562 JOHN K. MURLIN 



The net result would be that for each two molecules of alanin, yielding 

 2 molecules of NH 3 , one molecule of CO 2 would fail to appear in the ex- 

 pired air, hut would he eliminated as urea or water. Hence, for 6 mole- 

 cules of O 2 absorbed, only 5 would come back as CO 2 and the true R. Q. 

 of alanin would be 5/6 = 0.833. If all proteins were made up of ammo 

 acids as simply as this, the R. Q. for their combustion would be as easily 

 computed. The respiratory quotient of glycocoll would be 0.75; that of 

 leucin would be 0.73. But that of lysin containing two NH 2 groups and 

 requiring, therefore, one molecule of CO 2 for elimination of the N as 

 urea for each single molecule of the substance, would be only 0.71. The 

 more diamino acids contained in a protein, therefore, and the more leucin, 

 the lower would he the respiratory quotient. With gelatin, which con- 

 tains a high percentage of glycocoll, one might expect a somewhat higher 

 quotient than with casein which contains no glycocoll and a much larger 

 amount of leucin. Taking an example of a highly synthetized protein such 

 as 1-leucyl-triglycyl-l-leucyl-triglycyi-l-leticyl-octoglycyl-glycin, which was 

 put together by E. Fischer and whose exact chemical structure is there- 

 fore known, we find that 45 molecules of O 2 would be necessary to produce 

 complete combustion ; that molecules of CO 2 would be needed to remove 

 the OTI 2 in the form of (NH 4 ) 2 CO 3 ; and that when this ammonium car- 

 bonate breaks down by dehydration to form urea, none of the carbon would 

 return to the respiration and none of the oxygen would be available for 

 combustion. The R. Q. therefore would be 0.81. 



Taking the elementary analysis of protein of the human body and 

 adopting the percentages used by Magnus-Levy we get the following com- 

 position after making allowances for the elements which would appear 

 in the urine and the feces: C, 38.6 per cent; H, 4.24 per cent; O, 9.24 

 per cent. For the combustion of 100 grams of such protein, 127.6 gm. 

 O 2 in addition to that already present in the molecule would be needed 

 and 141.5 gm. C0 2 would be formed. Taking the ratio of the oxygen in 

 CO 2 (102.9 gm.) to the total oxygen required, the quotient is 0.807 or 



by the longer calculation '* * 2 X = 0.807. The respiratory 



1,27. b gm. OOo 11 



quotient of a complete protein such as is ordinarily used in rebuilding 

 the human tissues, but which, because it is not needed for this purpose, is 

 oxidized as completely as it is possible to oxidize protein in the body, is 

 thus approximately the same as that for alanin. We may think of this 

 amino acid as representing the type of fuel available when protein is 

 burned. 



Laulanie(c) in 1898 gave a very simple method of calculating the 

 thermal quotient for oxygen from the respiratory quotient. This method 

 is strictly applicable however only under conditions where the metabolism 

 of protein is entirely negligible, or is calculated independently and suitable 



