2012 Ch2-2 Bacterial Energetics [호환 모드] PDF

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Bacterial ểngetics...

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2.6 Energetics and Bacterial growth ○ The purpose of this section is to explore relationships

between reaction energetics and bacterial growth

2.6 Energetics and Bacterial Growth • When cells grow rapidly in the presence of nonlimiting concentrations of all factors required for growth, cells make the maximum investment of energy for synthesis.

dx a − ds =Y( ) − bx a dt dt dx a / dt xa Yn = =Y −b - ds / dt − ds / dt

[2.5] [2.6]

• When substrate and all other required factors are unlimited in amount, -dS/dt will be at its maximum, and the net yield (Yn) will approach the true yield (Y)

• However, when substrate is limited, the net yield decreases as a decrease in the rate of substrate utilization (eq 2.6)

• Equation 2.7 2 7 indicates that the net yield becomes zero ( X a = const.) const ) when the energy supplied through substrate utilization is just equal to m, the maintenance energy

Yn = 0, Y − b

xa = 0, − ds/dt

then h

− dS / dt b = =m Xa Y

[ 2.7 ]

2.6 How to relate growth to energetics • Battley(1987) provided us with relationships presented between energy production and cell yield( [2.37])

R = f e Ra + f s Rc − Rd

[2.37]

- This method is based on electron equivalents and differentiates between the energy portion of an overall biological reaction and the synthesis portion (The equation 2.37) (MaCatry, 1971, 1975; Christensen and McCarty 1975)

• A practical advantage of the approach : electron equivalents are easily related to measurements of widespread utility in practice environmental engineering practice - BOD (biochemical oxygen demand), COD` (calculated oxygen demand), COD (chemical oxygen demand) - One equivalent of oxygen demand (OD) = One equivalent of electron donor = 8g as O2

2.6.1 Free Energy of the Energy Reaction • Gibb’s free energy = H (enthalpy) – T · S(entropy) ' ΔGr0 : standard free energies corrected to pH 7

: All constituents are at unit activity except pH =7.0 [Ethanol ; 1mole 1mole, O2 ; 1 atm (partial pressure), pressure) CO2 ; 1 atm (partial pressure, liquid water ; solvent] 0

0' r

Ex Ethanol ( ΔGr = ΔG , because H+ is not a component of the equation) Ex.

2.6.1 Free Energy of the Energy Reaction 1. Aerobic oxidation (Oxygen as electron acceptor) Ex 1.1 The range of organic materials from Methane to glucose - from -96 to -120 kJ/e- eq (Δ 24) Ex 1.2 1 2 The range of inorganic materials from iron oxidation to hydrogen oxidation (fuel cell !) - from -5 to -119 kJ/e- eq (Δ 114)

2. Anaerobic oxidation (Carbon dioxide as electron acceptor) Ex 2.1 The range of organic materials from acetate oxidation to glucose oxidation - from -3.87 to -18.82 kJ/e- eq (Δ14.95) The great differences in reaction free energies for aerobic verse anaerobic and organic verse inorganic reaction have great effects on resulting bacterial yields

fe

vs. fs

2.6.1 Free Energy of the Energy Reaction ΔGr : reaction free energy for nonstandard concentrations v1 A1 + v2 A2 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ → vm Am + vm+1 Am+ 1 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +vn An

[2.40]

n

0 = ∑ vir Ai

[2.41]

The left of equation : negative

The right of equation : positive

vir = the stoichiometric coefficient for constituent Ai in reaction r n

ΔGr = ΔG + RT ∑ ln ai 0 r

i =1

ai = the activity of constituent Ai

[ 2.42]

2.6.1 Free Energy of Ethanol Oxidation (concentration correction) '

Ex 1. Ethanol ( ΔGr0 = ΔG 0 , because H+ is not a component of the equation) r

1 1 1 1 CH 3 CH 2 OH + O 2 = CO 2 + H 2O 4 12 4 6

[CO 2 ]1 / 6 [H 2 O]1 / 4 ΔG r = ΔG + RT ln [CH 3 CH 2 OH ]1 / 12 [O2 ]1 / 4 0 r

Condition : - Ethanol : 00.002M, 002M - O2 partial pressure : 0.21 atm - CO2 partial pressure : 0.0003 atm - Temp. Temp : 20 ◦C - All activity coefficients : 1.0

[2.43]

2.6.1 Free Energy of Ethanol Oxidation (concentration correction) '

Ex 1. Ethanol ( ΔGr0 = ΔG 0 , because H+ is not a component of the equation) r

[CO2 ]1 / 6 [ H 2 O]1 / 4 ΔGr = ΔG + RT ln [ CH 3 CH 2 OH ] 1 / 12[ O2 ]1 / 4 0 r

[2.43]

[0 .0003]1/ 6 [1]1/ 4 ΔGr = − 109.900 + 8.314(273 + 20) ln [0.002]1/12[0.21]1/ 4 = − 109.900 + 1.100 (~ 1%) = −111.000 J / e − eq or −111 kJ / e − eq

• Conclusion : - The reaction free energy, corrected for concentrations (1.1 kJ/eeq) that that are within aa typical range for biological systems systems of interest interest, is within 1 % of the standard free energy of -109.9 kJ/e-eq.

- Concentration corrections become important when the concentration of e- donor or acceptor is very low or Δ G 0 is < -10 kJ/e- eq r

2.6.1 Free Energy of the Energy Reaction

'

• In Table 2.2 and 2.3, only the ΔGr0 value is given. However, we need ΔG 0 value to get ΔG r . r • When pH is significantly different different from from 7, 7 then corrections for for the the effect of H+ concentration may be necessary. '

ΔGr0 = ΔGr0 + RTv H + ln[10 −7 ] '

ΔGr0 = ΔGr0 − RTvH + ln[10−7 ]

Tip

The

[ 2.44]

´ values in Tables 2.2 and 2.3 already have the pH correction for pH =7.0

2.7 Yield Coefficient and Reaction Energetics • The microbial yield from substrate utilization : 3 Step

(recall

Figure 1.13 Æ next slide)

- 1 Step : the energy reaction creates high-energy carriers, such as ATP - 2 Step : the energy carriers are “spent” to drive cell synthesis or cell maintenance - 3 Step : a certain amount of thermodynamic free energy is lost with each transfer

• How to compute the energy costs of cell synthesis and how to account for the energy lost in transfers (= How to estimate

and the true yield (Y) based on thermodynamic principles)

• For determining true yield (Y), the energy for maintenance is set to zero (Eq. 2-5), so that all energy is used for cell synthesis. dx a − ds [2.5] =Y( ) − bx a dt dt

2.7 How to estimate principles

and Y based on thermodynamic

- ΔG : the energy required to synthesize one equivalent of cells from a given s carbon source (when NH4+1 is N-source) -

th energy requ iiredd to t convertt th the carbon b source tto pyruvate t ΔG p : the ΔG pc : the energy required to convert the pyruvate carbon to cellular source

- First, we must determine the energy change resulting from the conversion of the carbon source to the common organic intermediates that the cell uses uses for synthesizing macromolecules macromolecules (use pyruvate as a representative intermediate) '

Δ Gp = 35 .09 − Δ Gc0 [2.45] -

35.09 kJ /e– eq. : free energy of half reaction for pyruvate synthesis (O-16, in Table 2.3) '

- ΔG0 : free energy of half reaction for the given e- donnor (=carbon source) c : For heterotrophic bacteria, the carbon source almost always is the e-

-

donor. donor Example, Carbon source (Acetate) Æ CO2 + HCO3- Æ pyruvate -27.4kJ 35.09kJ

2.7 How to estimate principles

and Y based on thermodynamic '

ΔG p = 35.09 − ΔGc0 -

[ 2.45]

In autotrophic reactions, inorganic carbon is used as the carbon source. In photosynthesis, the H+ and e- for reducing CO2 to form cellular organic matter comes from water. water

So we take ΔG 0' to equal to that for the water – oxygen reaction (- 78.72 kJ / e- eq) c ¼ O2 + H+ + e= ½ H2O (Table2-2)

ΔG p = 35.09 – (- 78.72) = 113.8 kJ/e- eq. - Second, Pyruvate carbon is converted to cellular carbon. The energy required here is

ΔG pc

= [an estimated value of 3.33kJ/g cell (McCarty, 1971)] x [5.65g/e- eq] < from Table 2.4, ammonium as Nitrogen source, (113.8/20 = 5.65 > = 18.8 kJ/e-eq

2.7 How to estimate

and Y based on thermodynamic principles

- Finally, the energy that is lost in the electron transfers is considered by including a term for energy-transfer efficiency (є)

ΔGs =

Δ Gp

εn

+

ΔG pc

ε

[2.46]

= the energy requirement for cell synthesis n = +1, when ΔG p is positive (energy is required in its conversion to pyruvate) < Ex. Conversion of acetate to pyruvate > More energy is required than the the thermodynamic thermodynamic amount amount n = -1, when ΔG p is negative (energy is released by its conversion to pyruvate) < Ex. Conversion of glucose to pyruvate > S Some off th the thermodynamic th d i energy iis llost. t

2.7 How much energy is needed to synthesize an equivalent of cells ?

•

Δ Gr

: free energy released per equivalent of donor oxidized for energy generation

A

equivalents of electron donor

:

AΔ Gr : the energy released by donor oxidized for energy generation •

•

When this energy is transferred to the energy carrier, a portion is lost through transfer inefficiencies. If transfer efficiency here is the same as that for transferring energy(є ), then the energy transferred to the carrier is A ε ΔGr An energy balance must be maintained around the energy carrier at steady state:

ΔG p

Aε ΔGr + ΔGs = 0

[2.47]

n ε A=

+

Δ G pc

ε

− εΔGr

[2.48]

2.7 How to estimate AεΔ Gr + ΔGs = 0

and Y based on thermodynamic principles

[2.47]

ΔGp

+

ΔG pc

n ε A= ε − εΔ Gr

[2.48]

• The equation 2.48 indicates that equivalents of donor used for energy production (A) per equivalent of cells formed. • It increases as the energy required for synthesis (numerator) from the given carbon source increases and as the energy released (denominator) by donor oxidation decreases.

Tip

- Under optimum conditions, є = 55 to 77 percent are typical - The

energy-transfer efficiency is key factor that needs to be assumed in solving l i E Equation ti 22.48 48 ( a є va lue l off 0.6 0 6 frequently f tl is i employed) l d)

2.7 Yield Coefficient and Reaction Energetics ΔG p n ε A=

+

ΔGpc

ε

− ε ΔGr

•

[2.48]

The equation 2.48 does not include energy of maintenance, the resulting value of A is for the situation of the true yield or Y .

f s0 =

A 1 and f e0 = 1 − f s0 = 1+ A 1+ A

[ 2.49]

A equivalents ; equivalents of e- donor which must be oxidized to supply energy in order to to synthesize an equivalent equivalent of of cells cells 1 equivalent ; part of the e- donor used for synthesis 1 +A ; equivalents of the total donor

2.3 Substrate partitioning and cellular yield

• •

Yn is less than Y because some of the electrons originally present in the substrate must be consumed for energy of maintenance. When considering net yield, the portion of e- used for synthesis is fs rather than fs0 : the portion of e- for energy production is fe rather than fe0

f e0 + f s0 = 1

fe + f s = 1

f s < f s0

f e > f e0

Example 2.10

Example 2.11

2.8 Oxidized Nitrogen Sources nitrogen as as an an inorganic inorganic • Microorganisms prefer to use ammonium nitrogen nitrogen source for cell synthesis because it already is in the (-Ⅲ) oxidation state (next slide) • But, when ammonium is not available for synthesis, many prokaryotic cells can use oxidized forms of nitrogen as alternative - nitrate(NO3-), nitrite(NO2-), dinitrogen(N2)

2.8 Oxidized Nitrogen Sources ΔGs =

ΔGp

ε

n

+

ΔGpc

ε

[2.46]

• The Δ G pc part of ΔGs depends on the nitrogen source, source since since a mole mole of of C5H7O2N contains different of electron equivalents -

ΔG pc = 18.8 (NH4+), 13.5 (NO3-), 14.5 (NO2-) , 16.4(N2) (unit : kJ/e- eq)

ΔG pc for NO3- = [an estimated value of 3.33kJ/g cell ] X [4.06g(=113.8/28)] = 13.5 kJ/e-eq

• Once A is computed from Equation 2.48, fs0 and fe0 are computed in the usual manner from Equation 2.49. Then the overall stoichiometric reaction is generated from the appropriate donor(Rd), acceptor(Ra) and synthesis(Rc) half-reactions ΔG p n A= ε

+

ΔG pc

ε

− εΔGr

[ 2.48]

fs0 =

A 1 and fe0 = 1 − fs 0 = 1+ A 1+ A

[2.49]

Example 2.13...