In the Paired End Library manual 3kb span (Oct. 2009 version), step 3.11 final library size selection, there is a bouble AMPure size selection on amplified paired end library. I understand the first ampure removes the larger fragment and the second ampure remove the small fragment in order to generate a sharp peak around 500 bp. But I couldn't understand the ratio it uses in the second ampure.
Based on AMPure beads titration, if the PE cutoff value is 0.7, x=70ul. I add 70ul AMpure beads to 100ul PCR product. At this ratio (0.7x), larger fragments (i.e. >600 bp) will bind to ampure bead. Supernatant contains everything <600 bp will be ampured for a second time to remove smaller fragment (let's say <500 bp). My question is on the second AMPure selection. Continue with my example, the supernatant of the 1st AMPure is 170ul, then I add 100ul EB, the total volume is 270ul. How much AMPure is required? Y=X+20=70+20=90ul. 90ul AMPure to 270ul DNA, the AMPure ratio is 0.33x. At such low ratio, how can DNA around 500bp bind to the AMPure beads? Remember at the titration, the ratio tested are between 0.5x to 1X. Even at 0.5x, 500bp bind is mostly removed.
Hope to get some answers soon. Many thanks!
Based on AMPure beads titration, if the PE cutoff value is 0.7, x=70ul. I add 70ul AMpure beads to 100ul PCR product. At this ratio (0.7x), larger fragments (i.e. >600 bp) will bind to ampure bead. Supernatant contains everything <600 bp will be ampured for a second time to remove smaller fragment (let's say <500 bp). My question is on the second AMPure selection. Continue with my example, the supernatant of the 1st AMPure is 170ul, then I add 100ul EB, the total volume is 270ul. How much AMPure is required? Y=X+20=70+20=90ul. 90ul AMPure to 270ul DNA, the AMPure ratio is 0.33x. At such low ratio, how can DNA around 500bp bind to the AMPure beads? Remember at the titration, the ratio tested are between 0.5x to 1X. Even at 0.5x, 500bp bind is mostly removed.
Hope to get some answers soon. Many thanks!
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