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1. Veins contain deoxygenated blood with increased CO2. Increasing PCO2 causes an increase in carbonic acid, decreasing pH (typically 7.3).
2. While you’ll learn more about it in the cardiac block, congestive heart failure is a condition in which the heart can no longer provide enough pressure to perfuse the system. Because of the blood pressure back up in the lungs, patients present with pulmonary edema, most apparent when laying down for long periods of time (edema moves with gravity). By the Starling’s Law, if there are no changes in the colloid osmotic pressure, then any changes in the hydrostatic pressure will cause an altered net fluid movement. Thus, 25mmHg – 15 mmHg = 10 mmHg.
With the congestive heart failure and pulmonary edema, frequently RBCs leak into the pulmonary space, the pigment of which yields hemosiderin. These RBCs are eaten by the resident phagocytes, the macrophages. These macrophages filled with hemosiderin are typically called “heart failure cells.”
The levels of CO2 in the alveoli closely match those in the arteries regardless of exercise (remember, your ABG calculates ARTERIAL blood gas concentrations, thus before your blood hits the capillaries in your CO2 producing muscles). However, there is always a difference between the alveoli and arterial: the A-a difference (10-15 mmHg), in which the PAO2 is always greater than the PaO2.
5. Remember: the output of the right ventricle roughly matches that of the left ventricle. Thus, the blood flow to the lungs roughly equals that of the systemic blood flow.
6. In emphysema, the patient has pockets of air that are not perfused with blood. Thus, these areas are a form of dead space (alveolar dead space). Note that the remaining answers are incorrect: anatomic dead space (conducting airways) are not increased, alveoli don’t “recruit”, and such a drastic increase in intrapleural fluid would cause other pulmonary problems not related to emphysema.
7. This patient has the classic signs of diabetic ketoacidosis, an ion gap metabolic acidosis. Note the low pH and low [HCO3-] (characteristics of metabolic acidosis). Since she is breathing rapidly in compensation, her PaCO2 is very low (20). Also, her ion gap is elevated: 144-(105+14) =25
8. In fixed obstruction of the major airway, the FVC (X-axis length) is normal. However, the flow rates are drastically reduced, since something is in the way. graph 5
9. In asthma, the FVC is slightly decreased (the chest is a normal size but has a difficult time fully filling). However, since the bronchioles leading to the lungs are constricted, the flow rates are drastically reduced. This causes a decreased FEV1/FVC ratio, causing
a concave “scoop” to appear on the slope. Note that administering bronchodilators mitigates the problem. graph 2
10. In restrictive lung diseases, the chest no longer expands properly. Thus, this causes a roughly equivalent decrease in FEV1/FVC, and the slope of the curve is normal. However, the FVC is drastically reduced. graph 4
In emphysema, the FEV1 and FVC are both reduced, with the FEV1 being more affected than FVC. Thus, the FEV1/FVC ratio is reduced, causing the “scooping” of the slope. COPD may or may not show significant bronchodilator response. graph 3
In this patient, excessive vomiting of gastric acid causes a large loss of H+. this, in turn, causes metabolic alkalosis (pH up), with increased bicarb and relatively normal PCO2.
13. Note that the PO2 is much lower than normal with increase PCO2 and low pH, suggesting respiratory acidosis. Using the Henderson-Hasselbalch equation, you’ll see that [HCO3-] is ~41.5, which is slightly elevated, indicating compensation. The patient cannot be breathing room air, because then the alveolar pO2 would be 12.5 mmHg. The patient must be breathing supplemental oxygen.
First, to calculate the alveolar-arterial difference: PAO2=PinspiredO2-PACO2/R = 74 mmHg. This value is much higher than the PO2, indicating some V/Q mismatch. Next, notice that this patient has some form of metabolic acidosis. Since her PCO2 is relatively high, this pattern most closely follows respiratory acidosis.
15. 7.32 = 6.1 + log([HCO3-]/(0.03 x PCO2)). [HCO3-]=30
16. Note that B-lactam drugs directly act on the bacterial cell wall without directly stimulating the immune system
A patient with a right to left shunt with normal lungs has a 20% shunt of her cardiac output. The blood leaving her pulmonary capillaries has an oxygen concentration of 20ml per 100 ml, and her arterial blood has a concentration of 19ml per 100ml. What is the oxygen concentration of her mixed venous blood (in ml per 100ml)?
(Qs/Qt) = (CcO2 - CaO2)/(CcO2 - CvO2) --> CVO2 = CcO2 -[ (Qt/Qs)(CcO2 - CaO2)]
Qs/Qt = 0.2
CVO2 = 20 mL/100 mL - [(5)*(20 mL/100 mL - 19 mL/100 mL)] = 15 mL/100 mL
A woman with iron deficient anemia sees you in your clinic. Assuming 98% sat, PaO2 = 100 mm Hg, and an O2 content of 15 ml/100ml, how many grams/100ml of hemoglobin does she have?
total O2 concentration mL O2/100 mL blood = (1.39 mL O2/g Hb) * Hb g/ 100 mL*O2 saturation) +(0.003 mL O2/(100 mL blood*mmHg) * PaO2 mm Hg -->
(total O2 concentration mL O2/100 mL blood - 0.003 mL O2/(100 mL blood*mmHg)( PaO2 mm Hg)/(1.39 mL O2/g Hb * O2 saturation) = Hb g/ 100 mL
(15 mL O2/100 mL - 0.003 mL O2/(100 mL blood*mm Hg) * 100 mm Hg) = 14.7 mL O2/100 mL blood
(1.39 mL O2/g Hb * 0.98 = 1.36 mL O2/g Hb
(14.7 mL O2/100 mL blood)/(1.36 mL O2/g Hb) = 10.8 Hb/100mL blood
A sixty year old male with Hodgkin’s lymphoma was recently treated with bleomycin as part of his chemotherapy regimen. A known side effect of bleomycin is pulmonary fibrosis, a condition that causes thickening of alveolar-air interface and reduced compliance. In this patient, the alveolar-air interface averages to roughly three times thicker, and he has 1/2 the gas exchange area compared to healthy lungs. How does the oxygen diffusion of this patient compare to that of a healthy person?
Vgas is proportional to A/T *D * (P1-P2)
" " 0.5A/3T * D (relatively constant compared to normal person) * (P1-P2)(relatively constant compared to normal person)
V gas is proportional to 0.5/3 of a normal person = 1/6
A patient in respiratory distress has an arterial PCO2 of 70 mm Hg. What percentage of oxygen must be given to this patient to return his alveolar PO2 to normal (100 mmHg)? HINT: Assume sea level humidified air.
PAO2 = (FiO2*(Pb - PH2O) - PACO2/R
assume at sea level R=0.8, Pb=760, PH2O=47, PACO2=PaCO2
(PAO2 + PACO2/R)(Pb-PH2O) = FiO2
100 mm Hg O2 + (70 mm Hg CO2/0.8)/(760 mm Hg -47 mm Hg) = .263
.263 * 100 = 26.3%
You want to measure the O2 concentration of mixed venous blood in the pulmonary artery without placing a catheter. Assume a pulmonary vascular resistance of 10 mm Hg / liter / min, mean pulmonary arterial and venous pressure of 45 and 5 mm Hg respectively, O2 consumption of 300 ml / min, and O2 concentration of arterial blood of 18 ml/100ml. Calculate the O2 concentration in ml * 100ml-1. HINT: First calculate blood flow.
Fick equation: VO2 = Q(CaO2 - CvO2) --> CaO2 (VO2/Q) = CvO2
where VO2 = O2 consumption, Q = bloodflow, CaO2 =arterial O2 concentration, CvO2=venous O2 concentration
PVR = (input - output pressure blood flow)/Q -->
Q = (45 mm Hg - 5 mm Hg)/(10 mm Hg/L/min) = 4 L/min --> 4 L/min (1000 mL/L) = 4000 mL blood/min
4000 mL blood/min = 40 * 100 mL blood/min
18 mL O2/100 mL - (300 mL O2/min)/(40 * 100 mL blood/min) = 10.5 mL O2/100 mL
Please refer to the attached figure.
Filling which alveoli with 100% oxygen will cause the biggest change on the O2 concentration of blood at position 5?
1. Note that the question asks for the biggest CHANGE. Adding oxygen to the alveolus with relatively little oxygen (A), would cause a greater change than the others. In terms of V/Q mismatch, A has a relatively low V/Q (low ventilation with normal perfusion). Thus, increasing ventilation would cause a drastic change in O2 concentration.
Figure for Problem Set 2.pdf 65 KB
Which numbered blood vessel has the highest PO2?
2. If perfusion is decreased, the capillary gases become much more equilibrated to the inhaled air (since the CO2 isn’t much released from poorly perfused capillaries). In terms of V/Q mismatch, C has a high V/Q (well ventilated, poorly perfused), which would cause it to closely match inhaled air (O2~140).2. If perfusion is decreased, the capillary gases become much more equilibrated to the inhaled air (since the CO2 isn’t much released from poorly perfused capillaries). In terms of V/Q mismatch, C has a high V/Q (well ventilated, poorly perfused), which would cause it to closely match inhaled air (O2~140).
Please refer to the attached figure.
Will supplying B with 100% O2 greatly affect O2 CONTENT at 5?
3. This question tests you on your understanding of O2 content. O2 content is the overall O2 in your blood (O2 bound to hemoglobin + dissolved O2). Since the majority of your O2 is bound to hemoglobin by 100 mmHg at saturating levels in good V/Q match, changing the inspired air (and thus the PO2) would have little effect. Simply, it won’t affect the amount of O2 bound to hemoglobin (it’s already saturated), and it offers only a very minimal increase in dissolved O2 in the blood (factor is 0.003 x PO2).
Figure for Problem Set 2.pdf 65 KB
Will supplying B with 100% greatly affect CO2 CONTENT at 5?
4. Since the CO2 dissociation curve is linear, changes in PCO2 changes CO2 content (unlike the sigmoidal oxygen binding curve). However, note that supplementation with 100% oxygen would not change the levels of PCO2 in the blood. Simply, normal
inspired air has almost no CO2 – thus, 100% oxygen would likewise have almost no CO2. Rather, the CO2 comes from the blood (40 mmHg in healthy adults). Thus, since the PCO2 does not change, the CO2 content will not change as well.
If blood flow to 4 were completely eliminated, what does that make C?Figure for Problem Set 2.pdf 65 KB
Remember: physiological dead space = anatomic dead space + alveolar dead space. Anatomic dead space is the conducting airways, not involved in gas exchange. Alveolar dead space occurs when there’s high ventilation but not perfusion (infinite V/Q).
If A is completely occluded, what does that make 1?Figure for Problem Set 2.pdf 65 KB
. Shunt, on the other hand, occurs when there’s high perfusion with no ventilation (0 V/Q). Make sure you understand V/Q mismatch, what it means when V/Q gets too high (moving towards dead space), or too low (towards shunt)!
7. Pressure = 4T/r. Thus, since the tension (T) is the same, the ratio is 1:2.
8. With surfactant, the surface tension (T) of two different sized spheres is different. Overall, smaller spheres with surfactant have less surface tension than larger ones (which is why smaller alveoli do not collapse into larger ones). Thus, since the values of T are different, you cannot simply use proportions to determine the ratio of pressures.
9. Adding in series increases resistance, adding in parallel decreases it. This property also applies to lung capillaries. Pressure = flow x resistance, whereas the blood flow to the heart roughly equals systemic flow. However, since the network of capillaries in the lungs is in parallel, it is a low resistance system (and thus low pressure).
10. a. Increasing parasympathetic tone constricts the airways, increasing resistance and moving the equal pressure point closer to the alveoli.
b. At smaller lung volumes, the lungs are less expanded (thus decreasing the recoil force). c. Likewise, in emphysema, a condition in which the alveolar septa are lost, recoil force is also lost causing a decrease in maximum flow rate.
d. The equal pressure point is where the intrapleural pressure equals the pressure of the releasing air. In this case then, any additional resistance past that point has minimal effects, since the pressure is already below the releasing pressure.
11. With positive pressure ventilation, the lungs are expanded. With lung expansion, radial traction opens larger extraalveolar vessels (a). Simultaneously, alveolar expansion constricts alveolar pulmonary capillaries (d), increasing vascular resistance (b). Finally, the positive intrathoracic pressure decreases venous return to the thorax (c). Since the pulmonary capillaries are constricted, this increases the likelihood of Zone 1, in which the alveolar pressure exceed that of blood flow.
12. While many of the interventions have been attempted, low tidal volume ventilation has been best shown to benefit patients in ARDS.
13. Peripheral receptors respond to PCO2, PO2, and pH. However, the carotid body plays a much more important role than the arotic body.
14. Since there is low atmospheric CO2, pretty much all the CO2 in the lungs comes from the patient. Thus, adding oxygen does little to help (a). Reducing the FRC and adding external dead space to “trap” the CO2 would not address the elevated CO2. Rather, ventilating the alveoli by increasing tidal volume would help “blow off” the excess CO2.
15. In ILD, lung inflammation (b) causes fibrosis, which thickens the alveolar-air interface. This reduces the diffusion capacity (a) and decreases compliance (a). However, both the FEV1 and FVC are reduced at roughly proportional levels – thus, patients in ILD (and other restrictive lung diseases) usually present with normal FEV1/FVC ratios (and no scooping of the slope in the flow-volume loops)
16. SCUBA diving would cause a normal lung volume (6L), and breath hold diving would cause a decreased lung volume (2L); thus, b and e are incorrect. Air embolism and pneumothorax is a risk on ascent (not descent, a, c).
17. As an adaptation, those living at high altitudes have high hematocrit to bind more oxygen (more erythropoietin – this is why many athletes do high altitude training). However, no matter how adapted you are, your PO2 is dependent solely on your inspired air. If there’s less PO2 in the air (eg. at high elevations), then there’s less dissolved in your blood. Even with this drawback, since the native Himalayan at 14,000 feet has higher hematocrit, they could very well have similar O2 content as your sea level blood, although their PO2 is drastically decreased. (Remember: O2 content includes both hemoglobin-bound and dissolved O2!)
18. Remember: hot exercising muscles need O2! All the factors do cause dissociation of oxygen – however, decreased pH (not increasing pH, such as with HCO3-) also causes dissociation.
B2 adrenergic agonists act on your sympathetic nervous system to open airways, but do not address the inflammatory cause. Corticosteroids work in many ways to reduce cytokines/apoptose inflammatory cells, antiIgE reduces mast cell degranulation, and leukotrienes come from mast cells.
20. If the volume flow rate and driving pressures are the same, then the resistances must be the same. In laminar flow, R = 8nl/pi*r^4. Thus:
Resistance (X) = Resistance (Y)
What drives expiration? In normal expiration, air simply flow from a high pressure to a low pressure (if your alveolar pressure is 5, and your mouth is 0, then the pressure gradient is 5). However, what about in forced expiration? Remember the higher pleural pressure, the more airway compression, causing the equal pressure point limitation. Thus, there’s no pressure gradient between your alveoli and mouth – rather, the pressure difference between your intrapleural pressure and alveoli/equal pressure point drives air out (38-30 = 8).
To calculate the climber’s alveolar PCO2, use the alveolar gas equation: PO2 = PIO2 – PCO2/R
37 = (253-47) * 0.21 – PCO2/0.8
PCO2 = 5 mmHg
To bring his PO2 to 100 mmHg, he would need: 100 = PIO2 – 5/0.8
PIO2 = 106.25 mmHg.
With the pressure being 253 mmHg still:
(253 – 47) x %O2 = 106.25 %02 = 51.6%
Note that as the climber breathes oxygen, he would most likely slow his ventilation and increase his PCO2 over time.
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