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

Solution

The experiment simplified:

• We have a protein of interest (POI) having (a) his-tag, (b) TEV protease site and (c) a fused GFP. Example like this:
• We have incubated our POI (with tags and infusions) his-tagged TEV protease.
• TEV protease has cleaved all the his tags and GFP from our POI. So now our protein is free of his tag and GFP.
• We pass the solution through Ni-NTA column that can strongly bind his tags (TEV protease has his tags, cleaved his tags with GFP will bind the column strongly)
• The flowthrough will collect the unbound fractions which we expect to be our cleaved protein of interest POI. If our protein non-specifically has mild interactions with the beads of the Ni-NTA column, it is expected to be simply washed out using low volume of wash buffer.
• imidazole solution is then applied with increasing concentrations to elute TEV protease and the cleaved his-tags with GFP.

1) The amount of GFP-POI in nanomoles

According to Beer-Lambers Law:

Absorbance = extinction coefficient x concentration x path length

0.655 (abs of uncleaved protein POI-GFP) =  63,300 x C X 1

C = 0.655 / 63300 = 1.03 x 10^-5M = 4.5uM = 1.03 x 10⁴ nM or 1.03 x 10⁴  nanomoles / L

1.03 x 10⁴ nanomoles / L means in 1L (1000ml) there are 1.03 x 10⁴ nanomoles of POI-GFP

1 ml would have 1.03 x 10⁴ / 1000 nmoles

or, 10ml would contain = (1.03 x 10⁴ / 1000) x 10 = 103 nanomoles of POI-GFP

2) The amount of TEVP added in nanomoles

Concentration added = 1mg/ml = 1 g/L

Molecular weight of TEVP = 29kDa = 29000 g/mol approx

Calculate to find concentration in nanomoles:

No. of moles = mass / molecular weight

= 1g / 29000g/mol

Molarity = no. of moles / L

= 3.4 x 10^-5 moles/L or 3.4 x 10⁴ nmoles/L

= ​​3.4 x 10⁴ nmoles / L

= (3.4 x 10⁴ ) nmoles / 10⁶ uL

= 3.4 x 10^-2 nmoles / uL

In 50ul, amount of TEVP = 3.4 x 10^-2 x 50 ul = 170 x 10^-2 nanomoles = 1.7 nmoles

3. Concentration of POI in namomoles

According to Beer-Lambers Law:

Absorbance = extinction coefficient x concentration x path length

0.283 (abs found in flowthrough + 2ml wash buffer, this fraction contains our protein) =  32,850 x C X 1

C = 0.283 / 32,850 = 8.6 x 10^-6M =8.6 uM = 8.6 x 10³ nM or 8.6 x 10³ nanomoles / L

8.6 x 10³ nanomoles / L means in 1L (1000ml) there are 8.6 x 10³ nanomoles of POI

1 ml would have 8.6 x 10³ / 1000 nmoles

or, 12ml (10ml flowthrough + 2ml wash buffer that eluted our protein) would contain = (8.6 x 10³ / 1000) x 12 = 103.2 nanomoles of POI

Calculate concentration of POI in mg/ml

0.283 (abs found in flowthrough + 2ml wash buffer, this fraction contains our protein) =  32,850 x C X 1

C = 0.283 / 32,850 = 8.6 x 10^-6M =8.6 x 10^-6 moles /L

Molarity = no. of moles / L

and, no. of moles = mass / molecular weight

or, mass = no of moles x molecular weight

= 8.6 x 10^-6 x 34000 Da

= 8.6 x 10^-6 g x 34000 g/mol = 2.92 x 10^-1 g

Therefore , 8.6 x 10^-6M = 2.92 x 10^-1 g / L or, 0.292 mg/ml

4. If we compare the amount of POI-GFP (before tag removal) and that of the eluted POI we can see they are same i.e, 103 nanomoles. This means 100% of TEV protease cleavage took place and the POI-GFP sample was pure.

5. Absorbance due to TEVP in the pooled fractions that contain it

The final imidazole eluted fractions contain both TEVP and GFP. Both their presence contribute to the Abs value 0.534.

Absorption is addidive for mixtures.

Concentration of GFP is same as that of our POI i.e, 8.6 x 10^-6M, as one POI molecule was tagged with exactly one GFP molecule (1:1 ratio).

Abs_total = Abs_TEVP+Abs_GFP

0.534 = E_TEVPC_TEVPL + E_GFPC_GFPL

0.534 = Abs_TEVP + (30,430 x 8.6 x 10^-6M x 1)

Abs_TEVP=0.534 - 0.262 = 0.272 (Answer)