Energetics of sediment microbes

(1)

(c) Heribert Cypionka, www.icbm.de/pmbio

Energetics of sediment microbes

Principle for writing and presentations:

Numbers in the text only when the reader should keep them in mind

Today:

Examples for comparison and

assessment of results - based on numbers

No numbers in the text

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A culture produced 3 mM sulfide

Heat production: 0.2 µW/g sediment

Sulfate reduction rate in a tidal flat: 50 nmol

^.

cm

^{-3 .}

d

^-1

Water: 5

^.

10

⁶

bacteria/ml

Mediterranean sediment: 3 µg DNA/g Doubling time of a culture: 10 h

Measured data

Measured data are not the result!

Data

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?

Questions

(3)

(c) Heribert Cypionka, www.icbm.de/pmbio Motivation

Bacterial cell

Size: 1 µm Ø, Volume: 4/3 π r

³

= 0.524

^.

10

^-15

l

(? pico or femto)

(0.5 µm Ø → 0.065

^.

10

^-15

l) Specific weight: 1.05 g/cm

³

, wet mass: 550

^.

10

^-15

g (0.5 µm Ø → 68

^.

10

^-15

g) Dry mass 20 %: 110

^.

10

^-15

g

Protein content 50 %: 55

^.

10

^-15

g

Turbidity (Bausch & Lomb-Photometer, 436 nm, depends on size):

OD 1 → 5

^.

10

⁸

cells/ml (Dv.) 0.05 - 0.2 mg dry mass/ml with OD

₄₃₆

= 1

Bacterial cell

(4)

Genome of prokaryotes

**3.6 * 10**

⁶

**base pairs (average, range 0.7 - 11 * 10**

⁶⁾

2 Bits per base, 1 MByte per bacterial genome (E.coli)

**4 * 10**

^-15

g DNA per bacterium (E.coli)

**15 * 10**

^-15

g RNA (90 % ribosomal) per bacterium (E.coli) Length of a DNA molecule: 1.4 mm (E.coli, man: ≈2 m)

Number of genes: 3000 (E.coli, man: ≈30 000)

Genome data

2407-2410

Meteor M40

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Sapropel S7, per cm ^-3 : 10 ⁸ bacteria, 3 µg DNA,

Meteor M40, Coolen et al. (2002) Science 296:2407-2410

Meteor data

Sapropel layer S7: 10

⁸

Bacteria cm

^-3

, 3 µg DNA per g dry Sediment

10

⁸

* 4

^.

10

^-15

= 0.4

^.

10

^-6

g DNA

Extraction efficiency, genome size, counting errors...?

No experiment is perfect. Data assessment and

interpretation should evaluate possible flaws and point to the reliable results.

Water content of the sediment? → decreasing values Specific weight of the sediment? → increasing values

Sapropel data

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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

Energetical classification

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) Calculation of free energies

- Use correct stoichiometry

- Use realistic protonation (H

⁺

, CO

₂

/HCO

₃^-

, HS

^-

/H

₂

S...) - Consider solublility (Fe

³⁺

, Fe

²⁺

, Mn

⁴⁺

...)

∆G

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Glucose oxidation with oxygen

C ₆ H ₁₂ O ₆ + 6 O ₂ → 6 CO ₂ + 6 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 : -917.2 (reactant: x -1) +917.2 O ₂ : 0 (reactant: x -6) 0 CO ₂ : -394.4 (product: x 6) -2366.4 H ₂ O : -237.2 (product: x 6) -1423.2

Sum: ∆G°'= -2872.4 kJ mol ^-1

∆G of glucose oxidation

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

Real concentrations

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∆G in a growth experiment C ₆ H ₁₂ O ₆ + 6 O ₂ → 6 CO ₂ + 6 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 )

= -2872.4 + 8.314 * 298/1000 * ln([0.1 ⁶ * 1 ⁶ )]/[0.01 * 0.2 ⁶ ])

= -2873.5

Growth experiment

0 100 200 300 400 500 600

0 5 10 15 20 25

Sulfate concentration [mM]

Sediment depth [cm]

sulfate [mM]

0 100 200 300 400 500 600 -250 -245 -240 -235 -230 -225

Free energy change [kJ]

free energy change

Abhängigkeit von ∆G von der Sulfatkonzentration (Wattsediment) Die vollständige Umsetzung Lactat mit Sulfat liefert unter

Standardbedingungen eine freie Energie ∆G

₀

von 254.4 kJ/mol Darstellung der Abhängigkeit der freien Energie von der Sulfatkonzentration

Annahme, dass die Konzentration aller Reaktionspartner gleich bleibt und sich nur die Sulfatkonzentration ändert

Antje Gittel

Look carefully on scales at the axes!

Tidal flat sediment

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Conditions ∆G (kJ/mol) Standard 152 [H

₂

]=0.001 83 [Sulfate]=0.01 140 [HS

^-

]=0.001 169

∆G of sulfate reduction under different conditions

Excel file for download:

deltaG-calculator.xls

Value of ∆G is essential because ATP phosphorylation requires 50 - 75 kJ/mol

∆G of sulfate reduction

Conditions ∆G (kJ/mol) Standard +18/-130 [H

₂

]=0.001 -50/-62

∆G of syntrophic ethanol degradation to methane + acetate

Excel file for download:

deltaG-calculator.xls

2 C

₂

H

₅

OH + CO

₂

→ 2 CH

₃

COO

^-

+ 2 H

⁺

+ CH

4

∆G°' = -112 kJ/mol

Although H

₂

is not visible in the reaction sum, its concentration modulates how much energy is available for the two syntrophic partners.

Syntrophic ethanol degradation

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Condition ∆G (kJ/mol) Standard -32 [ATP]=0.01,

[ADP]=0.001,

[Pi]=0.01 -49

∆G of intracellular ADP phosphorylation

Intracellular concentrations = ∆G

_biol.

download: deltaG-calculator.xls

∆G of ATPase

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 (0.0010.01)/(0.01 1) = 0.001

∆G

_biol.

= ∆G

₀

' + RT ln 0.001 = ∆G

₀

' -17 = -49 kJ/mol

∆G

_biol

= -50 kJ/mol

3.) For regeneration consumed: mostly about 75 kJ/mol ATP

Value of ATP

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ATP synthase: Reversible phosphorylation of ADP coupled to the transport of protons across the membrane

≈12.5 kJ/mol of protons

ATPase mechanism

Maintenance energy

≈ 4 mmol ATP g

^-1

(dry mass)

^-1

h

^-1

= 4800 J d

^-1

(g dry mass)

^-1

≈10

⁹

ATP cycles per bacterium and hour

1 - 10 mM ATP in the cytoplasm, cell volume 10

^-16

- 10

^-15

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 62:147-151 Harder J (1997) FEMS Microbiol Ecol 23:39-44

We do not understand survival in a population with a doubling time of 10 000 years

The key problem

Maintenance energy

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Sulfate reduction in a tidal flat

43 nmol Sulfate d

^-1

cm

^-3

How many cells can we expect to be responsible for this activity?

0 100 200 300 400 500 600

0 5 10 15 20 25

Sulfate concentration [mM]

sulfate [mM]

0 100 200 300 400 500 600 -250 -245 -240 -235 -230 -225

Free energy change [kJ]

free energy change

Antje Gittel Sulfate reduction rate

SRR in tidal flat

Assumptions

• 1 ATP per sulfate reduced (compare with : 5 ATP/O

₂

)

• Y

_ATP

= 10 g dry mass/mol ATP (experience)

• td = 24 h µ = 0.0289 h

^-1

• maintenance: m

_e

= 4 mmol ATP (g dry mass)

^-1.

h

^-1

1 µg cells (dry mass) reduce per day:

**24 * 4 = 96 nmol sulfate for maintenance**

Biosynthesis of 1 µg dry mass requires 0.1 µmol ATP or sulfate reduction

Sum: 196 nmol SO

₄^{2- .}

µg (dry mass)

^{-1 .}

d

^-1

(half for maintenance!)

With dry mass per cell of 110

^.

10

^-15

**g follows 9.1 * 10**

⁶

cells are in 1 µg

• Dry mass of a cell: 110

^.

10

^-15

g

**(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

SRR in tidal flat

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How much heat ist produced?

• C

₆

H

₁₂

O

₆

+ 6 O

₂

→ 6 CO

₂

+ 6 H

₂

O

∆G°' = -2872 kJ/mol

43 nmol SO

₄^{2- .}

cm

^{-3 .}

d

^-1

• C

₆

H

₁₂

O

₆

+ 3 SO

₄^2-

+ 3 H+ → 6 CO

₂

+ 3 HS

^-

+ 6 H

₂

O

∆G°' = -480 kJ/mol

43

^.

10

^{-9 .}

480 000 = 20.6 mJ

^.

cm

^{-3 .}

d

^-1

1 J = 1 W

^.

s

20.6 mJ

^.

d

^-1

= 0.24 µW

Heat production

How much heat ist produced?

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) 68

^.

10

^-15

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

Whales vs. bacteria

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(c) Heribert Cypionka, www.icbm.de/pmbio Motivation

Der E

lefant, das Riesentier, der braucht zwei Pfund

Klosettpapier....

Motivation

Exponential function instead

of simple 'rule of three'

calculation...

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- 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

₃^2-

+ H

₂

O → SO

₄^2-

+ HS

^-

+ H

+

∆G°' = -21.9 kJ/mol

Consortia carrying out anaerobic methane oxidation

CH

₄

+ SO

₄^2-

+ H

⁺

→ CO

₂

+ HS

^-

+ 2 H

₂

O ∆G°' = -21.0 kJ/mol or even:

CH

₄

+ SO

₄^2-

→ HCO

₃^-

+ HS

^-

+ H

₂

O ∆G°' = -16.2 kJ/mol

Back to the key problem:

What is the minimum energy required for sustaining life?

Harder (1997) FEMS Microbiol Ecol 23:39-44

Harder

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(c) Heribert Cypionka, www.icbm.de/pmbio 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 4 × 10¹⁴gC yr^-1and ocean-margin sites 1 × 10¹⁴gC yr^-1, minus C burial rate (5 × 10¹²gC yr^-1and 10 × 10¹²gC yr^-1for open- ocean and ocean-magins, respectively) is available for subsurface microorganisms. The efficiency of carbon assimilation of 0.529 was used to calculate the turnover times.

The turnover times of bacteria were in the range of 0.25 - 1.91 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 open- ocean 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 10³yrs given by Whitman et al. for the turnover time of the total prokaryotic biomass in subsurface sediments.

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 sub- seafloor biosphere identified as living bacteria. Nature 433:861-864

Where is the maintenance energy requirement?

Schippers

Energetics of sediment microbes