PULMONARY PHYSIOLOGY REVIEW - ANSWERS
Question 1.
The physiological dead space is:
The answer is C. - often increased in lung disease
The expired and arterial
PCO2 are used, not the arterial PO2.
Choice B is incorrect because the physiological dead space contains the
anatomic dead space, and may be substantially larger if there is inequality of
ventilation and blood flow within the lung. This is the reason why the
physiological dead space is frequently increased in lung disease (choice C)
Hypoxic pulmonary vasoconstriction:
The answer is B. - improves matching of ventilation and blood flow in some lung diseases.
Although the mechanism of hypoxic pulmonary vasoconstriction is not fully understood, we know that central nervous connections are not required because the phenomenon can be demonstrated in isolated lungs. Therefore, choice A is incorrect.
Choice B is correct. Suppose a lobe or lobule of lung is poorly ventilated because of partial bronchial obstruction. The resulting alveolar hypoxia will reduce the blood flow through the mechanism of hypoxic pulmonary vasoconstriction. The result is improvement in the matching of ventilation and blood flow.
Choice C is incorrect. Reducing the PO2 of the blood entering the lung results much less vasoconstriction than reducing the PO2 of alveolar gas.
Choice D is incorrect. Hypoxic pulmonary vasoconstriction is important in the perinatal period. When the newborn baby makes the transition from placental to air breathing, it is important for pulmonary vascular resistance to fall precipitously within a few seconds. As a consequence, pulmonary blood flow dramatically increases from its value of only about 15% of the cardiac output in utero. The increase in pulmonary blood flow is assisted by closure of both the ductus arteriosus and the foramen ovale.
Pulmonary vascular resistance increases:
The answer is A. - at high altitude.
Pulmonary vascular resistance increases at high altitude as a result of the global alveolar hypoxia. The exact mechanism is still unknown but is apparently a local effect on the smooth muscle of the pulmonary arterial wall. The increase causes right ventricular hypertrophy with characteristic ECG changes.
Pulmonary vascular resistance during space flight would, if anything decrease as blood flow becomes more uniform. Anemia would decrease viscosity - a term in the resistance formula during exercise, pulmonary capillaries would distend causing resistance to fall.
With exercise, pulmonary arterial pressure tends to rise causing recruitment and distension of pulmonary vessels leading to a fall in pulmonary vascular resistance.
In a subject with normal lungs, which of the following (acting alone) would have the greatest effect on O2 delivery (arterial o2 concentration x cardiac output)?
The answer is D. - hemoglobin reduced to 50% of normal.
Halving alveolar ventilation would double alveolar PCO2 and therefore reduce alveolar and arterial PO2 but not enough to halve arterial O2 content. Also, hypoxia would stimulate cardiac output via sympathetic stimulation of heart rate.
A 50% right to left shunt would also lower arterial O2 content but not by 50% since the shunted blood (venous blood) has a significant amount of oxygen (normally about 75% saturated).
Doubling inspired O2 - if you picked this choice …………@#$%!!!!.
Reducing hemoglobin concentration to half normal would reduce arterial O2 content (at any saturation) to half normal, therefore choice D is correct.
A patient is breathing air at sea level and has a respiratory exchange ratio of 1.0. The arterial blood values are:
PO2 90 mm Hg
PCO2 20 mm Hg
pH 7.30
These indicate that the:
The answer is E. - all of the above.
The alveolar PO2 is found from the alveolar gas equation:
PAO2 = PIO2 - PACO2/R
We can assume that the inspired PO2 is the sea level normal value of 149 mm Hg. Therefore, neglecting the small correction factor, the alveolar PO2 is 149 - 20/1 or 129 mm Hg. Thus the alveolar-arterial PO2 diffusion is 129 - 90 = 39 mm Hg.
The PCO2 of 20 mm Hg means that the patient is hyperventilating. The combination of the low PCO2 and low pH means that plasma bicarbonate concentration is reduced (Henderson-Hasselbalch equation
pH = pK’ = log
[HCO3-]/(PCO2 x a)
The data would fit a metabolic acidosis with partial respiration compensation.
In a patient with anemia and normal lungs:
The answer is D. - PO2 of mixed venous blood is reduced.
A patient with anemia and normal lungs typically has a normal arterial PO2. Thus, choice A is incorrect. Because the position of the oxygen dissociation curve is typically normal, the arterial oxygen saturation is also normal and thus, choice C is incorrect.
From the Fick principle:
O2 = x (CaO2 - CO2)
If the oxygen consumption and cardiac output are normal, the arterial-venous O2 concentration difference will also be normal. Thus, choice B is incorrect. As a matter of fact, cardiac output is sometimes reflexly increased in anemia, and if this occurs, the arterial-venous O2 concentration difference will be decreased.
Choice D is correct. Although the arterial PO2
is typically normal, the PO2 of mixed venous blood must fall. This
is because the venous oxygen concentration falls to a very low level as the
normal amount of oxygen is extracted, and thus the venous
Features of mild carbon monoxide poisoning include:
The answer is D. - decreased arterial O2 concentration.
Carbon monoxide poisoning causes some of the hemoglobin in the blood to be combined with CO to give carboxyhemoglobin. As a result, there is a decreased arterial O2 concentration (choice D is correct). However, it is important to realize that this does not decrease the arterial PO2. In this respect, carbon monoxide poisoning is similar to anemia where again the arterial PO2 is typically normal but the arterial oxygen concentration is reduced.
Choice B is not correct. There is an increase in the affinity of the hemoglobin for oxygen as evidenced by the leftward shift of the oxygen dissociation curve. Thus, carbon monoxide poisoning causes a fall in tissue PO2 for two reasons: first less oxygen is carried in the arterial blood, and second the unloading is impeded by the higher hemoglobin affinity.
Choice C is incorrect. Ventilation is unaffected in mild carbon monoxide poisoning; the chemoreceptors respond primarily to the arterial PO2 and, because this is normal, ventilation remains essentially unchanged.
The word “mild” is included in the stem of the question because with very severe carbon monoxide poisoning there may be brain damage which may alter ventilation.
The increase in ventilation which occurs immediately following ascent to high altitude:
The answer is D. - increases still further over the course in the next 1-3 days.
The increase in ventilation which occurs immediately following ascent to high altitude is caused by stimulation of the peripheral chemoreceptors by the arterial hypoxemia. Because of the very high blood flow per tissue mass of the carotid chemoreceptors, they essentially respond to arterial PO2 and therefore choice C is incorrect.
The increase in ventilation is partly inhibited by the resulting fall in PCO2 which causes a rise in pH of the CSF. This reduces the stimulation of the central chemoreceptors which works against the increased drive from the peripheral chemoreceptors. However, after a day or so, the pH of the CSF is reduced to some extent by the outward movement of bicarbonate and thus this inhibitory effect on the central chemoreceptors is diminished. Consequently, ventilation increases still further over the course of the 1-3 days.
The increase in ventilation is not caused by the reduced work of breathing the less dense air though it is true that the work of breathing is reduced to some extent at high altitude. If a subject is asked to breathe as fast and as deeply as he can for 15 seconds, it can be shown that the amount of air that is exhaled (measured at BTPS) is greater at high altitude than at sea level, but this does not cause increased ventilation.
In a patient with stable chronic lung disease, elevated PaCO2 and decreased PaO2:
The answer is D. - All of these.
Many patients with chronic lung disease have chronic hypoxemia and chronic CO2 retention with an elevated arterial PCO2. Typical values might be an arterial PO2 in the 40s or 50s and an arterial PCO2 in the 50s. Such a patient may have the pH of his CSF near the normal value of about 7.32. Although the pH may have fallen in the initial stages of hypercapnia, the arterial PCO2 is often stable at the high value for so long that the pH of the CSF returns to near normal as a result of an increase in CSF bicarbonate concentration. Therefore, choice A is correct
Such a patient has a reduced ventilatory response to CO2. Indeed this is one reason why there is chronic CO2 retention. Part of the reason for the reduced ventilatory response is the near normal value of the CSF pH. Another reason is that such a patient often has severe airway obstruction with a markedly increased airway resistance. In this case, even if a raised arterial PCO2 causes an increase in the number of impulses to the ventilatory muscles from the respiratory center, they are unable to increase ventilation in the normal way because of the airway obstruction. Thus, choice B is correct.
C is correct. In these patients, the hypoxic ventilatory drive from the peripheral chemoreceptors is often the most important stimulus to ventilation. This has important practical consequences. If such a patient is brought into the emergency room and given oxygen to relieve his arterial hypoxemia, his main stimulus to breathing may be abolished, an he may develop lethal CO2 retention and respiratory acidosis. The best way to manage these patients is to give relatively small increases in oxygen concentration (typically 24 to 28 %) and to monitor the arterial PO2, PCO2 and pH carefully for signs of additional CO2 retention and respiratory acidosis.
Choice D is correct. These patients typically have an almost fully compensated respiratory acidosis, but the arterial hydrogen ion concentration is usually slightly increased.
Back to Questions
Question 10.
Answer is C. The data given show that R
increases from 0.8 to 1. The alveolar PO2 equation states that
PAO2 = PIO2 - PACO2/R.
If PIO2 & PACO2 do not change, then changing R
from 0.8 to 1.0 results in an INCREASE of alveolar PO2.
PIO2
= (PB – PH2O) x FIO2 = (250 – 47) x.21 ≈
42.6 mm Hg
PAO2 = PIO2 –
(PaCO2/R) = 42.6 – (40/1) = 2.6 mm Hg.
PIO2 = (PB – 47)FIO2 at
sea level this is PIO2 = (760 – 47)0.21 = 149.7; so to keep
inspired PO2 at 149.7 on Everest summit, the required inspired O2
fraction is calculated from rearranging the equation for PIO2:
PIO2
= (PB – PH2O) x FIO2 to solve for FIO2
FIO2 = PIO2/(PB
– 47)/= 149.7/(250-47) = 0.74 or 74% oxygen.
Alveolar PCO2
= arterial PCO2 = 50 mm Hg. If R = 0.8, and oxygen consumption = 250 ml/min, then carbon dioxide
production can be calculated by rearranging the equation for R as follows:
R
= CO2/O2
or CO2 = R x O2 = 0.8 x 250 ml/min = 200 ml/min
To solve for A
A = (CO2/ PACO2) x 863 =
(200/50) x 863 = 3452 ml/min
The answer is D, 40 mm Hg.
The diffusing capacity of the lung for carbon
monoxide is defined as the volume of CO transferred into the blood per minute
per unit alveolar partial pressure of CO. In symbols this becomes:
DL = CO/PACO = 20/0.5 = 40
ml/min/mm Hg
Incidentally this value suggests that the subject
was exercising because the resting value is usually lower. The DLCO
increases on exercise because of recruitment and distension of pulmonary
capillaries and the increase in pulmonary capillary blood volume.
The answer is E.
Time = vol/flow = 200/6,000 ml/60 sec = 200 ml/100 ml/sec = 2.0 sec
A is the best answer
Because the reduction in mixed venous [O2]
from increased tissue O2 extraction will cause a sometimes-lethal
drop in systemic arterial [O2] and PaO2.
C is the best answer
Use of 100 % inspired O2 in differential
diagnosis of arterial hypoxemia.
Potential cause: Hypoventilation
By definition,
hypoventilation results in hypercapnia. Therefore, hypoxemia due to pure
alveolar hypoventilation will always be associated with hypercapnia. The
(A - a)PO2 difference in pure hypoventilation will be normal or
below normal (if on the steep part of the oxygen dissociation curve) as PAO2
will decrease according to the alveolar PO2 equation,
PAO2
= PIO2 - (PaCO2/R)
The hypoxemia of hypoventilation is readily removed by oxygen therapy simply by
increasing PIO2.
Potential cause: Diffusion Impairment
Oxygen therapy also readily removes hypoxemia due to diffusion impairment by
increasing the driving pressure for oxygen across the alveolar capillary
barrier. Note that in diseases which result in a thickening of alveolar
walls e.g., interstitial fibrosis, the ventilation/perfusion relationships are
also probably affected which may account for most of the hypoxemia at rest
while the diffusion impairment causes hypoxemia during exercise (due to
decreased transit time of red cells in capillaries of alveoli).
Potential cause: Ventilation-Perfusion Inequality
Oxygen administration also usually increases arterial PO2 but the
rise in PO2 depends on the severity of V/Q inequality. With
100 % oxygen administration, every alveolus that is ventilated eventually
washes out its nitrogen so that alveolar PO2 = [(PB - PH2O)
- (PCO2/R)] or about 550 mm Hg.
Potential cause: Shunt
The
only mechanism of hypoxemia where arterial PO2 remains well below
alveolar PO2 during 100 % oxygen breathing is V/Q = zero; i.e.,
venous admixture or anatomical shunt. Shunted blood does not "see"
the 100 % O2 and because of the shallow slope of the oxygen
dissociation curve at high values of PO2, the mixed blood leaving
the lungs generally has a subnormal O2 content, thus a low PO2.
There is, however, some useful gain of arterial PO2 over that
resulting from air breathing.
&nbs