Question

1. Based on concentration from Nanodrop, determine volume of RNA in µL needed for 5 µg. For example if concentration was 77.4 ng/µL, then:

(5 µg)(1000 ng/µL)(x µL / 91.6 ng) = _____ µL

Concentration from Nanodrop = 91.6 ng/µL

2. Since DNase I Reaction buffer is 10X and volume for reaction is 100 µL, how much 10X DNase I reaction buffer will you to need to obtain final concentration of 1X?

3. Based on volume RNA and 10X DNase I reaction buffer added, determine how much RNase free water you will need to add to reach final volume of 100 µl.

Answer 1

The amount of 10X DNase I reaction buffer for 100ul of reaction may be calculated as follows:

For 1ul 10X DNase I reaction buffer, 10units of buffer is present

For 100ul reaction volume, the amount of 10X DNase I reaction buffer needed will be= 100units/10X=10ul

Hence 10ul of 10X DNase I reaction buffer is to be added to a 100ul reaction volume.

Concentration of RNA from Nanodrop=91.6ng/ul

Needed concentration of RNA to add in the reaction is 5000ng (5ug)

The volume of RNA needed to be added will be:

1ul of RNA = 91.6ng

?ul of RNA= 5000ng

=> 1*5000/91.6=54.58ul of RNA needs to be added in the reaction to make it to a final concentration of 5000ng.

The amount of RNase free water added will be the amount required to make up to 100ul after adding RNA and 10X DNase I reaction buffer

Hence the reaction will contain the following amounts of each component:

RNA=54.58ul

10X DNaseI reaction buffer= 10ul

RNase free water=35.42ul

Total reaction volume=100ul

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