566 



JOHK R. MURLIH 



5. The Non-Protein Respiratory Quotient. It was but a step from 

 the method just given to a simpler calculation based upon a table giv- 

 ing the heat values of oxygen or carbon dioxid for the non-nitrogenous 

 combustion. 



The respratory exchange due to protein is thus given by Lusk (h). 



TABLE 7 



To oxidize these amounts of carbon and hydrogen would require 138.18 gm. 

 O 2 and there would be produced 152.17 gm. CO 2 . From which it may 

 be deduced that for every gram of nitrogen appearing in the urine from 

 meat there would be absorbed from the breath (138.18-^-16.28 =) 8.45 

 grams of oxygen, and there would be produced (152.17 -r- 16.28 =) 9.35 

 grams of carbon dioxid. Hence by multiplying the nitrogen elimination 

 in the urine whether of an hour or a day by these factors and subtracting 

 from the total oxygen absorbed and carbon dioxid eliminated the non- 

 protein respiratory quotient is obtained. 



By a method entirely analogous to that of Laulanie (page 562) it is 

 possible to learn the heat values of oxygen for each value of this respiratory 

 quotient. Zuntz and Schumberg (b) prepared a table setting forth these 

 values which is now widely employed. As reproduced here the heat values 

 of both oxygen and CO 2 per liter of the gas at and 760 mm. Hg may 

 be read off for any value of the non-protein R. Q. given to two places. 



It will be noted that the values for pure fat (R. Q. 0.71) and pure 

 carbohydrate (R. Q. 1.0) combustion differ but slightly from those of 

 Lefevre given in Table 2 (page 556). 



The calculation of the heat production from the respiratory exchange 

 and the nitrogen in the urine involves then the following steps : 



(1) Determination of total O 2 and CO 2 of respiration in grams. 



(2) " " " N in the urine. 



(3) Multiply IsT of urine by 8.45 = O 2 for protein. 



(4) " N " " " 9.35 = CO 2 " " 



(5) Subtract these values from total O 2 and CO 2 . 



(6) Convert to volume and get Non-prot. R. Q. 



