ELEMENTARY EXPERIMENTAL PHYSIOLOGY 185 



It will be sufficient if the student makes the determination in the 

 following way. He should collect a sample of expired air and analyse 

 it; then he should determine the average volume of air which he 

 breathes in a minute. The methods involved have been described in 

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

 as follows: 



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

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



7000 

 had, therefore, absorbed 21 - 16 = 5 x = 350 c.c. of oxygen and 



discharged 4 x -^-TTT- = 280 c.c. of carbon dioxide. His respiratory 

 quotient, the ratio of the volume of carbon dioxide discharged to the 

 volume of oxygen absorbed is -^r- 2 = J^ = ^ = 0'8. 



v/2 OD\) v 



There is a decrease of about -V i n 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 slightly larger 

 volume of inspired air contained the following volume of oxygen : 

 _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 : 



c j 20-94x80 01 10 

 Oxygen of inspired air= =21 '18 c.c. 

 /"'07 



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

 Respiratory quotient = TT^-^TTQ^ ^"- 



U._j O'Jo 



The respiratory quotient varies according to the nature of the food 

 which undergoes oxidation in the body; thus, for carbohydrates 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 0. 



C0 2 _6_ 

 2 ~6~ 

 Albumin (empirical formula) : 



C 72 H 119 N 18 22 S + 770 2 = 63C0 2 + 38H 2 + 9CO(NH 2 ) 2 + SO 3 . 



