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Consider a foreign gene with recognition sequence as GAATTC for EcoR1. Now suppose that it is being cut at two palindromic sequence to form sticky ends.

Here the sticky ends are formed such that its position is parallel to each other around the axis of DNA.

The plasmid DNA has a recognition sequence and is being cut at a site and as obvious would be cut with the sticky ends as counterparts. If we join the foreign DNA to this vector DNA the problem of combining the DNA arises. How will we cope with this problem or if I am going wrong can someone tell me where?

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    $\begingroup$ I am having trouble understanding your question. Inserting a fragment of DNA into a plasmid is pretty standard procedure, especially if the restriction sites are the same, just cut, purify, mix, and ligate. Can you clarify where you don't understand? $\endgroup$
    – user137
    Commented Oct 11, 2015 at 7:29
  • $\begingroup$ @user137 Let us consider a two dimensional case for this where the DNA is represented by two parallel lines where only one of the lines extended represents the strand with two sticky ends.In the plasmid DNA this would not be the case as the sticky ends are counterparts therefore opposite to each other at the site where it is cut off.This position of sticky ends of the plasmid DNA is not allowing the combination of sticky ends of the foreign DNA. $\endgroup$
    – Sikander
    Commented Oct 11, 2015 at 8:08
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    $\begingroup$ A drawing would help I guess. I can't visualize the issue. $\endgroup$
    – AliceD ♦
    Commented Oct 11, 2015 at 9:11
  • $\begingroup$ yeah it would certainly help.But I have some complications in doing so.I am not able to upload the image.I hope some symbols would help. $\endgroup$
    – Sikander
    Commented Oct 11, 2015 at 11:03
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    $\begingroup$ Instead of the ridiculous characters, try and use standard molecular biological notation. Sticky ends will either be a 5' overhang or a 3' overhang. EcoRI cuts between the G and the A and produces a 5' overhang of 5'-AATT...-3'. And what do you mean by "parallel to each other around the axis of DNA" If you are saying that your target gene has two EcoRI sites that flank it then the cut at the 3' end will still produce a 5'-AATT...-3' overhang, but on the complementary strand. If you cut your plasmid with EcoRI then it will pair complementary to those ends. $\endgroup$
    – AMR
    Commented Oct 11, 2015 at 13:15

2 Answers 2

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In the Google search engine I entered "ligation of sticky ends"; then I selected the "Images" tab. Here is the first result, which addresses your question exactly:

Annealing of complementary 5'-overhanging single-stranded ends on double-stranded DNA fragments

This originates from this website: AddGene

Based on your question it sounds like you understand the basic rules of forming Watson-Crick base pairs; but your diagram and question omits the innate polarity of the sugar-phosphate backbone in DNA and RNA. The convention in depicting double-stranded DNA is that the "top" strand (sometimes referred to as "Watson") is written in the 5'-to-3' direction, reading from left to right. Since the two DNA strands are anti-parallel, it follows that the "bottom" strand (maybe nobody refers to it as "Crick" any longer?) is written in the 3'-to-5' direction -- if you are reading from left to right. If you want to decode the biological information encoded in the bottom strand then you need to read it from right to left.

This is a long-winded way of saying that I usually note the 5' and 3'-ends on a drawing of DNA to emphasize this, and if I can figure out how to do so I will update the image.

Here is a slightly better version from Google:

Annealing of sticky ends

Which is from this site: A WordPress Blog

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  • $\begingroup$ my question is different.Now consider the one Watson strand of some foreign DNA. As one move from 5' end we get the sticky end 5'-AATT.After some locations let the sticky end start again as 5'-TTAA.Now when we combine this foreign DNA piece to the plasmid DNA it is not necessary that the sticky ends of foreign DNA just fall above their complimentary cut counterparts of the plasmid DNA to complete the helical structure. $\endgroup$
    – Sikander
    Commented Oct 11, 2015 at 15:56
  • $\begingroup$ You may be confused about palindromic DNA sequences. Perhaps the simplest explanation is this: The Eco RI enzyme recognizes and cuts this double-stranded sequence: 5'-GAATTC-3' This sequence is the same on the top strand AND on the bottom strand--if you read the sequence in the 5'-to-3' direction. All of the sticky ends generated after digestion will have the sequence 5'-AATT-3', and will therefore form Watson-Crick base pairs with each other. There is no such thing as a 5'-TTAA-3' sticky end generated by Eco RI. That would require a completely different enzyme. $\endgroup$
    – mdperry
    Commented Oct 11, 2015 at 17:52
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Assume this is your plasmid (N-Ñ denotes any complementary pair):

          enter image description here

And you have an insert which has EcoRI site at its 5' end.

When you do an EcoRI digestion of the insert and the plasmid you'll get these products.

  • Digested insert:

               enter image description here

  • Digested vector

          enter image description here

During ligation, insert will pair with the vector in this manner (black is the vector and blue is the insert):

           enter image description here

If you now want to ligate the other end you can use the HindIII site. However, if you have same restriction sites on both the ends (for instance EcoRI) of the insert then you will get two ligated products which will have opposite orientations of the insert.

I hope this clarifies your doubt.

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  • $\begingroup$ let us take a plastic model of DNA with sticky ends which are also complentary.I take another foreign plastic model of DNA with varying lengths which also have same sticky ends to combine with the former one.Now every length of DNA may not combine with it because of different orientation? We cannot twist it forcefully to combine. $\endgroup$
    – Sikander
    Commented Oct 12, 2015 at 14:46
  • $\begingroup$ @Sikander I can't really understand what you mean. Are you saying that there is a ssDNA region but only the first few residues are sticky. Ligation would happen only in the situation I mentioned above. You can draw a diagram to illustrate what you are asking. Use powerpoint to make a diagram, save the diagram as a picture and then post it in the question. $\endgroup$
    – WYSIWYG
    Commented Oct 12, 2015 at 14:56
  • $\begingroup$ In a test tube, where you have added two different pieces of DNA--with complementary sticky ends, if the DNA fragments are too short to circularize--like your plastic model, then what happens is that you form random multimers of the fragments, that all get ligated in a line. Eventually, when their cumulative length allows the two free ends to reach around and find each other, you will achieve intramolecular circularization. This can all be assayed be gel electrophoresis and electron microscopy. $\endgroup$
    – mdperry
    Commented Oct 12, 2015 at 17:31
  • $\begingroup$ @mdperry yeah! that is the possible answer I was looking for.You mean that there are many foreign fragments which will linearly combine till they get such orientation that their sticky ends are able to combine with their complementary in vector DNA. $\endgroup$
    – Sikander
    Commented Oct 12, 2015 at 17:55
  • $\begingroup$ @Sikander Yeah they can form multimers. Typically you should get a distribution of number of multimeric repeats. It also depends on concentration. In any case it would be great if you can add an illustration in your question to clarify the exact point that you are interested in. $\endgroup$
    – WYSIWYG
    Commented Oct 12, 2015 at 19:07

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