Question

1a

Your cell suspension is 6.3 x 10^8 cells /mL. Explain how to prepare a tube of 1 mL with cells at
2 x 10^5 cells/mL.

1b

Now that you have your tube of 1 mL with cells at 2 x 10^5/mL, explain how to add 10^4 cells to each
of 10 wells. Do you have enough cells?

Answer 1

1a) us e M1V1= M2V2

HERE, V1 is required , so V1 = M2V2/M1

V1 = 2×10^5 × 1 ml / 6.3× 10^8

V1 = 0.31 mcl , to be be disbsolved in water.

1b)

Put 0.05 ml in each well. Yes we have double the amount than what is required.

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