# Math in Clinical Chemistry

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Clinical Chemistry Note on Math in Clinical Chemistry, created by liebmansb on 07/10/2013.
Note by liebmansb, updated more than 1 year ago
 Created by liebmansb almost 11 years ago
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## Resource summary

### Page 1

1. Simple Dilution (Dilution Factor Method based on ratios) A simple dilution is one in which a unit volume of a liquid material of interest is combined with an appropriate volume of a solventliquid to achieve the desired concentration. The dilution factor is the total number of unit volumes in which your material will be dissolved. The diluted material must then be thoroughly mixed to achieve the true dilution. For example, a 1:5 dilution (verbalize as "1 to 5" dilution) entails combining 1 unit volume of solute (the material to be diluted) + 4 unit volumes of the solvent medium (hence, 1 + 4 = 5 = dilution factor). The dilution factor is frequently expressed using exponents: 1:5 would be 5e-1; 1:100 would be 10e-2, and so on. Example 1: Frozen orange juice concentrate is usually diluted with 4 additional cans of cold water (the dilution solvent) giving a dilution factor of 5, i.e., the orange concentrate represents one unit volume to which you have added 4 more cans (same unit volumes) of water. So the orange concentrate is now distributed through 5 unit volumes. This would be called a 1:5 dilution, and the OJ is now 1/5 as concentrated as it was originally. So, in a simple dilution, add one less unit volume of solvent than the desired dilution factor value.Example 2: Suppose you must prepare 400 ml of a disinfectant that requires 1:8 dilution from a concentrated stock solution with water. Divide the volume needed by the dilution factor (400 ml / 8 = 50 ml) to determine the unit volume. The dilution is then done as 50 ml concentrated disinfectant + 350 ml water. Top of page 2. Serial Dilution A serial dilution is simply a series of simple dilutions which amplifies the dilution factor quickly beginning with a small initial quantity of material (i.e., bacterial culture, a chemical, orange juice, etc.). The source of dilution material (solute) for each step comes from the diluted material of the previous dilution step. In a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step leading up to it.  Final dilution factor (DF) = DF1 * DF2 * DF3 etc. Example: In a typical microbiology exercise the students perform a three step 1:100 serial dilution of a bacterial culture (see figure below) in the process of quantifying the number of viable bacteria in a culture (see figure below). Each step in this example uses a 1 ml total volume. The initial step combines 1 unit volume of bacterial culture (10 ul) with 99 unit volumes of broth (990 ul) = 1:100 dilution. In the second step, one unit volume of the 1:100 dilution is combined with 99 unit volumes of broth now yielding a total dilution of 1:100x100 = 1:10,000 dilution. Repeated again (the third step) the total dilution would be 1:100x10,000 = 1:1,000,000 total dilution. The concentration of bacteria is now one million timesless than in the original sample.

3. Making fixed volumes of specific concentrations from liquid reagents: V1C1=V2C2 Method Very often you will need to make a specific volume of known concentration from stock solutions, or perhaps due to limited availability of liquid materials (some chemicals are very expensive and are only sold and used in small quantities, e.g., micrograms), or to limit the amount of chemical waste. The formula below is a quick approach to calculating such dilutions where:  V = volume, C = concentration; in whatever units you are working.   (stock solution attributes) V1C1=V2C2 (new solution attributes) Example: Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin (= C1) and you want to make 200 ul (= V2) of solution having 25 mg/ ml (= C2). You need to know what volume (V1) of the stock to use as part of the 200 ul total volume needed.V1 = the volume of stock you will start with. This is your unknown.C1 = 100 mg/ ml in the stock solutionV2 = total volume needed at the new concentration = 200 ul = 0.2 mlC2 = the new concentration = 25 mg/ ml By algebraic rearrangement: V1 = (V2 x C2) / C1V1 = (0.2 ml x 25 mg/ml) / 100 mg/mland after cancelling the units,V1 = 0.05 ml, or 50 ul So, you would take 0.05 ml = 50 ul of stock solution and dilute it with 150 ul of solvent to get the 200 ul of 25 mg/ ml solution needed. Remember that the amount of solvent used is based upon the final volume needed, so you have to subtract the starting volume form the final to calculate it.

Dilution

Concentration

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