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For the preparation of the hypotonic solution We calculate the grams needed to obtain a 0.075 Molar KCL solution. Considering that the molecular weight of the KCL is 74.55 gr / mol and that therefore 74.55 g in a liter of water equals 1 Molar we apply the proportion 74.55: 1 = X: 0.075 and we will have that X = 5.67 Then weighing 5.67 gr of KCl in 1 liter of water we will have a solution at 0.075 Molar We only needs 30 milliliters total for our experiment, we prepare 500 milliliters of it then we halve the grams of KCl to be weighed. 5.67 / 2 = 2.8 gr. We will use the remaining volume of excess KCl for subsequent experiments. The solution must be stirred to dissolve the solute in water The culture is transferred from the polypropylene test tube to a test tube with a conical and polystyrene bottom suitable for centrifugation. Finally, the centrifuge is set at 2100 revolutions per minute for 5 minutes and is started. The supernatant is eliminated with an automatic vacuum aspirator. Alternatively, aspiration can be done using a plastic pasteur pipette equipped with a teat. To progressively pass from the aqueous phase to the organic phase, we proceed with an intermediate phase which consists in adding a prefix (consisting of methanol: acetic acid in a ratio of 3 to 5) to the pre-existing aqueous phase, we prepare the prefix, consisting of 3 parts of methanol and 5 of acid by mixing 3 milliliters of methanol and 5 milliliters of acetic acid we will now dispense 400 microliters of prefix in each tube with a single-channel pipette in the pre-existing hypotonic solution. We use the same pasteur for the first three tubes because they are three replicates of the same sample and we proceed equally for the next three replicates A new centrifuge is always made at 2100 revolutions per minute for 5 minutes. Now we proceed to the removal of the hypotonic and the prefix to which we will add pure methanol which allows to completely dehydrate the crops and preserve them until the drop phase The tubes are closed and the cells resuspended. The chemical hood is turned off and the tubes are placed in the freezer pending the next dripping phase which could take place from a few days to a few months later. After at least half an hour in the freezer, the methanol cells are centrifuged again at 2100 revolutions per minute for 5 minutes A new centrifuge is always made at 2100 revolutions per minute for 5 minutes. The tubes are housed in the cruise and a centrifuge is made at 2100 revolutions per minute for 5 minutes. The tubes are housed in the cruise and the last centrifuge is always made at 2100 revolutions per minute for 5 minutes. 24.45 milliliters of medium are collected in two stages: 24 milliliters with the pipettor using a 25 milliliters serological tip and subsequently I will withdraw 450 microliters with the single-channel pipette In the preparation of the complete culture medium for each tube of soil the following are needed: RPMI 4, 0 75 milliliters for six tubes 24.45 total milliliters We need 10% fetal bovine serum, 0.5 milliliters for six tubes three milliliters in total 1% of antibiotics required 10,000 units per milliliter, 0.0 50 milliliters for six tubes 0.3 milliliters total finally 1.5% of phytohaemagglutinin 0.0 75 milliliters is required for six tubes 0.45 milliliters in total. At the 24th hour 3 cultures are treated with 6mM diepoxibutane, the tubes are taken from the tremostat, and brought under a biological hood. For this reason, 2.5 microliters of diepoxybutane are added to each tube, then they are closed and stirred. The same operations are carried out for the control pipes: these are brought under a biological hood, open, the vehicle is added, or 2.58 microliters of water, then they are closed and stirred. The tubes are put back in a thermostat at 37 ° C until the end of the growth of the crops always on a rotating plate. DEB preparation For the preparation the diepoxibutane to be added to the cultures at a concentration of 6µM starting from a 12M concentration. It is necessary to apply the formula initial concentration multiplied initial volume equal final concentration multiplied final volume from this 12 Molar multiplied initial volume equal to 6 micro Molar multiplied 5 milliliters, therefore 12 Molar multiplied initial volume equal 6 micro molar 5 multiplied 10 raised to minus 3 liters then 12 molar multiplied initial volume equal to 30 multiplied 10 raised to minus 6 multiplied 10 raised to minus 3 finally we will have that initial volume is equal to 30 multiplied 10 raised to minus 6 multiplied 10 raised to minus 3 all fraction 12 therefore initial volume is equal to 2.5 multiplied ten raised to minus 9 liters Being a very small volume, a dilution of 1 to 1000 is required Then 1 microliter of DEB at 12 moles plus 999 microliters of sterile water. This solution will have a final concentration of 12 millimolar