The following modules will allow you to practice converting between units and creating solutions. These are skills that are super useful in any lab setting, so the overall goal of this exercise is to help you feel comfortable when doing this type of lab math.

Each module will have an explanation and a series of questions. To check your answers, click the "Show Answers" button. If you have an incorrect answer and want to try again, no problem! Simply re-enter your new answer and it will automatically be checked. If you don't want to see the feedback immediately, re-click the "Show Answers" button to toggle off the feedback.

This module is based on the Web Exercise template created by the #PsyTeachR team at the University of Glasgow, based on ideas from Software Carpentry.

Getting comfortable with units

Volumes

This section will help you practice converting between units of volume. Milliliters are abbreviated as 'mL' and microliters as 'uL'. Milliliters and microliters are volumes with a base in liters: one milliliter is one thousandth of a liter. The prefix 'milli' means 'thousandth'. Therefore, 5 milliliters is equivalent to 5/1,000 L or 0.005 L. The prefix micro means 'millionth'; Therefore, 5 microliters is equivalent to 5/1,000,000 or 0.000005 L.

Now let's practice these conversions!

  • 100 mL is L

\[\mathbf{100ml * (\frac{1L}{10000ml}) = 0.1L}\]

  • 47 mL is L

  • 641 mL is L

  • 0.07 L is mL

  • 0.009 L is mL

  • 0.001 L is mL

  • 1 L is equivalent to

Because milliliters and microliters are on the same 'liters' base scale, this means they are related to one another. One milliliter is equivalent to one thousand microliters; conversely, one thousand microliters are equivalent to one milliliter. Therefore, 5 microliters are 5/1000 or 0.005 mL.

Let's practice these conversions!

  • 94 ul is mL

\[\mathbf{94ul * (\frac{1mL}{1000ul}) = 0.094mL}\]

  • 207 uL is mL

  • 601 uL is mL

  • 0.013 mL is uL

  • 0.004 mL is uL

  • 0.001 mL is uL

  • 1000 uL are equivalent to

Concentrations

A concentration is a description of the amount of a substance within a volume. In biology and chemistry, concentrations are reported as molarities (abbreviated M), or the number of moles of a substance per liter of solvent. Therefore, 1M is equivalent to 1 mole of sbustance in 1 L of solvent.

Let's practice!

  • 100 M is moles of substance in 1 L of solvent.

\[\mathbf{100M * (\frac{1mole}{1L}) = 100moles}\]

  • 24 M is moles of substance in 1 L of solvent.

  • 53 M is moles of substance in 1 L of solvent.

  • 6 M is moles of substance in 500 mL of solvent.

  • 17 M is moles of substance in 500 mL of solvent.

  • 8 M is moles of substance in 100 mL of solvent.

When working in a lab, the concentrations are usually in the micromolar range. Recall that the prefix 'micro' means millionth. Because Molarity is abbreviated as 'M', micromolar as 'uM'. A molarity of 1 is equivalent to 1,000,000 uM. On the other hand, 1000 uM is equivalent to 0.001 (or 1000/1,000,000) M. For these questions, you'll also have to convert between uM and mM as well as mM and M, so be sure to pay attention to the units.

Let's practice these conversions! Note: If your answer is 1000 or above, there is no need to include commas (i.e 9999 instead of 9,999).

  • 1000 uM is M

\[\mathbf{1000uM * (\frac{1M}{1,000,000uM}) = 0.001M}\]

  • 650 uM is mM

  • 3600 uM is M

  • 246 uM is mM

  • 19 uM is mM

  • 0.01 M is uM

  • 0.003 M is uM

  • 0.00298 M is uM

  • 0.273 M is mM

  • 0.641 M is mM

  • 500 uM is equivalent to

Combine your knowledge from the volumes and concentrations sections to answer the questions below:

  • 50 uM is micromoles (umol) of substance in 1000 uL of solvent.

\[\mathbf{1000uL * (\frac{1L}{1,000,000uL}) * (\frac{50umol}{1L})= 0.05umol}\]

  • 750 mM is moles of substance in 100 mL of solvent.

  • 15 M is moles of substance in 100 mL of solvent.

Making solutions

This section will help you practice the relationship between volume and concentration. Let's say you have a solution that has volume V1 and concentration C1. If you add more solvent to the solution, you've now created a new volume (V2) and the concentration has also changed as the solute has become more diluted (C2).

In other words,

  • C1 is the initial concentration of the solution.
  • V1 is the initial volume of the solution.
  • C2 is the final concentration of the solution after dilution.
  • V2 is the final volume of the solution.

The relationship between these variables is represented by the equation \[\mathbf{C_1*V_1 = C_2*V_2}\] To solve for V1, divide C2*V2 by C1. In other words:

\[\mathbf{V_1} = \frac{C_2*V_2}{C_1}\] Note that the same equation applies of the second volume has been concentrated rather than diluted -- the math remains the same!

Now let's practice: Your lab manager asks you to make 500mL of a 500uM solution. You have a stock solution that is 1mM and a jug of deionized water.

Before solving for V1, make sure your units for C1 and C2 are the same. Notice that the C1 is in mM, so let’s convert C2 to mM as well:

  • 500 uM is equivalent to

  • C1 =

  • C2 =

  • V1 =

  • V2 =

Now solve for V1:

  • V1 is mL

To create 500mL of a 500uM solution, add of the 1mM solution to of water.

Use the same approach to try the following problem:

Your lab manager asks you to make 100mL of a 100uM solution. You have a stock solution that is 250uM and a jug of deionized water.

  • C1 = uM

  • C2 = uM

  • V2 = mL

Now do the calculations to make the solution:

To create 100mL of a 100uM solution, add mL of the 250uM solution to mL of water.

Your lab manager asks you to make 15mL of a 0.5uM solution. You have a stock solution that is 100uM and a jug of deionized water.

To create 15mL of a 0.5uM solution, add uL of the 100uM solution to mL of water.

\[\mathbf{C_1*V_1 = C_2*V_2}\] \[\mathbf{V_1* (100uM) = (15mL)*(0.5uM)}\] \[\mathbf{V_1 = (\frac{15mL*0.5uM}{100uM}) = 0.075mL * (\frac{1000uL}{1mL}) = 75 {\mbox{ ul of the 100uM solution}}}\] \[\mathbf{15mL {\mbox{ total solution}} - 0.075mL {\mbox{ 100uM solution}} = 14.925mL {\mbox{ of water}}}\]

Your lab manager asks you to make 390mL of a 5uM solution. You have a stock solution that is 500uM and a jug of deionized water.

To create 390 mL of a 5 uM solution, add mL of the 500uM solution to mL of water.

Your lab manager asks you to make 46mL of a 0.7uM solution. You have a stock solution that is 100uM and a jug of deionized water.

To create 46 mL of a 0.7 uM solution, add uL of the 100uM solution to mL of water.

Your lab manager asks you to make 10 mL of a 25uM solution. You have a stock solution that is 1M and a jug of deionized water.

To create 10 mL of a 25 uM solution, add uL of the 1M solution to 10 mL of water.

This solution is problematic because

A solution to this conundrum is to

This could be achieved by

This is known as creating a serial dilution. We'll cover more in the next parts of this tutorial!

From concentrations to dilutions and serial dilutions

Dilutions

In some applications, concentrations of a solute is not given in terms of molarity but is instead provided as a ratio. This is most commonly seen in experiments involving antibodies, such as Western blots or immunohistochemistry. For example, in the methods section of a paper, you may see that the working concentration of an antibody written as 1:500. This ratio gives you information about how the antibody was diluted into solvent (typically a buffer) for use. While it's presented as a ratio, you can also think of it like a fraction depicting the dilution of the antibody in a volume of solvent:

\[\mathbf{\frac{1}{500} = {\displaystyle\frac{\mbox{Volume of antibody (in ul)}}{\mbox{Volume of solvent (in ul)}}}}\] 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 500ul; instead, it provides enough information so that other researchers could scale up (or scale down) by using a bit of math.

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, which is equal to 1 (indicating that it is undiluted).
  • V1 is the initial volume of the solution.
  • C2 is the final concentration of the antibody after dilution.
  • V2 is the final volume of the solution.

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

  • C1 = 1
  • V1 is the amount of antibody to be pipetted; this is what we will solve for in the equation.
  • C2 = 1:1000
  • V2 = 1000ul
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, 1ul of antibody would be added from the antibody stock solution to 1000ul of buffer.





Now let's practice how this ratio would be used. If the working concentration of an antibody was listed as 1:500,

  • uL of this antibody would be used if you wanted to create 3000ul of the buffer-antibody solution.

\[\mathbf{1* V_1 = (\frac{1}{500}) * 500 = 1uL}\]

  • uL of this antibody would be used if you wanted to create 1mL of the buffer-antibody solution.

  • uL of this antibody would be used if you wanted to create 3mL of the buffer-antibody solution.

  • uL of this antibody would be used if you wanted to create 55mL of the buffer-antibody solution.



Why would different volumes of buffer be used in experiments? The answer depends on several factors such as the type of sample and the cost of the antibody. For example, when performing a Western blot, the membrane has to be submerged (or covered) by the diluted antibody solution. This typically requires at least several milliliters of solution. However, antibodies can be quite expensive (i.e. hundreds of dollars for several microliters!), so scientists want to conserve their reagents. Therefore, determining the final volume of solution to use requires a delicate balance of technique and thriftiness.




Practicing dilutions

Not all antibodies are used at a 1:500 dilution. Each dilution has to be determined empirically (i.e. by doing the experiment). In the examples below, determine the volume of antibody that would be added when given a working dilution and a volume of buffer.

  • For an antibody used at a 1:10 dilution, use uL from the stock solution to create 1000ul of the buffer-antibody solution.

\[\mathbf{1* V_1 = (\frac{1}{10}) * 1000 = 100uL}\]

  • For an antibody used at a 1:250 dilution, use uL from the stock solution to create 1000ul of the buffer-antibody solution.

  • For an antibody used at a 1:1000 dilution, use uL from the stock solution to create 1000ul of the buffer-antibody solution.

  • For an antibody used at a 1:30 dilution, use uL* from the stock solution to create 1000ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth and round up (e.g. an answer of 10.38 would be reported as 10.4).

  • For an antibody used at a 1:51 dilution, use uL* from the stock solution to create 1000ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth and round.

  • For an antibody used at a 1:1 dilution, use uL* from the stock solution to create 1000ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth and round up.

  • For an antibody used at a 1:100 dilution, use uL* from the stock solution to create 500ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth and round up.

Serial dilutions

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 1ml (or 1000ul) 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 0.0066ul of the antibody stock solution into your buffer.

Perhaps you can see a problem ...

Most micropipettes cannot deliver a volume less than 0.5ul. 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 dilution of a substance in solution. A serial dilution is needed when the volume of antibody (or any solute) to be added is lower than what can be reliably measured/delivered. A serial dilution can be created at any concentration, but typically they are created at 1:500 or 1:1000. Therefore, to create a serial dilution, 1ul of antibody from the stock solution would be added to a specific volume of solvent (typically a buffer). This solution is then further diluted to reach the final working dilution.

To determine how much of a serial dilution to add to the final solution, we return to the basics of what we've learned with making a solution: \[\mathbf{C_1*V_1 = C_2*V_2}\]

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 solution.
  • Vi is the initial volume of the solution.
  • Cf is the final concentration of the antibody solution.
  • Vf is the final volume of the solution.

To continue with our example from above, 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 is the amount of antibody to be pipetted; this is what we will solve for in the equation.
  • C2 = 1:150,000
  • V2 = 1000ul

\[\mathbf{(\frac{1}{1000})*V_1 = (\frac{1}{150000})*1000}\] \[\mathbf{V_1 = 1000*(\frac{1}{150000}) * 1000 = 6.7}\] In other words, 6.7ul of antibody would be added from the antibody serial dilution solution to 1000ul 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.


Practicing serial dilutions

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

  • For an antibody used at a 1:10,000 dilution, use uL from the 1:1000 serial dilution to create 1000ul of the buffer-antibody solution.

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

  • For an antibody used at a 1:40,000 dilution, use uL from the 1:1000 serial dilution to create 1000ul of the buffer-antibody solution.

  • For an antibody used at a 1:125,000 dilution, use uL from the 1:750 serial dilution to create 500ul of the buffer-antibody solution.

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

  • For an antibody used at a 1:80,000 dilution, use uL from the 1:5000 serial dilution to create 6ml of the buffer-antibody solution.

  • For an antibody used at a 1:4000 dilution, use uL* from the 1:1000 serial dilution to create 1000ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:6000 dilution, use uL* from the 1:1000 serial dilution to create 500ul of the buffer-antibody solution. | NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:18000 dilution, use uL* from the 1:2000 serial dilution to create 10ml of the buffer-antibody solution. | NOTE: Report value to the nearest tenth.

  • For an antibody used at a 1:58000 dilution, use uL* from the 1:1200 serial dilution to create 15ml of the buffer-antibody solution. | NOTE: Report value to the nearest tenth.