Dyspnea 1 & 2 Quiz
Questions from our small group sessions to help study for Cardio/Resp
Questions and Answers
- 1.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through.Which way does air move through the hole in the thorax immediately after the hole opens?
- A.
The air moves out from the inward force from the chest wall
- B.
The air moves in from the negative pressure between the chest wall and the lung
- C.
The air moves out from the collapse of the lung
- D.
The air moves in from the negative pressure in the lung
Correct Answer
B. The air moves in from the negative pressure between the chest wall and the lungExplanation
When the knife is withdrawn, the lung seals but there is a gaping hole in the chest wall. This creates a communication between the outside environment and the space between the chest wall and the lung, known as the pleural cavity. Normally, the pressure in the pleural cavity is negative compared to the atmospheric pressure, which helps keep the lungs inflated. However, with the hole in the chest wall, air can pass through and equalize the pressure between the pleural cavity and the atmospheric pressure. Therefore, the air moves in from the negative pressure between the chest wall and the lung.Rate this question:
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- 2.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through.Will the lung collapse?
- A.
No. The tension from the pleural space will hold it in place.
- B.
No. The outward recoil from the chest wall holds the lung inflated.
- C.
Yes. The lung will collapse due to its elastic recoil.
Correct Answer
C. Yes. The lung will collapse due to its elastic recoil.Explanation
Yes, the lung will collapse due to its elastic recoil. When the knife cuts the lung, it causes a hole in the chest wall that allows air to pass through. As the lung has elastic properties, it will naturally recoil or shrink back to its original size. This recoil will cause the lung to collapse, leading to a condition called pneumothorax. Despite the vital structures being missed and little bleeding occurring, the presence of a gaping hole in the chest wall allows air to enter the pleural space, leading to lung collapse.Rate this question:
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- 3.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through. Will the circumference of his chest increase or decrease?
- A.
Increase because he is breathing heavily
- B.
Increase because of the outward recoil of the chest wall
- C.
Decrease because of the collapsed lung
- D.
Decrease because he is taking shallow breaths
Correct Answer
B. Increase because of the outward recoil of the chest wallExplanation
The correct answer is "increase because of the outward recoil of the chest wall." When the lung is cut and air can pass through the gaping hole in the chest wall, the air will accumulate in the pleural space between the lung and the chest wall. This accumulation of air creates positive pressure, causing the chest wall to expand or recoil outward. As a result, the circumference of the chest will increase.Rate this question:
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- 4.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through. With this left sided injury, will his trachea move to the left or right?
- A.
Left. The trachea shifts towards the side with more intrathoracic pressure.
- B.
Left. The trachea shifts towards the side with less intrathoracic pressure
- C.
Right. The trachea shifts towards the side with less intrathoracic pressure
- D.
Right. The trachea shifts towards the side with more intrathoracic pressure
Correct Answer
C. Right. The trachea shifts towards the side with less intrathoracic pressureExplanation
When there is a penetrating injury to the chest, such as in this case, the trachea tends to shift towards the side with less intrathoracic pressure. In this scenario, the lung is cut but seals when the knife is withdrawn, resulting in a gaping hole in the chest wall that allows air to pass through. This creates a pneumothorax, which leads to a decrease in intrathoracic pressure on the affected side. As a result, the trachea will shift towards the opposite side, which in this case is the right side.Rate this question:
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- 5.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through. To treat him, you place an occlusive dressing over the wound on his left chest and position him on his side. Would blood oxygen be highest if he were to lie on his left side or his right side?
- A.
Left to improve circulation to the wounded side
- B.
Left to improve perfusion to the side with ventilation
- C.
Right to improve perfusion to the side with ventilation
- D.
Right to improve circulation to the wounded side
Correct Answer
C. Right to improve perfusion to the side with ventilationExplanation
Blood oxygen would be highest if he were to lie on his right side. This is because the left lung is cut and there is a gaping hole in the chest wall that allows air to pass through. By lying on the right side, it helps to improve perfusion to the side with ventilation, allowing for better oxygenation of the blood.Rate this question:
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- 6.
A 27-year old male makes a minor transgression in a bar and a disgruntled patron plunges a knife into his left chest. Fortunately all vital structures are missed. The lung is cut but seals when the knife is withdrawn. There is little bleeding, however; there is a gaping hole in the chest wall that air can pass through. Someone places one end of a tube into his left pleural space and makes an airtight seal. The other end of the tube is placed into a pail of water on the floor with the tip 2cm below the surface. Why?
- A.
More than 2 cm of pressure wouldn't allow trapped air to escape
- B.
The underwater seal prevents air from going back into the pleural space
- C.
Bubbles are seen in the chamber when air is still escaping from the lung
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
The correct answer is "all of the above". Placing one end of the tube into the left pleural space and the other end into a pail of water creates an underwater seal. This seal prevents air from going back into the pleural space, allowing trapped air to escape. Bubbles are seen in the chamber when air is still escaping from the lung. This setup ensures that more than 2 cm of pressure doesn't build up, which could hinder the escape of trapped air.Rate this question:
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- 7.
A 45-year old female has episodic attacks of SOB, wheeze, and cough. You make a diagnosis of asthma, a disease characterized by high airways resistance. During the attack will the peak forced expiratory flow rate be increased or decreased?
- A.
Increased due to high airways resistance
- B.
Increased due to high pressure change
- C.
Decreased due to high airways resistance
- D.
Decreased due to high pressure change
Correct Answer
C. Decreased due to high airways resistanceExplanation
Flows will be reduced. Asthma symptoms of wheeze, cough and SOB are due to increased airways resistance (inflammation mucous, remodeling). Thus flow rates are determined by Flow = (pressure change from alveolus to mouth)/resistanceRate this question:
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- 8.
A 45-year old female has episodic attacks of SOB, wheeze, and cough. You make a diagnosis of asthma, a disease characterized by high airways resistance. What will happen to inspiratory flow rates?
- A.
Reduced more than expiratory flow
- B.
Reduced less than expiratory flow
- C.
Increased more than expiratory flow
- D.
Increased less than expiratory flow
Correct Answer
B. Reduced less than expiratory flowExplanation
Inspiratory flow will be reduced for the same reason, but to a lesser degree. the flow rate still depends on the pressure difference across the tube and the resistance of the tube. However, on inspiration the resistance of the tube is somewhat diminished by the surrounding negative pressure, helping to keep the intrathoracic tubes open. On expiration the positive pressure surrounding the tubes tends to collapse them and increase resistance.Rate this question:
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- 9.
A 45-year old female has episodic attacks of SOB, wheeze, and cough. You make a diagnosis of asthma, a disease characterized by high airways resistance. Will the amount of gas in the chest at the end of expiration of each tidal breath (the FRC) increase or decrease?
- A.
Increase
- B.
Decrease
- C.
Remain constant
- D.
A and C
- E.
B and C
Correct Answer
D. A and CExplanation
the end expiratory pressure (passive expiration) is the FRC. This volume represents the balance of forces working on lungs and chest wall. The FRC will likely remain constant BUT it could increase if some lung units have very long time constants (e.g. it takes a long time to empty the unit due to high resistance) and if breathing is fast. Such lung units will not have a chance to empty completely before the following breath. This can lead to 'gas trapping' in which case there is extra air in the chest just before the next breath. This is clinically very important as it makes it harder for the respiratory muscles to work and can have cardiovascular implications (reduced venous return)Rate this question:
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- 10.
A 45-year old female has episodic attacks of SOB, wheeze, and cough. You make a diagnosis of asthma, a disease characterized by high airways resistance. Will the amount of pleural pressure needed to make a tidal breath increase or decrease?
- A.
Increase because of gas trapping
- B.
Increase because of higher resistance
- C.
Decrease because of higher resistance
- D.
Decrease because of gas trapping
Correct Answer
B. Increase because of higher resistanceExplanation
the amount of pleural pressure needed to take the same breath depends on the resistance; higher resistance will require high pressure difference to generate the same flow. It will take more effort to generate more pleural negative pressure and this will likely be experienced as dyspnea.Rate this question:
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- 11.
A 59-year-old male with scleroderma-associated pulmonary fibrosis comes to your office. He has a chest x-ray that shows that hsi lung fields are more white than normal, indicating excessive scarring. Would you expect his lung volumes to be larger or smaller than normal?
- A.
Larger because of gas trapping
- B.
Larger because of stiffened lung tissue
- C.
Smaller because of reduced compliance
- D.
Smaller because of less inward recoil
Correct Answer
C. Smaller because of reduced complianceExplanation
Lung volumes would be smaller because of reduced compliance (increased stiffness) of lungs caused by the scarring. At FRC, the inward recoil of the lungs is greater than normal so FRC is smaller.Rate this question:
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- 12.
A 59-year-old male with scleroderma-associated pulmonary fibrosis comes to your office. He has a chest x-ray that shows that hsi lung fields are more white than normal, indicating excessive scarring. Would you expect his flow rates to be greater or less than normal?
- A.
Increased because of increased radial traction
- B.
Increased because of use of accessory muscles
- C.
Decreased because of less inward recoil
- D.
Decreased because of reduced compliance
Correct Answer
A. Increased because of increased radial tractionExplanation
Interestingly, the flow rates (for a given lung volume) may actually increase because of increased radial traction on airways due to the surrounding "stiff" lung parenchyma.Rate this question:
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- 13.
A 75-year-old woman develops the flu. The virus destroys the lining (epithelial cell) of her lung and serum leaks into the airspaces.will the amount of gas in her chest increase or decrease?
- A.
Increase because of gas trapping
- B.
Increase because of increased compliance
- C.
Decrease because of a loss of surfactant
- D.
Decrease because of decreased compliance
Correct Answer
C. Decrease because of a loss of surfactantExplanation
the volume of the chest will decrease because of the loss of pulmonary surfactant and therefore, an increase in surface tension. Recall pressure = 2T/R so more transalveolar pressure is required to keep alveoli open when surface tension increases. This effect is greatest for the smallest alveoli; so it is the small alveoli that tend to collapse if surface tension is high.Rate this question:
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- 14.
A 75-year-old woman develops the flu. The virus destroys the lining (epithelial cell) of her lung and serum leaks into the airspaces. She is placed on a mechanical ventilator so that volume can be delivered to her lungs through a tube inserted into her trachea. Suddenly, the pressure to obtain 500ml volume (breath) increases. What is the possible cause?
- A.
Increased resistance from a mucous plug
- B.
Increased resistance from bronchospasm
- C.
Increased resistance from a blocked tube
- D.
Reduced compliance from a pneumothorax
- E.
Reduced compliance from pulmonary edema
- F.
All of the above
Correct Answer
F. All of the aboveExplanation
The possible cause for the sudden increase in pressure to obtain 500ml volume (breath) could be due to all of the above reasons. Increased resistance from a mucous plug, bronchospasm, or a blocked tube can all contribute to increased pressure needed to deliver the desired volume to the lungs. Additionally, reduced compliance from a pneumothorax or pulmonary edema can also increase the pressure required for ventilation. Therefore, all of these factors combined could be the cause for the increased pressure in this scenario.Rate this question:
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- 15.
A 43-year-old woman was brought to the ER by a neighbour who found her unconscious. the patient had been in good health but despondent over the last several days. There was an empty pill bottle ont he table where she was found and a call to the pharmacist indicated that 30 tables of a sedative had been given to her one week before. On physical examination she was unresponsive to verbal or painful stimuli and had no deep tendon reflexes. BP 140/85, resp rate 6 and shallow, pulse 120. ABG 7.12/72/30/25. Calculate the A-a gradient during spontaneous respiration assuming a barometric pressure of 660 torr and R= 0.8
- A.
10mmHg
- B.
12mmHg
- C.
20mmHg
- D.
23mmHg
Correct Answer
A. 10mmHgExplanation
A-a gradient = 130-72(1.25)-30 = 10mm HgRate this question:
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- 16.
A 43-year-old woman was brought to the ER by a neighbour who found her unconscious. the patient had been in good health but despondent over the last several days. There was an empty pill bottle ont he table where she was found and a call to the pharmacist indicated that 30 tables of a sedative had been given to her one week before. On physical examination she was unresponsive to verbal or painful stimuli and had no deep tendon reflexes. BP 140/85, resp rate 6 and shallow, pulse 120. ABG 7.12/72/30/25....mechanical ventilation with RA 7.37/40/73/25. Describe the mechanism of her hypoxemia.
- A.
V/Q mismatch
- B.
Hypoventilation
- C.
Low inspired oxygen (FiO2)
- D.
Pure shunt
- E.
Diffusion defect
Correct Answer
B. HypoventilationExplanation
In this case, the A-a gradient is normal, there is no barrier to oxygen exchange and we can conclude that the total hypoxemia is due to alveolar hypoventilation.Rate this question:
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- 17.
51-year-old lady has been noted by her family physician to be increasingly somnolent, irritable, gaining weight and has massive swelling of her legs. She had smoked two packsof cigarettes a day for many years. ABG on admission to the to the ICU: 7.36/64/26/34 on room air, and 7.34/66/95/34 on 35% O2. 6 weeks later: 7.40/44/65/26. Calculate the A-a gradient breathing room air now and 6 weeks later.
- A.
Now: 24, six weeks later 10
- B.
Now: 10, six weeks later 24
- C.
Now: 26, six weeks later 12
Correct Answer
A. Now: 24, six weeks later 10Explanation
The A-a gradient is a measure of the difference in oxygen partial pressure between the alveoli and the arterial blood. A higher A-a gradient indicates impaired gas exchange in the lungs. In this case, the A-a gradient is 24 on admission and 10 six weeks later. This suggests that there has been an improvement in gas exchange over time. This improvement could be due to various factors, such as smoking cessation or treatment of any underlying lung disease.Rate this question:
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- 18.
51-year-old lady has been noted by her family physician to be increasingly somnolent, irritable, gaining weight and has massive swelling of her legs. She had smoked two packsof cigarettes a day for many years. ABG on admission to the to the ICU: 7.36/64/26/34 on room air, and 7.34/66/95/34 on 35% O2. 6 weeks later: 7.40/44/65/26. What is the mechanism for hypoxemia on admission?
- A.
Low inspired oxygen (FiO2)
- B.
Hypoventilation
- C.
V/Q mismatch
- D.
Pure shunt
- E.
Diffusion defect
- F.
B and C
- G.
B and D
Correct Answer
F. B and CExplanation
The patient's ABG values on admission show a normal pH, indicating that acid-base balance is not the cause of hypoxemia. The low PaO2 and normal PaCO2 suggest a problem with oxygenation rather than ventilation, ruling out hypoventilation as the sole cause. The presence of V/Q mismatch, indicated by the low PaO2 and normal A-a gradient on room air, suggests that there is a problem with the matching of ventilation and perfusion in the lungs. This, combined with the patient's history of smoking, suggests that both V/Q mismatch and hypoventilation contribute to the hypoxemia. Therefore, the correct answer is B and C.Rate this question:
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- 19.
51-year-old lady has been noted by her family physician to be increasingly somnolent, irritable, gaining weight and has massive swelling of her legs. She had smoked two packs of cigarettes a day for many years and is obese. ABG on admission to the to the ICU: 7.36/64/26/34 on room air, and 7.34/66/95/34 on 35% O2. 6 weeks later: 7.40/44/65/26. What are the factors that contributed to her high CO2?
- A.
COPD
- B.
Chest wall restriction from obesity
- C.
Sleep apnea
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
she may have COPD with associated ALVEOLAR hypoventilation: COPD causes a large dead space, high resistance to flow, and overinflation (with the respiratory muscles in a disadvantageous position).
She may have restriction of the chest wall due to obesity causing low TOTAL and ALVEOLAR ventilation. The fat may make it hard to expand the chest (low compliance leading to decreased volumes).
She could also have sleep apnea and consequent malfunction of the resp control centre and CENTRAL hypoventilation.
To test these, you would do pulmonary function tests, history of sleep apnea, sleep test with overnight monitoring of CO2 level.Rate this question:
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- 20.
51-year-old lady has been noted by her family physician to be increasingly somnolent, irritable, gaining weight and has massive swelling of her legs. She had smoked two packsof cigarettes a day for many years and is obese. ABG on admission to the to the ICU: 7.36/64/26/34 on room air, and 7.34/66/95/34 on 35% O2. 6 weeks later: 7.40/44/65/26. When the patient was placed on supplemental O2 (35%), the CO2 increased. Why would this happen?
- A.
High O2 depresses ventilatory drive in patients with severe lung disease
- B.
Hypoxic vasoconstriction creates more dead space
- C.
Increasing the amount of oxygen in the blood decreases the amount of Hb free for CO2
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
In some patients with severe lung disease, the hypoxic drive to ventilation becomes very important. These patients have chronic CO2 retention and the pH in their brain has returned to near normal in spite of a raised PCO2. Arterial hypoxemia becomes the chief stimulus to ventilation, and a high O2 mixture given to relieve hypoxemia, may depress the ventilatory drive. Therefore, when supplemental O2 is given, monitoring must be frequent. Oxygen is a drug.
Hypoxic vasoconstriction is the tendency for pulmonary vessels to constrict in areas on low PaO2, in an attempt to match ventilation and perfusion. The use of supplemental O2 may increase blood flow to diseased areas of the lung and subsequently drop the pulmonary pressure to good areas of the lung creating deadspace.
Deoxygenated blood (reduced HbO2) has a greater ability to carry CO2. Increasing the amount of oxygen in the blood reduces the amount of Hb available for CO2 binding, thus increasing dissolved CO2 - this is called the HALDANE EFFECTRate this question:
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- 21.
A 27-year-old white male was seen in the office because of breathlessness on exertion for several weeks. He owns 2 cockatiels and noted symptoms shortly after purchasing them. His chest x-ray revealed a diffuse increase in the interstitial markings throughout both lungs. ABG at rest on room air: 7.41/36/64, on 100% O2: 7.39/40/500, on exercise and room air: 7.56/24/38. What is the physiologic mechanism of his hypoxemia? (check all that may apply)
- A.
Low inspired oxygen
- B.
Hypoventilation
- C.
V/Q mismatch
- D.
Pure shunt
- E.
Diffusion defect
Correct Answer(s)
A. Low inspired oxygen
C. V/Q mismatch
E. Diffusion defectExplanation
The mechanism of his hypoxemia is not alveolar hypoventilation because his CO2 is normal. The dramatic rise in arterial oxygen when breathing 100% O2 excludes a right to left shunt . Since his hypoxemia increases strikingly with exercise, it is compatible with a diffusion abnormality. This is also compatible with a V/Q mismatch. The fifth cause of hypozemia is low FiO2. THis patient would have a higher PaO2 at sea level than in Calgary.Rate this question:
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- 22.
A 64-year-old male has been admitted for evaluation of genito-urinary symptoms and a recent GI hemorrhage. 2 days after admission he required a urinary catheter to permit bladder drainage. on his 5 hospital day he developed a shaking chill and his arterial blood pressure decreased to 80/60. At that time, the following lab data was obtained: Hb 80, ABG 7.32/30/50 SaO2 75%, lactate 7, blood cultures grew e.coli, JVP 2cm below right atrium, temperature 39. You do a CXR and find that there is diffuse patchy airspace disease consistent with ARDS (acute respiratory distress syndrome). Why does he have a low JVP?
- A.
Widely dilated vascular bed due to sepsis
- B.
Hypovolemia
- C.
Heart failure
- D.
A and B
Correct Answer
D. A and BExplanation
The patient has a low JVP (jugular venous pressure) because of both a widely dilated vascular bed due to sepsis and hypovolemia. Sepsis can cause vasodilation, leading to a decrease in systemic vascular resistance and subsequently a decrease in venous return to the heart. This can result in a low JVP. Additionally, the patient's recent GI hemorrhage suggests blood loss and hypovolemia, which can also contribute to a low JVP. The combination of sepsis-induced vasodilation and hypovolemia can further exacerbate the decrease in JVP.Rate this question:
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- 23.
A 64-year-old male has been admitted for evaluation of genito-urinary symptoms and a recent GI hemorrhage. 2 days after admission he required a urinary catheter to permit bladder drainage. on his 5 hospital day he developed a shaking chill and his arterial blood pressure decreased to 80/60. At that time, the following lab data was obtained: Hb 80, ABG 7.32/30/50 SaO2 75%, lactate 7, blood cultures grew e.coli, JVP 2cm below right atrium, temperature 39. You do a CXR and find that there is diffuse patchy airspace disease consistent with ARDS (acute respiratory distress syndrome). Assuming a cardiac output of 3.5L/min, calculate the delivery of oxygen to his vital organs
- A.
523mlO2/min
- B.
364mlO2/min
- C.
287ml O2/min
- D.
???
Correct Answer
C. 287ml O2/minExplanation
Delivery O2 = (Hb*1.34mlO2/gHb * SaO2*CO) + (dissolved O2*CO)
= [(80g/L*1.34mLo2/gHb*0.75) + (0.03mlO2*PaO2)] * CO
= 281.4 + 5.3 = 286.7ml/min
1.34mlO2 is the amount of O2 carried by one gram of fully saturated Hb
YOU WILL NOT HAVE TO DUPLICATE THIS QUESTION ON THE EXAM BUT YOU NEED TO KNOW THE DETERMINANTS OF O2 DELIVERY - note the small effect of dissolved O2 in the blood compared to the effect of O2 bound to HbRate this question:
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- 24.
A 64-year-old male has been admitted for evaluation of genito-urinary symptoms and a recent GI hemorrhage. 2 days after admission he required a urinary catheter to permit bladder drainage. on his 5 hospital day he developed a shaking chill and his arterial blood pressure decreased to 80/60. At that time, the following lab data was obtained: Hb 80, ABG 7.32/30/50 SaO2 75%, lactate 7, blood cultures grew e.coli, JVP 2cm below right atrium, temperature 39. You do a CXR and find that there is diffuse patchy airspace disease consistent with ARDS (acute respiratory distress syndrome). What approaches might be taken to improve tissue oxygen delivery?
- A.
Blood transfusion to increase Hb
- B.
Medicate to increase cardiac output
- C.
Give him oxygen to increase O2 saturation
- D.
Shift O2 dissociation curve to the left to carry more O2 (lower temp, correct acidosis)
- E.
All of the above
Correct Answer
E. All of the aboveExplanation
All of the above approaches might be taken to improve tissue oxygen delivery. Blood transfusion can increase the hemoglobin level, which can carry more oxygen. Medication can be given to increase cardiac output, improving the blood flow and delivery of oxygen to tissues. Giving oxygen can increase the oxygen saturation in the blood, ensuring an adequate supply of oxygen to tissues. Shifting the oxygen dissociation curve to the left by lowering temperature and correcting acidosis can enhance the release of oxygen from hemoglobin to tissues.Rate this question:
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- 25.
30 year old HIV positive male comes to ER for a 3-week history of increasing SOB, non-productive cough, fevers and night sweats. CXR shows diffuse interstitial disease. Physical exam reveals mottled extremities and his oropharynx appears blue. Resp rate is 30/min. What is the cause for his mottled hands and blue oropharynx?
- A.
Central cyanosis
- B.
Desaturated Hb because of hypoxemia
- C.
Desaturated Hb becasue of chronic respiratory acidosis (curve shifting)
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
The mottled extremities and blue oropharynx indicate central cyanosis, which is caused by a decrease in the oxygen saturation of hemoglobin. In this case, the patient's symptoms of increasing SOB, non-productive cough, fevers, and night sweats suggest a respiratory infection or inflammation. The diffuse interstitial disease seen on CXR further supports a respiratory pathology. The combination of these factors can lead to hypoxemia, resulting in desaturated hemoglobin. Additionally, chronic respiratory acidosis can cause a shift in the oxygen-hemoglobin dissociation curve, further contributing to desaturated hemoglobin. Therefore, all of the above options are potential causes for the mottled hands and blue oropharynx in this patient.Rate this question:
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- 26.
30 year old HIV positive male comes to ER for a 3-week history of increasing SOB, non-productive cough, fevers and night sweats. CXR shows diffuse interstitial disease. Physical exam reveals mottled extremities and his oropharynx appears blue. Resp rate is 30/min. ABGs: 7.5/28/42/22. What is the acid base disturbance?
- A.
Acute metabolic alkalosis with appropriate compensation
- B.
Acute respiratory alkalosis with inappropriate compensation
- C.
Chronic respiratory alkalosis with appropriate compensation
- D.
Acute respiratory alkalosis with appropriate compensation
Correct Answer
D. Acute respiratory alkalosis with appropriate compensationExplanation
for every decrease CO2 by 10 expect HCO3 to decrease by 4 (chronic) or 2 (acute). CO2 is down by 10, HCO3 is down by 2, therefore this is acute respiratory alkalosis with appropriate compensationRate this question:
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- 27.
30 year old HIV positive male comes to ER for a 3-week history of increasing SOB, non-productive cough, fevers and night sweats. CXR shows diffuse interstitial disease. Physical exam reveals mottled extremities and his oropharynx appears blue. Resp rate is 30/min. ABGs: 7.5/28/42/22. What are the possible causes of his hypoxemia? (calculate the A-a gradient and check all that apply)
- A.
FiO2
- B.
Hypoventilation
- C.
V/Q mismatch
- D.
Shunt
- E.
Diffusion barrier
Correct Answer(s)
C. V/Q mismatch
D. Shunt
E. Diffusion barrierExplanation
He has an A-a gradient of 53 indicating significant V/Q mismatch due to lung disease.Rate this question:
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- 28.
A 56-year-old banker has smoked 2ppd for 40 years and occasionally notes dyspnea on exertion. Three weeks ago, he noticed a reduced ability to concentrate and do arithmetic. His wife states that he has an occasional hand twitch and his colour does not look good. pH 7.35 PCO2 60, PO2 31. Determine his primary acid base disturbance and use the henderson hasselbach equation to determine whether or not the compensation is appropriate.
- A.
Primary chronic respiratory acidosis with appropriate compensation
- B.
Primary chronic respiratory acidosis with inappropriate compensation
- C.
Primary metabolic acidosis with appropriate compensation
- D.
Primary acute respiratory acidosis with appropriate compensation
Correct Answer
A. Primary chronic respiratory acidosis with appropriate compensationExplanation
[H+] = (24* [CO2])/[HCO3]
In the physiological range H+ and pH are linearly related. At a pH of 7.35, H+ is approximately 48, you can calculate HCO3 = 30.
In this case, CO2 increased by 20; HCO3 increasd by 6 consistent with compensated chronic resp acidosis.Rate this question:
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- 29.
A 56-year-old banker has smoked 2ppd for 40 years and occasionally notes dyspnea on exertion. Three weeks ago, he noticed a reduced ability to concentrate and do arithmetic. His wife states that he has an occasional hand twitch and his colour does not look good. pH 7.35 PCO2 60, PO2 31. Why is his oxygen low? (calculate A-a gradient and check all that apply)
- A.
Low inspired oxygen
- B.
Hypoventilation
- C.
V/Q mismatch
- D.
Pure shunt
- E.
Diffusion defect
Correct Answer(s)
B. Hypoventilation
C. V/Q mismatchExplanation
(A-a gradient: (660-47)*0.21 - (60/0.8) - 31 = 24 (elevated)
This is consistent with a diagnosis of smoking-related lung idsease, although further tests are required to establish the diagnosis. Therefore, hypoxemia is due to both hypoventilation and V/Q mismatch.Rate this question:
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- 30.
Ms. IA is a known drug addict. She is found unconscious in the street and brought to the ER. On physical exam she is unresponsive to deep pain. Resp rate is 4. ABG 7.20/72/33/27. What is the primary acid base problem?
- A.
Acute metabolic acidosis
- B.
Chronic respiratory acidosis
- C.
Acute respiratory acidosis
- D.
Chronic metabolic acidosis
Correct Answer
C. Acute respiratory acidosisExplanation
Compensation for respiratory acidosis is 10:1 for acute and 10:3 for chronic. CO2 is up by about 30 and HCO3 is up by 3, consistent with acute respiratory acidosis.Rate this question:
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- 31.
Ms. IA is a known drug addict. She is found unconscious in the street and brought to the ER. On physical exam she is unresponsive to deep pain. Resp rate is 4. ABG 7.20/72/33/27. What is the cause of hypoxemia?
- A.
Low inspired oxygen
- B.
Hypoventilaiton
- C.
V/Q mismatch
- D.
Pure shunt
- E.
Diffusion defect
Correct Answer
B. HypoventilaitonExplanation
the A-a radient (130-72/0.8)-33=7 indivating that the hypoxemia is due entirely to hypoventilation (normal A-a- gradient is less than 12)Rate this question:
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- 32.
Ms. V.J. is expecting her first baby in 6 weeks. She has noticed more and more shortness of breath and now has to pause halfway up a flight of stairs. pH 7.47, PCO2 24, PO2 90. What is the primary acid base disturbance?
- A.
Chronic metabolic alkalosis
- B.
Acute metabolic alkalosis
- C.
Chronic respiratory alkalosis
- D.
Acute respiratory alkalosis
Correct Answer
C. Chronic respiratory alkalosisExplanation
calculate the HCO3 = 24*24/34 = 17
the CO2 is decreased by 16, and HCO3 by 7, roughly 10:4, indicating chronic resp alkalosis.Rate this question:
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- 33.
Ms. V.J. is expecting her first baby in 6 weeks. She has noticed more and more shortness of breath and now has to pause halfway up a flight of stairs. pH 7.47, PCO2 24, PO2 90. What is the most likely cause of her SOB?
- A.
Low inspired oxygen
- B.
HypERventilation
- C.
V/Q mismatch
- D.
Pure shunt
- E.
Diffusion defect
Correct Answer
B. HypERventilationExplanation
the most likey cause is hyperventilation of pregnancy though to be due to hormonal influences on control of ventilation. One should ascertain that the A-a gradient is normal to be sure we are not missing a lung problem: A-a gradient is 10 which is normalRate this question:
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- 34.
A 32- year-old male developed acute hemorrhabe from duodenal ulcer and aspirated vomitus. He required intubation and ventilation and continuous gastric suction. After 8 hours: 7.64/28/85/30 ventilated with supplemental O2. Na 130, Cl 90, K 3.4, BUN 14. What is the acid base disturbance?
- A.
Respiratory alkalemia
- B.
Metabolic alkalemia
- C.
Respiratory acidosis
- D.
A and B
- E.
WHAT'S GOING ON HERE???!?!?!
Correct Answer
D. A and BExplanation
this is a mixed disturbance: metabolic alkalosis AND respiratory alkalosis. The mechanism is likely loss of gastric HCl due to suctioning and vomiting causing metabolic alkalosis. If the ventilator is set to high (high lung volumes or high rate) a resp alkalosis ensues.Rate this question:
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- 35.
60 year old male with COPD and cor pulmonale is taking diuretics for leg swelling. He develops pneumonia. In the ER he is found to be febrile, tachycardic with a BP of 70/40. ABGs 7.36/50/60/27 on 40% supplemental O2. Na 132, Cl 87, K 3.1 BUN 20, lactate 7.Calculate the anion gap, and the expected HCO3.
- A.
AG 18, HCO3 18
- B.
AG 16, HCO3 12
- C.
AG 23, HCO3 23
Correct Answer
A. AG 18, HCO3 18Explanation
At first glance, the values seem to indicate a compensated chronic resp acidosis with 10:3 compensation.
BUT!
Note the NORMAL pH and HIGH ANION GAP!
The AG=18, which indicates at least 6mmol/L of unmeasured anion (acid) so we would expect HCO3 to be 18. The HCO3 is actually 8 higher than the expected HCO3 of 18 - could be due to a contomitant primary metabolic alkalosis or could be due to compensation for chronic resp acidosis.Rate this question:
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- 36.
60 year old male with COPD and cor pulmonale is taking diuretics for leg swelling. He develops pneumonia. In the ER he is found to be febrile, tachycardic with a BP of 70/40. ABGs 7.36/50/60/27 on 40% supplemental O2. Na 132, Cl 87, K 3.1 BUN 20, lactate 7. Fortunately, you check the chart and a blood gas was done one week early at a "routine visit". 7.45/50/60/33 What is his acid base disorder?
- A.
High anion-gap primary metabolic acidosis
- B.
Primary chronic metabolic alkalosis
- C.
Primary chronic respiratory acidosis
- D.
All of the above
- E.
Brain...exploding...can't...think...
Correct Answer
D. All of the aboveExplanation
this patient has a chronic respiratory acidosis (due to COPD) and a primary metabolic alkalosis (likely due to diuretics). Since the last visit he has developed sepsis with associated lactic acidosis causing the third abnormality, a high anion gap metabolic acidosis. Thus this is a mixed acid base disorder. Multiple acid base disturbances are not uncommon in clinical practice.Rate this question:
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