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Question 1
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer: 50 ml oxygen/100 ml blood
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 2
Incorrect
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Breathing hyperbaric gas mixture
Correct Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Incorrect
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A 48-year-old man smoker presented to the doctor complaining of a persistent cough and shortness of breath. A chest X-ray indicated the presence of a right upper lung mass. Biopsy of the mass revealed the presence of pink cells with large, irregular nuclei. What is the most probable diagnosis?
Your Answer: Mesothelioma
Correct Answer: Squamous cell carcinoma
Explanation:Squamous cell carcinoma, is a type of non-small cell lung cancer that accounts for approximately 30% of all lung cancers. The presence of squamous cell carcinoma is often related with a long history of smoking and the presence of persistent respiratory symptoms. Chest radiography usually shows the presence of a proximal airway lesion. Histological findings include keratinisation that takes the form of keratin pearls with pink cytoplasm and cells with large, irregular nuclei.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 4
Correct
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Lung compliance is increased by:
Your Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 5
Incorrect
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The most likely cause of a low p(O2) in arterial blood is:
Your Answer: Hyperaemic hypoxia
Correct Answer: Hypoxic hypoxia
Explanation:Hypoxia is when the whole body or a region is deprived of adequate oxygen supply. Different types of hypoxia include the following:
– Hypoxic hypoxia, which occurs due to poor oxygen supply, as a result of low partial pressure of oxygen in arterial blood. This could be due to low partial pressure of atmospheric oxygen (e.g., at high altitude), sleep apnoea, poor ventilation because of chronic obstructive pulmonary disease or respiratory arrest, or shunts. The other types of hypoxia have a normal partial pressure of oxygen.
– Anaemic hypoxia occurs due to low total oxygen content of the blood, with a normal arterial oxygen pressure.
– Hyperaemic hypoxia occurs due to poor delivery of oxygen to target tissues, such as in carbon monoxide poisoning or methemoglobinemia.
– Histotoxic hypoxia results due to inability of the cells to use the delivered oxygen due to disabled oxidative phosphorylation enzymes.
– Ischaemic (or stagnant) hypoxia occurs due to local flow restriction of well-oxygenated blood, seen in cases like cerebral ischaemia, ischaemic heart disease and intrauterine hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 6
Incorrect
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A 47-year old-woman diagnosed with pancreatitis presented to the emergency department complaining of a worsening shortness of breath, fever, agitation and cough. Oxygen saturation was 67% in room air. Her respiratory status continued to deteriorate therefore she was intubated. She was admitted to the intensive care unit for management. Chest X-ray demonstrated bilateral perihilar opacities. The patient failed conventional treatment and died several days later. At autopsy, the lung shows growth of type 2 pneumocytes and thickened alveolar walls. What is the most probable diagnosis?
Your Answer: Cardiogenic pulmonary oedema
Correct Answer: Adult respiratory distress syndrome
Explanation:Acute (or adult) respiratory distress syndrome (ARDS) is a life-threatening lung condition characterised by a non-cardiogenic pulmonary oedema that leads to acute respiratory failure. The most common risk factors for ARDS include trauma with direct lung injury, sepsis, pneumonia, pancreatitis, burns, drug overdose, massive blood transfusion and shock. Acute onset of dyspnoea with hypoxemia, anxiety and agitation is typical. Chest X ray most commonly demonstrates bilateral pulmonary infiltrates. Histological changes include the exudative, proliferative and fibrotic phase. ARDS is mainly a clinical diagnosis.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 7
Correct
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The pleural cavity is the space between the two pulmonary pleurae which cover the lungs. What is the normal amount of pleural fluid?
Your Answer: 10 ml
Explanation:Pleural fluid is a serous fluid produced by the serous membrane covering normal pleurae. Most fluid is produced by the parietal circulation (intercostal arteries) via bulk flow and reabsorbed by the lymphatic system. The total volume of fluid present in the intrapleural space is estimated to be only 2–10 ml. A small amount of protein is present in intrapleural fluid. Normally, the rate of reabsorption increases as a physiological response to accumulating fluid.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 8
Incorrect
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During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer: Unchanged
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 9
Correct
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After a severe asthma attack, a 26-year-old woman is left in a markedly hypoxic state. In which of the following organs are the arterial beds most likely to be vasoconstricted due to the hypoxia?
Your Answer: Lungs
Explanation:Hypoxic pulmonary vasoconstriction is a local response to hypoxia resulting primarily from constriction of small muscular pulmonary arteries in response to reduced alveolar oxygen tension. This unique response of pulmonary arterioles results in a local adjustment of perfusion to ventilation. This means that if a bronchiole is obstructed, the lack of oxygen causes contraction of the pulmonary vascular smooth muscle in the corresponding area, shunting blood away from the hypoxic region to better-ventilated regions. The purpose of hypoxic pulmonary vasoconstriction is to distribute blood flow regionally to increase the overall efficiency of gas exchange between air and blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 10
Correct
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A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 11
Incorrect
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer: Decrease in 2,3-diphosphoglycerate (DPG)
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 12
Incorrect
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A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer: Decreased total lung capacity
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 13
Incorrect
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Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer: Decreased production of erythropoietin
Correct Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 14
Incorrect
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The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 10 mmHg
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 15
Incorrect
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Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: FEV1 of 75% (normal > 65%), arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Calculate his anatomical dead space.
Your Answer: 50 ml
Correct Answer: 100 ml
Explanation:Dead space refers to inhaled air that does not take part in gas exchange. Because of this dead space, taking deep breaths slowly is more effective for gas exchange than taking quick, shallow breaths where a large proportion is dead space. Use of a snorkel by a diver increases the dead space marginally. Anatomical dead space refers to the gas in conducting areas such as mouth and trachea, and is roughly 150 ml (2.2 ml/kg body weight). This corresponds to a third of the tidal volume (400-500 ml). It can be measured by Fowler’s method, a nitrogen wash-out technique. It is posture-dependent and increases with increase in tidal volume. Physiological dead space is equal to the anatomical dead space plus the alveolar dead space, where alveolar dead space is the area in the alveoli where no effective exchange takes place due to poor blood flow in capillaries. This physiological dead space is very small normally (< 5 ml) but can increase in lung diseases. Physiological dead space can be measured by Bohr’s method. Total ventilation per minute (minute ventilation) is given by the product of tidal volume and the breathing rate. Here, the total ventilation is 600 ml times 12 breaths/min = 7200 ml/min. The problem mentions alveolar ventilation to be 6000 ml/min. Thus, the difference between the alveolar ventilation and total ventilation is 7200 – 6000 ml/min = 1200 ml/min, or 100 ml per breath at 12 breaths per min. This 100 ml is the dead space volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 16
Correct
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The chest X-ray of a 72 year old patient reveals the presence of a round lesion containing an air-fluid level in the left lung. These findings are most probably suggestive of:
Your Answer: Lung abscess
Explanation:Lung abscesses are collections of pus within the lung that arise most commonly as a complication of aspiration pneumonia caused by oral anaerobes. Older patients are more at risk due to poor oral hygiene, gingivitis an inability to handle their oral secretions due to other diseases. Chest X-ray most commonly reveals the appearance of an irregularly shaped cavity with an air-fluid level.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 17
Correct
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Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 18
Correct
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What causes a reduction in pulmonary functional residual capacity?
Your Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 19
Correct
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Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer: Respiratory acidosis
Explanation:COPD leads to respiratory acidosis (chronic). This occurs due to hypoventilation which involves multiple causes, such as poor responsiveness to hypoxia and hypercapnia, increased ventilation/perfusion mismatch leading to increased dead space ventilation and decreased diaphragm function.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 20
Incorrect
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Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer: Is part of vital capacity
Correct Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 21
Correct
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. What is his alveolar ventilation?
Your Answer: 3000 ml/min
Explanation:Alveolar ventilation is the amount of air reaching the alveoli per minute. Alveolar ventilation = respiratory rate × (tidal volume – anatomical dead space volume). Thus, alveolar ventilation = 10 × (550 − 250) = 3000 ml/min.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 22
Correct
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How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your Answer: 500 ml
Explanation:Pulmonary circulation is the portion of the cardiovascular system which carries deoxygenated blood away from the heart, to the lungs, and returns oxygenated blood back to the heart. The vessels of the pulmonary circulation are very compliant (easily distensible) and so typically accommodate about 500 ml of blood in an adult man. This large lung blood volume can serve as a reservoir for the left ventricle, particularly during periods when left ventricular output momentarily exceeds venous return.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 23
Correct
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A 70-year-old male who has smoked since his teens complains of progressive shortness of breath and a persistent cough. He is diagnosed with COPD. Which of the following abnormalities is most likely to be present in his pulmonary function tests?
Your Answer: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 24
Correct
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A 45-year-old man complains of shortness of breath, cough and chest pain. Chest X ray revealed a perihilar mass with bronchiectasis in the left mid-lung. Which of the following is most probably associated with these findings?
Your Answer: Bronchial carcinoid
Explanation:Bronchial carcinoids are neuroendocrine tumours that arise from Kulchitsky’s cells of the bronchial epithelium. Kulchitsky’s cells belong to the diffuse endocrine system. Patients affected by this tumour may be asymptomatic or may present with symptoms of airway obstruction, like dyspnoea, wheezing, and cough. Other common findings are recurrent pneumonia, haemoptysis, chest pain and paraneoplastic syndromes. Chest radiographs are abnormal in the majority of cases. Peripheral carcinoids usually present as a solitary pulmonary nodule. For central lesions common findings include hilar or perihilar masses with or without atelectasis, bronchiectasis, or consolidation. Bronchial carcinoids most commonly arise in the large bronchi causing obstruction.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 25
Incorrect
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Abnormal breathing is noticed in a of victim of a road traffic accident, who sustained a head injury. The breathing pattern is characterised by alternate periods of waxing and waning tidal volumes with interspersed periods of apnoea. This breathing pattern is known as:
Your Answer: Agonal breathing
Correct Answer: Cheyne–Stokes breathing
Explanation:Cheyne-Stokes breathing is an abnormal breathing pattern with breathing periods of gradually waxing and waning tidal volumes, with apnoeic periods interspersed. It is usually the first breathing pattern to be seen with a rise in intracranial pressure and is caused by failure of the respiratory centre in the brain to compensate quickly enough to changes in serum partial pressure of oxygen and carbon dioxide. The aetiology includes strokes, head injuries, brain tumours and congestive heart failure. It is also a sign of altitude sickness in normal people, a symptom of carbon monoxide poisoning or post-morphine administration. Biot’s respiration (cluster breathing) is characterized by cluster of quick, shallow inspirations followed by regular or irregular periods of apnoea. It is different from ataxic respiration, which has completely irregular breaths and pauses. It results due to damage to the medulla oblongata by any reason (stroke, uncal herniation, trauma) and is a poor prognostic indicator. Kussmaul breathing, also known as ‘air hunger’, is basically respiratory compensation for metabolic acidosis and is characterized by quick, deep and laboured breathing. It is most often seen in in diabetic ketoacidosis. Due to forced inspiratory rate, the patients will show a low p(CO2). Ondine’s curse is congenital central hypoventilation syndrome or primary alveolar hypoventilation, which can be fatal and leads to sleep apnoea. It involves an inborn failure to control breathing autonomically during sleep and in severe cases, can affect patients even while awake. It is known to occur in 1 in 200000 liveborn children. Treatment includes tracheostomies and life long mechanical ventilator support.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 26
Correct
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Which of the following substances is most likely to cause pulmonary vasodilatation?
Your Answer: Nitric oxide
Explanation:In the body, nitric oxide is synthesised from arginine and oxygen by various nitric oxide synthase (NOS) enzymes and by sequential reduction of inorganic nitrate. The endothelium of blood vessels uses nitric oxide to signal the surrounding smooth muscle to relax, so dilating the artery and increasing blood flow. Nitric oxide/oxygen blends are used in critical care to promote capillary and pulmonary dilation to treat primary pulmonary hypertension in neonatal patients post-meconium aspiration and related to birth defects.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 27
Incorrect
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A 45-year-old-female is suspected to have a pulmonary mass. Supposing that she has a neoplasm, which of the following are most commonly found to involve the lung:
Your Answer: Squamous cell carcinomas in smokers
Correct Answer: Pulmonary metastases
Explanation:Lung metastases occur when a cancer started in another part of the body (primary site) spreads to the lungs. The lungs are among the most common site where cancer can spread due to its rich systemic venous drainage, almost every type of cancer can spread to the lung. The most common types of cancer that spread to the lung are breast, colorectal, kidney, testicular, bladder, prostate, head and neck cancers.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 28
Correct
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Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 29
Correct
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After a total colectomy and ileotomy, a 50-year old diabetic man who was a known case of diabetic nephropathy had persistent metabolic acidosis. The patient appeared well perfused, with normal vital signs and normal fluid balance. Investigations revealed:
Sodium = 132 mmol/l
Potassium = 6.6 mmol/l
Creatinine = 185 μmol/l (2.16 mg/dl)
Chloride = 109 μmol/l
8am cortisol = 500 nmol/l (18 μg/dl)
pH = 7.29, p(CO2) = 27 mmHg
p(O2) = 107 mmHg
standard bicarbonate = 12 mmol/l.
What is the likely causes of his acidosis?Your Answer: Renal tubular acidosis
Explanation:Acidosis here is due to low bicarbonate. The low p(CO2) is seen in compensation. The anion gap is normal, ruling out intra-abdominal ischaemia (which leads to metabolic acidosis). If it was a gastrointestinal aetiology, low potassium would be seen. The history of diabetic nephropathy predisposes to renal tubular acidosis. Type 4 (hyporeninaemic hypoaldosteronism) is associated with high potassium and is found in diabetic and hypertensive renal disease.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 30
Correct
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Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
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This question is part of the following fields:
- Physiology
- Respiratory
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SESSION STATS - PERFORMANCE PER SPECIALTY
