Week 5: More oxygen glucose deprivation work!

Day 21 and 22: 

Splitting and plating cells for oxygen glucose deprivation experiments:

I labelled four plates as follows:

My intention was to do a similar experiment to last week where I deprive c17.2 cells of glucose and oxygen by using artificial cerebrospinal fluid with oxygen vaccumed out of it. My plates were control, 1.5 hour oxygen/glucose deprivation, 3 hour oxygen glucose deprivation and 4 hour oxygen glucose deprivation. A 4 hour plate was implemented to ensure that damage to the cells was completed properly, seeing as there was little effect of cell damage when using the Hoechst stain.

The method I used was:

1μM is needed of each of the peptides for these experiments, therefore a 1/10 dilution of the stock of the peptide is required in epindorph tubes.

- The TAT and Myr-p110 were both 10mM stock solutions whilst TAT-p110 was 4mM stock solution as it would not dissolve properly previously.

- Obtain one more epindorph tube for each of the stock peptide preparations ie. 6 tubes in total, with one empty tube for each peptide. Labelling for the TAT tube: “1/10 TAT” (1mM). Labelling for the Myr-p110 tube: “1/10 Myr-p110” (1mM). Labelling for the TAT-p110 tubes: “1/10 TAT-p110” (1mM).

- Put 45 μl media into each of the tubes – the aim is to have 50 μl total volume by the end of the dilution.

- Take 5μl TAT Stock 10mM and place it into the 1/10 TAT tube. Vortex to ensure that the stock has dissolved properly.

Calculation of the volumes of each of the diluted peptides required to make up final concentration 1μM in each well for 500μl media:

Use the equation V = n/C

NOTE: there are 18 wells in total for each peptide. 6 wells per plate for each peptide per plate and there are 4 plates. Therefore 6x4 = 24 wells. We need to make enough solution of each peptide in for all of the total plates. Due to pipetting error, scale it up to 25 wells. 

For TAT and Myr-p110 (Both 10mM stock solutions): To make 1 μM final: Use 1mM dilution tube, ie. 1/10. 1mM is the same as 1000 μM.
Therefore: 1 μM/1000 μM = 1x10^-3.
1x10^-3 x 500 μl = 0.5 μl/500 μl media in ONE WELL.
For 25 wells: 0.5 μl / 500 μl x 20 = 12.5 μl peptide /12.5 ml media

For TAT-p110 (4mM Stock solution):

To make 1 μM final: Use 0.4mM dilution tube, ie. 1/10. 0.4mM is the same as 400 μM.
Therefore: 1 μM/400 μM = 2.5x10^-3.
2.5x10^-3 x 500 μl = 1.25 μl/500 μl media in ONE WELL. For 25 wells: 1.25 μl / 500 μl x 25 = 31.25 μl peptide /12.5 ml media

After 90 minutes:

- Take out the hypoxia chamber that had the 1.5 hr plate in. Aspirate out all of the artificial CSF.
- Place 500 μl media in the control row, 500 μl of 1 μM TAT in the TAT row, 500 μl Myr-p110 in the Myr-p110 row and 500 μl 1 μM TAT-p110 into the TAT-p110 row.
- place the plate into the incubator.

For the control plate:
- Aspirate out the old media,
- Place 500 μl media in the control row, 500 μl of 1 μM TAT in the TAT row, 500 μl Myr-p110 in the Myr-p110 row and 500 μl 1 μM TAT-p110 into the TAT-p110 row.
- place the plate into the incubator.

After 3 hours:
- Take out the hypoxia chamber that had the 3 hr plate in. Aspirate out all of the artificial CSF.
- Place 500 μl media in the control row, 500 μl of 1 μM TAT in the TAT row, 500 μl Myr-p110 in the Myr-p110 row and 500 μl 1 μM TAT-p110 into the TAT-p110 row.
- place the plate into the incubator.

After 4 hours:
- Take out the hypoxia chamber that had the 3 hr plate in. Aspirate out all of the artificial CSF.
- Place 500 μl media in the control row, 500 μl of 1 μM TAT in the TAT row, 500 μl Myr-p110 in the Myr-p110 row and 500 μl 1 μM TAT-p110 into the TAT-p110 row.
- place the plate into the incubator.

I plated the control plate with media and the p110 peptide only and it was not subjected to oxygen glucose deprivation at all. Once the 1.5, 3 and 4 hour oxygen glucose deprivation plates were taken out of their respective hypoxia chambers, then I also treated them with the peptide and left them incubated. I will analyse the results of this experiment tomorrow!