Help with Lab Math - preparing a gel (Oct/22/2008 )

I need help figuring out this - Im new at this. I need to prepare a 1.25% agarose gel in TAE buffer to run a sample that is 33ul. The gel box is 10x25 (cm) and the comb has teeth that are .5x.15 (cm). I'm supposed to be able to come up with: total amount of TAE and total amount of agar. Can someone show me how to figure this out?

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250ml of TAE + 3.125g agar +EtBr.

area of the gel box: 10x25cm= 250cm^2
Assume the gel has a depth of 1cm,
then 250cm^2 x 1cm= 250cm^3
1cm^3= 1ml
Therefore, the volume required is 250ml

make a 1.25% agarose gel in TAE buffer
250ml x 1.25%
=3.125g

Hope this may help

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Thank you so much! Do you always figure "about 1 cm" for the depth? I was trying to figure the depth out from the numbers I had. Someone in my class said that the gel depth "had to be able to fit the sample (33ul)". So, I didn't know how to figure out a depth that will be acceptable for 33ul.

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If your combs are 5 mm deep, the gel should also be about 6 mm deep. I would use a bit more than half the amount discussed above.

You need to know the comb width to determine the well volume.

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Assuming the teeth of the comb has a thickness of 0.1cm

the volume of the well is:

0.1cm x 0.5cm x 0.15cm = 1mm x 5mm x 1.5mm

=7.5mm^3

1cm^3 = 1ml= 1000ul

1cm^3= 10mm x 10mm x 10mm = 1000mm^3

therefore 1000ul= 1000mm^3
1ul= 1mm^3

Hence, the volume of the well is 7.5ul
I think you need a comb that has a wider teeth.


Hope this may help.

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