Q H+H+ H+H+ ADP ATP H+H+ H+H+ NAD NADH e – acceptor End product H+H+ H+H+ H+H+ H+H+ e – donor e – donor End product H+H+ H+H+ H+H+ H+H+ Cytoplasm Environment Electron Transport Chain e–e– e–e– e–e– Q QH 2 c 3+ c 2+ Courtesy Qusheng Jin

Electron donating reaction Electron accepting reaction oxidized reduced e–e–

e–e– Electron donating reaction Electron accepting reaction HCO 3 SO 4 2– (CH 2 O) n HS – –

Monod equation Microbial kinetics r+r+ r max = k[X] KDKD r max /2 [D][D] Biomass growth r + forward rate k rate constant [X] biomass concentration [D] donor concentration K D half-saturation constant Y growth yield

Qusheng Jin inks contract at Univ. Oregon Kinetic-Thermodynamic Theory Rate constant Biomass concentration Kinetics of donation, acceptance Thermodynamic factor Jin & Bethke, GCA (2005)

Key Assumptions Chemiosmotic model of ATP synthesis. Enzymatic catalysis of donating, accepting reactions: Single rate-limiting step. Reaction intermediates at steady state. Reactions proceed forward, backward.

Rate constant × biomass conc. Enzymatic catalysis of electron donating and accepting reactions Thermodynamic term: Energy  G r available in environment Energy  G ATP used to create m ATPs Average stoichiometric number  Rate of Microbial Respiration Microbial growth Y = growth yield (mg mmol –1 ) D = decay constant (s –1 ) Jin & Bethke (2002)

Thermodynamic Potential Factor Jin & Bethke, GCA (2005) ATPs produced Average stoichiometric number ADP phosphorylation Thermodynamic factor

T-C Monod Thermodynamically Consistent Rate Laws T-C Dual Monod r overall rate, r +  r - k rate constant [X] biomass concentration [D], [A] donor, acceptor conc. K D, K A half-saturation constants F T thermodynamic factor

Task 1 — Arsenate Reduction by B. Arsenicoselenatis Bacterium grows on arsenate in alkaline, saline bioreactor Use experimental data to estimate kinetic parameters What factors control rate of reaction?

Arsenate Reduction by Bacillus Arsenicoselenatis lactate acetate arsenate arsenite = F D = F A

B. arsenicoselenatis k + (mol mg –1 s –1 )7×10 –9 K D (molal –1 )10 KAKA 0.1 m2.5  G P (kJ mol –1 ) 50  4 Initial biomass (mg kg –1 )0.5 Y (mg mol –1 )5000 D (s –1 )0 Data correspond to reaction written for consumption of one lactate.

Time elapsed (hr) Concentration (mg kg –1 ) Biomass (mg kg –1 )

! ! ! ! ! ! ! ! ! ! › › › › › › › › › › 0 5 ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ œ œ œ œ œ œ œ œ œ œ Time elapsed (hr) Concentration (mmolal) Lactate AsO 4 −−− As(OH) 4 − CH 3 COO − Concentration (mmolal) Biomass (mg kg –1 )

Time elapsed (hr) Kinetic or thermo- dynamic factor Biomass (mg kg –1 ) Reaction rate (nmol kg –1 s –1 ) FDFD FTFT FAFA

Task 2 — Microbes in Aquifer Fermentation of organic matter supplies acetate to flowing water. SO 4 in recharge. Sulfate reducers compete with acetoclasts. What is distribution of microbial activity in aquifer?

H 2 S(aq) reacts with FeO(ox) in seds to form FeS(s). Seeds grow to form community Ca-HCO 3 + SO 4 Aquifer seeded with microbes Aquitards (fermenters) Acetate produced by fermentation. Respired by sulfate reducers Dismutated by methanogens Ac v x = 10 m yr –1 Microbes in Aquifer

Acetotrophic sulfate reducers Acetoclastic methanogens Competition for Acetate CH 3 COO – + SO 4  2 HCO 3 + HS – 2– – CH 3 COO – + H 2 O  CH 4 + HCO 3 –

SO 4 reducersMethanogens k + (mol mg –1 s –1 )10 –9 (a) 3×10 –9 (b) K D (mmolal)0.07 (a) 5 (b) K A (mmolal) (c) 0.2— m (d) 1½  G P (kJ mol –1 ) (e) 45  (f) 52 Y (mg mol –1 )4300 (a) 2000 (b) D (s –1 ) (g) 10 –9 (a) Ingvorsen et al. (1984); (b) Oude Elferink et al. (1994); (c) Ingvorsen et al. (1984), Pallud and Van Cappellen (2006); (d) Bethke et al. (2008); (e) Jin and Bethke (2009); (f) Jin and Bethke (2005, 2007); (g) Bethke (2008). Data correspond to reactions written for consumption of one acetate.

Concentration (μmol kg –1 ) X position (km) SO 4 reducers Methanogens CH 3 COO – CH 4 SO 4 2– flow Biomass (mg kg –1 )

×10 –9 10×10 – Reaction rate (mol kg –1 yr –1 ) X position (km) flow Thermodynamic factor, F T SO 4 reducers Methanogens SO 4 reducers m = 1,  = 5 Methanogens m = ½,  = 2

Growth rate should not be negative. For methanogens, at small m Ac So m Ac must be large enough that Extinction Vortex

Craig M. Bethke and Brian Farrell © Copyright 2016 Aqueous Solutions LLC. This document may be reproduced and modified freely to support any licensed use of The Geochemist’s Workbench® software, provided that any derived materials acknowledge original authorship.