Microbial Growth

Question 1

What factors can inhibit microbial growth?

Answer 1

Nutrient availability,
Toxin build up,
Quorum sensing- limits growth when population reaches its organisational quorum.

Question 2

Describe batch cultures.

Answer 2

This is a closed system culture of microorganisms that needs:
Specific nutrient types,
Temperature, pressure, aeration and other environmental conditions.
Only a few generations are allowed to grow before all nutrients are used up.
Batch cultures are the most simplest and most commonly used methods of fermentation.

Question 3

What are the two main batch culture methods?

Answer 3

Shake flasks,

Industrial batch reactors.

Question 4

What are the advantages and disadvantages of batch growth shake flasks?

Answer 4

The advantages are that it is easy to perform and has a relatively low cost. Also a large number of experiments can be run in parallel.
The disadvantages are that the experimental data can be difficult to interpret. Also the dynamic conditions used throughout the experiment can lead to experimental variability.

Question 5

What are the advantages and disadvantages of batch growth bioreactors?

Answer 5

The advantages are that the closed system means there is a low risk of contamination. Also certain conditions can be controlled, such as pH and oxygen availability.
The disadvantages are that dynamic conditions occur throughout the experiment which can lead to experimental variability.

Question 6

What are the phases of microbial growth?

Answer 6

Lag phase- in this phase the bacteria are adapting to the environment and to the nutrients they are surrounded with. The death of the microorganisms equals the growth rate.
Log phase- at this point there is active growth with the total number of cells doubling after each generation.
Stationary phase- at this rate the death rate equals the growth rate.
Death phase- at this point nearly all nutrients have been consumed and the pH is too low for survival so the death rate suppresses growth.

Question 7

What are the two main ways of measuring bacterial cell mass?

Answer 7

The two ways are indirect and direct.
Direct physical measurement of dry weight, wet weight or cell volume could be conducted. Or direct chemical measurement of a chemical component of the cells could be conducted.
Indirect measurement of chemical activity such as rate of oxygen production or consumption could be conducted.
Indirect optical density measurement.

Question 8

What methods could be used to measure bacterial cell numbers?

Answer 8

Direct microscope counts could be conducted using special slides known as counting chambers. One problem with this is that only dense concentrations can be counted.
Indirect viable cell counts, colonies counted are known as colony forming units with the number of these units being related to the viable number of bacteria present.
Indirect optical density measurements.

Question 9

How can the specific growth rate be determined from a graph?
What is specific growth rate?

Answer 9

The Cell growth phase is not linear but increases exponentially over time, therefore it cannot be considered as a rate of cells per hour. If a graph is plotted with loge of the cell concentration against time then a linear slope is created. The slope represents the specific growth rate.
Specific growth rate is the rate of change in biomass or cell concentration relative to biomass or cell concentration already present. Its units are hr^-1, min^-1.

Question 10

How is the specific growth rate related to doubling time?

Give two examples of organisms and their double times.

Answer 10

Td = (Loge2)/μ
∴μ = 0.693/Td
Escherichia coli: a doubling time of 17 min in glucose salts medium,
Streptococcus lactis: a doubling time of 26 min in milk medium.

Question 11

What is the Monod equation?

Answer 11

The Monod equation relates limiting nutrient concentration to a populations growth rate:
μ = μmax S/Ks + S
Where μ is the specific growth rate, μmax is the maximum specific growth rate, S is the substrate concentration and Ks is the Monod constant.

Question 12

What is Ks the measure of?

Answer 12

It is the Monod constant and it is the measurement of the affinity of an organism for a substrate.

Question 13

On a Lineweaver Burke plot what does the X intercept represent and what does the y intercept represent.
What is plotted on the y axis in this plot and what is plotted on the x axis?

Answer 13

The x intercept is -1/Ks
The y intercept is -/μmax
1/μ is plotted on the y axis and 1/S is plotted on the x axis.

Question 14

What is the batch culture growth theory equation and how can it be expressed graphically?

Answer 14

ΔX = μX.Δt
Where X is the concentration of biomass, t is the time and μ is the specific growth rate.
On integrating to give natural logarithms the equation becomes:
LnX = μt + LnXo. A graph of LnX against t would result in μ being the gradient and Xo being the y intercept.

Question 15

What is the yield?

Answer 15

It is the amount of biomass produced per amount of substrate. Biomass can be very specific and the units varied (g/g, g/mole etc).
Yield is calculated via the following equation:
Yield = biomass final - biomass initial/ substrate initial - substrate final

Question 16

When total conversion is assumed what would the equation for yield be?

Answer 16

Yield = biomass final/ substrate initial

Question 17

What is productivity?

Answer 17

This is the specific rate of product formation. It is also called the product quotients (qp). This is a rate so it is a function of time with units such as h^-1. In most cases it is: yield/time

Question 18

If the product quotient is constant throughout exponential growth what does this suggest?
If the product quotient is low through exponential growth but rises during the deceleration and stationary phases then what does this suggest?

Answer 18

If the product quotient is constant then it describes the production of growth linked primary products.
If the product quotient is low through exponential phases but rises during deceleration and stationary phases then it describes non-growth linked products.

Question 19

How can a culture be kept in the log phase?

Why might this be required?

Answer 19

A culture can be kept in the lop phase by actively feeding in nutrients. Also wastes and dead cells should be removed and the physical environment should be maintained. For a continuous culture this can be done via a chemostat.
A extended log phase may be required for production of growth-linked primary metabolites or products.

Question 20

Outline batch cultures and continuous cultures.

Answer 20

In a batch culture the culture environment is constantly changing, but it is a closed system. Growth, product formation and substrate utilisation all terminate after a certain time interval.
In a continuous culture fresh nutrient medium is continually supplied to a well-mixed culture with products and cells simultaneously withdrawn. Growth and product formation can be maintained for prolonged periods of time.

Question 21

What factors effect the bacterial population growth in continuous cultures?

Answer 21

The flow rate,
The relationship between the organisms growth rate and the limiting nutrient concentration,
The amount of bacteria produced per unit mass of nutrient,
Concentration of the limiting nutrient in the feed.

Question 22

The flow of medium into the chemostat is described by what?
Outline via equations how the net change in biomass per unit time can be obtained and how this changes under steady state conditions?

Answer 22

The dilution rate: D = F/V V = volume of vessel. F = flow rate Lh^-1
ΔX/Δt = μX - DX is the equation for the net change in biomass per unit time. When steady state conditions occur the biomass produced equals the biomass lost therefore: μX = DX so μ = X
Therefore under steady state conditions the specific growth rate of a culture is controlled by the dilution rate.

Question 23

Under steady state conditions within a chemostat, if D is increased what happens to μ?
Can D be increased indefinitely?

Answer 23

If D is increased then μ will also increase.
No, a chemostat works under a limiting substrate. As D increases the concentration of the limiting substrate increases. The dilution rate at which X = 0 is known as the critical dilution point. If the dilution is increased over this complete wash out of the cells occurs as the cells have insufficient time to double before being washed out of the reactor.

Question 24

Why were fed batch systems introduced and why are they being turned to now?

Answer 24

The fed batch system was introduced for bakers yeast production in the beginning of the last century, and for antibiotic production in the 1950s. It is being increasingly explored for the production of recombinant proteins.

Question 25

Describe the process outline for fed batch systems.

Answer 25

They are fed with a substrate solution so that the substrate component is growth limiting.
Substrate is often sugar rather than a complete medium.
The flow rate is low which means that the dilution is low.

Question 26

What are the main reasons for using fed batch systems?

Answer 26

Substrate limitation offers a tool for reaction rate control, as it avoids the production of toxic products.
Substrate limitation offers metabolic control, this is because it avoids catabolite repression. For example presence of glucose as a substrate prevents production of cAMP, which is required for the activit of CRP which can repress transcription of genes.

Question 27

What is microbial growth rate?
What is microbial generation?
What is generation time?

Answer 27

The change in cell number or cell mass per unit time.
This is the interval form the formation of two cell from one.
This is the time required for the cell population to double.