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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification 1

An evaluation of direct and indirect DNA quantification methods using chicken liver extracts

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Introduction The hypothesis of this experiment is that the DNA yield determined with the diphenylamine method will be larger than the yield obtained with the weighing method. Deoxyribonucleic acid (DNA) is a cellular polymer made up of deoxyribonucleotides which are joined together by phosphodiester bonds forming a sugar-phosphate backbone. DNA has the structure of a double helix in which the nucleotide bases of one DNA strand base pair with the bases of a second, antiparallel DNA strand (1). DNA encodes for the inheritable genetic information of all prokaryotic and eukaryotic cells and of many viruses (2). In eukaryotic cells DNA is found in the nucleus with some contained in mitochondria and also in chloroplasts of plants (3). Studying the physical and chemical properties of DNA as well as determination of the genetic information contained by DNA sequencing and the manipulation of DNA in genetic engineering requires the isolation and accurate quantification of DNA. DNA can be extracted from cells and quantified using various methods. Extraction methods employ a step in which cells are broken up using mechanical or chemical treatments, followed by several steps that serve to remove non-wanted cellular components such as lipids and proteins (4). DNA is then usually precipiated from the prepurified extract using ethanol followed by quantification (5). Several methods are available for the quantification of DNA. This study will compare and evaluate two such methods, a direct weighing method and an indirect diphenylamine method (6). The direct weighing method is fast and easy to perform as it determines the mass of ethanol precipitated DNA. However, if not removed, RNA is co-precipitated with DNA and contributes to the weight as both polymers have very similar physical properties (4). In contrast, the diphenylamine method uses a chemical reaction that is specific for DNA (6).

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Methods Experiments were carried out as described (7). All values were calculated using a pocket calculator. The curve was drawn using Excel.

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Results Determining the yield of DNA using the diphenylamine method required the generation of a DNA standard curve. For this, the absorbencies of four standard samples prepared in duplicates and treated with diphenylamine were determined (table 3). The standard deviations (SD) for the averaged absorbencies were small (0.3% to 6.9%, table 3). The averaged absorbencies were corrected for the average of the blank value (SD: 0.7% to 1.6%, table 3). The corrected, averaged standard absorbencies were plotted over the concentration and a line of best fit model was used to calculate a linear standard curve (y = 0.7278x + 0.0254, R² = 0.9905, figure 1). Using the weighing method, the dry mass or yield was determined to be 0.98% of the tissue mass (table 2). This value was obtained by first calculating the wet mass of the DNA precipitate obtained from ethanol precipitation of a 3 mL aliquot of the total extract (5.3 mL). When spooled onto a Pasteur pipette, the precipitate appeared yellowish, had a sticky consistency and the size of a small pea. The wet mass of the DNA precipitate 88 mg (table 1) was used to calculate the DNA wet mass per gram chicken liver as 155 mg and the dry mass or yield as 9.8 mg (table 2). The yield of DNA from one gram chicken liver was equivalent to 0.18% of the tissue mass (table 7). This was determined by first measuring the absorbencies of a concentrated (c), one in two (1/2) and one in five (1/5) diluted liver DNA samples, prepared in duplicates, after diphenylamine treatment (table 4). The absorbencies of duplicates were averaged and corrected for the blank absorbance (table 4). The errors for the corrected absorbencies (SD) were between 1.0% and 11.4% (table 4). Using the corrected values, the DNA concentrations were obtained from the standard curve (figure 1, table 5) and used to calculate the DNA concentrations before dilution (0.018 mg x mL-1 (1/5), 0.359 mg x mL-1 (1/2) and 0.454 mg x mL-1 (c), table 6). The DNA concentration obtained using the 1/5 diluted sample was much lower than the values for the other two samples and was not further taken into Page 1 of 10

SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification 50 51

account. The average of the concentrated and one in two diluted sample (table 6), was used to calculate the DNA amount per gram chicken liver as 1.8 mg (0.18% of the tissue weight, table 7).

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Discussion The DNA yield appeared 5.4 fold higher for the weighing method than the diphenylamine method (tables 2 and 7) supporting the hypothesis that the weighing method would result in a higher DNA yield per gram chicken liver than the diphenylamine method. The yield determined using the weighing method was 9.8 mg per gram chicken liver (table 2). The DNA appeared yellowish, sticky and had the size of a small pea. The yield determined using the diphenylamine method was 1.8 mg per gram chicken liver (table 7). The errors for the corrected absorbencies (SD) were between 1.0% and 11.4% (table 4). The DNA concentrations were obtained from the standard curve (figure 1, table 5) which had an R² = 0.9905. Given the differences between the two methods is it really worthwhile to use the weighing method to determine the DNA content? The answer should be ‘yes’ as it can give a quick and easy obtainable estimate of the DNA amount.

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification References (1)

Watson JD and Crick FHC, 1953, Molecular structure of nucleic acids, A structure for deoxyribonucleic acid, Nature Vol 171, pp 737-738. (2) Hershey AD and Chase M, 1952, Independent functions of viral protein and nucleic acid in growth of bacteriophage, Journal of General Physiology Vol 36, pp 39-56. (3) Timmis JN, Ayliffe MA, Huang CY and Martin W, 2004, Endosymbiotic gene transfer: organelle genomes forge eukaryotic chromosomes, Nature Reviews Genetics Vol 5, pp 123-135. (4) Grange JM, Fox KR and Morgan NL (eds) 2009, Genetic Manipulation : Techniques and Applications, Online ISBN: 9781444314144 , Wiley, Chichester. (5) Dale JW, Von Schantz M and Plant N, 2011, From Genes to Genomes : Concepts and Applications of DNA Technology 3rd edition, ISBN 978-0-470-68386-6, Wiley, Hoboken. (6) Burton K, 1956, A study of the conditions and mechanisms of the diphenylamine reaction for the colorimetric estimation of deoxyribonulceic acid, The Biochemical Journal Vol 62, pp 315323. (7) Molecular Biology of the Cell SCIE1106 Lab Manual 2014, University of Western Australia, pp 33-45. (8) Traganos F, Darzynkiewicz Z and Melamed MR, 1982, The ratio of RNA to total nucleic acid content as a quantitative measure of unbalanced cell growth, Cytometry Vol 2, pp 212- 218. (9) Mustafa S and Mittal A, 1982, Protein, RNA and DNA levels in liver and brain of starved catfish Clarias batrachus, Japanese Journal of Ichthyology, Vol 28, pp 396-400. (10) Maniatis T, Fritsch EF and Sambrook J, 1982 Molecular Cloning; A Laboratory Manual, Cold Spring Harbor Press, Cold Spring Harbor, NY. (11) Pace CN, Treviño S, Prabhakaran E and Scholtz JM, 2004, Protein Structure, Stability and Solubility in Water and Other Solvents, Philosophical Transactions: Biological Sciences Vol 359, pp 1225-1235.

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification APPENDIX 1 – Calculations The DNA was extracted from 1g chicken liver in 5.3 ml of buffer. Of this, only a 3 ml aliquot was used to determine the amount of DNA obtained using either the weighing method (calculation set 1) or the diphenylamine method (calculation set 2). Calculation set 1: Calculating the yield of DNA obtained from a chicken liver extract as determined using the direct DNA weighing method. . (I)

Calculating the wet mass of DNA of the precipitate obtained from 3 mL extract: Wet mass of DNA (3 mL extract) = Mass of tube and wet DNA – Mass of empty tube With the values obtained: Wet mass of DNA (3 mL extract) = 6.660 g – 6.572 g  Wet mass of DNA = 0.088 g = 88 mg The wet mass of DNA obtained from 3 mL of chicken liver extract was 0.088g or 88 mg.

(II)

Calculating the wet mass of DNA from one gram of chicken liver: =  With the values obtained:

= =

= 155.47 mg

The wet mass of DNA obtained from 1 gram of chicken liver extract was 155.47 mg. (III)

Calculating the dry mass yield of DNA from one gram of chicken liver using the DNA wet mass (m w) to DNA dry mass (md) relationship: 1 g wet mass of DNA (mw) is the equivalent of 63 mg dry mass of DNA (m d)  md = 0.063 x mw With the values obtained:  md = 0.063 x 155.47 mg  md = 9.79 mg The dry mass yield of DNA obtained from one gram of chicken liver using the direct weighing method was 9.79 mg. This was the equivalent of 0.98% of the tissue mass.

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification

Calculation set 2: Calculating the yield of DNA obtained from a chicken liver extract as determined using diphenylamine method. (IV)

Calculating the amount of DNA in DNA standard dilutions: -1

The concentration of the DNA standard solution was 1 mg x mL . The following volumes were used to make dilutions of the standard solution: Standard tube

TE buffer (ml)

1 mg/ml DNA solution (ml)

1

0.25

0

2

0.2

0.05

3

0.15

0.1

4

0.05

0.2

The formula DNA amount (mg) = C x V (C: concentration, V: volume) was used to calculate the DNA amount in the final volume of 0.25 mL. The following calculation for standard tube 2 shows an example of this calculation: -

DNA amount (mg) = 1 mg x mL 1 x 0.05 mL DNA amount (mg) = 0.05 mg The DNA amount in standard sample 2 was 0.05 mg.

(V)

Calculating the concentration of DNA in DNA standard dilutions: The formula C1 x V1 = C2 x V2 (C: concentration, V: volume) was used to calculate the concentration of the dilutions. The following calculation for standard tube 2 shows an example of this calculation: -1

1 mg x mL x 0.05 mL = C2 x 0.25 mL -1

 1 mg x mL x

= C2

 C2 = 0.2 mg x mL

-1

-1

The concentration of DNA in standard 2 was 0.2 mg x mL .

(VI)

Calculating the average absorbencies for DNA standard and chicken liver DNA samples: The average absorbencies were calculated according to the formula: Average absorbance (sample x) = The following example shows the calculation for the standard sample 2: Average absorbance (standard 2) =

= 0.2475

The standard deviation (SD) of the average absorbance was calculated as: SD = √

∑



with xsample representing the individual sample absorbance readings,  sample representing the corresponding sample absorbance average and n sample the number of absorbance readings of that sample. The average absorbance of the two DNA standard 2 samples was 0.2475  0.001.

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification

(VII)

Calculating the corrected average absorbencies for DNA standard and chicken liver DNA samples: The average absorbencies were corrected for the absorbance measured for a sample that did not contain DNA (blank). The absorbance of the blank reflected the absorbance of light at 595 nm due to chemical reactions taking place without DNA being present. The corrected absorbencies were calculated according to the formula: corrected absorbance (sample x) = absorbance (sample x) - absorbance (blank sample) The standard deviation of the corrected samples ( SD) was calculated as: SD = √(

) (

)

with SDsample representing the individual standard or liver DNA sample SD values, SD blank representing the SD value of the average of the standard DNA sample with a DNA concentration of ‘0’ (blank sample) and nblank and nsample the number of absorbance readings of that sample. As an example the corrected absorbance for standard 2 was 0.1895 0.003.

(VIII)

Calculating the DNA concentrations of undiluted chicken liver DNA samples from values obtained from the DNA standard curve: The concentration of the undiluted liver DNA samples was calculated using the following formula: -1

Concentrated DNA sample (mg x mL ) = The example below shows the calculation for the ½ diluted chicken liver sample: -1

Concentrated DNA sample (mg x mL ) =

-1

Concentrated DNA sample (mg x mL ) =

-1

= 2 x 0.179 mg x mL = 0.358 mg x mL

-1

-1

The DNA concentration of the ½ diluted DNA sample before dilution was 0.358 mg x mL .

(IX)

Calculating the average concentration of an undiluted chicken liver DNA samples: The average concentration was calculated using only the concentration (undiluted) values obtained for the undiluted and the 1/2 diluted sample as the concentration of the 1/5 diluted sample was too low and did not agree with the other two sample concentrations. The average undiluted sample concentration was calculated as follows: Average concentration =

mg x mL

-1

The following example shows the calculation for the standard sample 2: -1

Average concentration =

mg x mL = 0.407 mg x mL

-1

-1

The average concentration of the DNA samples was 0.407 mg x mL .

(X)

Calculating the standard deviation for the average DNA concentration: The standard deviation (SD) of the average DNA concentration was calculated in analogy to the SD calculation shown under calculations 2, step VI. The average concentration of the DNA samples was 0.407  0.067 mg x mL

-1

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification

(XI)

Calculating the DNA yield of one gram chicken liver as determined using the diphenylamine method: First the amount of DNA in 2.5 mL TE buffer was calculated according to the following formula: Average concentration (undiluted sample) x 2.5 mL (TE-buffer) = amount of DNA in TE buffer -1

0.407 mg x mL x 2.5 mL = amount of DNA in TE buffer amount of DNA in TE buffer = 1.02 mg This DNA amount was equivalent to the amount of DNA precipitated from 3 mL extract. Hence it can be used to calculate the yield according to the formula: =

=

With the values obtained: =

= 1.8 mg

The yield of DNA obtained from 1 gram of chicken liver extract was 155.47 mg as determined using the diphenylamine method. This is the equivalent of 0.18% of the tissue weight.

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification APPENDIX 2 – Tables and figures Table 1: Wet mass determination of a chicken liver DNA precipitate using a direct weighing method. DNA was extracted from 1g chicken liver in 5.3 ml of buffer. Three millilitre of the extract were used to isolate the DNA and to determine the wet mass of the DNA using a direct weighing method. First the mass of an empty sample tube was determined by weighing and then the mass of the sample tube plus wet DNA was determined. The wet mass of the DNA precipitate was calculated as the difference between the tube plus DNA and the empty tube (appendix 1, calculation set 1, step I).

Mass (g) of empty sample tube

Mass (g) of sample tube and wet DNA

Wet mass (g) of the precipitate

6.572 g

6.660 g

0.088 g

Table 2: Calculated wet and dry mass (yield) of DNA from one gram chicken liver from values obtained using the weighing method. The wet mass per gram chicken liver was calculated as shown in appendix 1 (calculation set 1, step II) using the wet mass determined with the weighing method from 3 mL extract (table 1). The dry mass (yield) per gram chicken liver was calculated as shown in appendix 1 (calculation set 1, step III).

Precipitate

Precipitate mass (g)

Precipitate mass as fraction (%) of tissue sample mass

Wet Dry (yield)

155.5 mg 9.8 mg

15.55 % 0.98 %

Table 3 : Absorbance values and calculated average absorbencies of DNA standard samples as determined via spectrophotometry using the diphenylamine method. The concentrations of the standard DNA samples were calculated as shown in appendix 1 (calculation set 2, steps IV and V). The absorbance values were obtained by measuring the light absorbance at 595 nm. The average absorbance readings represent the average of the duplicate absorbencies and were calculated as shown in appendix 1 (calculation set 2, step VI). The corrected average absorbencies (Average absorbance of standard minus blank) were calculated as shown in appendix 1 (calculation set 2, step VII). SD: Standard deviations for the average sample absorbance readings were calculated as shown in appendix 1 (calculation set 2, step VI). SD : SD value for the difference of a sample absorbance average minus the blank absorbance average. The SD  was calculated as shown in appendix 1 (calculation set 2, step VII).

Concentration of Standard DNA (mg x ml-1)

Absorbance reading duplicate 1

Absorbance reading duplicate 2

Average absorbance reading ± SD (SD%)

Corrected average absorbance ± SD (SD%)

0

0.0550

0.0610

0.0580 0.004 (6.9%)

0.2

0.2470

0.2480

0.0

0.2475 0.001 (0.4%)

0.1895 0.003 (1.6%)

0.4

0.3965

0.3985

0.3975 0.001 (0.3%)

0.3395 0.003 (0.9%)

0.8

0.6475

0.6515

0.6495 0.003 (0.5%)

0.5915 0.004 (0.7%)

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SCI1106 Molecular Biology of the Cell – lab 1: DNA extraction and quantification Table 4: Absorbance values and calculated average absorbencies of diluted liver DNA samples as determined via spectrophotometry using the diphenylamine method. The absorbance values were obtained by measuring the light absorbance at 595 nm. The average absorbance readings represent the average of the duplicate absorbencies and were calculated as shown in appendix 1 (calculation set 2, step VI). The corrected average absorbencies (Average absorbance of sample minus blank) were calculated as shown in appendix 1 (calculation set 2, step VII). SD: Standard deviations for the average sample absorbance readings were calculated as shown in appendix 1 (calculation set 2, step VI). SD: SD value for the difference of a sample absorbance average minus the blank absorbance average. The SD was calculated as shown in appendix 1 (calculation set 2, step VII).

Corrected average absorbance ± SD (SD%)

Dilution of liver DNA sample

Absorbance reading duplicate 1

Absorbance reading duplicate 2

Average absorbance reading ± SD (SD%)

1/1

0.4100

0.4140

0.4120 0.0028 (0.7%)

0.3560

1/2

0.2100

0.2180