Identifying the components of the mitophagy pathway after neonatal hypoxic ischaemic brain injury

Day 29 and 30 After my OGD plates were incubated for 1 and a half hours I treated them with 0.3, 1 and 3uM of TAT and TAT-p110 as follows: Calculation of the volumes of each of the diluted peptides required to make up final concentration of 0.3 Μm in 500 μl media: Use the equation V = n/C For TAT (10mM stock solutions): To make 0.3μM final: Use the 0.1mM dilution tube, ie. 1/100. 0.1mM is the same as 100μM. Therefore: 0.3 μM/100 μM = 3x10 -3 . 3x10 -3 x 500 μl = 1.5 μl/500 μl media in ONE WELL. For 15 wells: 1.5 μl / 500 μl x 15 = 22.5 μl/7.5 ml media. 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 15 wells: 0.5 μl / 500 μl x 15 = 7.5 μl/7.5ml media To make 3 μM final: Use 1mM dilution tube, ie. 1/10. 1mM is the same as 1000 μM. Therefore: 3 μM/1000 μM = 3x10 -3 . 3x10 -3 x 500 μl = 1.5 μl/500 μl media in ONE WELL. For

Day 27 and 28: Assays, secondary antibody incubation and splitting cells: The assays: I ran assays for all of the OGD plates for Hoechst, MTT and LDH and the results were as follows: Secondary antibody incubation for immunofluorescence: I had to incubate the coverslips with secondary antibodies against the TOMM20 (which would be an indicator of mitochondrial morphology and presence) and Drp1 (which would be an indicator of fission. TOMM 20 were going to appear red whilst Drp1 was going to appear green on the confocal microscope. I incubated the coverslips in darkness for the rest of the day. At the end of the day, I mounted the coverslips onto slides which contained DAPI which would mark the nucleus and this would appear blue on the confocal microscope. Mounting the coverslips onto the slides: The slides were labelled with their relevant characteristic and the cloverslips were placed on top. The coverslips were secured in place with nailvarnish to the microscop

Day 25 and 26: Result of my western blot:   LC3B was identified successfully in all of the lanes: The brightness of the LC3B in each of the lanes: In lane 2: 2620, In lane 3: 4130, In lane 364, In lane 5: 1390, In lane 6: 3690. Plating the cells for this week: Immunofluorescence: The plates I split the cells that I had incubated in the flask in order to put them in the immunofluoresnce plates. Seeing as there were 24 wells in total, I wanted to plate enough cells for 28 wells to make up for pipetting error. 30, 000 cells per plate were required therefore 30, 000 x 28 = 840, 000. I found that after splitting the cells i had 997, 600 cells/ml. Thus, 840, 000 / 997, 600 = 842ul required. The amount of media required: 750ul x 28 = 21ml media. I had to sterilise coverslips to attach the cells to coverslips for subsequent confocal #microscopy. After sterilising the coverslips with 70% ethanol, I carefully (without breaking them!) placed them back into the wells. I

Day 23, and 24 I split the cells which were present in the flask into two separate flasks for more examination next week on different experiments. After counting the cells, I found that there were 1.124 x 10 5 /ml cells were found in the cell sample and 350K cells were required in each of the 6 plates. Calculation for the volume of cells and media needed in each plate: 350, 000 x 8 = 2, 800, 000. 2, 800, 000/1, 124, 000 = 2.45 ml cells are required in the media ie. 2.5 ml cells. 3ml media is required in each plate, therefore 3ml x 8 = 24 ml media 1.124 x 10 5 /ml cells were found in the cell sample and 350K cells were required in each of the 6 plates. I plated the cells as follows: - I placed 3ml media in each of the plates, incubated them to allow them to settle for atleast 4 hours before the oxygen glucose deprivation. After 4 hours: - Harvest the No OGD and place it in its own epindorph tube. Leave plate 3 in the incubator because that needs to be harvested after

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 pepti

Day 20: Analysing the effect of Oxygen glucose deprivation on C17.2 cells upon the treatment of peptides TAT, Myr-p110, and TAT-p110   Aim: to assess the affect of OGD on C17.2 cells after 1.5 hours and 3 hours by assaying LDH, Hoechst and MTT. Reason for assaying LDH: LDH is lactate dehydrogenase which catalyses the conversion of pyruvate -> lactate during anaerobic respiration. Anaerobic respiration only involves the process of glycolysis which occurs only in the cytosol. Thus, if the peptides became cytotoxic to the cells, then the lactate dehydrogenase would leak out into the medium and therefore be detectable by increased absorbance and less transmission. Reason for assaying Hoechst: Hoechst specifically stains nuclei blue. If there are more nuclei present this means that less cells have died as a result of cytotoxicity. If there are less nuclei present this means that more cells have died as a result of cytotoxicity. It is required that we make a blank and then a trea