SIMPLE DILUTIONS

In many areas of the clinical laboratory, a specimen may have to be diluted so that it can be analyzed. In chemistry, the specimen may have a concentration outside of the linear range of the method or instrument used for analysis. Antibiotics are diluted in microbiology to determine the minimal inhibitory concentration for microorganisms. Serial dilutions are performed on serum in immunology to determine titers of antibodies. A poorly performed or interpreted dilution may lead to errors of analysis and subsequently affect patient treatment.

There are two parts to a dilution. The first part is the sample to be diluted; the second is the diluent used to perform the dilution. In Chapter 6 you will learn about solutions. In a solution there are also two parts: the solute (the part that is being placed into the solution) and the solvent (the liquid into which the solute is being diluted). The solute plus the solvent makes a solution. So for a dilution, the sample that is being diluted is the solute and the diluent is the solvent.

Many lyophilized quality control materials have an accompanying diluent to be used for reconstitution. This diluent may contain buffers necessary for the correct concentration of the analytes to be tested. If deionized water is used instead of the manufacturer’s supplied diluent, the control material will not be correctly buffered, and inaccurate results may occur. Therefore, it is essential that the proper diluent be used for reconstitution.

When a dilution is required, often it is referred to as a “1 to 10” or perhaps a “1 to 2” dilution. A 1 to 10 dilution means that for every 1 part of sample, there is a total of 10 parts of the solution. In a 1 to 10 dilution, 1 part of the sample is used and 9, not 10, parts of the diluent are used. If 10 parts of the diluent would be used, then there would be a total of 11 parts to the solution.

A simple dilution uses the following formula:

Sample volumeSample volume+Diluent volume

or

Sample volumeTotal volume

Notice that the sample volume is included in the denominator of the equation. This is because it is part of the total volume of the solution. Many dilution errors are made by forgetting this crucial fact.

Example 4–1

A 1 to 4 dilution of serum is to be made on a sample that is too high to measure on the chemistry analyzer for the creatinine method that is used. (When a result is too high for an instrument to measure it, the result is referred to as “outside of the linear range of the instrument.”) The total volume of the dilution is to be 100 μL. What volumes of serum and diluent (deionized water) are needed?

To solve this problem, use ratio and proportion.

Crossmultiplying the equation yields the following:

25.0 μL is the sample volume and 100.0 − 25.0 or 75.0 μL is the diluent volume.

Example 4–2

A 1 to 10 dilution must be prepared to make a total volume of 100.0 μL. How much serum must be used?

Again, use ratio and proportion to solve this problem.

Crossmultiplying the equation yields the following:

Therefore, 10 μL of serum will be used for the dilution. To determine the amount of diluent needed, subtract 10 from 100: 100 μL total volume −10 μL sample volume = 90 μL diluent volume.

Additional Examples

 A 1 to 20 dilution must be performed. How many parts of the sample must be used, and how many parts of the diluent must be used?

A 1 to 20 dilution consists of 1 part sample to 19 parts diluent.

 Using this 1 to 20 dilution, a total volume of 100 μL are needed. How is this dilution performed?

Using ratio and proportion, 5 μL of serum are used, and 95 μL of diluent are used to make a total volume of 100 μL. This is also 1 part serum to 19 parts diluent (5 × 19 = 95).

 A 1 to 4 dilution of a sample must be performed, with a total volume of 200 μL. How much serum would be needed?

Using ratio and proportion, 50 µL of serum would be needed and 150 µL of diluent would be used.

 If you had a total volume of 100 μL and you used 25 μL of serum to make your dilution, what was the value of the actual dilution you performed?

Now to add a little twist to it: determine the dilution that is performed given the sample volume and diluent volume.

Using ratio and proportion:

Crossmultiplying the equation yields:

Therefore, a dilution made with 25 μL in a total volume of 100 μL is a 1 to 4 dilution.

 If you had a total volume of 250 μL and you used 50 μL of serum to make your dilution, what would be the value of the actual dilution you performed?

Using ratio and proportion again, a dilution made with 50 μL in a total volume of 250 μL is a 1 to 5 dilution.

 If you had a total volume of 100 μL and you used 10 μL of serum to make your dilution, what would be the value of the actual dilution you performed?

The actual value of the dilution would be a 1 to 10 dilution.

DILUTION VARIATIONS

If you had a dilution where you took 0.5 parts sample to 9.5 parts diluent, what is the value of the actual dilution you performed?

The value would be a 1 to 20 dilution. At first, you might have thought that it is a 1 to 10 dilution; however, the dilution fraction’s numerator must be a whole number of at least 1.0. Therefore, the 0.5 has to be doubled to 1 and the denominator number of 10.0 (0.5 sample volume + 9.5 diluent volume) is doubled to 20.

DILUTED SPECIMEN VALUES

Once the specimen has been diluted and analyzed, the result obtained must be corrected for the dilution. A common preventable laboratory error is the failure to correct a diluted specimen’s result properly. An easy way to correct the result is to use a “factor.” This factor is the reciprocal of the dilution that was performed. In the first example and in the preceding example, a 1 to 4 dilution was performed. The factor for these dilutions is the reciprocal of 1 to 4, or 4.

1÷14OR114=4

The creatinine result obtained from Example 4–1 would be multiplied by 4 to obtain its true value.

Example 4–3

Using the creatinine example, suppose that the linear range of the creatinine assay the instrument could measure was from 0.1 mg/dL to 15.0 mg/dL. A patient sample is analyzed, and the result is too high to be measured on the instrument. A 1 to 5 dilution is performed on the serum and the reanalyzed diluted sample result is 5.0 mg/dL. The diluted sample result must be multiplied by the dilution factor to determine the concentration in the undiluted sample.

Therefore, the patient’s actual creatinine result is 25.0 mg/dL, not 5.0 mg/dL. Failure to perform this crucial step would mean the reporting of a creatinine level of 5.0 mg/dL instead of 25.0 mg/dL, which could be harmful to the patient.

Example 4–4

A patient’s glucose result is outside the linear range of the analyzer; 10.0 μL of serum are added to 90.0 μL of diluent and the diluted sample reanalyzed. The glucose value of the diluted sample is 75.0 mg/dL. What glucose value is reported to the physician, and what is the dilution factor that will be used to calculate this result?

The first step is to determine the dilution that was performed:

Next, the dilution factor is the reciprocal of the dilution that was performed, in this case 1 to 10.

The last step is to multiply the diluted result by the factor:

The technologist reports a glucose value of 750.0 mg/dL to the physician. This example demonstrates how critical it is to perform the dilution calculations correctly. In this example, a glucose value of 75.0 mg/dL is within normal range; a value of 750.0 mg/dL is life-threatening.

Additional Examples

 A patient’s AST value is outside of the linear range. The patient’s serum sample is diluted by taking 25 μL of serum and 75 μL of diluent. It is reanalyzed and the diluted value that is obtained is 400 IU. What is the actual true AST value?

The dilution was performed by adding 25 μL of serum to 75 μL of diluent in a 1 to 4 dilution (25100). The patient’s true AST value is 1600 IU (400 × 4).

 A different patient’s glucose sample needed to be diluted. An amount of 30 μL of serum is added to 150 μL of diluent. The diluted sample’s result was 145 mg/dL. What is the true glucose value?

The dilution that was performed was a 1 to 6 dilution. Therefore the patient’s actual glucose result is 145 × 6 or 870 mg/dL.

 A patient’s ALT value is outside of the linear range; 50 μL of serum was added to 200 μL of diluent. The diluted specimen was reanalyzed and a value of 350 IU was obtained. What is the patient’s true ALT value?

The dilution that was performed was a 1 to 5 dilution. Therefore, 350 × 5 = 1750, so the true ALT value is 1750 IU.

DILUTIONS VERSUS RATIOS

Learning about dilutions and ratios can be confusing for students because different textbooks refer to them in different ways. In this book, dilutions will be referred to with a slash, as in ½, whereas ratios will be referred to with a colon, as in 1:4. In a ratio, the numerator is the parts of the sample used, whereas the denominator is the parts or amounts of diluent used.

Radio:1:4=one part sample, four parts diluent

Contrast this with a dilution in which the numerator denotes the parts of the sample used but the denominator is the total parts (diluent plus the sample).

Dilution:14=one part sample, three parts diluent

In the above ratio example, a 1:4 ratio is a ⅕ dilution.

Another way of stating the difference between ratios and dilutions is that a dilution is formed by the ratio of the sample volume to the total volume. The ratio tells you how much of the sample and diluent are necessary to form the dilution. Using the above example, the 1:4 ratio tells you that you need one part sample to four parts diluent to make a ⅕ dilution.

Example 4–5

To clarify the difference between a ratio and a dilution, suppose 50 μL of serum was added to 100 μL of diluent. Consider the following questions:

What is the ratio of serum to diluent?

The ratio of serum to diluent is 50 parts serum to 100 parts diluent or a 1:2 (1 to 2) ratio.

What is the dilution in this example?

Fifty parts of serum were diluted into 100 parts of diluent. Using the dilution formula:

Therefore, 50 parts sample to 100 parts diluent is a 1 to 2 ratio, but a 1 to 3 dilution.

This is a critical distinction when calculating the original concentration as Example 4–4 demonstrates.

Additional Examples

 If 15 μL of sample was added to 90 μL of diluent, what is the ratio and what is the dilution?

15 μL added to 90 μL is a 1 to 6 ratio and a 1 to 7 dilution.

 If 30 μL of sample was added to 90 μL of diluent, what is the ratio and what is the dilution?

30 μL added to 90 μL is a 1 to 3 ratio and a 1 to 4 dilution.

 If 20 μL of sample was added to 80 μL of diluent, what is the ratio and what is the dilution?

20 μL added to 80 μL is a 1 to 4 ratio and a 1 to 5 dilution.

SERIAL DILUTIONS AND MULTIPLE DILUTIONS

Often in immunology or microbiology, serial dilutions are performed. An advantage of a serial dilution is that very large dilutions of the serum can occur with only a small amount of serum and diluent used. For example, suppose that a 11000 dilution of serum was needed for a test. Theoretically, 1.0 mL of sample and 999.0 mL of diluent could be used (or 0.5 mL sample and 499.5 mL diluent).

However, this is a large volume of diluent and sample. Using serial dilutions, much smaller sample and diluent volumes can be used. A serial dilution is simply a series of sequential dilutions of the sample until the final dilution is reached. By definition, a serial dilution uses the same dilution for each tube in the series. A multiple dilution series uses different dilutions for each tube. In both cases, the final dilution is the product of each individual dilution:

Final dilution=(dilution1)(dilution2)(dilution3),etc.

In immunology, serial dilutions are performed using the same dilution over and over again. If a serial dilution was composed of a series of 1 to 10 dilutions, that dilution series can be referred to as a “tenfold” dilution series. Common serial dilutions series used in immunology are ½, ¼, and 110. If reagents are to be added directly to serially diluted tubes, it is important to discard from the last tube in the series the same sample volume used for all other dilutions in the series. For example, if a tenfold dilution was performed with five tubes with a sample volume of 10 μL and a diluent volume of 90 μL, 10 μL from the fifth tube must be discarded. If the 10 μL are not removed from the last tube, its total volume would be 100 μL. However, all the other tubes have a total volume of 90 μL, as 10 μL are removed from each to be used in the next dilution in the series. By removing the 10 μL from the last tube, when additional reagents are added to the diluted tubes, all tubes in the reaction sequence will have the same volume.

In chemistry, samples may be serially diluted, or multiple dilution series may be used. In either a serial dilution or multiple dilution series, if the original concentration is known, then the concentration of each tube in the series can be calculated in one of two ways.

Example 4–6

A multiple dilution is performed on a sample to check the pipetting skills of a student clinical laboratory technician. Five tubes are used in the dilution. The concentration of the sample is 1650 mg/dL. The sample is diluted ⅕ (tube 1), rediluted ½ (tube 2), diluted again ¼ (tube 3), and then ⅕ (tube 4), and then diluted 110 (tube 5). After the student performs the dilution, the diluted specimen will be analyzed, and the student will be graded for precision of pipetting. What will be the diluted concentration in each tube? Figure 4–1 is a schematic of the multiple dilution.

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1. Immunohematology Laboratory
2. Quality Assurance and Quality Control in the Clinical Laboratory
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4. Calculations Associated with Solutions

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