The Elisa guidebook

Table 3

Relationship of Weights

	Relationship
Unit	to gram
Gram	1
Milligram		10¨C3
Microgram		10¨C6
Nanogram		10¨C9
Picogram		10¨C12
Femtogram		10¨C15
Attogram		10¨C18

The dilution required is 1/100. Divide this into the required final volume:

This is the volume of neat (undiluted sample) to be added into the required volume in microliters, which is therefore 100 µL.

In this example there would be little difficulty in using microliters. Thus, 10 mL/100 = 0.1 microliters. We would therefore add 100 µL of the neat sample to 9900 µL (9.9 mL) of diluent (final volume minus the volume of the added neat sample). The conversion of the 0.1 mL into the units of the micropipet would then have to made, i.e., 0.1 mL = 100 µL.

A slightly more complex calculation illustrates the benefit of initial conversion to microliters of all the volumes.

8.2¡ª

Making a 1/200 Dilution of a Neat Sample in a Final Volume of 4 mL

Convert the required volume to microliters = 4 mL = 4000 µL. The dilution factor is 200. Therefore, we need 4000 µL/200 µL = 20 µL of neat sample. (A check on such calculations should always be made.)

Thus, the dilution factor ¡Á

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volume of sample being diluted should equal the required volume. We therefore have 200 ¡Á 20 µL = 4000 µL (4.0 mL). Note here by that using milliliters, we would have 4/200 = 0.02 mL. This becomes a little more difficult to relate to the microliter setting of the micropipets.

8.3¡ª

Calculating a Diluted Sample

When a sample is already diluted, we have to include this in the calculation. Thus, we have an already diluted sample at 1/50. We require a final dilution of 1/1000 in a final volume of 20 mL.

This type of calculation causes most problems. Convert the required volume to microliters. The final volume required = 20 mL = 20,000 µL. The dilution factor required is 1000.

Now, assume the sample were not already diluted. Then we would add 20,000/1000 = 20 µL of neat sample. Therefore, since the sample is already diluted 1/50, we have to add more. The factor is determined by the known dilution factor (50), so we multiply the value for undiluted sample by this factor:

This may seem obvious, but taking a more complex dilution.

8.4¡ª

Dilution of a Final Volume

We require a 1/18,000 dilution in a final volume of 27 mL. We already have a dilution of the sample at 1/25. By converting to microliters we have a required final volume is 27,000 µL. Assume the sample were undiluted, then we would require 27,000/18,000 = 1.5 µL. Since it is already diluted then multiply the 1.5 µL by the dilution factor:

Check: 18,000 ¡Á 37.5/25 = 27,000 µL = 27 mL.

When high dilutions are needed in small volumes, it may be necessary to make up a limited dilution series to avoid wasting reagents.

8.5¡ª

Requiring a High Dilution

Direct addition of the undiluted sample would require 5000/100,000 = 0.05 µL. This volume is impossible to pipet. For most practical purposes, the pipetting of volumes <5 µL is not recommended. The high dilution just given can be achieved by two manipulations. There are alternatives, and these are governed by the availability of the reagent being diluted. If it is available in only small volumes, then the initial volume used to make the first dilution can be made small. Thus, a 1/100 dilution can be made into 1000 µL by adding 10 µL of neat to 990 µL of diluent. Calculating the amount of this 1/100 dilution

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to be diluted as required in a final volume of 5 mL, we have 5000/100,000 ¡Á 100 (the already produced dilution factor) = 5 µL. Thus, by adding an initial dilution into 1 mL of buffer, we can produce high dilutions using acceptable pipetting volumes.

8.6¡ª

Reviewing of Method for Dilution

1.Determine the final volume required and convert this to microliters.

2.Divide this by the dilution factor. This is the sample volume to be added to the final volume of diluent in microliters.

3.When there has been predilution of the sample, follow steps 1 and 2, and then multiply the predilution factor by the volume found in step 2.

4.When high dilutions are needed, make two or three dilutions of sample in small volumes.

8.7¡ª

Compensation of Volumes

Since we require a final volume (calculated in stage 1 of Subheading 8.6., then addition of sample will obviously increase the final volume. As an extreme example, we may require a 1/10 dilution of a sample in 1 mL. This means adding 100 µL to a final volume of 1 mL. Following the procedures in Subheading 8.1¨C8.6., we would end up with a volume of 1100 µL. Thus, this would not achieve an accurate 1/10 dilution¨Crather 1/11. The obvious course is to compensate for the addition of the sample volume by taking removing this from the final volume calculation. Thus, in the same example, the final volume required is 1000 µL and the sample volume calculated is 100 µL. Therefore we need to remove 100 µL of diluent to compensate for the extra 100 µL of sample added, so we need 900 µL of diluent plus 100 µL of sample to achieve a perfect 1/10.

The accuracy then of the dilution depends on removing the sample volume from the final volume. However, this depends on the actual volume to be added. As an example, we may require a final dilution of 1/100 in 1 mL. Thus, we need a final volume of 1000 µL and need to add 10 µL of sample. Here, we could compensate by removing 10 µL of diluent before adding the 10 µL of sample, but, in terms of accuracy, we can see that the difference between the dilutions with and without compensation are minimal. Hence, with compensation we have a dilution of 10/1000 = 1/100, and, without compensation we have a dilution of 10/1010 = 1/101. Effectively there is no practical difference.

8.7.1¡ª

When to Compensate

There can be no exact rule about when to compensate because there may be occasions in which the activity of a sample is affected by a very small difference in dilution. However, this is not usually true in ELISA and as a strong

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guideline I suggest that when the volume of the sample being added is >2% of the required volume, compensation should always be used. Thus, to make a 1/200 dilution, 5 µL can be added to 1000 µL (0.5%); 10 µL can be added to 1000 µL (1%); 20 µL can be added to 1000 µL (2%); 40 µL should be added to 960 µL.

9¡ª Pipets

As in all assays, accurate pipetting is vitally important to obtain consistent results. Examine the multichannel pipet. It can be of fixed volume (dispensing a fixed specified volume) or of variable volume. For variable-volume pipets, the volume delivered can be adjusted by turning the knob at the top of the pipet so that the volume read on the side of the pipet handle is altered. These are digital pipets: the volume is shown as a number on the side in microliters.

Practice setting up different volumes. Remember to note where the comma (denoting the decimal point) is. Thus, 200 = 200 µL, and 20.0 = 20 µL.

Some pipet volumes are altered using a vernier scale. Follow the instructions provided by the manufacturer. Generally these are not recommended because it is easier to make mistakes with them, and the scale tends to wear. Multichannel pipets are designed to deliver 4, 8, or 12 vol simultaneously, and therefore are ideal for microtiter plates. The pipets having 12 channels offer highest flexibility in that up to 12 channels can be utilized. Any number of the channels can be loaded with a tip (up to 12). This fact often confuses workers when they first encounter such pipets. Practice the pipetting action and putting on tips.

9.1¡ª

Pipetting Action

Remember always to use the pipet whose maximum volume is nearest to the volume you require. All pipets should be calibrated on a routine basis (every month). Techniques for performing calibrations can be obtained from commercial companies supplying the pipets. Some companies also provide a calibration service. Special calibration tips with precise volumes marked on the outside allow routine examination of the volumes being dispensed. Pipets should also be checked for damage.

9.1.1¡ª

Picking up Solutions

Press the button on top of the pipet to the first stop before you put tips in solution, and then place tips in solution. Release the button steadily. You will notice that a volume of liquid is taken up into each tip. Check that each tip has the same volume. If not, expel the liquid after noting which tip was ''low" in volume. Press that tip on harder. Repeat the pipetting.

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9.1.2¡ª

Dispensing Solutions

Put the points of the tips in the wells resting on the sides of the plastic, if possible. Press the knob to the first stop. Solution will be expelled, try to pull the tips out up the side of the wells. When you have finished, either pull the tips off by hand or press the tip ejector on the side of the pipet; this works with some tips but not others.

9.1.3¡ª Single-Channel Pipets

Single-channel pipets are used to deliver single volumes of solution, particularly for small volumes. These can be the vernier or digital type, fixed or variable volumes.

9.1.4¡ª Troughs

Reservoirs for liquid dispensing by multichannel pipets are used as reservoirs (containers) for the solutions in the ELISA for multichannel pipets. These can be homemade or commercial, and either single or multiple troughs. Some are only suitable for eight channels.

9.2¡ª

Pipetting Exercise

To get used to the pipets, perform some simple exercises in making dilution series in microtiter plates. The following materials are needed:

1.Multichannel pipet.

2.Trough.

3.Tips.

4.Solution of dye (e.g., phenol red in water, trypan blue).

5.Water.

Making a dilution series is highly important in immunoassay. In most cases, multiple dilution ranges can be made using the multichannel pipets. A dilution series is given next.

For a twofold dilution series, in 50 µL vol add 50 µL of the substance to be diluted to 50 µL of diluent, mix, transfer 50 µL of this dilution to another 50 µL of diluent, and repeat to the required range.

In microtiter plates, the required volume of diluent is dispensed in the wells with the multichannel pipet. The substance being titrated is added either at the starting dilution (in the test volume) to the first well and to the second well as an equal volume of the starting dilution plus the volume of diluent in the well, or to the first row only in the test volume with the sample at twice the required initial concentration into the test volume of diluent. The solution in the first or second row is then mixed using the multichannel pipet, and the test volume is

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transferred to the next well containing the test volume of diluent, using the pipetting action of the multichannel pipet.

The process is repeated over all wells, each containing the test volume of diluent. In most cases carryover of reagent can be ignored when the same tips are used for the dilution series. Thus any number of dilution series up to 12 rows (8 dilutions) or 8 rows (12 dilutions) can be prepared simply. Thus, for the exercise, the steps are as follows:

1.Add 50 µL of diluent (water) to each well of a plate.

2.Add 50 µL of dye solution to the first row (A¨CH) of the plate using a multichannel pipet.

3.Mix by pipetting action (first stop only used to avoid frothing).

4.Transfer 50 µL of diluted dye from row 2 to 3, and mix.

5.Transfer 50 µL from row 3 to 4, and mix.

6.Repeat to row 12.

We now have 8 rows (A¨CH) of the same dilution range of the dye. This is a twofold range (equal volumes transferred) from 1/2, 1/4, 1/8, and so on, to 1/4096. You can make any fold dilution range provided the volumes you transfer are effectively large enough to be pipetted and mixed after transfer.

Examine the dilution range to ensure that the rows appear as if they have the same volume and that there is a logical dilution effect along the rows. Discard the first attempt, throw dye into sink, wash the plate under the tap, blot the plate dry, and repeat.

Try the following dilution ranges:

1.A 3-fold dilution range: 50 µL vol (25 µL carried over into 50 µL diluent).

2.A 4-fold dilution range: 50 µL vol (approx 17 µL carried over into 51 µL).

3.A 5-fold dilution range: 60 µL vol (15 µL carried over into 60 µL).

4.A 10-fold dilution range: 100 µL vol (11 µL carried over into 110 µL).

5. A 5-fold dilution range: 100 µL vol (25 µL carried over into 100 µL).

9.2.1¡ª

Effect of Different Dilution Series

The choice of which dilution range to use depends on what activities are being titrated. Thus, if there is a large quantity of antibody in a serum, then a high dilution is necessary in order to assess this.

Preliminary experiments to assess ELISAs often involve the titration of reagents of unknown strengths. In such cases, the appropriate selection of dilution ranges is important. It is simple to make two-, three-, and fourfold dilution ranges, but what are the advantages? Table 4 gives the effective dilutions of a sample using the different ranges over eight wells (eight dilution steps) and illustrates the ranges of dilutions covered by each.

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Table 4
Dilutions Obtained Using Different Dilution Series
				Well number
	1	2	3	4		5	6	7	8
Range				Dilutions produced
Twofold range	2	4		8	16	32	64	128	256
Threefold range	3	9		27	81	243	729	2178	6534
Fourfold range	4	16		64	256	1024	4096	16384	65536

This shows that a simple adjustment of diluting range dramatically increase the dilution of samples. The three-or fourfold range is convenient in that the range titrates samples at relatively high concentration. Thus, low-titre samples might be observed but it is also useful in case samples have high titers. In the case of the twofold range a high-titer sample would show color across all the wells and no titer would be indicated. Preliminary experiments can be followed with the most suitable ranges according to the titer established as in Table 4.

10¡ª Molarities

To make up a 1 M solution of a compound we take the molecular weight of that compound in grams and dissolve this to I L (final volume) of liquid (usually distilled water).

Thus, the molecular weight of a particular compound must be calculated from the atomic formula or read off the reagent bottle.

Take a simple example of sodium chloride (NaCl). The molecular weight is 58.5. Therefore 58.5 g made up to 1 L in distilled water represents a 1 M solution. A 0.1 M solution would contain 58.5/10 = 5.85 g/L.

Often we do not require a large volume so that 100 mL of a 1 M solution of NaCl would contain 5.85 g of NaCl since 1 L of 1 M contains 58.5 g, and 100 mL = 1/10 L. Therefore, we require 10 times less NaCl.

Another way to calculate this is to always calculate the amount of chemical needed to give the required molarity per milliliter. Thus, 1 M = 58.5 g/L = 0.0585 g/mL. If we require 50 mL at 1 M, we therefore require 50 ¡Á 0.0585 g made up to 50 mL. This helps calculation of more difficult molarities.

For example, we require a 0.125 M solution of NaCl in 35 mL. Calculate the number of grams per milliliters at the required molarity: 1 M = 58.5 g/L and 0.125 M = 58.5/8 (0.125 M/1 M) = 7.35 g/L = 0.0073 g/mL. We require a final volume of 35 mL. Therefore, 35 ¡Á 0.0073 = 0.256 g = 2.57 g (rounded up).

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For 2 M solutions, we require twice the weight of that required for a 1 M solution, which is 117 g/L in the case of NaCl. If we require 17 mL of a 2 M solution of NaCl, 1 M NaCl = 58.5 g/L and 2 M NaCl = 117.0 g/L = 117/1000 =0.117 g/mL. We need 17 mL = 0.117 ¡Á 17 = 1.99 g.

Again, a fraction of a molarity is best calculated by assessing needs per mL for example, we require a 0.15 M solution of NaCl in 25 mL. Therefore, 1 M NaCl = 58.5 g/L; 1.5 M = 1.5 ¡Á 58.5 g/L = 87.75 g/L; and 0.15 M = 87.5/10 = 8.75 g/L = .0088 g/mL. Hence, we need 25 mL = 25 ¡Á 0.0088 g = 0.22 g.

Sometimes molarities are expressed in millimolar quantities, e.g., 100 mM, 10 mM, 30 mM. One millimolar = 1/1000 M. Thus, for NaCl, 58.5/1000 = 0.0585 g/L = 1 mM.

Hence, a 10 mM solution contains 0.585 g/L, and a 100 mM solution contains 5.85 g/L.

References

1.Harlow, E., and Lane, D., eds. (1988) Antibodies¡ªA Laboratory Manual, Cold Spring Harbor Laboratory Press, Cold Spring Harbor, NY.

2.Catty, D., ed. (1988) Antibodies¡ªA Practical Approach Volume I, IRL Press, Oxford, England.

3.Catty, D., ed. (1988) Antibodies¡ªA Practical Approach Volume II, IRL Press, Oxford, England.

4.(1982) Linscott's Directory of Immunological and Biological Reagents, Linscott, Mill Valley, CA.

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6¡ª

Practical Exercises

The aim of this chapter is to illustrate the principles of ELISA by (1) showing worked examples of each assay, including diagrams of plates and representational data from assays, (2) analyzing such data in terms of important rules that are learned at each stage, and (3) providing full working instructions for investigators to be able to perform each assay so that they obtain their own data to be analyzed.

This chapter can be used in several ways. First, researchers without access to reagents will obtain a working knowledge of ELISA through the examples. Second, it can be used in training courses in which reagents may be provided (as indicated in the text). Third, the information will be useful to investigators who have already had some experience with the technique but may have had difficulties in obtaining and analyzing data.

Remember that it is the application of the ELISA to specific problems, and not the methodology for its own sake, that is the most important reason the techniques should be mastered.

1¡ª

Test Schemes

You are already familiar with the concepts in ELISA, whereby an antigen binds to an antibody that can be labeled with an enzyme or, in turn can be detected with a species-specific antibody (enzyme labeled). All the ELISAs described are variations on this theme. Inherent in the methods of ELISA is that one of the reagents is attached to a solid phase, making the separation of bound (reacted) and unbound (nonreacted) reagents simple by a washing step. Before performing ELISA on disease agents, it is useful to train operators how to use reagents of defined reactivity, which are easily available and which provide security problems. An ideal system is to use an immunoglobulin (Ig) and, more particularly, an immunoglobulin G (IgG) as an antigen. Do not get confused since you have learned that the antibody population contains high levels of IgG acting as antibody. In the context of learning the principles, we are

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using IgG as an antigenic protein, because, (1) IgG from one animal species can be injected into another animal species so that a specific antiserum to that IgG is prepared; and, (2) Such antibodies can be labeled with enzyme, or detected with a second species-specific antibody labeled with enzyme.

Such reagents are defined, easy to standardize, stable, and available commercially. The particular IgG system chosen in this chapter involves the guinea pig, but similar tests can be performed with IgG from other species using the appropriate antispecies reagents. The systems described are analogous to the ones most commonly used to examine problems associated with diagnosis.

The schemes are described as symbols and as practical exercises in full as follows:

I	= solid-phase microtiter plate well
Ag	= antigen
Ag1, Ag2, etc.	= particular antigens highlighted
I-Ag	= antigen passively attached to solid phase
I-Ab, I-AB	= particular antibodies passively coated to wells
Ab	= antibody
AB	= antibody from a different species to Ab
AbX, AbY	= different antibodies identified by subscript letters
Anti-Ab	= antispecies specific antibody (species in which Ab was made)
Anti-Ab*E	= antispecies-specific antibody labeled with enzyme
W	= washing step
+	= addition of reagents and incubation step
S	= substrate/chromophore addition
R	= reading the test in spectrophotometer.

Many of the practical steps are similar. The conjugates used are all horseradish peroxidase (HRP) and the substrate/chromophore is H2O2/ortho-phenylenediamine (OPD). The following practical details are helpful:

1. Substrate/chromophore. The easiest method is to use commercial tablets that are preweighed. Also, citrate/phosphate tablets can be purchased (pH 5.0). There are commercial preparations of this substrate/chromophore that require addition of only water. I recommend 30 mg tablets, which make 75 mL of solution in buffer. Unused OPD solution (without added hydrogen peroxide) can be stored at ¨C20¡ãC but should be examined closely for discoloration on thawing. Use the completely made up solution as soon as possible. All liquids should be at optimal temperature. The hydrogen peroxide can be purchased as 3 or 6% solution and should be stored as instructed by suppliers. Tablets of urea/peroxide can also be obtained and used to make up a stock of hydrogen peroxide of defined strength rather than purchasing liquid that has certain transportation restrictions. Hydrogen peroxide should be added to the required concentration just before addition to wells; for example, add 5 µL of hydrogen peroxide (30% w/v) to every 10 mL of OPD

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solution (in citrate phosphate buffer pH 5.0) or 25 µL of 6% (w/v) hydrogen peroxide. It is imperative that the strength of the hydrogen peroxide be accurate.

2.Washing solution is phosphate-buffered saline (PBS) without addition of Tween-20. Washing requires the addition and emptying of wells four times.

3.Blocking buffer is PBS containing a final concentration of 1% bovine serum albumin (BSA) and 0.05% Tween-

20.This should be made in volumes necessary to complete tests as required, but can be stored at 4¡ãC if contamination is avoided, and should always be warmed and inspected for contamination before use.