Bacteria sorting ?

Gerhard Nebe-von-Caron Gerhard.Nebe-von-Caron at Unilever.com
Mon Mar 4 15:13:00 EST 2002


There is nothing wrong with the practical approach of volume measurement as
normally you do not know the exact conditions over the orifice. However, the
volume of a droplet should not be relevant to your question.

First of all you have to make sure your sheath is sterile and particle free at
the point of measurement. As you usually can't use antibacterials when sorting
bugs you have to be that little bit cleaner!

If you trigger on fluorescence you won't be able to see nonfluorescent bacteria
that will also be sorted. This is why I always emphasise in the tutorial the
importance of multichannel triggering as there are events where one gets
fluorescent signals but insufficient light scattering and the other way round.
Most bacteria pump out most dyes so appear unlabelled, depending on available
energy and the influx and efflux kinetics of the dyes used and might therefore
remain hidden if you trigger on fluorescence. For cultured organisms light
scatter tends to work fine as a trigger as your cells will have ample scatter
signals. If you still drown in noise you should worry about it's origins. If
the origin lives in the sheath - tough.

You have to make sure you reduce the flow rate below the levels of acceptable
coincidence. This means the event-flowrate or datarate, not the sample volume
flow rate which is a different matter. Also be aware that you are not able to
discriminate for doublets any more as you can not use pulse shape processing
(with very few exceptions), and for example a unlabelled cell stuck to a
labelled cell remains undetectable.

The question about the rate at which you detect a single cell in a droplet
leaves me a bit worried. The distribution of bacteria into single droplets is
based on Poisson statistics, but in a simple way if you oscillate the flow cell
at 30kHz and you have an eventrate of 30000 per second you will have an average
of 1 bacterium per droplet. If your eventrate is 60000 it will be 2 cells per
droplet on average. Thus the frequency distribution of bacteria into droplets
does not really depend on the size of the droplet, but mainly on the event
frequency relative to your droplet generation frequency. If however your
staining or triggering limits you to only detecting 10% of your bacteria you
will get the impression of a false eventrate e.g. 6000 cells per second or only
a cell in every 5th droplet whilst there are in real terms still 2 bacteria per
droplet. This method has been used by a couple of people for efficient
preenrichment sorts, having for example on average 4 cells per droplet.
Triggering on the rare fluorescent event to sort droplets enriches a rare event
from lets say 1 in 10exp6 to 1in4 which can be rerun fur purity quite easily.

Hope this helps

Regards

Gerhard Nebe-von-Caron

Research Scientist
Applied Science & Technology Group
SEAC - Safety and Environmental Assurance Centre
Unilever Colworth, Sharnbrook, Bedfordshire,  UK - MK44 1LQ

Tel: +44 (0)1234 264822, Fax: +44 (0)1234 222552
E- mailto:Gerhard.Nebe-von-Caron at unilever.com





-----Original Message-----
From:	Andy Johnson [SMTP:andy at brc.ubc.ca]
Sent:	Monday, March 04, 2002 9:15 PM
To:	Cytometry Mailing List
Subject:	Bascteria sorting ?


Hello

For those of you who know the true finer details about flow and fluid
dynamics ... can somebody tell me the calculation to work out the volume
size of droplets.  The mathamatical approach rather than collecting and
counting the volume.

We are using a 50 µm nozzle at 30psi (BD vantage) and trying to single cell
sort bacteria onto  96 well plate.

The first question is how large are the droplets (volume) ?

Can you determine the dilution factor of your samples (bacteria) ?

We are sorting bacteria and triggering on the fluorescence signal to reduce
background noise.  The questions is : at what rate of sample can we assume
that the instrument is detecting a single cell in a droplet ?

Looking at the numbers it would appear that non-fluorescent bacteria must
be in the same droplets, but I may be assuming something incorrectly.

Any help appreciated and further information available if it will help you
with the solution !

Andy


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