Cell fixation and SRB staining

Cell fixation (proliferation)

Repeat previous step 4 times.

Ensure plates are dry and return the plates to the teaching assistant. They will be stored per group in a drawer until tomorrow when you’re doing the next procedure.

Protein extraction

What is the function of the 1% acetic acid solution?

1. To degrade proteins by changing the pH
   Correct = no
2. To wash away all excess SRB solution
   Correct = yes
3. To dissolve cellular proteins
   Correct = no
4. To kill all cells by changing the pH
   Correct = no

What is the correct way of disposing SRB after removal from the plate?

1. Remove the SRB withe the suction needle in to the vacuum pump.
   • This is not correct. The whole vacuum pump will turn pink if we do so. Never use the suction needle to remove SRB!
2. In a tube the regular waste bin.
   • This is not correct. SRB is a chemical. You may not throw it in the regular waste bin.
3. In a tube in the biohazard waste bin.
   • So close, but not entirely correct.
4. Collect in the original tube for reuse.
   • Correct! You can reuse the SRB solution.

Why is it necessary to wash the plate with 1% acetic acid?

1. The pH of acetic acid ensures that SRB stays bound to proteins during the wash steps.
   • This is correct! SRB binds to proteins under mild acidic conditions. We just want to remove the unbound SRB. By washig with 1% acetic acid, we remove the unbound parts, but not the bound ones.
2. The pH of acetic acid ensures that the cells stay securely fixed to the well walls
   • This is incorrect. Remeber what we did in the TCA step.
3. The pH of acetic acid denatures the proteins in the cells, making them more susceptible to SRB staining.
   • This is incorrect. If acetic acid would make the cells beter susceptible to SRB, would it be useful to add this after staining with SRB?
4. The pH of acetic acid washes off all the SRB staining off the cell.
   • This is incorrect. If it washes off all the SRB staining, wat is left to measure?

Which of the following statements is true in regards to the function of adding Tris?

1. SRB reacts with Tris resulting in the pink colour through which the protein concentration is measured.
   • This is incorrect. There is no conversion reaction between Tris and SRB.
2. The basic nature of Tris releases SRB from it’s bond with proteins resulting in the pink colour through which the protein concentration is measured.
   • This is correct. SRB wil be extracted using basic conditions like Tris.
3. Under the pH of Tris the colour of SRB turns pink making it able to measure protein concentration.
   • This is incorrect. SRB is already a pink solution.
4. Tris is added in order to dissolve any remaining cell fragments which are not proteins and may disrupt the absorbance measurement
   • This is incorrect. Remeber the TCA step. During this step we removed already all the remaining cell fragments and fixated the cells.

What does the SRB solution do?

1. Stain only cellular membrane proteins
   Correct = no
2. Stain cellular nuclei
   Correct = no
3. Stain cellular membranes
   Correct = no
4. Stain all cellular proteins
   Correct = yes

Feedback if correct: Indeed! SRB solution stains all cellular proteins.

Feedback if incorrect: This is incorrect. Read the background information for this technique again and choose another option!

Which process in the signalling pathway is inhibited by the inhibitors in your experiments?

1. Protein interaction
   Correct = no
2. Phosphorylation
   Correct = yes
3. Receptor activation
   Correct = yes
4. Ligand binding
   Correct = no

Feedback if correct: Correct! The phosphorylation of proteins in the Akt or Erk pathways will be inhibited. That is the reason why we will execute a Western Blot. It gives us information about the phosphorylation of these pathways.

Feedback if a: Think about which process will be inhibited.

Feedback if incorrect: Does the inhibitor work before the receptor or after the receptor?

On which signalling pathway do the inhibitors have an effect on?

1. Lapatinib inhibits MEK, which is involved with the ERK signalling pathway.
   Correct = no
2. MK-2206 inhibits PI-3K, which is involved with the ERK signalling pathway.
   Correct = no
3. Lapatinib inhibits EGFR, which is involved with Akt signalling and the ERK signalling.
   Correct = yes
4. Selumetinib inhibits MEK, which is involved with the ERK signalling pathway.
   Correct = yes
5. Lapatinib inhibits Akt, which is involved with the ERK signalling pathway.
   Correct = no
6. MK-2206 inhibits MEK, which is involved with the Akt signalling pathway.
   Correct = no

Feedback if correct: Correct! The inhibitors inhibit the phosphorylation of the MEK and Akt proteins which affect their own signalling pathway and the gene expression of proteins involved with migration, proliferation and resistance.

Feedback if incorrect: Two of the options are correct. Look up which inhibitor affect the ERK and Akt pathway and how they affect these pathways? The information can be found in the provided datasheets.

Before starting an experiment with live cells it is important to check your cells with a microscope. Select everything that you should check before starting your experiment:

1. Bacterial or fungi infection
   Correct = yes
2. Is there equal confluency in each well
   Correct = yes
3. The outer wells
   Correct = no
4. Color of the medium
   Correct = yes
5. The effect of the inhibitors
   Correct = no
6. Are cells homogenously distributed in one well
   Correct = yes
7. Are the cells attached to the bottom?
   Correct = yes
8. Shape of the cells
   Correct = yes
9. Floating cells in the medium
   Correct = yes

Feedback if correct: That is correct. As you can see, there are a lot of factors that could affect your experiment. It is important to write down every irregularity when checking your cells. When your data don’t match your expectations, you can trace back the cause by looking it up in your notes.

Feedback if c: What is inside the outer wells?

Feedback if e: Inhibitor is not always present when the cells should be checked.

For the combi inhibitor condition, the working concentration is twice the final concentration compared to the single inhibitor condition. What is the reason for that?

1. The cell density is higher in these wells, so the concentration inhibitor should also be higher.
   • Incorrect: If seeded correctly, each well has approximately the same cell density.
2. Because the inhibitor will compete with the other inhibitor. The concentration should be doubled so that the final concentration is the same.
   • Incorrect: The inhibitors don’t compete in solution.
3. We want to see what the effect is of double concentration in combination with two inhibitors.
   • Incorrect: Having a higher concentration is possible in an experiment. For instance, a dose-response experiment is often used in research. But in this experiment, the end concentration should be the same to answer your research question.
4. The working concentration is twice as high, because it will be diluted again in the well. The final volume in the well should be considered in the calculation.
   • Correct: The working concentration is twice as high, because it will be diluted again with the other inhibitor solution. The final volume in the well is twice as high as the volume you add. This should be considered when making your working solutions. This way, the final concentration of the inhibitors in your single and combi condition would be the same.

Above you can see your plate layout for the plate you will use for SRB and IF (1 plate each). Since you need similar conditions for both experiments, you can make inhibitor volumes for both plates together. Calculate the volume needed of inhibitor and RPMI complete medium to make your solution for the lanes with “Inhibitor A” to fill up both plates (note that both cell lines will be treated with this inhibitor).

The stock concentration is 2 mM, the target concentration is 1µM and each well can hold 100 µL of liquid. On top of the amount that you need to fill all wells, add 6 wells of volume to your calculation to compensate for pipetting errors.

Note: this is an example calculation for practice, you will not do this during the actual experiment, for that you will find novel instruction in the “protocol” tab.

We need __d__ µL of stock solution

We need __c__ µL of RPMI

• How many wells are exposed to “Inhibitor A”?
• The “combi” condition needs a different concentration of “inhibitor A”. Don’t include these wells in your calculation
• You have two plates that you need to expose.
• Use formula: C1 x V1 = C2 x V2

Feedback if d > 2998.5 and d < 3000 and c = 1.5: Very good! you calculated it correctly.