Q) A 50 ug/mL solution of pure DNA will give an absorbance of 1.0 reading at 260
ID: 1016392 • Letter: Q
Question
Q) A 50 ug/mL solution of pure DNA will give an absorbance of 1.0 reading at 260 nm. Based upon this information, CALCULATE the concentration of DNA in your purified sample. Also, DETERMINE the % amount of DNA in your original leaf tissue on a mass percentage basis. For this latter calculation, assume that all the DNA in your original share of the lysis buffer was present in the final purified DNA sample.
Absorbance of my purified DNA at 260 nm = 0.027.
Total volume in cuvette = 420 uL (400 uL dH2O + 20 uL purified DNA)
Mass of spinach leaves used to prepare this DNA sample = 0,75g
600 uL of Lysis Buffer per 100 mg of original wet of leaf tissue was used to make the unpurified DNA solution.
Volume of Lysis Buffer added to the spniach leaves and liquid N2 = (600 uL)/(100 mg) = 6 uL/mg * 750 mg = 4500 uL.
Volume of spinach leave solution given to each group = 600 uL.
Purification process involed the addition of 140L of Precipitation buffer, 420 L of Isopropanol, 300L of sterile dH2O, 150L of DNA binding solution, 300L of absolute ethanol, 1300L of wash buffer and 100L of elution buffer at 65°C.
After purification, volume of purified DNA obtained = 100 uL.
I was able to determine the concentration by multiplying abs260(0.027) * 50 ug/mL * dilution factor(420/20) = 28.35 ug/mL
But I dont know how to find the % amount of DNA in the original leaf tissue. My friend got 8.02 * 10^-3 but he's not sure if he's right.
Explanation / Answer
Concentration of DNA in your purified sample = 28.35 ug/mL
After purification, volume of purified DNA obtained = 100 uL.
Hence, mass of the DNA = 28.35 * 100 * 10^ -3 ug = 2.835 ug
Mass of spinach leaves used to prepare this DNA sample = 0.75g
So, % amount of DNA in your leaf tissue (mass percentage basis) = 2.835 * 10^-6 g/0.75 g * 100% = 3.78 * 10^ -4 %
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