Appendix I: Exponential Numbers and the Metric System

Transcription

1 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 = Likewise: 1 = = = ,000 = ,000 = ,000 = ,000,000 = ,000,000 = ,000,000 = ,000,000,000 = 10 9 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 = = = = 10 2 Likewise: 0.1 = = = = = = = = = 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. 113

2 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, = x 10 3 Practice: Express the following numbers in scientific notation: (a) 578 Answer: 5.78 x 10 2 (b) Answer:. 1.0 x 10-2 (c) 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: Use scientific notation to solve the following problem: 50 x 250 Solution: Rewrite each number in scientific notation: (5 x 10 1 ) x (2.5 x 10 2 ). Multiply the first two numbers: 5 x 2.5 = 12.5 Add the exponents: 10 1 x 10 2 = The answer is: 12.5 x 10 3, which can also be written as 1.25 x The same steps are used whenever multiplying exponential numbers, even with numbers less than one. Example: 0.5 x 0.25 = (5 x 10-1 ) x (2.5 x 10-1 ) = 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: x 8000 = (1.25 x 10-1 ) x (8 x 10 3 ) = 10 x 10 2 = 1.0 x

3 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 10 3 / 5 x 10 2 ) Divide the first numbers by themselves: 2.5 / 5 = 0.5. Subtract the bottom exponent from the top exponent: 10 3 / 102 = The answer is 0.5 x 10 1 or 5.0. The same rules apply if the exponents are negative. Example: / 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 10 1 / 3 x 10-1 ) = 0.5 x 10 2 = 5.0 x 10 1 = 50 Subtracting a negative number from a positive number is the same as adding a positive number to a positive number. Example: / 25 = (1.25 x 10-1 / 2.50 x 10 1 ) = 0.5 x 10-2 = 5.0 x 10-3 = 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. 115

4 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 Volume Length 1 gram (g) = 10 3 milligrams (mg) = 10 6 micrograms (µg) = 10 9 nanograms (ng) 1 liter (l) = 10 3 milliliters (ml) = 10 6 microliters (µl) 1 ml = 1 cubic centimeter 1 meter (m) = 10 3 millimeters (mm) = 10 6 micrometers (µm) = 10 9 nanometers (nm) 116

5 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, 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 = C V 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: 117

6 C 1 V 1 = C 2 V 2 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 C 1 to concentration C 2 and you need a certain volume V 2 of the solution, you can use the equation: V1 = C2V2 C1 = (mg / ml) ml (mg / ml) = 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 C 1, C 2, and V 2. To answer the question, plug the known numbers into the equation shown above. ( V1C1) = ( V 2C 2) (? ml 1M) = ( 10ml 0.01M) 10 ml 0.01M? ml = ( ) = 0.1ml ( 1M) Therefore, you would need to add 0.1 ml of 1 M Tris to 9.9 ml of H 2 O for a total volume of 10 ml of 0.01 M Tris. Alternatively, you can use this equation to calculate the final concentration (C 2 ) of a solution that has been diluted from volume V 1 to volume V 2 : C2 = C1V1 V 2 = (mg / ml) ml ml = mg / 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 H 2 O. Solution: Note that the dilution of 0.1 ml into 9.9 ml gives a final volume of 10 ml. Thus, you know V 1, V 2, and C 1. To determine C 2 simply plug the known numbers into the equation shown above. 118

7 ( V1C1) = ( V2C 2) ( 0.1 ml 2mg / ml) = ( 10ml? mg / ml) 0.1 ml 2 mg / ml? mg / ml = ( ) = 0.02 mg / ml= 20 µg / ml ( 10 ml) Diluting Cell Suspensions Concentrations of bacteria are often in the range of 10 8 to 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 10 5 cells/ml. Concentration = Numberof Particles Volume or C = P V Rearranging yields: V = P C In this case: V = 300cells cells / ml =10 3 ml = 0.001ml Therefore, to spread 300 cells you would need 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 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: C = P V = 300 cells 0.1 ml = 3000 cells ml = 3x103 cells ml Call this plating concentration C 2 and the original cell suspension concentration C 1 (= 3 x 10 5 cells/ml). For convenience, assume the final volume needed (V 2 ) is 10 ml. Recall that C 2 = P 2 /V 2. However, if some cells are taken from the original suspension (P 1 ) 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, P 2 = P 1. By substituting C 1 V 1 = C 2 V 2, you can solve for V 1, the volume of the original suspension as follows: 119

8 V 1 = C 2 V 2 C1 = (3 103 cells / ml)(10ml) ( cells / ml) = 0.1ml Thus, if 0.1 ml of the original cell culture (3 x 10 5 cells/ml) is diluted to 10 ml, this will give a final working concentration of 3 x 10 3 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 10 8 cells/ml, a typical concentration. In this case, V 1 would equal ml (=10-4 ml) if V 2 were 10 ml. V 1 = (3 103 cells / ml)(10ml) ( cells / ml) =10 4 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 dilution fluid = Volume added Total volume e.g., Dilution = 0.1ml 0.1ml + 9.9ml = 0.1ml 10 ml = It is also useful to define the inverse of the dilution, the dilution factor (DF). Thus, if: Dilution = 1/100 Dilution factor = 100 Dilution = 1/n Dilution factor = n In a serial dilution, the total dilution is the product of the individual dilutions. 120

9 Dilution = = = = 10 6 The final dilution factor is the inverse of this, Because DF = V 2 / V 1, we can write the basic dilution formula (C 1 V 1 = C 2 V 2 ) as: In our example, C 1 = 3 x 10 8 (original concentration) and C 2 = 3 x 10 3 (concentration needed for plating). Thus: Thus, we need a series of dilutions in which the product of the dilution factors is 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 10 5 DF. At this point, we have gone from 3 x 10 8 to 3 x 10 3 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? 121

10 # of DF C 1 Volume colonies plated a ? 0.1 ml b? x 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. 122

11 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). 123

12 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 y axis label If applicable, units must be included. y axis ordinate origin of axes Joan Student, Bio 121, Sect. C Name of Laboratory Exercise Date of Assignment legend data points line connect-the-dots x axis abscissa x axis label If applicable, units must be included. Figure 1 This figure shows the basic format for a line graph. 124

13 Title of Graph Time (minutes) connect-the-dots line of best fit Figure 2 This figure shows a sample linear graph with line of best fit. 125

14 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 x- axis 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. 126

15 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) Number of Colonies per milliliter (phr ) Table 1. This table contains the example set of data used for Figures 3 and

16 3000 Linear Graph of Survivors as a Function of UV Exposure Time of UV Exposure (seconds) Figure 3. This figure shows a sample linear graph with a connect-the-dots curve. 128

17 Semi-logarithmic Graph of Survivors as a Function of UV Exposure Number of Colonies per ml Time of UV Exposure (seconds) Figure 4. This figure shows a basic semi-logarithmic graph. 129

18 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 Ultraviolet Visible Infrared Wavelength 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. 130

19 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. 131

20 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. 132

21 log I o I = ε c l Where I o 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 coefficient 1 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, ε 420 refers to the molar absorption coefficient at 420 nm. The log I o /I is the amount of light absorbed by the solution and therefore it is called the optical density (OD) or absorbance (A). A = ε c l 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 c l 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 ε l 1 Sometimes called an "extinction" coefficient because the absorption "extinguishes" light passing through the solution. 133

22 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. 134