Question

Use the following information for all questions in this exercise.

Protein information:

TEVP: Extinction coefficient: 32,290 M^-1cm^-1, MW: 28.6 kD

GFP-POI: Extinction coefficient: 63,300 M^-1cm^-1, MW: 65.9 kD

GFP: Extinction coefficient: 30,430 M^-1cm^-1, MW: 32.1 kD

POI: Extinction coefficient: 32,850 M^-1cm^-1, MW: 33.8 kD

A protein sample of 10 mls, A₂₈₀ 0.655, containing a His-tagged GFP-linked protein with TEV site was treated with 50 ?ls of 1 mg/ml His-tagged TEV protease, then applied to a Ni-NTA column. We will assume that the TEV protease was complete with its reaction. The following represent the pooled fractions from each elution.

Initial flow through + 2 mls column wash: 12 mls, A₂₈₀ = 0.283

Remaining column wash and low Im buffer had no significant absorbance  

High Im buffer: 6 mls, A₂₈₀ = 0.534

Questions:

1) What is the amount of GFP-POI in nanomoles?

- I know we use Beer's Law; however, I do not know what absorbance value we should be using.

2) What is the amount of TEVP added in nanomoles?

3) What is the concentration of the POI (in mg/ml)

4) Based on this concentration, convert the POI to nanomoles and compare to the amount of GFP-POI. If they are the same, then we can assume the initial pool of GFP-POI was pure. If they are different, then there must have been an impurity present. Is your sample pure? Answer yes or no:

5) What is the absorbance due to TEVP in the pooled fractions that contain it?

*Note: More info on topic - High imidazole buffer should elute both GFP and TEVP.

Please help, and thank you!

Answer 1

Answer 1.

A = E C l

Where,

A = 0.655 (protein sample containing His-tagged GFP-POI)

E = 63,300 M-1cm-1

L = 1 cm (standard cuvette path length)

A = E C l

0.655 = 63300 x C x 1

C = 0.655 / 63300

C = 0.0000103475 M

C = 347.5 nM

The observed concentration is for 1 ml sample.

Total volume is 10 ml so, concentration will be C = 347.5 nM x 10 ml = 3475 nM

Answer 2.

Amount used of 1 mg/ml His-tagged TEV protease : 50 ?l

So, 1 mg/ml = 1 ug/ ul

Therefore, by adding 50 ul in solution, 50 ug protease is added.

Convert into moles using molecular weight: 28.6 kD using the formula :

m[g] = Q[mol] x Mw[kDa] x 103

The obtained concentration using the formula is 1.75 nmoles.

Answer 3.

A = E C l

Where,

A = 0.283 (protein sample containing His-tagged GFP-POI)

E = 32,850 M-1cm-1

L = 1 cm (standard cuvette path length)

A = E C l

0.283 = 32850 x C x 1

C = 0.283 / 32850

C = 0.000008614 M

The observed concentration is for 1 ml sample.

Total volume is 12 ml so, concentration will be C = 0.000008614 M x 12 ml = 0.000103 M

Converting into mg/ml

Convert into mg/ml using molecular weight: 33.8 kD using the formula :

m[g] = Q[mol] x Mw[kDa] x 103

The obtained concentration using the formula is 3480 mg/m