Question

suppose you were instructed to add .2ml of sample (tube A) to 9.8ml of diluent, but instead added 2.0ml of sample.

a) what was the intended dilution and what was the actual dilution in the tube (tube B)?

b) transfer 1ml of the culture from tube B into a tube containing 20ml of diluent (tube C). Mix well, transfer 3ml of the culture from tube C into a tube containing 5 ml of diluent (tube D). Is it a dilution series or serial dilution?

Answer 1

a) Intended amount of sample from tube A = 0.2 mL, Amount of diluent in tube B = 9.8 mL, Actual amount of sample added from tube A = 2.0 mL

Therefore, intended dilution in tube B = (0.2/10) = 2/100 = 1/50

Therefore, actual dilution in tube B = (2/11.8) = 20/118 = 10/59

b) After transferring 1 mL of culture from tube B into tube C containing 20 mL of diluent, the dilution in tube C = 1/21

But, the actual dilution in tube C = (10/59) x (1/21) = 10/1239

Now, after transferring 3 mL of the culture from tube C into tube D containing 5 mL of diluent, the dilution in tube D = 3/8

But, the actual dilution in tube D = (10/1239) x (3/8) = 10/3304

In, serial dilution, dilution factor is constant & the final total dilution is a product of each individual dilution in the series. Clearly this is not in our case. So, here it is in dilution series.