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148 PHYSIOLOGY CASES AND PROBLEMS

Case 26 Monoxide Poisoning

Herman Neiswander is a 65-year-old retired landscape architect in northern Wisconsin. One cold January morning, he decided to warm his car in the garage. Forty minutes later, Mr. Neiswander's wife found him slumped in the front seat of the car, confused and rapidly. He was taken to a nearby emergency department, where he was diagnosed with acute carbon monoxide poisoning and given 100% 02 to breathe. An arterial blood sample had an unusual cherry-red color. The values obtained in the blood sample are shown in Table 3-12.

TABLE 3-12 Mr. Neiswander's Arterial Blood Gases

Pao, (arterial P02) 660 mm Hg (normal, 100 mm Hg, room air)

Paco, (arterial Pc02) 36 mm Hg (normal, 40 mm Hg) % saturation 50% (normal, 95%-100%) rg QUESTIONS hum_ 1. In healthy people, the percent 0 2 saturation of in arterial blood is 95%-100%. Why was Mr. Neiswander's 0 2 saturation reduced to 50%?

2. What percentage of the heme groups on his hemoglobin were bound to carbon monoxide (CO)?

3. Draw a normal 0 2-hemoglobin dissociation curve, and superimpose the 02-hemoglobin dissociation curve that would have been obtained on Mr. Neiswander in the emergency department. What effect did CO poisoning have on his 0 2-binding capacity? What effect did CO poisoning have on the affinity of hemoglobin for 02?

4. How did CO poisoning alter 02 delivery to Mr. Neiswander's tissues?

5. What was the rationale for giving Mr. Neiswander 100% 0 2 to breathe?

6. In healthy people breathing room air, arterial P 02 (Pa02) is approximately 100 mm Hg. Mr. Neiswander had a Pa02 of 660 mm Hg while breathing 100% 02. flow is a value of 660 mm Hg possible? [Hint: There is a calculation that will help you to determine whether this value makes sense. For that calculation, assume that Mr. Neiswander's respiratory quotient (CO, production/02 consumption) was 0.81

7. What is an A-a gradient? What physiologic process does the presence or absence of an A-a gradient reflect? What was the value of Mr. Neiswander's A-a gradient while he was breathing 100% 02? What interpretation can you offer for this value? n 150 PHYSIOLOGY CASES AND PROBLEMS

pi ANSWERS AND EXPLANATIONS 1. Mr. Neiswander's percent 02 saturation was only SO% (normal, 95%-100%) because CO occu- pied 02-hinding sites on hemoglobin. In fact, CO binds avidly to hemoglobin, with an affinity that is more than 200 times that of 02. Thus, heme groups that should be bound to 02 were instead bound to CO. Hemoglobin that is bound to CO is called carboxyhemoglobin and has a characteristic cherry-red color.

2. Because the percent saturation of 0 2 was 50%, we can conclude that the remaining 50% of the heme sites were occupied by CO.

3. In the presence of CO, the 02-hemoglobin dissociation curve is altered (Figure 3-11). The maximum percent saturation of hemoglobin by 0 2 was decreased (in Mr. Neiswander's case, to 50%), resulting in decreased 0 2-binding capacity. A left shift of the curve also occurred because of a conformational change in the hemoglobin molecule caused by binding of CO. This conformational change increased the affinity of hemoglobin for the remaining bound 02.

Carbon monoxide poisoning Figure 3-11 Effect of carbon monox- ide on the 02-hemoglobin dissociation curve. Po, partial of . (Reprinted with permission from Costanzo LS: BRS Physiology, 3rd ed. Baltimore, Lippincott Williams &

25 50 75 100 Wilkins, 2003, p 143.)

4. 02 delivery to the tissues is the product of blood flow (cardiac output) and 02 content of the blood, as follows:

02 delivery = cardiac output x 02 content of blood

The 02 content of blood is the sum of dissolved 0 2 and 02 bound to hemoglobin. Of these two components, 02-hemoglobin is, by far, the most important. In Mr. Neiswander's case, 02 deliv- ery to the tissues was significantly reduced for two reasons: (1) CO occupied 02-binding sites on hemoglobin, decreasing the total amount of 0 2 carried on hemoglobin in the blood. (2) The remaining heme sites (those not occupied by CO) bound 0 2 with a higher affinity (consistent with a left shift of the 0 2-hemoglobin curve). This increase in affinity made it more difficult to unload 02 in the tissues. These two effects of CO combined to cause severe 0 2 deprivation in the tissues ().

5. Mr. Neiswander was given 100% 02 to breathe for two reasons: (1) to competitively displace as much CO from hemoglobin as possible arid (2) to increase the dissolved 0 2 content in his blood. As you have learned, dissolved 0 2 normally contributes little to the total 02 content of blood. However, in CO poisoning, the 0 2-carrying capacity of hemoglobin is severely reduced (in this RESPIRATORY PHYSIOLOGY 151

case, by 50%), and dissolved 0, becomes, by default, relatively more significant. By increasing the fraction of 0 2 in inspired air to 100% (room air is 21% 0 2), the Po, in Mr. Neiswander's alveolar gas and arterial blood will be increased, which will increase the dissolved 0 2 content (dissolved 02 = P02 x of 02 in blood).

6. While Mr. Neiswander was breathing 100% 02, the measured value for Pao, was strikingly high (660 mm Hg). Because pulmonary capillary blood normally equilibrates with alveolar gas, arterial Po, (Pao,) should be equal to alveolar P02 (PA02). Therefore, the question that we really need to answer is: Why was the PA02 660 mm Hg?

The is used to calculate the expected value for PA °, (as described102 in Case 20). For the alveolar gas equation, we need to know the values for Po, of inspired air PAcoz, and respiratory quotient. Pro, is calculated from the barometric pressure (corrected for water vapor pressure) and the fraction of 0 2 in inspired air (F1 02 ). In Mr. Neiswander's case, Frog is 1.0, or 100%. PAc02iS equal to Pac02, which is given. The respiratory quotient is 0.8. Thus: Pio, (PB — PH 2O) x

= (760 mm Hg — 47 mm Hg) x 1.0

= 713 mm Hg

PACO2 PA 02 = Proz R

= 713 mm Hg 36 mm Hg 0.8 = 668 mm Hg

From this calculation, we know that when Mr. Neiswander breathed 100% 0 2, his alveolar P02 (PA02) was expected to be 668 mm Hg. Assuming that his systemic arterial blood was equili- brated with alveolar gas, the measured Pao, of 660 mm Hg makes perfect sense.

7. The A—a gradient is the difference in Po, between alveolar gas ("A") and arterial blood ("a"). In other words, the A—a gradient tells us whether equilibration of 02 between alveolar gas and pulmonary capillary blood has occurred. If the A—a gradient is zero or close to zero, then perfect (or nearly perfect) equilibration of 0 2 occurred, as is normally the case. Increases in the A—a gradient indicate a lack of equilibration, as with a ventilation— (V/Q) defect (e.g., obstruc- tive disease), when a defect is present (e.g., fibrosis), or with a right-to-left cardiac shunt (i.e., a portion of the cardiac output bypasses the and is not oxygenated).

Mr. Neiswander's PAo, was calculated from the alveolar gas equation (see Question 6), and his Pao, was measured in arterial blood. His A—a gradient is the difference between the two values:

A—a gradient = PA02 Pa02 = 668 mm Hg — 660 mm Hg = 8 mm Hg

This small difference between the Po, of alveolar gas and the Po, of arterial blood implies that pulmonary capillary blood equilibrated almost perfectly with alveolar gas. In other words, CO poisoning caused no problems with V/Qmatching or 02 diffusion. 152 PHYSIOLOGY CASES AND PROBLEMS

,=:Key topics

A—a gradient Alveolar gas equation Carbon monoxide (CO) poisoning

Diffusion of 02 Left shift of the 0 2—hemoglobin dissociation curve 0 2—hemoglobin dissociation curve Right-to-left cardiac shunts Ventilation—perfusion (V/0) ratio