MSP Problem Set #7 ANSWER KEY

Respiratory – Mechanics

1. If you were a water molecule who got sucked into someone’s nose, what would you see on your journey from the nostril to an alveolus. In as much detail as you can muster, describe the path you take, how fast you are going at different points on your trip, what the scenery looks like (the changes in the cells of the surface of the airway), and with your x-ray vision note the underlying structures that keep the tunnel open. Also, at what point do you see the O2 molecules leaving the lumen and the CO2 molecules entering the lumen. (Assume that, along with having x-ray vision, you can see in the dark, you have eyes, you have a brain, and that you have a language -- even though you are a water molecule. This is Hollywood.)

after the terminal bronchioles, the air way starts to be used for gas exchange – before that, the air way is just for conduction and humidification (dead space)

The following are some ideas for the rest of the question:

THE PATH / THE SPEED / THE SURFACE / THE UNDERLYING STRUCTURE
nose / fast – single tube / pseudo-stratified, ciliated, columnar epithelium / bone, cartilage
pharynx / fast – single tube / same / Bone, cartilage
larynx / fast – single tube / ?? / cartilage
trachea / fast – single tube / ?? / cartilage
main bronchi, lobar bronchi, segmental bronchi, small bronchi / fast –airway generation 1-10, cross sectional area starts to increase – there is still the same amount of air , so velocity drops / columner, ciliated / firm cartilaginous support
bronchioles / very slow – large cross sectional area / cuboidal / not supported by cartilage – depend on elastic recoil of the alveolar septa -connective tissue fibrils, interstitial cells – fibroblasts,
terminal bronchioles / generation 16- total cross sectional area is large- very, very slow / cuboidal / same
respiratory bronchioles / bulk flow ceases and diffusion takes over as the dominant mechanism of ventilation / cuboidal to flat to flat between the alveoli / same
alveolar ducts / same / same as alveolus / same
alveolar sacs / same / same as alveolus / same
alveolus / same / Squamous -- Type I – squamous, polar, incapable of dividing; Type II -- cuboidal, junction, microvilli, synth/storage/ dischrge of surfactant, responsible for growth and repar; Type III -- rare cone shaped, unknown function; alveolar macs – ; mast cells -- / same

2. Being a thoughtful water molecule, you wonder what caused your decent into the far reaches of the lung.

a) What is the simple physical principle that caused you to move ?

pressure gradient and conductance (a path to follow)

b) Explain what generates this moving force (which muscles and how do they work).

muscles of inspiration:

  • diaphragm - contraction causes downward piston-like movement lowering intrathoracic pressure – you might want to talk about the orientation of the fibers (appositional action and insertional action)
  • intercostals – parasternal are primary muscles of inspiration, external are accessory muscles of inspiration
  • scalenes – now considered to be primary muscles of inspiration
  • sternocleidomastoids – accessory muscles of inspiration

c) What will cause you to be pushed out of the lung?

expiration is passive in normal breathing but can be helped by internal intercostals, rectus abdominus, and external oblique in forced expiration

d) What transmits the energy of these moving muscles into a driving force for air movement?

The movement of the muscles causes a drop in the intrathoracic pressure which is transmitted to the parietal pleura which –through a negative pleural pressure is transferred to the visceral pleura, and then to the alveolus. Pressure drops in the alveolus and the higher pressure air at the mouth is forced inward. (below, see diagram from the notes)

You might want to bring up the transpulmonary pressure (PTP) at this point. A better word for this is distending pressure – it is the pressure that causes the lung to expand and it is the difference between the pressure in the alveolus and the intrapleural space. See diagram on page 14.

PTP = Palv - Ppl

Remember - Intrapleural pressure is negative with respect to atmospheric pressure in inspiration and at the end of a normal passive exhalation -- it is lung elasticity vs. elastic properties of the chest wall

3. Explain the three curves in the picture at the right. What significance does the slope of each curve have? What factors are involved in the elastic properties of the chest wall and the lung? Where is FRC on the curve?

These curves are all the volume of each component with respect to distending pressure if the system were at equilibrium. In other words it is the amount of pressure needed to keep the component at that volume. For Pw and Pl (pressure of chest wall and pressure of lung respectivly) it is as if they were independent of each other. Prs (pressure of respiratory system) is the sum of Pw and Pl. It might help to explain this as a spring – The chest wall has its equilibrium position at 60% TLC. if you compress the chest it is going to want to pop back out (that’s why it takes a negative pressure to keep it in). If you expand the chest, the chest wall will want to drop back to it’s equilibrium position (that’s why it takes a positive pressure to keep it distended). With respect to the lung, its equilibrium position is zero volume, so it always wants to drop pull back to its zero position (that’s why it takes a positive pressure to keep it distended. For the Prs curve (what actually happens) both the lung and the chest wall are working at the same time. At FRC they balance each other out. That is why the distending pressure is zero.

The slope of the curves are significant because the represent the compliance of the lungs

The factors that go into the elastic properties of the lung and chest wall are as follows:

Lung: lung elastic tisue (elastin and collagen in the alveolar walls and bronchial tree); Surface tension( fluid tends to pull the lung back in); Surfactant (reduces surface tension so that the force to pull the lung back doesn’t overpower the ability to distend the lung)

FRC is the point where the distending pressure is zero. FRC is crucial - it is the equilibrium position in which the recoil forces of the lung and chest wall are equal and opposite.

  1. You have been assigned to a preceptorship with a pulmonologist for the year. One of your first patients happens to be none other than Christopher Reeve.
  1. As you may know, Mr. Reeve was paralyzed from the neck down in a horseback riding accident. Let’s say he suffered a partial paralysis of his diaphragm and other muscles of respiration. Discuss the effects of Mr. Reeve’s injury on the static and dynamic forces involved in respiration.

The point: To make them think about the forces inflating and deflating our lungs. We’re trying to derive this answer directly from the information in their notes, not from our PPD stuff. Static forces include the recoil properties of lung elastic tissue, the surface forces at the air fluid interface, and the compliance of the chest wall. These forces sum to make up the intrapleural pressure – lung recoil and surface tension working to decrease lung volume, and the chest wall working to expand it. Since they don’t require muscle activity, they would be unaffected in the case of diaphragmatic weakness. Dynamic forces are the ones we actually need to breathe. During inspiration, they include the diaphragm (as well as the internal parasternal intercostals, external intercostals, and scalenes). Expiration, however, is due to the lung elastic recoil and only muscle action involved is relaxation of the chest wall. Mr. Reeve will have trouble breathing air in, but no problem exhaling whatever does enter his lungs. His TLC and VC will be reduced, but his FRC and RV should be normal.

  1. The pulmonologist asks you to 1.) use the axes below to draw Mr. Reeve’s flow-volume relationship during a normal (not forced) exhalation both before and after his accident. In addition, he asks you to 2.) predict what you might find after testing Mr. Reeve’s Functional Residual Capacity, Vital Capacity, and Total Lung Capacity, then requests that you classify Mr. Reeve’s ventilatory defect as obstructive or restrictive.


1)

See tracings for normal (before accident) and restrictive (after accident). Because of his weakened muscles of inspiration, his maximum volume would be much less than that of what he was able to achieve before the accident. However, because normal exhalation is a passive process that does not rely on muscle strength, once he gets up to maximum flow, at each given volume his flow rates (before and after the accident) should be identical. This can be seen by the similar slopes of the before accident and after accident curves. Be sure to note the weird orientation of high and low on the x axis – students may find this confusing.

Also, you may wish to discuss how a time-volume graph during a forced expiration might look. (see the “restrictive disease” graph on page 25) Again, with the restrictive defect the maximum volume would be decreased. Due to loss of strength of the muscles of forced expiration, the slope of his forced expiration curve would probably be a little less steep than the “normal” graph.

2)Remember, the FRC reflects the equilibrium between the inward and outward static forces. Since these forces are not dependent on muscular action, we would expect to see no change in FRC in the paralyzed Mr. Reeve. Since VC and TLC both take into account respiratory activity, however, they would be reduced in a patient with a weakened diaphragm.

3)This is a good time to discuss what we see in spirometry in the two disease patterns. This one would be restrictive – decreased total lung capacity and decreased vital capacity. We’re not sure if they have had this lab yet, but if so, you can discuss how FEV1 and FVC would both be decreased. Note that many restrictive lung diseases (such as pulmonary fibrosis) are characterized by a normal or even increased FEV1/FVC ratio. Here, however, because he is suffering from neuromuscular weakness, his FEV1/FVC ratio would be decreased.

Compare this with an obstructive disease (such as emphysema) where the FEV1, FVC and the FEV1/FVC ratio are always reduced. Also mention about the increased FRC (“barrel chest”) and TLC found in emphysema due to increased lung compliance.

  1. You then accompany the pulmonologist on hospital rounds and see a patient with Isolated Diaphragm Paralysis. You observe that as he inhales his abdomen moves inward as his chest wall moves out. You notice that this seems “paradoxical.” When your preceptor asks to explain this interesting finding, you dazzle him by explaining:

Because his diaphragm is paralyzed, his inhalation is solely driven by his other respiratory muscles. He is able to use these muscles to expand his chest cavity. However, unlike the normal case, his diaphragm is passively raised in response to the drop in thoracic pressure. This causes a subsequent drop in abdominal pressure and the moving in of the abdominal wall.

  1. Your next patient is Herman, a 60 year-old man who has been smoking about a pack a day for thirty years.
  1. He has recently been diagnosed with emphysema. In the last few months he has noticed that he can’t get all of the air out of his lungs when he expires. One of his friends who has had some medical training mentioned something about Herman’s airways collapsing. What’s going on?

It seems that the emphysema has caused a loss of elastic recoil in the lungs. The intra-mural pressure in the lung is opposed outside of the airway by the Pleural Pressure (Ppl) and Static Elastic Recoil (Pst). As the Pst drops, it takes less distance for the intramural pressure to drop to a point where the pressure inside the airway is equal to the pressure outside of the airway are equivalent (the Equal Pressure Point). The problem in emphysema is that this equal pressure point occurs right next to the alveoli in the terminal bronchioles. The terminal bronchioles are not supported by cartilage as the rest of the airway is, so it collapses, causing air to be “trapped” in the alveoli. Hence, Herman’s feeling that he can’t blow all of his air off.

See diagram on page 19.

b. Using Nitrogen washout, the functional residual capacity of Herman’s lung is calculated to be approximately 2.75L. Assuming the following information: Expiratory Reserve Volume = 1.5L and Vital Capacity = 3.75 L, calculate Herman’s total lung capacity. Also, explain the Nitrogen washout.

TLC = Residual Volume (FRC-ERV) + Vital Capacity
TLC = 1.25 + 3.75 = 5L

In Nitrogen washout, a subject inhales 100% O2 until all of the N2 is washed out of the lungs. The volume of the expired air and the concentration of Nitrogen in that air are both measured. This allows us to calculate the volume of exhaled Nitrogen (Total Air Volume * N Concentration). This is equivalent to the volume of N2 that resided in the lungs prior to the inhalation of pure O2. Assuming you know the concentration of Nitrogen in the lungs before the test, you can determine the FRC using the following equation: FRC * [N]lungs = Volume of exhaled Nitrogen


c. What happens to the compliance of emphysemic lungs? What about the recoil? Which of the three curves on graph “C” most accurately depicts the pressure volume relationship seen in Herman’s lung?

Compliance is increased. Recoil is decreased. Line 3 on graph “C” represents lungs that are more compliant, which would be consistent with emphysema. Specifically, we see a larger change in volume for any given change in pressure.