Hi.
1.Pipette 0.9mlsterile saline (0.9% NaCl w/v) into each of six Eppendorftubes and label them.We recommend making a diagram in your notebook showing the procedure and the meaning of your labels.2.Resuspend the cells in the originalsuspension. Remove 100μl with the pipette and fresh tip and transfer it to one of the tubes with saline.Then thump or vortex the tube to suspend the cells. 3.Repeat the procedure serially, using a fresh tip for each of the remainingtubes, so that each tube contains 1/10ththe number of cells as the previous one.
4.Plating of the dilutions: resuspend thecells in each tube by thumping it before taking a sample, then pipette 100μl onto the center of an LB plate starting with the highest dilution (10-6). Take the glass spreader to a Petri dish with 95% ethanol, removeit and light it in a flame and, after it burns, cool it down by touching the agar on the outer most are of the plate. After it has cooled down, spread the cells over the surface with the spreader. Repeat with the 10-5and 10-4dilutions.5.Incubate the cells in an incubator at 37°C overnight.If your original culture had 1x109cells per ml, predict how many colonies you expect to see on each plate-10-4
I have this protocol and I want to calculate the colonies in plate 10-4. Does anyone know the result?
The predicted number of colonies can be calculated from the original equation used for standard plate count by serial dilution:
Original sample count (CFU/mL)= CFU/(D *V (mL)).
Where CFU are the number of colonies grown on the agar plate, D is the dilution used, and V is the volume of sample plated on the agar.
By rearranging the equation, the predicted number of colonies on agar plate inoculated from your 10^-6 dilution is as follows:
No. colonies = Original sample count * D* V
= 10^9 * 10^-6* 0.1 mL
= 100 cfu
Since you have plated a volume of 0.1ml (100 uL) from each tube.
So, idealy, 100 colonies are expected to be seen on the plates inoculated with 0.1 mL from the 10^-6 dilution. But this is just a theoretical calculation. Real count might be more or less different.
Similarly, 1000 colonies from the 10^-5 dilution, and 10000 colonies from the 10^-4 dilution are expected. Note that colonies on plates from the 10^-5 and 10^-4 will not be possibly counted because colonies will be extremly crowded and may even be seen as a lawn. Plates with more than 300 colonies or less than 25 are not statistically reliable.
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