@9 Questions 2 min per question.

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 Oxidation states

1/Pm) During the exponential growth of a bacterial batch culture the development of the following parameters over time is characteristic: biomass concentration X (g.L), the growth rate dX/dt (g/L/h) and the specific growth rate u (h-1). The characteristic development is that …
#...X and dX/dt increase exponentially while u (h-1) stays constant.
During exponential it is not only the biomass concentration (g/L) that increases exponentially but also the rate at which new biomass is formed (g/L/h). The reason for this is that the rate of new biomass formation depends linearly on the existing biomass concentration. As this increases exponentially, so must the rate of growth (g/L/h). If the exponentially increasing rate of growth is divided by the exponentially increasing biomass concentration, this will result in the specific growth rate (growth rate per amount of biomass present), which by cancelling g/L results in h-1 as its units. This scheme that the rate of the reaction is proportional to the concentration of a substrate or product is described by first order kinetics and is of fundamental importance for the understanding of bioprocesses.
...X and dX/dt increase exponentially while u increases linearly.
No.
...X increases exponentially while dX/dt and u increase linearly.
No.
...X increases exponentially while dX/dt and u remain constant.
No.
...X increases exponentially;dX/dt increases linearly and u remains constant.
No.
...X increases linearly while the dX/dt and u increase exponentially.
No.
No.
No.
@2 Growth, Doubling time
2/Pm)The doubling time of an exponentially growing batch culture can be determined using semi-logarithmic plots. Either the log of the biomass concentration is plotted over time or the biomass concentration is plotted against time on semi-logarithmic graph paper. What is the doubling time (h) of the culture if at time 5h the biomass concentration was 5g/L and at time 11h it was 20g/h.
#The doubling time is 3h.
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. This represents the doubling time.
The doubling time is 6h.
Incorrect.
The doubling time is 4h.
Incorrect
The doubling time is 2h.
Incorrect.
The doubling time is 6h.
Incorrect.
The doubling time is 12h.
Incorrect.
The doubling time is 25h.
Incorrect.
The doubling time is 5h.
Incorrect.
@3 Growth, specific growth rate
3/Pm) The 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).
Correct. Refer to study guide and CBLA.
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).
Incorrect.
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. It 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.
Incorrect.
...require less energy for maintenence.
Incorrect.
@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.