Growth - Overview
Microbial Growth –Overview of terms:
exponential growth
u
td
productivity
Substrate limitation of metabolism
Link between metabolism and growth
And 22
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
And 33
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
And 44
Growth - Overview
Microbial Growth III –
Maintenance coefficient
Microbial Growth IV
Growth in batch culture
Microbial Growth V
Chemostat
Microbial Growth VI
Determining growth constants
Biomass retention
And 55
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.
And 66
1.3 Carbon source
The 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 source
Nitrogen 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.
And 77
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.
And 88
OUR- Growth medium for microbes
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)
And 99
Growth- Overview of Energy Metabolism
=Dissimilation
simplifying FAD and ATP genration in TCA
glucose
glucolysis
TCA
cycle
2 ATP + 12 NADH
1 ATP  3 H+
Cell
ETC
each NADH  9 H+
O2
ATP
synthase
Overall:
38 ATP
allowing growth
And 1010
Growth- Simplified Scheme of Energy
preservation as ATP
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
And 1111
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.
And 1212
Growth- Exponential
Multiplication by binary fission:
0 min
30 min
60 min
1, 2, 4, 8, 16, 32 → exponential
And 1313
Growth- Exponential
(split… split… split )
The resulting seqeunce in numbers is exponential ( 2, 4, 8, 16, 32)
And 1414
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
And 1515
Growth- Exponential
(split… split… split )
Not only biomass (X) increases exponentially, but also the
rate at which it is produced.
Thus: dX/dt ˜ X , dX/dt = factor • X
The proportionality factor is μ the specific growth rate
dX/dt = μ * X
(g/L.h)
Integration gives
Take ln of both sides
(h-1) (g/L)
Xt = Xoeμt
ln(Xt) = ln(Xo) + μt
μ = (In(Xt) – In(Xo))
t
μ indicates how much more biomass is produced
And 1616
-1
per biomass present (g/h/g) = (h )
not examinable:
dX/dt = μ * X
dx = u *n X * dt
dx/X = u * dt
∫dx/x = ∫ udt
∫1/x * dx = ∫ udt
lnx = ut +c
x = e ut+c
x = e ut *ec
for t= 0: x=xo
xo = e ut *ec
Hence:
x = e ut *xo
17
Growth- Estimation of u from single interval
InXt – ln Xo
μ=
t
=
In 8 – In 2
90 min
2.77 – 0.693
=
90 min
=
2.077
90 min
= 0.023 min-1
= 1.3847 h-1
Doubling time =
In 2
μ
= 0.5 h
=
0.69
1.38 h-1
And 1818
Growth- Estimation of doubling time from
semilog plot
When plotting the log
of cell mass versus time
a straight line is obtained.
The slope of the line reveals
the doubling time.
The specific growth rate
can be calculated from the
doubling time by:
Advantage of plot: averaging
out, avoiding outliers
And 1919
Growth- Limitation and growth phases
Growth in batch
culture can not
continue forever
Typical industrial
growth curve incl.:
X
(g/L)
Time (h)
preparation time
(clean, sterilise, fill)
lag phase
log phase
stationary phase
decay phase
And 2020
Growth- Limitation and growth phases
lag
log
And 2121
Growth- Productivity in industrial batch cultures
Most important to
industry:productivity of the
process (g.L-1.h-1).
maximum
productivity
X
(g/L)
total
productivity
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)
Time (h)
Choice depends on cost
of operation and product
And 2222
Growth- Substrate Limitation
substrate saturation
Substrate
(g/L)
substrate limitation
Time (h)
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)
And 2323
Growth- Substrate Limitation
substrate
saturation
v
(g/L/h)
Substrate uptake rate at
different substrate
concentrations is
important
substrate
limitation
Substrate (g/L)
And 2424
Growth- Michaelis Menten model
What is the relationship
between substrate
concentration (S) and
its uptake rate (v) ?
vmax
(h-1)
v
(h-1)
v = vmax
substrate
limitation
kM
S (g/L)
S
------S + kM
Described by MichaelisMenten kinetics (standard
biochemistry knowledge)
And 2525
Growth- Michaelis Menten model
Growth- Relationship between Michaelis Menten
kinetics and and Monod kinetics
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)
And 2626
Growth- Michaelis Menten model
Growth- Michaelis Menten model
µ
(h-1)
What is the relationship
between substrate
concentration (S) and
its uptake rate (v) ?
µmax
(h-1)
v = vmax
substrate
limitation
kS
S (g/L)
S
------S + kM
Described by MichaelisMenten kinetics (standard
biochemistry knowledge)
And 2727
Growth- Michaelis Menten model
Growth- Relationship between Michaelis Menten
kinetics and and Monod kinetics
•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
v = vmax
S
------S + kM
µ=Y*v
S
µ = Y * vmax -------S + kS
And 2828
Growth- Michaelis Menten model
Substrate limitation of microbial growth
The two curves are described
by two properties:
µmax
(h-1)
The maximum specific growth
rate obtained with no substrate
limitation (umax (h-1))
µ
(h-1)
substrate
limitation
kS
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 (g/L)
And 2929
Growth- Michaelis Menten model
Substrate limitation of microbial growth
S
µ = µmax * ---------S + kS
µ
(h-1)
Typically there are low
and high substrate
specialists and
ecological “substrate
niches” for the specialists
to outcompete each other
kS kS
Substrate (g/L)
And 3030
Growth- Michaelis Menten model
Substrate limitation of microbial growth
S
µ = µmax * ---------S + kS
µmax
(h-1)
µ
(h-1)
substrate
limitation
kS
Substrate (g/L)
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.
And 3131
Growth- Michaelis Menten model
Substrate limitation of microbial growth
S
µ = µmax * ---------S + kS
µ
(h-1)
There is also room for
medium substrate
“allround” specialists
kS kS
Substrate (g/L)
And 3232
Growth- Michaelis Menten model
Substrate limitation of microbial growth
S
µ = µmax * ---------S + kS
µ
(h-1)
With the same ks the
organism with a higher
µmax will always win.
kS
Substrate (g/L)
And 3333
Growth- Michaelis Menten model
Substrate limitation of microbial growth
S
µ = µmax * ---------S + kS
µ
(h-1)
With the same same
umax the organism with a
lower ks will always win.
kS
Substrate (g/L)
And 3434
Growth- Michaelis Menten model
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
And 3535
Growth- Michaelis Menten model
Microbial Growth
Comparison of μmax and kS for competition under
Substrate limitation
Which of the two growth constants influences to a larger extent
The growth of an organism under substrate limitation (substrate
Concentration approaches zero)
Approach 1.
μmax • S
μ=
ks + S
For S approaching zero the μmax term approaches zero.
Thus it appears that μ would be mainly influenced by kS
(Textbook explanation).
And 3636
Growth- Michaelis Menten model
Microbial Growth
Comparison 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?
μmax(B) • S
μmax(A) • S
μ(B) =
μ(A) =
ks(B) + S
ks(A) + S
To compare growth rate of strain A and B: μ(A) = μ(B)
μmax(A) • S
ks(A) + S
=
μmax(B) • S
ks(B) + S
μmax(A)
ks(A) + S
=
μmax(B)
ks(B) + S
And 3737
Growth- Michaelis Menten model
2
1 + 0.1
=
1.82
>
1
0.5 + 0.1
1.67
At all substrate concentrations μmax is more important than kS
And 3838
Growth- Michaelis Menten model
Microbial Growth
Dependence 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 ceased Substrate
Growth ceased
inhibition
because of
because of lack
endproduct
of substrate
inhibition
And 3939
Growth- Yield Coefficient
Microbial Growth
Dependence of Biomass concentration on substrate used
(Yield Coefficient) - Intro
In the absence of inhibition the
biomass formed is correlated to
the substrate used (X)
X (g/L)
The correlation factor is the
Yield Coefficent (dimensionless, X/S)
[S] (g/L)
Typical Y for aerobes on glucose: 0.4 to 0.5
And 4040
Growth- Yield Coefficient
Microbial Growth
Yield Coefficient – Role in Establishing Correct Mass Balance
The biomass yield coefficient is essential to establish a
complete mass balance in a fermentation:
E.g. Substrate + Oxygen → Products + Biomass
The empirical formula for biomass must be known:
CH1.8O0.5N0.2
And 4141
Growth- Yield Coefficient
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 from
Gluconate into cell growth resulting in increased
acetate/ethanol ratio.
And 4242
Growth- Yield Coefficient
Microbial Growth
YS =
Significance of Special Yield Coefficients
X (g/L) - Only works for same substrate, pathway
S (g/L)
X (g/L)
S (mol/L)
Molar yield coefficient
-Works only for aerobes and for same
X (g/L)
ATP/O2
YO2 =
O2 (mol/L)
+ works for unknown or complex substrates
(e.g. cornsteep liquor, wastes
And 4343
Growth- Yield Coefficient
Microbial Growth
Significance of Special Yield Coefficients
Ye =
X (g/L)
Mole of reducing
equivalents respired
Similar to YO2 but works
also for other electron
acceptors
X (g/L)
YkJ =
kJ of heat of combustion
X (g/L)
YN, YP =
Mole of N or P
Works also for fermenting
Bacteria, and unknown
Substrates and pathways
For scientific purposes under
N or P limitation
And 4444
Growth- Yield Coefficient
Microbial Growth
YATP
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 ATP
consumption 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
And 4545
Growth- Yield Coefficient
Microbial Growth
YATP
Comparison of YS and YATP for glucose fermenting bacteria
YATP
ATP
Yield
YS
Organism
gX/
(mol ATP/
mol substance) mol ATP
Streptococcus lactis
19.5
2
9.8
Lactobacillus plantarum
18.5
2
9.4
Saccharomyces cerevisiae 18.5
Zymomonas mobilis
9
Aerobacter aerogenes
29
E. coli
26
2
1
3
3
9.4
9
9.6
8.6
The literature valuefor YATP is given as 10.5 g biomass/ mol
ATP (Baushop and Elsden 1960)
And 4646
Growth- Yield Coefficient
Microbial Growth
Calculation and Inconsistencies of YATP
ATP gained per mole of substrate can be estimated for
bacteria 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 of
expressing the yield coefficient it can also vary:
1. Not constant for all microbes (4.7 to 21) in rich media
2. Experimental YATP < theoretical YATP (30 g/mol ATP)
3. Low YATP on minimal media
4. 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
And 4747
Growth- Yield Coefficient
Microbial Growth
Maintenance Coefficient
1 mole of ATP generated during catabolism allows
• theoretically → synthesis of 32 g cells
• in praxis
→
10.5 g cells
The maintenance coefficient (ms) is the reason for
2/3 being “wasted”
1. Substrate transport into cell (e.g. against diffusion gradient)
2. Osmotic work
3. Motility
4. Intracellular pH
5. Replacement of thermally denatured proteins (↑ T → ↑ ms)
6. Leakage of H+ ions across membrane (uncoupling)
S (mg)
ms =
X (mg) • time (h)
ms influences Y, μ and the metabolic activity of the cells and
And 4848
is thus important to be considered in bioprocesses.
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/ h
mostly used: g substrate / g biomass / h = (h-1)
And 4949
Growth- maintenenace
Effect of Maintenance Coefficient on Growth Rate
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)
And 5050
Growth- maintenenace
Effect of maintenance coefficient on growth
rate
Effect of mS on Y?
Ymax
Y
(gX/
gS))
Y is the observed
yield coefficient.
The maximum yield
coefficient Ymax is
approached only
when u = umax
Specific Growth rate u (h-1)
Ymax is one of four
growth constants
And 5151
Growth- maintenenace
Effect of maintenance coefficient on growth
rate
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:
S
µ = µmax * -------- mS *Ymax
S + kS
And 5252
Growth- maintenenace
Effect of maintenance coefficient on growth
rate
ms the
respiration
u = (v – mS) *Ymax
S
u = ( vmax * -------- - mS) * Ymax
S + kS
activity used
to just stay
alive
S
µ = Ymax * vmax * -------- - mS *Ymax
S + kS
S
µ = µmax * -------- - mS *Ymax
S + kS
substrate uptake
And 5353
Growth- maintenenace
Effect of Maintenance Coefficient (mS) on growth Rate
ignoring mS
S
µ = µmax * ---------S + kS
µ
(h-1)
0
S(g/L)
including mS
S
µ = µmax * -------- - mS *Ymax
S + kS
And 5454
Effect of Maintenance Coefficient (mS) on growth Rate
Sm = critical substrate
concentration  growth is
zero
µ
(h-1)
mS.kS.Ymax
sm= -----------------umax- ms.Ymax
0
S(g/L)
Sm
And 5555
Growth- maintenenace
Effect of Maintenance Coefficient (mS) on growth Rate
The negative specific
growth rate (µ) observed in
the absence of substrate
(when S = 0)
(cells are starving, causing
loss of biomass over time)
µ
(h-1)
0
S(g/L)
is the decay rate mS*Ymax
- mS*Ymax
And 5656
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.
And 5757
Growth- maintenenace
How to obtain microbial growth constants
from chemostat runs
S
µ = µmax * -------- mS *Ymax
S + kS
Formulae used to determine ms and Ymax
1
ms
1
--- = --- + -----Y
D
Ymax
And 5858
Growth- maintenenace
How to obtain microbial growth constants
from chemostat runs
Formulae used to determine ms and Ymax
1
ms
1
--- = --- + -----Y
D
Ymax
_____1_______
ms
1
Y = --- + -----D
Ymax
Ymax = Y - u/ms
Y = Ymax + u/ms
And 5959
Growth- maintenenace
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)
And 6060
Transition from Batch to Chemostat
1. 1 Drop of Inflow of feed medium  increases X by a small amount
2. 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 slightly
4. 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)
And 6161
Effect of Increasing the Dilution rate
1. 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 :
3.  More substrate per time is supplied
4.  bacteria grow faster
5.  bacteria are washed out faster
6.  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
And 6262
Chemostat 4
Effect of loading rate increase (R = X * D)
1
Double F 
and D 
doubles μ
Almost doubles R
Affects X only a little
SR =
2 g/L h
SR =
2 g/L
SR =
2 g/L
h
Double SR 
doubles X
doubles R
Affects μ only a little
SR =
2 g/L
h
SR =
4 g/L
h
AndD6363
X and μ can be set independently, X by SR and μ by
Steady State Concentration
Key features of steady
states 1:
SR
X
•Inflow rate = Outflow rate
•Dilution rate= specific
growth rate u
R
•S limitation of growth
•X stays constant over
wide range of D
S
D
Dcrit
Effect of Dilution rate on chemostat
steady state concentrations
X= biomass, S= substrate,
SR= substrate in Reservoir R=productivity
•If D approaches umax 
washout (Dcrit)
•Beyond (Dcrit) S = SR
And 6464
Steady State Concentration
Key features of steady
states 2:
SR
X
•Open system, time factor
excluded
•Allows to study microbial
behaviour at constant
growth rate
R
•µ of culture can be
controlled by changing D
S
D
Dcrit
Effect of Dilution rate on chemostat
steady state concentrations
X= biomass, S= substrate,
SR= substrate in Reservoir R=productivity
•D  S  µ but not
X (because of washout)
•How can X be
controlled?
And 6565
Steady State Concentration
Key features of steady
states 3:
SR
X
•X can be controlled by
SR (dotted line using
more dilute feed)
•Doubling SR doubling
of X and of R (to a point)
R
•The level of X also
depends on Y
S
D
Dcrit
Effect of Dilution rate on chemostat
steady state concentrations
X= biomass, S= substrate,
SR= substrate in Reservoir R=productivity
And 6666
Steady State Concentration
Key features of steady
states 4:
SR
X
•R
=X *D
•g/L/h = g/L * h-1
•R is largely a function of
D until washout occurs
R
•R can be increased by:
S
D
Dcrit
Effect of Dilution rate on chemostat
steady state concentrations
X= biomass, S= substrate,
SR= substrate in Reservoir R=productivity
•D   µ but not X
•SR  X but not  u
•Such control does not
exist in batch culture. And 6767
Steady State Concentration
Productivity R :
SR
•R
=X *D
•g/L/h = g/L * h-1
X
•As R =D*X and u=D
R
R=
S
D
Dcrit
D.Y
SR – kS.D
-------------µmax -D
•R can be increased by
•D, Y, SR
•kS
•note: ms is ignored for high D
and Y is assumed to be ~Ymax
•Where is the max chemostat
And 6868
productivity?
Steady State Concentration
Max chemostat
productivy (Rm):
SR
X
Rm
•D at which R= Rm is
called Dm
•Dm can be calculated
from growth constants:
R
Dm = µmax. (1-
kS/(SR- kS)
S
D
Dcrit
•Rm = X* Dm
Note: at Dm some substrate wastage occurs 
Less conversion efficiency than at lower D
Operator can aim for: high productivity or high conversion efficiency
(of SX)
And 6969
Dm is dangerously close to Dcrit (especially when ks is low)
X
(g/L) chemostat
batch
maximum
productivity
SR
Chemostat productivity
•no lag phase,
•no preparation phase
batch
total
productivity
Time (h)
•but only highly
concentrated cell
suspensions with close to
exponential growth 
•very high productivity
compared to batch total or
maximum productivity.
And 7070
X
(g/L) chemostat
batch
maximum
productivity
SR
batch
total
productivity
Time (h)
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.
And 7171
Steady State Concentration
Key features of steady
states 2:
X
S
•R is largely a function of
D until washout occurs
•µ of culture can be
controlled by changing D
R
•X can be controlled by
changing SR (
D
Dcrit
Effect of Dilution rate on chemostat
steady state concentrations
X= biomass, S= substrate, R=productivity
And 7272
Chemostat 8
Productivity (R) –
Comparison Chemostat / Batch Culture
Batch culture
R=
Xmax - Xo
tprep + tlog
t prep = preparation time since last run
(cleaning, sterilizing, lag phase etc.)
t log = running time under full growth
R = DX (1 - tprep / tlog)
Chemostat
• for long runs R = DX
The productivity of chemostats can be several times higher
than in batch culture
And 7373
Chemostat 10
Oxygen limitation
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.
S-limited O2-limited S-limited
Chemostat under oxygen limitation
X = Y (SR 1
D ks )( kLa +1 )
μmax -D
D
1) Low D  no O2 limitation
2
X
2)  D  O2 limitation  μ<D   X
3)  X   O2 limitation
3
4
D
And 7474
Chemostat
The Critical Dilution Point
The critical dilutino point Dc results in washout of biomass
(X = 0, S = SR)
 μmax

 Dc
μmax SR
Dc =
 ks

 Dc
Ks + SR
SR little affect on Dc
Neglecting ms
Dilution rates > μmax can be used to determine μmax:
If D > μmax  logarithmic biomass washout
In Xt -In Xo
μmax =
+D
t
μmax = (In Xt - In Xo) t + D
X
Washout kinetics
Fixed D
D > μmax
And 7575
Time
Chemostat
Advantages and disadvantages compared to batch
Chemostat
+  prodctivity
+ constant requirement for cooling, O2 transfer, labour, etc
Heat requirements for batch cultures
Heat
Cool
Heat
And 7676
How to obtain microbial growth constants
from chemostat runs
1.
2.
3.
4.
5.
6.
7.
Run a chemostat at equilibrium making sure that
oxygen is not limiting
Note down the substrate in the reservoir (SR)
Calculate the Dilution rate from the Flowrate and the
reactor Volume: D= F/V. check that units cancel to h-1
Note down the steady state values for S, X and DO
Determine the observed yield coefficient Y by dividing
the biomass formed (X) by the amount of substrate
degraded (SR-S)
Change the dilution rate by about 10% steps upward
and downward. Wait for equilibrium each time.
Tabulate your values of X, S, SR, DO for each D in a
spreadsheet
And 7777
How to obtain microbial growth
constants from chemostat runs
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
And 7878
How to obtain microbial growth
constants from chemostat runs
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
And 7979
1/y
Graphical determination
of 2 growth constants
from chemostat steady
state data
mS
1/Ymax
1/µ
Dcrit
Effect Double inverse plot of Y and u
from chemostat runs (u=D)
And 8080
The relationship between
the rate of substrate
uptake (= OUR) and
growth rate
qS
some activity will be used
exclusively for
maintenance without
growth.
1/Ymax
ms
If qS is > ms  growth
will occur.
µ
Relationship between specific
metabolic acitity (qS) and specific
growth rate
qS = v = substrate uptake rate / X
And 8181
Microbial Growth
Ms 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
+S
Exogenous Respiration
q
e.g.
SPOUR
Endogenous Respiration
Time
For aerobic organisms the maintenance coefficient can be
And 8282
measured via the endogenous respiration rate.
• Formulas not to be used:
S
R = X* µmax * -------- - mS *Ymax
S + kS
And 8383
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)
And 8484
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)
And 8585
Chemostat productivity (R = X*D)
can be increased by
• 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 metabolites (produced after
growth)
And 8686
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 h1)
•  highest product formation rate of primary metabolites
•  highest substrate degradation rate
•  highest oxygen uptake rate
And 8787
Chemostat
Advantages 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)
And 8888
Chemostat
Main 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.
And 8989
Chemostat
Main 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
And 9090
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.
And 9191
Chemostat
Biomass 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
And 9292
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
And 9393
Chemostat
Biomass Feedback 2
The most simple idea to retain biomass: Filter
Feed
Chemostat
Filter
Outflow
Because the filter clogging (fouling)
simple bacterial filters cannot be
used
Cross flow filtration, and filter
capillaries are used for the
separation of expensive products
And 9494
Chemostat
Biomass Feedback 2
Biomass feedback by internal sedimentation of biomass
→ Flocculation
necessary (Stroke’s Law)
And 9595
Chemostat
Biomass Feedback 3
External biomass feedback in activated sludge treatment
Feed
Outflow
Settler
Reactor
Biomass Recycle
Biomass feed back: Ú X, Ú R, Ú Dc, Æ S
proportionally to the efficiency of feedback
And 9696
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)
And 9797
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
And 9898
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-ReactSettle-Decant
• Problems: mass transfer
(e.g. oxygen), channeling
Inflow
Outflow
And 9999
Practical Biomass retention
via biomass feedback
• Centrifuging of recycle
liquid
• Membrane filtration of
recycle liquid
• Flocculation
• Gravity settling of
flocculated biomass
Inflow
Outflow
Recycle
(Feedback)
And 100100
Steady State Concentration
Effect of biomass
feedback (here 3 fold):
Dotted line no feedback:
•Washout occuring early
SR
X
•3-fold Feedback
approximately:
•3*X 3*R  1/3* S
•allows 1/3 reactor size to
do same work
R
S
D
Dcrit
•Feedback essential for
pollutant removal. Can be
used 100-fold  100-fold
smaller treatment plant
•Note: same assumed
feed concentration (SR)
And 101101
Effects of growth constants on steady state concentrations
of biomass and substrate in a chemostat as a function of dilution rate (xaxis)
Effect of ms
Effect of decrease ks
Effect of increased Y
Effect of increased μmax
And 102102