Guyton hall resp , Respiratory system pretest , Respiratory Physiology Mini 1

In a maximal expiration, the total volume expired is (A) tidal volume (TV) (B) vital capacity (VC) (C) expiratory reserve volume (ERV) (D) residual volume (RV) (E) functional residual capacity (FRC) (F) inspiratory capacity (G) total lung capacity

27. The answer is B. The volume expired in a forced maximal expiration is forced vital capacity, or vital capacity (VC).

Effects of large intrapulmonary shunt and patient with metabolic acidosis?

A large intrapulmonary shunt will cause arterial oxygen tension to decrease. This will increase respiratory drive causing a respiratory alkalosis, which will result in a low PCO2 and a high pH. The low oxygen tension may also produce a lactic acidosis, which may attenuate the rise in pH produced by hyperventilation. A patient with metabolic acidosis will have a lower than normal pH, which will stimulate the chemical and peripheral chemoreceptors. The ensuing rise in alveolar ventilation will lower PCO2 and slightly increase PO2.

Which one of the following statements characterizes pulmonary compliance? a. It decreases with advancing age b. It is inversely related to the elastic recoil properties of the lung c. It increases in patients with pulmonary edema d. It is equivalent to DP/DV e. It increases when there is a deficiency of surfactant

B. Pulmonary compliance, defined as the ratio of change of lung volume to the change in pressure required to inflate the lung (∆V/∆P), is an index of lung distensibility. It decreases in patients with pulmonary edema and interstitial fibrosis and increases in patients with emphysema and in persons of advancing years. The stiffer the lung, the lower the pulmonary compliance. Surfactant lowers the surface tension in the alveoli and makes the lung more distensible.

The oxygen consumption of the respiratory muscles is decreased by a. A decrease in lung compliance b. A decrease in airway resistance c. An increase in the rate of respiration d. A decrease in the production of pulmonary surfactant e. An increase in tidal volume

B. Respiratory muscles consume oxygen in proportion to the work of breathing, which can be divided into resistance work and elastic work. Resistance work includes work to overcome tissue as well as airway resistance. A decrease in the amount of pulmonary surfactant would decrease lung compliance and increase the elastic work of breathing. An increase in respiratory rate increases the work of breathing.

Which one of the following will be greater than normal in a patient with a low V/Q ratio? a. PaCO2 b. PaO2 c. A-a gradient d. Oxygen dissolved in blood e. Oxygen combined with hemoglobin

C. Capillary blood draining from areas of the lung with low V/Q ratios has low oxygen saturations. Capillary blood draining from areas of the lung with high V/Q ratios does not have oxygen saturations higher than normal because, under normal circumstances, the blood is fully saturated. When blood from areas of low V/Q ratios (with less than normal saturations) combines with blood from high V/Q ratios (with no more than normal saturations), the saturation of the combined blood is less than normal. The low saturation of the blood causes the PO2 of the arterial blood to be lower than normal. In the lungs, the alveolar oxygen tension is higher in areas of high V/Q ratio and lower in areas of low V/Q ratio. When the air from the two regions combines, it produces a normal PO2. Because the alveolar PO2 is normal and the arterial PO2 is less than normal, the A-a gradient is greater than normal.

At the end of a quiet inspiration, intra-alveolar pressure is normally a. 240 cmH2O b. 24 cmH2O c. 0cmH2O d. 14 cmH2O e. 140 cmH2O

C. During inspiration, the pressure in the alveoli becomes negative and gas is drawn into the lung by the difference in pressure between the alveoli and the atmosphere. At the peak of a normal inspiration, this pressure difference is only a few centimeters of water. At the end of inspiration, the pressure in the alveoli becomes equal to the atmospheric pressure, and flow ceases. During expiration, the intra-alveolar pressure rises above atmospheric pressure and gas is expelled from the lungs. The intra-alveolar pressure again equals atmospheric pressure at the end of expiration.

Anemic patients have what blood gas partial pressures?

Normal arterial PO2 and reduced arterial oxygen content

High altitude residence sometimes results in right heart failure. What is the cause of this type of heart failure?

Pulmonary HTN caused by alveolar hypoxia

At what point during the tidal breath is the alveolar PCO2 at its highest value?

The alveolar PCO2 rises and falls with each breath. As CO2 diffuses into the alveoli from the blood, the alveolar PCO2 rises. The PCO2 falls when fresh air is added to the alveolar gas. The highest PCO2 occurs just before fresh air is inhaled. The first gas entering the alveoli during inspiration is actually alveolar gas because the conducting airways are filled with alveolar gas at the end of expiration. Fresh air doesn't enter the alveoli until the alveolar gas has been emptied from the dead space. In normal tidal breathing, dead space volume is approximately 30% of tidal volume. Therefore, the alveolar PCO2 continues to rise for the first one-third of each breath and reaches its highest value at the point labeled A.

In the transport of CO2 from the tissues to the lungs, which of the following occurs in venous blood? (A) Conversion of CO2 and H2O to H+ and HCO3- in the red blood cells (RBCs) (B) Buffering of H+ by oxyhemoglobin (C) Shifting of HCO3- into the RBCs from plasma in exchange for Cl- (D) Binding of HCO3 to hemoglobin (E) Alkalinization of the RBCs

The answer is A. CO2 generated in the tissues is hydrated to form H+ and HCO3 - in red blood cells (RBCs). H+ is buffered inside the RBCs by deoxyhemoglobin, which acidifies the RBCs. HCO3- leaves the RBCs in exchange for Cl- and is carried to the lungs in the plasma. A small amount of CO2 (not HCO3-) binds directly to hemoglobin (carbaminohemoglobin).

Which one 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 e. Lung compliance

A. The alveoli at the apex of the lung are larger than those at the base so their compliance is less. Because the compliance is reduced, less inspired gas goes to the apex than to the base. Also, because the apex is above the heart, less blood flows through the apex than through the base. However, the reduction in airflow is less than the reduction in blood flow,so that the V ̇/Q ̇ ratio at the top of the lung is greater than it is at the bottom. The increased V ̇ / Q ̇ ratio at the apex makes PaCO2 lower and PaO2 higher at the apex than they are at the base.

The activity of the central chemoreceptors is stimulated by a. An increase in the PCO2 of blood flowing through the brain b. A decrease in the PO2 of blood flowing through the brain c. A decrease in the oxygen content of blood flowing through the brain d. A decrease in the metabolic rate of the surrounding brain tissue e. An increase in the pH of the CSF

A. The central chemoreceptors are located at or near the ventral surface of the medulla. They are stimulated to increase ventilation by a decrease in the pH of their extracellular fluid (ECF). The pH of this ECF is affected by the PCO2 of the blood supply to the medullary chemoreceptor area as well as by the CO2 and lactic acid production of the surrounding brain tissue. The central chemoreceptors are not stimulated by decreases in PaO2 or blood oxygen content but rather can be depressed by long-term or severe decreases in oxygen supply.

In areas of the lung with lower than normal V/Q ratios, the a. Capillary CO2 tension is lower than normal b. Pulmonary vascular resistance is higher than normal c. Alveolar O2 tension is higher than normal d. Water vapor pressure is higher than normal e. Gas exchange ratio is higher than normal

B. Proper gas exchange requires an appropriate matching of ventilation and blood flow referred to as the V ̇ / Q ̇ ratio. If perfusion (Q ̇ ) exceeds ventilation (V ̇ ), the amount of oxygen diffusing into the capillary blood and the amount of car- bon dioxide diffusing out of the capillary blood are reduced. The result is a lower than normal PO2 and a higher than normal CO2. The lower PO2 will cause vascular resistance to rise above normal.

Which of the following changes occurs during strenuous exercise? (A) Ventilation rate and O2 consumption increase to the same extent (B) Systemic arterial PO2 decreases to about 70 mm Hg (C) Systemic arterial PCO2 increases to about 60 mm Hg (D) Systemic venous PCO2 decreases to about 20 mm Hg (E) Pulmonary blood flow decreases at the expense of systemic blood flow

The answer is A. During exercise, the ventilation rate increases to match the increased O2 consumption and CO2 production. This matching is accomplished without a change in mean arterial PO2 or PCO2. Venous PCO2 increases because extra CO2 is being produced by the exercising muscle. Because this CO2 will be blown off by the hyperventilating lungs, it does not increase the arterial PCO2. Pulmonary blood flow (cardiac output) increases manyfold during strenuous exercise.

Which person would be expected to have the largest A-a gradient? (A) Person with pulmonary fibrosis (B) Person who is hypoventilating due to morphine overdose (C) Person at 12,000 feet above sea level (D) Person with normal lungs breathing 50% O2 (E) Person with normal lungs breathing 100% O2

The answer is A. Increased A-a gradient signifies lack of O2 equilibration between alveolar gas (A) and systemic arterial blood (a). In pulmonary fibrosis, there is thickening of the alveolar/pulmonary capillary barrier and increased diffusion distance for O2, which results in lack of equilibration of O2, hypoxemia, and increased A-a gradient. Hypoventilation and ascent to 12,000 feet also cause hypoxemia, because systemic arterial blood is equilibrated with a lower alveolar PO2 (normal A-a gradient). Persons breathing 50% or 100% O2 will have elevated alveolar PO2, and their arterial PO2 will equilibrate with this higher value (normal A-a gradient).

If an area of the lung is not ventilated because of bronchial obstruction, the pulmonary capillary blood serving that area will have a PO2 that is (A) equal to atmospheric PO2 (B) equal to mixed venous PO2 (C) equal to normal systemic arterial PO2 (D) higher than inspired PO2 (E) lower than mixed venous PO2

The answer is B. If an area of lung is not ventilated, there can be no gas exchange in that region. The pulmonary capillary blood serving that region will not equilibrate with alveolar PO2, but will have a PO2 equal to that of mixed venous blood.

When a person is standing, blood flow in the lungs is (A) equal at the apex and the base (B) highest at the apex owing to the effects of gravity on arterial pressure (C) highest at the base because that is where the difference between arterial and venous pressure is greatest (D) lowest at the base because that is where alveolar pressure is greater than arterial pressure

The answer is C. The distribution of blood flow in the lungs is affected by gravitational effects on arterial hydrostatic pressure. Thus, blood flow is highest at the base, where arterial hydrostatic pressure is greatest and the difference between arterial and venous pressure is also greatest. This pressure difference drives the blood flow.

A 12 year old boy has a severe asthmatic attack. He experiences rapid breathing and becomes cyanotic. Arterial PO2 is 60 mmHg and his PCO2 is 30 mmHg. Which of the following statements about this patient is most likely to be true? A. Forced expiratory volume 1/FVC is increased B. V/Q ratio is increased in the affected areas of his lungs (C) His arterial PCO2 is higher than normal because of inadequate gas exchange (D) His arterial PCO2 is lower than normal because hypoxemia is causing him to hyperventilate (E) His residual volume (RV) is decreased

The answer is D. The patient's arterial PCO2 is lower than the normal value of 40 mm Hg because hypoxemia has stimulated peripheral chemoreceptors to increase his breathing rate; hyperventilation causes the patient to blow off extra CO2 and results in respiratory alkalosis. In an obstructive disease, such as asthma, both forced expiratory volume (FEV1) and forced vital capacity (FVC) are decreased, with the larger decrease occurring in FEV1. Therefore, the FEV1/FVC ratio is decreased. Poor ventilation of the affected areas decreases the ventilation/perfusion (V/Q) ratio and causes hypoxemia. The patient's residual volume (RV) is increased because he is breathing at a higher lung volume to offset the increased resistance of his airways.

A 49-year-old man has a pulmonary embolism that completely blocks blood flow to his left lung. As a result, which of the following will occur? (A) Ventilation/perfusion (V/Q) ratio in the left lung will be zero (B) Systemic arterial PO2 will be elevated (C) V/Q ratio in the left lung will be lower than in the right lung (D) Alveolar PO2 in the left lung will be approximately equal to the PO2 in inspired air (E) Alveolar PO2 in the right lung will be approximately equal to the PO2 in venous blood

The answer is D. Alveolar PO2 in the left lung will equal the PO2 in inspired air. Because there is no blood flow to the left lung, there can be no gas exchange between the alveolar air and the pulmonary capillary blood. Consequently, O2 is not added to the capillary blood. The ventilation/perfusion (V/Q) ratio in the left lung will be infinite (not zero or lower than that in the normal right lung) because Q (the denominator) is zero. Systemic arterial PO2 will, of course, be decreased because the left lung has no gas exchange. Alveolar PO2 in the right lung is unaffected.

A healthy 65-year-old man with a tidal volume (TV) of 0.45 L has a breathing frequency of 16 breaths/min. His arterial PCO2 is 41 mm Hg, and the PCO2 of his expired air is 35 mm Hg. What is his alveolar ventilation? (A) 0.066 L/min (B) 0.38 L/min (C) 5.0 L/min (D) 6.14 L/min (E) 8.25 L/min

The answer is D. Alveolar ventilation is the difference between tidal volume (TV) and dead space multiplied by breathing frequency. TV and breathing frequency are given, but dead space must be calculated. Dead space is TV multiplied by the difference between arterial PCO2 and expired PCO2 divided by arterial PCO2. Thus: dead space = 0.45 × (41 - 35/41) = 0.066 L. Alveolar ventilation is then calculated as: (0.45 L - 0.066 L) × 16 breaths/min = 6.14 L/min.

V/Q mismatch is partially compensated by local reflexes in the pulmonary vasculature and in the airways. Which are these reflexes?

Vasoconstriction due to low alveolar PO2; bronchoconstriction due to low alveolar PCO2

176. In the maximal expiratory flow - volume curves below , curve A would be typical of which of the following clinical presentations ? Airflow ( min / L ) vs Lung volume ( liters ) curve A is 8.25 to 5 (image description slope up then curves down same slope inversed and then at 7 it the angle flattens to 5) a . A 75 - year - old man who has smoked two packs of cigarettes per day for 60 years . His breath sounds are decreased bilaterally and his chest x - ray shows flattening of the diaphragm . b . A 68 - year - old man who presents with a dry cough that has persisted for 3 months . His chest x - ray shows opacities in the lower and middle lung fields . The man states that he was exposed to asbestos for approximately 10 years when he worked in a factory in his 30s . c . A 57 - year - old woman with pulmonary fibrosis who presents to the emergency room with shortness of breath . d . An 84 - year - old woman with a

a . A 75 - year - old man who has smoked two packs of cigarettes per day for 60 years . His breath sounds are decreased bilaterally and his chest x - ray shows flattening of the diaphragm . Cigarette smoking is the major cause of COPD . In obstructive lung diseases , the increase in airway resistance causes a decrease in expiratory flow rates and " air trapping , " which results in an increased residual volume , and thus total lung capacity . This hyperinflation pushes the diaphragm into a flattened position . Asbestosis and pulmonary fibrosis are restrictive lung diseases , in which curve C would be the typical MEFV curve . Decreased effort would decrease flow rates during the effort - dependent portion of a MEFV curve , but not during the effort - independent portion

177. A 14 - year - old adolescent girl presents with a lump in the neck . Fine needle aspiration biopsy reveals acinic cell carcinoma of the parotid gland . During the parotidectomy , there is compression injury of the glossop haryngeal nerve . As a result , which of the following respiratory reflexes will be impaired ? a . Aortic baroreceptor reflex b . Carotid body chemoreceptor reflex c . Hering Breuer inflation reflex d . Irritant airway reflex e . Juxta pulmonary capillary ( J ) receptor reflex

b . Carotid body chemoreceptor reflex The afferent pathway from the carotid body chemoreceptors is the Hering nerve , a branch of cranial nerve IX , the glossopharyngeal nerve . The vagus nerve constitutes the afferent pathway from the aortic baroreceptors , the J receptors , the irritant airway receptors , and the rapidly adapting stretch receptors mediating the Hering - Breuer inflation reflex .

164. A 34 - year - old woman presents in the emergency department with tachypnea and shortness of breath of acute onset . The history reveals that she has been oral contraceptives for 9 years . A lung scan demonstrates a perfusion defect in the left lower lobe . Which of the following occurs if the blood flow to alveolar units is totally obstructed by a pulmonary thromboembolism ? a . The V / Q ratio of the alveolus equals zero . b . The P0₂ of the alveolus will be equal to that in the inspired air . c . The Po₂ of the alveolus will be equal to the mixed venous P0₂ . d . There will be an increase in shunt ing ( venous admixture ) in the lung . e . There will be a decrease in alveolar dead space .

b . The P0₂ of the alveolus will be equal to that in the inspired air . A pulmonary thromboembolism results in areas of the lung that are ventilated , but not perfused , yielding V / Qra ratios of infinity and an increase in alveolar dead space . When the VIO ratio equalso , the PAO₂ of the affected alveoli will be the same as that in the humidified inspired air because atmospheric air enters the alveoli via the process of ventilation , but no gas exchange takes place because the alveoli are not perfused . Areas of the lung that are perfused but not ventilated constitute areas of shunting ( venous admixture ) , characterized as a VIÒ ratio equal to 0 , and having PAO₂ values that equilibrate with the mixed venous blood .

167. A 6'3 " tall , 140 - lb , 20 - year - old man was watching television when he felt pain in his shoulder blades , shortness of breath , and fatigue . His father noticed how pale he was and took him to the emergency department The physical exam revealed decreased tactile fremitus , hyperresonance , and diminished breath sounds . A chest x ray revealed a 55 % pneumothorax of the right lung , which was attributed to rupture of a bleb on the surface of the lung . What changes in lung function occur as a result of a pneumothorax ? a . The chest wall on the affected side recoils inward . b . The intrapleural pressure in the affected area equals to atmospheric pressure . c . The trachea deviates away from the affected lung d . There is hyperinflation of the affected lung . e . The V / Oratio on the affected side increases above normal .

b . The intrapleural pressure in the affected area equals to atmospheric pressure . When air enters the pleural space due to interruption of the pleural surface through either the rupture of the lung or a hole in the chest wall , the pressure in the pleural space becomes atmospheric , the lung on the affected side collapses because of the lung's tendency to recoil inward , and the chest wall on the affected side recoils outward . With collapse of the lung , the VIO ratio on the affected side decreases . The trachea shifts toward the affected lung in a spontaneous pneumothorax and away from the affected lung in a tension pneumothorax .

In an acclimatized person at high altitudes, oxygen delivery to the tissues may be adequate at rest because of a. An increase in hemoglobin concentration b. The presence of an acidosis c. A decrease in the number of tissue capillaries d. The presence of a normal arterial PO2 e. The presence of a lower-than-normal arterial PCO2

A. At a high altitude, arterial PO2 is low because of the low barometric pressure. Ventilation increases owing to stimulation of the carotid body chemoreceptors, and PaO2 increases somewhat, although it does not return to a normal, sea-level value. Hyperventilation causes a decrease in PaCO2, resulting in a respiratory alkalosis that may become fully compensated with time via renal mechanisms. A lower-than-normal PaCO2 would shift the oxyhemoglobin dissociation curve to the left, thereby increasing the affinity of hemoglobin for oxygen. Chronic hypoxia at high altitudes or in disease states stimulates the increased release of erythropoietin from the kidneys, which increases the hematocrit and thereby increases the hemoglobin concentration of the blood. Thus, at the same lowered PaO2, the oxygen content of blood and oxygen delivery to tissues increase. Chronic hypoxia also stimulates increased capillary growth in tissues.

During the early stages of an asthmatic attack, a. Arterial carbon dioxide tension decreases b. The equal pressure point moves toward the mouth c. Lung compliance increases d. Airway resistance decreases e. Arterial oxygen tension increases

A. During the early stages of an asthmatic attack, the increased airway resistance makes it difficult to inhale and exhale rapidly, and therefore each breath is slower and deeper. However, the accompanying dyspnea usually increases alveolar ventilation, and, as a result, PaCO2 decreases. Arterial oxygen tension usually decreases, despite normal alveolar oxygen levels, because of the increased V/Q mismatch that accompanies asthma. This hypoxia may also cause alveolar ventilation to increase. The equal pressure point moves toward the lung because of the increased respiratory effort. Acute asthmatic attacks do not produce any direct change in lung compliance. However, air trapping will cause functional residual capacity (FRC) to increase, and the larger lung volumes will lead to a decrease in lung compliance.

Which one of the following is higher at total lung capacity than it is at residual volume? a. Anatomical dead space b. Maximum static inspiratory pressure c. Lung compliance d. Airway resistance e. Alveolar pressure

A. Expanding the lungs lowers the intrathoracic pressure, which causes the airways to expand. Expanding the airways increases their volume and, therefore, the volume of the anatomical dead space. Expanding the airways also causes a decrease in airway resistance. Maximum static inspiratory pressure is maximal at function residual volume and decreases as lung volume is increased. Lung compliance is decreased; that is, the lungs are stiffer at high lung volumes. Alveolar pressure is atmospheric at residual volume and at total lung capacity.

Hyperventilation in response to a stressful situation leads to a. A decrease in the blood flow to the brain b. An increase in the activity of the central chemoreceptors c. A decrease in pH of the arterial blood d. An increase in the resistance of the pulmonary blood vessels e. A decrease in the excitability of nerve and muscle cells

A. Hyperventilation, by definition, reduces the arterial PCO2 below normal, that is, to a value less than 40 mmHg. Low arterial PCO2 in the blood perfusing the brain causes vasoconstriction of the cerebral vasculature, leading to a lowering of blood flow to the brain. The low arterial CO2 will increase the pH of the arterial blood (i.e., produce a respiratory alkalosis) and decrease the activity of the central chemoreceptors. Pulmonary vasculature, unlike the systemic vas- culature, is dilated by low PCO2 and high PO2 and constricted by high PCO2 and low PO2. The respiratory alkalosis will cause Ca2+ to bind to plasma proteins, lowering the concentration of ionized Ca2+. Low concentrations of ionized Ca2+ increase the excitability of nerve and muscle membranes, causing them to contract spontaneously or with minimal stimulation. The increased contractile activity of skeletal muscle produced by low concen- tration of ionized Ca2+ is called tetany

During normal spontaneous breathing the respiratory parameters of a person is measured as: Tidal volume (VT) = 500 ml and Respiratory frequency (FR) = 16 breaths/min. After a thorax injury breathing of this patient becomes very painful. To reduce the magnitude of chest excursion by breathing and thus reducing pain, the physician decides to ventilate him artificially (using a ventilator) with a smaller tidal volume but a larger frequency than his spontaneous breathing. His VT has been set to 250 ml and his FRto 32 breaths / min. How will the ventilation of this person be correctly described after transition of his breathing to artificial ventilation? A. Hypoventilation B. Hyperventilation C. Normoventilation

A. Remember that alveolar ventilation and PaCO2 are inversely related. Alveolar ventilation has decreased therefore PaCO2 has increased. When PaCO2 >40 it is hypoventilation.

1. What tends to decrease airway resistance ? A ) Asthma B ) Stimulation by sympathetic fibers C ) Treatment with acetylcholine D ) Exhalation to residual volume

B ) stimulation of sympathetic fibers A decrease in airway resistance is due to an increase in the diameter of the airway . Asthma causes bron choconstriction , which is prevented by ß - agonists . Sympathetic stimulation of the airways results in a relaxation of airways , decreasing resistance Acetyl choline is a bronchoconstrictor , increasing resistance . With low lung volumes there is a collapse of the air ways , leading to decreased diameter and increased resistance .

Which of the following conditions causes a decrease in arterial O2 saturation without a decrease in O2 tension? a. Anemia b. Carbon monoxide poisoning c. A low V/Q ratio d. Hypoventilation e. Right-to-left shunt

B. A rise or fall in the oxygen saturation of hemoglobin is caused by a decrease in the arterial O2 tension, which can result from a low V/Q ratio, hypoventilation, or an anatomical right-to-left shunt. Because carbon monoxide (CO) competes with O2 on the hemoglobin molecule, CO poisoning can cause a decrease in hemoglobin's saturation even when the PO2 is normal. In anemia, oxygen saturation is unchanged but the concentration of hemoglobin is less than normal, reducing the oxygen content.

When a person ascends to a high altitude, alveolar ventilation increases. Alveolar ventilation continues to increase over the next several days because a. The central chemoreceptors become more sensitive to low oxygen tensions b. The peripheral chemoreceptors increase their firing rate c. The plasma concentration of 2,3-DPG increases d. The pH of the cerebrospinal fluid decreases e. The oxygen-carrying capacity of hemoglobin increases

B. Hyperventilation occurs as soon as a person is exposed to low atmospheric pressures at high altitudes. The rise in pH attenuates the chemoreceptor response to low oxygen ten- sions. When the pH of the cerebrospinal fluid returns to normal, the chemoreceptors are restored to their normal responsiveness, and they are able to produce an even greater increase in ventilation. The plasma concentration of 2,3-DPG and, therefore, the oxygen-carrying capacity of hemoglobin increase with prolonged exposure to high altitudes. These changes increase the ability of the cardiovascular system to deliver oxygen to the tissues.

A 25 year old man has been suffering from malabsorption and steatorrhoea since his childhood. He is thin and has a frequent productive cough. In recent years he developed bronchiectasis associated with chronic retention of bronchial secretions and dysfunction of mucociliary transport. He is diagnosed with cystic fibrosis that developed into chronic obstructive lung disease (lung emphysema). As his disease worsens, he experiences difficulty breathing during expiration. His exhalation takes longer and he tries to help himself by lip pursing during forced expiration. What will most likely be the cause of lengthening his exhalation in this condition? A. A reduced alveolar pressure due to decreases in lung compliance B. A reduced alveolar pressure due to decreases in lung recoiling force C. A shift in the airway equal pressure point towards mouth D. A decrease in airway opening pressure

B. In emphysema you are increasing compliance, decreasing recoil force. Pursing lips during forced expiration will lengthen his exhalation because the patient is increasing the pressure at the mouth to decrease pressure in the alveolar region.

Airway resistance is lowest a. During a forced expiration b. At the total lung capacity c. At the residual volume d. During vagal stimulation e. When breathing gas with low oxygen

B. Most of the airway is within the thoracic cavity, and, therefore, the intrathoracic pressure affects airway diameter and resistance. The intrathoracic pressure is most negative at the total lung capacity. The negative thoracic pressure increases airway diameter and decreases airway resistance. During a forced expiration or at the residual volume, the intrathoracic pressure is positive, compressing the airways and increasing their resistance. The vagus nerve constricts airway smooth muscle. Low oxygen will have little effect on airway tone.

6. A 28 - year - old on oral contraceptives develops tachypnea and reports dyspnea . A ventilation perfusion scan is ordered to check for pulmonary thromboemboli . Which of the following best explains why , as she takes in a normal inspiration , more air goes to the alveoli at the base of the lung than to the alveoli at the apex of the lung ? A. The alveoli at the base of the lung have more surfactant B. The alveoli at the base of the lung are more compliant C. The alveoli at the base of the lung have higher V / Q ratios D. There is a more negative intrapleural pressure at the base of the lung E. There is more blood flow to the base of the lung

B. The alveoli at the base of the lung are more compliant During inspiration , when all alveoli are subjected to essentially the same alveolar pressure , more air will go to the more compliant alveoli in the base of the lung . Because the lungs are essentially " hanging " in the chest , the force of gravity on the lungs causes the intrapleural pressure to increase ( become less negative ) at the base of the lungs compared to the apex ( more negative intrapleural pressure ) . This also causes the alveoli at the apex of the lung to be larger than those at the base of the lung . Larger alveoli are already more inflated and are less compliant than smaller alveoli . Because of the effect of gravity on blood , more blood flow will go to the base of the lung . Ventilation is about three 3 times greater at the base of the lung , but flow is about 10 times greater at the base than at the apex of the lung ; therefore , the V / Q ratio is lower at the base than at the apex in a normal lung .

During a normal inspiration, more air goes to the alveoli at the base of the lung than to the alveoli at the apex of the lung because a. The alveoli at the base of the lung have more surfactant b. The alveoli at the base of the lung are more compliant c. The alveoli at the base of the lung have higher V/Q ratios d. There is a more negative intrapleural pressure at the base of the lung e. There is more blood flow to the base of the lung

B. There is a negative, or sub- atmospheric, intrapleural pressure (PIP) between the lungs and the chest wall due to the tendency of the chest wall to pull outward and the tendency of the lungs to collapse. Because the lungs are essentially "hanging" in the chest, the force of gravity on the lungs causes the PIP to be more negative at the top of the lung. This also causes the alveoli at the apex (top) of the lung to be larger than those at the base (bottom) of the lung. Larger alveoli are already more inflated and are less compliant than smaller alveoli. During inspiration, when all alveoli are subjected to essentially the same alveolar pressure, more air will go to the more compliant alveoli. Because of the effect of gravity on blood, more blood flow will go to the base of the lung. This does not appreciably affect lung compliance. Ventilation is about three times greater at the base of the lung, but flow is about 10 times greater at the base than at the apex of the lung. Therefore, the V ̇ / Q ̇ ratio is lower at the base than at the apex in a normal lung.

Complete transection of the brainstem above the pons would a. Result in cessation of all breathing movements b. Prevent any voluntary holding of breath c. Prevent the central chemoreceptors from exerting any control over ventilation d. Prevent the peripheral chemoreceptors from exerting any control over ventilation e. Abolish the Hering-Breuer reflex

B. Transection of the brain- stem above the pons would prevent any voluntary changes in ventilation by cutting the pathways from the higher centers. Breathing would continue because the pontine-medullary centers that control rhythmic ventilation would be intact. Inputs to the brainstem from the central and peripheral chemoreceptors that stimulate ventilation and from lung stretch receptors that inhibit inspiration (Hering-Breuer reflex) would also be intact and these reflexes would be maintained.

A young skier with normal pulmonary function (minute volume 4 L; pulmonary blood flow 5 L/min) who is recovering from a tibial fracture suddenly develops right-sided chest pain and tachypnea. Embolic occlusion of the right pulmonary artery is suspected. Which of the following alveolar gas measurements would immediately confirm the diagnosis? PO2 (mmHg) a. 125 b. 125 c. 100 d. 80 e. 80 PCO2 (mmHg) a. 60 b. 20 c. 40 d. 20 e. 60

B. Under normal conditions, the V/Q ratio in both lungs is the same, so that mixed alveolar gas in both lungs has the same PO2 and PCO2. The gas from the normal lung will be approximately normal (PO2 = 100 mmHg, PCO2 = 40 mmHg). While the gas from the occluded lung, which now represents alveolar dead space, will resemble tracheal gas (PO2 = 150 mmHg, PCO2 = 0 mmHg). When the gas from the two lungs mixes, the average alveolar O2 becomes 125 mmHg and the average PCO2 becomes 20 mmHg.

A patient with inadequate surfactant will have a relatively normal a. FEV1 b. FVC c. FEV1/FVC d. MVV e. V/Q ratio

C. Surfactant acts to diminish the surface tension of the lung. In the absence of surfactant, the lungs are more difficult to expand. Patients with a restrictive lung disease cannot expand their lungs as easily as normal, and, therefore, their total lung capacity is lower than normal. Because they have a lower total lung volume, the amount of gas they can exhale in 1 s (the FEV1) is less than normal. Similarly, their forced vital capacity (FVC) is lower than normal. However, because they have a greater than normal recoil force, they can exhale a normal or even fraction of their TLC in 1 s. Therefore, their FEV1/FVC is normal or even slightly higher than normal. The greater recoil reduces the amount of gas they can inhale with each breath so their maxi- mum voluntary ventilation (MVV) is less than normal.

The basic respiratory rhythm is generated in the a. Apneustic center b. Nucleus parabrachialis c. Dorsal medulla d. Pneumotaxic center e. Cerebrum

C. The basic respiratory rhythm originates from spontaneous rhythmic discharge of inspiratory neurons located in the respiratory center in the dorsal medulla. This basic rhythm can be modified by many factors, including voluntary control of breathing. In the pons are the apneustic center and the pneumotaxic cen-ter (which is located in the nucleus parabrachialis). These modify and regularize the basic respiratory rhythm to produce adequate breathing.

The equal pressure point would be most likely to move closer to the mouth with an increase in a. Airway resistance b. Lung compliance c. Lung volume d. Expiratory effort e. Airway smooth muscle tone

C. The equal pressure point moves further away from the lungs if the recoil force is increased and moves closer to the lungs when the intrapleural pressure is increased. Increasing the lung volume expands the alveoli and makes their recoil force greater. This moves the equal pressure point toward the mouth. If airway resistance increases, for example, by increasing airway smooth muscle tone, then a greater expiratory effort and consequently a greater intra- pleural pressure will be necessary to expel the gas from the lungs.

An increase in pulmonary blood flow during exercise a. Causes alveolar oxygen tension to decrease b. Causes the V/Q ratio at the top of the lung to increase c. Causes pulmonary arterial resistance to decrease d. Causes diffusion capacity to decrease e. Causes arterial oxygen saturation to decrease

C. The increase in blood flow through the pulmonary circulation during exercise increases the diameter of the pulmonary vessels and therefore decreases their resistance. The decrease in resistance can occur because the pulmonary vessels are very compliant and are large enough to prevent an increase in pulmonary artery blood pressure when cardiac output increases. The increased blood flow to the lung evens out the V/Q ratio, decreasing it at the top of the lung and increasing it at the bottom. Making the V/Q ratio more similar throughout the lung may decrease some of the shunting effect that normally occurs at the base of the lung where the V/Q ratio is quite low and may actually increase the oxygen saturation of arterial blood. The increased blood flow will also increase the surface area for respiratory exchange and therefore increase the diffusing capacity.

Which one of the following will decrease in a person with ventilation- perfusion (V/Q) abnormalities? a. Anion gap b. Arterial pH c. Arterial carbon dioxide tension d. Alveolar-arterial (A-a) gradient for oxygen e. Alveolar ventilation

C. V/Q mismatches will cause arterial oxygen levels (PaO2) to decrease. Decreased PaO2 will stimulate the peripheral chemoreceptors, which, in turn, will increase alveolar ventilation and decrease PaCO2. The decreased PaCO2 will cause a respiratory alkalosis (increasing pH). Hypoxemia will also cause lactate levels to rise, increasing the anion gap (and blunting the rise in pH). The fall in PaO2 causes the A-a gradient to rise.

3. A healthy , 25 - year - old medical student participates in a 10 - kilometer charity run for the American Heart Asso ciation . Which muscles does the student use ( contract ) during expiration ? A ) Diaphragm and external intercostals B ) Diaphragm and internal intercostals C ) Diaphragm only D ) Internal intercostals and abdominal recti E ) Scaleni F ) Sternocleidomastoid muscles

D ) Internal intercostals and abdominal recti Contraction of the internal intercostals and ab dominal recti pull the rib cage downward during ex piration . The abdominal recti and other abdominal muscles compress the abdominal contents upward toward the diaphragm , which also helps to eliminate air from the lungs . The diaphragm relaxes during ex piration . The external intercostals , sternocleidomas toid muscles , and scaleni increase the diameter of the chest cavity during exercise and thus assist with inspiration , but only the diaphragm is necessary for inspiration during quiet breathing .

Which of the following would normally be less in the fetus than in the mother? a. PaCO2 b. Pulmonary vascular resistance c. Affinity of hemoglobin for oxygen d. PaO2 e. Arterial hydrogen ion concentration

D. Because fetal hemoglobin (hemoglobin F) is chemically different from adult hemoglobin in that it has two a and two g chains instead of two a and two b chains, it has a greater affinity for oxygen. This is advantageous in the placental exchange of O2 from maternal blood (PaO2 = 100 mmHg) to fetal blood (PaO2 = 25 mmHg). PaCO2 is about 2 to 3 mmHg higher in the fetus than in the mother. Arterial H+ concentration is about the same in both. Pulmonary vascular resistance is high in the fetus, shunting blood away from the lungs. It decreases at birth and remains low.

During normal spontaneous breathing the respiratory parameters of a person is measured as: Tidal volume (VT) = 500 ml and Respiratory frequency (FR) = 16 breaths/min. After a thorax injury breathing of this patient becomes very painful. To reduce the magnitude of chest excursion by breathing and thus reducing pain, the physician decides to ventilate him artificially (using a ventilator) with a smaller tidal volume but a larger frequency than his spontaneous breathing. His VT has been set to 250 ml and his FR to 32 breaths / min. What will most likely be the status of vasculature of this person after transition of his spontaneous breathing to artificial ventilation? A. Systemic vasoconstriction B. Pulmonary Vasodilation C. Maintained systemic vascular tone D. Pulmonary vasoconstriction E. Maintained pulmonary vascular tone

D. In the pulmonary system this will induce vasoconstriction

PVR is decreased A. At low lung volumes B. With decreased CO C. By breathing low oxygen D. With increased pulmonary artery pressure E. At high lung volumes

D. Increased pulmonary artery pressure

A 25 year old man has been suffering from malabsorption and steatorrhoea since his childhood. He is thin and has a frequent productive cough. In recent years he developed bronchiectasis associated with chronic retention of bronchial secretions and dysfunction of mucociliary transport. He is diagnosed with cystic fibrosis that developed into chronic obstructive lung disease (lung emphysema). How will lip pursing helps this patient to exhale? A. It decreases pressure inside the airways B. It shifts the equal pressure point closer to alveolar region C. It causes alveolar pressure to increase during forced exhalation D. It prevents a negative transmural pressure to occur across the airways E. It causes alveolar pressure to decrease during forced exhalation

D. The lip pursing will cause an increase in pressure at the mouth that will prevent negative transmural pressure from occuring across the airways.

A 42-year-old woman with severe pulmonary fibrosis is evaluated by her physician and has the following arterial blood gases: pH = 7.48, PaO2 =55 mm Hg, and PaCO2 = 32 mm Hg. Which statement best explains the observed value of PaCO2? (A) The increased pH stimulates breathing via peripheral chemoreceptors (B) The increased pH stimulates breathing via central chemoreceptors (C) The decreased PaO2 inhibits breathing via peripheral chemoreceptors (D) The decreased PaO2 stimulates breathing via peripheral chemoreceptors (E) The decreased PaO2 stimulates breathing via central chemoreceptors

D. The patient's arterial blood gases show increased pH,decreased PaO2, and decreased PaCO2. The decreased PaO2 causes hyperventilation (stimulates breathing) via the peripheral chemoreceptors, but not via the central chemoreceptors. The decreased PacO2 results from hyperventilation (increased breathing) and causes increased pH, which inhibits breathing via the peripheral and central chemoreceptors.

The percentage of hemoglobin saturated with oxygen will increase if a. The arterial PCO2 is increased b. The hemoglobin concentration is increased c. The temperature is increased d. The arterial PO2 is increased e. The arterial pH is decreased

D. The percentage of hemoglobin saturated with oxygen depends on the level of PO2 in the blood and on other factors that affect the position of the oxyhemoglobin dissociation curve. Factors that shift the curve to the left, such as a decrease in PCO2, an increase in pH, or a decrease in temperature, would increase the percentage of hemoglobin saturated with oxygen as would an increase in PO2, provided that the percentage of saturation was not already at 100%. At a given PO2, increasing the concentration of hemoglobin would not affect the percentage of saturation but would increase the oxygen content of the blood.

During normal spontaneous breathing the respiratory parameters of a person is measured as: Tidal volume (VT) = 500 ml and Respiratory frequency (FR) = 16 breaths/min. After a thorax injury breathing of this patient becomes very painful. To reduce the magnitude of chest excursion by breathing and thus reducing pain, the physician decides to ventilate him artificially (using a ventilator) with a smaller tidal volume but a larger frequency than his spontaneous breathing. His VT has been set to 250 ml and his FR to 32 breaths / min. What will most likely be the alveolar ventilation and dead space ventilation of this person after transition of his breathing to artificial ventilation? A. Increased alveolar ventilation above normal B. Equal alveolar and dead space ventilations C. Decreased dead space ventilation below normal D. Decreased alveolar ventilation below normal E. Unchanged dead space ventilation

D. When you increase frequency, alveolar ventilation will increase, but a decrease in Vt will only impact the alveolar region (not the dead space) so that decrease in Vt will all decrease alveolar ventilation below normal.

2. The pleural pressure normal 56 - year - old woman is approximately -5 cm H₂O during resting conditions immediately before inspiration ( i.e. , at functional re sidual capacity [ FRC ] ) . What is the pleural pressure ( in cm H₂O ) during inspiration ? A ) +1 B ) +4 C ) 0 D ) -3 E ) -7

E ) -7 The pleural pressure ( sometimes called the in trapleural pressure ) is the pressure of the fluid in the narrow space between the visceral pleura of the lungs and parietal pleura of the chest wall . The pleural pressure is normally about -5 cm H₂O immediately before inspiration ( i.e. , at FRC ) when all of the respi ratory muscles are relaxed . During inspiration , the volume of the chest cavity increases and the pleural pressure becomes more negative . The pleural pres sure averages about -7.5 cm H₂O immediately before expiration when the lungs are fully expanded . The pleural pressure then returns to its resting value of -5 cm H₂O as the diaphragm relaxes and lung volume returns to FRC . Therefore , the intrapleural pressure is always subatmospheric under normal conditions varying between -5 and -7.5 cm H₂O during quiet breathing .

Because the small airways produce a small fraction of the airway resistance, their obstruction is difficult to detect. Which one of the follow- ing changes is indicative of small rather than large airway obstruction? a. An increase in maximum airflow during a forced expiration b. A decrease in the maximum expiratory pressure that can be generated c. An increase in the turbulent airflow produced during inspiration d. A decrease in the forced vital capacity e. An increase in the closing volume of the lung

E. Although it may seem paradoxical, about 80% of the total resistance to airflow occurs in the large- and medium-sized airways. Turbulent flow, which increases the pressure necessary to cause gas to flow, is more likely to occur in larger airways. In patients, an increased resistance to airflow usually indicates the presence of disease of the large airways.

The clinical sign of cyanosis is caused by a. An increase in the affinity of hemoglobin for oxygen b. A decrease in the percent of red blood cells (hematocrit) c. An increase in the concentration of carbon monoxide in the venous blood d. A decrease in the concentration of iron in the red blood cells e. An increase in the concentration of deoxygenated hemoglobin

E. Cyanosis is the blue color of the skin produced by desaturated hemoglobin. Cyanosis appears when 5 g of hemoglobin per 100 mL of blood are desaturated. For a per- son with a normal hemoglobin concentration of 15 g/100 mL, cyanosis appears when one-third of the blood is desaturated. For a person with polycythemia (a higher than normal concentration of hemoglobin), cyanosis may appear when only one-fourth of the hemoglobin is desaturated. This individual may not be hypoxic because of the high concentration of saturated hemoglobin. On the other hand, a person with anemia (a lower than normal concentration of hemoglobin) may have a significant portion of the hemoglobin desaturated without displaying cyanosis. This individual will not appear cyanotic but may be hypoxic.

Hypoxemia produces hyperventilation by a direct effect on the (A) phrenic nerve (B) J receptors (C) lung stretch receptors (D) medullary chemoreceptors (E) carotid and aortic body chemoreceptors

E. Hypoxemia stimulates breathing by a direct effect on the peripheral chemoreceptors in the carotid and aortic bodies. Central (medullary) chemoreceptors are stimulated by CO2 (or H+). The J receptors and lung stretch receptors are not chemoreceptors. The phrenic nerve innervates the diaphragm, and its activity is determined by the output of the brain stem breathing center.

Venous blood contains more CO2 than arterial blood despite the same PCO2. What is responsible for this?

Haldane effect: O2 is affecting the affinity of Hb for Co2/H+ Bohr: CO2 and H+ are affecting the affinity of Hb for O2

Swimmers hyperventilate because:

Increases the time before carotid and central PCO2/pH receptors increase their firing

A 38-year-old woman moves with her family from New York City (sea level) to Leadville Colorado (10,200 feet above sea level). Which of the following will occur as a result of residing at high altitude? (A) Hypoventilation (B) Arterial PO2 greater than 100 mm Hg (C) Decreased 2,3-diphosphoglycerate (DPG) concentration (D) Shift to the right of the hemoglobin-O2 dissociation curve (E) Pulmonary vasodilation (F) Hypertrophy of the left ventricle (G) Respiratory acidosis

The answer is D. At high altitudes, the PO2 of alveolar air is decreased because barometric pressure is decreased. As a result, arterial PO2 is decreased (<100 mm Hg), and hypoxemia occurs and causes hyperventilation by an effect on peripheral chemoreceptors. Hyperventilation leads to respiratory alkalosis. 2,3-Diphosphoglycerate (DPG) levels increase adaptively; 2,3-DPG binds to hemoglobin and causes the hemoglobin-O2 dissociation curve to shift to the right to improve unloading of O2 in the tissues. The pulmonary vasculature vasoconstricts in response to alveolar hypoxia, resulting in increased pulmonary arterial pressure and hypertrophy of the right ventricle (not the left ventricle).

The pH of venous blood is only slightly more acidic than the pH of arterial blood because (A) CO2 is a weak base (B) there is no carbonic anhydrase in venous blood (C) the H+ generated from CO2 and H2O is buffered by HCO3- in venous blood (D) the H+ generated from CO2 and H2O is buffered by deoxyhemoglobin in venous blood (E) oxyhemoglobin is a better buffer for H+ than is deoxyhemoglobin

The answer is D. In venous blood, CO2 combines with H2O and produces the weak acid H2CO3, catalyzed by carbonic anhydrase. The resulting H+ is buffered by deoxyhemo- globin, which is such an effective buffer for H+ (meaning that the pK is within 1.0 unit of the pH of blood) that the pH of venous blood is only slightly more acid than the pH of arterial blood. Oxyhemoglobin is a less effective buffer than deoxyhemoglobin.

A person with a ventilation/perfusion (V/Q) defect has hypoxemia and is treated with supplemental O2. The supplemental O2 will be most helpful if the person's predomi- nant V/Q defect is (A) dead space (B) shunt (C) high V/Q (D) low V/Q (E) V/Q = 0 (F) V/Q = ×

The answer is D. Supplemental O2 (breathing inspired air with a high PO2) is most helpful in treating hypoxemia associated with a ventilation/perfusion (V/Q) defect if the predominant defect is low V/Q. Regions of low V/Q have the highest blood flow. Thus, breathing high PO2 air will raise the PO2 of a large volume of blood and have the greatest influence on the total blood flow leaving the lungs (which becomes systemic arterial blood). Dead space (i.e., V/Q = ∞ has no blood flow, so supplemental O2 has no effect on these regions. Shunt (i.e., V/Q = 0) has no ventilation, so supplemental O2 has no effect. Regions of high V/Q have little blood flow, thus raising the PO2 of a small volume of blood will have little overall effect on systemic arterial blood.

170. A 26 - year - old man training for a marathon reaches a workload that exceeds his anaerobic threshold . If he continues running at or above this workload , which of the following will increase ? a . Alveolar ventilation b . Arterial pH c . Paco₂ d . Plasma [ HCO3 ] e . Firing of the central chemoreceptors

a . Alveolar ventilation During exercise , minute ventilation and alveolar ventilation increase linearly with carbon dioxide production up to a level of about 60 % of the maximal workload . Above that level , called the anaerobic threshold , muscle lactate spills into the circulation causing a metabolic acidosis , characterized by a decrease in pH and [ HCO3 ] The increased [ H ] stimulates the peripheral ( not central ) chemoreceptors to increase alveolar ventilation more proportionally than the increase in carbon dioxide production , resulting in a decrease in Paco₂ .

163. A 68 - year - old woman with pulmonary fibrosis presents with a complaint of increasing dyspnea while performing activities of daily living . She is referred for pulmonary function testing to assess the progression of her disease . Which of the following laboratory values is consistent with her diagnosis ? a . Decreased diffusing capacity of the lung b . Increased residual volume c . Decreased forced expiratory volume exhaled in 1 second ( FEV₁ ) / forced vital capacity ( FVC ) d . Increased lung comp liance e . Increased airway resistance corrected for lung volume

a . Decreased diffusing capacity of the lung In pulmonary fibrosis , the diffusing capacity of the lung is decreased due to an increase in the thickness of the diffusional barrier , as predicted by Fick law of diffusion . Pulmonary fibrosis is characterized by a decrease in lung compliance and an increase in lung elastic recoil ( " stiff ' lungs ) , which results in findings typical of a restrictive impairment . Pulmonary function test values characteristic of a restrictive impairment include a decrease in all lung volumes and capacities and a ratio of the FEV , to the total FVC that is normal or increased . Airway radius is decreased , and thus airway resistance is increased , at lower lung volumes , but in restrictive disorders , the airway resistance is normal when corrected for lung volume in contrast to obstructive disorders , in which an increased airway resistance is a hallmark of the functional impairment .

178. A 30 - year - old woman is admitted to the emergency department with dyspnea tachycardia , confusion , and other signs of hypoxia . The following laboratory data were obtained while the patient was breathing room air : Pao₂ = 67 mm Hg Sao2 = 90 % Paco₂ = 60 mm Hg PvO₂ = 30 mm Hg pH = 7.27 Svo₂ = 55 % [ HCO3 ] = 26 mEq / L Vo₂ = 350 mL / min [ Hb ] = 15 g % Cao2 - CVO₂ = 7 mL 0₂ / 100 mL Which of the following is the most appropriate classification of the patient's hypoxia ? a . Hypoxic hypoxia ( hypoxemia ) b . Anemic hypoxia c . Stagnant ( hypoperfusion ) hypoxia d . Histotoxic hypoxia e . Carbon monoxide poisoning

a . Hypoxic hypoxia ( hypoxemia ) Alveolar hypoventilation ( as evidenced by the higher - than - normal value of Paco₂ ) is a type of hypoxic hypoxia or hypoxemia ( as evidenced by the decreased Pao₂ ) . Anemic hypoxia is characterized by a decreased concentration of hemoglobin ( anemia ) or a reduction in the saturation of hemoglobin with oxygen ( SaO₂ ) expected for a given Pao2 , as would occur in carbon monoxide poisoning or methemoglobinemia . Stagnant hypoxia is characterized by a decreased cardiac output ; in this patient , cardiac output , calculated as ( Voy / Cao , - Cvo , ) is 5 L / min , which is normal . In histotoxic hypoxia , oxygen extraction is impaired , and thus CaO₂ - CvO₂ would be less than normal and SvO₂ would be greater than normal

173. A 58 - year - old woman experiences an acute exacerbation of asthma , which causes her breathing to become labored and faster . As a result , which of the following changes in airflow is expected ? a . Flow in the trachea and upper airways will become more laminar . b . The pressure gradient required for airflow will increase . c . The resistance to airflow will decrease . d . The resistance to airflow will increase linearly with the decrease in airway radius . e . Reynolds number will decrease .

b . The pressure gradient required for airflow will increase . An increased velocity of airflow will increase turbulent airflow , as predicted by an increased . Reynolds number . Resistance to turbulent airflow exceeds that for laminar airflow , and thus the pressure gradient required for airflow increases when flow is turbulent . Because the velocity of airflow is greatest in the trachea and large airways , the predisposition to turbulent airflow is greater in the central than in the peripheral airways . Airway resistance varies inversely with the fourth power of airway radius , according to Poiseuille law .

165. A 150 - lb patient scheduled for abdominal surgery is sent for preoperative evaluation and testing . His chest x - ray is normal , and pulmonary function results on room air show the following : Tidal volume 600 mL Respiratory rate = 12 / min Vital capacity = 5000 mL Pao₂ = 90 mm Hg Paco₂ = 40 mm Hg PECO₂ = 28 mm Hg The volume of the patient's physiological dead space , determined by applying the Bohr equation , equals which of the following ? a . 0.3 mL b . 150 mL c . 180 mL d . 420 mL e . 7200 mL

c . 180 mL Physiological dead space is the volume of the respiratory tract that is ventilated but not perfused by the pulmonary circulation . The Bohr equation for determination of the ratio of the physiologic dead space ( VD ) to the tidal volume ( VT ) is : VD / Vt = Paco₂ - PECO₂ / PaCo₂ VD / Vt = 40-28 / 40 = 0.3 VD / Vt X Vt = VD 0.3 x 600 mL = 180 mL Physiological dead space volume is equal to the sum of the anatomic dead space and the alveolar dead space . Anatomic dead space , which represents the volume of the conducting airways ( nose to the terminal bronchioles ) , can be measured using the Fowler technique , but it is often estimated as 1 mL per pound of body weight . Alveolar dead space represents the volume of alveoli that are ventilated but not perfused . Because there is normally no alveolar dead space , physiologic dead space volume approximates anatomic dead space volume in persons with normal lung function .

161. A group of third - year medical students accompanied a medical mission team to Peru , South America . After arriving at the airport in Bolivia , they hiked to a remote mountain village in the Andes at an elevation of 18,000 ft . With a barometric pressure of 380 mm Hg at this altitude , what would be the resulting Po₂ of the dry inspired air ? a . 160 mm Hg . b . 100 mm Hg . c . 80 mm Hg d . 70 mm Hg e . 38 mm Hg

c . 80 mm Hg According to Dalton's law , the partial pressure of a gas is the product of the fractional composition of the gas and the total pressure of the gaseous mixture . Oxygen constitutes approximately 21 % of dry atmospheric air . Therefore , the partial pressure of O₂ in dry atmospheric air equals the fractional concentration of oxygen ( FIO₂ ) times the atmospheric ( barometric ) pressure . At sea level , the barometric pressure is 760 mm Hg , yielding a PIO₂ of 160 mm Hg . At high altitude , the barometric pressure decreases in proportion to the decreased weight of the air above it . At an elevation of 18,000 ft in the Peruvian Andes , the barometric pressure is 380 mm Hg . yielding a PIO₂ of 80 mm Hg . Once inside the respiratory tract , the inspired air becomes warmed and humidified . The partial pressure of H₂ O vapor is temperature dependent rather than concentration dependent and at body temperature ( 37 ° C ) it is 47 mm Hg . The presence of H₂ O vapor reduces the partial pressure of the other gases in the atmosphere , and the PH₂ O must be subtracted from the total barometric pressure before multiplying by the fractional concentration of a gas to yield the partial pressure of the gas . Thus , at sea level , the humidified PIO₂ in the conducting airways is 0.21 ( 760-47 ) or 150 mm Hg , whereas , at 18,000 ft , the humidified , tracheal Po₂ would be 0.21 ( 380-47 ) or 70 mm Hg .

175. A newborn of 28 weeks of gestation develops respiratory distress syndrome . Mechanical ventilation on 100 % O₂ with 10 cm H₂O of positive end - expiratory pressure ( PEEP ) does not provide sufficient oxygenation . After porcine surfactant is instilled via a fiberoptic bronchoscope , the Paco₂ , fraction of inspired oxygen ( FIO₂ ) , and shunting improve impressively . The improvements in respiratory function occurred because surfactant increased which of the following ? a . Alveolar surface tension b . Bronchiolar smooth muscle tone c . Lung compliance d . The pressure gradient needed to inflate the alveoli e . The work of breathing

c . Lung compliance Pulmonary surfactant increases lung compliance by lowering alveolar surface tension . As a result , the pressure gradient needed to inflate the alveoli decreases , as does the work of breathing Although surfactant replacement therapy has proven to be beneficial in respiratory distress syndrome of the newborn , surfactant replacement therapy is not currently recommended in acute respiratory distress syndrome based on clinical evidence against efficacy of the therapy .

171. A medical student waiting for her first patient interview at the clinical skills center becomes very anxious and increases her rate of alveolar ventilation . If her rate of CO₂ production remains constant , which of the following will decrease ? a . pH b . Pao2 c . Paco₂ d . V/Q e . Alveolar - arterial Po₂ difference

c . Paco₂ Because the dead space air does not participate in gas exchange , the entire output of CO₂ in the expired gas comes from the alveolar gas . Accordingly , alveolar ( and arterial ) PCO₂ can be expressed in terms of CO₂ output and alveolar ventilation according to the following equation : PACO₂ = Paco₂ = VCO₂ / VA Thus , an increase in alveolar ventilation at a constant rate of carbon dioxide production will lower PACO₂ and Paco₂ . Hyperventilation increases PAO2 and PaO2 , with no change in the alveolar - arterial Po₂ difference . The VIO will be normal or increased .

162. A 28 - year - old man is admitted to the emergency department with multiple fractures suffered in a car accident . Arterial blood gases are ordered while the patient is breathing room air . After the first - year resident obtains an arterial blood sample from the patient , the glass plunger slides back , drawing an air bubble into the syringe before it is handed to the blood gas technician for analysis . How does exposure to room air affect the measured values of Po₂ and PCO₂ in arterial blood ? a . The measured values of both Pao , and Paco₂ will be higher than the patient's actual values . b . The measured values of both Pao , and Paco₂ will be lower than the patient's actual values . c . The measured PaO₂ will be higher and the measured Paco , will be lower than the patient's actual blood gas values . d . The measured Pao₂ will be lower and the measured Paco , will be higher than the patient's

c . The measured PaO₂ will be higher and the measured Paco , will be lower than the patient's actual blood gas values . Room air contains 21 % O₂ and 0.04 % CO₂ , yielding a PIO₂ of 160 mm Hg and a PICO₂ of 0.3 mm Hg . Thus , if a sample of arterial blood is equilibrated with room air , the measured PaO₂ will have an inaccurately high reading and the Paco , will have an inaccurately low reading . For this reason , collecting an " anaerobic " blood sample is critical in blood gas analysis . Also , once the sample is obtained , the syringe should be placed in a container of crushed ice to prevent any metabolism by the red blood cells , which can also affect the accuracy of the readings . In addition to being certain that an air bubble is not left in the syringe , it is best to use a glass rather than a plastic syringe because the arterial pressure will pump the blood sample into a glass syringe without requiring aspiration , and glass is more impermeable to the diffusion of gases than plastic . A plastic syringe is permissible if one is drawing the blood sample from an arterial line rather than doing an arterial " stick " and if the sample is promptly analyzed .

169. A 125 - lb , 40 - year - old woman with a history of nasal polyps and aspirin sensitivity since childhood presents to the emergency department with status asthmaticus and hypercapnic respiratory failure . She requires immediate intubation and is placed on a mechanical ventilator on an FIO₂ of 40 % , a control rate of 15 breaths per minute , and a tidal volume of 500 mL . Which of the following is her approximate alveolar ventilation ? a . 375 ml / min b . 3500 mL / min c . 5250 mL / min d . 5625 mL / min e . 7500 mL / min

d . 5625 mL / min Alveolar ventilation Va equals the tidal volume ( V t) minus the dead space volume ( Vd) times the breathing frequency (f ) . The dead space volume can be estimated as 1 mL / lb of body weight . VA = ( Vt - VD ) xf VA = ( 500 mL - 125 mL ) x 15 breaths per minute VA = 5625 mL / min

172. A 36 - year - old man with a history of AIDS and Pneumocystis infection presents to the emergency department with severe respiratory distress . The patient is placed on a ventilator at a rate of 16 , tidal volume of 600 mL , and FIO₂ of 1.0 . An arterial blood sample taken 20 minutes later reveals a Po₂ of 350 mm Hg , a Pco₂ of 36 mm Hg , and a pH of 7.32 . At a barometric pressure of 757 mm Hg , and assuming a normal respiratory exchange ratio ( R ) of 0.8 , the patient's alveolar oxy gen tension is approximately which of the following ? a . 105 mm Hg b . 355 mm Hg c . 576 mm Hg d . 665 mm Hg e . 712 mm Hg

d . 665 mm Hg The alveolar air equation is used to calculate the PAO₂ . PAO₂ = PIO₂ - ( Paco₂ / R ) Given the barometric pressure of 757 mm Hg , Fio₂ = 1.0 ( 100 % O₂ ) , Paco₂ 36 mm Hg and R = 0.8 , then PAO₂ = ( 1.0 ) ( 757-47 ) - ( 36 / 0.8 ) = 710-45 = 665 mm Hg Pneumocystis is an opportunistic fungal pulmonary pathogen that is an important cause of pneumonia in immunocompromised hosts . HIV patients with a CD4 + cell count below 200 / μL have an increased risk of developing Pneumocystis pneumonia . According to new and still evolving nomenclature , Pneumocystis carinii is the name of the organism derived from rats , and Pneumocystis jiroveci is the name of the organism derived from humans .

CO poisoning, what blood finding is expected?

decreased arterial O2 concentration

174. A 27 - year - old woman at 30 weeks of gestation goes to the obstetrician for a prenatal visit . During the visit , she expresses concern that she has been breathing faster than usual . Lab results revealed the following Pao₂ = 105 mm Hg Tidal volume = 480 mL Paco₂ = 30 mm Hg Respiratory rate = 30 breaths / min pH = 7.47 R = 0.8 [ HCO3 ] = 20 mEq / L [ Hb ] = 12 g % PECO₂ = 18 mm Hg Anion gap = 12 mEq / L Based on the data , what conclusions can you draw about the level of the patient's alveolar ventilation ? a . Alveolar ventilation exceeds her minute ventilation . b . Alveolar ventilation is inadequate due to rapid , shallow breathing c . Alveolar ventilation is less than her dead space ventilation d . Alveolar ventilation matches the increased CO₂ production during pregnancy . e . Alveolar ventilation is greater than normal .

e . Alveolar ventilation is greater than normal . Alveolar ventilation is the volume of air entering and leaving the alveoli per minute . Alveolar ventilation is less than the minute ventilation ( minute volume ) because the last part of each inspiration remains in the conducting airways and does not reach the alveoli . The minute ventilation is the product of tidal volume and respiratory rate ( 14,400 mL / min ) . Alveolar ventilation cannot be measured directly but must be calculated by subtracting dead space ventilation from minute ventilation . The ratio of the physiological dead space volume to the tidal volume ( VD / VT ) can be calculated using the Bohr equation ( Paco₂ PECO₂ / Paco₂ ) , and then multiplied by the VT to yield the dead space volume , which when multiplied by the respiratory rate yields the dead space ventilation ( 5760 mL / min ) . Thus , alveolar ventilation in this patient is 8640 mL / min . The adequacy of alveolar ventilation is determined by the alveolar air equation , which states that the Paco₂ is approximately equal to the rate of carbon dioxide production divided by the rate of alveolar ventilation . At a normal rate of alveolar ventilation , Paco₂ is in the normal range of 35 to 45 mm Hg . Assuming a constant rate of carbon dioxide production , a decrease in alveolar ventilation ( hypoventilation ) causes a higher Paco₂ than normal ( ie , > 45 mm Hg ) and a rate of alveolar ventilation that is greater than normal hyperventilation ) " blows off " excessive CO₂ causing Paco , to decrease below normal ( ie , < 35 mm Hg ) . Thus , in this patient , the Paco₂ of 30 mm Hg indicates that she is hyperventilating . If her increase in alveolar ventilation matched an increased carbon dioxide production , then Paco , would be in the normal range .

168. An insulation worker presents with a chief complaint of dyspnea on exertion . Pulmonary function test is consistent with a restrictive impairment . His arterial PO₂ is normal at rest but hypoxemic during exercise stress testing . Which of the following is the most likely explanation for the decline in the patient's Pao , during exercise compared with rest ? a . A decreased partial pressure gradient for O₂ diffusion during exercise b . A decreased surface area for diffusion during exercise c . An increase in hemoglobin's affinity for O₂ during exercise resulting in more oxy gen being transported as oxyhemoglobin and less in the dissolved state d . An increased uptake of oxygen from the blood by exercising skeletal muscles e . An underlying diffusion impairment coupled with a decrease in pulmonary capillary transit time during exercise

e . An underlying diffusion impairment coupled with a decrease in pulmonary capillary transit time during exercise The time course for oxy gen transfer across the alveolar - capillary ( A - C ) membrane is shown in the graph below . At the entry of the pulmonary capillary , the partial pressure of oxygen starts at the Po₂ of the mixed venous blood , about 40 mm Hg , and rises fairly rapidly , reaching equilibration with the alveolar Po₂ within about 0.25 of a second , or about one - third of the time the blood is in the pulmonary capillary at a normal resting cardiac output ( called the pulmonary capillary transit time or erythrocyte transit time = 0.75 s ) . Once equilibration occurs and the Po2₂ in the pulmonary capillary blood equals the alveolar Po₂ , there is no partial pressure gradient ( AP ) for further oxygen transfer across the A - C membrane . The time between equilibration and when an erythrocyte leaves the pulmonary capillary is referred to as pulmonary capillary reserve time ( -0.5 s at a resting cardiac output ) . Because of the pulmonary capillary reserve time at rest , even individuals with mild - to - moderate diffusion impairment may have sufficient transfer of oxygen across the A - C membrane by the time an erythrocyte leaves the pulmonary capillary , such that the blood exits with a Po₂ in the normoxic range . More severe diffusion impairments , on the other hand , may slow down oxygen transfer to the extent that equilibration of pulmonary capillary and alveolar Po₂ is never reached and the blood leaving the pulmonary capillary is hypoxemic . During exercise , as cardiac output increases , pulmonary capillary transit time decreases to as fast as -0.25 s . When diffusing capacity is normal , the decreased pulmonary capillary transit time does not decrease oxygen transfer across the A - C