Making use of information from a vector map
If the DNA being mapped is a recombinant plasmid, additional information may be derived from a map of the vector used to clone the DNA in addition to information that might be derived from a gel. For example, assume that cloning was done with the 500 bp vector mapped below.
Fig. 6. Cloning vector cut in its MCS
Restriction sites Y, M1, and M2 are sites in the MCS of the vector (Fig. 6). We know that sites in an MCS are always unique on the vector, otherwise the vector would be cut in unintended places when the MCS sites are used in cloning. So we know that there are no other Y sites on the vector. Information provided with the vector map also indicates that, while not present in the MCS, the X site is also unique on the vector. The information provided with the vector also states that there are no Z sites on the vector. During the creation of the recombinant plasmid, the vector was cut at the M1 and M2 sites on the MCS, removing the small 5 bp fragment between them (Fig. 6, right). The insert was then ligated to the M1 and M2 sticky ends to form a recombinant plasmid larger than 500 bp.
This recombinant plasmid was subjected to single digests of restriction enzymes X, Y, and Z. It was also digested with double digests of X+Y and Y+Z. The digests were electrophoresed, resulting in the gel of Fig. 7 (band positions are not to scale). Note that the absolute distances traveled by the bands are not necessarily accurate. The fragment sizes (in bp) are given next to the bands - in a real gel they would have to be calculated from a standard curve by comparison with molecular mass standards that are not shown on the gel.
Fig. 7 Hypothetical gel of single and double digests of a plasmid
At first, this gel looks hopelessly complex. However, if the problem is attacked systematically, it is not too difficult. First, it can be seen immediately that the plasmid is cut two times by enzyme X, two times by enzyme Y, and once by enzyme Z. Because both X and Y cut twice, their relationship is likely to be more complex than the relationship between Y and Z. (Because no X+Z digest was done, it would not be easy to determine the relationship between X and Z restriction sites directly.) Let's start by mapping the X+Y pieces.
Neither the 860 bp nor the 140 bp Y digest fragments are present in the X+Y digest. That means that both of these Y fragments must have been cut by X into smaller pieces. The 800 bp X+Y fragment can only have come from the 860 bp Y fragment - it is too large to be a piece of the 140 bp Y fragment. The other three X+Y fragments could possibly be fragments of the 140 bp Y fragment. However, because we know that the 860 bp Y fragment was cut to produce the 800 bp fragment, we know that the remainder of the 860 bp Y fragment must be 860-800=60 bp long. Thus, of the three small X+Y fragments, we know that the 60 bp fragment cannot have been formed from the 140 bp X+Y fragment because we have already accounted for the 60 bp fragment as the partner of the 800 bp X+Y fragment. So the remaining two small X+Y fragments (30 bp and 110 bp) must be the fragments of the 140 bp fragment. Comfortingly, their masses add up to 140 bp, so our logic is supported.
It is very important to note the logic that was used in the previous paragraph. It is tempting to use trial and error to simply guess at pairs of double-digest fragments that might possibly add up to larger fragments. But because we knew that the 800 bp X+Y fragment could not logically have come from the 140 bp Y fragment, we know that it and its partner via subtraction, the 60 bp X+Y fragment, must be fragments of the 860 bp Y fragment. By elimination, the 30 bp and 110 bp fragments must be fragments of the 140 bp Y fragment. There is no uncertainty about the pairing of X+Y fragments. From these pairings, we can infer that the Y fragments looked like this (not to scale) before they were cut:
|--------------------------|------------| |----------------------------------------------------------|----------------|
Y 110bp X 30bp Y Y 800bp X 60bp Y
Making use of information from the vector map
We are now in a position to see how these two fragments are related to the vector map that was introduced at the beginning of this section. From that map, we know that there should be a 60 bp fragment that would form in an X+Y digest. From the gel inferences that we made above, we can see that there is a 60 bp fragment that is a part of the 860 bp Y digest fragment. That 60 bp fragment must be the same one that we know about from the vector map. So we are now in a position to overlay what we have learned from the gel onto what we know from the vector map.
First we know that the total size of the recombinant plasmid must be 1000 bp (the sum of all of the fragments in any lane in the gel). We know from the vector map that the vector portion of the recombinant plasmid is 495 bp, so the insert must be about 1000-495=505 bp. We can summarize this information in the map of Fig. 8.
Fig. 8. Map using information from the vector
Because we know from the gel that the 800 bp X+Y fragment is adjacent to the 60 bp X+Y fragment, we can infer that the Y restriction site is 375 bp into the insert from the M2 vector/insert boundary (from the vector map we know that M2 is 425 bp from the X restriction site, therefore the 800 bp fragment extends 800-425=375 bp into the insert).
At this point we do not have sufficient information to know how the 140 bp Y fragment is oriented on the map. Either the 30 or the 110 bp fragment could be adjacent to the 60 bp X+Y fragment. In order to differentiate between these two possibilities we need to utilize information from the comparing the X+Y fragments to the fragments of the X single digest using an approach similar approach to the one we used when we compared the X+Y fragments to the Y single digest fragments. In this case it is clear. Neither the 800 nor 110 bp X+Y fragments could have resulted from cutting the 90 bp X fragment - they are both too large. Therefore, they must have been fragments of the 910 bp X fragment. By elimination the 30 and 60 bp X+Y fragments must be fragments of the 90 bp X fragment. That logic indicates that the X fragments must have been oriented like this before they were cut:
|------------|--------------------| |----------------------------------------------------------|----------------------|
X 30bp Y 60bp X X 800bp Y 110bp X
Comparison of these fragments to the developing map tells us that 30 bp X+Y fragment must lie next to the 60 bp fragment that we know about from the vector map. Since the M1 vector/insert boundary is 10 bp from the Y restriction site, this 30 bp fragment must extend 20 bp into the vector. This new information allows us to add the remaining X restriction site to our map (Fig. 9):
Fig. 9. Map with inferred second X position
The remaining task is to place the Z restriction site. Because the140 bp band in the Y+Z digest is the same size as the 140 bp band in the Y digest, we can assume that those fragments are actually one and the same. That means that the 860 bp Y fragment was cut by Z into the 600 and 260 bp pieces. There are two possible orientations of this cut (Fig. 10):
Fig. 10. Possible positions of the Z restriction site.
If the gel were the only source of information, it would not be possible to decide which of these two orientations is the correct one. However, because we have additional information from the vector map, we can falsify one of the two possibilities. We know from the vector map that there are no Z restriction sites on the vector. Thus any possibility which suggests that the Z restriction site must be located in the vector must be incorrect. The alternative on the right above suggests that the Z restriction site is 200 bp from the X restriction site that forms one side of the 60 bp fragment. However, we know that the M2 vector/insert boundary is 425 bp from that X site. So the alternative on the right must be wrong. The orientation on the left must be correct. With this information, we now have the complete map:
Fig. 11. Final map.
Lessons
There are several important points that you should have learned from this example:
1. Basic information about the total size of the recombinant plasmid and the number of times that each restriction enzyme cuts can be determined directly from the number and size of fragments present on the gel.
2. Inferring which double digest fragments must have come from single digest fragments is the most powerful way to determine which double digest fragments are adjacent to each other.
3. It is best to overlay information onto a map that contains information from the vector because information about the vector can eliminate some alternatives.
Many students attempt to solve this kind of problem by trial and error. Although it is possible to determine a solution by that method, it takes much longer and in the end there is a high degree of uncertainty about whether the map is correct and whether there might be other possible correct solutions. If instead the map is built based entirely on careful logical deductions, one can be confident that the final map must be correct.
You should also be aware that this example is somewhat unrealistic because the sum of small fragment sizes always added up exactly to the masses of larger fragments. In a problem where the fragment sizes are determined by measuring distances traveled by bands and application of those distances to a formula derived from a standard curve, the sums of smaller fragments often only approximately match the masses of larger fragments. The error in bp is larger for large fragments because a small error in measuring the distance traveled by the band corresponds to a larger error than a similar error in measuring the distance travelled by smaller bands.
Practice
For practice: Draw what the final complex map in the example above would have looked like if the cloning vector had the map shown below instead of the vector map given in the example.
