INVESTIGATIONS ON THE METABOLISM OF MILCH COWS. 813 



If the dog is fed with both fat and starch in addition to protein, it is 

 not possible to calculate the amount of fat and of starch broken down 

 in the body by determining the amount of carbon dioxid excreted in 

 addition to the nitrogen in the urine. In this case the amount of oxy- 

 gen consumed must be known in addition. This may be calculated. It is 

 not probable that the animal body can use protein when fed in connec- 

 tion with fat and starch more economically than protein from its tis- 

 sues when fasting. Voit found that the organic matter excreted in the 

 urine of a fasting dog was made up of 25.5 per cent carbon, 6.4 per cent 

 hydrogen, 34.4 per cent nitrogen, and 33.7 per cent oxygen. If 100 gm. 

 of protein were broken down in the body 10 gm. of nitrogen would be 

 found in the urine, together with carbon, hydrogen, and oxygen. The 

 data for calculating the amount of oxygen required for the combustion 

 of protein and the amount of carbon dioxid produced from protein are 

 therefore available. Suppose that 100 gm. protein is consumed and 

 4G.6 gm. organic matter excreted in the urine. Then — 



100 gin. protein consumed =53.0 gin. C, 7 gm. H, 16 gm. N, 23.0 gm.0, 1 gm.S 



46.6 gm. organic material in urine =11.9 gm. C, 3 gm. H, 16 gm. N, 15.7 gm.O, 



Remainder =41.1 gm. C, 4 gm. H, 7.3 gm.O, 1 gm.S 



The O, H, and S will be oxidized to C0 2 , H 2 0, and S0 3 . The oxida- 

 tion of 41.1 gm. (J would produce 150.7 gm. C0 2 . This would require 

 109.6 gm. O; 4 gm. H would produce 3G gm. H 2 and require 32 gm.O; 

 1 gm. S would produce 2.5 gm. S0 3 , requiring 1.5 gm. O, or a total of 

 143.1 gm. of oxygen. Deducting from this amount 7.3 gm. O furnished 

 by the food in excess of the amount excreted in the urine, leaves 135.8 

 gm. of oxygen as the amount which must be taken by the body from, the 

 inspired air. As noted above, the expired air contained 150.7 gm. C0 2 . 

 One liter of carbon dioxid weighs 1.90633 gm. and 1 liter of oxygen 

 1.43003 gm. at 760 mm. mercury pressure. The respiratory quotient — 

 that is, the ratio of carbon dioxid produced to oxygen consumed — 

 would be as follows: 



CQ 2 _ 150.7 -r- 1.96633 Q S()7 

 2 135.8 4- 1.43003 _ * 



If 17 gm. of nitrogen were excreted in the urine, as mentioned above, 

 then the amount of carbon dioxid produced by the combustion of pro- 

 tein in the body would be as follows : 



17 X 16 5 ° 7 == 16(U gm> °° 2 ( contaiuin S 43 ' 7 ^m. O). 

 The amount of oxygen necessary would be: 

 17 x 135.8 



16 



= 144.3 gm. O. 



Supposing the animal took 500 gm. of O from the air and excreted 

 160 gm. C0 2 in the breath, the amount of carbon which must have 



