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# Agarose Gel Electrophoresis question?

## Agarose Gel Electrophoresis question? - Biology Forum

### Agarose Gel Electrophoresis question? - Biology Forums. Ask questions and discuss the study of Biology. If you have Biology questions from your homework ask them here!

#1
03-11-2011, 04:09 PM
Pipette Filler

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Agarose Gel Electrophoresis question?

I have been given a question to answer and am unsure if I am making the right calculations. The question is:

A 4-cutter restriction enzyme such as Hae lll recognises a particular sequence of four base pairs. The probability that any given set of 4 bases is the Hae lll recognition site is 1/4 x 1/4 x 1/4 x 1/4. So, what would you expect to be the average fragment size when human DNA is cut with Hae lll?

So far, I have made the following calculations:

There are 6,000,000,000 base pairs in human DNA and 1/4^4 is 0.00390625.

So, 0.00390625 x 6,000,000,000= 23437500

Am I on the right lines? If anyone could give me some help with this question it would be great.Thakyou in advance.
#2
03-11-2011, 09:03 PM
M.D/Ph.D

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Re: Agarose Gel Electrophoresis question?

23,437,500 would be the number of times Hae III cuts the human genome, isn't it? I'm not completely sure, but if that's the case then if you divide the genome size (6,000,000,000) between the number of cuts (23,437,500), you'll have the average size of the resulting fragments:

--> 6,000,000,000 / 23,437,500 = 256 bp

So the answer would be 256 bp

Hope I'm right. xD
 The Following 2 Users Say Thank You to luisillo For This Useful Post: admin (03-11-2011), Emmalouise (03-12-2011)
#3
03-13-2011, 12:19 AM
Pipette Filler

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Re: Agarose Gel Electrophoresis question?

On that note: I am confused about how I would go about performing a similar calculation to deduce the average fragment size that a 6-cutter restrication enzyme like Hind III would produce?
#4
03-14-2011, 09:59 PM
M.D/Ph.D

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Re: Agarose Gel Electrophoresis question?

Considering the above example, if the recognition site is of 6 bp then the probability of cut would be: 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4, or (1/4)^6. That's is if the restriction sequence is completely nucleotide specific.

Considering that, a 6 bp restriction site enzyme would generate an average fragment lenght of 4096 bp.

It's more a probability issue than a molecular biology one.
 The Following User Says Thank You to luisillo For This Useful Post: biomedicine.girl (03-19-2011)

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