RESPIRATORY PHYSIOLOGY: PROBLEMS
Respiratory Ventilation: Answers

RESPIRATORY VENTILATION PROBLEMS:  Answers

1. A patient has a VC of 4.8 L, an ERV of 1.2 L, and an FRC of 2.5 L. What are his RV, TLC, and IC? Are these volumes within the normal range?

Answer: Using formulas or diagram:

FRC = ERV + RV so RV = FRC - ERV
TLC = VC + RV
IC = TLC - FRC

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Values are normal for young adult male of "average" size.

Question: What if FRC = 4.0 L and other values are as above?
Answer: RV = 2.8 L, TLC = 7.6 L. What might cause such values? Gas trapping, such as in emphysema.

2. After recovery from surgery involving cardiopulmonary bypass, patients often require positive pressure ventilation to assist their breathing. In a certain patient, a pressure of 30 cm H2O is needed for a tidal volume of 0.9L. Calculate respiratory system compliance, and decide whether the result of your calculation indicates the respiratory system is stiffer than normal, more compliant than normal, or within the normal range.

Note: Explained cardiopulmonary bypass. Also, positive pressure ventilation is accomplished -- endotracheal tube, pump or intermittent valve on air tank. Used for lung inflation and ventilation.

Answer: C = delta-V / delta-P = 0.9 L / 30 cmH2O = 0.03 L/cmH2O. Is much less than normal (0.1 L/cmH2O).

Question: Why might it be so low? Answer: Alveolar collapse (atelectasis); interference with ability to produce surfactant.

3. For the following three patients, indicate the classification of the ventilatory deficiency -- restrictive lung disease or obstructive -- and whether the condition appears acute or chronic. (Note: Isoproterenol is a bronchodilator which is effective in reducing airway resistance in acute conditions but is less effective in chronic obstructive disease.)

PATIENT A: Male, age 62, whose work was coating pipes with an asbestos compound. Smoked 30 cigarettes daily. Chest x-ray showed a massive shadow in the left lower lobe.

  Predicted Value Measured Value Percent Predicted Measured after Isoproterenol
VC 3.1L 0.9L 29% 0.9L
FEV-1 2.5 0.8 32% 0.85
FRC 3.2 1.6 50% 1.6
TLC 5.3 2.2 42% 2.2
FEV-1/VC >80% 88%   94%

Answer: Restrictive disease, as seen from very low lung TLC; assuming normal respiratory effort, implies pulmonary compliance is low. No obstructive disease, as FEV1 / VC is normal. Note minimal effect of Isoproterenol. What is the action of Isoproterenol?. Bronchodiator because beta-2 agonist.

Question: What might have caused this condition. Data suggest pulmonary fibrosis (asbestosis) and lung cancer (massive shadow, cigarettes, asbestos).

PATIENT B: Male, age 50. Smoked 2 packs of cigarettes a day for past 30 years. Chronic cough with sputum. Progressively increasing dyspnea.

  Predicted Value Measured Value Percent Predicted Measured after Isoproterenol
VC 4.1L 4.1L 100% 4.2L
FEV-1 3.3 1.4 42% 1.5
FRC 2.4 3.4 142% 3.4
TLC 5.0 6.0 120% 6.0
FEV-1/VC >80% 34%   36%

Note: Explain meaning of "dyspnea".

Answer: No restrictive disease (large TLC). Obstructive disease, low FEV1/VC; chronic, because not much improved with bronchodilator. Has bronchitis (chronic cough with sputum); may also have emphysema.

Question: How can TLC be larger than normal while VC is normal. Answer: Larger RV, due to gas trapping on expiration.

PATIENT C: Male, age 28. Recurrent attacks of wheezing since early twenties, precipitated by anxiety.

  Predicted Value Measured Value Percent Predicted Measured after Isoproterenol
VC 4.9L 4.0L 82% 4.5L
FEV-1 3.9 1.5 39% 3.0
FRC 3.6 4.0 110% 3.4
TLC 6.0 6.6 110% 6.6
FEV-1/VC >80% 38%   67%

Note: Wheezing is sound due to turbulence in airways; characteristic of airway obstruction.

Answer: No restrictive disease (large TLC). Obstructive disease indicated by low FEV1 / VC; acute because largely relieved by isoproterenol.

Note: Typical picture of asthma.

4. Below are shown alveolar pressure, intrathoracic pressure, and lung volume for a typical respiratory cycle of a normal person at rest.

 

A. Explain the shape of the intrapleural and alveolar pressure curves.

B. How would these curves be expected to change during heavy exercise?

C. How would these curves be expected to change in a person with severe restrictive disease (e.g. fibrosis)?

D. How would these curves be expected to change in a person with severe emphysema?

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Answer:

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Explain these curves:

Muscles inspiration contract (which? Mainly diaphragm, also external intercostals). This lowers intrapleural pressure, which is transmitted into the alveoli. Since the alveolar pressure is now negative (below atmospheric), air flows into the lungs, increasing their volume. The volume increase causes the lung to "pull back" more, causing the alveolar pressure to return toward zero (atmospheric). Finally, the volume will stop increasing, so inspiration ends.

When the inspiratory muscles stop contracting at the beginning of expiration, the stretched lungs now "pull back" harder than the chest wall pulls "out", so the alveolar pressure becomes positive and the lungs empty until the system reaches balance again. Note that this is quiet breathing at rest, so active contraction of the expiratory muscles is not involved.

Curves in exercise:

Larger intrapleural pressure swings (more vigorous contraction of respiratory muscles); intrapleural pressure less negative in expiration (active contraction of muscle during expiration - internal intercostals and abdominal wall); larger alveolar pressure swings; larger volume changes and faster flow rates.

Restrictive lung disease (fibrosis):

Low compliance ("stiff" lungs), so "pull back" harder at rest position. Results in lower (more negative) intrapleural pressure, smaller swings in alveolar pressure (more of the work of the respiration muscles goes into stretching the lungs, less into moving air), smaller tidal volume. Usually increased respiratory rate to compensate for low tidal volume and maintain alveolar ventilation.

Emphysema:

Loss of lung tissue leads to lower static lung compliance. Result: larger lungs at rest position and less negative intrapleural pressure. Inspiration is normal but in expiration, higher intrapleural pressure leads to airway narrowing (reduced expiratory flow rate) and my cause airway collapse (gas trapping).

5. A bright lad decides to lengthen his short snorkel with a 3 meter hose so he can dive deeper. Should you notify the life guard?

Answer: Yes.

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If use large tube (vacuum cleaner hose), very high dead space reduces alveolar ventilation.

If use small tube to reduce dead space, high resistance makes it impossible to move air fast, reducing total ventilation.

Suppose use large hose, breath in through mouth and out through nose (why does this solve the dead space and resistance problems?). Still notify the life guard because 300 cmH2O pressure on chest will prevent any inspiration (maximum inspiratory effort can only generate about 100 cmH2O water pressure).

How to go deep than 20-30 cm? (1) Hold breath. (2) Scuba (Self-Contained Underwater Breathing Apparatus) - tank of compressed air and valve; valve adjusts pressure at mouth to be the same as pressure on chest permitting normal breathing.

Problem of "bends" (decompression sickness) in scuba diving. High air pressure necessary to balance high water pressure causes nitrogen to dissolve in body tissues, especially fat tissue. If ascend slowly, nitrogen diffuses out of fat and is lost from lungs. But if ascend rapidly, nitrogen becomes a supersaturated solution in the blood and forms small bubbles (nitrogen emboli) particular at sites of mechanical vibration (e.g. joints). The emboli block tissue blood flow, causing pain and, if severe, tissue necrosis. Note use of hyperbaric chamber to recompress (to redissolve nitrogen) and then decompress slowly.