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Centrifugation - how to compensate lower g?

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3 replies to this topic

#1 Trof

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Posted 28 March 2012 - 06:50 AM

I have a protocol saying I need 15 minutes/20 000g to pellet lysate debris.
But my rotor can do 12 500 maximum, and my tubes can take only 9 500. For safety I will go with 9 250g.

I know it should be possible to compensate by increasing the time, but I have no idea what is the proportion. When I have 9 250 instead of 20 000, should I double it? Triple it? Is there some formula for that?

Our country has a serious deficiency in lighthouses. I assume the main reason is that we have no sea.

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#2 mdfenko

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Posted 28 March 2012 - 07:46 AM

it's not exactly a linear relationship but, especially since rcfs and times used are generally more than necessary, just increase time by the ratio of the original rcf to the rcf that you will use.

in your case, 20000/9250=2.162, 2.162x15min=32.43min.

in actuality, the differences are dependent on k factor of the rotor (at the speed used) and sedimentation coefficient of the material being pelleted.
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#3 Astilius

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Posted 28 March 2012 - 08:04 AM

mdfenko's nailed it. I was going to say that sedimentation rate is liable to be the biggest contributer to the non-linear relationship here but given solutions that have no unusual characteristics (high viscosity for example) the rule of thumb conversion as he's detailed is going to be more than adequate.

If in doubt run a pilot test and have a looksee.

Edited by Astilius, 28 March 2012 - 08:06 AM.

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#4 Trof

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Posted 28 March 2012 - 09:21 AM

I see. Sedimentation coeficient should come to my mind.
Anyway thanks a lot.

Our country has a serious deficiency in lighthouses. I assume the main reason is that we have no sea.

I never trust anything that can't be doubted.

'Normal' is a dryer setting. - Elizabeth Moon






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