Question

Lab 5: Practice Dilution Problems

You have a stock solution that contains 24 mg/ml of protein. You want to make a set of standards with the following concentrations:

                  8 mg/ml

                  4 mg/ml

                  2 mg/ml

You need a minimum of 300 ul of each standard.

Using the technique of serial dilution, explain how you would make your standards by filling in the chart below.

	Standard 1	Standard 2	Standard 3
concentration	8 mg/ml	4 mg/ml	2 mg/ml
dilution factor/ from
ml of concentrate
ml of diluent
final dilution of stock
final volume/tube

        

        

You have a stock solution that contains 200 mg/ml of protein. You want to make a set of standards with the following concentrations :

                  100 mg/ml   12.50 mg/ml

                  50 mg/ml    6.25 mg/ml

                  25 mg/ml    3.125 mg/ml

You need a minimum of 300 ul of each standard.

Using the technique of serial dilution, explain how you would make your standards by filling in the chart below.

	Std 1	Std 2	Std 3	Std 4	Std 5	Std 6
concentration
dilution factor / from
ml of concentrate
ml of diluent
final dilution of stock
final volume/tube

Dilute the following making sure you end up with a minimum volume of at least 10 ml.

stock conc	desired conc	dilution factor	volume stock	volume diluent
25 mg/ml	5 mg/ml
30 mg/ml	2 mg/ml
50 mg/ml	0.5 mg/ml
100 mg/ml	25 mg/ml
5 mg/ml	0.1 mg/ml
3 mg/ml	1.5 mg/ml
20% (v/v)	0.5% (v/v)
10x	1x
5x	1x
25% (w/v)	5% (w/v)
106 cells/ml	104 cells/ml
200 ppm	1 ppm
30% (w/v)	20% (w/v)
100% (v/v)	75% (v/v)

Answer 1

Say we need to make 2 mg/ml of the solution when stock available is 8 mg/ml. We can either directly implement,

S₁V₁=S₂V₂, where S₁ is the initial strength available, V₁ is the volume of this solution that should be taken to make V₂ volume of the solution of strength S₂ . Or we can go about this gradually, meaning we create first 4 mg/ml by diluting 8mg/ml with an equal volume, then further dilute 4 mg/ml by an equal volume to get 2 mg/ml. When a series of dilutions are done we call the method as serial dilution, it prevents us from having to measure very smalls volumes as V₁ .

S₁V₁=S₂V₂

24* V₁= 8 * 0.5

V₁= 0.167 ml, so we take 0.167 ml of the stock and add 0.333 ml of the diluent

For simplicity, I have taken 0.5 ml (500 microlitres) rather than 0.3 ml for calculations