If PAO 2 normally averages 100 mmHg, why is average PaO 2 =95 mmHg?? 1. V/Q differences from apex to base 2. Shunt To understand both influences we must.

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If PAO 2 normally averages 100 mmHg, why is average PaO 2 =95 mmHg?? 1. V/Q differences from apex to base 2. Shunt To understand both influences we must remember: - arterial O2 O2 content is a function of the contributing sources’ relative volumes and O2O2 contents. - the relationship between PO 2 and O2 O2 content in the presence of Hb is NOT LINEAR.

O 2 Content (ml/dl blood) PO 2 (mmHg)

PAO 2 = FIO 2 (P B - P H 2 O ) - PACO 2 FIO 2 + (1-FIO 2 ) R 0.21 *(747-47) (1-0.21) (range = =100 V/Q Matching Alveolar Gas Equation

O 2 Hb Saturation (%) PO 2 (mmHg) % O 2 O 2 Hb dissociation curve V/Q Matching

Cv O 2 Q T Cc' O 2 Q S Ca O 2 Q T Q T Cc' O 2 - Cv O 2 Q S Cc' O 2 - Ca O 2 V/Q Matching %Sat O 2 can be used in place of C If breathing 100% O 2, the shunt fraction can be approximated as 1% of the cardiac output for every 20 mmHg PAO 2 -PaO 2 difference.

P O 2 =40 C O 2 =15 P O 2 =100 C O 2 =20 P O 2 =54 C O 2 =17.5 P O 2 =40 C O 2 =15 P O 2 =40 C O 2 =15 P IO 2 =150 P AO 2 =100 Shunt Fraction= = 0.5 Ok….so let’s put him On 100% oxygen!

P O 2 =40 C O 2 =15 P O 2 =40 C O 2 =15 P O 2 =660 C O 2 =21 P O 2 =64 C O 2 =18 P O 2 =40 C O 2 =15 P IO 2 =700 P AO 2 =660 Shunt Fraction= = 0.5

P O 2 =40 C O 2 =15 P O 2 =50 C O 2 =17 P O 2 =100 C O 2 =20 P O 2 =64 C O 2 =18.5 P O 2 =40 C O 2 =15 P IO 2 =150 P AO 2 =100 P AO 2 =50 Shunt Fraction= = 0.3 Ok…. (a little more tenuously) Let’s put him on a little more oxygen???

P O 2 =40 C O 2 =15 P O 2 =85 C O 2 =19 P O 2 =235 C O 2 =21 P O 2 =100 C O 2 =20 P O 2 =40 C O 2 =15 P IO 2 =285 P AO 2 =235 P AO 2 =85 Shunt Fraction= = 0.17

Case Study The following data is obtained from a man with smoke inhalation injury who is breathing 100% oxygen: PaO2 190 mmHg PaCO2 36 mmHg SaO259% COHb40% pH *(747-47) (1-1) = 664 mmHg PAO 2 - PaO 2 = 474 Qs = 474/20 = 23.7%

Case Study A patient presents with pneumonia which involves the entire left lung, sparing the right. The following data is obtained on ambient air PaO2 52 mmHgSaO275% PaCO2 39 mmHgSmvO260% On 50% oxygen, the data obtained are: PaO2 65 mmHgSaO280% PaCO2 35 mmHgSmvO260% 0.21 *(747-47)- 39 * = 100 mmHg Shunt Fraction= = *(747-47)- 35 * = 311 mmHg Shunt Fraction= = 0. 47

Case Study In one lung anesthesia, only one lung (referred to as the dependent lung) is ventilated, while the non-dependent lung is not ventilated. Blood flow to the non-ventilated lung becomes shunt flow. This is in addition to any shunt flow through the dependent lung. During such a procedure the following data were obtained: mvO2 content15 ml/dl aO2 content19 ml/dl Assuming that oxygen content of blood leaving well ventilated regions of the dependent lung is 20 ml/dl, the calculated shunt fraction is Shunt Fraction= = 0.20

A patient sitting upright in bed is on positive pressure ventilation that maintains a positive end-expiratory pressure of 8 cm H 2 0 (~2 inches). You should be able to discuss the following questions concerning this patient based on this information alone: Provide a rationale for either an increase or decrease in the patient’s pulmonary arterial blood pressure after being placed on PEEP. Why might you consider putting a flow-directed pulmonary arterial (Swan- Ganz) catheter under fluroscopic guidance in this patient? Case Study