ADVANCED EXPERIMENTAL PHYSIOLOGY 139 



a minute. The methods involved have been described in previous 

 chapters. From the data obtained a calculation can be made as 

 follows : 



The man breathes 7 litres per minute, and the composition of the 

 expired air was 16 per cent, oxygen and 4 per cent, carbon dioxide ; 



he had, therefore, absorbed 21 - 16 = 5 X = 350 c.c. of 



100 



7000 



oxygen and discharged 4 X - = 280 c.c. of carbon dioxide. His 



1UU 



respiratory quotient, the ratio of the volume of carbon dioxide dis- 



C*O 280 4 

 charged to the volume of oxygen absorbed is 2 = - = - =0-8. 



O 



There is a decrease of about -^ in the volume of the expired air as compared 

 with the inspired air, when both are measured at and 760 mm. ; the deficit 

 is due to the absorption of a small quantity of oxygen which does not reappear 

 in combination with carbon as carbon dioxide, but passes out of the body in 

 other products of oxidation. The increased proportion of nitrogen in the 

 expired air must be taken into account when the respiratory quotient is 

 calculated from volumetric analysis ; thus for every 100 c.c. of expired air the 

 lightly larger volume of inspired air contained the following volume of oxygen : 



Q _ 20-94 x Nitrogen of expired air 



79-07 



The respiratory quotient, therefore, in a case in which the percentages of 

 nitrogen, oxygen and carbon dioxide are 80, 16 and 4, would be correctly 

 calculated as follows : 



Oxygen of inspired air = 2Q ' 94 x 8( = 21-18 c.c. 



Oxygen absorbed =21-18 16 = 5-18 c.c. 



PT) 4 



Respiratory quotient = - 2 = ^ - = 0-77. 

 O 2 5*18 



The respiratory quotient varies according to the nature of the 

 food which undergoes oxidation in the body ; thus, for carbohy- 

 drates it is 1, for protein 0-8, and for fat 0-7. The following formulae 

 represent the oxidation of these different substances :-- 



Dextrose : C 6 H 12 6 + 60 2 = 6C0 2 + 6H 2 O. 



CO, = 6_ 1 

 2 "6" 



Albumin (empirical formula) : 

 C 72 H 112 N 18 22 S + 77O 2 = 63C0 2 + 38H 2 O + 9CO(NH 2 ) 2 + S0 3 . 



CO* = ?3 = . 8 2 



2 " 77 " 



Olein : C 3 H 5 (C 18 H3 3 O a )3 + 800 2 = 57CO, + 52H..O. 



The respiratory quotient in a living organism is the resultant of 

 various chemical changes and may be complicated by the forma- 



