Parameters of rumen fermentation in a continuously fed sheep: evidence of a microbial rumination pool.

The feed and feces of a continuously fed sheep were analyzed for carbon, hydrogen, and nitrogen, with oxygen as the remainder. The daily feed-feces weight difference was used as the reactant in an equation representing the rumen fermentation. The measured products were the daily production of volatile fatty acids (VFA), CH(4), CO(2), and ammonia. The carbon unaccounted for was assumed to be in the microbial cell material produced in the rumen and absorbed before reaching the feces. The ratio of C to H, O, and N in bacteria was used to represent the elemental composition of the microbes formed in the rumen fermentation, completing the following equation:C(20.03)H(36.99)O(17.406)N(1.345) + 5.65 H(2)O --> C(12)H(24)O(10.1) + 0.83 CH(4) VFA + 2.76 CO(2) + 0.50 NH(3) + C(4.44)H(8.88)O(2.35)N(0.785) microbial cells absorbed With C arbitrarily balanced and O balanced by appropriate addition of water, any error is reflected in the H. The H recovery was 98.5%. The turnover rate constant for rumen liquid equilibrating with polyethylene glycol (PEG) was 2.27 per day. Direct counts and volume measurements of the individual types of bacteria and protozoa in the rumen were used to calculate the total microbial cell volume in the rumen, not equilibrating with it. The dry matter in the rumen (582 g) and the nitrogen content (12.05) of the microbes in the rumen were estimated, the latter constituting 85% of the measured N in the rumen. Calculations for rumen dry matter and nitrogen turning over at the PEG rate introduce big discrepancies with other parameters; a rumination pool must be postulated. Its size and composition are estimated. Arguments are presented to support the view that dry matter and some of the microbes, chiefly the protozoa, do not leave the rumen at the PEG rate. One experiment with the same sheep fed twice daily showed significantly less production of microbial cells than did the continuous (each 2 hr) feeding. Analysis of the microbial cell yield suggests that, on the basis of 11 mg of cells per adenosine triphosphate molecule, a maximum of six adenosine triphosphate molecules could have been formed from each molecule of hexose fermented.