05 Nucleic acid quantification

99 Calculations for Molecular Biology and Biotechnology. DOI: © 2010 Elsevier Inc. All rights reserved. 10.1016/B978-0-12-375690-9.00005-X 5.1  �Quantification�of�nucleic�acids�� by�ultraviolet�(uv)�spectroscopy Any experiment requiring manipulation of a nucleic acid most likely also requires its accurate quantification to ensure optimal and reproducible results. The nitrogenous bases positioned along a nucleic acid strand absorb ultraviolet (UV) light at a wavelength of 260 nm; at this wavelength, light absorption is proportional to nucleic acid concentration. This relationship is so well characterized that UV absorption is used to accurately determine the concentration of nucleic acids in solution. The relationship between DNA concentration and absorption is linear up to an absorption at 260 nm (A260) of 2 (Figure 5.1). For measuring the absorption of a nucleic acid solution in a spectrophotometer, most molecular biology laboratories will use quartz cuvettes with a width through which the light beam will travel 1 cm. Therefore, all discussions in this chapter assume a 1 cm light path. Nucleic acid quantification Chapter5 n  Figure 5.1  The concentration of DNA and absorbance at 260 nm is a linear relationship up to an A260 value of approximately 2. 0 0 0.5 1 1.5 2 2.5 3 50 100 µg DNA in 1 mL A bs or ba nc e at 2 60 n m 150 200 250 100 CHAPTER 5 Nucleic acid quantification 5.2  �determining�the�concentration�� of�double-stranded�dna�(dsdna) Applications requiring quantitation of double-stranded DNAs (dsDNAs) include protocols utilizing plasmids, viruses, or genomes. Quantitation is typically performed by taking absorbance measurements at 260, 280, and 320 nm. Absorbance at 260 nm is used to specifically detect the nucleic acid component of a solution. Absorbance at 280 nm is used to detect the presence of protein (since tryptophan (Trp) residues absorb at this wavelength). Absorbance at 320 nm is used to detect any insoluble light- scattering components. A spectrophotometer capable of providing a scan from 200 to 320 nm will yield maximum relevant information (Figure 5.2). For nucleic acids purified from a biological source (as opposed to those made synthetically), calculating the ratio of the readings obtained at 260 and 280 nm can give an estimate of protein contamination. Pure DNA free of protein contamination will have an A260/A280 ratio close to 1.8. If phenol or protein contamination is present in the DNA prep, the A260/A280 ratio will be less than 1.8. If RNA is present in the DNA prep, the A260/A280 ratio may be greater than 1.8. Pure RNA preparations will have an A260/A280 ratio close to 2.0. At 260 nm, DNA concentrations as low as 2 mg/mL can be detected. A solution of DNA with a concentration of 50 mg/mL will have an absorb- ance at 260 nm equal to 1.0. Written as an equation, this relationship is 1 50260 A unit of dsDNA g DNA/mL  A bs or ba nc e Wavelength (nm) 220 0 0.3 240 260 280 300 320 n  Figure 5.2  A typical spectrophotometric scan of double-stranded DNA (dsDNA). Maximum absorbance occurs at 260 nm. Absorbance at 230 nm is an indication of salt in the sample. If the sample is optically clear, it should give a very low reading at 320 nm, which is in the visible wavelength region of the spectrum. 101 Absorbance and optical density (OD) are terms often used interchange- ably. The foregoing relationship can also be written as a conversion factor: 50 1 0 . g DNA/mL OD The following problems use this relationship. problem�5.1  From a small culture, you have purified the DNA of a  recombinant plasmid. You have resuspended the DNA in a volume  of 50 L. You dilute 20 L of the purified DNA sample into a total  volume of 1000 L distilled water. You measure the absorbance of this  diluted sample at 260 and 280 nm and obtain the following readings:  A260  0.550; A280  0.324. a)  What is the DNA concentration of the 50 L plasmid prep? b)  How much total DNA was purified by the plasmid prep procedure? c)  What is the A260/A280 ratio of the purified DNA? solution�5.1(a) This problem can be solved by setting up the following ratio:   x      .    g DNA/mL OD gDNA/mL OD0550 50 1 0.      x    ( .   )(     ) .    gDNA/mL OD gDNA/mL OD  0 550 50 1 0     x  27 5.    gDNA/mL   This  amount  represents  the  concentration  of  the  diluted  DNA  solution  that was used for the spectrophotometer. To determine the concentration  of DNA in the original 50 mL plasmid prep, this value must be divided by  the dilution factor. A 20 mL sample of the plasmid prep DNA was diluted  into water for a total diluted volume of 1000 mL.   27 5 20 1000 27 5 1000 20 .   .    g DNA/mL gDNA/mL        27 500 20 1375      g DNA/mL gDNA/mL   5.2  Determining the concentration of double-stranded DNA (dsDNA) 102 CHAPTER 5 Nucleic acid quantification Therefore, the original 50 L plasmid prep has a concentration of 1375 g  DNA/mL. To bring this value to an amount of DNA per L, it can be multi- plied by the conversion factor, 1 mL/1000 L:   1375 1 1000 1 35       .      g DNA mL mL L g DNA/ L    Therefore, the 50 L plasmid prep has a concentration of 1.35 g DNA/mL. solution�5.1(b) The total amount of DNA recovered by  the plasmid prep procedure can  be calculated by multiplying  the DNA concentration obtained earlier by  the volume containing the recovered DNA:   1375 1 1000 50 1375 50 1000 68 75     ( )( ) .      g DNA mL  mL   L L g DN    AA   Therefore,  the  original  50 L  plasmid  prep  contained  a  total  of  68.75 g  DNA. However, since 20 L was used for diluting and reading in the spec- trophotometer,  only  30 L  of  sample  prep  remains.  Therefore,  the  total  remaining amount of DNA is calculated by multiplying the remaining vol- ume by the concentration:   30 1 35 40 5  .   .     L g DNA L g DNA    Therefore,  although  68.75 g  of  DNA  were  recovered  from  the  plasmid  prep procedure, you used up some of  it  for spectrophotometry and you  now have 40.5 g DNA remaining. solution�5.1(c) The A260/A280 ratio is   0 550 0 323 1 703 . . .   5.2.1  �using�absorbance�and�an�extinction�� coefficient�to�calculate�double-stranded�� dna�(dsdna)�concentration At a neutral pH and assuming a GC DNA content of 50%, a DNA solu- tion having a concentration of 1 mg/mL will have an absorption in a 1 cm light path at 260 nm (A260) of 20. For most applications in molecular 103 biology, GC content, unless very heavily skewed, need not be a consid- eration when quantitating high-molecular-weight dsDNA. The absorption value of 20 for a 1 mg DNA/mL solution is referred to as DNA’s extinction coefficient. It is represented by the symbol E or e. The term extinction coefficient is used interchangeably with absorption constant or absorp- tion coefficient. The formula that describes the relationship between absorption at 260 nm (A260), concentration (c) (in mg/mL), the light path length (l) of the cuvette (in centimeters), and the extinction coefficient at 260 nm (E260) for a 1 cm light path is A E260 260 lc This relationship is known as Beer’s Law. Since the light path, l, is 1, this equation becomes A E260 260 c Rearranging the equation, the concentration of the nucleic acid, c, becomes c A E  260 260 problem�5.2  A DNA solution has an A260 value of 0.5. What is the DNA  concentration in g DNA/mL?  solution�5.2 The answer can be obtained using the equation for Beer’s Law.   c A E  260 260     c   0 5 20 0 025 . . mg DNA/mL     0 025 1000 25. mg DNA/mL g mg gDNA/mL      Therefore, the sample has a concentration of 25 g DNA/mL. 5.2  Determining the concentration of double-stranded DNA (dsDNA) 104 CHAPTER 5 Nucleic acid quantification problem�5.3  A DNA solution has an A260 value of 1.0. What is the DNA  concentration in g DNA/mL?  solution�5.3 The answer can be obtained using the equation for Beer’s Law.   c A E  260 260     c   1 0 20 0 05 . .  mg DNA/mL     0 05 1000 .  mg DNA/mL   g mg 50  g DNA/mL      Therefore, the solution has a concentration of 50 g DNA/mL. Notice that  this value is the one described earlier.   1 50260   unit of dsDNA g DNA/mLA     5.2.2  �calculating�dna�concentration�� as�a�millimolar�(mM)�amount The extinction coefficient (E260) for a 1 mM solution of dsDNA is 6.7. This value can be used to calculate the molarity of a solution of DNA. problem�5.4  A solution of DNA has an absorbance at 260 nm of 0.212.  What is the concentration of the DNA solution expressed as millimolarity?  solution�5.4 This problem can be solved by setting up a relationship of ratios such that  it is read ‘1 mM is to 6.7 OD as x mM is to 0.212 OD.’   1 6 7 0 212  m OD  m OD M x M . .      ( )( . ) . 1 0 212 6 7  m  OD OD  m M x M      0 03.  mM x   Therefore,  a  DNA  solution with  an A260  of  0.212  has  a  concentration  of  0.03 mM. 105 problem�5.5  A solution of DNA has an absorbance at 260 nm of 1.00.  What is its concentration expressed as millimolarity?  solution�5.5 This problem can be solved using ratios, with one of those ratios being the  relationship of a 1 mM solution of dsDNA to the extinction coefficient 6.7.   x M M m  OD  m  OD1 00 1 6 7. .      x M M M m  m  OD  OD  m ( )( . ) . . 1 1 00 6 7 0 15=   Therefore, a solution of dsDNA with an A260 of 1.00 has a concentration of  0.15 mM. This relationship,   1 0 0 15260. .   of dsDNA  mA M   has frequent use in the laboratory. problem�5.6  A solution of DNA has a concentration of 0.03 mM. What is  its concentration expressed as pmol/L?  solution�5.6 A 0.03 mM solution, by definition, has a concentration of 0.03 millimoles per  liter. A series of conversion factors is used to cancel terms and to transform  a concentration expressed as millimolarity to one expressed as pmol/L:   0 03 1 1 10 1 10 306 9. mmol L L L pmol mmol  pmol/ L         Therefore, a 0.03 mM DNA solution has a concentration of 30 pmol DNA/L. 5.2.3  �using�picogreen®�to�determine�� dna�concentration Determining DNA concentration by measuring a sample’s absorbance at 260 nm, though a common practice and, for many years, the gold standard of methods for DNA quantification, can be prone to inaccuracies that result from the contribution that such contaminants as salt, protein, nucleotides 5.2  Determining the concentration of double-stranded DNA (dsDNA) 106 CHAPTER 5 Nucleic acid quantification (nts), and RNA can make to the absorbance value. In addition, a solution of DNA having a concentration less than 2 g/mL cannot be quantified reli- ably by measuring its 260 nm absorbance. More recently, however, fluores- cent dyes have been used as a tool for nucleic acid quantification, the best example of which is PicoGreen® from Life Technologies. PicoGreen offers the advantages that it is specific for dsDNA, only fluoresces when bound to dsDNA, and can detect dsDNA at concentrations as low as 25 pg/mL. Quantification of dsDNA using PicoGreen requires that a standard curve of known concentrations be made using the same reagents prepared for the unknown sample. The DNA used for the standard curve is typically pre- pared from bacteriophage  or calf thymus and is diluted (in TE) in con- centrations from 1 ng/mL to 1000 ng/mL. The DNA dilutions are combined with PicoGreen reagent, allowed to incubate for several minutes at room temperature, and then read on a spectrofluorometer that excites the samples at a wavelength of 480 nm and reads their emission intensity at 520 nm. A standard curve presents the samples’ fluorescence (y axis) vs. DNA con- centration (x axis). An equation describing that curve can then be used to calculate the DNA concentration of an unknown sample based on its fluo- rescence, as demonstrated in the following problem. problem�5.7  A dilution series of calf thymus DNA is assayed for its  DNA content using PicoGreen. The following results are obtained. (These  values represent the concentrations of DNA in the assay tubes.)   dna�concentration�� (ng/ml) fluorescence 1 5572 2.5 6945 5 9245 10 13 820 25 27 585 50 50 520 100 99 710 250 234 952 500 450 210 750 700 025 1000 920 110 A sample of human DNA using PicoGreen reagent generates fluorescence  of 28 795 units. What is its concentration?  107 solution�5.�7 We will generate a standard curve using Microsoft Excel, determine the line  of best fit’s (the regression line’s) equation, and then use that equation to  calculate the concentration of the unknown human DNA sample. The pro- tocol for using the Excel graphing utility can be found in Appendix A. In the Excel spreadsheet, enter ‘ng/mL (x)’ in the column A, row 1 box and  ‘Fluorescence (y)’ in the column B, row 1 box. Fill the data in for both col- umns. The spreadsheet should look similar to Figure 5.3. Plot the above data using the ‘XY (Scatter)’ chart type. When a trendline is  added according to the instructions in Appendix A, the chart will appear  on the spreadsheet as shown in Figure 5.4. 5.2  Determining the concentration of double-stranded DNA (DSDNA) n  Figure 5.3  The values of fluorescence for each diluted sample of calf thymus DNA used for the standard curve assay in Problem 5.7, as entered in an Excel spreadsheet. n  Figure 5.4  The regression line and equation for Problem 5.7 data, as calculated in Microsoft Excel. 108 CHAPTER 5 Nucleic acid quantification The regression equation for this line is, therefore, y  916.1x  4652.3. We  now calculate the concentration of the unknown sample using this equa- tion. The unknown sample generated a fluorescence value of 28 795. This  is the y value in the equation. We then solve for x to give us the DNA con- centration in ng/mL:   y x x x       916 1 4652 3 28 795 916 1 4652 3 916 1 28 795 4652 3 916 1 . . . . . . . xx x    24 142 7 24 142 7 916 1 26 35 . . . .   Therefore, the concentration of the unknown sample in the assay tube is  26.35 ng/mL. note: If a fluorescence value for an unknown sample falls outside of that  covered by the standard curve and outside of  the  linear  range of detec- tion, the sample should be diluted if too high or concentrated if too low  so that a reliable measurement can be obtained. 5.3  �determining�the�concentration�� of�single-stranded�dna�(ssdna)�� molecules 5.3.1  �single-stranded�dna�(ssdna)�concentration� expressed�in�g/ml To determine the concentration of single-stranded DNA (ssDNA) as a g/ mL amount, the following conversion factor is used: 1 33 OD of ssDNA g/mL  problem�5.8  Single-stranded DNA isolated from M13mp18, a derivative  of bacteriophage M13 used in cloning and DNA sequencing applications,  is diluted 10 L into a total volume of 1000 L water. The absorbance  of this diluted sample is read at 260 nm and an A260 value of 0.325 is  obtained. What is its concentration in g/mL?  109 solution�5.8 This problem can be solved by setting up a ratio, with the variable x rep- resenting  the  concentration  in g/mL  for  the diluted  sample. The equa- tion  can  be  read  ‘x g/mL  is  to  0.125 OD  as  33 g/mL  is  to  1 OD.’  Once  x  is  obtained,  the  concentration  of  the  stock  DNA  can  be  determined  by multiplying  the  concentration  of  the  diluted  sample  by  the  dilution   factor.   x  g/mL  OD   g/mL  OD0 125 33 1.      x   g/mL   g/mL   g/mL       0 125 33 4 125 . .   Therefore,  the  concentration  of  the  diluted  sample  is  4.125 g/mL.  To  determine  the  concentration  of  the M13mp18  DNA  stock  solution,  this  value must be multiplied by the dilution factor:   4 125 1000 10 4125 10 412 5 .         .        g/mL L L g/mL g/mL      Therefore, the stock of M13mp18 DNA has a concentration of 412.5 g/mL. 5.3.2  �determining�the�concentration�� of�high-molecular-weight�single-stranded�� dna�(ssdna)�in�pmol/l The concentration of high-molecular-weight ssDNA can be expressed as a pmol/L amount by first determining how many micrograms of the ssDNA are equivalent to one pmol. To do this, we use the average molecu- lar weight of a deoxynucleotide in a DNA strand. For ssDNA, it is taken to be 330 daltons. This value is then used as a conversion factor to bring the concentration of ssDNA expressed as g/mL to a concentration expressed in pmol/L. 5.3  Determining the concentration of single-stranded DNA (ssDNA) molecules The unit dalton is defined as 1⁄12 the mass of the carbon-12 atom. It is used  interchangeably  with  ‘molecular  weight,’  a  quantity  expressed  as  grams/ mole. That is, there are 330 g of nt per mole of nt. 110 CHAPTER 5 Nucleic acid quantification problem�5.9  A stock of M13mp18 DNA has a concentration of   412.5 g/mL. What is this concentration expressed in pmol/L?  solution�5.9 The cloning vector M13mp18 is 7250 nts in length. To express this as g/ pmol, the following relationship is set up, in which a series of conversion  factors is used to cancel terms:   7250 330 1 10 1 10 2 39 6 12 nts  g/mol nt   g g mole pmol   g/pmo        . ll   Therefore, 2.39 g of a 7250 nt-long ssDNA molecule is equivalent to 1 pmol.  This value can now be used to convert g/mL to pmol/L:   412 5 1 2 39 1 1000 0 17 .     .       .      g mL pmol g mL L pmol/ L     Therefore, the M13mp18 DNA stock has a concentration of 0.17 pmol/L. 5.3.3  �expressing�single-stranded�dna�(ssdna)� concentration�as�a�millimolar�(mM)�amount The extinction coefficient (E260) for a 1 mM solution of ssDNA is 8.5. This value can be used to determine the millimolarity concentration of any ssDNA solution from its absorbance. problem�5.10  A 1 mL sample of ssDNA has an absorbance of 0.285.  What is its mM concentration?  solution�5.10 The  following  relationship  can  be  used  to  determine  the  millimolarity  concentration.   x M M  .     .   m OD m OD0 285 1 8 5      x M M M    .   .   .    m ( m )( OD) OD m  1 0 285 8 5 0 03   Therefore, a solution of ssDNA with an absorbance of 0.285 has a concen- tration of 0.03 mM. 111 5.4  oligonucleotide�Quantification 5.4.1  optical�density�(od)�units Many laboratories express an amount of an oligonucleotide in terms of opti- cal density (OD) units. An OD unit is the amount of oligonucleotide dis- solved in 1.0 mL giving an A260 of 1.00 in a cuvette with a 1 cm light path length. It is calculated by the equation OD units oligonucleotide volume dilution factor  ( ) ( ) ( )A260 problem�5.11  Following its synthesis, an oligonucleotide is dissolved in  1.5 mL of water. You dilute 50 L of the oligonucleotide into a total volume  of 1000 L and read the absorbance of the diluted sample at 260 nm.   An A260 of 0.264 is obtained. How many OD units are present in the 1.5 mL  of oligonucleotide stock?  solution�5.11 Using the formula just given, the number of OD units is   OD units  mL  0 264 1 5 1000 50 . . L L         396 50 7 92.  OD units   Therefore, the 1.5 mL solution contains 7.92 OD units of oligonucleotide. 5.4.2  �expressing�an�oligonucleotide’s�� concentration�in�g/ml An A260 reading can be converted into a concentration expressed as g/mL using the extinction coefficient for ssDNA of 1 mL/33 g for a 1 cm light path. In other words, a solution of ssDNA with an A260 value of 1.0 (1.0 OD unit) contains 33 g of ssDNA per milliliter. Written as an equation, this relationship is 1 33 OD unit g ssDNA/mL  5.4  Oligonucleotide quantification 112 CHAPTER 5 Nucleic acid quantification problem�5.12  In Problem 5.11, a diluted oligonucleotide gave an A260  reading of 0.264. What is the concentration of t