I can not understand how to deal with the analysis of gel electrophoresis! I run

Question

I can not understand how to deal with the analysis of gel electrophoresis!

I run five kinds of sample; double digested plasmid DNA (by Ava2 +Pvu2) in lane1, single digest by Pvu2 in Lane3, single digest by Ava2 in lane4, pUC 19 control in lane6, and DNA Ladder in lane7.

and made this semi-log grhaph of X axisis: traveled distance (mm) VS base pairs (by using DNA laddar)

By using them, I calculated the size of double digest sample( by Ava2 + Pvu2) as ; 4000, 2100, 1500, 200.

As the gel figure shows, In lane 1, there are five bands. However, in this lab, I used two enzyme Ava2+Pvu2 to digest PUC 19.

digesting pUC19 by Ava and Pvu must produce 4 flagments ( these sizes are expected to be 1209, 933,322,222 bp, besause the total bp of pUC19 is 2686bp)

However, in the experiment, I got five bands ( in lane 1) ! I can not understand why there is extra band produced eventhough it seems two enzymes work well ( I think this case is not imcomplete-digestion, because if it was, there would be less than 4 bands produced).

Here, I have some question.

Q1: I am required to find the anti-log of the value in order to record the bp, How Can I find anti-log value? Please show me how to do it.

Q2: How can I explain there ere 5 bands ( 4 bands + 1 additonal unexpected band) in lane 1.

Q3: total bp of flagments in lane1 is 7800. This number is too high compeared with pUC 19's bp( 2686=around 27kb). The total bp of this sample should be within 2686~. Why is it too high? Does the number change to be with in 2686 after I caluculate anti-log of them?

I am comfused and can not find correct wat to deal with this problem, pelase help me.

Explanation / Answer

Q1. Here, anti log value will not be required. From the migration distance of the 5th band draw a line onto the best fit line to find its molecular size. I do not understand why only 4 values have been calculated when there are 5 bands seen in the gel. Kindly calculate the value of the 5th band as well, so that it would give an exact idea of the band sizes and it could be determined if the interfering band is of a genomic DNA or an average contamination., or a plausible mutation in the MCS region.

Q2. There maybe various reasons for the presence of an extra band which might include - some kind of contamination by an extra E.coli . Since it is obvious here that it cannot be a partial digest as the size of the partial digest will not be greater than the size of the entire genomic DNA , the reason is not a partial or incomplete digest. There is a high chance that the sample had some kind of contamination either via another genomic DNA or another E.coli cell that contained another plasmid.

Q3. The average size of the 1st lane is way more than the size of the plasmid . This is a clear indication that the interfereing band is of a genomic DNA. However, if it is of the genomic DNA , there would be more interferences I.e. Extra bands. Thus it can also be assumed that it might be because of a mutation, thus giving a much greater size than the original size of the plasmid.

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