1 Laboratory math and experimental design

Experimental design review

Good experiments start with good planning. The analysis of genetic and molecular information involves increasingly sophisticated tools and methodologies, many of which are advancing at a fast pace. To prepare you for this research, we need to start with the basics of experimental design.

• Observations are records of quantitative or qualitative data. In other words, data are things that a person can see or measure; for example, you can measure the concentration of something with a spectrophotometer.
• Independent variable(s) are factors that the investigator deliberately changes or manipulates. These are usually graphed on the X axis.
• The dependent variable is measured in the experiment (and usually graphed on the Y axis). As the biologist changes a given independent variable, those factors that depend on the independent variable will change. Thus, the biologist measures the dependent variables in the attempt to understand the relationship of independent and dependent variables. A negative control treatment can be no treatment at all, or a treatment that should not cause an effect.
• A positive control treatment is a treatment that is known to have an effect under standard conditions. Including negative and positive controls helps you know your experimental design is effective.
• Standardized variables are those variables that may cause changes in the dependent variables but which a scientist attempts to keep equal across all treatments. In other words, this is something the investigator is not experimenting with at this time. For example, in a plant growth experiment, a biologist might need to standardize the amount of water given to plants while investigating how amounts of light affect growth.

Lab Math Review

Biologists learn chemistry for a reason: all of life is delicately controlled chemical reactions. Biologists need to exactly measure the amount of a chemical or a solute in solution. For example, knowing the exact amount of glucose in the bloodstream before and after eating can be an indication how well a human processes sugar and if they are at risk for diabetes.

In the lab, you may see three different ways of expressing concentrations: molarity (M), percent (%) and times concentration (X). The only one you might see in daily life is the X – and if you have followed the directions to dilute frozen OJ, you have diluted a 4-5X stock to 1X to drink. Percentages are based on a weight to volume or volume to volume measure based on 100 ml of solvent (typically water). So, a 95% ethanol solution is 95 ml of ethanol with 5 ml of water (solvent). Molarity is the most important concentration expression in molecular labs since it allows us to work with precise numbers of moles or molecules in solution (1M= 1FW / 1L). Once we know the concentration of a stock, we can dilute to our desired concentration using this equation.

There is very little that you have to memorize for lab math. However, there are a few things you DO need to just know – and from these you can figure out just about everything else.

Units.  International units have special prefixes to indicate orders of magnitude, so that scientific notation is not always necessary.  A Kg is 1000 grams.  We use a lot of small quantities in molecular biology labs, so you will need to use the following prefixes:

Mole:   6.02 x 10²³ atoms (or molecules)

The weight in grams of one mole of a substance is whatever 6.02 x 1023 molecules of it weighs – the larger the molecule, the more one mole of it will weigh.The formula weight gives you the number of grams that is in one mole of the substance.

(Generally formula weights are on the bottle, but you can calculate them from the formula by summing the atomic weights of the components)

The abbreviation for mole is mol (not M, see below)

Molarity – very important in solution chemistry, since it describes a concentration.

The molarity is the number of moles per liter. Its abbreviation is M

A 1 molar solution (1M) is 1 mol/L (note that 1 mmol/1mL is also 1 M!)

A 1 mM solution is 1 mmol/L (or 1 umol/mL)

Note:  Do not confuse amount (mol) with concentration (M)

%  means parts of a hundred (per cent)

1% means one part out of one hundred:  1 g/100g  (w/w)

Since in solutions our solvent is frequently water (with a density of 1 g/mL), we often express the solution as   % w/v, or g/100 mL (i.e. 20% sucrose = 20 g sucrose/100 ml (final volume)

If the solute comes in liquid form, it may also be v/v  (20 mL + 80 mL H2O, 100 mL final vol)

How to solve lab math problems

Most of the problems that you solve in the lab involve the preparation of solutions.  I solve at least 95% of practical problems in the lab using either Dimensional Analysis or the CV=CV equation.

• 1) when you have solved the problem, think about the answer: DOES IT MAKE SENSE?
• 2) If it does, great – still check it: try to solve it another way, or go backwards!

Dimensional Analysis

This is a very useful technique.  Look at what you have, where you want to be, and what you need to know to get there, and set up the necessary conversion factors to get you there.

Consider:  How many meters in a mile?

What you want to find out is the # meters = 1 mile.

The only conversion between metric and English distance that I have committed to memory is 2.54 cm/in.  I do also know that there are 5280 ft in a mile, and 12 inches in a foot, and a cm is 1/100 of a meter.

Start with the original question and then arrange the conversion factors so that the units cancel each other out:

Keep your wits about you when you do this, especially on the “easy” metric conversions (sometimes people slip and put 100 meter per cm, which makes the answer hugely wrong). Once you cancel out the units, you will be left with meters, and then just crank the math.

You follow the same strategy to do solution chemistry.  Consider:

How many grams of NaCl do you need to make 200 mL a 0.5 M solution.? Or:

#  grams NaCl = 200 mL of a 0.5 M solution

First recognize that when you set up a problem and say “of” it is a signal to multiply.   Recast the M as mol/L and you will see that if you do multiply, you can cancel out your volume terms (just change the 200 mL to 0.2 L, or 1 L to 1000 mL)

Amount of NaCl (mol) = 200 mL x [latex]\frac{0.5 mol}{1000 mL}[/latex]

Now you need to find the number of grams in that many moles using the formula weight.  (For NaCl, the FW is 58.4, so there are 58.4 g in one mole) Arrange this so that moles cancel and leave you with grams:

Then just do the arithmetic! To prepare this solution, don’t forget that the NaCl takes up space in the final solution. So to make 200 mL of this solution, you cannot add 200 mL water. Instead, you dissolve the solute in ~50% of the final volume (in this case, 100 mL), and then add solvent until the final volume is reached (determined by a graduated cylinder or volumetric flask).

Image from LibreTexts, CC-BY-SA

The C1V1 equation   … or actually    C1V1 = CfVf           (f = final)

This equation is used for making dilutions.  We often make up concentrated versions of solutions and then dilute them as needed because it is easier and more accurate than weighing out small amounts of dry reagents all the time. This works because concentration x volume = amount.   A solution that is 4 M  is  4 mol /L, if you multiply that by a volume in L, you will be left with an amount.  If you have 10 mL of 4 M NaCl, you have 4 mol/L x .01 L or .04 moles.

To set up a C1V1 equation, consider the question:   How much (volume) of a stock solution do I need to get the volume and concentration that I want (what is the amount of the first solution that I need to get to my dilution?)

You want 300 mL of a 0.14 M solution of NaCl, and you have a stock solution of 4 M NaCl.

How many mL of 4 M NaCl do I need to give me 300 mL of 0.14 M?

And write :
(x mL) (4 M)  =  (300 mL) (0.14M)      (“of” tells you to multiply)

Solve it:
Since all the units match, you can now just ignore them and solve, and x will be in mL

4 x = 300 x 0.14;  x = 42/4  =  10.5 mL.  Add this to 289.5 mL and you have your 0.14 M solution.

Once you have solved the equation, you should do two things.

1. Ask yourself if it makes sense.  If for example, you got a volume larger than the original volume you had, perhaps you set things up wrong!  Check!
2. Plug the solution back into the equation, and solve for one of the other terms.

If your units do not match, convert them:   How many mL of a 0.2 M solution do I need to make 5 L of a 20 mM solution?

Set it up:         (x mL) (0.2 M) = (5 L )(20 mM)     But you cannot do the math yet; the units are not the same, so fix that:

(x L) (200 mM) = (5 L )(20 mM)    Now they match, so do the math

The biggest mistake people make with the C1V1 equation is when they try to use an amount instead of a concentration.  THIS DOES NOT WORK!

 A couple of special applications for this useful C1V1 tool:

One thing the C1V1 is very helpful for is when you must adjust the concentration of a solute that is contributed from more than one source.  Here is an example:

Let’s say you are doing a restriction enzyme digest in two steps.

In your first step, you have a 20 uL reaction that has (among other things) 50 mM NaCl.  After an hour, you want to add a different enzyme and increase the volume to 40 uL, and also increase the salt to 100 mM (because the 2nd enzyme works better at the higher salt).  Clearly, if you just add 20 uL of enzyme and water to the first reaction, you will dilute the salt to 25 mM, so you will have to add some concentrated salt as well.   But how much?  Let’s say the stock is 1 M.

The question:  How much 1 M NaCl do I add to 20 uL of 50 mM NaCl so that when I make the final 40 uL it will be 100 mM salt?

The equation:  (x) (1 M) + (20 uL) (50 mM) =  40 uL (100 mM)   [note that you have a term for each salt source]

Fix the units: (x) (1000 mM) + (20 uL) (50 mM) =  40 uL (100 mM)

OK, now they are all the same, so that means that x will be in uL, and you can now ignore the other units and crank the numbers:

1000 x + 20 x 50 = 40 x 100

1000 x + 1000 = 4000

1000 x = 3000,      so x = 3 uL.

Modified from a document created By K. Cawley, Portland Community College