BIOS E-162 Human Pathophysiology I Problem Set 2 Due Nov.23, 2009

Transcription

1 BIOS E-162 Human Pathophysiology I Problem Set 2 Due Nov.23, 2009 Note that the last question is for grad students only. Undergrads and Grads: Answer all parts. 1. Marie has just been hired for a job in a factory. Because of the nature of the emissions, Marie will have to wear a full face mask with an attached air filter while she is on the factory floor. She wants to see what it s going to feel like to breathe with the mask, so she puts it over her face and starts to breathe. Immediately before she put on the mask, her tidal volume was 400 ml, her frequency of breathing was 10 breaths per minute, and her P a CO 2 was 40 mmhg. The total volume of the air between the mask and her face is 100 ml (HINT: this is the new added dead space). Answer the following questions about Marie s breathing pattern before and after putting on the mask. You can assume that Marie s anatomical dead space is 100 ml. a) What was Marie s minute ventilation before she put the mask on? Her alveolar ventilation? Minute ventilation is tidal volume x frequency (0.4 L x 10 = 4.0 L/min) Alveolar ventilation is (V T -V d ) x f = ( ) x 10 = 3.0 L/min b) Assuming that Marie does not change her breathing pattern after putting on the mask, what will her new alveolar ventilation be (in L/min)? Her new dead space is 0.1L (anatomical dead space) L (space between mask and face) = 0.2L. Alveolar ventilation is (V T -V d ) x f = ( ) x 10 = 2.0 L c) Again, assuming that Marie does not change her breathing pattern after putting on the mask, what will her new P a CO 2 be (in mmhg) once a steady state is reached? PaCO2 will be the same as P A CO2. Previously her P A CO2 was 40 mmhg. P A CO2 is inversely proportional to alveolar ventilation. Her alveolar ventilation was reduced to two thirds of what it was originally, so if she doesn t change her breathing pattern, her new PaCO2 will be 3/2 times what it was (1.5 x 40 = 60 mmhg). In fact, Marie does change her pattern of breathing when she puts the mask on. Describe how this change comes about. (You will need to consult the general respiration notes about the control of breathing to be able to answer this question). The increase in PaCO 2 (see section c above) will result in more H + ion being generated both in the blood and in the cerebrospinal fluid (because CO 2 combines with water to form carbonic acid which dissociates into H + and HCO 3 - ions. The H + in the blood will stimulate peripheral chemoreceptors in the aorta and carotid arteries. The H + in the cerebrospinal fluid will stimulate central chemoreceptors in the brainstem. Both will act on the

2 respiratory control center in the medulla to increase the frequency and magnitude of respiratory muscle contractions, leading to increased ventilation. Undergrads and Grads: 2. A relatively new treatment for asthma is a drug called omalizumab, which is an antibody against IgE. Based on what you know about the early response to allergen inhalation in an asthmatic, explain why omalizumab should be effective in preventing bronchoconstriction that occurs following allergen inhalation. The early response consists of bronchoconstriction caused by products such as histamine and leukotrienes released from mast cells. Both are potent contractors of airway smooth muscle. Allergen binding to IgE on mast cells is what activates the mast cells to release these products. IgE antibodies might be effective in inhibiting allergic bronchoconstriction because they bind circulating IgE, preventing it from being able to bind to mast cells. It s also possible that the IgE antibodies could bind to the IgE on the mast cells and block it from being able to interact with allergen. Undergrads and Grads: Answer all parts. 3. Examine the results of the pulmonary function testing obtained on Mr. X, a 65 year old man, who has a30 year history of smoking 2 packs a day. VC: 4.0 L RV: 3.5 L FRC: 4.5 L FEV 1.0 : 2.4 L A normal man of Mr. X's age, sex, and height has the following pulmonary function tests: VC: 4.0 L RV: 2.0 L FRC: 3.2 L FEV 1.0 : 3.4 L TLC: 6.0 L a) What is Mr. X's TLC? TLC = VC + RV = = 7.5 L b) Based on what you have learned in class, what do you think is Mr. X's problem? The FEV 1 /FVC ratio is 2.4/4 = 60% (well below 80%), so Mr. X has some kind of obstructive disease. His TLC is substantially above what is predicted for his age, gender, and height. An increase in TLC in conjunction with the history of cigarette smoking suggests that Mr. X has emphysema.

3 c) Draw the lung pressure volume curve of a normal subject. Draw Mr. X's pressure volume curve on the same graph. Mr. X s PV curve should be shifted up (higher TLC) and to the left. d) What is the reason for Mr. X s airway obstruction. The primary reason for the airway obstruction in emphysema is loss of the tethering effect of the lung parenchyma on the airways. Undergrads and Grads: 4. Mr. Smith has pneumonia. Most of the left lung is affected and is completely full of pus and fluid, whereas the right lung is clear. Mr. Smith s PaO 2 is only 60 mmhg. a) Because you have been taking NSCI E-162, an idea occurs to you. You tell the staff treating the patient to turn him so that he is lying on his right side. You do measurements of blood gases after the patient has been lying like this for a while, and you find that his PaO 2 improves. Explain why. Mr. Smith has a shunt. His left lung is still receiving plenty of blood flow, but there is no ventilation to this part of the lung. Remember from our discussions of V/Q abnormalities that blood flow depends on gravity, the lowest parts of the lung get the most blood flow. The ideal for Mr. Smith would be to have more of his blood flowing to the part of the lung that is being ventilation, i.e. the right lung. If we have him lie on his right side, i.e. with his right lung down, more blood with go there and less will go to the left lung, where it won t be oxygenated. b) You also decide to have Mr. Smith breathe a gas mixture containing 40% O2, 60% N2, instead of regular room air. Do you think this will result in a substantial improvement in Mr. Smith s PaO 2? Why or why not? The left lung (the bad one) will never see the extra O2 that the patient is now breathing, so breathing even 100% O2 will still not result in any additional O2 entering the circulation from the bad lung. On the good side, the alveolar PO2 in the air in the good lung will increase, so the PO2 in the blood that goes through that side will also increase. However, the actual content of O2 coming from the blood that goes through that side will change very little, because the PO2 was already quite high, high enough that the hemoglobin was already fully saturated with O2 and no more can be added (except a very small amount that is physically dissolved in the blood). So the O2 content of the blood mixed from the right and left sides will change minimally, and consequently, the PaO2 will change very minimally.

4 Undergrads and Grads: 5. Pulmonary edema caused by left heart failure is usually self-limiting whereas pulmonary edema induced by exposure to toxins or viruses is often not. Explain why. Pulmonary edema caused by left heart failure is usually self-limiting because as fluid leaks out into the interstitial spaces in the lungs, the proteins in the capillary become more concentrated, there is an increase in the oncotic pressure, which tends to pull the fluid back into the capillary. Pulmonary edema induced by exposure to toxins or viruses is usually the result of damage to the endothelial cells that line the pulmonary capillaries. Once the integrity of the endothelial barrier is destroyed, proteins start leaking into the interstitial spaces, so the oncotic pressure of the interstitium starts to increase, further increasing fluid flux into the interstitium. This kind of edema is extremely dangerous and can result in respiratory failure. Grad students only: 6. Read this assigned article: Banauch GI, Alleyne D, Sanchez R, Olender K, Cohen HW, Weiden M, Kelly KJ, Prezant DJ. Persistent hyperreactivity and reactive airway dysfunction in firefighters at the World Trade Center. Am J Respir Crit Care Med Jul 1;168(1): Accessing the article: This article is available on Pubmed. To access this article, go to and find the article. You will find a link to a free full text version of the article. Summary and Questions: The article describes changes in pulmonary function in firefighters who responded to the World Trade Center disaster. 1. The article describes reductions in the FEV 1 /FVC ratio after WTC exposure. What do FEV 1 and FVC measure? What is the significance of a decrease in this ratio? FEV1 is the amount of air that is expired during the 1 st second of a forced expiratory maneuver from total lung capacity. It is a measure of how fast air can be expelled from the lungs. FVC is the total amount of air that can be expired during a forced expiratory maneuver. It is mostly a measure of the size of the lungs and is affected by height, gender and age, but can also be impacted by lung disease. The ratio FEV1/FVC reflects how fast air can be expelled normalized for the size of the lungs. A decrease in the FEV1/FVC is indicative of airway obstruction, narrowing of the airways. 2. Methacholine is a relatively stable analog of acetylcholine, a constrictor of airway smooth muscle. What is the significance of the methacholine hyperreactivity in the group of highly exposed firefighters?

5 Constriction of airway smooth muscle, as can occur following inhalation of methacholine, results in airway narrowing, usually measured as a change in the FEV1. Hyperreactivity means that the subject had large declines in FEV1 in response to low concentrations of methacholine, concentrations that do not affect FEV1 in normal subjects. Such a change indicates that the subject will likely experience airway obstruction in response to stimuli (particles, allergens, gases, exercise in cold air) that do not typically affect the airways of normal subjects. This is a characteristic feature of asthma: all asthmatics are hyperreactive, although not all individuals with hyperreactivity necessarily have asthma. 3. Describe the nature of the exposure following the WTC disaster. The article notes that Firefighters reported that except for their self-contained breathing apparatus (lasting only 8 15 minutes), respirators were generally not available during the first days post-collapse and, when available, were rarely worn. What is the public health message here. Environmental data indicated a largely particulate exposure. The size of the particles was such that most should have deposited in the nose and upper airways (mouth pharynx), but some were in the respirable range and clearly deposited in the lungs resulting in a syndrome that included cough, airway obstruction, and airway hyperreactivity, i.e. a syndrome that resembled asthma. The articles also notes that Firefighters reported that except for their self-contained breathing apparatus (lasting only 8 15 minutes), respirators were generally not available during the first days post-collapse and, when available, were rarely worn. Wearing of masks with air filters should have reduced a large amount of the particulate exposure and prevented the induction of reactive airways disease, an asthma-like condition that resulted in more respiratory symptoms and more and longer medical leave for symptoms. The article highlights the need to provide respiratory protection, especially in at the beginning of any future such disasters, since it was during this time period when most of the exposures that worsened lung function occurred.