19 Questions N301: Mid-Semester Test
@19 Questions N301: Mid-Semester Test
@1 Oxygen transfer 21) The rate of oxygen transfer from air to water is linearly proportional to the…
air flow rate of the aeration equipment.
the transfer of oxygen will increase with the air flow rate but by no means proportionally.
stirring speed of the reactor.
the transfer of oxygen will increase with the stirring speed but by no means proportionally.
#oxygen saturation deficit (cS - cL) of the water.
The saturation deficit is the driving force for oxygen transfer. In fact the proportionality factor between oxygen transfer rate and saturation deficit is given by the kLa value.
actual oxygen concentration (cL).
On the contrary the higher the actual dissolved oxygen concentration, the slower the OTR.
the power input for aeration into the reactor.
the transfer of oxygen will increase with the power input for aeration speed but by no means proportionally.
absolute temperature (K).
There is a relationship between cS and temperature but not necessarily between OTR and temperature.
number of bacteria suspended.
Bacterial numbers don't affect the OTR in a systematic way.
optical density of the solution.
Optical density has no known effect on OTR.
@2 Electron balance 1
2) How many moles of oxygen are needed for the complete oxidation of ethanol to CO2 ?
1
Incorrect.
2
Incorrect.
#3
Ethanol represents 12 reducing equivalents (electrons available) which requires 12/4 moles of oxygen.
4
Incorrect.
1.25
Incorrect.
1.5
Incorrect.
2.5
Incorrect.
3.5
Incorrect.
@3 Electron balance 3
3) How many moles of oxygen are needed for the complete oxidation of malic acid (COOH-CH=CH-COOH) to CO2 ?
1
Incorrect.
2
Incorrect.
#3
Malic Acid represents 12 reducing equivalents (electrons available) which requires 12/4 moles of oxygen.
4
Incorrect.
1.25
Incorrect.
1.5
Incorrect.
2.5
Incorrect.
3.5
Incorrect.
@4 Electron balance 4
4) How many moles of nitrate (NO3-) (for ammonification to NH3) are needed to allow for the complete oxidation of succinic acid (COOH-CH=CH-COOH) to CO2 ?
1
Incorrect.
2
Incorrect.
#1.5
Succinic Acid represents 12 reducing equivalents (electrons available) which requires 12/8 moles of oxygen.
4
Incorrect.
1.25
Incorrect.
3
Incorrect.
2.5
Incorrect.
3.5
Incorrect.
@5 Electron balance 3
5) Assuming an oxygen saturation concentration of 8 mg/L, what is the maximally possible elemental sulfur oxidation rate (mmol/L/h) to sulfate (SO4--) of a bioreactor with a kLa of 80 h-1 ?
120
Incorrect.
100
Incorrect.
60
Incorrect.
200
Incorrect.
#240
Max OTR = kLa * cs= 160 mg/L/h = , S oxidation to SO4-- produces 6 electrons while O2 accepts 4 per mol. Thus 6 * 160 / 4 mmol of S can be oxidized.
125
Incorrect.
150
Incorrect.
270
Incorrect.
@6 Chemostat
6) A bioprocess operator can control the specific growth rate in a chemostat precisely by…
#the flowrate of medium provided the reactor volume is known.
The flowrate is proportional to the dilution rate provided the reactor volume is not changed and the dilution rate is identical to the specific growth rate. Hence, yes the flowrate allows to control the specific growth rate precisely.
the concentration of substrate in the reservoir.
No. This allows to change the concentration of biomass but not of the specific growth rate.
the air flow rate to the reactor.
Growth is not necessarily linked to the air flow rate.
the kLa.
Growth is not really related to the specific mass transfer coefficient.
the ks value.
The half saturation concentration is a growth constant of the organism and not a parameter that the operator can control.
using biomass feedback.
No, biomass feedback can prevent washout but does not allow adjusting the specific growth rate of the microbes.
determining the exact time of harvesting the cells.
No, a chemostat is not usually "harvested" at a certain time.
@7 Chemostat
7) A bioprocess operator can control the concentration of biomass in a typical chemostat precisely by…
#the concentration of substrate in the reservoir (feed vessel).
Over wide spreads of dilution rates the amount of biomass present at steady state in a chemostat is proportional to the amount of substrate in the reservoir.
the flowrate of medium provided the reactor volume is known.
No, the flowrate and dilution rate determine the growth rate but not the
the air flow rate to the reactor.
Growth is not necessarily linked to the air flow rate.
the kLa.
Growth is not really related to the specific mass transfer coefficient.
the ks value.
The half saturation concentration is a growth constant of the organism and not a parameter that the operator can control.
genetically engineered mutants
It is not established that this can control the amount of biomass present. Usually mutants produce less biomass than wild types.
determining the exact time of harvesting the cells.
No, a chemostat is not usually "harvested" at a certain time.
@8 Microbial Growth
8) Under some circumstances the specific growth rate of a microorganism can be negative. This can happen when …
#the maintenance requirement for growth (maintenance coefficient) is higher than the substrate flux available to the cell.
This negative growth is also called death rate and can be measured by decreases of biomass. Initially storage compounds are used up and cells start shrinking. Later on the cells die and become potential food for the surviving cells ('cells chewing themselves up')
the kS value is higher than the umax value.
These two values cannot be compared as their units do not correspond.
the dilution rate of a chemostat is higher than the maximum specific growth rate (umax).
This will result in washout of cells but not negative growth.
the umax value is higher than the kS value.
These two values cannot be compared as their units do not correspond.
the dilution rate of a chemostat is higher than the actual specific growth rate (u).
This is a non-steady state condition resulting in lowering of the biomass concentration and is not typical for chemostats.
the actual growth rate is less than half that of the maximum specific growth rate (kS).
These two values cannot be compared as their units do not correspond.
@9 Chemostat
9) Single out the only INCORRECT answer. Compared to batch cultures chemostats…
#are better suitable for industrial processes as there is a smaller risk of contamination.
The problems with contamination from inside (mutation) and outside to chemostat cultures are the main drawback.
cause bacteria to grow under substrate limitation.
In batch culture bacteria grow under substrate saturation
provide a higher productivity.
Because of less preparation, cleaning, filling and lag time
involve less labor for operation.
are more easily automated.
have got uniform temperature requirement.
are easy to monitor
product inhibition is unlikely to occur.
@10 Chemostat
10) In a chemostat …
#microbial growth is substrate limited.
Yes, as soon as more feed is added
the productivity increases exponentially with the dilution rate.
the critical dilution rate depends strongly on the microbial maintenance coefficient
the productivity is always proportional to the dilution rate.
the microbial half saturation constant (kS) is less important than in batch culture.
the production of secondary metabolites is particularly effective.
substrate inhibition problems are more likely than in batch culture.
product inhibition is unlikely to occur.
@11 Denitrification
11) Per mol of nitrate converted, the denitrification (conversion of NO3- to N2) reaction require a suitable electron…
#donor providing 5 electrons.
Yes, the nitrogen in nitrate changes its oxidation state from +5 to 0.
donor providing 4 electrons.
No.
donor providing 8 electrons.
No.
donor providing 6 electrons.
No.
acceptor accepting 4 electrons.
No.
acceptor accepting 6 electrons.
No.
acceptor accepting 8 electrons.
No.
acceptor accepting 5 electrons.
No.
@12 Chemostat
12) Biomass recycle or biomass feedback in a continuous culture (e.g. chemostat) has the following effects
It decreases the potential productivity.
Biomass feedback will build up higher levels of biomass at the same dilution rate. The productivity is the product of biomass and dilution rate (R= X*D). Hence a higher productivity is expected.
#It allows the operator to run the system at higher dilution rates and thus higher productivity.
Yes by retaining the microbes in the reactor at dilution rates D higher than the maximum specific growth rate (umax), a higher productivity can be achieved as R=D*X
It increases the substrate concentration in the reactor.
Yes, as there are more microbes in the reactor they will have a higher substrate uptake activity compared to reactors without feedback.
It increases the yield coefficient.
By retaining the microbes in the reactor for longer
It lowers the risk of contamination from outside and inside (mutation).
There is no established reason why keeping the microbes in the reactor for longer should reduce the risk of contamination.
It results in increased specific growth rate of the culture.
The growth rate of the organisms in substrate-limited cultures is largely determined by the substrate concentration. However, biomass feedback will actually lower the substrate concentration.
It optimises oxygen transfer to the system.
There is no clear relationship between biomass recycle and oxygen usage. If at all ther
@13 Yield coefficient
13) A chemostat steady state culture is fed with a 36 mM acetate (CH3-COO-) solution. The substrate steady state concentration in the reactor is 4 mM and the biomass concentration 4 g/L. What is the current molar yield coefficient ?
#125
The molar Y is the biomass formed per mol of substrate degraded. Substrate degraded is 0.036M - 0.004M = 0.032M. Thus 4g/ 0.032M =125 g/mol
100
No.
75
No.
16
No.
0.16
No.
12.5
No.
1.25
No.
64
@14 Maintenance coefficient
14) Select the only INCORRECT answer. The yield coefficient Y…
is a "growth constant" independent of the growth rate
Y.
is likely to increase with temperature because of protein denaturation.
Y.
#has the same units as the kS value.
kS value is a substrate concentration (g/L) at which half maximum growth is obtained. While the maintenance coefficient is the rate of substrate used per g of cells per time (units are time-1).
has a noticeable effect on chemostat cultures at very low dilution rates
No.
is responsible for negative growth under substrate starvation.
No.
is likely to be higher under extreme conditions (e.g. pH,).
This is because it costs energy to maintain a correct internal pH against a very different external pH.
is expected to be higher for actively moving cells than for passive or attached cells.
Movement costs energy which is no longer available for growth.
@15 kLa
15) Select the only INCORRECT answer. The kLa value
has the units 1 over time.
Y.
is the specific mass transfer coefficient for oxygen.
Y.
can be obtained mathematically from an aeration curve.
Y.
can be determined by monitoring the oxidation of a sulfite solution.
This is the so called sulfite oxidation technique.
can be determined if the OUR and the steady state oxygen concentration is known.
Y.
describes the aeration capacity of a bioreactor.
Y.
#is dependent of the dissolved oxygen concentration.
This is the only incorrect statement. It is the OTR that is dependent on the oxygen concentration.
@16 Competition
16) An organism with a medium kS and medium umax value…
#can find its "ecological niche" at medium substrate concentrations.
Y.
will always be out-competed by organisms with a lower kS value.
N.
will always be overgrown by organisms with a higher umax value.
N.
will be particularly competitive in batch cultures.
N.
can be determined if the OUR and the steady state oxygen concentration is known.
Y.
describes the aeration capacity of a bioreactor.
Y.
#is dependent of the dissolved oxygen concentration.
This is the only incorrect statement. It is the OTR that is dependent on the oxygen concentration.
@17 Growth
17) In a batch culture Proteus vulgaris grew at of 0.2 h-1 . In a 10 L chemostat at a flow rate of 200mL/h, P. vulgaris left 4 mg/L of substrate undegraded. Which of the values below is closest to its half saturation constant kS ?
#40
10
20
30
50
5
2.5
80
@18 Degradation
18) The full oxidation of benzene (simple non-substituted aromatic ring) to CO2 can reduce how much NAD+ to NADH ?
#15
The reduction of NAD+ to NADH requires 2 electrons. A benzene ring contains 6 carbons with alternating double/single bonds. Hence the sum formula is (CH)6. Each carbon has an oxidation state of -1 and will donate 5 electrons when oxidized to CO2 (oxidation state of carbon of +4). 5 * 6 = 30 electrons and thus 15 NADH.
10
N
20
N
12.5
N
14
N
28
N
8
N
12
N
@L18 Conversion rate
19) After stopping the airflow to a fed batch system converting propanol (CH3-CH2-CH2OH) into propanoic acid (CH3-CH2-COOH), the dissolved oxygen concentration dropped from 5 to 2 mg/L within 30 seconds. What is closest conversion rate of the alcohol to the organic acid, neglecting the biomass yield ?
#11.25 mmol/h
OUR measured was 6 mg/L/min = 360mg/L/h =11.25 mmol/L/h. Alcohol to carboxylic acid donates 4 electrons, thus 1 mol of alcohol converted per mol of oxygen. Thus 11.25 mmol/L/h are converted.
15.5 mmol/h
Incorrect.
9.75 mmol/h
Incorrect.
7.15 mmol/h
Incorrect.
2.4 mmol/h
Incorrect.
55.5 mmol/h
Incorrect.
28.2 mmol/h
Incorrect.
8.25 mmol/h
Incorrect.
@ Maintenance coefficient
20 The maintenance coefficient of a microbe…
#can be determined by obtaining values of biomass produced per substrate consumed at various growth rates.
plays a particularly important role under substrate saturation (i.e. batch).
#is constant at all growth rates but relatively more important at slow growth.
determines how much ATP can be produced per mol of a substrate.
is identical to the decay rate.
#is equal to the growth rate at the critical substrate concentration.
can be obtained from a double reciprocal plot of 1/growth rate versus 1/substrate.
is the minimum (or critical) oxygen requirement of aerobic microbes.
@8 Microbial Physiology
NADH…
is the principal energy source or microorganisms.
#is the reduced form of NAD+.
is the oxidized form of NAD+.
#carries 2 more electrons than NAD+.
carries 2 less electrons than NAD+.
#is a principal product from glycolysis of glucose to pyruvate.
is an energetic advantage to all organisms.
can only be produced under aerobic conditions.
@ Oxygen transfer + Stoichiometry
An aerobic microbial culture oxidizes glycerol (CH2OH-CHOH-CH2OH) to acetic acid (CH3-COOH) and CO2 at a constant rate of 2 mmol/glycerol/L/h in a chemostat. What oxygen mass transfer coefficient (kLa) in h-1 is needed for the supply of sufficient oxygen and to allow a minimum oxygen concentration (cL) of 2 mg/L. Assume that the oxygen saturation concentration (cS) is 8 mg/L.
#16
glycerol (14/3) to acetate (8/2) produces 6 electrons, thus needs 1.5 moles of oxygen. Thus needed are 3 mmol/oxygen/L/h = 96 mg/L/h kLa=96/(8-2mg/L) = 96/6
12
Incorrect.
8
Incorrect by a factor of 2.
32
Incorrect by a factor of 2.
40
Incorrect.
10
Incorrect.
20
Incorrect.
9
Incorrect.
@4 Electron balance 2
How many moles of oxygen are needed for the complete oxidation of acetic acid (CH3-COOH) to CO2 ?
1
Incorrect.
#2
Acetic acid represents 8 reducing equivalents (electrons available) which requires 8/4 moles of oxygen.
3
Incorrect.
4
Incorrect.
1.25
Incorrect.
1.5
Incorrect.
2.5
Incorrect.
3.5
Incorrect.
@ Oxygen transfer + Electron balance
In the production of acetic acid (CH3-COOH) from ethanol (CH3-CH2OH) the current oxygen steady state concentration of the reactor is 2 mg/L. What is the current acetic acid production rate (mmol/L/h) when the kLa of the system is 40 h-1 and the oxygen saturation concentration is 8 mg/L?
#7.5
OTR=kLa*(8-2), OTR=40*6, OUR=240 mg/L/h = 7.5 mmol/L/h of O2
2
Incorrect.
15
Incorrect.
2.5
Incorrect.
5
Incorrect.
1
Incorrect.
12
Incorrect.
10
Incorrect.
@13 Chemostat
The efficiency of chemostat processes can be improved by biomass recycle (biomass feedback). This is due to allowing
the build-up of increased biomass concentration.
operation at dilution rates higher than the maximum specific growth rate.
operation at very high dilution rates the buildup of increased biomass and vhigher biomass concentration and allowing very high dilution rates.
a more complete conversion of the substrate into product.
savings in aeration costs
@16 Yield coefficient
The yield coefficient of a microbe…
#can vary with the temperature, pH and other conditions,
determines the substrate concentration at which half of the full growth is achieved.
#would be strongly affected by respiration uncouplers.
is constant throughout a batch culture.
#can vary for different substrates.
#is relatively constant when comparing biomass produced per mol of ATP
#increases in general with the specific growth rate of the culture.
decreases with increasing growth rate of the culture.
@ Yield coefficient
The yield coefficient Y…
is a growth constant for a particular strain and does not depend on the growth rate.
Ymax, the maximum Yield coefficient is the growth constant.
#typically increases with the specific growth rate of the microbial culture.
as faster growing cultures require relatively less substrate for maintenance purposes.
typically decreases with the specific growth rate of the microbial culture.
No.
is independent of the medium composition
No.
is independent of the temperature
No.
is higher in chemostats than in batch culture.
because microbes don't need to spend as much energy on producing all the building blocks for new cells.
is typically lower at higher temperatures as cells need to counteract protein denaturation.
It costs the cells energy to counteract protein denaturation and this energy expenditure deviates reducing power (NADH) and ATP from cell synthesis.
depends on the pH value
Like other