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Thread: PhysioRespiratory

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    A patient's airway pressure is being measured while he is breathing into a spirometer. Which of the following lung volumes would be associated with an airway pressure of +30 cm H2O? A. Functional residual capacity
    B. Minimal volume
    C. Residual volume
    D. Tidal volume
    E. Total lung capacity






    The correct answer is E. For this question, you need not worry about the actual value of the airway pressure. There is only one listed answer that would produce a positive airway pressure, and that is total lung capacity. To measure the airway pressure, a patient inspires or expires from a spirometer, and then relaxes while his airway pressure is measured. It is easy to determine whether the airway pressure would be negative or positive by practicing on yourself. Inhale to total lung capacity and relax. You will feel the sensation of wanting to blow air out--this creates a positive pressure in your airways. Any volume above functional residual capacity (FRC) will create a positive airway pressure, and any volume below FRC will create a negative airway pressure.
    The functional residual capacity (choice A) is the volume of air that remains in the lung after a normal expiration. The FRC is the equilibrium volume when the elastic recoil of the lungs is balanced by the tendency of the chest to spring out. Because this is the volume when the patient is "at rest," the airway pressure is zero. Breathe out normally (to reach FRC) and notice that there is no pressure in your airways.

    Minimal volume (choice B) can only be achieved with an excised lung. It is the volume of air remaining in an excised lung that is maximally deflated. It is smaller than the residual volume because the chest wall is not there to help draw the lung open.

    The residual volume (choice C) is the volume of air remaining in the lungs after maximal expiration. At volumes less than FRC, like residual volume, the airway pressure would be less than 0 cm H2O. Exhale all the way (to residual volume), and relax--you will feel the sensation of wanting to draw air in--this creates a negative pressure in your airways (like a vacuum).

    Tidal volume (choice D) is the volume of air that is inhaled or exhaled with each normal breath.





    A medical student volunteers to have his lung volumes and capacities measured for his organ physiology laboratory class. He is connected to a spirometer containing a known concentration of helium. He is instructed to breathe several times until the helium has equilibrated between the spirometer and his lungs. He is then instructed to exhale as much air as he possibly can. Calculations are made to determine the amount of air remaining in his lungs, which is called the A. expiratory reserve volume
    B. functional residual capacity
    C. inspiratory capacity
    D. inspiratory reserve volume
    E. residual volume
    F. tidal volume
    G. vital capacity






    The correct answer is E. There are two ways to arrive at the correct answer to this question. The first is to simply remember the definition of residual volume (RV): the amount of air remaining in the lungs after maximal exhalation. The second way is to recall that the helium dilution technique described above is used to measure functional residual capacity (FRC) and RV, which narrows the reasonable option choices to B and E only. All of the other volumes and capacities can be directly measured with spirometry because they are blown into the spirometer. Only FRC and RV represent amounts of air that remain in the lungs.
    Expiratory reserve volume (choice A) is the volume expelled by an active expiratory effort after passive expiration.

    Functional residual capacity (choice B) is defined as the amount of air remaining in the lungs after passive expiration.

    Inspiratory capacity (choice C) is the maximal amount of air inspired after a passive expiration.

    Inspiratory reserve volume (choice D) is the amount of air inspired with a maximal inspiratory effort over and above the tidal volume.

    Tidal volume (choice F) is the amount of air that is inspired (or expired) with each normal breath.

    Vital capacity (choice G) is the largest amount of air that can be expired after a maximal inspiratory effort.



    A 12-year-old girl has been severely anemic (Hb = 6.0 g/dL) for several months. Which of the following is most likely to be decreased during resting conditions? A. 2,3-diphosphoglycerate
    B. Arterial PO2
    C. Cardiac output
    D. Mixed venous PO2
    E. Red blood cell H+ concentration






    The correct answer is D. Oxygen delivery to the tissues is approximately equal to cardiac output x hemoglobin concentration x the amount of oxygen extracted from the blood (oxygen extraction = arterial oxygen content - venous oxygen content). Some of the compensatory mechanisms that come into play in an anemic person with a low hemoglobin content include the following: increased levels of 2,3 diphosphoglycerate (2,3-DPG; choice A), increased cardiac output (choice C), and increased red blood cell H+ concentration (choice E). The increase in 2,3-DPG and red blood cell H+ concentration cause the oxygen-hemoglobin dissociation curve to shift to the right, which facilitates unloading of oxygen in the tissues.
    The arterial PO2(choice B) and arterial oxygen saturation of hemoglobin are independent of hemoglobin concentration. However, a decrease in the hemoglobin concentration of the blood causes a proportionate decrease in the oxygen-carrying capacity of the blood. (Think of a glass beaker containing a solution of hemoglobin. If a gas having a PO2 of 100 mm Hg is bubbled through the solution, the PO2 of the solution will be 100 mm Hg regardless of the hemoglobin concentration. However, if the concentration of hemoglobin in the beaker is doubled, the beaker will now contain twice as much oxygen.) Because each gram of hemoglobin can normally carry a total of 1.34 g oxygen, the arterial oxygen content of this patient can be calculated as follows: 1.34 mL O2/g Hb x 6.0 g Hb/dL blood = 8.04 mL O2/dL blood (normal = 20 mL O2/dL blood). When the oxygen content of the arterial blood is decreased to only 8 mL O2 /dL, the loss of oxygen to the tissues will cause the venous oxygen tension to fall to lower-than-normal levels during rest and to very low levels whenever exercise is attempted.



    A healthy 42-year-old woman with a history of anxiety attacks sits in the hospital waiting room as her 3-year-old daughter undergoes open heart surgery for a septal malformation. The woman experiences a feeling of suffocation and is obviously hyperventilating. She informs her husband that she feels faint and has blurred vision. Which of the following is most likely to relieve the symptoms caused by hyperventilation? A. Breathing a 10% oxygen/90% nitrogen mixture
    B. Breathing 100% nitrogen
    C. Breathing in and out of a plastic bag
    D. Intravenous administration of bicarbonate
    E. Lying down






    The correct answer is C. Hyperventilation associated with states of anxiety can lead to feelings of faintness, suffocation, tightness in the chest, and blurred vision. Individuals undergoing such an attack may not be aware of overbreathing. The anxious, hyperventilating woman is "blowing off" carbon dioxide, which lowers her arterial PCO2. Many of the symptoms associated with anxiety attacks are probably caused by a decrease in cerebral blood flow secondary to low arterial PCO2. Recall that carbon dioxide is a major regulator of cerebral blood flow, i.e., carbon dioxide dilates the brain vasculature, and conversely, the vasculature constricts when carbon dioxide levels are low. The decrease in cerebral blood flow leads to cerebral hypoxia, which is probably responsible for the fainting and blurred vision. An attack may be terminated by breathing in and out of a plastic bag because this can increase carbon dioxide levels in the blood. Inhaling a 5% carbon dioxide mixture would also be effective.
    Breathing a mixture of 10% oxygen/90% nitrogen (choice A) or 100% nitrogen (choice B) can decrease oxygen delivery to the brain and thereby worsen the symptoms caused by hyperventilation.

    Hyperventilation results in hypocapnia (low PCO2), which causes alkalosis (high blood pH). Bicarbonate (choice D) should not be administered to an alkalotic patient.

    The feelings of faintness and blurred vision resulting from hyperventilation are not relieved by lying down (choice E).



    A 62-year-old man with a 50-year-history of cigarette smoking has a complete work-up in a pulmonary function laboratory. The table below shows pulmonary volumes and capacities obtained using simple spirometry and helium washout techniques.
    Lung Volumes Values
    Functional residual capacity 5.0 L
    Inspiratory reserve volume 1.5 L
    Inspiratory capacity 2.0 L
    Vital capacity 3.5 L

    What is the total lung capacity of this patient?


    A. 6.5 L
    B. 7.0 L
    C. 7.5 L
    D. 8.0 L
    E. 8.5 L






    The correct answer is B. The total lung capacity is the sum of the functional residual capacity and inspiratory capacity. The easiest approach to this type of question is to construct a simple spirogram (see below) and fill in the values provided in the table. This approach eliminates the need to memorize formulas. The missing volume or capacity can be easily determined from the spirogram. The total lung capacity, functional residual capacity, and residual volume are often increased in the emphysematous lungs of patients with a long-term history of cigarette smoking.

    A inspiratory reserve volume
    B expiratory reserve volume
    C inspiratory capacity
    D vital capacity
    E residual volume
    F functional residual capacity
    G tidal volume
    H total lung capacity





    An inexperienced resident physician is asked to draw arterial blood gases from a patient with severe chronic obstructive pulmonary disease, who is not yet on oxygen therapy. The resident attempts to draw blood from the femoral artery at the groin, but actually draws blood from the femoral vein. When compared to the arterial sample that should have been obtained, this venous sample will show which of the following differences?

    p02 pC02 pH
    A. Decreased Increased Decreased
    B. Decreased Increased Increased
    C. Increased Decreased Decreased
    D. Increased Decreased Increased
    E. Increased Increased Increased







    The correct answer is A. The tissues use oxygen, so venous blood has decreased p02 compared to arterial blood. The tissues produce C02, which makes carbonic acid, and thus causes the venous blood to be slightly more acidic (decreased pH) than arterial blood, although both values can still be within normal limits because of the effects of blood buffering. Arterial puncture for blood gas collection is considered sufficiently difficult that some hospitals only allow physicians to collect the blood. At other institutions, specially trained non-physician personnel may be allowed to do the blood collections. Remember that in the groin, the femoral vein lies medial to the femoral artery, and the femoral nerve lies lateral to the artery.




    A premature infant has initially good APGAR scores, but several hours after birth develops severe respiratory distress, which improves when the baby is placed in an incubator with increased oxygen tension. The neonatologist explains to the parents that the problem is that the neonate's immature lungs are unable to synthesize enough secretions to keep the alveoli open. The most important constituent of these secretions is a lipid containing which of the following? A. α-ketoglutarate
    B. Arachidonic acid
    C. Choline
    D. Proline
    E. Retinol






    The correct answer is C. The baby has respiratory distress syndrome of the newborn. This occurs when immature lungs are unable to synthesize adequate amounts of surfactant to lower the surface tension within the lungs and prevent alveolar collapse. The most important constituent of surfactant is dipalmitoylphosphatidylcholine, which is formed when a diacylglycerol (dipalmitoylglycerol) condenses with phosphated choline derived from CDP-choline. Other constituents of surfactant include phosphatidylglycerol, apoproteins, and cholesterol.
    α-ketoglutarate (choice A) is a citric acid cycle intermediate.

    Arachidonic acid (choice B) is the lipid precursor of prostaglandins and leukotrienes.

    Proline (choice D) is an imino (rather than amino) acid found in proteins, notably collagen.

    Retinol (choice E) is one form of vitamin A, and is used in the light-sensitive visual cycle in the eye.



    A term neonate is born after a long, difficult delivery. The baby has an APGAR score of 3, so arterial blood is drawn for blood gas studies 3 minutes after delivery. Arterial blood gas studies show a PO2 of 10 mm Hg, PCO2 of 27 mm Hg, and pH of 7.09. Which of the following is the best interpretation of these studies? A. Markedly decreased PCO2, suggesting hyperventilation
    B. Markedly decreased pH, suggesting acidosis
    C. Markedly decreased PO2, suggesting respiratory failure
    D. Markedly increased pH, suggesting alkalosis
    E. Within normal limits






    The correct answer E. This is something of a trick question, but it is included to illustrate a specific point of which you should be aware. The biochemistries of neonates, especially in the first minutes to hours of life, can be strikingly different from those of adults. Specifically, the normal range of arterial blood pH at birth is 7.11 to 7.36; the normal range of PO2 at birth is the strikingly low 8 to 24 mm Hg; and the normal range of PCO2 at birth is the low 27 to 40 mm Hg.
    PCO2(choice A) is normally lower for infants than adults.

    pH (choices B and D) may vary over the first few hours of life in infants who subsequently do well from 7.09 to 7.50 (a range broader than reported "normals" in the question stem, but still not necessarily clinically significant).

    PO2(choice C) can be very low at birth but comes up during the first day of life, when it may still normally be as low as 54 mm Hg.



    A medical student, whose baseline alveolar PCO2 level was 40 mm Hg, begins to voluntarily hyperventilate for an experiment during his respiratory physiology laboratory. If his alveolar ventilation quadruples and his CO2 production remains constant, approximately what will be his alveolar PCO2? A. 4 mm Hg
    B. 10 mm Hg
    C. 20 mm Hg
    D. 80 mm Hg
    E. 160 mm Hg






    The correct answer is B. When you hyperventilate, CO2 is blown off. The amount of CO2 blown off is inversely proportional to alveolar ventilation. This is shown by the alveolar ventilation equation:
    VA = VCO2/PACO2, where

    VA = alveolar ventilation

    VCO2 = CO2 production

    PACO2 = alveolar PCO2

    So, if VCO2 remains the same, and VA quadruples, PACO2 must decrease by 4 fold; 40 mm Hg decreases to 10 mm Hg.



    A native to the Peruvian Andes is being studied at his home elevation of 15,000 ft. He is found to have an arterial PO2 of 40 mm Hg, an O2 saturation of 75% , and an arterial O2 content of 20 mL O2/dL blood. An increase in which of the following makes it possible for the arterial O2 content to be "normal" even though the arterial PO2 and O2 saturation are far below the typical values for a person living at sea level? A. Capillary density
    B. Hematocrit
    C. Hyperventilation
    D. Pulmonary diffusion capacity
    E. Ratio of ventilatory capacity to body mass






    The correct answer is B. In full acclimatization to hypoxia, the hematocrit can increase from a normal, sea level value of 45% to an average of 60% to 65% with a proportional increase in blood hemoglobin levels. This increased level of hemoglobin in the blood makes it possible to carry normal amounts of oxygen even though the arterial PO2 and oxygen saturation are far lower than the sea level normal. Each gram of hemoglobin can carry 1.34 mL O2. With a normal blood hemoglobin concentration of 15 g/dL and an O2 saturation of 97%, the arterial O2 content would be 15 x 1.34 x 0.97 = 19.5 mL O2/dL (not counting dissolved oxygen). In the acclimatized Andean native with a blood hemoglobin concentration of 20 g/dL and oxygen saturation of 75%, the O2 content would be 20 x 1.34 x 0.75 = 20.1 mL O2/dL (not counting dissolved oxygen).
    A high capillary density (choice A) facilitates the exchange of nutrients (which includes oxygen) and metabolites between the blood and tissue spaces via diffusion, but does not affect arterial O2 content.

    Hyperventilation (choice C) can lower alveolar PCO2 and therefore raise the alveolar PO2. This is a powerful mechanism for raising the arterial PO2, but it does not affect the O2 content at a given PO2.

    The pulmonary diffusion capacity (choice D) increases at high altitude, which ensures equilibrium of PO2 between alveolar air and pulmonary capillary blood despite the lower-than-normal oxygen pressure gradient. However, this increase in pulmonary diffusion capacity does not affect the O2 content at a given PO2.

    Andean natives have a large chest in proportion to body size, giving a high ratio of ventilatory capacity to body mass (choice E). The large lungs of Andean natives can reduce the physical effort required to maintain adequate oxygenation, but this is independent of the oxygen-carrying capacity of their blood.





    A pulmonologist is testing a patient's lung volumes and capacities using simple spirometry. Which of the following lung volumes or capacities cannot be measured directly using this technique? A. Expiratory reserve volume
    B. Functional residual capacity
    C. Inspiratory reserve volume
    D. Tidal volume
    E. Vital capacity






    The correct answer is B. The functional residual capacity is the amount of air left in the lungs after a normal expiration. Because this volume cannot be expired in its entirety, it cannot be measured by spirometry. Essentially, lung volume that contains the residual volume, which is the amount of air remaining after maximal expiration (e.g., functional residual capacity and total lung capacity), cannot be measured by spirometry. These volumes can be determined using helium dilution techniques coupled with spirometry or body plethysmography.
    The expiratory reserve volume (choice A) is the volume of air that can be expired after expiration of a tidal volume.

    The inspiratory reserve volume (choice C) is the volume of air that can be inspired after inspiration of a tidal volume.

    Tidal volume (choice D) is the amount of air inspired or expired with each normal breath.

    Vital capacity (choice E) is the volume of air expired after a maximal inspiration.



    A normal, healthy, 25-year-old man lives at the beach. His twin brother has been living in a mountain cabin for the past 2 years. Which of the following indices would be expected to be higher in the man living at sea level? A. Diameter of pulmonary vessels
    B. Erythropoietin production
    C. Mitochondrial density in a muscle biopsy
    D. Renal bicarbonate (HCO3-) excretion
    E. Respiratory rate






    The correct answer is A. A number of physiologic changes occur in a person living at high altitude. The diminished barometric pressure at high altitude causes alveolar hypoxia and arterial hypoxia. Pulmonary vasoconstriction occurs in response to alveolar hypoxia; therefore, the diameter of the pulmonary vessels would be greater in the brother living at sea level. All the other choices describe physiologic processes that would be enhanced by living at high altitude.
    Increased erythropoietin production (choice B), caused by arterial hypoxia, leads to increases in hematocrit in people living at high altitude.

    Mitochondrial density increases (choice C) in people chronically exposed to the hypoxemia caused by living at high altitude.

    At high altitudes, the ventilation rate increases, causing a respiratory alkalosis. The kidney then compensates by increasing the excretion of HCO3- (choice D).

    Increasing the rate of respiration (choice E) is a very useful adaptation to the hypoxic conditions of high altitude. The primary stimulus is the hypoxic stimulation of peripheral chemoreceptors.



    A 56-year-old woman with a 75-pack-year history of smoking cigarettes visits her physician because of shortness of breath. The physician sends her to a pulmonary function laboratory for a complete work-up. The table below shows pulmonary volumes and capacities obtained using simple spirometry and helium washout techniques.

    Lung volumes Amount
    Functional residual capacity 4.5L
    Inspiratory reserve volume 1.5L
    Inspiratory capacity 2.0L
    Vital capacity 3.0L

    What is the residual volume of this patient?


    A. 1.5 L
    B. 2.0 L
    C. 2.5 L
    D. 3.0 L
    E. 3.5 L
    F. 4.0 L






    The correct answer is E. Problems of this type are best approached by drawing a spirogram (see below) and filling in the values provided in the table. This approach eliminates the need to memorize formulas for the various pulmonary volumes and capacities because these become obvious when the spirogram is examined. In this particular problem, it is first necessary to calculate the total lung capacity (TLC) by adding the functional residual capacity and inspiratory capacity (TLC = 4.5 L + 2.0 L = 6.5 L). The residual volume is then calculated as the difference between TLC and vital capacity, i.e., 6.5 L - 3.0 L = 3.5 L. The total lung capacity, functional residual capacity, and residual volume are often increased in the emphysematous lungs of patients with a long-term history of cigarette smoking.
    A inspiratory reserve volume
    B expiratory reserve volume
    C inspiratory capacity
    D vital capacity
    E residual volume
    F functional residual capacity
    G tidal volume
    H total lung capacity




    A 72-year-old man visits his physician because of a cough that has persisted for several months. The man thinks he was exposed to asbestos for 4 years in his late fifties at an insulation factory. A chest X-ray shows mottling. The table below shows pulmonary volumes and capacities obtained using simple spirometry and helium washout techniques.

    Lung volumes Amount
    Functional residual capacity 2.0L
    Inspiratory reserve volume 1.5L
    Inspiratory reserve capacity 2.0L
    Vital capacity 3.3L
    Tidal volume 0.5L

    What is the total lung capacity of this patient?


    A. 2.0 L
    B. 3.0 L
    C. 4.0 L
    D. 5.0 L
    E. 6.0 L






    The correct answer is C. The total lung capacity is the sum of the functional residual capacity and inspiratory reserve capacity. Problems of this type are best approached by drawing a spirogram (see below) and filling in all values provided in the table. This approach eliminates the need to memorize formulas for the various pulmonary volumes and capacities because these become obvious when the spirogram is examined.
    The presence of asbestos in the lungs can cause an interstitial process that slowly develops into diffuse pulmonary fibrosis after a long latent period. Pulmonary fibrosis decreases lung compliance and increases the elasticity of the lungs. Patients with asbestosis usually have a decrease in total lung capacity, functional residual capacity, and residual volume.







    Arterial oxyhemoglobin 97%
    Arterial partial pressure of oxygen 100 mm Hg
    Arterial oxygen content 97 mL oxygen/L blood
    Mixed venous oxyhemoglobin 50%
    Mixed venous partial pressure of oxygen 35 mm Hg
    Mixed venous oxygen content 47 mL oxygen/L blood
    Cardiac output 6 L/min


    A 25-year-old man weighing 150 pounds has lost both kidneys to renal disease and undergoes hemodialysis therapy 3 days each week. He becomes severely anemic (hemoglobin = 7.0 g/dL) because his kidneys are no longer producing erythropoietin. The data above were obtained from the patient during resting conditions. What is his resting oxygen consumption?

    A. 100 mL/min
    B. 150 mL/min
    C. 200 mL/min
    D. 250 mL/min
    E. 300 mL/min






    The correct answer is E. Oxygen consumption can be calculated using the Fick equation as follows: oxygen consumption = cardiac output x (arterial O2 content - venous O2 content). Therefore, oxygen consumption = 6 L/min x (97 mL O2/L - 47 mL O2 /L) = 6 L/min x (50 mL O2/L) = 300 mL O2 /min. Note that each liter of blood loses 50 mL of oxygen as it passes through the tissues. Because 6 liters of blood pass through the tissues each minute, and because each liter of this blood loses 50 mL of oxygen, a total of 300 mL of oxygen are used by the tissues each minute.


    Which of the following would shift the oxygen-hemoglobin dissociation curve to the right? A. Carbon monoxide poisoning
    B. Decreased PCO2
    C. Decreased pH
    D. Decreased temperature
    E. Decreased 2,3-DPG






    The correct answer is C. The loading of O2 is facilitated when the oxygen dissociation curve shifts to the left, and the unloading of O2 is facilitated when the oxygen dissociation curve shifts to the right. A good way to remember the conditions that promote dissociation of O2 is to think of exercising muscle, which has decreased pH (choice C) because of the accumulation of lactic acid, increased PCO2 (compare with choice B) because of the increased rate of aerobic metabolism, increased temperature (compare with choice D), and increased 2,3-DPG (2,3-diphosphoglycerate; compare with choice E) because of increased glycolysis.
    Carbon monoxide poisoning (choice A) left-shifts the oxygen dissociation curve, which interferes with the unloading of O2. Carbon monoxide also strongly binds to available sites on hemoglobin.
    " VALUE HAS A VALUE ONLY IF ITS VALUE IS VALUED "
    Never Let Student Die In Your Heart When It Dies You Want Remain A Doctor But You Will Be A Technician

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    thanks

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    Thank you very much

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