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MCB 253
Appendix I: Exponential Numbers and the Metric System
It is often necessary to deal with very large or very small numbers in molecular
biology. Exponential numbers expressed in scientific notation provide a shorthand
method for writing large and small numbers and for doing calculations with them.
Every number can be expressed as the product of two numbers, the second one
being some power of ten.
For example, the number 100 can be expressed as 10 x 10 = 102. Likewise:
1 = 100
10 = 101
100 = 102
1,000 = 103
10,000 = 104
100,000 = 105
1,000,000 = 106
10,000,000 = 107
100,000,000 = 108
1,000,000,000 = 109
The exponent to which ten is raised is the number of zeroes present in writing out
the number in the ordinary way.
Almost the same rules are used to express numbers less than one. For example, the
number 0.01 can be expressed as:
0.01 =
1
1
1
1
×
=
= 2 = 10 −2
10 10 100 10
Likewise:
0.1
0.01
0.001
0.0001
0.00001
0.000001
0.0000001
0.00000001
0.000000001
=
=
=
=
=
=
=
=
=
10-1
10-2
10-3
10-4
10-5
10-6
10-7
10-8
10-9
In this case, the number of the negative exponent to which ten is raised is equal to
the number of digits to the right of the decimal point.
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All numbers can be expressed in scientific notation. In scientific notation, numbers
that are not some exact power of ten are written with the first integer only to the left
of the decimal point and the other numbers to the right of it, multiplied by some
power of ten.
Example: Write the number 1,234.0 in scientific notation.
Answer: 1, 234.0 = 1.234 x 103
Practice: Express the following numbers in scientific notation:
(a) 578
Answer: 5.78 x 102
(b) 0.010
Answer:. 1.0 x 10-2
(c) 0.000267
Answer: 2.67 x 10-4
Calculations with exponential numbers
Addition and subtraction are both done by rewriting all the numbers involved so
that their exponents are raised to the same power of ten. The first part of each
number is then added (or subtracted). The power of ten stays the same. If the sum
no longer has one integer to the left of the decimal place, the number can simply be
rewritten with a new power of ten. Multiplication with exponential numbers is a
little more involved. First, rewrite each number in scientific notation. Second,
multiply the first numbers. Third, simply add the exponents.
Example:
Solution:
Use scientific notation to solve the following problem: 50 x 250
Rewrite each number in scientific notation: (5 x 101) x (2.5 x 102).
Multiply the first two numbers: 5 x 2.5 = 12.5
Add the exponents: 101 x 102 = 103.
The answer is: 12.5 x 103, which can also be written as 1.25 x 104.
The same steps are used whenever multiplying exponential numbers, even with
numbers less than one.
-1
Example:
0.5 x 0.25 = (5 x 10-1) x (2.5 x 10 ) = 12.5 x 10-2= 1.25 x 10-1 =0.125
The rules also apply when multiplying numbers greater than one by numbers less
than one. Simply express the numbers in scientific notation, multiply the first
numbers, add the exponents, and then express the answer in correct scientific
notation. (Remember: when adding a negative number to a positive number,
subtract the negative number from the positive number).
Example:
0.125 x 8000 = (1.25 x 10-1) x (8 x 103) = 10 x 102 = 1.0 x 103
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Division with exponential numbers is similar to multiplication. First, rewrite the
problem in scientific notation. Second, divide the first numbers. Third, subtract
the bottom exponent from the top exponent. Finally, express the answer in correct
scientific notation.
Example:
Solve the following problem: 2500 / 500 = ?
Solution:
Rewrite it as follows: (2.5 x 103 / 5 x 102)
Divide the first numbers by themselves: 2.5 / 5 = 0.5.
Subtract the bottom exponent from the top exponent: 103 / 102 = 101.
The answer is 0.5 x 101 or 5.0.
The same rules apply if the exponents are negative.
Example: 0.015 / 0.3 = (1.5 x 10-2 / 3 x 10-1) = 0.5 x 10-1 = 5.0 x 10-2
Remember, when subtracting one negative number from another negative number,
you end up adding the second number to the first number.
Example: 15 / 0.3 = (1.5 x 101 / 3 x 10-1) = 0.5 x 102 = 5.0 x 101 = 50
Subtracting a negative number from a positive number is the same as adding a
positive number to a positive number.
Example: 0.125 / 25 = (1.25 x 10-1 / 2.50 x 101) = 0.5 x 10-2 = 5.0 x 10-3 = 0.005
To subtract a positive number from a negative number, add the positive number to
the negative number and express the answer as a negative. By following the rules
described above, both multiplication and division may be combined in the same
problem.
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Units of Measurement and Dimensional Analysis
As you do the various calculations necessary in the lab, you will generally be
working with numbers that have units identifying what they are a measure of. In
order to do your calculations correctly, you must be able to manipulate the units
along with the numbers so that your answer has the correct units. Checking to
make sure that the units cancel out to give the answer in the correct units is called
dimensional analysis.
Common units
Mass
1 gram (g)
= 103 milligrams (mg)
= 106 micrograms (µg)
= 109 nanograms (ng)
1 liter (l)
= 103 milliliters (ml)
= 106 microliters (µl)
= 1 cubic centimeter
Volume
1 ml
Length
1 meter (m)
= 103 millimeters (mm)
= 106 micrometers (µm)
= 109 nanometers (nm)
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Appendix II: Calculation of Dilutions
Concentration
Concentration is the amount per volume, and it is commonly expressed in several
different ways:
1. Grams per liter (g/l). Note that g/l is the same as milligrams per
milliliter (mg/ml).
2. Moles/liter (M or molar). A 1 M solution of compound X is equivalent
to the molecular weight of compound X made up to 1 liter volume with
water. For example, to prepare a 1 M solution of sodium chloride (NaCl)
which has a molecular weight of 58.45, 58.45 g of NaCl is dissolved in
water and the volume made up to 1 liter.
3. Percent solution for solids = grams per 100 milliliters of solution (g/100
ml), e.g. 10 g/100 ml = 10% solution 0.1 g/100 ml = 0.1% solution
4. Percent solution for two liquids when one is water. The volume of the
second liquid divided by the total volume times 100% describes the
solution. For example, 70 ml ethanol per 100 ml total is 70% ethanol.
5. Number of cells per milliliter (cells/ml)
It is often necessary to dilute solutions to obtained a desired concentration. In order
to make the proper dilutions, you must be able to determine the concentration of a
solution after dilution or the volumes needed to obtain the desired dilution. A few
simple formulas allow you to solve any one-step dilution problem, where:
C = concentration
P = number of particles (expressed in mass or number)
V = volume
DF = dilution factor
Concentration is defined by the equation: C =
P
V
Thus, if P = mg and V = ml, then C = mg/ml
or if P = cells and V = ml, then C = cells/ml.
This equation can be rearranged two ways.
P = CV
V = P/C
(e.g., if C = mg/ml and V = ml, then P = (mg/ml) x (ml) = mg)
(e.g., if P = g and C = g/l, then V = g/(g/l) = liters)
For all dilutions: P1 = P2. That is, the number of particles does not change during
dilution -- only the volume changes. The subscript 1 indicates a solution before
dilution and the subscript 2 indicates a solution after dilution:
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C1V1 = C2V2
This is a fundamental equation for solving dilution problems. It can be solved for
any one of the variables if the other three are known. It applies both to chemical
solutions and to cell suspensions. Thus, if you want to dilute a solution at
concentration C1 to concentration C2 and you need a certain volume V2 of the
solution, you can use the equation:
V1 =
C 2V 2 (mg / ml) ml
=
= ml
C1
(mg / ml)
Example: Given a 1 M stock solution of Tris buffer prepare 10 ml of 0.01 M Tris.
Solution: The question you need to answer is, "what volume of 1 M Tris do you
need to dilute to give a total volume of 10 ml of 0.01 M Tris". Thus, you know C1,
C2, and V2. To answer the question, plug the known numbers into the equation
shown above.
(V 1C1) = (V 2C 2)
(? ml × 1M ) = (10ml × 0.01M )
(10 ml × 0.01M) = 0.1ml
∴? ml =
(1M )
Therefore, you would need to add 0.1 ml of 1 M Tris to 9.9 ml of H2O for a total
volume of 10 ml of 0.01 M Tris.
Alternatively, you can use this equation to calculate the final concentration (C2) of
a solution that has been diluted from volume V1 to volume V2:
C2 =
C1V 1 (mg / ml) ml
=
= mg / ml
V2
ml
Example: What would be the final concentration if 0.1 ml of a 2 mg/ml stock
solution of Ampicillin is added to 9.9 ml H2O.
Solution: Note that the dilution of 0.1 ml into 9.9 ml gives a final volume of 10
ml. Thus, you know V1, V2, and C1. To determine C2 simply plug the known
numbers into the equation shown above.
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( V 1C1) = (V 2C 2 )
(0.1 ml × 2mg / ml ) = (10ml × ? mg / ml)
(0.1 ml × 2 mg / ml ) = 0.02 mg / ml= 20 µ g / ml
∴? mg / ml =
(10 ml )
Diluting Cell Suspensions
Concentrations of bacteria are often in the range of 108 to 1010 cells per ml. Thus,
in order to plate no more than 300 cells, the optimal number of cells that should be
spread on a plate, it is first necessary to dilute the cell suspension. For example,
assume that you have a bacterial suspension with a concentration of 3 x 105
cells/ml.
Concentration =
Numberof Particles
P
or C =
Volume
V
Rearranging yields: V =
In this case:
V=
P
C
300cells
= 10 −3 ml = 0.001ml
5
3×10 cells / ml
Therefore, to spread 300 cells you would need 0.001 ml (= 1 µl). This is too small
a volume to plate directly because the serological pipettes used for plating are not
able to measure such a small volume and 0.001 ml cannot be spread uniformly
across the surface of an agar plate. Therefore, the original cell suspension must be
diluted so that a larger volume (100 µl or 0.1 ml is ideal) can be plated. To
determine the volume of the original suspension that needs to be diluted, first
decide on the concentration of cells that you need for plating. For example, if you
want to spread 100 µl containing no more than 300 cells, you would need a cell
concentration of:
P 300 cells 3000 cells 3x103 cells
C= =
=
=
V
0.1 ml
ml
ml
Call this plating concentration C2 and the original cell suspension concentration C1
(= 3 x 105 cells/ml). For convenience, assume the final volume needed (V2) is 10
ml. Recall that C2 = P2/V2. However, if some cells are taken from the original
suspension (P1) and diluted with sterile dilution fluid (saline or water), the number
of cells will not change during the dilution – only the volume in which the cells
are suspended changes. In other words, for all cell dilutions, P2 = P1. By
substituting C1V1 = C2V2, you can solve for V1, the volume of the original
suspension as follows:
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C2 V 2 (3 × 103 cells / ml)(10ml)
V1 =
=
= 0.1ml
5
C1
(3 × 10 cells / ml)
Thus, if 0.1 ml of the original cell culture (3 x 105 cells/ml) is diluted to 10 ml, this
will give a final working concentration of 3 x 103 cells/ml, which is suitable for
plating. Dilution of 0.1 ml to 10.0 ml is a 100-fold or 10-2 dilution. This dilution
factor is widely used in diluting bacterial cultures for plating because it is
convenient to do in the lab.
C1V1 = C2V2
This formula can be used to determine how to make any of the single-step
dilutions of cells, media, or chemical solutions necessary in the lab.
A single-step dilution is not sufficient in most cases involving bacteria because the
original cell concentrations are too high. For example, consider an original cell
concentration of 3 x 108 cells/ml, a typical concentration. In this case, V1 would
equal 0.0001 ml (=10-4 ml) if V2 were 10 ml.
V1 =
(3 × 103 cells / ml)(10ml)
−4
= 10 ml
8
(3 × 10 cells / ml)
Again, this is too small a quantity to pipette. Because one dilution is insufficient,
the answer is to make an ordered series of dilutions of the original cell suspension,
a serial dilution, to bring it to a concentration that can be easily plated.
Dilution =
Volume added
Volume added
=
Volume added + Volume dilution fluid
Total volume
e.g., Dilution =
0.1ml
0.1ml
1
=
=
0.1ml + 9.9ml 10 ml 100
It is also useful to define the inverse of the dilution, the dilution factor (DF).
Thus, if:
Dilution = 1/100
Dilution = 1/n
Dilution factor = 100
Dilution factor = n
In a serial dilution, the total dilution is the product of the individual dilutions.
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Dilution =
1
1
1
1
1
1
1
×
×
= 2 × 1 × 3 = 6 = 10 −6
100 10 1000 10 10 10
10
The final dilution factor is the inverse of this, 106.
Because DF = V2 / V1, we can write the basic dilution formula (C1V1 = C2V2) as:
In our example, C1 = 3 x 108 (original concentration) and C2 = 3 x 103
(concentration needed for plating). Thus:
Thus, we need a series of dilutions in which the product of the dilution factors is
105. There are an infinite number of possible dilutions, but it is easiest to work in
powers of 10 and, for convenience, to limit the dilutions to 1/10 or 1/100.
This dilution series will produce the 105 DF. At this point, we have gone from 3 x
108 to 3 x 103 cells/ml. Using this formula, we can calculate any of the parameters
necessary to determine cell concentration or to develop a plating scheme. For
example, how would you determine the missing numbers in the tables below?
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a
b
# of
colonies
132
?
DF
C1
106
104
?
3.5 x 106
Volume
plated
0.1 ml
0.1 ml
Answers:
The above calculations assume that all of the cells plated will be viable and give
rise to colonies. In fact, cell viability is rarely, if ever, 100%, so final cell
concentrations are best determined by counting the actual number of colonies that
appear on the plates (assuming each viable cell gives rise to a colony). This is
clearly important in experiments in which you measure cell killing. Unfortunately,
this method is time-consuming, requiring about 24 hours after the platings are
done to determine the cell number.
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Appendix III: Graphing Data
The graphical presentation of data is a very important part of scientific analysis. The key
components of a graph are the same whether you are drawing a graph by hand or using a
computer drawing application. Every instructor has certain expectations with regard to
how graphs are constructed. Make certain that you understand exactly what your
teaching assistant requires.
Linear Graphs
A line graph should be well organized, neat, and accurate. Figure 1 shows the basic
format for a line graph.
• Use an entire 8.5" x 11" piece of paper for the graph. This means that the graph
will need to be expanded to fill the page. Grid paper should be used for a linear
graph.
• The title of the graph should be placed at the top of the page. In the upper right
corner should be your name and other pertinent information about the assignment.
Example:
"Title of Graph"
Joan Student, MCB 151, Section C
Experiment 3: Name of Experimental Exercise
Date of Assignment
• Place the origin of axes approximately one inch in and to the right of the bottom
left corner and approximately one inch up from the bottom of the sheet. The origin
is where the x and y axes meet.
• Draw the two axes. It is customary to make the y-axis approximately 1 and 1/2
times as long as the x-axis. The x-axis is the horizontal axis and the y-axis is the
vertical axis. The x-axis is often called the abscissa, and the y-axis is called the
ordinate. The independent variable is most often placed on the x-axis and the
dependent variable is most often placed on the ordinate.
• Divide each axis at regular intervals using as much of the axis as possible. Number
the division marks clearly.
• Label each axis and always include units if applicable.
• All data points must be shown on the graph. When more than one curve is plotted
on the same graph, use different types of lines or curves (e.g. solid, dashed, dotted)
or use different shapes for the data points (e.g. dots, squares, triangles).
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• Some lines are best drawn by "connecting the dots," while others are most accurate
when a "line of best fit" is drawn. Figure 1 shows a line drawn by "connecting the
dots." If the data points are not in a straight line, in many cases it is better to fit a
curve to the data. The line of best fit can be determined statistically, but for the
purposes of this course, the line of best fit can be drawn by attempting to place the
line equidistant between the data points. The line can be drawn with an equal
number of points on each side of the line or with approximately equal distances
between the line and all the points on either side of the line (fig 2).
• If more than one line is to be drawn on a graph, a legend can be used to key out the
different lines.
Example:
dotted line = 1-year data
dashed line = 5-year data
solid line = 10-year data
grid
title of line graph
Joan Student, Bio 121, Sect. C
Name of Laboratory Exercise
Date of Assignment
14
13
12
11
y axis label
If applicable, units
must be included.
legend
10
9
data points
8
7
6
5
4
y axis
ordinate
3
line
connect-the-dots
2
1
1
2 3
4
5 6 7
8 9 10
origin of axes
x axis
abscissa
x axis label
If applicable, units must be included.
Figure 1
This figure shows the basic format for a line graph.
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Title of Graph
14
12
10
8
6
4
2
0
0
1
2
3
4
5
Time (minutes)
connect-the-dots
line of best fit
Figure 2 This figure shows a sample linear graph with line of best fit.
125
6
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Semi-logarithmic Graphs
The use of semi-logarithmic (semi-log) graph paper is an important tool in graphing
scientific data that extends over a wide range of values. Semi-log graph paper provides
the range needed to graph data points on a single sheet of paper by compressing the scale
for large numbers and expanding the scale for very small numbers. This is done by
choosing a logarithmic scale for one axis (usually the y axis) while leaving the other axis
(usually the x axis) in the linear format. Semi-log paper comes with different numbers of
cycles, and the span of the data will determine what size is needed. Three-cycle semilogarithmic paper has been chosen for this example. A set of data (Table 1) is graphed on
linear paper and produces a curve (Figure 3). The same set of data, graphed on semi-log
paper, produces a straight line (Figure 4). When extrapolation from a graph is necessary,
a straight line is needed.
A semi-logarithmic graph should be well organized, neat, and accurate. Specific
instructions for creating such a graph follow.
• Use an entire 8.5" x 11" piece of paper for the graph. This means that the graph
will need to be expanded to fill the page. Semi-logarithmic paper should be used
for a semi-log graph.
• The title of the graph should be placed at the top of the page. In the upper right
corner should be your name and other pertinent information about the assignment.
Example:
"Number of Survivors as a Function of UV Exposure"
Joan Student, MCB 151, Section C
Experiment 4: UV Damage and Mutagenesis
Date of Assignment
• The origin of axes is determined for you on the semi-log paper. The origin is where
the x and y axes meet.
• The two axes have been drawn for you. It is customary to make the y-axis
approximately 1 and 1/2 times as long as the x-axis. The x-axis is the horizontal
axis and the y-axis is the vertical axis. The x-axis is often called the abscissa, and
they axis is called the ordinate. The independent variable should be placed on the xaxis and the dependent variable should be placed on the ordinate.
• Divide the x-axis at regular intervals using as much of the axis as possible. Number
the division marks clearly.
• The y-axis has been divided into cycles. A cycle is a series of non-uniformly
spaced lines which are often numbered with the single digits 1–9 (for each cycle).
With 3-cycle semi-log paper you will notice that the numerical labels on the y-axis
are based on multiples of 10 and the line spacing sequence repeats itself 3 times.
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When labeling the y-axis, each cycle should increase by the number the cycle
started with. For example, if you decided to begin the first cycle with 1 you
wouldn’t need to change the labeling, the next number would be 2, then 3, then 4,
etc. until you got to 10. Since 10 starts the second cycle each pre-printed number
will need to have a 0 added to it so that each division increases by 10; the 2
becomes 20, the 3 becomes 30, and so forth until you reach 100. Now, since 100
begins the third cycle each pre-printed number will need to have two zeros added to
it so that each division will increase by 100. You may find that the first cycle needs
to begin with a number that is higher or lower than 1 depending on your range of
numbers. Just make sure you write in the appropriate amount of zeros at each cycle
so that your scale is clearly represented. The use of a logarithmic y-axis is
commonly preferred over graphing the log of the y component of the ordered pair
and the x component on linear graph paper.
• Label each axis and always include units if applicable.
• All data points must be shown on the graph. When more than one set of ordered
pairs is plotted on the same graph, use different types of lines (e.g. solid, dashed,
dotted) or use different shapes for the data points (e.g. dots, squares, triangles).
• Figure 3 shows a line drawn by "connecting the dots." The data points are not in a
straight line in this instance, so it is better to fit a curve to the data. Figure 4 is by
its nature a straight line. The most frequent purpose of a semi-logarithmic graph is
to generate a linear relationship from which data can be extrapolated.
Sample sheets of semi-log paper can be found at the end of your lab manual. Remember
to make copies of these sheets so that they can be used more than once.
Time of UV Exposure
(seconds)
0
2
4
6
8
10
Number of Colonies
per milliliter (phr –)
8600
2600
790
240
70
20
Table 1. This table contains the example set of data used for Figures 3 and 4.
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Linear Graph of Survivors as a Function of UV
Exposure
3000
2600
2500
2000
1500
1000
790
500
240
70
20
0
2
4
6
8
10
Time of UV Exposure (seconds)
Figure 3. This figure shows a sample linear graph with a connect-the-dots curve.
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Semi-logarithmic Graph of Survivors as a Function of UV Exposure
10000
Number of Colonies per mL
2600
1000
790
240
100
70
20
10
2
4
6
8
Time of UV Exposure
(seconds)
Figure 4. This figure shows a basic semi-logarithmic graph.
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Appendix IV: Spectrophotometry
Spectrophotometry is the measurement of light absorption or transmission
through a solution.
For example, the concentration of bacteria can be determined by measuring light
scattering in a spectrophotometer. Spectrophotometry is also useful for measuring the
absorption of light by specific compounds. Because different compounds absorb light
at different wavelengths, and the amount of light absorbed is proportional to the
concentration of the compound, spectrophotometry provides a sensitive, qualitative and
quantitative analysis of the components of solutions. In addition, it is usually a nondestructive method of analysis. Thus, spectrophotometry is one of the most valuable
analytical techniques available to biochemists:
i) Unknown compounds may be identified by their characteristic absorption of
ultraviolet, visible, or infrared light;
ii) Concentrations of known compounds in solutions may be determined by
measuring the light absorption at one or more wavelengths;
iii) Enzyme-catalyzed reactions can often be assayed by measuring the
appearance of a product or disappearance of a substrate with a unique absorption
spectrum.
Various types of photometers are used in biochemistry and microbiology.
Colorimeters and spectrophotometers measure the amount of light absorbed by
solutions. Turbidimeters and nephelometers measure the light scattered by
suspensions. Fluorimeters measure the fluorescence produced by absorbed light.
These instruments can use a very wide range of electromagnetic radiation from the far
ultraviolet to the far infrared portion of the spectrum.
Region
Wavelength
Ultraviolet
Visible
Infrared
100 nm-400 nm 400 nm-800 nm 800 nm-100 µm
When light is passed through a solution containing a molecule, the molecule
only absorbs certain specific wavelengths of light. For example, o-nitrophenol (ONP)
strongly absorbs blue light and allows yellow light to pass through. Thus, solutions of
ONP appear yellow. Protein solutions do not absorb visible light and thus appear to be
uncolored. However, proteins strongly absorb specific wavelengths of ultraviolet light.
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Since different molecules absorb light at different wavelengths, the wavelengths of
light absorbed by a solution of an unknown molecule may help identify the molecule.
A plot of the relative amount of light absorbed by a compound as a function of
wavelength is called an absorption spectrum. For example, the absorption spectrum of
ONP is shown below.
Figure 11.5: Absorption Spectrum of ONP.
The amount of light absorbed is proportional to the concentration of molecules
in the solution. Therefore, spectrophotometry can also be used to determine the
concentration of a molecule. The most accurate way to determine the concentration of
a molecule is to assay the absorption of monochromatic light (a single wavelength) at a
wavelength the compound strongly absorbs. The optimal wavelength to use is
determined from the absorption spectrum. For example, the wavelength absorbed
maximally by ONP is 420 nm.
Determining the light absorbed by a molecule in solution could be complicated
if the molecule is dissolved in a solvent which absorbs light. To eliminate this
problem, the absorption of an unknown in solution is corrected for the absorption of the
solvent. This is done by setting the spectrophotometer to read zero for a cuvette
containing solvent only (a "blank"). Any absorption due to the solvent will then be
automatically "subtracted" from the sample readings.
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Spectrophotometers
A spectrophotometer is an instrument used to measure the amount of light of a
given wavelength that passes through a sample. The essential components of a
spectrophotometer are shown below.
Figure 11.6: Components of a Typical Spectrophotometer.
A tungsten lamp emits white light that consists of a mixture of all wavelengths. The
light is focused on a mirror, then the beam of white light passes through a
monochromator (a prism or grating) which divides the light into different wavelengths.
The desired wavelength is selected and passed through the sample chamber. The
solution is placed in a clear tube (called a "cuvette") which is inserted into the sample
chamber. A photoelectric detector measures the light that passes through the solution
and converts the signal to an electric current that is translated to an optical density
reading on the meter. The greater the amount of light absorbed or scattered by the
solution, the greater the optical density recorded.
Beer's Law. The fraction of the incident light that is absorbed by a solution depends
on the thickness of the sample, the concentration of the absorbing compound in the
solution, and the chemical nature of the absorbing compound. Light absorption
follows an exponential rather than a linear law. The relationship between
concentration, length of the light path through the solution, and the light absorbed by a
solution are expressed mathematically by the following equation.
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I
log o = ε c l
I
Where Io is the intensity of the beam of light entering the solution and I is the intensity
of light transmitted by the solution, ε is the molar absorption coefficient1 of the
molecule, c is the concentration of the molecule in moles per liter (M) and l is the
length of the optical path through the solution in centimeters (i.e. the width of the
cuvette). The width of the cuvette is almost always 1 cm. The units of ε are liter/(mole
x cm) or M-1 cm-1. Since the amount of light absorbed by a molecule is not the same
at all wavelengths (see the absorption spectrum of ONP shown above), a molecule will
have a characteristic ε for each wavelength. Therefore, the absorption coefficient is
usually followed by a subscript that indicates the wavelength. For example, ε420 refers
to the molar absorption coefficient at 420 nm.
The log Io/I is the amount of light absorbed by the solution and therefore it is
called the optical density (OD) or absorbance (A).
A = εcl
Absorbance is a linear function of concentration. Thus, a plot of optical density versus
concentration should yield a straight line. However, often Beer's Law is not valid for
highly concentrated solutions: at high concentrations the absorbance is not linearly
proportional to the concentration. Usually quantitative spectrophotometric
measurements are done under conditions where Beer's Law is valid since this allows
the concentration to be derived from the optical density by a simple linear standard
curve. Note that ε equals the OD observed when a 1 M solution is analyzed in a
photometric cell with a light path length of 1 cm.
ε=
A
cl
Why is it useful to determine ε? Once you have determined the ε for a specific
compound, it is possible to calculate the concentration of that compound in a solution
from the absorbance by simply rearranging the equation:
c=
A
εl
1Sometimes
called an "extinction" coefficient because the absorption "extinguishes" light
passing through the solution.
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Appendix
MCB 253
This approach is often used to calculate the number of molecules of product produced or
substrate consumed in an enzymatic reaction by measuring the absorbance of the solution
before and after the reaction.
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