Estimation of arterial PCO2 in the elderly Article (Web of Science)


  • Arterial PCO2 (PaCO2), determined directly in the radial artery, was compared with indirect estimates of PCO2 in six elderly men (mean age 73.8 yr). Estimates of PaCO2 included arterialized venous PCO2 (PavCO2); end-tidal PCO2; mean alveolar PCO2, calculated by using a reconstruction of the alveolar oscillation in PCO2 and accounting for the presence of dead space (time-weighted mean for PCO2 throughout the respiratory cycle); and values calculated by using the empirical formula developed by Jones et al. (N. L. Jones, D. G. Robertson, and J. W. Kane. J. Appl. Physiol. 47: 954–960, 1979), which incorporates end-tidal PCO2 and tidal volume (PaCO2 derived from end-tidal PCO2 and VT). Measurements were made at rest and during cycle ergometry at 25 and 50 W while the subjects breathed various gas mixtures (euoxic-eucapnic, hypoxic-eucapnic, hyperoxic-eucapnic, and hyperoxic-hypercapnic). The mean differences between the estimates and the actual PaCO2 at rest and in 25- and 50-W exercise were as follows: PavCO2, 0.3 +/- 0.7 (SD), -0.1 +/- 0.7, and 1.8 +/- 1.2 Torr; end-tidal PCO2, 2.9 +/- 1.7, 4.0 +/- 3.1, and 3.7 +/- 3.2 Torr; time-weighted mean of alveolar PCO2, 2.6 +/- 1.9, 3.3 +/- 3.1, and 3.6 +/- 3.8 Torr; and PaCO2 derived from end-tidal PCO2 and VT, 2.4 +/- 1.3, 1.3 +/- 3.0, and 0.6 +/- 2.9 Torr. It is concluded that mean PavCO2 agreed most closely with mean PaCO2 both at rest and in exercise. All methods of deriving PaCO2 using measurements from the respired gases overestimated arterial values at rest. Of the noninvasive techniques, mean estimates calculated using the regression equation developed by Jones et al. corresponded most closely with PaCO2 in exercise.


  • St Croix, C. M.
  • Cunningham, D. A.
  • Kowalchuk, J. M.
  • McConnell, A. K.
  • Kirby, A. S.
  • Scheuermann, Barry W
  • Petrella, R. J.
  • Paterson, D. H.

publication date

  • 1995

published in

number of pages

  • 7

start page

  • 2086

end page

  • 2093


  • 79


  • 6