illustrate how you would make 4 serial 4fold dilutions of an

illustrate how you would make 4 serial 4-fold dilutions of an original stock soluton. Indicate for each tubem including original stock solution, the original stock solution, the following information: a) what is the n-fold for each tube? b) what is the relative concent concentration for each tube? c) what is the actual concentration in each tube if the original concentration=1M? d) if the original tube has 333 cells per mL, how many cells per mL are in each dilution tube? e) If you pipetted 0.25mL of the second dilution tube unto a cuvette, how many cells would there be in the cuvette?

Solution

I) As we are serially diluting, the serial n factor for each tube will be 4 four, 16 fold and 64 fold ( wrt to stock) respectively. ( serial dilutions get multiplied - this is by definition)

2) The relative concentration of each the three tubes wrt the stock will be 0.25X, 0.0625X, 0.015625X, assuming concentration of stock is X.( n- fold simply means a dilution of 1/n- this is also by definition)

3) if the original concentration is 1 M, this means there is 1 mole of solute in 1 liter or 1000 ml of solution. For first four-fold dilution, we have to take one part of the stock to three parts of the solvent such that overall we have four parts. Assuming we take 1 ml of solution and add it to 3 ml of solvent, we will now get a solution that has (1/1000)/4 moles/ml or 0.00025 moles/ml or 0.25 moles/litre or simpy 0.25 M. We can see therefore that this matches clearly what we have calculated in part 2 of the question, because, actually a four-fold dilution is nothing but a 1/4 dilution. Therefore the next two tubes will have concentrations of 0.0625M and 0.015625M respectively.

d) Again, we have to simply divide by 4 each time, so the first tube will have 100 cells/ml, the second will have 25 cells/ml and the third approximately 6.25 or 6 cells/ml ( cells cannot be a fraction).

Another way to visualize practically or understand n-fold dilutions is as follows: - ( in this case four-fold)

First tube: 1 ml of stock + 3 ml of solvent = 400 cells / (3+ 1) ml of solution or 400/4 = 100 cells/ml

Second tube: 1 ml of tube one + 3 ml of solvent = 100 cells/ (3+ 1) ml of solution or 100/4 = 25 cells/ml

Third tube: 1 ml of tube two + 3 ml of solvent = 25 cells / (3+1) ml of solution = 6.25 or 6 cells/ml

Always remember, four fold means we want final parts of the dilution to be four. The easiest way to do this is taken one part stock plus 3 parts of solvent. This can also be scaled up to 2 parts stock plus 6 parts solvent and so on.

5) As we have calculated in the previous question, the concentration of the second dilution tube was 100cells/ml.

therefore in 0.25 ml there are 0.25 x 100 or 25 cells present.