Question

You started with a cell count of 5.2 times 10^5 cells in a total volume of 3ml. You need lo seed a six-well plate with the following numbers. Calculate the volume of cell suspension you will need to transfer into 1 well. Each letter represents a different scenario. a. 1.2 times 10^4 cells/mL and up to 2 mL/well b. 2.5 times 10^5 cells/well and up to 2 mL/well c. 3 times 10^4 cells/well and up to 2 mL/well

Answer 1

Given, total number of cell count in total volume of 3ml is 5.2x10⁵

Number of cell count in 1ml volume is 5.2x10⁵ / 3

=1.73 x 10⁵

a. 1.2 x 10⁴ cells/ mL and upto 2mL/well

By using the Unitary method

volume of cell suspension required= number of cell count/ number of cell count in 1mL

=1.2 x 10⁴ / 1.73 x 10⁵

=0.0694 mL

  

Therefore, volume to be seeded for 2mL/well= 2x 0.0694mL

=0.1388mL

b. 2.5 x 10⁵cells/ well and upto 2mL/well

By using the Unitary method

volume of cell suspension required= number of cell count/ number of cell count in 1mL

=2.5 x 10⁵ / 1.73 x 10⁵

=1.45mL

Therefore vulume to be seeded / well in 2mL is 1.45mL

c. 3 x 10⁴cells/ well and upto 2mL/well

By using the Unitary method

volume of cell suspension required= number of cell count/ number of cell count in 1mL

=3 x 10⁴ / 1.73 x 10⁵

   =0.173mL

Therefore vulume to be seeded / well in 2mL is 0.173mL