Energetics of sediment microbes

Transcription

1 Energetics of sediment microbes Principle for writing and presentations: Numbers in the text only when the reader should keep them in mind No numbers in the text Today: Examples for comparison and assessment of results - based on numbers 1

2 Measured data Data A culture produced 3 mm sulfide Sulfate reduction rate in a tidal flat: 50 nmol. cm -3. d -1 Heat production: 0.2 µw/g sediment Water: bacteria/ml Doubling time of a culture: 10 h Mediterranean sediment: 3 µg DNA/g Measured data are not the result! Questions Questions - How is the free energy in natural environments calculated? - How much ATP is required to double a cell? - How much energy does a cell need for survival? 2

3 Bacterial cell Bacterial cell Size: 1 µm Ø, Volume: 4/3 π r 3 = l (? pico or femto or atto) (0.5 µm Ø l) Specific weight: 1.05 g/cm 3 Wet mass: g 0.5 µm Ø g) Dry mass 20 %: g Protein content 50 %: g Turbidity (Bausch & Lomb-Photometer, 436 nm, depends on cell size): mg dry mass/ml with OD 436 = 1 OD cells/ml (Dv.) Genome of prokaryotes Genome data 3.6 * 10 6 base pairs (average, range * 10 6 ) 2 Bits per base, 1 MByte per bacterial genome (E.coli) Length of a DNA molecule: 1.4 mm (E.coli, humans: 2 m) Number of genes: 3000 (E.coli, man: ) 4 * g DNA per bacterium (E.coli) 15 * g RNA (90 % ribosomal) per bacterium (E.coli) 3

4 Meteor M Meteor data Sapropel S7, per cm -3 : 10 8 bacteria, 3 µg DNA, Meteor M40, Coolen et al. (2002) Science 296:

5 Sapropel data Sapropel layer S7: 10 8 Bacteria cm -3, 3 µg DNA per g dry Sediment 10 8 * = g DNA Water content of the sediment? decreasing values Specific weight of the sediment? increasing values Extraction efficiency, genome size, counting errors...? No experiment is perfect. Data assessment and interpretation should evaluate possible flaws and point to the reliable results. Energetical classification Energetical classification of processes: free (utilizable) energy G G < 0: exergonic, thermodynamically spontaneously possible G = 0: reversible, thermodynamically in equlilibrium G > 0: endergonic, not spontaneously reacting 5

6 Calculation of free energies G The free energy ( G) of a chemical reaction (at constant pressure and temperature) is easily calculated from tabulated enthalpies of formation ( G f ) G = Σ G f (Products) - Σ G f (reactants) - Use correct stoichiometry - Use realistic protonation (H +, CO 2 /HCO 3-, HS - /H 2 S...) - Consider solublility (Fe 3+, Fe 2+, Mn 4+...) Glucose oxidation with oxygen C 6 H 12 O O 2 6 CO H 2 O Enthalpies of formation under standard conditions ([ ]: 298 K, reactants 1 mol/l in water [gas 1atm], [']: ph=7) in kj/mol C 6 H 12 O 6 : (reactant: x -1) O 2 : 0 (reactant: x -6) 0 CO 2 : (product: x 6) H 2 O : (product: x 6) G of glucose oxidation Sum: G '= kj mol -1 6

7 Real concentrations Consideration of real concentrations G = G 0 + RT ln(c P /c E ) - Multiply concentrations, if more than 1 reactant or product - Stoichiometric factors go to the exponent Growth experiment G in a growth experiment C 6 H 12 O O 2 6 CO H 2 O [Gluc] = 10 mm, [O 2 ] =0.2 atm, [CO 2 ]= 0.1 atm, [H 2 O] = '1' G = G 0 + RT ln(c products /c reactants ) = * 298/1000 * ln([0.1 6 * 1 6 )]/[0.01 * ]) =

8 Dependence of G on the concentration of sulfate (tidal flat sediment) Complete oxidation of lactate with sulfate yields under standard conditions: G 0 = kj/mol Assumption that all concentrations except sulfate remain constant!? Tidal flat sediment Sulfate concentration [mm] Free energy change [kj] Sediment depth [cm] Sediment depth [cm] sulfate [mm] free energy change Antje Gittel Look carefully on scales at the axes! G of sulfate reduction under different conditions G of sulfate reduction Conditions G (kj/mol) Standard 152 [H 2 ]= [Sulfate]= [HS - ]= Value of G is essential because ATP phosphorylation requires kj/mol Excel file for download: deltag-calculator.xls 8

9 G of syntrophic ethanol degradation to methane + acetate Syntrophic ethanol degradation 2 C 2 H 5 OH + CO 2 2 CH 3 COO H + + CH 4 G ' = -112 kj/mol Conditions G (kj/mol) Standard +18/-130 [H 2 ]= /-62 Although H 2 is not visible in the reaction sum, its concentration modulates how much energy is available for the two syntrophic partners. Excel file for download: deltag-calculator.xls G of intracellular ADP phosphorylation G of ATPase Condition G (kj/mol) Standard -32 [ATP]=0.01, [ADP]=0.001, [Pi]= Intracellular concentrations = G biol. download: deltag-calculator.xls 9

10 Value of ATP Value of ATP 1.) Textbook (standard conditions) ATP + H 2 O ADP + P i G 0 ' = -32 kj/mol 2.) In the cell: [ATP] 10 mm, ADP 1 mm, [P i ] 10 mm, [H 2 O]=1 product/reactant ratio is (0.001*0.01)/(0.01 * 1) = G biol. = G 0 ' + RT ln = G 0 ' -17 = -49 kj/mol G biol = -50 kj/mol 3.) For regeneration consumed: mostly about 75 kj/mol ATP ATPase mechanism 12.5 kj/mol of protons ATP synthase: Reversible phosphorylation of ADP coupled to the transport of protons across the membrane 10

11 Maintenance energy The key problem Maintenance energy 4 mmol ATP g -1 (dry mass) -1 h -1 = 4800 J d -1 (g dry mass) ATP cycles per bacterium and hour 1-10 mm ATP in the cytoplasm, cell volume l 6 * 10 5 ATP molecules per cell 1 Cycle per sec for every ATP molecule Seitz H-J, Cypionka H (1986) Arch Microbiol 146:63-67 Müller RH, Babel W (1996) Appl Environ Microbiol 62: Harder J (1997) FEMS Microbiol Ecol 23:39-44 We do not understand survival in a population with a doubling time of years Sulfate reduction in a tidal flat SRR in tidal flat Sulfate reduction rate Sulfate concentration [mm] Free energy change [kj] Sediment depth [cm] Sediment depth [cm] sulfate [mm] free energy change Antje Gittel 43 nmol Sulfate d -1 cm -3 How many cells can we expect to be responsible for this activity? 11

12 Assumptions SRR in tidal flat 1 ATP per sulfate reduced (compare with : 5 ATP/O 2 ) Y ATP = 10 g dry mass/mol ATP (experience) td = 24 h µ = h -1 maintenance: m e = 4 mmol ATP (g dry mass) -1. h -1 Dry mass of a cell: g 1 µg cells (dry mass) reduce per day: 24 * 4 = 96 nmol sulfate for maintenance (43/196)* 9.1 * 10 6 = 2 * 10 6 cells with a doubling time of 24 h and standard maintenance energy requirement would reduce 43 nmol sulfate per day Biosynthesis of 1 µg dry mass requires 0.1 µmol ATP or sulfate reduction Sum: 196 nmol SO µg (dry mass) -1. d -1 (half for maintenance!) With dry mass per cell of g follows 9.1 * 10 6 cells are in 1 µg How much heat ist produced? Heat production C 6 H 12 O O 2 6 CO H 2 O G ' = kj/mol C 6 H 12 O SO H+ 6 CO HS H 2 O G ' = -480 kj/mol 43 nmol SO cm -3. d = 20.6 mj. cm -3. d -1 1 J = 1 W. s 20.6 mj. d -1 = 0.24 µw 12

13 How much heat ist produced? Whales vs. bacteria 10 7 cells in 1 cm 3 produce 0.2 µw (fast bacteria might produce 100 times more) Wet weight of a bacterial cell (0.5 µm Ø) g 68 * 10-8 g wet cells produce 0.2 µw 0.68 g produce 0.2 W 0.68 kg produce 200 W 68 kg (man) would produce 20 kw (we produce 2500 kcal per day = 120 W) 68 t (medium-sized whale) would produce 20 MW Back to the key problem: What is the minimum energy required for sustaining life? - 1 ATP per substrate? - 1 proton per substrate? - Organisms with extremely low G: Sulfate reducers carrying out thiosulfate disproportionation (Bak and Cypionka, 1987) S 2 O H 2 O SO HS - + H + G ' = kj/mol Consortia carrying out anaerobic methane oxidation CH 4 + SO H + CO 2 + HS H 2 O G ' = kj/mol or even: CH 4 + SO 2-4 HCO 3- + HS - + H 2 O G ' = kj/mol 13

14 What is the minimum energy required for sustaining life? Harder (1997) FEMS Microbiol Ecol 23:39-44 Harder Axel Schippers, Lev N. Neretin, Jens Kallmeyer, Timothy G. Ferdelman, Barry A. Cragg, R. John Parkes and Bo B. Jørgensen (2005) Prokaryotic cells of the deep subseafloor biosphere identified as living bacteria. Nature 433: Schippers The turnover times of bacteria were in the range of yrs, both, for the open ocean and for the ocean-margin sites. Higher turnover times for living bacterial biomass of 7 yrs for ocean-margin and 22 yrs for openocean sediments were calculated from the global estimates of carbon flux available for the subsurface bacterial community and the total living bacterial biomass. All these values are comparable to turnover times of prokaryotes in soil and aquatic habitats and are considerably lower than the value of 1-2 x 10 3 yrs given by Whitman et al. for the turnover time of the total prokaryotic biomass in subsurface sediments. The turnover time of living Bacteria was calculated by dividing the carbon flux available for the subsurface community by the total number of living Bacteria estimated as described above separately for the open-ocean and ocean-margin sites. We assumed that 1 % of the total primary production in both, the open-ocean gc yr -1 and ocean-margin sites gc yr -1, minus C burial rate ( gc yr -1 and gc yr -1 for openocean and ocean-magins, respectively) is available for subsurface microorganisms. The efficiency of carbon assimilation of was used to calculate the turnover times. Where is the maintenance energy requirement? 14