- I personally have not had the experience of <30 ug cRNA with 5 ug
total RNA. (I adhere rather painstakingly - almost mule-headedly - to
the protocol, but anyone might too, if one was operating with $5000
worth of chips/reageants at one go. ;P)
- Also, the amount used for the IVT reaction, even for starting amount
of 5 ug total RNA, is only about 10 uL of the 12 uL from cDNA
synthesis
reaction/cleanup, so one can assume that there would be even more cRNA
synthesized if I used the entire cDNA synthesis reaction.
- I am curious - why is there a need to assume equal amounts of
starting
total RNA, if the loading amount of cRNA is the same? Wouldn't
normalization manage any minor differences in loading cRNA? The amount
loaded on the HGU133A chip is fixed at about 10 ug cRNA (although the
Affymetrix protocol makes you jump through hoops to calculate that)
(since the hybridization amount is 15 ug, but only 200 uL out of 300
uL
of hybridization mixture is used for the GeneChip)
Regards,
Min-Han
-----Original Message-----
From: James MacDonald [mailto:jmacdon@med.umich.edu]
Sent: Monday, October 06, 2003 10:33 AM
To: vos@stats.ox.ac.uk; Tan, MinHan
Cc: bioconductor@stat.math.ethz.ch
Subject: RE: [BioC] Pooling in microarray studies
What you need to have is 16 ug of fragmented cRNA to hybridize to the
chip. If you start with 10 ug, there is a very high probability that
you
will end up with at least 16 ug at the end of the molecular biology
step. However, if you only start with 8 ug, you are less likely to end
up with the magical 16 ug, but it is still highly probable. We
recently
did something like 8 chips with only ~5 ug for each, and had 16 ug at
the end for all of them.
Jim
James W. MacDonald
Affymetrix and cDNA Microarray Core
University of Michigan Cancer Center
1500 E. Medical Center Drive
7410 CCGC
Ann Arbor MI 48109
734-647-5623
>>> Wiesner Vos <vos@stats.ox.ac.uk> 10/06/03 10:02AM >>>
Thanks for the replies Min-han and Jim.
I suspected that this may be the case (keeping the
papers in mind about comparing expression measures
on the Dilution data eg. Cope et al.)
I suspected that the use 8ug would be perhaps
not make much of a difference if you use
an expression measure like for example
RMA that seems to be reasonably independent of
the actual amount of RNA on a chip. Thanks
again for your help.
Wiesner J. Vos
Department of Statistics
University of Oxford
1 South Parks Road
OX1 3TG
United Kingdom
On Mon, 6 Oct 2003, Tan, MinHan wrote:
> - Perhaps it is an policy issue with your microarray core facility,
but
> the standard Affymetrix GeneChip Technical Manual recommends 5 - 20
ug
> total RNA.
>
> - I routinely use 5 ug total RNA with no apparent problems.
>
> - So yes, you should be able to do single sample hybridizations.
>
> Regards,
> Min-Han
>
>
>
> -----Original Message-----
> From: Wiesner Vos [mailto:vos@stats.ox.ac.uk]
> Sent: Monday, October 06, 2003 9:27 AM
> To: bioconductor@stat.math.ethz.ch
> Subject: [BioC] Pooling in microarray studies
>
>
>
> I have question arising to the pooling of mRNA
> samples. Someone approached me about the
> following problem:
>
> The study wants to use Affymetrix chips to study
> changes in expression between a group of treated
> mice and a group untreated mice. There are 10 mice
> in each group. It is only possible to extract
> 8 ug of RNA from each mouse, not enough for one chip. (According to
the
> experimenters they require 10 ug per
> chip) So it is not possible to use biological
> replicate chips for each individual mice. Now the issue
> is whether to perhaps pool the RNA in each group
> and carry out analysis on technical replicates from the pooled
> samples.
>
> As I understand it pooling may reduce the precision, with
> the risk that one or few samples can dominate the outcome, and that
> averaging over single sample hybridisations is perhaps safer than
using
> pooled samples. However in this case you cannot do single sample
> hybridisations.
>
> I was wondering if the following approach is an acceptable
compromise
to
> retain at least some information on the between sample variation in
each
> group:
>
> Mix the RNA from 2 different mice on a single chip to get 5
> hybridisations, where the hybridisation on each chip is from the mix
of
> the RNA samples of two mice? I though that this may enable you to
some
> extend if all the mice are behaving similarly. Ofcourse one would
not
be
> able to distinguish between the behaviour of the two mice relating
to
> the same chip. Or is it better to accept that you do not have enough
RNA
> to hybridize the sample for each individual to a separate chip and
pool
> the samples and accept the risk that one sample may dominate the
> outcome? The best solution did not seem obvious (to me at least!)
>
> Any comments will be much appreciated.
>
> Wiesner
>
>
>
> Wiesner J. Vos
> Department of Statistics
> University of Oxford
> 1 South Parks Road
> OX1 3TG
> United Kingdom
>
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