I have very limited knowledge when it comes to DNA fingerprinting. I know about the technique gel electrophoresis and how everything is carried out. But I don't know how to read the result or understand that. Can anyone please explain me how to read the result [profile of bands]?

Because of my lack of knowledge, I can't solve problems like this one. Please explain me using this one so that I can get a clear idea about DNA profiling. This seems like home-work problem but the thing is I want to understand the technique of restriction mapping by reading the result of restriction digestion on electrophoresis and my intention is to understand the topic. Please Help.

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And I can't describe how I tried to solve this question because I don't have any kind of knowledge about this technique [Restriction Mapping] so my efforts are very silly to mention and I have improper way of approach to this as I don't know how to do the Restriction Mapping.

But I am given the answer which I tried to make sense by working on it but I just couldn't help it. It seems easy after seeing the Map but I don't get the relation between the profile of bands and the Map so if I am given a different problem I will fail to come up with a reasonable answer.

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What I am trying to say is, by seeing this map and working on it I get to agree that this is the correct mapping for the data given but I have some doubts, like, why we keep 3Kb distance between BamHI and EcoRI sites. Working out in some another way will end up with wrong answer. How these inticacies are understood? By reading the profile of bands [data given] right? So that's what I am asking for you to explain to me. I don't know anything about it.

Thank you always.

  • 2
    $\begingroup$ This is homework, one way or another. Please let us know what you tried to solve your problems and exactly what the bottle neck is. $\endgroup$
    – AliceD
    Commented Oct 9, 2015 at 11:18
  • $\begingroup$ Homework questions are off-topic on Biology unless you have shown your attempt at an answer. For more information see our homework policy. $\endgroup$
    – MattDMo
    Commented Oct 9, 2015 at 13:49
  • $\begingroup$ @MattDMo I would let this question go. As you can see from my description of how to go from the gels to a restriction maps, these types of HW are extremely difficult if you don't have a really good handle on what you are doing. $\endgroup$
    – AMR
    Commented Oct 9, 2015 at 20:48
  • $\begingroup$ @AliceD I remember getting this pretty easily in school and then I would have a line of my classmates asking me to explain how to solve these. These are not trivial problems and they are not something that you can look up, and unless can work it out, it is very easy not to even know where to begin. I would cut the OP some slack. This is probably one of the harder Molecular Bio questions that can be asked at the undergraduate level. $\endgroup$
    – AMR
    Commented Oct 9, 2015 at 20:51
  • $\begingroup$ @AMR, after edits this question is great. No worries. I retracted my vote and +1'd $\endgroup$
    – AliceD
    Commented Oct 9, 2015 at 22:03

2 Answers 2


How you would determine the plasmid map from just the restriction digest data

The thought process behind doing this is a bit involved.

  1. From the question you are told that the Gene is 8kb and does not contain an EcoRI site inside of the gene
    • I say not inside because the ends of the gene were made with EcoRI cuts, but digesting in EcoRI will only produce the 8kb fragment of the gene
  2. The vector is 6kb and you know that they have EcoRI sites at either end, but you are not told what other cut sites there are.
  3. When you digest the plasmid in BamHI you get 3 fragments, meaning that your plasmid has 3 BamHI sites somewhere in it.
  4. You know that your total plasmid size is 14kb (easiest way to know is just look and see that they tell you that there is an 8kb gene and a 6kb vector). Looking at the Double Restriction Digest, you can see that the five fragments add up to 14kb, which implies that there are only one of each fragment in the plasmid, and they are produced by 5 separate cuts.
    • This is an important piece of information because this tells you that you do not have 2 fragments of the same size making up different parts of the plasmid.
    • Remember that when you do a restriction digest on a circular piece of DNA, the number of fragment you get equals the number of cuts (restriction sites) in the plasmid. So because there were 5 fragments, you have 5 separate restriction sites
    • From the Double Digest data, you also know that the smallest fragment of DNA that you have is 1kb and the largest is 4kb.
  5. Since the 4kb site is common to both the BamHI only and the the Double Digest, as since you do not have a 0.5kb or 1.5kb fragment (the remainders if the Double Digest 4kb fragment had been produced from the 4.5kb or 5.5kb fragments in the BamHI digest) in the Double Digest, you know that the 4kb fragment is the same in both the BamHI digest and the Double Digest, and is produced by 2 BamHI cuts.
    • You can now work on roughly placing where the 4kb fragment should be in the map.
    • If you try to place that 4kb fragment in the vector, then you would either generate a 2kb fragment, or something less than 2kb and a very small fragment.
    • Since you do not have a 2kb fragment or fragments smaller than 1kb, you know that it cannot be in the vector region. Also in order to get a 2kb fragment from the 6kb vector the 4kb fragment would have needed to be cut once with EcoRI and from the description, above, we worked out that the 4kb fragment is produced by two BamHI cuts
    • It could have been that the vector was 1kb + 4kb + 1kb, however from point 4 you determined that you only had one of each fragment, so you can't have two 1kb fragments.
  6. All of that tells you that it has to be in the 8kb gene portion, though you need some more information to determine exactly where.
    • Remember that you do know that the 4kb fragment was produced by 2 BamHI cuts, which means that it cannot be at the very end of the gene region, as that would require it to be cut once with EcoRI
    • This also tells you that you have at least two BamHI cut sights in the gene.
  7. What I would do here is draw two circles, the first having an 8kb arc and a 6kb arc, then draw second circle and make three cuts in it.
    • From 6 we worked out that the 4kb piece will overlap somewhere in the 8kb area, that it cannot be at an end and that there need to be 2 BamHI restriction sites within the gene.
    • You also have to consider the 4.5kb and 5.5kb parts from the BamHI digest. Since 4kb plus either of those two are greater than 8kb, then you know that the cut that separates 4.5kb from 5.5kb will be somewhere in the 6kb vector. This is how you know that the third BamHI site is in the vector.
  8. Now that you have determine that the 4kb fragment is in the gene you can start to work out the other pieces.
  9. You had 2 fragments from the BamHI digest that were 5.5kb and 4.5kb and in the double digest you have fragments that are 1.0kb, 2.5kb, 3.0kb, and 3.5kb.
    • Since you have already determine that that 4kb piece is the same for both digests, then at this point you can tell yourself that the 1.0kb and 3.5kb fragments made up the 4.5kb fragments and the 2.5kb and 3.0kb fragments made up the 5.5kb fragment.
  10. Since you know that there has to be two EcoRI cuts that make the 6kb fragment, you are going to need two fragments that add up to 6kb.
    • The only way to get this with the four fragments that you have from the double digest is with the 2.5kb piece and the 3.5 kb piece.
    • So you have now determined the exact position of one of the BamHI cut sites. It is in the Vector, 2.5kb from one of the EcoRI sites and 3.5kb from the other EcoRI sites.
  11. Now that you determined where one of the sites is, you know that the 2.5kb fragment was paired with the 3.0kb fragment, so this tells you that there is a BamHI site 3kb away from the EcoRI site in the Gene.
  12. From there you can connect the 4.0kb piece to the end of the 3.0kb piece because you know that the 3.5kb piece has to be attached to the 1kb piece and cannot be attached to the 3kb piece.
  13. All that is left is to close the circle. You know that you have 1.0kb more of the gene that needs to be place and this will connect the 7.0kb that you determine of the gene in step 11 to the 3.5kb fragment in the vector. You have that 1.0kb fragment, and that is the one.

So that is how you determine how to construct the plasmid map from the restriction digests.

What the distances between the cut sites tell you

The cut sites are determined by the sequence of the DNA that the Restriction Enzyme recognizes.

  • EcoRI recognizes GAATTC
  • BamHI recognizes GGATCC

You will notice that these are palindromic sequences, so 5'-GAATTC-3' is 3'-CTTAAG-5' on the complimentary strand.

All that the distances are telling you is that in the areas in between the cut sites, the sequences GAATTC and GGATCC do not appear. If they did, then the Restriction Endonuclease would recognize that position and cut the DNA, and you would have an additional cut site at that location.

Assuming that you were given the plasmid map, this should be the answer.

The answer should be C. The 1kb and 3kb fragments are the DNA fragments from the gene that are generated from the EcoRI / BamHI digest. So one end of the fragment was produced with EcoRI and the other end of the fragment was produced with BamHI

The 4kb fragment is the result of BamHI restriction alone. So both ends of the fragment were produced by BamHI so it cannot be the answer to a fragment produced by EcoRI / BamHI restriction.

If they are saying that they also want answer A as well, then their question is very poorly worded, in my opinion.


The short answer:

After the Eco digest, you have two pieces, (which means a single cut) one 6 kb which is all vector, one 8 kb, which is all gene. Then you digest those with another enzyme, but the pieces that came from the vector should add up to 6, and the pieces that came from the gene should add up to 8. The 2.5 and 3.5 long pieces must be the cut up vector, the rest must be the cut up gene. There is no other way to group pieces of those lengths into 6 kb and 8 kb parent fragments.

To find the order of them all:

Basically, you have the 5 smaller fragments, and you want to put them in the right linear order, like this:

enzyme - DNA of some size - enzyme - DNA of some size - enzyme ...etc.

We know that there is 6 kb Eco RI fragment, and THEN we cut with Bam, so there have to be Bam fragments that add up to 6, and the rest will add up to 8. In this set that HAS to be the 3.5 and 2.5 which are part of the 6kb fragment that you get when cutting with Eco alone. So we know part of the sequence :

Eco - 2.5 kb - Bam - 3.5 kb - Eco - ??? Bam ??? Bam ??? Eco

Since there are only two, order doesn't matter.

Now, if the 4.0 kb fragment was adjacent to the 3.5 kb fragment, the result of cutting that with Bam alone would be 7 kb. There is no fragment that size, so that's not right. Same with the 3.0 kb fragment. So, it must be the 1.0 kb fragment, because that would yield the 4.5 kb fragment that we do see.

Eco - 2.5 kb - Bam - 3.5 - Eco - 1.0 - Bam ??? Bam ??? Eco

Again, if the 4.0 were adjacent to the 2.5 kb fragment, the Bam alone digest would have a 6.5 kb fragment, which it doesn't. So the 4.0 fragment must be in the middle, with the 3.0 fragment adjacent to the 2.5 fragment, which yields the 5.5 kb digestion Bam alone digestion product that we observe.

Eco - 2.5 kb - Bam - 3.5 kb - Eco - 1.0 kb - Bam - 4.0 kb - Bam - 3.0 kb - Eco

  • 1
    $\begingroup$ You cannot assume the 2.5 and 3.5kb fragments make up the 6kb fragment right away. You have to get to it by a process of elimination that rules out the 4kb and 2 X 1kb fragments from being a possibility. $\endgroup$
    – AMR
    Commented Oct 9, 2015 at 21:01
  • $\begingroup$ We know the whole vector is 14 kb. We know the 5 fragments visible add up to 14. Therefore, there is only one fragment of each size. $\endgroup$
    – swbarnes2
    Commented Oct 12, 2015 at 17:06
  • $\begingroup$ Your answer makes the point without explaining how you get there. You can't make the assumption without ruling out the 4kb and 1kb two times first. You did that as a matter of course, because you are familiar with how to do it. Someone who doesn't understand needs that explained to them. You also don't mention that there are 5 fragments that add up to 14kb in the initial part of your answer and you can't assume that the OP will have picked up on that or understand the significance. $\endgroup$
    – AMR
    Commented Oct 12, 2015 at 18:47
  • $\begingroup$ The other answer already existed when I wrote mine, and it lays out why each band must represent only a single fragment of that size. $\endgroup$
    – swbarnes2
    Commented Oct 12, 2015 at 22:04

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