Lung Mechanics
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You are called to examine a premature early born infant. He is suffering from breathlessness and is cyanotic. Your diagnosis is "infant respiratory distress svridrome, (IRDS). After treatment with artifrciaI surfactant, the situation is under control. What did surfactant do?
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It decreased the lung compliance
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It decreased surface tension in the lung
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It increased the force of respiratory muscles
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It Increased the activity of inspiratory motor neurons
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It increased the concentration of erythrocytes in the blood
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This quiz on 'Lung Mechanics' assesses key concepts in respiratory physiology, focusing on lung compliance, pressures, and disease states. It aims to enhance understanding of lung function and disorders, crucial for medical students and professionals.
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- 2.
Alveolar surfactant acts to INCREASE pulmonary
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Surface tension
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Compliance
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Airway resistance
Correct Answer
A. ComplianceExplanation
Alveolar surfactant acts to increase pulmonary compliance. Compliance refers to the ability of the lungs to stretch and expand during inhalation. By reducing surface tension within the alveoli, surfactant decreases the forces that work against lung expansion, allowing for easier and more efficient breathing. Therefore, an increase in pulmonary compliance means that the lungs are more flexible and have a higher capacity for expansion.Rate this question:
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- 3.
Contraction of the abdominal muscles is important in
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Normal (quiet) inspiration.
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Forced (maximum) inspiration
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Normal (quiet) expiration
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Forced (maximum) expiration.
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None of the above
Correct Answer
A. Forced (maximum) expiration.Explanation
During forced (maximum) expiration, the contraction of the abdominal muscles plays a crucial role. This contraction increases the intra-abdominal pressure, which helps to push the diaphragm upward and compress the thoracic cavity. As a result, more air is forcefully expelled from the lungs, allowing for a deeper and more complete expiration. In contrast, during normal (quiet) expiration, the diaphragm relaxes and the abdominal muscles are not actively involved in the breathing process. Therefore, the correct answer is forced (maximum) expiration.Rate this question:
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- 4.
A patient has a skeletal abnorrnalltv (Kyphoscohosis). You are interested to verify the compliance of his chest wall. You are measuring changes in lung volume by a spirometer. Which pressure change you have to measure in addition to DV to be able to determine the compliance of his chest wall?
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Changes In alveolar pressure (D,PA)
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Changes in intra pleural pressure (D,Ppl)
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Changes In airway pressure (D,Paw)
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Changes in alveolar 02-partial pressure (D,PAO2)
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Changes in inspired O2-partial pressure (D,PIO2)
Correct Answer
A. Changes in intra pleural pressure (D,Ppl)Explanation
To determine the compliance of the patient's chest wall, you need to measure changes in intra pleural pressure (D,Ppl). Compliance is a measure of how easily the chest wall expands and contracts, and it is influenced by changes in intra pleural pressure. By measuring the changes in intra pleural pressure, you can assess the elasticity and flexibility of the chest wall, which will help determine its compliance. Changes in alveolar pressure, airway pressure, alveolar O2-partial pressure, and inspired O2-partial pressure are not directly related to measuring the compliance of the chest wall.Rate this question:
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- 5.
The compliance of the respiratory system IS changing with lung volume. At which lung volume has the respiratory systsm It 's greates compliance?
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FRC
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RV
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TLC
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Minimal lung volume (below RV)
Correct Answer
A. FRCExplanation
The respiratory system's compliance refers to its ability to stretch or expand. Compliance is highest at the functional residual capacity (FRC), which is the volume of air remaining in the lungs after a normal exhalation. At this point, the lungs are neither fully expanded nor fully deflated, allowing for optimal flexibility and ease of breathing. Therefore, FRC is the lung volume at which the respiratory system has its greatest compliance.Rate this question:
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- 6.
A patient inhales 2 L of air and holds his breath. At the beginning of inspiration his lung transmural pressure is + 5 cm H20. During breath holding his lung transmural pressure is + 10 cm H20 . What is his lung compliance in L / cm H20 ?
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0.1
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0.2
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0.3
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0.4
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0.5
Correct Answer
A. 0.4Explanation
The lung compliance can be calculated by dividing the change in lung volume by the change in lung transmural pressure. In this case, the change in lung volume is 2 L (since the patient inhales 2 L of air) and the change in lung transmural pressure is 5 cm H20 (from +5 cm H20 to +10 cm H20). Therefore, the lung compliance is 2 L / 5 cm H20, which simplifies to 0.4 L / cm H20.Rate this question:
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- 7.
Breathing is associated with changes in intra pleural pressure (PPI)' During which of the following respirations can this pressure become positive (greater than atmospheric pressure) .
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During normal inspiration
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During normal expiration
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During forced inspiration of a large volume (3 L)
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During forced expiration of a large volume (3 L)
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During breath holding at FRC
Correct Answer
A. During forced expiration of a large volume (3 L)Explanation
During forced expiration of a large volume (3 L), the pressure in the intra pleural space can become positive (greater than atmospheric pressure). This is because during forced expiration, the diaphragm and other respiratory muscles contract forcefully, causing a rapid decrease in lung volume. This increase in pressure within the lungs can lead to a positive intra pleural pressure, which helps to push air out of the lungs.Rate this question:
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- 8.
A Patient is suffering form obstructive lung disease. His chest wall and respiratory system compliances are measured as: Chest wall compliance (Ccw)= 200 ml / cm H20 Respiratory system compliance (Crs)= 100 ml/ cm H20 What is Lung Compliance (Cl) in ml / cm H20
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50
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100
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200
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300
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400
Correct Answer
A. 200Explanation
1/Crs = 1/Cl + 1/Ccw therefore 1/Crs - 1/Ccw = 1/Cl
1/100 - 1/200 = 1/Cl = 1/200 thererore Cl = 200Rate this question:
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- 9.
A premature baby with difficulty breathing probably has :
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An increased FRC
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An increased TLC
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A decreased RV.
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A and C.
Correct Answer
A. A decreased RV.Explanation
A premature baby with difficulty breathing is likely to have a decreased RV (residual volume). This is because the RV is the volume of air remaining in the lungs after a maximum exhalation, and a decreased RV indicates that the baby is not able to fully exhale and clear the lungs properly. This can be a result of underdeveloped lung function in premature babies, leading to difficulty in breathing and inadequate lung ventilation.Rate this question:
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Current Version
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Mar 21, 2023Quiz Edited by
ProProfs Editorial Team -
Nov 11, 2011Quiz Created by
Chachelly
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