Double digests
Unfortunately, we cannot tell from the gel of Example A (above) how the fragments or cut sites are related to each other. If we simultaneously digest the plasmid with enzyme X and enzyme Y, there are two possibilities: Y could cut within the small fragment produced by enzyme X (Fig. 3, left), or Y could cut within the large fragment produced by enzyme X (Fig. 3, right).
Fig. 3 Two possible outcomes of a double digest of enzymes X and Y
These two possibilities would produce different band patterns on gels (Fig. 4). The left and right gels correspond to the left and right possible outcomes in Fig. 3, respectively.
Fig. 4 Band patterns for double digests illustrated in Fig. 3. Bands present in both double and single digests are colored red.
On these gels, you should observe two important principles:
1. When the band representing a fragment produced by a single digest of a certain restriction enzyme (e.g., X) is also present when a second restriction (e.g., Y) is added in a double digest, then that second enzyme does not cut within the fragment. In the scenario on the left, the 700 bp fragment from the X digest is not cut by enzyme Y, so its 700 bp band is present on the gel in both the X and X+Y lanes. In the scenario on the right, Y does not cut the small (300 bp) X digest fragment, so it is present in both the X and X+Y lanes.
2. When a fragment produced by a single digest of a certain restriction enzyme (e.g., X) is cut upon the addition of a second enzyme (e.g., Y), the band which represented it in the single digest lane will not be present in the double digest lane. Rather, two or more smaller bands, having masses that add up to the mass of the larger single digest band, will appear in the double digest lane and will not have counterparts in the single digest lane. For example, in the left scenario, the 300 bp X fragment, whose band is visible in the X single digest lane, was cut by enzyme Y into smaller 200 and 100 bp fragments that appear in the double digest lane. The same phenomenon can be observed with the 700 bp fragment in the right scenario. Thus far, we have only considered X digest fragments being acted upon by Y. However, the same logic applies to the Y fragment being acted upon by X. The single 1000 bp Y digest fragment is cut into three smaller pieces by X. These pieces appear in the double digest lane and their masses sum up to 1000 bp in both scenarios.
Working backwards from a gel to a map
In the example shown here, we have the benefit of knowledge of the position of the restriction sites on the plasmid and can use that information to predict the bands that would be seen on a gel. However, in real life we often have the reverse situation. We have only the pattern of bands on a gel. We must use that pattern to determine the size of DNA fragments that were formed from cutting the plasmid and to infer the relative positions of those fragments on the plasmid before it was cut.
To draw these inferences, we use the converse of Principle 2 discussed above. If there are only two bands which are present in a double digest lane that do not have corresponding bands in a single digest lane, then the fragments represented by those bands must be common fragments of a larger band in the single digest lane and the mass of that larger fragment must be the sum of the masses of the smaller bands. In addition, we can infer that if those two double digest fragments are common fragments of a larger fragment, then they must lie adjacent to each other on the map. This seems trivial in the current example where there are only three fragments, because in that case every fragment is adjacent to every other fragment. However, this is very useful information in a more complicated situation where there are more than three fragments in a double digest, particularly if one of the two fragments has already been positioned on the map through previous inferences.
If we had not had previous knowledge of the plasmid map and only had access to the banding patterns on the gels, we could use the following process to deduce the map. In the gel on the left, we note that the 100 and 200 bp X+Y fragments do not have corresponding bands in the X single-digest lane. So they must have formed by fragmentation of a larger band which is about 300 bp. That band must be the lower band in the X single digest lane, because we can already account for the upper band (it was the 700 bp fragment which was not cut by enzyme Y). We can diagram the 300 bp fragment like this:
|----------------------------|----------------|
X 200bp Y 100bp X
The 700 bp fragment looks like this:
|--------------------------------------------------------------------------------------------------------------------------|
X 700bp X
Both fragments have X "sticky ends" and if the ends of the 700 bp fragment were ligated to the ends of the 300 bp fragment, the map shown at the top left of the previous page would result. You can repeat this exercise using the gel on the right to infer that the 200 and 500 bp fragments are pieces of a larger 700 bp fragment that has an X sticky end on both ends.
Practice
For practice, try to draw the restriction map for this gel:
Fig. 5. Practice gel
