Download Bio301 final exam 2005 with model answers

Murdoch University
2005 Bio 301
Industrial and Environmental Microbiology
Final test.
Student No:………………………………
Student name:………………………………..
30 Total Marks
Section 1: 20 marks
Answer 10 out of the following 12 questions. Each question carries
2 marks
1.1. Oxygen transfer
The oxygen transfer from air to solution is affected by a number of factors, amongst
them are temperature and the amount of oxygen that is already dissolved in solution.
Explain in your own words how a change in temperature and a change in dissolved
oxygen concentration affects the oxygen transfer rate.
The oxygen transfer rate is determined by the kLa and oxygen saturation deficit
(OTR = kLa * (cS-cL).
Effect of change in temperature:
The temperature affects the oxygen solubility. With increasing temperature the solubility of oxygen
in water decreases. The oxygen solubility is given by the oxygen saturation concentration cS. As the
oxygen transfer rate is driven by the saturation deficit (the difference between current oxygen
concentration cL and the cS (CS-cL)) an increase in temperature would lower this driving force by
lowering the cS value. Hence one could expect for the same cL that the oxygen transfer rate
decreases at an increased temperature. However in line with Arrhenius law of reaction rates in
general being increased by temperature one would expect that the kLa value would increase with
increasing temperature. The combined effect of lowered solubility and increase kLa is difficult to
predict.
Effect of DO on oxygen transfer rate.
The saturation deficit (cS-cL) is the driving force for the
oxygen transfer rate (OTR = kLa * (cS-cL). In fact the
transfer rate of oxygen is proportional to the saturation
deficit and inverse proportional to the actual oxygen
concentration (cL), see figure below. If cL is at saturation
(cL=cS) then the OTR is zero, while it is at a maximum
level (called Standard OTR) when cL is zero.
OTR
cL
lefe
1.2. Oxygen uptake
Online monitoring of the oxygen uptake rate of aerobic processes is a very powerful
tool for process monitoring and process optimisation. Provide precise and complete
instructions to an operator of a bioreactor how to set up the reactor for online OUR
monitoring.
2005 Bio301 final test 1/11
To establish a reactor with online OUR monitoring the dissolved oxygen concentration is determined
by an online dissolved oxygen probe. The air-supply to the reactor must be constant such that an
oxygen steady state concentration (equilibrium concentration) is established. The kLa of the system
must be known to determine the OUR (which , under steady state equals the OTR) from the kLa,
the oxygen steady state concentration cL and the satuation concentration cS : OUR = OTR = kLa (cssL).The kLa can be obtained by a short interruption of the oxygen supply and monitoring the drop in
DO over time (OUR). This, typically linear drop represents the OUR of the steady state system. By
using the steady state DO concentration as cL the kLa can be obtained from KLa = OUR / (cS-cL)
Alternatively the kLa can be obtained beforehand with no culture in the reactor but using the same
aeration conditions and plotting an aeration curve after deoxygenation with sulfite catalysed by cobalt
ions. Here the kLa is obtained as the slope from a plot of the respective OTR at the different DO levels.
1.3. Growth competition
(a) List the four growth constants of microbes that allow the prediction of their
growth. (b) Give suitable units of the growth constants. (c) Explain for each growth
constant, whether a higher value or a lower value would prove a competitive
advantage against a competing organisms and why.
The four growth constants are:
maximum specific growth rate umax (h-1, or g cells produed per g cells present per h)
substrate half saturation constant ks or kM (g/L)
maintenance coefficient mS (h-1, or g of substrate used per g of cells per h)
maximum yield coefficient Ymax (dimensionless, or g of cells produced per g of substrate degraded)
The umax gives the maximum specific rate with which an organism can grow, when no substrate
limitation occurs. A higher umax value allows an organism to grow faster at any substrate
concentration and hence will be a competitive advantage
The ks value gives the substrate concentration at which half the maximum substrate uptake (or half the
maximum specific growth rate) is reached. A bacterium with a high kS value needs high substrate
concentrations to grow close to maximum specific growth rate, while an organism with a low kS value
can grow close to umax even at low substrate concentrations. Hence it is a competitive advantage to
have a lower not a higher ks value. At very high substrate concentrations the Ks value is not important
while it is critical for substrate limiting conditions such as in a chemostat.
the mS value gives the rate of specific substrate uptake that is necessary to obtain zero growth by just
maintaining the cell maintenance requirements. If the substrate concentration and hence the substrate
uptake rate is less than that, growth will be negative and the death rate is higher than the growth rate.
An organisms with a high maintenance requirement (mS value) will have a competitive disadvantage
because it grows slower due to the use of an increase fraction of the substrate taken up for maintenance
rather than growth.
The Ymax value gives the maximum yield coefficient observed when the culture is growing without
substrate limitation. An organism with a higher Ymax value will grow faster than an organism with a
lower Ymax value as it produces more biomass (and hence faster growth) for the same amount of
substrate consumed
.
2005 Bio301 final test 2/11
1.4. Chemostat
Plot the typical trends of specific growth rate, substrate concentration and productivity
of a chemostat as a function of (a) dilution rate (b) time. Indicate in your plot where
the critical dilution rate and the maximum productivity is.
Under chemostat operation the specific growth rate u is identical to the Dilution rate (see plot) until the
critical dilution rate is reached at which point further growth is not maintainable.
The productivity R increases initially linearly with the dilution rate, but with increasing D there is
undegraded substrate and hence less biomass present in the reactor deviating R from being proptional
to D.
The substrate concentration is typically very low but when approaching the critical dilution rate Dcrit
the substrate concentration tend to be that of the reactor inflow (reservoir)
As the chemostat is a steady state system with all variables being constant over time, u, R and S remain
constant (see fig).
Rmax
u
u
u
R
R
S
D
S
Dcrit
Time
1.5. Product formation, stoichiometry of bio-reactions
Explain the differences in primary and secondary metabolites. Which ones are suited
to chemostats and batch cultures and why ?
Primary metabolites are essential by-products from the energy metabolism and are always produced as
long as substrate is consumed and growth occurs. The production rate of primary metabolites is in
parallel with the specific growth rate. PM can be produced under chemostat or batch conditions.
Examples are ethanol or lactic acid.
Secondary metabolites are compounds that are not essential for metabolism and growth and are
typically produced after growth during the stationary phase (e.g. some antibiotics, or plant essential
oils). When plotting the specific product formation rate and the specific growth rate over time there are
two separate phases - the growth phase (tropho phase )and the product formation phase (idio phase).
Secondary metabolites can not be produced by chemostat cultures as the product is not formed during
growth while a chemostat by definition requires the microbes to grow, otherwise the are washed out.
SM can only be produced by batch cultures or sequential reactor setups.
1.6. Fermentation pathways
Outline two of the following glucose fermentation pathways: heterolactic acid
fermentation, propionic acid fermentation, butyric acid fermentation, butanol/acetone
fermentation, homoacetogenesis. Explain how they differ in terms of which electron
acceptor is used for reducing equivalents (NADH) from glycolysis and point out the
differences in ATP gain.
2005 Bio301 final test 3/11
1.7. Organic acid production
Explain the significance of anaplerotic sequences (replenishing the TCA cycle) in
some biotechnological processes. Give examples (bioprocess and substrate used).
Name the key enzymes involved.
The TCA cycle has acetate as the input and CO2 and NADH as the key outputs. In its cyclic operation
the C2 compound acetate fuses with oxalo-acetate (OAA) (C4) to form citrate (C6). Citrate is then
gradually oxidised and decarboxylated to form OAA again. If either citrate or other metabolites
between citrate and OAA are removed from the cycle for biosynthetic or other purposes the cycle
cannot continue because of a lack of OAA the starting and finishing compound.
Anaplerotic sequences are short biochemical paths that provide metabolites to the TCA cycle to
replenish it when TCA intermediates are removed from the cycle when they are excreted (e.g. citric
acid) or used for the production of other metabolites such as glutamate from alpha keto glutarate. In the
biotech production of citric acid, vast amounts of citrate are removed from the cycle and excreted into
the medium. In order to provide an alternative path for OAA formation, Pyruvate (10-3) is
carboxylated to OAA (10-4). The enzyme catalysing this reaction (pyruvate carboxylase) is the key
anaplerotic sequence for industrial citric formation. In much the same way does the malic enzyme
represent an anaplerotic sequence as it provides a precursor for OAA by reductive carboxylation of
pyruvate (10-3) to malate (12-4) which is then converted to OAA in the usual TCA way. The malic
enzyme is the responsible anaplerotic sequence in the biotechnological production of glutamate. If
pyruvate is not available (e.g. the substrate is acetate, ethanol, fats or hydrocarbons) resulting in the
formation of acetate but not pyruvate (beta-oxidation), then the glyoxylate cycle can function as an
anaplerotic sequence.
1.8. Nitrogen removal
Simultaneous nitrification and denitrification is a novel way or nitrogen removal from
wastewater. Explain what the difficulties are to have both processes running in
parallel and how SND can be accomplished.
Nitrification involves the aerobic oxidation of ammonia, the key inorganic nitrogen compound in waste
water to nitrate or nitrite. It is catalysed by strictly aerobic bacteria that use oxygen as the electron
donor for ammonium oxidation. Denitrification is an anaerobic process that relies on the oxidation of
organic material using nitrate or nitrite as the electron acceptor and reducing it to N2 gas. This reaction
is carried out by aerobic bacteria but in the absence of oxygen, as oxygen inhibits nitrate reduction and
is used as the preferred electron acceptor. To accomplish nitrogen removal from ammonia to N2 both
reactions need to operate. In traditional processes this is done sequentially by providing aerobic
conditions followed by anaerobic conditions. In the process of SND both reactions are aimed to be
operating at the same time. This requires the provision of sufficient oxygen to enable nitrification while
leaving the oxygen concentration low enough to avoid oxygen inhibition of denitrification. Via precise
DO control, or oxygen oscillations SND can be accomplished. It is believed that biofilms and bacterial
flocs enhance SND by providing micro-conditions or aerobic zones at the outside and anaerobic zones
at the inside.
1.9. Modelling
Explain in a few clear steps how the Michaelis-Menten model expressed by the
following equation: v = vmax * S/(kS +S) can be used in a spreadsheet to predict
(model) the change in S over time. v = actual reaction rate, vmax = maximum reaction rate, S =
substrate concentration, kS = substrate half saturation constant
2005 Bio301 final test 4/11
The rate equation is used to determine the current v from the current S and the constants ks and umax.
Then the S converted is determined by multiplying v (mg/L/sec) with the iteration interval chosen (a
short time of say 1 sec). Then the new substrate concentration at the end of the interval is determined
by subtracting the S converted from the previous S. This process is iteratively repeated, starting to
calculate a new rate from the new substrate concentration. By filling down cells, two or more data
columns can be generated to plot S versus time.
In detail:
Title 4 columns ”Time, “v”, “S deg”, and “S current”. The first data cell of each column represents
conditions before the reaction. Enter 0(time) , 0 (v), 0 (S deg) and 100 (S current).
Rate data.. Enter the rate equation v = vmax * S/(kS +S) into second data cell of column “v” and use as S
the substrate concentration from the first data line (100 mg/L).
Time data. In the second data cell of the “S deg” column, calculate the amount of substrate degraded
over the time interval (e.g. ti =1 sec) by multiplying the rate (mg/L/s) with the time interval (sec). In
the second data cell of the “S current” calculate the new substrate concentration by subtracting the
substrate degraded (cell 2 of “S deg”) from the previous substrate concentration (cell 1 of “S curr”)
Time data.In the “time” column, add the time interval for the overall simulation (1 sec) to the previous
time and fill down a column of data for time.
Iteratively Repeat the process. Fill down the commands (equations of the cells above with their relative
referencing to other cells) to produce a table with model data ready for plotting.
1.10. Process monitoring and control
A proportional- integral- differential controller (PID) can be used to control important
process parameters such as dissolved oxygen concentration or pH to a desired
setpoint. Explain how the three different elements (P, I, D) work and give at least one
advantage and one disadvantage of each element.
For proportional control the error (difference between set-point and actual value) is used to produce
control action, in proportion to the error size. A large error will cause a strong corrective action and a
small error will cause little action (e.g. increase in air flow rate). Exclusively proportional control can
work, but has the tendency to overshoot.
The differential controller exclusively considers the current trend of the signal (slope, obtained from
two measurements over an interval). If the trend is zero then it will initiate no change, even if the
current error is large. Differential control by itself is inadequate for process control.
A combination of P and D is a better controller as for example the trend towards the set-point plus the
current distance from the set-point allows predicting linearly when the set-point will be hit.
The integral controller considers the sum of errors (product of error size and duration of errors) over
the past and suggests corrective action on the basis of overall errors (either high or low) of the past. If
the sum of high and low errors of the past are equal, then no corrective action is taken independent how
far away from the set-point the current signal is. The combination of P and I can work for some
processes.
Problems:
P overshoots as it does not consider current trends
I inadequate control if the current error is large but the sum of previous errors about equals zero due to
high and low errors cancelling.
D inadequate control when current signal is constant but far away from setpoint
A time analogy for PID of present, past and future, respectively can be used to visualise the PID
controller.
2005 Bio301 final test 5/11
1.11. Biogeochemistry
Give 3 examples of microbial processes that use anaerobic bacterial respiration. Point
out
 the electron donor and electron acceptor of the reactions.
 the environmental significance of the process
 what specifically makes the process a “respiration”
1. Bacterial sulfate reduction
 Simple organic substances such as organic acids or alcohols  sulfate
 Sulfate reduction is generally and undesired process as its endproduct sulfide is toxic,
explosive, corrosive and malodorous. sulfate reducing bacteria can cause microbial
corrosion of steel by involving in the consumption of electrons via cathodic hydrogen from
the steel surface.
 It is a respiration because ETC is used for the transfer of electrons from electron carriers to
sulfate. In the process ATP is generated
2. Bacterial denitrification
 Organic substances  nitrate to N2
 Nitrate reduction is a desired process in the nutrient removal from waste water, converting
nitrate to harmless N2 gas. In agricultural soils nitrate reduction can deplete soils of fertilisers
and is an undesired process.
 It is a respiration because ETC is used for the transfer of electrons from electron carriers to
nitrate. In the process ATP is generated
3. Methane formation
 Hydrogen, acetate, methanol  to CO2 reduction to methane. More strictly speaking it is a
methyl respiration as only the last reduction step during. But it is also termed CO2 reduction
or CO2 respiration.
 It is a respiration because ETC is used for the transfer of electrons from electron carriers to the
co-factor F420 bonded the methyl group resulting in methane formation. In the process ATP is
generated.
 Methane is a valuable renewable fuel that can be harvested from anaerobic digestion of
organic wastes. It involves different groups of bacteria producing mainly acetate and hydrogen
and the methane producing bacteria as the terminal electron accepting species
4. Fumarate respiration
Propionic bacteria using lactate or sugar as electron donor and using fumarate formed during the
methyl-malonyl pathway as electron acceptor
2005 Bio301 final test 6/11
1.12. Methanogenesis, Anaerobic Digestion
Explain the significance of hydrogen production and consumption (interspecies
hydrogen transfer) in anaerobic digestion .
H2 : High turnover rate (30%) of all electrons. But low pool size (low concentrations).
H2 thermodynamically inhibits its own production from the
2005 Bio301 final test 7/11
Section 2: Short answer questions 4 marks
Answer 4 out of the following 6 questions:
2.1. Under which conditions are reducing equivalents (e.g. reduced electron carriers)
an energetic advantage or disadvantage for bacterial cells? Give examples.
advantage for respiring cells as they can make ATP via respiration and electron
transport chain
Disadvantage for fermenting bugs as they need to manufacture a suitable
electron acceptor (e.g. pyruvate in lactic acid fermentation, acetaldehyde in
ethanolic fermentation)
2.2 Competition. An organism with a maximum specific growth rate µmax of 0. 2 h-1
is cultivated under substrate limiting conditions (substrate concentration of 0.8 mM)
resulting in a much slower than maximum specific growth rate of 0.05 h-1 . What is
its substrate half saturation constant (kS) ?
u = umax * S /(KS+S)
0.05 h-1 = 0.2 h-1 * 0.8 mM / (Ks + 0.8 mM)
2.3 In a continuous culture bioprocess how can you control the biomass concentration
(X) and the specific growth rate (µ) respectively ?
X by the substrate in the feed (reservoir, SR)
u by the dilution rate ( flowrate/volume)
2005 Bio301 final test 8/11
2.4 A fermenting organism degrades 1 mole of glucose to 1 mole of acetate, 44.8 l of
hydrogen and unknown amounts of ethanol and CO2. Establish a stoichiometric
fermentation balance.
1 (CH2O)6 + 3 H2O  1 CH3-COO- + 2 H2 + CH3-CH2OH + 2 HCO3- + 3
H+
glucose has 24 electrons, acetate +2 H2 have 8 +4 electrons = 12 electrons. Hence
1 ethanol is missing for the full electron balance.
To balance carbon 2 mol of CO2 must be formed (bicarbonate)
To balance O add 3 water
To balance H add 3 H+
2.5 How much methane gas may theoretically be produced per mole of the following
substrates (assuming 100 % conversion efficiency): acetate, lactate, propionate,
butyrate, ethanol, glucose, butanol ?
1 mole of CH4 represents 8 electron equivalents
ac = 8/8 =1
lac=12/8=1.5
glucose = 24/8=3
`
2.6 Write comments on the significance of poly hydroxy alkanoates (PHA) such as
poly-hydroxy-buryate (PHB) in microbial bioprocesses. Point out the precursors for
synthesis of PHB.
2005 Bio301 final test 9/11
Section 3: Calculations 6 marks
Answer 2 out of the following 3 questions:
3.1 Bioenergetics
The anaerobic conversion of butyrate (CH3-CH2-CH2-COO-) into two moles of
acetate (CH3-COO-) and hydrogen gas is an important reaction but is energetically
close to the equilibrium. Calculate the driving force DG of this reaction when the
concentrations of reactants are:
butyrate (1 mM),
acetate (100 mM),
hydrogen partial pressure (100 ppm = 0.0001 bar)
The values of DeltaG of formation of the reactants are as follows: butyrate
-352.63 kJ/mol, acetate: 369.41 kJ/mol, H2: 0 kJ/mol, H+: 0 kJ/mol, H20:
237.178 kJ/mol
3.2 Growth constants calculation:A continuous aerobic process (chemostat) was
used to convert a medium containing 90 g/L of methanol CH3-CH2OH) into biomass
for Vitamin B12 production (35 mg B12/ g of dry biomass). The biomass
concentration and the residual substrate concentration in the reactor (27 L working
volume) were monitored as a function of the dilution rate D and are displayed below:
D
h-1
0.015
0.038
0.075
0.150
0.375
0.500
0.625
0.682
0.732
Acetate
g/L
0.01
0.02
0.05
0.11
0.42
0.84
2.10
4.21
16.89
Biomass
g/l
19.51
19.68
19.73
19.75
19.70
19.61
19.33
18.87
16.08
Determine the 4 growth constants (maximum specific growth rate (µmax), the
substrate half saturation constant kS, the maximum growth yield coefficient (Ymax)
and the maintenance coefficient (mS) of the organism used. Use the graph paper
supplied if needed. Do not forget the units when stating your results.
3.3 An aerobic chemostat is operated at an oxygen steady state concentration of 2.5
mg/L. After interruption of the airflow the DO decreased every second by 0.2 mg/L.
The reactor was run at room temperature so the oxygen saturation concentration was 8
mg/L. What is the current productivity of acetic acid (CH3-COOH) from this reactor
considering that ethanol (CH3-CH2OH) is the substrate? If the kLa of the reactor is
not changed, what is the maximum possible acetate production rate that this reactor
would allow, if the bacteria were much more active (e.g. by increasing the substrate
concentration in the feed inflow).
2005 Bio301 final test 10/11
OUR = 0.2 mg/sec = 720 mg/L/h = 720/32 mmol/L/h
ethanol (12 e) to aceate (8 e-) needs one mole of O2 hence ethanol conversion rate
is also 720/32 mmol/L/h
kLa = OUR / (cs-cL)
kLa = 720 mg/L (8 – 2.5 mg) = 720/5.5
Standard OTR = kLa * cs = (720/5.5 ) * 8 mg/L/h
2005 Bio301 final test 11/11