Souther blot and restriction mapping

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Mistersnuffle
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Souther blot and restriction mapping

Post by Mistersnuffle » Sat Oct 04, 2014 11:17 am

Hi all,

I would greatly appreciate any help with constructing a restriction map from Southern Hybridisation data. I can construct restriction maps from fragment data obtained through endonuclease digestion and separation by standard gel electrophoresis, however, Southern Blot data only shows those fragments that a probe has hybridised to and I am struggling to find the correct approach. The parting words from lecturer were 'It's just a matter of logic, consider that fragments may overlap'. The question is as follows:

Genomic DNA was digested and run on a gel. The gel was blotted and used for Southern hybridisation. The probe was a 2.4 kb XhoI fragment. Using the southern data, determine the restriction map for genomic DNA.

Data:
Enzyme Approximate Band Sizes

E ------------- 6.0 ----- 1.8
X ------------- 2.4
B ------------ >10 ----- 3.6
XE ------------ 1.7 ----- 0.7
XB ------------ 2.3 ----- 0.1
BE ------------ 2.0 ----- 1.6 ----- 0.2

Any help with this would be greatly appreciated.

Thank you.

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JackBean
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Post by JackBean » Wed Oct 08, 2014 9:57 am

What I did was to draw several lines each with marking the placeswhere the enzymes must cut and space in between

X must be Xho obviously as it gives exactly one band of the same size

so
X + E gives |-------0,7----|-----------1,7----------|
X + B gives |------2,3--------------------|---0,1--|
B + E gives |-----------2,0---------|------------1,6-------|----0,2------|

if you compare XE and XB, you can get either 1.6 fragment in between (with my positioning) or 0.6 fragment with the opposite position of fragments in XE (i.e. |--1,7--|--0,7--| ).
Now you just have to put it together:
X---0,7---E-------1,6--------B--0,1--X
(you know the X must be on the borders and thus E and B cut in between)
Now it's just matter of trying where B and E may cut to fit the pattern of single-digest
E has 1,8 fragment which coincides with 1,6+0,1+0,1 so E cuts 0,1 to the right from the right X site and 5,3 to the left from left X site
B has 3,6 fragment which is made together from 1,6 (BE), 0,7 (XE) and 1,3 (invisible in BX digest, but combination with previous fragment visible as 2,0 fragment in BE digest)
so the final map looks like this:
E-------4,0-----B---1,3---X---0,7---E------1,6--------B--0,1--X--0,1--E-------------->9,8-------------B
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zeeshan002
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Post by zeeshan002 » Tue May 05, 2015 10:33 am

Genomic DNA was digested and run on a gel. The gel was blotted and used for Southern hybridisation. The probe was a 2.4 kb XhoI fragment. Using the southern data, determine the restriction map for genomic DNA.

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