7
$\begingroup$

I am learning RNA-seq analysis. I always encounter this phase "poly(A)+ RNA". After searching, I got this: "Most messenger RNAs contain a poly(A) tail, while structural RNAs do not. Poly(A) selection therefore enriches for messenger RNA. The technique has proved essential for construction of cDNA libraries."

Does it mean that when constructing the cDNA, we always use the ones with a A's tail? Then other RNAs like tRNA are filtered and not used to get their cDNA? I do find some RNAs records on UCSC where microRNAs are included.

When we say poly(A)+ RNA library, what is the difference between normal one? Or normal RNA library is a poly(A)+ RNA library? Thanks.

$\endgroup$
1
  • 1
    $\begingroup$ Note that this isn’t always done: there are RNA-seq protocols which do intentionally not enrich for mRNA and hence don’t do poly(A) selection. After all, we might be interested in other RNA as well (but this is much rarer since most RNA-seq is done as a quantification of gene expression). $\endgroup$ Commented Apr 25, 2012 at 13:21

1 Answer 1

11
$\begingroup$

One of the primary reasons to use poly(A) selection is to eliminate the massive amount of ribosomal RNA (rRNA) present in the samples. The alternative is to use ribosomal RNA depletion kits / techniques to remove as much rRNA as possible before sequencing. Without rRNA depletion a large proportion (~60-80%) of the reads would map to rRNAs.

poly(A) selection also reduces the amount of pre-mRNA in the samples, which reduces the number of intron aligned reads.

$\endgroup$
5
  • $\begingroup$ Thanks, GWW. So if we are not saying "poly(A) RNA", the reads we get will be largely from rRNAs? And for poly(A) selection, is it going to remove other RNAs like microRNA? $\endgroup$
    – Joy
    Commented Apr 25, 2012 at 5:01
  • $\begingroup$ @CaiShaojiang, yes it will not bind to miRNAs $\endgroup$
    – bobthejoe
    Commented Apr 25, 2012 at 8:02
  • 1
    $\begingroup$ @bobthejoe miRNA can be (usually is) polyadenylated as well. $\endgroup$ Commented Apr 25, 2012 at 13:23
  • 2
    $\begingroup$ @KonradRudolph: Not exactly. The pri-miRNA is polyadenylated, but what is (eventually) called the miRNA isn't, so ... $\endgroup$ Commented Apr 25, 2012 at 21:40
  • $\begingroup$ @Steve Correct. Unfortunately, I oversimplified. $\endgroup$ Commented Apr 25, 2012 at 22:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .