Mike - Biological Separation Processes Flashcards

Question 1

Q

How does yield vary with the number of steps in a process?

Answer

A

Product yield decreases with increased number of steps.

Question 2

Q

What are the 4 filtration classifications?

Answer

A

Micro
Nano
Ultra
Reverse osmosis

Question 3

Q

What are the different membrane performance models?

Answer

A

General equations
Resistance model
Film model
Gel polarisation model
Osmotic pressure model

Question 4

Q

How May membranes be classified?

Answer

A

Structure: symmetric, asymmetric
Configuration: flat, tubular, hollow fibre
Material: organic, inorganic
Surface charge: positive, negative, neutral

Question 5

Q

How do symmetric and asymmetric membranes differ?

Answer

A

Symmetric: also called homogeneous. A cross section shows a uniform porous structure

Asymmetric: in a cross section, you can see 2 different structures - a thin, dense layer and a porous support layer

Question 6

Q

What are the different modes of filtration?

Answer

A

Unstirred frontal (“dead-end”) filtration

Stirred frontal (“dead-end”) filtration

Cross-flow filtration

Question 7

Q

What does the configuration and module of a membrane refer to?

Answer

A

Configuration: geometric form given to the synthetic membranes

Module: name of the devices supporting one of more membranes

The module seals and isolates the different streams. The geometry and specific fluid movement through the confined space characterises each module. The type of flux, transport mechanism, and the membrane surface phenomena depend on the module design.

Question 8

Q

What are the main types of membrane module?

Answer

A

Tubular
Flat sheet
Spiral wound
Hollow fibre

Question 9

Q

What are features of a flat-configured membrane?

Answer

A

The active layer is flat.

Synthesised as a continuous layer.

They’re used in plate-and-frame and spiral wound filters.

They offer high surface area : volume ratio

Question 10

Q

What do plate and frame membrane systems consist of?

Answer

A

Layers of membranes separated by corrugated structural sheets, alternating layers with feed material flowing in and retentate flowing out in one direction whilst permeate flows in the other.

Question 11

Q

How do tubular, spiral-wound, and hollow fibre membrane modules differ?

Answer

A

Highest to lowest:

Surface area - HF, SW, T

Flux - SW, T, HF

P Loss - T, SW, HF

Replacement difficulty - HF, SW, T

Question 12

Q

What materials are used in membrane modules?

Answer

A

Organic:
Made of polymers
Low cost
Problems can arise with their mechanical or chemical resistance, temperature, pH, solvent and pressure handling.

Inorganic:
Mainly made from metal oxides / ceramics, glass, carbon, or metal.
5-10 times more expensive than organic materials
High chemical resistance and can withstand high temperatures
Low selectivity
Fragile

Question 13

Q

How are synthetic membranes prepared?

Answer

A

Sintering / fusion
Casting
Leaching
Stretching
Nucleation track

Question 14

Q

What are the advantages of membrane processes?

Answer

A

Processing can be at modest (even at low) temperatures.
Chemical and mechanical stresses can be minimized.
No phase change is involved (energy demand is modest).
In many cases, selectivity is good.
Concentration and purification may be achieved in one step.
Equipment can be easily scaled up either in batch or continuous
operations, or in a closed system for effective containment.

Question 15

Q

What are the disadvantages of membrane processes?

Answer

A

Concentration polarisation
Membrane fouling
Particle interactions (aggregation)
Low membrane lifetime
Limited economies of scale (6/10ths rule does not apply)

Question 16

Q

Properties of reverse osmosis:

Answer

A

Hyperfiltration - offers very high resistance.

Very popular and used for filtering ions and low molecular weight species (MW < 200)

Pore size < 1 nm

Water molecules freely pass (0.2 nm)

Pressure difference acts as the driving force

The operating pressure is between 10-25 bar and 40-80 bar for brackish and seawater respectively.

Average flux: 5-40 L/m2h

Membranes mostly made of CA or PA

Configuration in spiral-wound or hollow fibre.

Typical conversion is between 10-30%

Flow is limited by concentration polarisation

Rejection is approx. 99%

Question 17

Q

What industries is RO used in?

Answer

A

Waste water treatment
Drinking water
Food industry
Seawater desalination (30-40% of market)
Biotech

Question 18

Q

Properties of nanofiltration, NF:

Answer

A

Typical pore size of 2nm (between UF and RO)

Separation mainly due to electrostatic interaction and size exclusion. Rejects neutral molecules by size exclusion and multivalent salts by electrical charge.

Driving force is pressure difference. Moderate pressure of 15 bar.

Average flux: 20-80 L/m2h

Question 19

Q

What industries is NF used in?

Answer

A

Water treatment

Food industry

Drug production

Metal recovery

Question 20

Q

Properties of ultrafiltration, UF:

Answer

A

Microporous membranes with pore size between 1 and 50 nm.

Rejects particles from 15 - 2000 A (polymers, proteins, and colloids) - Mr of 5000 to 5*10^6 Da

Driving force is pressure difference

Operates under 1 - 10 bar.

Average flux: 5 - 200 L/m2h

Separation mostly by size exclusion

Any configuration

Significant fouling and pore plugging

Question 21

Q

What industries is UF used in?

Answer

A

Food
Wastewater
Textiles

Question 22

Q

Properties of dialysis for separation:

Answer

A

Separates ions and species of low Mr (~<100 Da)

Ionic membranes

Concentration gradient acts as the driving force.

Slow and low selectivity

Question 23

Q

Properties of microfiltration for separation:

Answer

A

Very popular and used

Pore size between 0.05 and 10 um

Rejects particles between 0.2 and 10 um (bacteria, fragmented cells, or colloids) Mw > 3*10^5.

Pressure difference acts as the driving force

Low operation pressure (0.2 - 3.5 bar)

Average flux: > 200 L/m2h

Question 24

Q

Properties of liquid membranes:

Answer

A

Consist of a liquid barrier between two phases, not yet used in industry

The driving for is chemical potential and/or concentration

There are 2 configurations: Emulsion (ELM) and Supported liquid membranes (SLM)

Potential applications include: removal of cations in solution, selective separation of gases, recovery of acid or basic compounds, organic compound separation in complex mixtures.

Question 25

Q

What are the advantages and disadvantages of liquid membranes?

Answer

A

Pros:
high flows due to transport velocity in liquids
Selective separation due to presence of specific reagents
Pumping effect due to carrier equilibrium
Small quantities of solvent needed

Cons:
Low stability of emulsions in ELM
Leaching out of organic phase from the pores of an SLM.
ELM - low practical interest
SLM -

Question 26

Q

How is retention (or rejection) of a membrane calculated?

Answer

A

R = (Cf - Cp)/Cf

= 1 - Cp / Cf

Where:
Cf : solute concentration in the feed
Cp : solute concentration in the permeate
R = 100% - complete retention in the feed, ideal semi-permeable membrane
R = 0% - solute and solvent pass through the membrane freely, no separation

Question 27

Q

How is the selectivity factor, a(A/B) calculated?

Answer

A

a(A/B) = (ya/yb)/(xA/xB)

Where:
yA and yB: concentrations of components A and B in the permeate

xA and xB: concentrations of components A and B in the feed

Question 28

Q

What does MWCO stand for?

Answer

A

The normal molecular weight cut-off (MWCO).

The MWCO is normally defined as the molecular weight of a solute for which R = 0.9.

Question 29

Q

How is flux calculated for MF and UF processes?

Answer

A

Jv = dP/(mu*Rm)

Where:
Jv is flux
P is pressure
mu is dynamic viscosity
Rm is hydrodynamic resistance (/m)

Question 30

Q

What is the general membrane equation?

Answer

A

J = dP/ [mu*(Rm + Rc + Rf)]

J: membrane permeation rate (flux expressed as volumetric rate per unit area)
dP: pressure difference applied across the membrane (transmembrane pressure)
Rm: resistance of the membrane,
Rc: resistance of layers deposited on the membrane (filter cake, gel foulants)
Rf: “resistance” of the concentration polarization film layer
mu: viscosity of the permeate solution

Question 31

Q

What are the problems with increased fouling and reduced material flux through a membrane?

Answer

A

Fouling - irreversible reduction of flux throughout time

Pore size reduction by irreversible adsorption of compounds

Pore plugging

Formation of a gel layer over the membrane surface (cake)

Question 32

Q

What is considered when designing a membrane process?

Answer

A

Performance
Cost

Material
Number of states
Pore size
Module type
Control strategy
Energy consumption

Question 33

Q

What information is required before making a final decision regarding a membrane process?

Answer

A

The effect of concentration, pressure, and crossflow rate on flux
The rejection characteristics of the membrane
The effect of temperature on flux and rejection
The rates of fouling
Of the cleaning regime to be adopted
Expected operational lifetime of the membrane

Question 34

Q

What are the stages of a membrane process design procedure?

Answer

A

Select the membrane and operation module
Predict module performance
Consider fouling
Plant design (batch, feed and bleed, single pass configurations?)

Question 35

Q

What is a single pass membrane configuration?

Answer

A

When there are several parallel trains of membrane models in series. The number of trains is progressively reduced in order to maintain the crossflow rate above the minimum value required to keep concentration polarisation and fouling under control

Question 36

Q

How may we approach improving/enhancing flux through membranes?

Answer

A

Via modification of the hydrodynamics above the membrane surface (e.g. turbulent flow, changes in channel geometry, rotating membranes, back-flushing, pulsed flow)

By modifying the properties of the membrane surface (can be permanent modifications e.g. dynamic changes)

Via modification of the electrical forces of solute-solute and the solute-membrane surface.

Question 37

Q

Why must membrane material be considered when cleaning?

Answer

A

Polymeric membranes are usually supplied coated with a preservative (such as glycerine),

Once this preservative is removed, interactions will take place between the feed and the polymer, and the membrane will behave differently.

Whilst of great operational significance, permeate flux is in itself a poor indicator of surface condition.

In addition to pore size (or molecular weight cut off) hydrophobicity, roughness and charge are also crucial in determining membrane filtration performance.

Try to understand the interactions between cleaning agents, surfaces and deposits (many food based deposits swell on contact with caustic agents)

Select an appropriate cleaning agent based on material compatibility, and a knowledge of the deposit composition, structure and location.

Cleaning agent temperature and concentration are often more important that transmembrane pressure (TMP) and cross flow velocity (CFV) in controlling the cleaning process.

Question 38

Q

How can temperature influence membrane cleaning?

Answer

A

Higher cleaning agent temperatures reduce viscosity, and speed up cleaning reactions such as dissolution, peptide bond hydrolysis etc. Performance improvements can be dramatic even moving from 30 - 50°C. However, too high a temperature can sometimes inhibit the cleaning of membranes.

Try to clean at a lower TMP than that used during the filtration process. High pressures force the partially cleaned deposit into membrane pores. If possible, clean at zero transmembrane pressure.

Surfactant micelles often change shape approaching the cloud point. This can lead to dramatic changes in viscosity and therefore cleaning performance.

Question 39

Q

What is the basis of chromatography?

Answer

A

Chromatography is a laboratory technique for the separation of a mixture.
The mixture is dissolved in a fluid called the mobile phase, which carries it through a system on which a material called the stationary phase is fixed.

The individual components are retained by the stationary phase differently and separate from each other while they are running at different speeds through the column with the eluent. At the end of the column they elute one at a time. During the entire chromatography process the eluent is collected in a series of fractions.
The composition of the eluent flow can be monitored and each fraction is analyzed for dissolved compounds, e.g. by analytical chromatography, UV absorption, or fluorescence.

Question 40

Q

What are the 3 ways of eluting (removing via washing) an ionically-bound protein?

Answer

A

Change in pH

Increase the salt concentration to swamp the charge (can be at the same pH)

Include a completing ion having the same charge but which has a higher affinity for the exchanger

Question 41

Q

Name the action principle and separation method for:

Adsorption chromatography

Ion-exchange chromatography

Molecular sieve chromatography

Affinity chromatography

Answer

A

Adsorption chromatography:
Action by - surface binding
Separation by - molecular structure

Ion-exchange chromatography:
Action by - ionic binding
Separation by - surface charge

Molecular sieve chromatography:
Action by - size exclusion
Separation by - molecular size and shape

Affinity chromatography:
Action by - biospecific adsorption/desorption
Separation by - molecular structure

Question 42

Q

What is the stationary phase in chromatography?

Answer

A

The stationary phase or adsorbent in column chromatography is a solid. The most common stationary phase for column chromatography is silica gel silica gel, followed by alumina.

Cellulose powder has often been used in the past. Also possible are ion exchange chromatography, reversed-phase chromatography (RP), affinity chromatography or expanded bed adsorption (EBA). The stationary phases are usually finely ground powders or gels and/or are microporous for an increased surface area.

Question 43

Q

What does the eluent refer to (regarding chromatography)?

Answer

A

The mobile phase.

The eluent / mobile phase is either a pure solvent or mix of different solvents, chosen so that the retention factor of the compound of interest in 0.2 to 0.3.

Question 44

Q

What is the effect of a faster flow of the eluent (mobile phase in chromatography)?

Answer

A

It minimises the time needed to run a column, thus minimising diffusion and resulting in better separation.

Question 45

Q

What are the 3 main isotherms that can be used to describe the binding dynamics of column chromatography:

Answer

A

Linear

Langmuir

Freundlich

Question 46

Q

When does the linear isotherm occur, regarding chromatography?

Answer

A

When the solute concentration needed to be purified is very small relative to the binding molecule.

Thus, equilibrium can be defined as:

q* = Kc*

Where:
q* - loading
K - eq constant
c* - target molecule conc’ at equilibrium

Question 47

Q

When does the Freundlich isotherm occur, regarding chromatography?

Answer

A

When the column can bind to many different samples in the solution that need to be purified. (Empirical correlation)

Since many different samples have different binding constants to the beads, there are many different K’s, thus the Langmuir model is not good here.

q* = Kcm*

Where:
q* - loading
K - eq constant
c*m - target molecule conc’ at equilibrium

Question 48

Q

What does the Langmuir isotherm show?

Answer

A

The Langmuir isotherm relates the coverage or adsorption of molecules on a solid surface to gas pressure or concentration of a medium above the solid surface at a fixed temperature.

θ is the fractional coverage of the surface, P is the gas pressure or concentration, α is a constant. The model this has one adjustable parameter, compared to the two adjustable parameters in Freundlich.
The constant α is the Langmuir adsorption constant and increases with an increase in the binding energy of adsorption and with a decrease in temperature.

Question 49

Q

What are favourable and unfavourable isotherms?

Answer

A

Both Langmuir and Freundlich are examples of “favourable” isotherms i.e. a plot of q* vs c* has a convex shape such that q»c at low concentrations. This allows effective recovery from dilute streams and minimises the unadsorbed losses in an adsorption process.

An unfavourable isotherm is concave in shape i.e. q»c only at high liquid phase concentrations such that effective adsorptive concentration from dilute streams is not possible.

Question 50

Q

A suspension of hydrated beads in water can be considered to consist of two immiscible aqueous phases.

What does the ability of a protein to partition between two phases depend on?

Answer

A

The relationship between the hydrodynamic radius of the protein and the pore size in the beads.

The partition coefficient,

Kp = cg/cl

Question 51

Q

Regarding chromatography, what is the retention time?

Answer

A

The time between sample injection and an analyte peak reaching a detector at the end of the column is termed the retention time (tR).

Each analyte in a sample will have a different retention time.

The time taken for the mobile phase to pass through the column is called tM.

Question 52

Q

What is the retention factor used for?

How is it found?

Answer

A

the retention factor, k’, is often used to describe the migration rate of an analyte on a column (also called the capacity factor).

The retention factor for analyte A is defined as;

k’A = (tR - tM)/tM
Where:
tR - retention time
tM - time for mobile phase to pass through the column

Question 53

Q

What does the Theoretical Plate Model of Chromatography consider?

Answer

A

The plate model supposes that the chromatographic column is contains a large number of separate layers, called theoretical plates.

Separate equilibrations of the sample between the stationary and mobile phase occur in these “plates”.

The analyte moves down the column by transfer of equilibrated mobile phase from one plate to the next.

Plates serve as a way of measuring column efficiency, either by stating the number of theoretical plates in a column, N, or by stating the plate height; the Height Equivalent to a Theoretical Plate (the smaller the better).

Question 54

Q

What is HETP?

How may it be found?

Answer

A

Height equivalent to a theoretical plate (HETP), used to convert empirically the number of theoretical trays to packing height.

If the length of the column is L, then HETP = L / N

The number of theoretical plates that a real column possesses can be found by examining a chromatographic peak after elution.

The number of theoretical plates, N, is a dimensionless number, which is related to the ratio between the retention time, tr, and the width of the peak containing the compound. If the peaks are reasonably symmetric, it can be assumed that they are Gaussian in shape. In this case, N can be found empirically.

Question 55

Q

What is the relationship to find the number of theoretical plates for a chromatography column via the half-height method?

Answer

A

N = 5.54(Ve/W1,2)^2

Using a Gaussian shaped peak curve.

Where:
N - number of theoretical plates
Ve - elution volume or retention time (tr) [ml, sec, or cm]
h - peak height
w1/2 - width of the peak at half the peak height [ml, sec, or cm]

Question 56

Q

What is the relationship to find the number of theoretical plates for a chromatography column via the tangents to the peak method?

Answer

A

N = 16*(Ve/Wb)^2

Where:
N - number of theoretical plates
Ve - elution volume or retention time (tr) [ml, sec, or cm]
h - peak height
Wb - width of the peak at the base line [ml, sec, or cm]

Question 57

Q

How is the peak asymmetry factor found, considering chromatograms?

Suggest an appropriate value for As:

Answer

A

As = b/a

Where:
As - peak asymmetry factor
b - distance from the point at peak midpoint to the trailing edge (measured at 10% of peak height)
a - distance from the leading edge of peak to the midpoint (measured at 10% of peak height)

As should be less than 1.25 for an efficient separation.

Question 58

Q

What is the Van Deemter equation?

What is it used for?

What are the different terms?

Answer

A

HETP = A + B/u + Cu

Used for establishing plate height for column chromatography.

Where:
u - average velocity of the mobile phase
A - eddy diffusion factor
B - Longitudinal diffusion factor
C - resistance to mass transfer

Question 59

Q

Why is eddy diffusion considered in column chromatography?

Answer

A

The mobile phase moves through the column which is packed with stationary phase.

Solute molecules will take different paths through the stationary phase at random. This will cause broadening of the solute band, because different paths are of different lengths.

Question 60

Q

Why is longitudinal diffusion considered in column chromatography?

Answer

A

The concentration of analyte is less at the edges of the band than at the centre. Analyte diffuses out from the centre to the edges. This causes band broadening. If the velocity of the mobile phase is high then the analyte spends less time on the column, which decreases the effects of longitudinal diffusion.

Question 61

Q

Why is resistance to mass transfer considered in column chromatography?

Answer

A

The analyte takes a certain amount of time to equilibrate between the stationary and mobile phase.

If the velocity of the mobile phase is high, and the analyte has a strong affinity for the stationary phase, then the analyte in the mobile phase will move ahead of the analyte in the stationary phase.

The band of analyte is broadened. The higher the velocity of mobile phase, the worse the broadening becomes.

Question 62

Q

How can chromatography column operation be improved?

Answer

A

It is often found that by controlling the capacity factor, k’, separations can be greatly improved. This can be achieved by changing the temperature (in Gas Chromatography) or the composition of the mobile phase (in Liquid Chromatography).

The selectivity factor, a, can also be manipulated to improve separations.

When a is close to unity, optimising k’ and increasing N is not sufficient to give good separation in a reasonable time. In these cases, k’ is optimised first, and then a is increased by one of the following procedures:
Changing mobile phase composition
Changing column temperature
Changing composition of stationary phase
Using special chemical effects (such as incorporating a species which complexes with one of the solutes into the stationary phase)

Question 63

Q

Give examples of natural and synthetic support materials used for adsorption and chromatography:

Answer

A

Natural:
Agarose - a galactose containing polysaccharide derived from sea weed. It has a high gel strength and can be cast into beads of different sizes.

Dextran - a linear polymer which can be cast into beads and crosslinked using epichlorhydrin. The most common forms are used for size based separations.

Cellulose - another polymer of glucose. This is not widely used for proteins as its pore structure is ill defined and it has poor flow properties in packed beds.

Synthetic:
Trisacryl - gels have good biological stability, are hydrophilic and can be cast in a range of pore sizes

Polyacrylamide - have good biological stability but are mechanically weak.

Hydroxylalkyl Methacrylate - Synthetic gels with good chemical and physical properties.

Question 64

Q

What are the two methods used to prepare a chromatography column?

Answer

A

The dry method and the wet method.

Dry method:
The column is first filled with dry stationary phase powder, followed by the addition of the mobile phase, which is flushed through the column until it is completely wet, and from this point is never allowed to run dry.

Wet method:
A slurry is prepared of the eluent with the stationary phase powder and then carefully poured into the column.
Care must be taken to avoid air bubbles.
A solution of the organic material is pipetted on top of the stationary phase.
Eluent is slowly passed through the column to advance the organic material on top of the column.

Answer 65

A

Loading - to load the adsorbent by adsorbing the required product(s).

Washing - to wash the adsorbent to remove, or at least reduce the quantity of, non-adsorbed contaminants.

Elution - to change conditions such that the binding interaction is displaced and the product can be recovered from the adsorbent.

(Regeneration - in the case of ion-exchange supports an acid or base treatment to ensure that the charge groups have the correct counter ion prior to the next loading cycle.)

The requirements and constraints imposed vary between these process steps and it is important that the choice of a resin for a given application is not based solely on loading conditions.

Answer 66

A

A = Fp / J

Where Fp is the permeate flowrate and J is the flux.

Flux can be found from the relationship:
J = (ΔP - π)/(μ*Rm)

(ΔP - transmembrane pressure / π - osmotic pressure / Rm - hydrodynamic resistance)

Answer 67

A

(i) Choose a membrane area
(ii) Calculate corresponding flux
(iii) From polarisation equation determine Cm
(iv) From Cm determine osmotic pressure
(v) Determine operating pressure needed via…

J = (ΔP - π)/(μ*Rm)

(ΔP - transmembrane pressure / π - osmotic pressure / Rm - hydrodynamic resistance)

Answer 68

A

The loading equation…

q* = q.max * C* /(kd +C*)

Can be rearranged into the Lineweaver-Burk form…

1/q* = 1/q.max + kd/(q.maxC)

Slope: kd/q.max
Y-int: 1/q.max

Answer 69

A

Isotherms where a plot of q* vs c* has a convex shape such that q»c at low concentrations.

This allows effective recovery from dilute streams and minimises the unadsorbed losses in an adsorption process.

Both Langmuir and Freundlich are examples of “favourable” isotherms

Answer 70

A

Isotherms where the q* vs c* plot is concave in shape i.e. q»c only at high liquid phase concentrations.

Effective adsorptive concentration from dilute streams is not possible.

Answer 71

A

Chemical properties:

Charge
Hydrophobicity
Biological function
Surface chemistry

Physical properties:
- Size

Answer 72

A

Chemical:

the requirements are similar for all separation options
a single core structure can be used for most separations given appropriate derivatisation
the base core structure resin should be inert (regarding interactions with the material to be rejected)
(i. e. hydrophilic, uncharged and no aromatic groups which could lead to unwanted (i.e. non specific) hydrophobic interactions).

Physical:

pore structure
particle size and geometry
mechanical stability

Answer 73

A

Adsorption is via a metal ion mediated coordination complex between protein and support.

Not as specific as affinity adsorption.

Is capable of combining high selectivity with tight binding.

Properties can be controlled by choice of metal ion (Cu most common)

Answer 74

A

The basis of ion exchange adsorption is provided by the ability of two substances to dissociate in solution and to exchange ions.

Three ion exchange groups are commonly used in protein separations and can be chemically substituted onto beaded resin materials (CM, DEAE, and SP)

These are characterised in terms of the ion exchanged, and the strength of acid or base required to prevent their dissociation, hence there are “strong” and “weak” examples of both anion and cation exchangers. In practice the pH stability of proteins precludes the use of most “strong” ion exchangers.

Answer 75

A

Porous bead supports (despite constraints)

A compromise between kinetics and capacity is needed…

Non porous supports: rapid kinetics but low capacity and high pressure drop in packed columns
Porous materials: higher capacities, but at the expense of slower binding kinetics and reduced mechanical stability.

Answer 76

A

Biological recognition and extreme selectivity, demonstrated by:

Antibody/antigen
Enzyme/substrate
Lectin/sugar
Neurotransmitter/receptor
Specific binding proteins

As the mixture of proteins is passed through the chromatography column, those proteins that have a binding site for the immobilised substrate will bind to the stationary phase, while all other proteins will be eluted in the void volume of the column.

Answer 77

A

Solely on the physical dimensions rather than the chemistry of the molecules to be separated.

Uses a porous hydrophilic inert support usually chemically cross-linked dextran with a tightly defined pore size distribution.

Separation must be conducted in a column or packed bed configuration

Separation is based on differences in the partition coefficients between molecules of different size between the liquid and the gel phase.

The sample is applied as a discrete band and is washed through the column using a suitable buffer .

Smaller molecules are able to access a larger fraction of the gel volume and move more slowly than material which is completely excluded. This means that the degree of separation obtained is in part a function of column length.

Answer 78

A

Non porous supports: rapid kinetics but low capacity and high pressure drop in packed columns

Porous materials: higher capacities, but at the expense of slower binding kinetics and reduced mechanical stability.

Answer 79

A

The term reversed-phase describes the chromatography mode that is just the opposite of normal phase, namely the use of a polar mobile phase and a non-polar [hydrophobic] stationary phase.

Answer 80

A

Expanded bed adsorption

Answer 81

A

Identify three from: ion exchange, affinity, hydrophobic, or size exclusion.

In each case the bases for the interaction i.e. charge, bio-specificity, balance of hydrophobic/hydrophilic regions and size should be identified.

It should be pointed out that while the others are suitable for both zonal and frontal applications, size exclusion is only suitable for zonal separations

Answer 82

A

Supports for proteins are typically porous.

Moreover, they should be:

Chemically and biochemically inert
Hydrophilic to minimise non specific interactions
Mechanically stable to support high flow rates
In the case of porous supports pore size distributions should be compatible with the size of the molecule to be adsorbed.

Answer 83

A

Compromises:

Mechanical stability must be set against chemical and biochemical suitability. Mechanical stability restricting the maximum length of an unsupported bed leading to the use of stacked columns.

Kinetic advantages of non-porous supports must be set against capacity constraints.

Surface area of adsorbent must be set against pressure drop.

Low capacities requiring large bed volumes and large diameter columns requiring sophisticated inlet configuration.

Problems of presence of particulates causing column blocking leading to development of expanded bed systems.

Collectively these problems provide the incentive for the investigation of novel configurations e.g. membrane and sponge based adsorbents.

Answer 84

A

Linear

Langmuir

Freudlich

Answer 85

A

It describes the migration rate of an analyte on a column.

k’A = (t R - tM )/ tM

Where:
tR - retention time (time between sample injection and an analyte peak reaching a detector at the end of the column)

tM - time taken for the mobile phase to pass through the column

Answer 86

A

Selectivity is a measure of the ability of the chromatographic system to distinguish between sample components.

This factor can be visualized as the distance between two chromatographic peaks.

a = k ‘B / k ‘A
When calculating the selectivity factor, species A elutes faster than species B. The selectivity factor is always greater than one.

Answer 87

A

The resolution of a elution is a quantitative measure of how well two elution peaks can be differentiated in a chromatographic separation.

It is defined as the difference in retention times between the two peaks, divided by the combined widths of the elution peaks.

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Mike - Biological Separation Processes Flashcards

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