Respiratory Flashcards
Regarding the lungs, which of the following statements is correct?
A Ventilation/perfusion (VQ) quotient is directly proportional to gas exchange
B Ventilation is greatest in middle zone
C Perfusion is greatest at the base
D Ventilation/perfusion (VQ) quotient is inversely proportional to gas exchange
C
Explanation
Ventilation is greatest in the lower zones of the lungs. Base of the lungs has less negative pressure (-2.5 vs. -10mmhg) and therefore a small resting volume and expands well on inspiration. Ventilation perfusion (VQ) will affect gas exchange in different ways: A high VQ means more ventilation and less blood flow and the arterial blood PO2 will approach that of inspired air. With a low VQ, the arterial PO2 will approach that of mixed venous blood.
Regarding ventilation, which of the following options is correct?
A Increased respiratory rate (RR) decreases physiological dead space
B Lung units with a high Ventilation/perfusion (VQ) quotient have decreased alveolar minute ventilation
C Anatomical dead space is 1ml/kg
D At the end of inspiration, the chest wall relaxes and the alveoli recoil back to their original size
D
Explanation
During inspiration, the intercostal muscles contract and the diaphragm moves down. The intrapulmonary pressure decreases as the alveoli are pulled open and the lung expands. The pressure inside the lungs reduces compared to the outside atmospheric pressure and air moves into the lungs. At the end of inspiration, the intercostal muscles and the diaphragm relax, (and after that) the elastic properties of the alveoli recoil back to their original size, the intrapulmonary pressure increases above atmospheric pressure and air moves out of the lungs
Anatomical dead space is 2ml/kg (150ml). Physiological dead space is the volume of gas that does not eliminate CO2.
An increased respiratory rate will increase the volume of the dead space. If you breathe rapidly and shallow, lets say a TV of 150ml and a respiratory rate of 40/min. Your pulmonary ventilation (total exchange of air between the lungs and ambient air-minute ventilation) will be 6000ml/min but your alveolar ventilation (total exchange of air that reaches the alveoli and is available for gaseous exchange with the blood per unit time) will be 0. Therefore your dead space increases dramatically (over ventilation). Dead space also depends on the size and posture of the patient.
PV= TV X RR.
AV= (TV-Dead space) X RR
Normal: PV= 500 X 12= 6000ml/min. AV=(500-150) X 12 =4200ml/min.
Abnormal breathing PV=150 X 40=6000ml/min. AV=(150-150) x 12= 0 Massive increase in dead space
Lung units with high V/Q quotients, have high ventilation (decreasing PCO2 and increasing O2). An increased ventilation will increase tidal volume and thus increase your minute ventilation.
Extra: Physiological dead space is the sum of anatomical dead space (air that does not reach the alveoli) and alveolar dead space (air that enters poorly or nonperfused alveoli); increasing respiratory rate does not improve ventilation efficiency because dead space is increased. Conversely, decreasing respiratory rate and increasing tidal volume has been shown to improve ventilation efficiency via alveolar recruitment and distension, thus reducing alveolar dead space.
The physiological effects of slow breathing in the healthy human
Marc A. Russo et al
Which of the following statements is true in relation to airway resistance?
A It decreases while breathing through the nose
B It is independent of lung volumes
C It is equal in inspiration and expiration
D It increases with forced expiration
D
Explanation
Airway resistance will rise as the lung volume decreases. Airway resistance is not equal in both expiration and inspiration and airway resistance will increase with nasal breathing (halving the size of the tube increases the resistance 16 fold).
With regard to lung compliance, which of the following statements is correct?
A It decreases with age
B It changes in inspiration compared with expiration
C It is independent of lung volume
D It decreases with emphysema
B
Explanation
Compliance is slightly greater when measured during deflation then when measured during inflation.
Compliance = change of volume divided by change of pressure.
Compliance increases with age and emphysema and is decreased in pulmonary congestion and interstitial pulmonary fibrosis.
Compliance of a human lung is approximately 200ml/cm water.
Regarding the Respiratory Quotient (RQ), which of the following options is correct?
A RQ brain > 0.95
B RQ fat = 0.95
C RQ Carbohydrates= 0.95
D RQ liver > 0.95
A
Explanation
The RQ of
-Fat is 0.7
- Carbohydrates is 1.00
- Brain is 0.97-0.99
Respiratory Quotient (RQ) is the steady state ratio of CO2 production to O2 production in metabolism. Carb RQ=1 (H and 2O in same proportion as water)
Lipid RQ= 0.7 indicating more oxygen is required (due to ratios)
RQ also can be calculated for individual organs by comparing arterio-venous alterations in CO2 and O2: e.g. Brain 0.97-0.99 hence principal source is carbs (but small amount of other is used) RQ should be distinguished from R (Resp exchange ratio) which is ratio of CO2/O2 at any point and can be influenced by ventilation (e.g. R=2.0 in strenuous exercise/acidosis and <1 in alkalosis)
What is the oxygen pressure in the bronchi at an altitude where barometric pressure is 500 mmHg, breathing 30% O2?
A 129 mm Hg
B 85 mm Hg
C 135 mm Hg
D 105 mm Hg
C
Explanation
Alveolar gas equation
PAO2 = O2 % x (atmospheric pressure-47) - (pCO2 [arterial pressure of cardon dioxide]/0.8)
PAO2=0.3(500-47) is equal to 135.9mmHg
Note: Because the question said bronchi, the amount of CO2 is very little. In inspiration CO2 levels at your mouth and main bronchi are negligible. As you go down the bronchial tree into the alveoli, CO2 starts to make up a concentration. Therefore if the question asks for the pressure in the alveoli, you will need the CO2 pressure (to complete the alveolar gas equation). In this case, because it is in the bronchi the CO2 can be ignored.
Vapour water pressure is 47mmHg
Note: This question has been asked with slightly different values: breathing room air- 20.93%- allowed to use 21%
0.21 X (500-47)=95mmHg
Without a calculator: what is a fifth (1/5) of 453 (500-47)=90.6 choose the option 95mmHg. (option given were 85mmHg, 95mmHg, 115mmHg and 125mmHg)
If compliance of the lung is 30mL/cmH20 and the average tidal volume is 600mL, the pressure change per breath is?
A 12 cm H20
B 10 cm H20
C 18 cm H20
D 20 cm H20
D
Explanation
Compliance is the volume changes per pressure change. C=V/P, therefore P=V/C, P=600/30 which is equal to 20 cm H20
Note: With this question, the volume has been left in ml and not converted to L. However, if you converted it litres, then the answer would still be 20 cm H20.
600ml=0.6L and 30ml=0.03L. Therefore-0.6/0.03=20
There is a similar question in the data base where the conversion to litres has been done. Please be aware of this. It may be prudent to do the question both ways
Given that the intrathoracic pressure changes from 5 cmH2O to 10 cmH2O with inspiration and a tidal volume (TV) of 500 mls, what is the compliance of the lung?
A 10 L/cmH2O
B 0.1 L/cmH2O
C 1 L/cmH2O
D 100 L/cmH2O
B
Explanation
Compliance is the volume change per pressure change.
C=V/P
C=L/cmH2O
Therefore;
C=0.5/5 which equals 0.1
Extra: This is sneaky, remember to convert it to the L/cmH20 rather than keeping it in ml/cmH20. Very sneaky question
Residual volume in a 70kg man most closely approximates which of the following values?
A 3.0 litre
B 1.0 litre
C 2.0 litre
D 4.0 litre
B
Explanation
Residual volume is approximately 1.2L
RV= amount of air remaining in the lungs after maximum expiration; keeps the alveoli inflated between breaths and mixes with fresh air on the next inspiration
The source for the RV amount is from the physiology textbook- Ganong edition 25
TV ~500ml (-750ml)
IRV ~2000ml
ERV ~1000ml
RV ~1300ml
VC ~3500ml
IC ~2500ml
FRC ~2500ml
TLC ~5000ml
Permanent high altitude is associated with all of the following changes except?
A Increased alveolar ventilation
B Increased arterial blood HCO3
C Increased pulmonary artery pressure
D A low PaCO2
B
Explanation
Permanent high altitude states result in the kidneys excreting HCO3, which compensates for the alkalosis caused by hyperventilation. This is the most important feature of acclimatization to high altitude. The mechanism of hyperventilation is hypoxic stimulation of the peripheral chemoreceptors. The resulting low arterial PCO2 and alkalosis tends to inhibit this increase in ventilation. After 2-3 days the arterial blood pH is returned to normal by the renal excretion of HCO3. The brakes on ventilation are thus reduced and respiration may increase further. Arterial 2,3 diphosphoglycerate (DPG) also increases.
Extra: The body’s response to high altitude includes the following: ↑Erythropoietin → ↑hematocrit and hemoglobin ↑2,3-DPG (allows ↑ release of O2 and a right shift on the Hb-O2 disassociation curve) ↑renal excretion of bicarbonate (use of acetazolamide can augment for treatment) Chronic hypoxic pulmonary vasoconstriction (can cause Right Ventricular Hypertrophy)
Extra:
A explanation on increased pulmonary artery pressure; https://www.ncbi.nlm.nih.gov/books/NBK555925/ “ Many organ systems utilize vasodilation to enhance oxygen delivery in the setting of hypoxia. In contrast, the lung responds to hypoxia with increased pulmonary vasoconstriction. Blood is shunted away from poorly oxygenated lung zones towards healthy alveoli in an effort to minimize V/Q mismatch. This intuitively makes sense in the setting of lobar pneumonia or other focal disease processes involving the lung. In high altitude pulmonary hypertension, all lung fields experience the same degree of hypoxia. Significant vasoconstriction in all parts of the lung and elevated pulmonary artery pressures lead to pulmonary hypertension. The resultant increase in pulmonary pressure forces fluid into the lungs, causing acute pulmonary edema. “
With regard to the distribution of pulmonary blood flow, which of the following options is correct?
A Hypoxia leads to pulmonary dilation
B Typically, there is a zone at the apex which is not perfused
C The mean pulmonary arterial pressure is 8 mmHg
D In some areas, flow is determined by the arterial/alveolar pressure difference
D
Explanation
Only in diseased lungs is there an area which is not perfused. The mean pulmonary arterial pressure is 15mmHg. Hypoxia leads to pulmonary constriction.
With regard to pulmonary gas exchange, which of the following statements is correct?
A The diffusion rate for CO2 is double that of O2
B Transfer of nitrous oxide is perfusion limited
C Diffusion is inversely proportional to the partial pressure gradient
D Transfer of O2 is diffusion limited
B
Explanation
Diffusion is proportional to the partial pressure gradient.
The diffusion rate for CO2 is twenty times faster than O2.
The transfer of O2 is perfusion limited. The rise in partial pressure when O2 enters a red blood cell is great. Under typical resting conditions , the capillary PO2 virtually reaches that of alveolar gas when the red cell is about one third of the way along the capillary. Under these conditions, O2 transfer is perfusion limited. However, in some abnormal circumstances when the diffusion properties of the lung are impaired, for example, because of thickening of the blood -gas barrier, the blood PO2 does not reach the alveolar value by the end of the capillary, and now there is some diffusion limitation as well.
When nitrous oxide moves across the alveolar wall into the blood, no combination with Hb takes place. As a result, the blood has nothing like the avidity for nitrous oxide that it has for CO, and the partial pressure rises rapidly. The partial pressure of nitrous oxide has reached that of alveolar gas by the time the red cell is only one tenth of the way along the capillary. After this point , almost no nitrous oxide transferred. Thus the amount of gas taken up by the blood depends entirely on the amount available blood flow and not at all on the diffusion properties of the blood-gas barrier. The transfer of nitrous oxide is perfusion limited
When the red cell enters the capillary, CO (carbon monoxide) moves rapidly across the extremely thin blood-gas barrier form the alveolar gas into the cell. As a result, the content of CO in the cell rises. However, because of the tight bonds that form between CO and Hb within the cell, a large amount of CO can be taken up by the cell with almost no increase in partial pressure. Thus, as the cell moves through the capillary, the CO partial pressure in the blood hardly changes, so that no appreciable back pressure develops , and the gas continues to move rapidly across the alveolar wall. It is clear that the amount of CO that gets into the blood is limited by the diffusion properties of the blood-gas barrier and not by the amount of blood available. The transfer of CO is therefore said to be diffusion limited
A permanent inhabitant at 4,500 feet has adapted by which of the following mechanisms?
A Increased ventilation
B A high alveolar PO2
C A decreased 2,3, diphosphoglycerate (DPG)
D A normal HCO3-
A
Explanation
A permanent inhabitant at 4,500 feet will have a lower alveolar PO2 relative to sea level. By hyperventilating, he can increase his O2. He will have an increased 2,3 diphosphoglycerate (DPG). He must also increase ventilation, as this is the most important feature in acclimatization. He will have a low arterial HCO3, which the body creates by renally excreting the bicarbonate allowing the patient to hyperventilate.
What is the PO2 of alveolar air with a CO2 of 64, breathing room air at sea level and a respiratory exchange ratio of 0.8?
A 70
B 35
C 52
D 74
A
Explanation
Alveolar gas equation PAO2= (atmospheric pressure – vapour pressure) X inspired oxygen percentage - PaCO2/0.8
Therefore;
PAO2= (760-47) X 0.21 – 64/0.8 which equals 69 mmHg
Note: if you include the correction fraction of 2mmhg, the answer could be 71-72 as well.
The respiratory exchange ratio is the flow of CO2 molecules across the alveolar membrane per minute divided by the flow of O2 molecules across the membrane per minute
When walking at a steady pace the increase in respiratory rate is due to?
A Increased pH
B Decreased PO2
C Increased CO2
D None of the above
D
Explanation
The pH, PCO2 and P02 remain constant during moderate exercise. An abrupt rise in ventilation at the start of exercise is due to psychic stimuli and afferent impulses from the joints, muscles and tendons. The gradual increase is humoral even though PCO2, PO2 and pH remain constant in moderate exercise. The increase in ventilation is proportionate to the increase in O2 consumption but the mechanism responsible for stimulation is still the subject of much debate.
What is the residual volume left in the lung after forced maximal expiration?
A 1Litre
B 3 Litres
C 2 Litres
D 0.5 Litres
A
Explanation
The volume expelled by an active expiratory effort after passive expiration is the expiratory reserve volume (ERV), and the air remaining in the lungs after a maximal expiratory effort is the residual volume (RV). Men=1.2 and women=1.1 litres
Note: I do not think that the question is referring to ONE lung. Hopefully the answers will not give options which can create confusion
Compliance of the lung is reduced by all the following with the exception of?
A Emphysema
B Fibrosis
C Consolidation
D Alveolar oedema
A
Explanation
Compliance is also increased in the normal ageing lung
In control of ventilation, the medullary chemoreceptors respond to changes by which of the following?
A H+ concentration
B O2 tension
C CO2 tension
D HCO3 concentration
A
Explanation
The central chemoreceptors are surrounded by brain extracellular fluid and respond to the changes in H+ concentration. An increase in H+ concentration stimulates ventilation and a decrease inhibits it. Although the BBB is relatively impermeable to H+ and HCO3- ions, CO2 molecule diffuses easily across it. When the blood PCO2 rises, CO2 diffuses into the CSF from the cerebral blood vessels liberating H+ ions that STIMULATE the chemoreceptors.
The Haldane effect refers to which of the following options?
A The dissociation constant for the bicarbonate buffer system
B The carriage of O2 according to Henry’s law
C The chloride shift that maintains electrical neutrality
D The increased capacity for deoxygenated blood to carry CO2
D
Explanation
Henry’s law refers to the amount of dissolved oxygen, which is proportional to the pressure of oxygen. At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
The chloride shift which maintains electrical neutrality (in the plasma) is the Hamburger effect. Cl- shifts into the RBC as HCO3- shifts out (part of the Haldene effect)
The Haldane effect is a property of haemoglobin. Deoxygenation of the blood increases its ability to carry carbon dioxide, and for Hydrogen ions; this property is the Haldane effect. Conversely, oxygenated blood has a reduced capacity for carbon dioxide.
After residing at an increased altitude for some time, which of the following occurs?
A pH of arterial blood is returned to near normal
B Hypoxia stimulates the aortic body over the carotid body
C Nitrogen levels are increased
D Alkalosis and a low pCO2 prevents an increase in ventilation
A
Explanation
At high altitude, the atmospheric pressure is reduced thus for the same partial pressure of O2 the alveolar PAO2 is reduced. Hypoxaemia drives hyperventilation by stimulation of peripheral chemoreceptors. The resulting low arterial PCO2 and alkalosis tends to inhibit this increase in ventilation, but after a day or so, the CSF pH is brought partly back by movement of bicarbonate out of the CSF, and after 2 or 3 days, the pH of the arterial blood is returned nearer to normal by renal excretion of bicarbonate. These brakes on ventilation are then reduced, and it increases further. In addition, there is now further evidence that the sensitivity of the carotid bodies to hypoxia increases during acclimatization.
At an altitude of 18,000 feet there is much less of the atmosphere pressing down. The concentration of nitrogen, oxygen, and other gases is half the concentration at sea level. The air is less dense. This is often referred to as thin air, with molecules more spread out because of the reduced pressure.
Which is true regarding the pulmonary hypoxic vasoconstriction?
A The stimulus response curve of this response is linear
B Occurs in response to low alveolar PO2
C It occurs in response to a low pulmonary arterial PO2
D Nitric Oxide has shown to be a potent vasoconstrictor
B
Explanation
Hypoxic pulmonary vasoconstriction occurs in response to a low alveolar PO2 and not in response to a low pulmonary arterial blood PO2. The precise mechanism of this response is not entirely known. The stimulus response curve is non linear, when the alveolar PO2 is altered in the region above 100mmHg, little change in vascular resistance is seen. When the alveolar PO2 is reduced below 70mmHg, marked vasoconstriction take place. And at a very low PO2, the local blood flow may be almost abolished. NO (nitric oxide) a endothelium derived vasoactive substance is a relaxing factor for blood vessels.
The haemoglobin dissociation curve moves to the right under which of the following circumstances?
A Alkalosis
B Raised CO2
C Low temperature
D Decrease in 2,3 DPG
B
Explanation
A rise in temperature, an decrease in the pH (acidosis), an increase in 2,3 DPG (an end product of red cell metabolism) and a rising pCO2 all shift he haemoglobin dissociation curve to the right.
Remember: an exercising muscle is acidic, hypercarbic and hot, and it benefits from an increased unloading of O2 from its capillaries. I.e. a right shift.
This is known as the Bohr Effect: a decrease in blood pH or an increase in blood CO2 concentration will result in haemoglobin proteins releasing their loads of oxygen (right shift) and a decrease in carbon dioxide or increase in pH will result in haemoglobin picking up more oxygen (left shift)
Which zone of the lung is sometimes NOT perfused?
A Zone 3
B Zone 1
C Zone 2
D Zone 4
B
Explanation
Zone 1 is not observed in the normal healthy human lung. In normal health pulmonary arterial (Pa) pressure exceeds alveolar pressure (PA) in all parts of the lung. It is generally only observed when a person is ventilated with positive pressure or haemorrhage. In these circumstances, blood vessels can become completely collapsed by alveolar pressure (PA) and blood does not flow through these regions. They become alveolar dead space.
Zone 2 is the part of the lungs about 3 cm above the heart. In this region blood flows in pulses. At first there is no flow because of obstruction at the venous end of the capillary bed. Pressure from the arterial side builds up until it exceeds alveolar pressure and flow resumes. This dissipates the capillary pressure and returns to the start of the cycle. Flow here is sometimes compared to a starling resistor or waterfall effect.
Zone 3 comprises the majority of the lungs in health. There is no external resistance to blood flow and blood flow is continuous throughout the cardiac cycle. Flow is determined by the Ppa-Ppv difference (Ppa - Ppv), which is constant down this portion of the lung. However, transmural pressure across the wall of the blood vessels increases down this zone due to gravity. Consequently the vessels walls are more stretched so the calibre of the vessels increases causing an increase in flow due to lower resistance.
Zone 4 can be seen at the lung bases at low lung volumes or in pulmonary oedema. Pulmonary interstitial pressure (Pi) rises as lung volume decreases due to reduced radial tethering of the lung parenchyma. Pi is highest at the base of the lung due to the weight of the above lung tissue. Pi can also rise due to an increased volume of ‘leaked’ fluid from the pulmonary vasculature (pulmonary oedema). An increase in Pi causes extra alveolar blood vessels to reduce in calibre; in turn causing blood flow to decrease (extra alveolar blood vessels are those blood vessels outside alveoli). Intra alveolar blood vessels (pulmonary capillaries) are thin walled vessels adjacent to alveoli which are subject to the pressure changes described by zones 1-3. Flow in zone 4 is governed by the arterio interstitial pressure difference (Pa − Pi). This is because as Pi rises, the arterial calibre is reduced, thereby increasing resistance to flow. The Pa/Pv difference remains unchanged since Pi is applied over both vessels.
The ventilation/perfusion ratio is higher in zone #1 (the apex of lung) when a person is standing than it is in zone #3 (the base of lung) because perfusion is nearly absent. However, ventilation and perfusion are highest in base of the lung, resulting in a comparatively lower V/Q ratio.
Extra: another way the question may be worded:
In which zone of the lung is capillary pressure sometimes zero?
What happens initially to pCO2 and pO2 with ventilation/perfusion (V/Q) mismatch?
A pCO2 increases, pO2 decreases
B pCO2 unchanged, pO2 decreases
C pCO2 decreases, pO2 decreases
D pCO2 increases, pO2 increases
B
Explanation
Ventilation/perfusion inequality has a much greater effect on pO2 than pCO2.
The CO2 dissociation curve is linear in working range. Chemoreceptor stimulation increases ventilation and CO2 output especially in lung areas with high V/Q ratios.
The O2 dissociation curve however is nonlinear. Areas with high V/Q ratio add little O2 with increased ventilation. Whereas low V/Q ratio areas have lower pO2 close to mixed venous blood and overall pO2 is reduced. Hypoxia cannot be corrected by increased
An increase in pO2 reduces stimulation of aortic bodies. Which of the following also reduces stimulation?
A Decreased pH
B Decreased pCO2
C Increased pCO2
D Hypoventilation
B
Explanation
Peripheral chemoreceptors are rapidly responding and are located in the carotid and aortic bodies. They respond to decreased arterial pO2. There is a minor response to increased pCO2. Decreased pH causes a carotid response only.
Hyperventilation is the response to an acid load applied to the body. What causes this response?
A Ventral medulla
B Pre-Botzinger neurons
C Aortic bodies
D Pons
A
Explanation
The most important receptors involved in the minute-by-minute control of ventilation are those situated near the ventral surface of the medulla in the vicinity of the exit of the 9th and 10th nerves. In animals, local application of H+ or dissolved CO2 to this area stimulates breathing within a few seconds. At one time, it was thought that the medullary respiratory centre itself was the site of action of CO2, but it is now accepted that the chemoreceptors are anatomically separate.
Extra:
The central chemoreceptors, located on the ventral aspect of the medulla, are activated by an increase in CO2 or acidity.4 The best known effects of central chemoreceptor activation are increases in ventilation. Indeed, studies have suggested that central chemoreceptor activation, depending on conditions, mediates 50% to 90% of the ventilatory response to hypercapnia.
https://www.sciencedirect.com/topics/neuroscience/central-chemoreceptors
