Question

Calculation: dilute the rna sample by 100 times, and the reading of od260 is 0.5. what is the concentration of rna in the original sample? if we need 2 ug of total rna for the reaction, how much volume should be taken from the original sample?

Answer 1

The concentration of RNA in the original sample is 25   (µg/ml) and the volume taken from the original sample is 0.023 ml.

RNA yield can be assessed using various methods including absorbance (optical density), agarose gel electrophoresis, or the use of fluorescent RNA-binding dyes.

RNA concentration is estimated by measuring the absorbance at 260nm, adjusting the A260 measurement for turbidity, multiplying by the dilution factor, and using the relationship that an A260 of 1.0 = 50µg/ml pure RNA.

Concentration (µg/ml) = A260 reading × dilution factor × 50µg/ml

                                     = 0.5 × 1 × 50

                                     = 25   (µg/ml)

RNA yield (µg) = RNA concentration × total sample volume (ml)

Total sample = RNA concentration - RNA yeild (µg)

                     = 25 µg - 2 µg

                       = 23 µg

The total yield is obtained by multiplying the RNA concentration by the final total purified sample volume. = 0.023 ml

                   

However, RNA is not the only molecule that can absorb UV light at 260nm. Since DNA also has a great absorbance at 260nm, and the aromatic amino acids present in protein absorb at 280nm, both contaminants, if present in the RNA solution, will contribute to the total measurement at 260nm. Additionally, the presence of guanidine will lead to a higher 260nm absorbance. This means that if the A260 number is used for the calculation of yield, the RNA quantity may be overestimated.

Learn more about RNA here:-brainly.com/question/28073947

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