Question

You are given a sample of RNA diluted 1:5. You read the OD260 for be 0.45. How much of the non-diluted RNA would be needed to obtain 5ug (micro-grams) of total RNA?

A. 5 ml

B. 111 ul

C. 11.1 ul

D. 55.5 ul

E. 1.1 ul

Answer 1

(D) 55.5 ul *** Given RNA diluted - 1:5 00260 - 0.45 We Knowa For a 1-cm equals 1.0 Pathlength, the optical density at 260nm for the RNA = 40 ug/ml 100 = 40 ug me 004500 will be = 40 ugime x0.45 40x0.45= 18 ug/ml So, " Concentration of diluted RNA at OD 0.45 I will be 18 uglme we know, the RNA is 5 times diluted. To make, the concentration of diluted RNA Equal to undiluted or Non-diluted RNA, we will Multiply concentration of diluted RNA with 5 Factor. concentration 1 of diluted dilution RNA = 18 ug Ime Factor = 5 " so, Concentration conc RNA = 18 x5 of id. Non-diluted 1. = gouglme CSScanned with CamScanner

concentration of Chi Non-diluted RNA = gougl me Now, we want calculate volume of Non-diluted RNA to obtain Sug of total RNA. So, here we can assume Stock solution is Concentrated Non-diluted RNA Stock solution = go ug/ml elie uci= noug Vi = ? V we have to calculate. Vi is the volume of Von-diluted RNA solution to obtaing Seeglme solution. 4. Here working solution will be suglme RNA solution, Coq=5 ug , vg = 1 me 1. By using Formula - CiUFCqva Гc - Go5 Via? Lea=bug 90 MgX Vi= 5 ugxime Lugalme me ", - sugxime = o. 555 me 05.55Y ден Im ooo en 0:55 5 ml - Com 55.5 del Volume of Non diluted RNA solution Required to obtain sug of total RNA (V.) = 55.5 ul. CS$lanned with CamScanner