How do you determine the DNA sequence by looking at the picture of gel?

Question

And why are there more fragments in the "small" bands of the gel then there are in the "long" ones.
Why can this technique of DNA sequencing only be used for small fragments of DNA?
Thank you VERY MUCH!!!

Answer 1

There are different methods for sequencing.
One is based on doing a PCR using a mix of the four dNTPs+ an analogue of one of them that terminates the reaction. E.g. you use a primer that will anneal upstream of the sequence that you want to sequence and let the polymerase elongate using dATP, dTTP, dCTP, dGTP and analogue of dATP.
In this case, when the polymerase has to add an A it will either use normal dATP or the analogue. If it uses normal dATP then the reaction will continue and the produced fragment wil grow in size. If the analogue is incorporated then the elongation of that molecule stops at that point.

Depending on the ratio of normal dATP and its analogue you have a certain probability that the analogue is incorporated.
Let's assume that the polymerase would be building the sequence
TTAGGACCAT
Then it would produce TTA and stop if it incorporates the analogue at position 3. Thus the probability that this product forms is equal to the probability that the analogue is incorporated(let's call that pAana).
If it incorporates the normal dATP it continues producing
TTAGGA and stops if it incorporates the analogue at position 6.
Thus, in order to produce TTAGGA you need the combined probability (p) of incorporating normal A (let's call that pAnorm) at 3 position and an analogue at position 6. So the probability to have this product is p=pAnorm*pAana which is smaller than pAana since 0
In order to get the TTAGGACCA fragment you would need to have the combined probability of having normal A at positions 3 and 6 and anologue at 9 so it is p= pAnorm *pAnorm *pAana, and so on

As you realize the probability that a certain product forms reduces as its desired size increases. Thus the frequency of their formation drops and you end up with fewer molecules of large size compared to that of small size. That's why you can't use it for big DNA fragments: the probability that the analogue is incorporated only very near the end of the sequence becomes extremely small.

In order to find the sequence you need to repeat the experiment three more times, each time using one base analogue.
Then you run the gel which has 4 lanes (one from the experiment with the A analogue, one with the T analogue, one with the C and one with the G analogue).

In the gel the smaller fragments (which correspond to the begining ot the sequence) run faster, thus they will be at the bottom of the gel. Thus you start reading from bottom to top, based on which lane you see a band. E.g. for the sequence
I gave you (TTAGGACCAT), you have 10 bases in total so the gel would be

size of fragm .. .A .. T .. C .. G
10 .. .. .. .. .. .. .. .. .. .. *
9 .. .. .. .. .. .. .. .. ..*
8 .. .. .. .. .. .. .. .. .. .. .. .. .. *
7 .. .. .. .. .. .. .. .. .. .. .. .. .. *
6 .. .. .. .. .. .. .. .. ..*
5 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. *
4 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. *
3 .. .. .. .. .. .. .. .. ..*
2 .. .. .. .. .. .. .. .. .. .. . *
1 .. .. .. .. .. .. .. .. .. .. . *


where * corresponds to the band on the gel

Answer 2

you must review Fred.Sanger's work on wikipedia! (' http://en.wikipedia.org/wiki/Dideoxy_termination ')
and maxam & gilbert Method for DNA Sequencing...!