Microbial Growth – Overview of terms: exponential growth u td productivity

Microbial Growth –Overview of terms:exponential growthutdproductivitySubstrate limitation of metabolismLink between metabolism and growth

Growth - Overview

Microbial Growth I

1) Energy metabolism overview: glycolysis, TCA cycle, respiration chain, ATP synthase

2) Growth medium components, energy, carbon, nitrogen, phosphorus sources , minerals, trace elements, buffer

3) Growth rate, specific growth rate, exponential growth, semilog plot, maximum and total productivity, lag phase

Growth - Overview

Microbial Growth II –

1) Substrate limitation

2) Michaelis Menten model of substrate dependent substrate uptake rate vmax, km

3) The yield coefficient connects the Michaelis Menten Model to the Monod model of substrate dependent growth rate, umax, ks

4) Yield coefficient measurements

5) Yield coefficient is not constant

6) Maintenance coefficient

7) Pirt model contains 4 growth constants, ms, Ymax

Growth - Overview

Microbial Growth III –Maintenance coefficient

Microbial Growth IVGrowth in batch culture

Microbial Growth VChemostat

Microbial Growth VIDetermining growth constantsBiomass retention

Growth - Overview

1 Growth Medium Ingredients

1.1 The rationale of media recipes

Bacterial cells typically grow by cell division into two daughter cells. To do this they require a suitable growth medium. Growth media recipes in the literature vary widely and it can be confusing to students to discriminate between essential ingredients and replaceable ones. Rather than blindly following recipes, it would be more useful for microbiologists to be able to design own media, or to modify or optimise exiting media. For this it is useful to understand the generic microbial growth requirements. The ingredients of a typical growth medium satisfy a number of principal needs of growing cells by providing a source of Energy, carbon, nitrogen, phosphate, sulfur, minerals, trace elements, vitamins, growth factors, buffer capacity.

1.2 Energy source

Microbial growth (assimilation) is an endergonic process and requires energy input for the conversion of ingredients from the growth medium into biomass. This energy is derived from the energy source component of the growth medium. Typically an energy source consists of a suitable electron donor and electron acceptor. It is the transfer of electrons from the electron donor (a redox couple that of a more negative potential than that of the electron accepting redox couple) to the acceptor that liberates energy which is conserved as ATP. The ATP is typically generated during this electron transfer via electron transport phosphorylation (electron carriers, electron transport chain, proton gradient, ATP synthase).

For most bacteria the electron donor is an organic compound being oxidised to CO2 and the electron acceptor is oxygen, which is supplied by allowing air access to the growht container (e.g. petri dishes or shake flasks). However many bacteria use inorganic reduced chemical species as the electron donor, such as ferrous iron, sulfide, ammonia, or hydrogen gas which supply electrons by being oxidised to ferric iron, sulfate, nitrite or protons, respectively. Electron accepting species alternative to oxygen are can be ferric iron, nitrate, sulfate, CO2, etc. which accept electrons by being reduced to ferrous iron, N2, sulfide and methane respectively. When supplying specific electron acceptors it needs to be considered that the presence of multiple potential electron acceptors can cause mutual inhibition. For example the presence of oxygen inhibits the reduction of nitrate, sulfate and CO2 while the presence of nitrate inhibits the reduction of sulfate and CO2. The use of external electron acceptors other than oxygen is still a respiration (anaerobic respiration).

If a suitable electron donor and electron acceptor is provided this enables the bacteria to generate ATP by harnessing the energy liberated from the flux of electrons from electron donor to electron acceptor.

1.3 Carbon sourceThe carbon source for microbial growth is typically an organic compound and is often identical to the

electron donor. For example an aerobic bacterium growing by oxidising sugar to CO2 for ATP generation will also use the same sugar as a starting material for biomass synthesis. The sugar may be partly degraded via glycolysis to pyruvate or even to Acetyl-CoA to use parts of the TCA cycle for biomass synthesis purposes (compare to pathway of glutamate formation later in this text). The ratio of carbon that is used as energy source (catabolism, dissimilation) or as carbon source for biomass synthesis (anabolism, assimilation) determines the bacterial yield coefficient. Aerobic bacteria degrading a sugar use about 40 to 50% of the carbon for assimilation and the rest for energy generation. Hence a yield coefficient of 0.4 to 0.5 (g of microbial cells formed per g or carbon source used) is commonly observed.

In rare but interesting cases no formal carbon source needs to be provided to the growth medium. This is the case for autotrophic bacteria. Similar to plants and algae, autotrophic bacteria use CO2 as the carbon source which can be obtained from the air supply. However the additional supply of CO2 via a bicarbonate buffer (HCO3

- + H+ H2CO3 CO2 + H2O) helps the growth of autotrophs by minimising CO2 limiting conditions. Examples of autotrophic bacteria with industrial and environmental significance include: nitrifying bacteria that use NH3 as electron acceptor and Fe2+ and sulfur oxidising bacteria. These bacteria of the genera Nitrosomonas and Thiobacillus are used for nitrogen removal from wastewater and bioleaching of ores respectively.

1.4 Nitrogen sourceNitrogen is next to carbon, hydrogen and oxygen (the latter two are sourced from water) the

quantitative most important element. Nitrogen is needed for the synthesis of enzymes and other proteins. In minimal media it is typically supplied as ammonium or nitrate salts, in rich media it is supplied as an organic nitrogen source (e.g. part of yeast extract or peptone based media). With nitrate as the nitrogen source the bacteria need to first reduce the nitrate to ammonia (assimilative nitrate reduction) followed by ammonia assimilation into biomass.

Some bacteria are capable of using the relatively inert (triple bond) N2 from air as the nitrogen source. Hence selective media for such nitrogen fixing bacteria do not include any nitrogen source.

1.3 Phosphate source

The next most important element for microbial growth is phosphorous. In contrast to the other elements in biomass, phosphorus does not undergo oxidation or reduction and stays in its phosphate status (hence phosphate source). Phosphate needs to be present in all microbial growth media, without it growth cannot occur. Phospholipids and ATP are two examples of essential phosphate containing cell components. Rich media produced from biomass hydrolysates (e.g. yeast extract or peptone) contain organic phosphate compounds.

Many media use manifold more phosphate than necessary for biosynthetic purposes (e.g. about 50 mM). Here phosphate serves as the buffer species for pH control. While phosphate buffers are typically recommended to be added in two components, the more acidic phosphate (KH2PO4) and the more basic phosphate (K2HPO4) to result in a precise pH of the final media, it can also be provided by other phosphate sources followed by adjustment of the pH to a precise setpoint by any suitable acid or base. This pH adjustment will result in producing the same ratio of hydrogen and dihydrogen phosphate as suggested by the original recipe.

1.4 Sulfur source

Sulfur is needed for protein synthesis and hence essential to all growth media. For aerobic media sulfur is added as sulfate, which the bacteria reduce (assimilatory sulfate reduction) to sulfide prior to assimilation into amino acids. Alternative sulfur sources are sulfide (for anaeroboic media) or organic sulfur sources such as yeast extract or cysteine.

Components of Growth Medium:Energy source (electron donor and acceptor)C-source (e.g. sugar)N-source (e.g. NaNO3)P-source (e.g. KH2PO4)other minerals (e.g. Na+Mg+, SO42-,Trace elements (e.g. Co, Mn, Fe, etc)Vitamins (e.g. cyanocobalamin)Buffer (e.g. carbonate or phosphate buffer)

OUR- Growth medium for microbes

glucose

TCA cycle

ETC

ATPsynthase

1 ATP 3 H+

glucolysis

2 ATP + 12 NADH

each NADH 9 H+

Overall:38 ATPallowing growth

Cell

O2

Growth- Overview of Energy Metabolism=Dissimilation

simplifying FAD and ATP genration in TCA

Important Quantities:

ATP-synthase: 3H+ 1 ATP

ETC: 1 NADH 3*3 = 9 H+

2 NADH reduce 1 O2

1 NADH = 2 electron equivalents

1 O2 accepts 4 electron equivalents

glycolysis: 1 glucose 12 NADH

1 glucose 12*9=108 H+ = 36 ATP

+ 2 ATP from glycolysis via substrate level phosphorylation = 38 ATP

Growth- Simplified Scheme of Energy preservation as ATP

Minor corrections not needed for exams:

During TCA cycle not only NADH is produced but also FAD.

FAD translocates only 2 H+ rather than 3 hence 2 less ATP.

However TCA also generates 2 ATP not mentioned in

simplified balance.

Growth- Simplified Scheme of Energy preservation as ATP

Multiplication by binary fission:

0 min

30 min

60 min

1, 2, 4, 8, 16, 32 → exponential

Growth- Exponential

The resulting seqeunce in numbers is exponential ( 2, 4, 8, 16, 32)

Growth- Exponential(split… split… split )

The resulting sequence in numbers is exponential ( 1,2, 4, 8, 16, 32).

Not only the biomass (X) increases exponentially but also the rate at which it is produced (calculate from above)

growth rate is NOT constant in batch culture (similar to OTR not being constant) needed: a constant that describes the speed of binary fission (similar to kLa in oxygen transfer)

Plotting the growth rate as a function of time will reveal


Not only biomass (X) increases exponentially, but also the rate at which it is produced.

The proportionality factor is μ the specific growth rate

dX/dt = μ * X(g/L.h) (h-1) (g/L)

Thus: dX/dt X , dX/dt = factor • X˜

Xt = XoeμtIntegration gives

Take ln of both sides

μ indicates how much more biomass is producedper biomass present (g/h/g) = (h-1)

ln(Xt) = ln(Xo) + μt

μ = (In(Xt) – In(Xo))t


not examinable:dX/dt = μ * Xdx = u *n X * dtdx/X = u * dt∫dx/x = ∫ udt∫1/x * dx = ∫ udtlnx = ut +cx = e ut+c

x = e ut *ec

for t= 0: x=xo

xo = e ut *ec

Hence:x = e ut *xo

Growth- Estimation of u from single interval

= 0.023 min-1

= 1.3847 h-1

μ =InXt – ln Xo

t

=In 8 – In 2

90 min

=2.77 – 0.693

90 min

=90 min

2.077

Doubling time = =In 2

μ 0.69

1.38 h-1

= 0.5 h

When plotting the log of cell mass versus timea straight line is obtained.

The slope of the line revealsthe doubling time.

The specific growth ratecan be calculated from thedoubling time by:

Advantage of plot: averagingout, avoiding outliers

Growth- Estimation of doubling time from semilog plot

Growth in batch culture can not continue forever

Typical industrial growth curve incl.:

preparation time (clean, sterilise, fill)lag phaselog phasestationary phasedecay phase

X(g/L)

Time (h)

Growth- Limitation and growth phases

loglag

Growth- Limitation and growth phases

Most important to industry:productivity of the process (g.L-1.h-1).

Productivity is the overall product (here biomass X) concentration produced per time required.

The process can be stopped for maximum productivity or maximum product concentration (total productivity)

Choice depends on cost of operation and product

X(g/L)

Time (h)

maximum productivity

total productivity

Growth- Productivity in industrial batch cultures

In most environmental and many industrial bioprocesses (e.g. chemostat), the growth rate is limited by substrate availability.

Substrate uptake rate at different substrate concentrations is important (limitation and saturation of substrate)

Substrate(g/L)

Time (h)

substrate saturation

substrate limitation

Growth- Substrate Limitation

Substrate uptake rate at different substrate concentrations is important

v(g/L/h)

Substrate (g/L)

substrate saturation

substratelimitation

Growth- Substrate Limitation

What is the relationship between substrate concentration (S) and its uptake rate (v) ?

Described by Michaelis-Menten kinetics (standard biochemistry knowledge)

v = vmax -------kMS +S

v(h-1)

S (g/L)

substratelimitation

kM

vmax(h-1)

Growth- Michaelis Menten model


Michalis Menten Model: predicts substrate uptake fromsubstrate concentration

Monod Model: predicts specific growth rate from substrate concentration

Under substrate limitation: Substrate concentration

Substrate uptake rate (SUR) ATP production rate

rate of producing new cells (u)

Growth- Relationship between Michaelis Menten kinetics and and Monod kinetics


What is the relationship between substrate concentration (S) and its uptake rate (v) ?

Described by Michaelis-Menten kinetics (standard biochemistry knowledge)


µ(h-1)

S (g/L)

substratelimitation

kS

µmax(h-1)



µ = Y * vmax --------kSS +S

•What is the relationship between specific growth rate (µ) and specific substrate uptake rate (v)

•Relationship is given by the yield coefficient Y (g of X formed per g of S degraded).

•v= substrate uptake rate (SUR) but can also be OUR

•Note: unlike µmax and kS, Y is not a true growth constant.

kS and kM are equivalent

µ = Y * v


Growth- Relationship between Michaelis Menten kinetics and and Monod kinetics


The two curves are described by two properties:

The maximum specific growth rate obtained with no substrate limitation (umax (h-1))

and the half saturation constant (Michaelis Menten constat), giving the substrate concentratation at which half of the maximum u is reached (ks (g/L)).

Substrate limitation of microbial growth

µ(h-1)

Substrate (g/L)

substratelimitation

kS

µmax(h-1)



µ = µmax * ----------kSS +S

µ(h-1)

Substrate (g/L)kS

Typically there are low and high substrate specialists and

ecological “substrate niches” for the specialists to outcompete each other

kS



µ = µmax * ----------kSS +S

To be most competitive against other microbes a low ks value and a high umax value are important.

This simplified growth model only uses 2 out of 4 growth constants.

µ(h-1)

Substrate (g/L)

substratelimitation

kS

µmax(h-1)



µ = µmax * ----------kSS +S

µ(h-1)

Substrate (g/L)kS

There is also room for medium substrate “allround” specialists

kS



µ = µmax * ----------kSS +S

µ(h-1)

Substrate (g/L)kS

With the same ks the organism with a higher µmax will always win.



µ = µmax * ----------kSS +S

µ(h-1)

Substrate (g/L)kS

With the same same umax the organism with a lower ks will always win.


Conclusions – substrate limitation

• Substrate limitation slows down metabolism• Slowed metabolism slows growth (how? via Y!)• The limitation effect can be quantified (S/(S+kS))• The quantifier term has values between 0 and 1• e.g. if S=kS then u is half of umax• different microbes have different kS• competition between microbes is determined by kS and

umax

• What is missing -- maintenance, death, Ymax


Microbial GrowthComparison of μmax and kS for competition under

Substrate limitation

Which of the two growth constants influences to a larger extentThe growth of an organism under substrate limitation (substrateConcentration approaches zero)

Approach 1.

μ =μmax • Sks + S

For S approaching zero the μmax term approaches zero.Thus it appears that μ would be mainly influenced by kS(Textbook explanation).


Microbial GrowthComparison of μmax and kS for competition under

Substrate limitation

Approach 2.

Question is doubling of μmax (strain A) or halving of kS (strain B)having a larger effect on μ?

μ(A) =μmax(A) • Sks(A) + S

μ(B) =μmax(B) • Sks(B) + S

To compare growth rate of strain A and B: μ(A) = μ(B)

μmax(A) • Sks(A) + S

μmax(B) • Sks(B) + S

=

μmax(A)

ks(A) + Sμmax(B)

ks(B) + S=


21 + 0.1

10.5 + 0.1

=

1.82 > 1.67

At all substrate concentrations μmax is more important than kS


Microbial GrowthDependence of Biomass concentration on substrate used

(Yield Coefficient) - Intro

Final X in several batch cultures with increasing [S]

X (g/L)

Substrate Concentration (g/L)

Growth ceasedbecause of lackof substrate

Growth ceasedbecause ofendproductinhibition

Substrateinhibition

Growth- Yield Coefficient

Microbial GrowthDependence of Biomass concentration on substrate used

(Yield Coefficient) - Intro

X (g/L)

[S] (g/L)

In the absence of inhibition thebiomass formed is correlated tothe substrate used (X)

The correlation factor is theYield Coefficent (dimensionless, X/S)

Typical Y for aerobes on glucose: 0.4 to 0.5


Microbial GrowthYield Coefficient – Role in Establishing Correct Mass Balance

The biomass yield coefficient is essential to establish acomplete mass balance in a fermentation:

E.g. Substrate + Oxygen → Products + Biomass

The empirical formula for biomass must be known:

CH1.8O0.5N0.2


E.g. Gluconate degradation by Klebsiella

a. By resting Cells (non growing): Ideal biocatalyst

1 gluconate → 1.5 acetate + 0.5 ethanol + 2 formate

b. By growing cells:

1 gluconate + 0.174 NH3 + 0.04 H2O →1.4 acetate + 0.3 ethanol + 1.7 formate + 0.87 CH1.8O0.5N0.2 Thus: Growing cells incorporate 14.5 % of carbon fromGluconate into cell growth resulting in increasedacetate/ethanol ratio.


YS =X (g/L)S (g/L)

X (g/L)S (mol/L)

YO2 =X (g/L)

O2 (mol/L)

Microbial GrowthSignificance of Special Yield Coefficients

- Only works for same substrate, pathway

Molar yield coefficient

-Works only for aerobes and for sameATP/O2

+ works for unknown or complex substrates(e.g. cornsteep liquor, wastes


YN, YP =X (g/L)Mole of N or P

YkJ =X (g/L)kJ of heat of combustion

Ye =X (g/L)

Mole of reducingequivalents respired

Microbial GrowthSignificance of Special Yield Coefficients

Similar to YO2 but worksalso for other electronacceptors

Works also for fermentingBacteria, and unknownSubstrates and pathways

For scientific purposes underN or P limitation


Microbial GrowthYATP

YATP =X (g/L)

Mole of ATP

+ completely comparable between different physiological types+ can compare efficiency of growth for aerobes and anaerobes- requires knowledge about how much ATP is gained

Note: Moles of ATP generated can be estimated for many pathways:e.g. glycolysis to pyruvate 2 ATPconsumption of 1 mole of O2 ~ 2 NADH ~ 4 electrons 6 ATP

For rich media where all cell building blocks are provided (e.g. Yeast Extract): YATP = 10.5 g/mol


Comparison of YS and YATP for glucose fermenting bacteria

Microbial GrowthYATP

Organism YS ATP Yield(mol ATP/mol substance)

YATPgX/mol ATP

Streptococcus lactisLactobacillus plantarum

Saccharomyces cerevisiaeZymomonas mobilis

Aerobacter aerogenesE. coli

19.518.5

18.59

2926

22

21

33

9.89.4

9

8.69.6

9.4

The literature valuefor YATP is given as 10.5 g biomass/ molATP (Baushop and Elsden 1960)


Microbial GrowthCalculation and Inconsistencies of YATP

ATP gained per mole of substrate can be estimated forbacteria growing in rich media from Ys if the YATP is known(e.g. 10.5 g/ATP)E.g. If Ys = 21 g/mole of substrate → about 2 mole of ATP generated per mol of substrate

Although the YATP is more consistent than any other way ofexpressing the yield coefficient it can also vary:1.Not constant for all microbes (4.7 to 21) in rich media2.Experimental YATP < theoretical YATP (30 g/mol ATP)3.Low YATP on minimal media4.YATP dependent on growth conditions (ease of life)5.↑ temperature → ↓ YATP (thermal denaturation of proteins)6.Unsuitable growth conditions → ↓ YATP

7.Likely Reason for less cells formed: Higher ATP usage for cell maintenance rather than cell growth Maintenance coefficient


Microbial GrowthMaintenance Coefficient

1 mole of ATP generated during catabolism allows• theoretically → synthesis of 32 g cells• in praxis → 10.5 g cellsThe maintenance coefficient (ms) is the reason for2/3 being “wasted”1.Substrate transport into cell (e.g. against diffusion gradient)2.Osmotic work3.Motility4.Intracellular pH5.Replacement of thermally denatured proteins (↑ T → ↑ ms)6.Leakage of H+ ions across membrane (uncoupling)

ms influences Y, μ and the metabolic activity of the cells andis thus important to be considered in bioprocesses.

ms =S (mg)

X (mg) • time (h)

Growth- maintenenace

Effect of Maintenance Coefficient on Growth Rate

What is maintenance coefficient?The energy supply rate needed to maintain the life functions of a non growing cell.

Units?strictly speaking: mol ATP/ cell/ hmostly used: g substrate / g biomass / h = (h-1)


What does the maintenance coefficient (mS) affect?

• ms is the reason for Y not being constant. ms Y hence”: ms u (compare slide) Why? Because some substrate is taken up just for maintenance, not for growth.

• Effect is more apparent in slow growing cultures than in fast growing cultures.

• Slow growing cultures can have a very low Y .

mS (gS/gX/h) * Ymax (gX/gS) = Decay rate (h-1)


Effect of Maintenance Coefficient on Growth Rate

Effect of maintenance coefficient on growth rate

Effect of mS on Y?

Y is the observed yield coefficient.

The maximum yield coefficient Ymax is approached only when u = umax

Ymax is one of four growth constants

Y(gX/gS))

Specific Growth rate u (h-1)

Ymax



mS and Ymax can be combined:

mS (gS/gX/h) * Ymax (gX/gS) = Decay rate (h-1)

The Pirt equation of growth includes all four growth constants:

µ = µmax * -------- - mS *YmaxS + kS

S



µ = Ymax * vmax * -------- - mS *YmaxS + kS

S


u = ( vmax * -------- - mS) * Ymax S + kS

S

u = (v – mS) *Ymax

substrate uptake

µ = µmax * -------- - mS *YmaxS + kS

S

ms the respiration activity used to just stay alive

Effect of Maintenance Coefficient (mS) on growth Rate

µ = µmax * ----------kSS +Sµ

(h-1)

S(g/L) including mS0

µ = µmax * -------- - mS *YmaxS + kS

S

ignoring mS

µ(h-1)

S(g/L)

Sm = critical substrate concentration growth is zero

Sm

0

sm= ------------------umax- ms.Ymax

mS.kS.Ymax



µ(h-1)

S(g/L)

The negative specific growth rate (µ) observed in the absence of substrate(when S = 0) (cells are starving, causing loss of biomass over time)

is the decay rate mS*Ymax

- mS*Ymax

0


Points about ms

ms more heat produced (e.g. uncoupler)

• when S= 0 and u is negative (decay rate) then any oxygen uptake is via endogenous respiration.


How to obtain microbial growth constants from chemostat runs

Formulae used to determine ms and Ymax

1 ms 1Y D Ymax--- = --- + ------

µ = µmax * -------- - mS *YmaxS + kS

S


Y = --- + ------


Formulae used to determine ms and Ymax

1 ms 1Y D Ymax--- = --- + ------

Y = Ymax + u/ms Ymax = Y - u/ms

ms 1 D Ymax

_____1_______


Chemostat 1

How to increase growth of microbial culture?

1. Increase initial substrate concentration problem: substrate inhibition

2. Add more substrate during growth (Fed-batch) problem: endproduct inhibition

3. Replace old medium including endproducts by fresh medium at given intervals (semi continuous)

4. Automate semicontinuous culture by applying constant inflow and outflow (continuous culture, chemostat)

Transition from Batch to Chemostat 1. 1 Drop of Inflow of feed medium increases X by a small amount2. 1 Drop of reactor volume flows out removes biomass (X)3. If the Feed drop allows more X to grow than is taken out by the

reactor drop X in reactor will increase slightly4. Repeat the loop and there will be gradual increase in X

5. However, as the X increases each drop of outflow will contain more X which is removed from the reactor rate of out put increases

6. as a consequence the biomass will reach a certain steady state level.

7. The flow rate can be varied but has little effect on the level of biomass concentration.

8. The specific flowrate (i.e. flowrate (L/h) per reactor volume (L) is called the Dilution rate

Assumptions: Reactor volume stays constant(Use learning activity on chemostat to visualise how it works)(use bioprosim simulation to demonstrate chemostat versus batch)(Could use a modelling spreadsheet to demonstrate the steady state)

Effect of Increasing the Dilution rate1. After a steady state is reached from continuously adding and

removing drops (continuous dilution rate) a steady state is reached consisting of a constant high biomass level (X) and a very low substrate level (S) because the biomass degrades the substrate down to limiting concentrations.

2. If the dilution rate is doubled :33 More substrate per time is supplied33 bacteria grow faster33 bacteria are washed out faster33 steady state is reached again with a

• slightly higher substrate steady state concentration,• slightly lower biomass level • bacteria growth rate being twice as high• bacteria growth rate u compensating for the Dilution rate D

Chemostat 4Effect of loading rate increase (R = X * D)

X and μ can be set independently, X by SR and μ by D

1Double F and D doubles μAlmost doubles RAffects X only a little

Double SR doubles Xdoubles RAffects μ only a little

SR = 2 g/L h

SR = 2 g/L

h

hh

SR = 2 g/LSR =

2 g/L

SR = 4 g/L

Effect of Dilution rate on chemostat steady state concentrationsX= biomass, S= substrate,

SR= substrate in Reservoir R=productivity

X

S

D

Ste

ady

Sta

te C

once

ntra

tion Key features of steady

states 1:

•Inflow rate = Outflow rate

•Dilution rate= specific growth rate u

•S limitation of growth

•X stays constant over wide range of D

•If D approaches umax washout (Dcrit)

•Beyond (Dcrit) S = SR

R

Dcrit

SR



X

S

D

Ste

ady

Sta

te C

once

ntra


states 2:

•Open system, time factor excluded

•Allows to study microbial behaviour at constant growth rate

•µ of culture can be controlled by changing D

D S µ but not X (because of washout)

•How can X be controlled?

R

Dcrit

SR



X

S

D

Ste

ady

Sta

te C

once

ntra


states 3:

•X can be controlled by SR (dotted line using more dilute feed)

•Doubling SR doubling of X and of R (to a point)

•The level of X also depends on Y

R

Dcrit

SR



D

Key features of steady states 4:

•R = X * D•g/L/h = g/L * h-1

•R is largely a function of D until washout occurs

•R can be increased by:

D µ but not X

SR X but not u

•Such control does not exist in batch culture.

Dcrit

X

SSte

ady

Sta

te C

once

ntra

tion

R

SR

Productivity R :

•R = X * D•g/L/h = g/L * h-1

•As R =D*X and u=D

•R can be increased byD, Y, SRkS

•note: ms is ignored for high D and Y is assumed to be ~Ymax

•Where is the max chemostat productivity?

R = D.Y --------------µmax -DSR – kS

.D

X

S

D

Ste

ady

Sta

te C

once

ntra

tion

R

Dcrit

SR

X

S

D

Ste

ady

Sta

te C

once

ntra

tion Max chemostat

productivy (Rm):

•D at which R= Rm is called Dm

•Dm can be calculated from growth constants:

•Rm = X* Dm

R

Dcrit

SR

Dm = µmax. (1- kS/(SR- kS)

Rm

Note: at Dm some substrate wastage occurs Less conversion efficiency than at lower DOperator can aim for: high productivity or high conversion efficiency (of SX)Dm is dangerously close to Dcrit (especially when ks is low)

Chemostat productivity

•no lag phase,

•no preparation phase

•but only highly concentrated cell suspensions with close to exponential growth

•very high productivity compared to batch total or maximum productivity.

SR

batchmaximum productivity

batchtotal productivity

X(g/L)

Time (h)

chemo-stat

Comparison of productivity batch vs chemostat:•Chemostats:

•no lag phase,

•no preparation phase

•but only highly concentrated cell suspensions with close to exponential growth

•very high productivity compared to batch total or maximum productivity.

SR

batchmaximum productivity

batchtotal productivity

X(g/L)

Time (h)

chemo-stat

Effect of Dilution rate on chemostat steady state concentrations

X= biomass, S= substrate, R=productivity

X S

D

Ste

ady

Sta

te C

once

ntra


states 2:

•R is largely a function of D until washout occurs

•µ of culture can be controlled by changing D

•X can be controlled by changing SR (

R

Dcrit

Chemostat 8Productivity (R) –

Comparison Chemostat / Batch Culture

Batch culture

R = Xmax - Xo

tprep + tlog

R = DX (1 - tprep / tlog) Chemostat

• for long runs R = DX

t prep = preparation time since last run (cleaning, sterilizing, lag phase etc.)

t log = running time under full growth

The productivity of chemostats can be several times higher than in batch culture

Chemostat 10

The classical chemostat theory is based on the concept of substrate limitation.

In practice the oxygenation capacity of a bioreactor may become limiting. This results in the deviation of the classical chemostat.

Chemostat under oxygen limitation

X = Y (SR - )( +1 )D ksμmax -D

kLaD

1) Low D no O2 limitation

2) D O2 limitation μ<D X

3) X O2 limitation

S-limited

1

2

3

4

X

Oxygen limitation

D

S-limitedO2-limited

ChemostatThe Critical Dilution Point

The critical dilutino point Dc results in washout of biomass (X = 0, S = SR)

Dc = μmax SRKs + SR

Neglecting ms

μmax Dc ks DcSR little affect on Dc

μmax = + DIn Xt -In Xo

t

Dilution rates > μmax can be used to determine μmax:If D > μmax logarithmic biomass washout

μmax = (In Xt - In Xo) t + D

Washout kineticsFixed DD > μmax

X

Time

ChemostatAdvantages and disadvantages compared to batch

Chemostat+ prodctivity+ constant requirement for cooling, O2 transfer, labour, etc

Heat requirements for batch cultures

Heat HeatCool


1. Run a chemostat at equilibrium making sure that oxygen is not limiting

2. Note down the substrate in the reservoir (SR)

3. Calculate the Dilution rate from the Flowrate and the reactor Volume: D= F/V. check that units cancel to h-1

4. Note down the steady state values for S, X and DO

5. Determine the observed yield coefficient Y by dividing the biomass formed (X) by the amount of substrate degraded (SR-S)

6. Change the dilution rate by about 10% steps upward and downward. Wait for equilibrium each time.

7. Tabulate your values of X, S, SR, DO for each D in a spreadsheet


7. Tabulate your values of X, S, SR, DO for each D in a spreadsheet

8. Considering that at steady state u=D, produce additional columns that calculate 1/Y and 1/u

9. Plot 1/Y against 1/u and obtain the linear equation with the slope. Make sure you are clear about the units of both axes and the slope

10. Obtain the ms and Ymax constants from the slope and the intercept respectively

11. Alternatively using simultaneous equations allows you to calculate the constants from the equation


12. Tabulate the inverse of the substrate concentration present in the chemostat (S that determines u)

13. Tabulate 1/(u +ms*Ymax)14. Use the double inverse plot of the two items

above. This is similar to the Lineweaver Burk Plot

15. Read 1/u from the Y axis intercept and -1/ks from the x axis intercept

16. All 4 of the growth constants are now obtained

Effect Double inverse plot of Y and u from chemostat runs (u=D)

1/µ

1/y

Graphical determination of 2 growth constants from chemostat steady state data

mS

Dcrit

1/Ymax

Relationship between specific metabolic acitity (qS) and specific

growth rateqS = v = substrate uptake rate / X

µ

q SThe relationship between the rate of substrate uptake (= OUR) and growth rate

some activity will be used exclusively for maintenance without growth.

If qS is > ms growth will occur.

1/Ymax

ms

Microbial GrowthMs at [S] = 0

• Even if [S] = 0, some metabolism is required to stay alive.

• The substrate used from either energy stores (e.g. PßHB)or biomass itself.Thus:Net growth = total growth – biomass consumed

Endogenous Respiration

q

Time

+ SExogenous Respiration

For aerobic organisms the maintenance coefficient can bemeasured via the endogenous respiration rate.

e.g.SPOUR

• Formulas not to be used:

R = X* µmax * -------- - mS *YmaxS + kS

S

A chemostat is the preferred culture method …

• A chemostat is the preferred culturing method…• …when the process wants to select for the fasted growing strain

(single cell protein, degradation of pollutants)• …when substrate limitation (e.g. substrate toxicity) is desired• …contamination or mutation does not play a role (e.g. extreme

conditions of temperature, pH, etc.)• …for studying metabolic behavior at specified conditions (e.g. pH,

cell density, substrate concentration, product concentration, specific growth rate) (remember u and X can be set constant separately by selecting D and SR)

• … for studying effects of shocks (e.g. toxic substances) or minute disturbances of equilibrium (pH change, DO change)

Chemostats are not suitable for…

• production of recombinant products (tendency of backmutation to the wild strain “contamination from inside”)

• aseptic cultures with high tendency of contamination (continuous sterile supply of feed and harvest is difficult)

• where traditional methods play a role• where there is a need for changing conditions (e.g.

preventing respiratory deficient mutants in brewing, feast and famine regime

• the production of secondary metobolites (produced after growth)

Chemostat productivity (R = X*D) can be increased by

• production of recombinant products (tendency of back-mutation to the wild strain “contamination from inside”)

• aseptic cultures with high tendency of contamination (continuous sterile supply of feed and harvest is difficult)

• where traditional methods play a role• where there is a need for changing conditions (e.g.

preventing respiratory deficient mutants in brewing, feast and famine regime)

• the production of secondary metabolites (produced after growth)

Chemostat productivity (R = X*D) can be increased by

• maximising the dilution rate (D)• maximising the biomass concentration in the reactor (via

substrate in the feed (SR) or via organism with high Ymax)

• Highest biomass productivity (g of cells produced L-1 h-1)

highest product formation rate of primary metabolites highest substrate degradation rate highest oxygen uptake rate

ChemostatAdvantages and disadvantages compared to batch

+ less labour+ facilitates automation+ constant output+ easy for problem monitoring+ - selects for best growing strain- more susceptible to contamination from outside longer running times higher probability continuous sterilisation difficult- contamination from inside (back mutation to wild type)- breaking tradition of batch processes (e.g. beer)

ChemostatMain applications

1. Where the fastest growing strain is required, e.g. single cell protein (SCP), effluent, strain selection (e.g. higher affinity).

2. Generating functional understanding (not empirical observation) e.g. process optimisation, research.

2.1. Time is excluded every part of a batch process can be studied over extended periods. E.g. to investigate problem; set chemostat to exactly these conditions.

2.2. Permits to vary μ by changing D under otherwise constant conditions.

ChemostatMain applications

2.3. Study of environmental changes (pH, temperature, salinity, [S] etc.) at constant μ

2.4. Allows to study growth and metabolism under substrate limitation (ecological studies, pollution, waste treatment)

2.5. Allows to study effect of 1 parameter only

3. Batteries of small research chemostats to complement to sophisticated industrial batch reactor

Why not continuous beer brewing?

1. Contamination from inside and outside (e.g. respiratory deficient mutants (RMD))

2. Start up time until state gives low quality beer

3. Although HRT is 5 to 10 times less than for batch process, only small economic benefit is achieved in comparison with the cleaning condition and packaging of the product.

ChemostatBiomass Feedback

Justification for biomass feedback• High [X] is essential for high metabolic activity

-ds/dt = μ X/Y

• Usually X can be controlled by SR in reaction

• For waste water treatment (e.g. activated sludge) SR can not be controlled and is very low

• Biomass feedback can keep X high, S extremely low and achieve high R

Idea

Prevent X from being washed out

Technique

Physically retain or return biomass resulting in longer biomass (solid) retention time SRT than liquid retentiontime (HRT)

Options

• Immobilization of cells (e.g. fixed bed reactors, fluidised bed reactors, rotating disk contractor (RDC))

• Internal recycle

• Eternal recycle

ChemostatBiomass Feedback 2

The most simple idea to retain biomass: Filter

Because the filter clogging (fouling) simple bacterial filters cannot be usedCross flow filtration, and filter capillaries are used for the separation of expensive products

Feed

ChemostatFilter

Outflow


Biomass feedback by internal sedimentation of biomass

→ Flocculation necessary (Stroke’s Law)


External biomass feedback in activated sludge treatment

Biomass feed back: Ú X, Ú R, Ú Dc, Æ S proportionally to the efficiency of feedback

Feed

Reactor

Biomass Recycle

Outflow

Settler

Retaining biomass in a chemostat

• Preventing biomass from washout• allows buildup of higher biomass concentration

• when is this useful?• When feed concentration (SR) can not be concentrated

(eg. feed toxicity or fixed feed (e.g. waste water))• When only very little biomass growth will be obtained

(e.g. mineral media, autotrophs, extreme conditions such a bioleaching)

Retaining biomass in a chemostat

• How can biomass be retained?

• In theory a filter that allows liquid outflow but no biomass outflow works

• In practice: Filter fouling

• Alternatives (Cross flow filtration,

Inflow

Outflow

Practical Biomass retention inside the reactor

• Biofilm reactors• (a) fixed bed reactor

(trickle reactor)• (b) fluidised bed reactor• (c) sludge blanket reactor

(settling biomass flocs) • (e.g. sequencing batch

reactor (Feed-React-Settle-Decant

• Problems: mass transfer (e.g. oxygen), channeling

Inflow

Outflow

Practical Biomass retention via biomass feedback

• Centrifuging of recycle liquid

• Membrane filtration of recycle liquid

• Flocculation• Gravity settling of

flocculated biomass

Recycle(Feedback)

Inflow

Outflow

X

S

D

Ste

ady

Sta

te C

once

ntra

tion

Effect of biomass feedback (here 3 fold):Dotted line no feedback:•Washout occuring early

•3-fold Feedback approximately:•3*X 3*R 1/3* S•allows 1/3 reactor size to do same work

•Feedback essential for pollutant removal. Can be used 100-fold 100-fold smaller treatment plant

•Note: same assumed feed concentration (SR)

R

Dcrit

SR

Effects of growth constants on steady state concentrationsof biomass and substrate in a chemostat as a function of dilution rate (x-axis)

Effect of ms Effect of decrease ks

Effect of increased Y Effect of increased μmax

Documents

Microbial Growth – Overview of terms: exponential growth u td productivity