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Biochemical Engineering > Biochem Week 10 > Flashcards

Biochem Week 10 Flashcards

1
Q

What does a turbidostat measure?

A

The absorbance and turbidity of the culture in a growth culture and automatically regulates it to maintain a preset cell density

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2
Q

How does the specific growth rate in a turbidostat culture compare to the maximum growth rate?

A

It is very close to the maximum growth rate

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3
Q

How is the specific growth rate controlled?

A

It is controlled by the rates of internal cellular reactions and expressed in optical density of culture biomass

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4
Q

How is the optical density measured?

A

By photometers. Incident light is scattered by the culture and the transmitted light is detected and measured by the photometer. When it reaches a certain level, pump turned on to return turbidity to required

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5
Q

What is the aim of the turbidostat?

A

To keep the culture turbidity constant by manipulating the feed of the medium.

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6
Q

What happens if turbidity too great?

A

The feed rate of the medium is increased to dilute the turbidity back to setpoint.

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7
Q

What happens if turbidity too small?

A

The feed rate falls so that the growth culture can restore the turbidity.

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8
Q

What is the major problem for the optical surfaces of detectors?

A

They are easily fouled by growth, microbial biofoams, foam or precipitates from medium. In practice, turbidostat operates for short amount of time but control of turbidity eventually is unreliable.

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9
Q

What word is interchangable with nutrient?

A

Substrate

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10
Q

Define the specific growth rate and the specific production rate?

A 
mu = r (x) /X specific growth rate (hr-1)
pi = r (p)/X specific production rate (hr-1)
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11
Q

3 things that the substrate is used for

A
  1. Growing biomass (with yield Y (x/s) = -r(x)/r(s) )
  2. Making product (with yield Y (p/s) = -r(p)/r(s) )
  3. Maintaining the cell alive in its environment
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12
Q

Define ‘maintenance’

A

Maintenance includes all substrate expenses that are required for e.g. DNA chemistry, protein synthesis, transport against osmotic gradients

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13
Q

When can the rate of consumption of substrate be linked to that of cell growth and product formation?

A
  1. The appropriate yields are included

2. An allowance is made for ‘maintenance’

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14
Q

As there is no flow in or out of a batch reactor, what are the rates equal to?

A

The rates equal the derivative of concentrations with respect to time.
r(x) = dX/dt
mu = r(x)/X = [dX/dt]/X —> specific growth rate
And the balance for the product:
pi = [dP/dt]/X —> specific rate of production of product

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15
Q

The four phases of biomass concentration.

A
  1. lag phase
  2. growth phase
  3. stationary phase
  4. death phase
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16
Q

What is a limiting substrate?

A

A limiting substrate is one which is present at relatively low levels of abundance so that change in its concentration may affect growth rate.
This can be used to advantage for controlling growth rate or cell cycle or enzyme expression

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17
Q

The Monod Model

A

mu = mu(max) x S/(S+KS)
S is the concentration of limiting substrate
Ks is the order of magnitude of concentration of nutrient that becomes limiting

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18
Q

What occurs if S is greater than Ks and if S is less than Ks in the Monod model?

A

If S»Ks then mu~mu(max)

If S<

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19
Q

Is Ks specific?

A

Yes, Ks is specific to a given LIMITING nutrient. Usually the limiting nutrient is glucose but sometimes it is oxygen.

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20
Q

The bioprocess can be separated into two different sections.

A
  1. Bioreactor.
    Streams include the feed (medium and O2) and the cell innoculum. Product stream is the broth
  2. Separation Train
    The broth becomes separated from debris, additive, spent buffers, waste products.
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21
Q

The stages in bioproduct recovery.

A
  1. Solids or oil removal produced by cell disruption to protect downstream operations from cells or concentrate biomass.
  2. Initial isolation/ concentration.
    This is to concentrate product and reduce volume as much as possible. This saves capital and small operating costs downstream
  3. Primary purification
    To get near pure product and remove undesirable by-products. To meet specs or protect expensive final purification apparatus.
  4. Final purification
    Meet specs if these are stringent
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22
Q

What is cell disruption?

A

The breaking up of cell walls or cell membranes so as to get the intercellular product out.

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23
Q

Examples of techniques of cell disruption on small scale and their disadvantages on a large scale.

A

Addition of chemicals - the chemicals must be removed at a later stage
Ultrasound - it is expensive

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24
Q

When is cell lysis and whole cell isolation used?

A

Cell lysis is used when an organelle or a DNA molecule is required and whole cell isolation is used for a lymphocyte or bacterial cell is desired.

25
Q

Preferred technique of cell disruption on a large scale

A

The preferred technique on a large scale is “homogenising” by pressurising the broth and jetting through a narrow orifice in the product line. This induces large shear stress that can tear cells apart.

26
Q

Alternative method for cell disruption on a large scale

A

Agitator mill in which abrasive beads are added to the broth and removed later.

27
Q

How is a secondary protein structure created?

A

Hydrogen bonds create secondary structures within peptides e.g. alpha helices, beta sheets
There are corkscrew-like twists and pleated folds in the polypeptide chain.

28
Q

How is a tertiary protein formed?

A

Further folding of the protein. It is a complex three-dimensional shape with multiple twists and bends in polypeptide chain. It is based on the interaction of the side chains with each other.

29
Q

How is a quartenary protein formed?

A

Two or more polypeptide chains are bonded together. Folded proteins can assemble together. e.g. haemoglobin

30
Q

The structure of antibodies

A

Depending on highly variable amino acid assembly at tips of Y-branches, they can bind to substance (‘antigens’) with very high specificity.

31
Q

Advantages of microbial vs. plant and mammalian cells biologics production platform.

A

Compared to mammal, risk of contamination is low
It is less expensive and less labour intensive
Production is found to be higher in microbes
Microbial microorganisms are easier to engineer genetically

32
Q

Disadvantages of microbial vs. plant and mammalian cells biologics production platform

A

In case of protein product, folding may not be proper, lack of disulfide bonds are seen when compared to animals and plant cells.
Post translational modification doesn’t take place in microbes which leads to change in protein of interest and immune response in humans.
The computation of the post translational processing is advanced.
The proteins may be inclusion bodies - meaning that they are insoluble in the proteins which makes separation more difficult

33
Q

How does sedimentation separate material?

A

Separation by size and density. It is gravity-led, there is a settler tank, there is vast throughputs, it is very common in wastewater works.

34
Q

How does filtration separate material?

A

Separation by size and molecular weight.

35
Q

What is filtration?

A

It is sweeping the medium tangentially to the surface of a membrane and it will keep the cake thin. It allows for continuous operation.

36
Q

What is the operating mode of filtration in which the medium is swept tangentially to the membrane surface?

A

It is known as ‘cross flow’. ‘Dead-end’ is without tangential flow.

37
Q

Describe the cakes in dead-end and cross-flow

A

Cross-flow gives a looser and thinner cake. Dead-end gives a thick and compressed cake.

38
Q

What is the biggest problem associated with filtration?

A

Fouling especially with proteins depositing onto the membrane

39
Q

Why is the separation train expensive?

A
  • A product of high purity is required for the pharmaceutical industry for example
  • The proteins are very fragile
40
Q

What happens if the disulphide bonds in proteins break?

A

Then the protein will become denatured and lose it’s shape. The disulphide bond will be broken by light for example.

41
Q

What makes the proteins fragile?

A

If the disulphide bonds break.

42
Q

How does the denaturation change with the pH?

A

As the pH decreases then the level of denaturation will increase.

43
Q

Why is post-translational processing required for proteins?

A

The newly formed proteins are not functional yet and require further alteration

44
Q

The path of a newly synthesised protein in a cell

A

Endoplasmic reticulum –> Golgi apparatus –> Endosome (quality check) then either a lysosome or out to the cell

45
Q

What are the steps of a separation train and by what methods are they achieved?

A
  1. Removal of solids and oils - centrifugation/ sedimentation
  2. Initial cell isolation to increase the concentration and reduce operating costs - filtration (cross flow vs. dead flow)
  3. Primary purification steps - CHROMATOGRAPHY
  4. Final purification steps - if stringent standards are required. Crystallisation.
46
Q

What are three types of chromatography?

A
  1. Size exclusion chromatography
  2. Ion exchange chromatography
  3. Affinity chromatography
47
Q

What is the basis of chromatography?

A

The speed that the protein travels through the chromatography column is dependent on the interactions of the protein between the solid and liquid medium

48
Q

How does the speed of the protein in the column depend on the strength of the interaction with the solid medium?

A

The stronger the interaction of the protein with the solid medium then the slower the protein travels with the column.

49
Q

How does size exclusion chromatography work?

A

The smaller proteins will go into holes of solid medium. Strong interaction with the solid medium and so will travel faster. Those that are larger will not travel as quick through the medium.

50
Q

How does ion exchange chromatography work?

A

The solid medium is positively charged. The larger the negative charge on the protein then the stronger the interaction with the solid medium. This means that the protein will travel more slowly through the column. The ions are eventually detached by a varying flowrate.

51
Q

Which molecules does affinity chromatography usually apply to?

A

Antibodies

52
Q

How does affinity chromatography work?

A

Antibodies bind to specific antigens. The solid phase is the antigens that the antibodies bind to. The proteins that are not antibodies will pass straight through and the antibodies will remain attached to the solid medium. This is expensive for the solid medium.

53
Q

Is the oxygen concentration high or low in water?

A

Low

54
Q

How does the oxygen concentration change with temperature?

A

The oxygen concentration decreases with an increasing temperature. This means that it may be more desirable to use a lower temperature to increase the oxygen concentration.

55
Q

What is the equation for the oxygen transfer rate?

A

OTR = kLA(Cs - Cl)
where kL is the local liquid transfer coefficient
A is the area
Cl is the dissolved oxygen concentration in the local liquid
Cs is the concentration dissolved in the liquid-gas interface

56
Q

The mass balance for oxygen

A

VdC/dt = FinCin - FoutCout - VmuX - OTR + VkLa(Cs - Cl) - V.OUR

57
Q

Methods for:

  • Removing the cells from a suspended cell solution
  • Increasing the concentration of cells
  • Primary purification steps
  • Further purification
A
  • Centrifugation (or Microfiltration or ultrafiltration)
  • Solvent extraction
  • Chromatography (packed bed adsorption)
  • Chromatography (affinity chromatography)
58
Q

Four limitations in the transfer of oxygen

A
  1. Transfer of oxygen through the boundary of the bubble
  2. Transfer of oxygen through the medium - if the medium is poorly stirred
  3. Transfer of oxygen through the boundary of the biomass
  4. Transport through the biomass

Biochemical Engineering (14 decks)

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