to inspiratory muscles to initiate respiration.

 

152. Intravenous lactic acid increases ventilation. The receptors responsible for this effect are located in the
A. medulla oblongata.
B. carotid bodies.
C. lung parenchyma.
D. aortic baroreceptors.
E. trachea and large bronchi.
.

Ans. B. Increase in lactic acid causes ↓Ph. Though both peripheral (carotid bodies) and central (medulla) chemoreceptors respond to Ph, hydrogen ions in systemic blood can not cross blood brain barrier. So, intravenous lactic acid will have effect only on peripheral (carotid) chemoreceptors.

 

153. Spontaneous respiration ceases after
A. transection of the brain stem above the pons.
B. transection of the brain stem at the caudal end of the
medulla.
C. bilateral vagotomy.
D. bilateral vagotomy combined with transection of the brain
stem at the superior border of the pons.
E. transection of the spinal cord at the level of the first thoracic segment.
.

Ans. B. The “dorsal respiratory center” (located in dorsal medulla) is the main drive for rsepiration because it initiates inspiration. So, transaction of the brain stem at the caudal end of medulla would result in cease of signals to inspiratory muscles from medulla, resulting in cease of respiration.

 

154. The following physiologic events that occur in vivo are listed in random order:(1) decreased CSF pH;(2) increased arterial Pco2 ;(3) increased CSF Pco2 ;(4) stimulation of medullary chemoreceptors;(5) increased alveolar Pco2 . What is the usual sequence in which they occur when they affect respiration?
A. 1, 2, 3, 4, 5
B. 4, 1, 3, 2, 5
C. 3, 4, 5, 1, 2
D. 5, 2, 3, 1, 4
E. 5, 3, 2, 4, 1

.

Ans. D. [5]In hypoventilation, alveolar Pco2 increases as all the Co2 in alveoli can not be expired. [2] Due to increased Pco2 in alveoli, Co2 in the blood can not move out into the alveoli, so arterial Pco2 increases. [3] As Co2 can cross the blood brain barrier, it passes from blood to CSF [1] As Co2 is acidic, increased Co2 in CSF decreases CSF Ph. [4] ↓Ph stimulates medullary chemoreceptors to increase respiratory rate.

.

155. The following events that occur in the carotid bodies when
they are exposed to hypoxia are listed in random order:
(1) depolarization of type I glomus cells;(2) excitation of
afferent nerve endings;(3) reduced conductance of hypoxia sensitive K+ channels in type I glomus cells;(4) Ca2+ entry into type I glomus cells;(5) decreased K+ efflux.
What is the usual sequence in which they occur on
exposure to hypoxia?
A. 1, 3, 4, 5, 2
B. 1, 4, 2, 5, 3
C. 3, 4, 5, 1, 2
D. 3, 1, 4, 5, 2
E. 3, 5, 1, 4, 2
.

Ans. E. [3] Hypoxia decreases the conduction of K+ channels. [5] So, K+ efflux is decreased, and K= remains inside the cells. [1] As K+ efflux causes repolarisation, blocking K+ efflux causes depolarization of the cells.[4] Depolarization of the cells result in Ca2+ influx.

[2] This initiates an impulse that is transmitted to afferent nerves.

 

156. njection of a drug that stimulates the carotid bodies would be expected to cause
A. a decrease in the pH of arterial blood.
B. a decrease in the Pco2 of arterial blood.
C. an increase in the HCO3– concentration of arterial blood.
D. an increase in urinary Na+ excretion.
E. an increase in plasma Cl– .
.

Ans. B. The carotid bodies contain peripheral chemoreceptors. Upon stimulation, the chemoreceptors increases respiratory rate (hyperventilation). The ↑RR increases removal of Co2 from the blood, so the blood Ph decreases.

157. Variations in which of the following components of blood or CSF do not affect respiration?
A. Arterial HCO3– concentration
B. Arterial H+ concentration
C. Arterial Na+ concentration
D. CSF CO2 concentration
E. CSF H+ concentration

.

Ans. C. Na+ has no effect on chemoreceptors, so Na= do not affect respiration.

.

158. Maximum reduction in the vital capacity is likely to be seen During anaesthesia in

A. Kidney position

B. Lateral position

C. Lithotomy position

D. Prone position

E. Supine position

.

Ans. C

.

159. In quiet breathing, the principal inspiratory muscle is

A diaphragm

B. Internal intercostal

C. Levator costae

D. Pectoralis major

E. Scalenus anterior

.Ans A

.

160. Clinically, central cyanosis is evident when reduced haemoglobin in the circulation is more than

A. 1 oin/d1

B. 2 gm/dl

C. 3 gm/di

D. 4 gm/di

E. 5 gm/di

.

Ans E

.

161. Pulse oxymeter may give false reading if

A. Patient is dark skin

B. Increased conc. of HB-A

C. Increased conc. of HB-F

D. Aortic stenosis

.

Ans. A

.

162. A 60 years old patient of emphysema presents with dyspnea of sudden Onset with chest pain. On examination, pulse 110/min; bp 105/65 mmhg. Trachea shifted towards right, hyper resonance percussion note, and Decreased breath sounds on left side. The investigation which will be Most helpful in the diagnosis is:

A. Arterial blood gases

B. Cardiac enzymes

C. Chest x-ray

D. E.c.g.

E. Sputum routine e4mination

.

Ans. C

.

163. During the transport of carbondioxide in the blood, the Amount of carbondioxide dissolved in the plasma at 45 mmhHg is Following:

A. 0.3 %

B. 2.7 %

C. 23 %

D. 7.5 %

E. 70%

.

Ans. B

.

164. Basement membrane of lungs alveoli:

A. Comprises of type ii collagen

B. Contributes in the formation of surfactant

C. Has chondroitin sulphate as a constituent component

D. Is a mucopolysaccharide

E. Separates epithelium from endothelium in alveolar wall

.

Ans. E. Layers of respiratory membrane include:

.

  1. Surfactant
  2. thin layer of fluid-water
  3. Alveolar epithelium
  4. Epithelial basement membrane
  5. Interstitial space
  6. Capillary basement membrane
  7. Capillary endothelium

    .

    .

    1.jpg

    .

    .

     

    165. Maximum reduction in the vital capacity is likely to be seen during anaesthesia in

    A. Kidney position

    B. Lateral position

    C. Lithotomy position

    D. Prone position

    E. Supine position

    .

    Ans. C

    .

    166.Underlying mechanism for breathlessness in this patient is:

    A. Alveolar capillary block

    B. Depression of respiratory center

    C. Low blood count

    D. Low oxygen pressure

    E. Ventilation – perfusion mismatch

    .

    Ans. E

    .

    166. Normal parameters for Spo2 reading in a healthy person is

    A. 60-70%

    B. 70-80%

    C. 80-90%

    D. 85-90%

    E. 90-100%

.

Ans. E. Sp02 can be broken down into the following components: ‘S’ indicates saturation; P indicates pulse. Normal range: 90 – 100%

.

167. Compared with adult hemoglobin, fetal hemoglobin has a

A. Different beta- chain

B. Higher p1/2 for oxygen

C. Lower bohr effect

D. Lower affinity for 2,3-bisphosphoglycrate

E. Lower life span

.

Ans. C

.

Haldane effect

.

Deoxygenation of the blood increases its ability to carry carbon dioxide; this property is the Haldane effect. Conversely, oxygenated blood has a reduced capacity for carbon dioxide.

.

BOHR EFFECT

.

Hemoglobin’s oxygen binding affinity is inversely related both to acidity (Hydrogen ions concentration) and to the concentration of carbon dioxide. That is to say, a decrease in blood pH which leads to an increase in blood CO2 concentration will result in hemoglobin proteins releasing their load of oxygen.

.

DIFFERENCE BETWEEN HALDANE AND BOHR EFFECT

.

The simplest way to differentiate the two effects is to identify which molecule is the cause of the change.

.

The Haldane effect describes how oxygen concentrations determine hemoglobin’s affinity for carbon dioxide. For example, high oxygen concentrations enhance the unloading of carbon dioxide. The converse is also true: low oxygen concentrations promote loading of carbon dioxide onto hemoglobin. In both situations, it is oxygen that causes the change in carbon dioxide levels.

.

The Bohr effect, on the other hand, describes how carbon dioxide and H+ affect the affinity of hemoglobin for oxygen. High CO2 and H+ concentrations cause decreases in affinity for oxygen, while low concentrations cause high affinity for oxygen.

.

To further illustrate the difference, it might help to look at specific examples. In the lungs, when hemoglobin loaded with carbon dioxide is exposed to high oxygen levels, hemoglobin’s affinity for carbon dioxide decreases. This is an example of the Haldane effect.

.

In active muscles, carbon dioxide and H+ levels are high. Oxygenated blood that flows past is affected by these conditions, and the affinity of hemoglobin for oxygen is decreased, allowing oxygen to be transferred to the tissues. Because we are looking at the situation from the perpsective of carbon dioxide changing oxygen affinity, this is an example of the Bohr effect.

.

So, haldane effect describes the loading of O2 in lungs and bohr effect describes the release of O2 in tissues.

.

EFFECT OF FETAL HEMOGLOBIN ON BOHR EFFECT

.

Fetal hemoglobin has high affinity for oxygen as compared to adult hemoglobin. It means that fetal hemoglobin do not release oxygen easily, so it has lower bohr effect.

.

EFFECT OF HALDANE AND BOHR EFFECT ON OXYGEN-HB DISSOCIATION CURVE

.

  • BohR effect shift the curve to the Right.
  • Haldane effect shifts the curve to the left.

.

168. Left shift of oxygen-Hb curve is called:

a. Bohr effect

b. Haldane effect

c. Chloride shift

d. Dead space

.

Ans. B

.

169. During aclimatization at high attitude, there is

A. A marked increase in p 50 value

B. Acidosis

C. An increase in pulmonary ventilation

D. Decreasesd erythropoietin secretion

E. Pulmonary vasodilation

.

Ans. C

.

.

Pgas= Fgas  (Patm Pwater)

.

Where Pgas= Partial pressure of the gas, Fgas = Concentration (fraction) of the gas, Patm = Atmospheric pressure, Pwater = Partial pressure of water vapors = 47 mmHg

.

This equation shows that partial pressure of oxygen in the inspired air is directly proportional to atmospheric pressure. At high altitude, the atmospheric pressure is low. So, partial pressure of oxygen in the inspired air is low. As Po2 of alveoli and that of pulmonary capillaries blood are in equilibrium with each other, ↓Po2 in alveoli results in ↓Po2 in the blood. ↓Po2 stimulates peripheral chemoreceptors, which results in hyperventilation (i.e., ↑Pulmonary ventilation).

.

170. During deep sea diving, which of the symptom of acute Oxygen deficiency may prove fatal?

A. Disturbances of vision

B. Dizziness

C. Irritability

D. Nausea

E. Seizures

.

Ans. E

.

171. Pulmonary artery pressure is increased

A. By nitrites

B. By sympathetic stimulation

C. In essential hypertension

D. In exercise

E. In hypoxia

.

Ans. E. Hypoxia causes vasoconstriction in the lungs, resulting in pulmonary hypertension.

.

172. Increasing alveolar ventilation increases the blood ph because

A. It activates neural mechanism that removes acid from the blood

B. It decreases the pco2 of the blood

C. It increases the p02 of the blood

D. It makes haemoglobin a strong acid

E. The increased muscle work generates more carbon dioxide

.

Ans. B. Hypoventilation causes respiratory acidosis due to accumulation of Co2 in the body, while hyperventilation causes respiratory alkalosis due to ↓Co2 in the body.

.

173. Out of the following only one is not vapour at room

Temperature

A. Carbon dioxide

B. Ether

C. Methane

D. Nitrous oxide

E. Oxygen

.

Ans. E

.

174. A person reveals ph 7.35-7.45, paco2 35-45 hg, Pao2>70 mmhg, Hco3 22-26 meq/l, base excess 2.0 to 2.0meq/l and cao2 16-22ml02/dl. The Individual is most likely breathing

A. At high altitude

B. At sea level

C. During the last trimester of pregnancy

D. In a closed space (vessel)

E. Under high atmospheric pressure

.

Ans. B

.

175. Lung compliance increases in

A. Bronchial obstruction

B. Decreased intrapleural pressure

C. Increased elastic recoil

D. old age

E. Stiff lung

.

Ans. D. Lungs of an infant are more elastic than that of an old man. As elastic recoil of the lungs opposes lungs expansion and hence, decreases lungs compliance, so lungs of an old man are more compliant than that of an infant.

.

176. A capnometer is used for measuring end tidal Co2. The other Possible use of this device is

A. Indicator of air embolism

B. Indicator of cardiac arrhythmias

C. Measuring oxyhaemoglohin concentration

D. Measuring saturation of carbon dioxide

E. U!travoiletanalyser

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Ans. A. Capnometer is a monitoring device that measures and numerically displays the concentration of carbon dioxide in exhaled air.

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http://www.zoll.com/uploadedImages/Public_Site/Products/R_Series_Defibrillators/capnogram.jpg

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177. Charle’s gas law is applicable to the volume of a gas

A. At a constant temperature

B. At constant pressure

C. For a non-ideal gas

D. For a specific gas

E. With absolute pressure

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Ans. B. Charle’s law states that “At a fixed pressure, the volume of a gas is proportional to the temperature of the gas.” This law describes how a gas expands as the temperature increases; conversely, a decrease in temperature will lead to a decrease in volume.

178. Vasodilation at high altitudes occurs due to

A. NO

B. Increase 02

C. Increase CO2

D. Increase HCO3

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Ans. C

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179. Neonates are most likely deficient in which cells

A. Pneumocytes

B. Surfactant

C. Lymphocytes

D. Eisinophits

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Ans. B

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180. Respiratory centre is sensitive to

A. CO2

B. 02

C. K+

D. H+

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Ans. A

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181. Person exposed to asbestose. In order to know his Inspiratory Reserve voiurne(IRY), what Combination is necessary?

A. T.V + E.R.V

B. T.V + V.C

C. T.V + R.V

D. T.V + F.R.C

E. T.V + I.R.V

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Ans. E

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182. Person during respiration uses accessory muscles and nasal flaring, which type of respiration He has

A. Labor breathing

B. Shallow breathing

C. Periodic breathing

D. Chyne stokes breathing

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Ans. A. Labored respiration or labored breathing is an abnormal respiration characterized by evidence of increased effort to breathe, including the use of accessory muscles of respiration, stridor, grunting, or nasal flaring.

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183. In respiratory tract exchange of gasses occur at

A. Respiratory bronchioles

B. Alveolar duct

C. Alveoli

D. Alveolar ac

E. Alveolar duct

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Ans. C

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184. During forced expiration, the most likely statement regarding the physiology of Respiratory system

A. Bronchi collapse during expiration

B. There is dynamic collapse of respiratory bronchioles due to thin walls

C. The volume after forced expiration is called tidal volume

D. Vital capacity is the sum of inspiratory plus expiratory reserve volumes

E. None of the above is correct

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Ans. B

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185. The substance which will cross blood brain barrier first

A. CO

B. CO2

C. 02

D. Blood

E None of them

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Ans. B. Co2 has very high diffusion capacity, and it can diffuse across the membranes very rapidly.

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186. Which statement is correct about cyanosis

A. when deoxygenated herisnembin exceeds 2 mg

B. When deoxygenated hemoglobin exceeds 5 mg

C. When deoxygenated hemoglobin exceeds 10 mg

D. When carboxyhernoglobin is present nr blood

E. None of them is correct

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Ans. B

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187. REGARDING TIDAL VOLUME

A. Helium dilution is the method to measure it

B. Is part of minute ventilation

C. Increase in restrictive lung disease

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Ans. B. Minute ventilation (alveolar ventilation) = (tidal volume – Dead space) X Respiratory rate

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188. A PATIENT WITH BRONCHIAL ASTHMA AND BRONCHOSPASM, WHAT IS CORRECT

A. Decreased Residual volume

B. Dec FEV, Inc Vital capacity

C. Inc Tidal volume

D. Dec Tidal volume

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Ans. B

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189. A person has developed type II alveolar cells damage resulting in low surfactant. What will Happen in this pt

A. Decrease recoil-capacity of lung

B. Decrease in compliance

C. Decrease in pulmonary resistance

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Ans. B

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190. A person is performing exercise. What will have been decrease in its upper lobe

A. Blood flow

B. Ventilation

C. V/Q ratio

D. Pulmonary resistance

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Ans. C

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191. FEV1/FRV In normal person is

A. 60

B. 70

C. 80

D. 85

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Ans. C

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192. Increased ventilation at base of lungs is due to

A. Increased V/Q ratio there

B. Base contain more alveoli

C. Presence of Wide segmental bronchioles

D. There is more blood flow at base

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Ans. D

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193. Both central and peripheral chemorecptors respond to:

A. Dec ph and inc Pco2

B. Dec o2

C. Inc ph and dun Pco2

D. Bp

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Ans. A

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194. Oxygen hb dissociation curve shifts to left due to?

A. Increased col

B. Decreased ph

C. At high altitude

D. Carbon mono oxide poisoning

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Ans. D

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195. In hypoxemic state, respiratory centre is stimulated through?

A. Aortic sinus

B. Carotid body

C. Arterial pco2

D. Carotid sinus

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Ans. B

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196. During exercise a man develops tachypnea due to stimulation of respiratory center. Which is causative factor for this stimulation?

A. Decreased 02

B. Increased CO2

C. Increased H ions

D. Decreased HCO3

E. Decreased function of lungs

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Ans. B

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197. During exercise, what happens

A. Decrease HR

B. Increase pulmonary compliance

C. Decrease pulmonary compliance

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Ans. B

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198. If a person cannot excrete CO2, which of the following will be present in him?

A. Decreased PCO2

B. Elevated PH

C. Decreased HCO3

D. Decreased P02

E. Alkaline urine

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Ans. D. If a person can not exchange Co2, it means he is having hypoventilation. Hypoventilation causes ↑Pco2 and ↓Pco2.

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Table Hypoventilation Vs. Hyperventilation

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Pco2

Po2

Ph

Hypoventilation

 ↓ (respiratory acidosis)

Hyperventilation

 ↑ (respiratory alkalosis)

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199. Treatment of hyperbaric oxygen is best explained as treatment of

A. Asthma

B. Type 2 respiratory failure

C. Decompression sickness

D. Carbon monoxide poisoning

E. Gas gangrene

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Ans. C

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200. If alveolar ventilation doubles and co2 production Remains constant, the arterial pco2

A. Does not change

B. Doubles

C. Increases four-fold

D. Increases sixteen – fold

E. Is halved

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Ans. E

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201. Surfactant in the lung

A. Decreases the compliance of lungs

B. Increases by cigarette smoking

C. Is produced by type 1 alveolar epithelial cells

D. Its molecules move apart during expiration

E. Lowers the surface tension of the fluid lining the alveoli

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Ans. E

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