COS 30-8
A cost:benefit analysis of lignocellulose decomposition based on energetic tradeoffs

Tuesday, August 6, 2013: 10:30 AM
L100J, Minneapolis Convention Center
Daryl L. Moorhead, Environmental Sciences, University of Toledo, Toledo, OH
Gwenaëlle Lashermes, Fractionnement des AgroRessources et Environnement, Institut National de la Recherche Agronomique, Reims, France
Robert L. Sinsabaugh, Department of Biology, University of New Mexico, Albuquerque, NM
Michael N. Weintraub, Environmental Sciences, University of Toledo, Toledo, OH
Background/Question/Methods

Despite its ambiguous composition, the acid insoluble fraction of litter isolated by proximate carbon analysis remains one of the best predictors of decay rate. Although it clearly consists of more than lignin, its representation in the lignocellulose index (LCI=lignin/[lignin+holocellulose]) continues to serve as a common index to litter quality with many models reducing decay rates as LCI increases. The energetic cost of lignin decay appears to be too high for a net energy gain from its products, but patterns of lignin decay and activities of oxidative enzymes responsible for it suggest that degrading lignin may provide access to other, biochemically shielded resources, including energy-rich compounds. Earlier studies have shown that lignin (C3) decay rate can be expressed as a first-order equation (dC3/dt = k3*C3), given a decay rate coefficient (k3) that is a linear function of LCI (k3=m3*LCI+b3), starting at a LCI = 0.4 (LCImn) and reaching a maximum (k3max) at LCI = 0.7 (LCImx). Studies have also shown that energy rich compounds, like holocellulose (C2), decay more slowly with increasing LCI (e.g., dC2/dt=k2*C2, wherein k2=m2*LCI+b2). 

Results/Conclusions

If we assume that LCImx represents the condition at which net energy yield from dC2/dt matches the cost of dC3/dt, then the relationship between k2 and k3 is defined by both LCI and this energetic tradeoff. Moreover, if LCImx is stable, then k2 = k3max at LCImx. If the efficiency of C-acquisition from C2 is e2, and the cost of C3 degradation is e3, then e3 = e2*(LCImn-LCImx)/LCImx. The slope of k2 with LCI (m2) must be opposite that of the change in k3 with LCI (m3) and steeper by the ratio of e2/e3. Thus k2 = m3*e2/e3*(LCI+LCImx) + k3max. This solution can be used to calculate the energetic yield of dC2/dt+dC3/dt given LCI, for example, the energy gained by microbial activity that increases litter LCI.