NORMAL PKOCESSES OF ENERGY METABOLISM 561 



per cent O. One hundred grams of such fat would require 288.6 gm. 

 O 2 in addition to that already present in the molecule for complete con- 

 version to CO 2 and H 2 O. There would be produced 280.5 gm. CO 2 . The 



relationship of - is ; and this divided by -, the molecular 



\J 2 ^oo.o 32 



weight, or multiplied by - would give the respiratory quotient = 0.706. 



A slightly simpler calculation, as noted above, is to determine the weight of 

 O 2 necessary to form CO 2 (in this case 204.0 grams) and divide this 



directly by the weight of total O 2 required; thus: - - = 0.706. 



288.6 



The respiratory quotient of all food fats is in the neighborhood of 

 0.71. The same is true also of body fat. Hence whether pure body fat 

 or pure food fat were being burned, the R. Q. would be approximately 0.71. 

 As a matter of fact, this quotient is probably never actually produced 

 under normal conditions; for there is always some protein being de- 

 stroyed, and, since under the conditions of high fat combustion, whether 

 from starvation or excessive fat ingestion this small am'ount of protein 

 is readily oxidized, there is a mixed quotient contributed in small part 

 by the oxidation of protein and in large part by the oxidation of fat. On 

 the assumption that the protein quotum of energy production is 15 per. 

 cent and the other 85 per cent is from fat, Magnus-Levy estimates that 

 the actual respiratory quotient should be 0.722, while if the remaining 

 85 per cent is produced from carbohydrate, the quotient should be 0.971. 



The respiratory quotient of proteins will, of course, depend upon the 

 exact formula employed ; but since all proteins are made up of amino acids, 

 the exact relationship can best be appreciated by starting with a single 

 amino acid. If alanin is given to an animal, it will be oxidized after deam- 

 ination, as follows : 



CH 3 . CHNH 2 . COOH + HOH == CH 3 . CHOH . COOH + NH 3 

 CH 3 .CHOH.COOH + 3 O 2 == 3 CO 2 + 3 H 2 O 



The respiratory quotient of this reaction would be 1.0 since the volume 

 of O 2 is just equal to the volume of CO 2 produced. But the NH 3 is not 

 yet disposed of. It cannot remain in the body as NH 3 and it cannot be 

 eliminated as a gas, for the lungs are not permeable to NH 3 even if it 

 could be carried in the blood as gas. Actually, the NH 3 will unite with 

 the CO 2 to form ammonium carbonate, thus : 



2 KH 3 + C0 2 + H 2 = (NH 4 ) 2 CO 3 

 Later, this is converted to urea, thus: 



NH 2 \ 



CO 3 2 H,O = CO 



NH 2 / 



