Note: Content of this page is modified from the Cell Biology Laboratory Manual by Dr. William H. Heidcamp, Biology Department, Gustavus Adolphus College
Cell biology deals with things which are relatively small. The units of measurement typically used are the micron at the light microscope level, and the nanometer at the electron microscope level. For molecular measurements, the norm is the Angstrom. These units are defined within the following table:
Measure Symbol Relative Length Exponential Notation
|Measure||Symbol||Relative Length||Exponential Notation|
|Micrometer or micron||µ||.000001||10⁻⁶|
From this table it is apparent that:
- 10 Å = 1 nm
- 1000 nm = 1 mm
- 10 mm = 1 cm
Not apparent are that:
- 1 inch = 2.54 cm = 25.4 mm = 25,400 µ = 25,400,000 nm
- 1 mm = 0.04 inches
Volumes are measured relative to a liter, with the most commonly used measurements, the milliliter and the microliter.
The following table gives the relative volumes:
|Measure||Symbol||Relative Volume||Exponential Notation|
- 1,000 µl = 1 ml
- 1 ml = 1 cm3
- 1 gallon = 3.8 liters
- 1 quart = 0.95 liters
- 1 liquid ounce = 29.6 ml
- 1 drop of water = 50µ l
The most common measurements of weight at the gram, milligram and microgram.
|Measure||Symbol||Relative Weight||Exponential Notation|
NOTE: the information below regarding Molarity and Percent Consenctration is for general reference and will not be required knowledge for our first lab project.
Most concentrations used throughout cell biology are those of a solute disolved or suspended within a solvent, and in most cases the solvent is water. There are two general methods of identifying the concentration of a solution; as molarity or as a percent. Molarity is based on the number of moles of solute in the solvent, while percent is based on the number of parts, either grams (for a solid solute) or milliliters (for a liquid solute). Molarity equals the number of moles of solute in 1 liter of solution. A mole is equal to the gram molecular weight (or formula weight) of the solute. Sodium Chloride (NaCl), for example has a formula weight of 58.43 (22.98 for Na and 35.43 for Cl). Thus, if 58.43 grams of NaCl are dissolved in 1 liter of water, the result would equal a 1 molar solution of NaCl. This is
designated as 1 M NaCl, or as simple M NaCl. We often deal with solutions of less volume than 1 liter, and the following should be noted:
1 M NaCl = 58.43 grams / liter = 58.43 mg/ml = 58.43
µ U g/µ U l.
A 0.002 M NaCl solution contains 0.002 moles of NaCl or 0.1168 grams (0.002 x 58.43) in one liter of solvent. Note that molar is abbreviated as M, but that there is no abbreviation for moles. The 0.002 M solution contains 0.002 moles (or 2 millimoles) or solute in one liter. A 0.002 M solution would contain 0.001 moles of solute in a half liter.
- The number of moles = Volume (in liters) x Molar Concentration
- The number of millimoles = Volume (in ml) x Molar Concentration
Note that chemical equations are always balanced via moles. Moreover, note that for dilutions of known concentrations, one can use the simple formula:
- Molarity x Volume = Molarity x Volume
If you have a 0.002 M solution of NaCl and you wish to obtain 100 ml. of a 0.001 M solution,
- 0.002 M x Needed Volume = 0.001 M x 100 ml
- Needed Volume = 0.001 M x .1 liters / 0.002 M = .050 liters= 50 ml.
- Measure 50 ml of the 0.002 M solution, and dilute it to 100 ml with the solvent (usually water, or an appropriate buffer).
Molarity is appropriate for use when chemical equations are to be balanced. When we deal with physical properties of solutions, molarity is not as valuable as a similar measurement of concentration, molality. For colligative properties of solutions (freezing point depression, boiling point elevation, osmotic pressure, density, viscosity), there is a better correlation between the property and molality. Molality (designated with a lower case m) is equal to the number of moles of solute in 1000 gm of solvent. At first this may not appear any different from molarity, since a ml of water equals 1.0 gm. Indeed, for dilute solutions in water, there is little or no practical difference between a molar solution and a molal solution. In concentrated solutions, with temperature fluctuations and with changes in solvent, there is appreciable difference.
A 2 m (2 molal) solution of sucrose contains 684.4 gm of sucrose (twice the molecular weight or two moles of sucrose)dissolved in 1000 gm (approximately 1 liter) of water. The weight of this solution is 684.4 gm + 1000 gm or 1684.4 gm. This solution (2 m sucrose) has a density of 1.18 gm/ml or 1180 gm/liter.
Since there are 1684.4 gms, division by the density (1180 gm/liter) would indicate that there are 1.43 liters of solution. That is, 684.4 gm of sucrose dissolved in 1000 gm of water would yield 1.43 liters of solution. This solution would contain 2 moles of sucrose, however and would have a molarity = 2 moles/1.43 liters or 1.40 M. So, a 2 m sucrose solution equals a 1.4 M sucrose solution.
In the example above of 2 m sucrose, there were 684.4 gm of sucrose in the final solution which weighed 1684.4 gm (684.4 gm sucrose + 1000 gm water). The percent of sucrose on the basis of weight is therefore 684.4/1684.4 x 100, or 40.6%.
There are three means of expressing concentration in the form of a percent figure:
- Percent by weight (w/w); gm solute / 100 gm solvent
- Percent weight by volume (w/v); gm solute /100 ml solvent
- Percent by volume (v/v); ml solute / 100 ml solution
For dilute solutions, these differences are not significant, but at higher concentrations, they are. Chemists (when they use Percent designations) usually use w/w. Biochemists and physiologists more often use w/v. Both use v/v if the solute is a liquid. It is important to distinguish among these alternatives.
Using ethanol as an example, consider a 20% solution of ethanol in water, mixed according to the three designations of w/w, w/v and v/v.
- w/w would contain 20 g of absolute ethanol mixed with 80 gm of water to yield a 20% (w/w) solution.
- w/v would contain 20 g of absolute ethanol mixed with water to form a final volume of 100 ml.
- v/v would contain 20 ml of absolute ethanol diluted to 100 ml with water. The three solutions are not the same. First, the density of alcohol is not equal to that of water, and thus conversion of g to ml is not equivalent. A 20% (w/w) solution of ethanol, for example, has a density of 0.97 g/ml and 20 gm of ethanol plus 80 gm of water would have a volume of 103 ml. The % (w/v) for this solution would be 20 gm ethanol / 103 ml, or 19.4% (w/v). Similarly, absolute ethanol has a density of 0.79 gm/ml and thus 20 ml of ethanol would weigh 15.8 gm. A 20% (v/v) solution would contain 15.8 gm of ethanol in 100 ml and be a 15.8% (w/v) solution. So, for ethanol:
- 20% (w/w) = 19.4 % (w/v)
- 20% (w/v) = 20.0 % (w/v)
- 20% (v/v) = 15.8 % (w/v)
- In cell biology, the most common use of Percent solution is as (w/v). In practice, these are simple solutions to mix. For a 20% (w/v) sucrose solution, for example, simply weigh 20 gm of sucrose and dissolve to 100 ml with water.
- Unless specifically stated otherwise, solutions lacking the appropriate designation should be assumed to be (w/v).