Dilutions

On this page:

AMOUNT vs CONCENTRATION

PERCENTAGES/PROPORTIONS/FRACTIONS

DILUTION FACTORS - Important formulae

EXAMPLES OF DILUTION PROBLEMS

SERIAL DILUTIONS

LINEAR DILUTION SEQUENCES

INTRODUCTION

A very important concept in biological science is how to adjust the concentration of solutions to required levels. Why do we need to do this? To adjust concentrations and volumes of solutions to within working limits i.e. so that you do not have to use megalitres of solvents or weigh out microgram quantities of materials, both of which would be impractical.

AMOUNT vs CONCENTRATION

An amount of something is a quantity i.e.a mass (grams) or a volume (litres). A concentration, on the other hand, is a mass in a specified volume (g/l or mol/l etc).

If you have a solution of NaCl with a concentration of 10 mg/ml and you have 100 mls of this solution in one container and 200 mls in another container, then the concentration of the salt is the same in each but the amount of dissolved salt in the first container is less than in the second container. The first container has 10mg/l x 100ml= 1000 mg salt, but the second container has 10 mg/ml x 200ml = 2000mg salt. If you take 10 mls (an aliquot) from each container, both lots of 10 mls have the same concentration and contain the same amount of salt.

A solution is simply a vehicle for the materials dissolved within it. You can manipulate amounts in solution as you would if they were solids.

PERCENTAGES/PROPORTIONS/FRACTIONS

Before we get on to dilutions, an understanding of the concept of proportion is essential.

Percentages

Percentage is a proportion of the total expressed as a value out of 100. It is calculated by taking the observed value, dividing it by the total and multiplying by 100 i.e.

For example, in a differential white blood cell count, 100 white cells are counted and assigned to 5 different categories

neutrophils 63
lymphocytes 22
monocytes 9
eosinophils 5
basophils 1

total 100

percent neutrophils = 63/100 x 100

= 63%

If the total is not 100, the principle is the same eg if 540 people are tested for their ABO blood group:

A 189
B 81
AB 54
O 216

Total 540

Percent group O = 216/540 x 100

= 40%

Proportion

Alternatively, if you know that 35% of the population have blood group A, then how many people in your test group would be expected to have group A, i.e. what proportion of the total would be group A?

= 189 people out of 540

Fractions

Fractions (numbers less than 1 but greater than 0) represent a proportion of a whole number and can be expressed as a percentage. Eg

50% equals 50 out of 100 or 50/100 which equals ½

Useful fractions to remember:

50% = ½
25% = ¼
20% = 1/5
33^1/3% = 1/3
10% = 1/10
12^1/2% = 1/8

Fractions and dilution factors are often used interchangeably but be sure you are certain of the meaning e.g.

A 1 in 2 dilution is often written as ½ and means that equal quantities of solution and diluent are mixed, resulting in a solution with half the original concentration. In this case the solution and diluent are mixed in equal proportions.

DILUTION FACTORS

The dilution factor is how much you wish to dilute a solution and can be calculated in 2 ways.

The dilution factor is the proportion of the new solution which is composed of the original solution (sample).

The volume of diluent that must be added to the sample is calculated by:

Diluent volume = total volume – sample volume

Remember the above two equations and you will be able to work out any dilution problem.

EXAMPLES OF DILUTION PROBLEMS

1. You are given the concentration of a starting solution (10mg/ml of NaCl), the concentration of the desired diluted solution (1 mg/ml of NaCl) and the required volume of the diluted solution (100 mls). How do you produce the right volume with the correct concentration?

Dilution Factor = 1 mg/ml/10 mg/ml = 1/10 (units cancel out in this case)

What is the sample volume (how many mls of 10mg/ml NaCl)?

sample volume/total volume = dilution factor = 1/10

Total volume = 100 mls

sample volume/100 = 1/10 hence

sample vol. = 100 x 1/10 = 10 mls

Diluent volume = total vol (100) - sample vol (10) = 90 mls

So you take 10 mls of the stock solution and add 90 mls of diluent (water) to make 100 mls of concentration 1mg/ml NaCl from a stock solution of 10mg/ml.

2. You want to make a 1 in a 100 dilution of a bacterial culture. What volumes do you use? You want 10 mls of the diluted culture.

Dilution factor = 1/100 (given)

sample volume/total volume = 1/100

total volume = 10 mls

sample volume = 1/100 x 10 = 0.1 ml

diluent volume = total – sample = 10.0 - 0.1 = 9.9 mls.

So you add 0.1 mls of culture to 9.9 mls of diluent (saline)

If the above original culture contained 2x10⁸ bacteria per ml and you diluted it 1 in 100, what is the concentration of the diluted culture?

Dilution factor = 1/100 = concentration of diluted culture/original concentration = conc dilute/2x10⁸

1/100 x 2x10⁸ = concentration dilute culture

concentration dilute culture = 2x10⁸/10²

Concentration diluted culture = = 2x10^8-2 = 2x10⁶ bacteria/ml

The volumes are irrelevant, but remember the units!

SERIAL DILUTIONS

Serial dilutions are a series of dilutions carried out consecutively i.e. one dilution is made from the previous one and so on. They can be serial 1 in 2 (doubling) or serial 1 in 10 or any other dilution factor. They represent an exponential series of dilution factors where the concentration of each dilution produced is a multiple of the next, with the multiple the same between each member of the series. They are used when the concentrations desired are an exponential sequence and not an arithmetic one.

Antibody titrations – serial 1 in 2 or doubling dilutions

All tubes contain 1 ml of diluent. 1 ml of serum is added to the first tube making a 1 in 2 dilution. 1 ml of this is removed and added to the second tube making a 1 in 4 dilution. 1 ml of this is removed and added to the third tube making a 1 in 8 dilution and 1 ml of this is removed and added to the fourth tube to make a 1 in 16 dilution. This can continue until the desired number of dilutions have been made, the excess 1 ml from the final tube being discarded. The dilutions are 1 in 2 dilutions because equal volumes of sample and diluent are mixed in each tube.

BACTERIAL VIABLE COUNT – SERIAL 1 IN 10 DILUTIONS

Each tube would contain 0.9 mls of diluent and 0.1 ml of culture would be added to the first tube, making a 1/10 or 10^-1 dilution. 0.1 ml of this dilution would be added to the next tube making a 1/100 or 10^-2 dilution and so on. In a viable count 0.1mL of each dilution is plated out and grown. Colonies are counted and from this information, the original culture concentration can be calculated. E.g.

Dilution factor Number of Colonies
1/10 or 10^-1 Confluent growth
1/100 or 10^-2 Confluent growth
1/1000 or 10^-3 >300 colonies
1/10,000 or 10^-4 285
1/100,000 or 10^-5 31
1/1,000,000 or 10^-6 3
1/10,000,000 or 10^-7 No growth

To calculate the starting concentration of bacteria, determine the concentration per ml for the plates which had countable numbers of colonies. In the above case, the 10^-4 plate is the most accurate. The number of colonies on the plate represents the number of bacteria in 0.1 ml (plated volume), so to determine bacteria per ml, multiply the number of colonies by 10. This gives a concentration of 2850 bacteria per ml which is the diluted concentration in the equation given at the start of this section. Dilution factor is 1/10⁴.

concentration of diluted sample/concentration original = DILUTION FACTOR

rearrange the equation:
concentration of original solution = concentration dilute/dilution factor = 2850/10^-4 = 2850 x 10⁴ per ml

= 2.85 x 10⁷ bacteria/ml

LINEAR DILUTION SEQUENCES

These sequences occur in assays such as colorimetric assays or antibiotic sensitivity tests where there are a number of standards, which are a linear or arithmetic sequence and not an exponential series.

If the standards are an exponential series e.g. 0, 1, 10 100, 1000 mg/ml then serial dilutions can be used.

If the standards are a linear series eg 0, 2, 4, 6, 8, 10 mg/ml then serial dilutions cannot be used. In this case a number of individual dilutions must be calculated for each standard.

Examples

1. Gentamicin assay using a stock solution of 100 μg/ml. Standards required are:

10, 8, 6, 4, 2, 1 μg/ml

You require 10 mls of each standard.

10 μg/ml Standard

Dilution factor = concentration dilute/original concentration = 10/100 = 1/10

Sample volume = total volume x dilution factor

Sample volume = 1/10 x 10 = 1 ml

Diluent volume = total volume – sample volume = 10-1 = 9 ml

For 10 μg/ml standard you add 1 ml of stock to 9 mls of diluent.

8 μg/ml standard

Dilution factor = 8/100

Sample volume = 8/100 x 10 = 0.8 ml stock solution

Diluent volume = 10 – 0.8 = 9.2 mls

For 8 μg/ml standard you add 0.8 ml of stock to 9.2 mls of diluent.

Continuing this process for the other standards, you can produce the following protocol:

Get into the habit of drawing up this type of table for your dilution protocols. It is easy to see exactly what must be done and it minimises pipette usage. For the above protocol you only need two pipettes, one for aliquoting the diluent and one for the stock solution. In contrast, with a serial dilution you need a new pipette for every tube to avoid carryover.

One feature of the above problem is the volumes used can be relatively accurately measured with either graduated pipettes or automatic pipettes, because the smallest volume is 100 μl (0.1ml).

If the concentration of the gentamicin stock solution in the above example was changed to 1 mg/ml, what would happen to the protocol?

The dilution factor for the 10 μg/ml standard then becomes

= 1/100

Sample volume becomes 0.1 ml and diluent volume 9.9ml

The entire protocol changes to

The volumes for the stock solution can be measured reasonably accurately with an autopipette, but volumes like 9.99 mls are very difficult to measure accurately. You are mixing a small volume with a very large one and this always leads to inaccuracy.

The answer is to predilute the stock solution.

In this case you would predilute the stock solution 1 in 10 to give a concentration of 100 μg/ml and then follow the protocol in example 1 using the diluted stock.

When a predilution is required, dilute the stock to the nearest multiple of 10 above the highest standard required. If the highest standard is a multiple of 10, dilute the stock to that value.

2. In a protein assay, the stock protein solution is 80g/l. The standards range is 0.1 to 0.8 g/l.

Dilute original stock solution 1 in 80 to give an intermediate stock of 1 g/l

Following the principles above: