76 PRINCIPLES OF ANIMAL NUTRITION. 
As a matter of fact, however, it is not easy to determine satis- 
factorily the proportion of the respiratory exchange due to the 
proteids, both because the nitrogenous products of their meta- 
bolism are numerous and occur in varying proportions in the urine, 
and because we may not always be justified in assuming complete 
oxidation of the non-nitrogenous residue. Computations of the 
nature indicated above, therefore, must be accepted with some 
reserve. 
A simpler case, and one which has been extensively investigated, 
is the nature of the increased metabolism arising from muscular 
exertion. As we shall see in a succeeding chapter, such exertion 
causes a marked increase in the respiratory exchange while pro- 
ducing at most but aslight effect upon. the proteid metabolism. 
If we neglect altogether this latter effect, the ratio between the in- 
crements of carbon dioxide and oxygen will indicate whether the 
additional material consumed during the performance of the work 
consisted of fat or carbohydrates or a mixture of the two, of course 
on the same assumption as before, viz., that subStantially only 
these two classes of substances are available in the schematic body. 
For example, in an investigation by Zuntz, cited on a subsequent 
page, the performance of one kilogram-meter of work of draft 
by a dog caused the following increments in the respiratory ex- 
change: 
ORY Pen isa eae laane nate nea 1.6704 c.c. 
Carbon dioxide........................ 1.4670 “ 
Respiratory quotient.................. 0.878 
Assuming, as above, that these amounts arise from the oxida- 
tion of fat and carbohydrates only, let x equal the amount of oxy- 
gen consumed in the oxidation of fat and 1.6704—z the amount 
consumed in the oxidation of carbohydrates. Since the respira- 
tory quotient of fat is 0.7069, the x cubic centimeters of oxygen 
would yield 0.7069z cubic centimeters of carbon dioxide, while the 
1.6704—z2 cubic centimeters of oxygen used to oxidize the carbohy- 
drates would yield an equal volume of carbon dioxide. We there- 
fore have— 
0.70692x + (1.6704—x) =1.4670, 
whence x=0.6939. 
