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An Absorbing Markov Chain approach to understanding the microbial role in soil carbon stabilization


The number of studies focused on the transformation and sequestration of soil organic carbon (C) has dramatically increased in recent years due to growing interest in understanding the global C cycle. While it is readily accepted that terrestrial C dynamics are heavily influenced by the catabolic and anabolic activities of microorganisms, the incorporation of microbial biomass components into stable soil C pools (via microbial living cells and necromass) has received less attention. Nevertheless, microbial-derived C inputs to soils are now increasingly recognized as playing a far greater role in stabilization of soil organic matter than previously believed. Our understanding, however, is limited by the difficulties associated with studying microbial turnover in soils. Here, we describe the use of an Absorbing Markov Chain (AMC) to model the dynamics of soil C transformations among three microbial states: living microbial biomass, microbial necromass, and C removed from living and dead microbial sources. We find that AMC provides a powerful quantitative approach that allows prediction of how C will be distributed among these three states, and how long it will take for the entire amount of initial C to pass through the biomass and necromass pools and be moved into atmosphere. Further, assuming constant C inputs to the model, we can predict how C is eventually distributed, along with how much C sequestrated in soil is microbial-derived. Our work represents a first step in attempting to quantify the flow of C through microbial pathways, and has the potential to increase our understanding of the microbial role in soil C dynamics.

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This work was financially supported by the DOE Great Lakes Bioenergy Research Center (DOE BER Office of Science DE-FC02-07ER64494), USDA-CSREES and NSF-DMS 0906497. We would like to thank Dr. R. Jackson for his help with the proposed idea, Drs. C. Xu and J. Zhu for the discussions on the earlier stage of this study. We would also like to thank the editor and two anonymous reviewers. The manuscript is much improved because of their inputs.

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Correspondence to Chao Liang.



Prove why T = [ m / ( m + n )] ( Lx + Ly   2)  + [ n / ( m + n )](Dx + Dy   2) holds:

Step 1: we first show that the expected number of transitions from state L to state D is the (1; 2)-th element of the fundamental matrix N minus one, i.e. Ly − 1.

Let C(s) = 1 if the C starts from L and enters D in the s-th step, C(s) = 0 otherwise. In order to calculate the expectation of C(s), we need to know the probability of the event that C starts from L and enters D in the S-th step, denoted by p (S) LD . In other words, the expectation is

$$ M[C(S)] = \sum\limits_{S = 1}^{\infty } {p_{LD}^{(S)} } $$

By the Markov Chain rule, we know p (S) LD equals to the (2, 3)-th element of PS. Furthermore, we know it also equals to the (1, 2)-th element of QS by the canonical form of Pt. Hence, we have

$$ M[C(S)] = \sum\limits_{S = 1}^{\infty } {p_{LD}^{(S)} } = \sum\limits_{S = 1}^{\infty } {Q_{12}^{(S)} } = Ly - 1 $$

The last equality follows from Eq. 6

$$ {\mathbf{N}} = \left( {{\text{I}} - {\text{Q}}} \right)^{ - 1} = {\text{I}} + {\text{Q}} + {\text{Q}}^{ 2} + {\text{Q}}^{ 3} + \ldots $$

Step 2: We next calculate the expected number of transitions that the C transit from the state L or D to the air state. We consider two cases: (i) C starts from L; (ii) C starts from D.

For the case (i), the expected number of transitions from L to L (D) is Lx − 1 and Ly − 1, respectively, based on the results in step 1. For the case (ii), the expected number of transitions from D to L (D) is Dx − 1 and Dy − 1, respectively, based on the results in step 1. Then, the expected number of total transitions equal to

P(C starts from L) × (Lx + Ly − 2) + P(C starts from D) × (Dx + Dy − 2) = 

$$ \left[ {m/\left( {m + n} \right)} \right]\left( {Lx + Ly - 2} \right) + \left[ {n/\left( {m + n} \right)} \right]\left( {Dx + Dy - 2} \right) $$

This completes the whole proof.

Prove why L(stable) = f · Lx and D(stable) = f · Ly holds:

We first have the following relations:

$$ \begin{aligned} & {\text{Z}}^{(0)} = \left[ {{\text{L}}^{(0)} ;{\text{D}}^{(0)} } \right] \\ & {\text{Z}}^{( 1)} = {\text{Z}}^{(0)} {\text{P}} + \left[ {{\text{f}};0} \right] \\ & {\text{Z}}^{( 2)} = {\text{Z}}^{( 1)} {\text{P}} + \left[ {{\text{f}};0} \right] \\ & \ldots \\ \end{aligned} $$

where (L(0); D(0)) are the initial distribution of (L; D), respectively. Based on the above relation, we can prove that

$$ \left[ {{\text{L}}^{{({\text{k}})}} ;{\text{D}}^{{({\text{k}})}} } \right] = \left[ {{\text{L}}^{(0)} ;{\text{D}}^{( 0)} } \right]{\text{Q}^{k}} + \left[ {f;0} \right]\left( {I + \sum\limits_{i = 1}^{k - 1} {Q^{i} } } \right) $$

through mathematical inductions. Since our model is the AMC, Qk → ∞ as k → ∞. Then we have

$$ \left[ {{\text{L}}^{{({\text{k}})}} ;{\text{ D}}^{{({\text{k}})}} } \right] \to \left[ {f;0} \right]\left( {I + \sum\limits_{i = 1}^{k - 1} {Q^{i} } } \right) \to \left[ {f;0} \right]N $$

by the equation that N = (I − Q)−1 = I + Q + Q 2 + 

Hence we have L(stable) = f · Lx and D(stable) = f · Ly as k approaches infinity.

This completes the whole proof.

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Liang, C., Cheng, G., Wixon, D.L. et al. An Absorbing Markov Chain approach to understanding the microbial role in soil carbon stabilization. Biogeochemistry 106, 303–309 (2011).

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