8 Questions 2.3 Minutes Per Question. N251: Mid-Semester Test 2000 Could Be About 40 Questions

@8 Questions 2.3 minutes per question. N251: Mid-Semester Test 2000 Could be about 40 questions with 1 min for each. Could be simple-medium-advanced for each topic. Topics could include: oxidation states. stoichiometry. kinetics. thermodynamics. water. sugar. fats. Would make 7 times 3 is 21. Say 3 on sugars and fats would be 11 times 3 is 33 looks OK. Other categories could include: Comprehension. Facts. Problems solving

@1 Oxygen Demand

1/Pm) The pollution strength of organic material is typically measured as the oxygen demand. What is the theoretical molar oxygen demand of butanol CH3-CH2-CH2-COOH (moles of O2needed for the complete oxidation to CO2 per mole of butanol)?
#5
Butanol represents 20 electron equivalents. 1 mole of oxygen accepts 4 electron equivalents. Hence 5 moles of oxygen are needed for the complete oxidation to CO2
6
No.
4
No.
20
No.
4.5
No.
6.2
No.
5.5
No.
10
No.

@2 Oxygen Demand

2/Pm) The pollution strength of organic material is typically measured as the oxygen demand. What is the theoretical molar O2 demand of oxalic acid COOH-COOH (moles of oxygen needed for its complete oxidation to CO2)?
#0.5
Oxalic acid consists of two carboxyl groups, each representing 1 electron equivalent. The two electrons can be accepted by o.5 mole of oxygen. In terms of oxidation states, the oxidation state of oxalic acid changes from +3 for each carbon to +4. The two electrons allow 0.5 mole of O2 to change its oxidation state from zero to -2 (both atoms).
1
No.
2
No.
4
No.
5
No.
6
No.
2.5
No.
1.5
No.

@3 Oxygen Demand

3/Pm) The pollution strength of organic material is typically measured as the oxygen demand. What is the theoretical molar oxygen demand of valeric acid, a saturated aliphatic carboxylic acid with 5 carbon atoms?
#6.5
Valeric acid (CH3-CH2-CH2-CH2-COOH) represents 7 + 3*6 + 1 = 26 electrons. 1 mole of oxygen accepts 4 electron equivalents. Hence 6.5 moles of oxygen are needed for the complete oxidation to CO2
6
No.
7
No.
5
No.
5.5
No.
7.5
No.
8
No.
10
No.

@4 Oxygen Demand

4/Pm) Sulfate reducing bacteria can oxidize ethanol (CH3-CH2OH) to acetic acid (CH3-COOH) using sulfate (SO42-) as the electron acceptor. How many moles of sulfate are being reduced to sulfide (H2S) per mole of ethanol converted to acetate ?
#0.5
The oxidation of ethanol to acetate donates 4 electrons (OS of alcohol group of –1 changes to OS of carboxyl group of +3). As the sulfur in sulfate changes its oxidation state from +6 to –2 it can accept 8 electrons. Hence 0.5 moles of sulfate are needed per mole of ethanol oxidized to acetate.
1
No.
1.5
No.
2
No.
2.5
No.
3
No.
3.5
No.
4
No.

@5 Oxygen Demand

5/Pm) Denitrifying bacteria using nitrate (NO3-) as the electron acceptor (reduction to N2). How many moles of a pentose (C5 sugar; (CH2O)5 ) can denitrifying bacteria oxidize per mole of nitrate reduced ?
#4
The N in nitrate changes its oxidation state from + 5 to zero (in the element N2) thus accepting 5 electrons. Each carbon group in the sugare has an oxidation state of zero (CH2O). Thus each carbon will give 4 electrons when oxidized to CO2 where the carbon oxidation state is +4. The five carbons in the sugar will donate 5* 4 = 20 electrons to 4moles of nitrate.
5
No.
6
No.
20
No.
4.5
No.
6.2
No.
5.5
No.
10
No.

@6 Oxygen Demand

6/Pm) The conversion of succinate (HOOC- CH2-CH2-COOH) to fumarate (H))C-CH=CH-COOH) is...
#...an oxidation donating 2 electrons.
The change from the CH2-CH2 group to the CH=CH group changes the oxidation state in both carbons from -2 to -1. Thus each carbon donates 1 electron in the oxidation of succinate to fumarate.
... a reduction accepting 2 electrons.
No.
...an oxidation accepting 2 electrons.
No.
...an oxidation donating 4 electrons.
No.
...a reduction accepting 1 electrons.
No.
...a reduction donating 4 electrons.
No.
...an oxidation accepting 4 electrons.
No.
...an oxidation donating 1 electrons.
No.

@7 Chemostat O2

7/Pm) Bacteria of the species Geobacter can respire ferric iron (Fe3+) to ferrous iron (Fe2+) rather then O2 to water. How many moles of ferric iron are needed to accept all electrons from glycerol (CH2OH-CHOH-CH2OH) oxidation to CO2 ?
#14
Glycerol represents 5+4+5 = 14 electrons. The oxidations states of each carbon change from -1 0 -1 to +4 +4 +4, again accounting for 14 electrons donated. As each ferric iron accepts only 1 electron to be reduced to ferrous iron, 14 are needed to accept electrons from glycerol.
12
Incorrect.
4
Incorrect
8
Incorrect.
16
Incorrect.
2
Incorrect.
10
Incorrect.
18
Incorrect.

@8 Chemostat O2

8/Pm) What kLa in h-1 (assume cS=8mg/L) is needed to allow a chemostat to operate at a dilution rate of 0.2 h-1 with a feed of 0.1 mol/L of methanol (CH3OH). The methanol should be completely oxidized to CO2. Ignore biomass production. The minimum oxygen concentration in the reactor should be 2 mg/L.
#160
The specific substrate input rate (also called loading rate) to the chemostat is given by 20 mmoles/L/h. This requires a molar oxygen input of 1.5 mol per mol of methanol input (6 electrons donated by methanol over 4 accepted by oxygen) equalling 30 mmoles/L/h = 960 mg/L/h as the OTR needed. This OTR should be achieved while maintaining 2 mg/L of D.O. kLa= OTR * (cS-cL) = 960 mg/L/h / (8-2 mg/L) = 160 h-1.
120
Incorrect.
64
Incorrect
200
Incorrect.
60
Incorrect.
80
Incorrect.
180
Incorrect.
50
Incorrect.
@3 Growth, specific growth rate
3/Pm) This conditions in this question are identical to the previous one. However this time you are asked to calculate the specific growth rate u (h-1) rather than the doubling time (h). What is the specific growth rate (h-1) of the culture if at time 5h the biomass concentration was 5g/L and at time 11h it was 20g/h.
#The specific growth rate is between 0.22 and 0.23 h-1.
Over the period of 6 hours (11-5h) the biomass doubled twice (from 5 to 10 g/L and from 10 to 20 g/L). Thus it doubles once every 3 hours, which represents the doubling time. By dividing ln 2 (about 0.693) by the doubling time (3h) the specific growth rate can be calculated (about 0.231 h-1).
The specific growth rate is between 0.11 and 0.12 h-1.
Incorrect.
The specific growth rate is between 1.1 and 1.2 h-1.
Incorrect
The specific growth rate is between 0.15 and 0.16 h-1.
Incorrect.
The specific growth rate is between 0.25 and 0.26 h-1.
Incorrect.
The specific growth rate is between 0.9 and 1.0 h-1.
Incorrect.
The specific growth rate is between 0.05 and 0.055 h-1.
Incorrect.
The specific growth rate is between 0.75 and 0.80 h-1.
Incorrect.
@4 Type of batch culture termination.
4/Cs) When the biomass and product concentration of a batch fermentation process is plotted as a function of time; the rate of product formation generally (i.e. primary metabolites) increases exponentially and proportional to the biomass concentration. However towards the end of the batch culture when the stationary phase is approached; the way the rates of biomass and product formation slow down indicates what became inhibiting to the organisms. A limitation by the energy/carbon source results in...
#...a sudden stop of biomass and product formation.
The energy source that is usually also the carbon source (such as a sugar) is typically depleted rather suddenly. This is due to the low kS values of microbes for the substrate. It means that they can keep taking up the substrate at maximum rates until the substrate concentration has reached very low levels (mg/L). As a result the sudden depletion of end products results in a sudden stop of growth and an equally sudden stop of the production of primary metabolites that are directly derived from the substrate.
...a gradual stop of biomass and product formation.
Incorrect. With no energy source there will be no growth and no formation of primary metabolites.
...a gradual stop of biomass formation and linear continuation of product formation.
Incorrect. With no energy source there will be no growth and no formation of primary metabolites.
...an immediate stop in product formation and a linear continuation of growth.
Incorrect. With no energy source there will be no growth and no formation of primary metabolites.
...a sudden stop in biomass formation and linear continuation of product formation.
Incorrect. With no energy source there will be no growth and no formation of primary metabolites
...a sudden stop in biomass formation and linear continuation of product formation.
Incorrect. With no energy source there will be no growth and no formation of primary metabolites.
Incorrect.
Incorrect.
@5 Growth constants
5) There are four growth constants that describe the growth of micro-organisms in bioreactors or the environment. These are:
#The maximum specific growth rate (umax); the half saturation constant (kS); the maintenance coefficient (mS) and the maximum yield coefficient (Ymax).
Incorrect.
The specific growth rate (u); the half saturation constant (kS); the maintenance coefficient (mS) and the maximum yield coefficient (Ymax).
Incorrect.
The specific growth rate (u); the maximum saturation constant (kmax); the maintenance coefficient (mS) and the saturation yield coefficient (yS).
Succinic Acid represents 12 reducing equivalents (electrons available) which requires 12/8 moles of oxygen.
The specific growth rate (u); the saturation constant (kS); the maintenance coefficient (mS) and the yield coefficient (yS).
Incorrect.
The metabolic quotient (qS), the oxygen saturation deficit (cS-cL), the maximum specific growth rate (umax), the half satauration constant (kS).
Incorrect.
The maximum specific growth rate (umax); the half saturation constant (kS); the maintenance coefficient (mS) and the metabolic coefficient (qS).
Incorrect.
The maximum specific growth rate (umax); the half saturation constant (kS); the oxygen uptake rate (OUR) and the metabolic coefficient (Ymax).
Incorrect.
Incorrect.
@6 ks value
6/Fs) One of the microbial growth constants is the kS value.
#It’s units are g/L. It provides the substrate concentration at which the specific growth rate is half of its maximum value (umax).
Yes. It is also called the half-saturation constant.
It’s units are h-1. It provides half of the maximum specific growth rate (umax).
Incorrect.
It’s units are g/L/h and provides half of the maximum growth rate.
Incorrect.
It is dimensionless. It provides the amount of substrate needed to produced 1 g of biomass.
Incorrect.
It’s units are g/L. It provides the minimum substrate concentration needed to maintain cellular activities.
Incorrect.
Incorrect.
Incorrect.
Incorrect.
@7 Competition
7) Compared to an organism with a low kS value; an organism with a high kS value will...
#...grow relatively slowly at low substrate concentrations.
Yes. A high kS value means that the organism has half of its maximum growth rate at a reasonably high substrate concentration. At low substrate concentrations (e.g. less than the kS value) the growth rate of the organism will be severely restricted.
...be very efficient in taking up low substrate concentrations.
Incorrect.
...produce substantially more biomass per g of substrate.
Incorrect. You relate to the yield coefficient.
...grow substantially faster at high substrate concentrations.
Incorrect.
...grow substantially slower at high substrate concentrations.
Incorrect.
...produce substantially less biomass per g of substrate.
Incorrect.
...grow relatively fast at high substrate concentrations.
No, the change of oxidation states of carbon atoms indicates a redox reaction.
...require less energy for maintenence.
No, the change of oxidation states of carbon atoms indicates a redox reaction.
@8 Monod calculation
8/Fa) According to the Monod model; what will be the specific growth rate (in h-1) of an organism with a kS value of 0.2 mg/L and the capability to grow with a maximum specific growth rate (umax) of 0.6 h-1 when the current substrate concentration is 0.1 mg/L?
#0.1
The Monod model predicts that the specific growth rate (u) is determined by : u = umax *S / (kS +S). Here 0.6 h-1 * 0.1 mg/L / (0.2+0.1 mg/L) = 0.06 mg/L/h / 0.3 mg/L = 0.1 h-1.
0.01
Incorrect.
2
Incorrect.
0.5
Incorrect.
0.02
Incorrect.
0.2
Incorrect.
0.3
Incorrect.
0.12
Incorrect.
@8 Yield coefficient
8/Ps) Substrate inhibition in a batch culture becomes apparent by...
#...no growth occurring at all as the substrate concentration inhibits right from the beginning.
Yes. At the beginning the substrate concentration is highest. If it is at inhibitory levels no growth can occur.
...growth being linear rather than exponential.
Incorrect.
...growth occurring but when more substrate would be added this would not result in further growth.
Incorrect.
...a slowing down of exponential growth leading to stationary growth.
Incorrect.
...negative growth after the stationary phase leading to the decay phase.
Incorrect.
...a slightly lower final biomass concentration than without substrate inhibition.
Incorrect.
...a slower specific growth rate during exponential phase.
Incorrect.
Incorrect.
@9 ATP yield
9/Fs) During the fermentation of glucose in a rich medium via the Emden Meyerhof Parnass pathway of glycolysis an anaerobic bacterium can produce 2 moles of ATP per mol of glucose. How much biomass would you expect to be formed per mol of glucose ?
#20 -22 g.
Per mol of ATP 10.5 g of biomass can be formed (gro3). Hence per mol of glucose it would be twice that: 21 g.
10 - 11 g.
No, the water content is not the relevant parameter.
10 - 11 mg.
Incorrect.
21 mmol.
No. Biomass cannot be expressed in mmol; as it is not a pure substance.
10.5 mol.
No. Biomass cannot be expressed in mmol; as it is not a pure substance.
20 mol.
No. Biomass cannot be expressed in mmol; as it is not a pure substance.
55 g.
Incorrect.
12.5 g.
Incorrect.
@10 Lipds vs Carbohydrates
10/Cs) In comparison with carbohydrates …
#lipids are more reduced.
Correct.
lipids are more polar.
Incorrect.
lipids are more stable.
Incorrect.
lipids contain more oxygen atoms.
Incorrect.
lipids are more soluble
Incorrect.
contain less double bonds.
Incorrect.
contain more hydroxyl groups.
@11 Denitrification
11) In comparison to a glycosidic bond linking two sugar molecules the ester bond of fats
#forms between a triple alcohol and
Correct.
#contains a double bonded oxygen
No.
does not release water when being formed.
The ester bond forming between a fatty acid and an alcohol also releases water.
does not involve hydroxyl groups.
No, glycosidic and ester bonds involve the reaction with a hydroxyl group.
does not form a stable bond.
No.
always requires water as a reactant.
No.
can not be hydrolysed enzymatically.
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