In the previous unit, we learned about w/v calculations and serial dilutions. In this unit, we'll practice how to create an antibody solutions of a specific concentration.

In some applications, the concentration of a solute isn’t given in molarity, but instead as a ratio. This is especially common in experiments involving antibodies, such as Western blots, immunohistochemistry, or ELISA. For example, in the methods section of a scientific paper, you might see an antibody listed at a 1:500 dilution. This ratio tells you how the antibody was diluted into a solvent (usually a buffer) for use in the experiment.

Even though it’s presented as a ratio , you can also think of it as a fraction that describes the proportion of antibody in the total volume. A 1:500 dilution means 1 part antibody and 499 parts buffer, for a total of 500 parts. So, if you're making 1 mL of this dilution, you'd mix 2 μL of antibody with 998 μL of buffer (because 1/500 × 1000 μL = 2 μL).

\[\mathbf{\frac{1}{500} = {\displaystyle\frac{\mbox{Volume of antibody (in μl)}}{\mbox{Total volume of solvent (in μl)}}} = {\displaystyle\frac{\mbox{2 μl antibody}}{\mbox{1000 μl total (998 μl buffer + 2 μl antibody)}}}}\] These ratios are always provided in the most simple terms (i.e. with "1" as the numerator). This does not mean that the final volume of buffer used in the experiment was 500 μl; instead, it provides enough information so that other researchers could scale up (or scale down) by using a bit of math.


Why are antibody concentrations given as ratios rather than molarity?

Unlike small molecules, antibodies are large, complex proteins that are often sold as stock solutions with unknown or variable molarity. Their functional effectiveness in a given assay (like binding a target protein) is often determined experimentally rather than through precise molar concentrations. Because of this, labs typically report how well an antibody works at a certain dilution ratio, rather than needing the exact number of moles per liter.

If you are provided with a dilution ratio and a final volume, you can determine how much antibody to add through the same principles that we used before when making solutions:

  • C1 is the initial concentration of the antibody (also called the stock solution), which is equal to 1 (indicating that it is undiluted).
  • V1 is the volume of the stock solution that will be used to create the working solution; this is typically what you are solving for in these types of problems
  • C2 is the final concentration of the antibody after dilution.
  • V2 is the final volume of the working solution.

Let's practice with an example. If an antibody is to be used at 1:1000 in 1000 μl of buffer, then

  • C1 = 1 (stock solution)
  • V1 = the amount of antibody to be pipetted; this is what we will solve for in the equation.
  • C2 = 1:1000
  • V2 = 1000 μl
Therefore, the amount of antibody to be added would be \[\mathbf{C_1*V_1 = C_2*V_2}\] \[\mathbf{1*V_1 = (\frac{1}{1000})*1000}\] \[\mathbf{1*V_1 = (\frac{1}{1000}) * 1000 = 1}\] In other words, 1 μl of antibody would be added from the antibody stock solution to 1000 μl of buffer.





Let's practice this concept:

  • How many μl of antibody would you take from a stock solution if you wanted to create 3000 μl of a 1:500 working solution? uL

\[\mathbf{1* V_1 = (\frac{1}{500}) * 3000 = \mbox{6 μL}}\]

  • How many μl of antibody would you take from a stock solution if you wanted to create 1 ml of a 1:800 working solution? μL

  • How many μl of antibody would you take from a stock solution if you wanted to create 1000 μl of a 1:250 working solution? μL

  • How many μl of antibody would you take from a stock solution if you wanted to create 260 μL of a 1:500 working solution? μL

  • How many μl of antibody would you take from a stock solution if you wanted to create 100 μL of a 1:500 working solution? μL NOTE: Report value to the nearest tenth and round up (e.g. an answer of 10.38 would be reported as 10.4).

  • How many μl of antibody would you take from a stock solution if you wanted to create 900 μL of a 1:5000 working solution? uL NOTE: Report value to the nearest tenth and round up (e.g. an answer of 10.38 would be reported as 10.4).

Some antibodies are used at very small working dilutions. For example, let's say that an antibody you're interested in using for an experiment has a working concentration of 1:150,000 and that you want to create 1 ml (or 1000 μl) of the solution. Based on the math we've practiced above, this means that you would need to add

\[\mathbf{V_1 = (\frac{1}{150000}) * 1000 = 0.0066}\] or 0.0066 ul of the antibody stock solution into your buffer.

Perhaps you can see a problem ...

Most micropipettes cannot accurately deliver a volume less than 0.5 μl. Therefore, pipetting this small of a volume would not work. To do this experiment, you would need to create a serial dilution of the antibody stock that can then be further diluted to be used in the experiment.

A serial dilution is a stepwise process of diluting a substance in solution. It’s often used when the volume of solute needed (such as an antibody) is too small to be measured accurately with standard lab equipment. Serial dilutions allow you to start with a more measurable volume and dilute in stages to reach very low, precise concentrations.

For example, antibodies are often used at dilutions like 1:5000 or 1:10000, but directly measuring 0.05 μL or 0.01 μL for a final volume can be unreliable if not impossible. Instead, you can create a serial dilution:

  1. First, add 1 μL of concentrated antibody to 1000 μL of buffer to make a 1:1000 intermediate dilution.
  2. Then, further dilute this intermediate dilution to reach the desired final working dilution. For example, place 50 μl of the 1:1000 intermediate dilution into 450 μl buffer (1:10 dilution) to create a 1:10000 final dilution.

To make it easier to visualize, you may want to think in terms of the initial and final solution:

\[\mathbf{C_i*V_i = C_f*V_f}\] In other words,

  • Ci is the initial concentration of the antibody serial dilution
  • Vi is the volume of the serial dilution solution that will be used to create the working solutions.
  • Cf is the final concentration of the antibody working solution.
  • Vf is the final volume of the working solution.

Let's illustrate this with an example. Imagine that you have an antibody that has a working concentration of 1:150,000 and that you want to create 1ml (or 1000ul) of the solution. To make this possible, you've created a 1:000 serial dilution of the antibody.

  • C1 = 1:1000
  • V1 = the amount of the serial dilution to be pipetted; this is what we will solve for in the equation.
  • C2 = 1:150,000
  • V2 = 1000 μl

\[\mathbf{(\frac{1}{1000})*V_1 = (\frac{1}{150000})*1000}\] \[\mathbf{V_1 = 1000*(\frac{1}{150000}) * 1000 = 6.7}\] In other words, 6.7 μl of antibody would be added from the antibody serial dilution solution to 1000 μl of buffer to reach a final working concentration of 1:150,000.



NOTE: Serial dilutions can be made at any dilution. Some scientists like to determine the serial dilution by first considering how much of the serial dilution they would like to pipette (i.e. setting V1 to a specific volume). C1 is then calculated and the serial dilution is made.


Now let's put what we've learned into practice!

  • Let's say you have an antibody that is used at a 1:10,000 dilution. You create a 1:1000 serial dilution. How many μl of the serial dilution would be used to create 1000 μl of the working dilution? μL

\[\mathbf{(\frac{1}{1000})* V_1 = (\frac{1}{10,000}) * 1000}\] \[\mathbf{ V_1 = (\frac{1}{10,000}) * 1000 * 1000 = \mbox{100 μL}}\]

  • Let's say you have an antibody that is used at a 1:125,000 dilution. You create a 1:750 serial dilution. How many ul of the serial dilution would be used to create 500 μl of the working solution? μL

To create a 1:750 serial dilution, use of an undiluted solution to create of the buffer-antibody solution.

  • Let's say you have an antibody that is used at a 1:80,000 dilution. You create a 1:5000 serial dilution. How many μl of the serial dilution would be used to create 6ml of the working solution? μL

  • For an antibody used at a 1:1000 dilution, how many μl would you use from a 1:1000 serial dilution to create 1000 μl of the working solution? μL NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:4000 dilution, how many μl would you use from a 1:1000 serial dilution to create 500 μl of the working solution? μL NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:270000 dilution, how many μl would you use from a 1:2000 serial dilution to create 10 ml of the working solution? μL NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:55000 dilution, how many μl would you use from a 1:1200 serial dilution to create 15 ml of the working solution? μL NOTE: Report value to the nearest tenth.

In the Lab Math series, you have practiced the following concepts:

  1. Concentrations and unit conversions
  2. Creating solutions
  3. Creating antibody solutions

Ready to put it all together? Click here to go to the final module.





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