Field Flow Fractionation

Question 1

What is the RI signal proportional to?

Answer 1

RI-signal is proportional to concentration

Question 2

What is the MALS signal proportional to?

Answer 2

MALS-signal is proportional to
concentration*molar mass.

Question 3

What is MALS?

Answer 3

Multi angle light scattering

Question 4

What is RI?

Answer 4

Refractive Index

Question 5

Population 2 has a low MALS-signal but a high RI-signal and has a
relatively short elution time. What does this tell us about population 2?

Answer 5

The RI-signal is proportional to concentration while the MALS-signal is proportional to
concentration*molar mass. This tells us that the concentration in this fraction is relatively high
but the molar mass is relatively low as the MALS-signal is considerably lower than for
populations 3 and 4. Another way of viewing it is that the elution times are shorter than for 3
and 4 and thus it would correspond to something relatively small in comparison to 3 and 4 but
at a non-negligible concentration.

Question 6

When analyzing high molar mass biopolymers (such as amylopectin) with AF4, overloading of
sample in the separation channel may cause problems and impair resolution. Hence, the
injected amount should be carefully optimized before analysis commences. Which concentration is most optimal for analysis?

Answer 6

As it is desirable to maximize S/N then the highest injected mass that does no influence
retention would be the optimal choice.

Question 7

Suggest an explanation to why overloading increasing potential problem for high molar
mass branched biopolymers than for those of relatively low molar mass.

Answer 7

It is more likely that problems with inter-molecular
interaction arises if the polymers are large and even more so if they are branched as this may
possibly lead to entanglements of polymer chains from different molecules. Another way of
viewing this potential problem is that the viscosity increases in the sample zone due to the
large volume that is occupied if the molecules are larg

Question 8

What happens if you overload a sample?

Answer 8

If a too large mass is injected the sample may not
be able to relax (i.e. find its equilibrium distance from the accumulation wall) simply as there
are to many molecules/particles present which leads to sample interaction and influences
diffusion (Brownian motion). In turn, this can cause tailing or fronting and impaired separation

Question 9

How do you determine molar weight given data for kc/R(theta) and sin2(theta/2)?

Answer 9

Plot kc/R(theta) on y axis and sin2(theta/2) on x axis. Intercept will give you the inverse of molar mass (=1/Mw)

Question 10

How do you determine radius of gyration given P(theta) values?

Answer 10

Using the formula P(theta) = 1 - (16pi2n2/3lambda2)r2*sin2(theta/2)

Question 11

What is the elution order in AF4? And how would bovine serum albumin dilute with monomers, dimers and trimers, and what would the tail be?

Answer 11

As the elution order in AF4 goes from small to large, the order would be monomer, dimer, trimer and the tail is likely higher
oligomers or aggregates.

Question 12

Explain what you think will happen to the fractogram if you would increase the cross flow rate

Answer 12

By increasing the crossflow rate retention time will increase as the sample
zone will be more compressed towards the accumulation wall (and hence experience lower
flow rate along the channel). For the same reason, we also expect that the resolution between
the oligomers would increase as the velocity variation over the sample zone (away from the
accumulation wall) would decrease.
Secondary answer: It should be noted that if we increase the crossflow excessively then the
higher retention (longer residence time in the channel) could however cause a increase in
longitudinal zone broadening, leading to “smeared” out peaks and loss of resolutio

Question 13

Explain what you think will happen to the fractogram if you decrease the channel out flow rate

Answer 13

The decrease will lead to longer retention times as the flow rate along the
channel is lower and hence analyte migration is slower.
Secondary answer: The peak height could increase as the analytes will elute a smaller volume
(lower dilution at the channel outlet). This could however be compensated for by increased
zone broadening due to the longer residence time in the channel (see answer above in i.)
which will depend on the actual flow settings

Question 14

Explain what you think will happen to the fractogram if you change to a twice as long channel

Answer 14

The decrease will lead to longer retention times as the flow rate along the
channel is lower and hence analyte migration is slower.
Secondary answer: The peak height could increase as the analytes will elute a smaller volume
(lower dilution at the channel outlet). This could however be compensated for by increased
zone broadening due to the longer residence time in the channel (see answer above in i.)
which will depend on the actual flow settings

Question 15

Explain what you think will happen to the fractogram if you change to a twice as thick channel height

Answer 15

This change would lead to an increase in retention according to:
tr = w2/6VDqct0
Where retention time is strongly dependent on w. It arises for the difference in steepness och
the velocity gradients in the parabolic flow profile along the channel.

Question 16

Suggest an approach to develop an AF4-method taking into consideration channel type, flow
settings, injected amount

Answer 16

Things to consider are
-Selection of channel thickness (i.e. thick or thin).
* Selection of flows and possibly crossflow gradients.
* Injected amounts.
* Optimization of crossflow and possibly crossflow gradients.
* Is unwanted interaction with the accumulation wall observed? What can be done to
try to eliminate/reduce it?

Question 17

What does retention depend on?

Answer 17

Field induced velocity (cross flow velocity) and the diffusion coefficient of analyte

Question 18

How is the diffusion coefficient related to size?

Answer 18

Something large will have a lower diffusion and be more compressed. Will therefore elute later. Something small will have a higher diffusion and be less compressed. Will therefore elute faster.

Question 19

What is the retention level RL?

Answer 19

RL = tr/t0 = wU/6D = wQ / 6AD

Question 20

What does dn/dc describe?

Answer 20

How the refractive index increases with increasing solute concentration

Question 21

What is isotropic light scattering?

Answer 21

Rayleigh scattering with no angular dependence of scattered light intensity. Valid when size of scatter is much smaller than wavelength (proteins)

Question 22

What is anisotropic light scattering?

Answer 22

Rayleigh-Gans-Debye RGD scattering. Angular dependence of scattered light intensity. Valid when size of scatter is larger than wavelength. For polymers and nanoparticles. If we want lower wavelength go with x-ray

Question 23

What is P(theta)?

Answer 23

The form factor, describes how scattered light varies with angle. Affected by the rg.

Question 24

When is rg = a?

Answer 24

If layer is sufficiently thin and it is hollow inside.

Question 25

When is rg = sqrt(3/5)?

Answer 25

It is the average distance of the centre to mass.

Question 26

What is an explanation of the light scattering signal being low?

Answer 26

It shows the concentration. Small particles scatter light less.

Question 27

What is a reasonable retention level RL?

Answer 27

At least 5 and up to 40

RL = tr/t0

Question 28

How do you change the retention level?

Answer 28

By changing the cross flow rate and outflow rate (Qc/Qout)

Question 29

How can you create a gradient?

Answer 29

You make a cross flow gradient

Question 30

What is the benefit of AF4 over SEC?

Answer 30

AF4 can separate larger things so you can look at a much larger range. Also can change the membrane yourself and it is cheaper.