TP12_Restriction_mapping2\TP12_Restriction_mapping2
Restriction Mapping
At some point in a cloning project it will be necessary to construct a restriction map of a plasmid.
This will involve manipulating restriction digest fragments into a circular map. There is a systematic approach to constructing a map, which is illustrated in the sample problem below. Once you have mastered the sample problem (plasmid pRIT450), complete at least pRIT451 (Question 1) and your personal question (Question 1).
Working with actual restriction fragments from a gel is a bit messier than these problems because of experimental error in determining fragment sizes. Sometimes the mathematical approach used here needs to be supplemented with some common sense.
The circular plasmid pRIT450 has two restriction sites each for the restriction enzymes PvuII, EcoRV, and BtrI. From the data below, determine the size and restriction map of the resulting plasmid. The fragment sizes are given in kilobasepairs(kb):
PvuII 7.0 4.0
EcoRV 6.0 5.0
BtrI 8.9 2.1
PvuII + EcoRV 4.3 3.3 2.7 0.7
PvuII + BtrI 6.1 2.8 1.2 0.9
EcoRV + BtrI 5.0 2.1 2.1 1.8
The first task is to determine the size of the plasmid. We can observe that all fragment length for each enzyme or enzyme combination is sums to 11 kb in the table above. For example PvuII 7 + 4 =11; EcoRV: 6 + 5 = 11 and BtrI: 8.9 + 2.1 = 11). Since all digestions give the same total sum, we can say with some confidence that the total size is 11 kb and that there are probably no distinct fragments with the same size except for 2.1 fragment indicated in the last row.
It is important to remember that there is not necessarily only one solution, but the suggested solution has to fit the observed fragment sizes.
The first step is to locate two enzymes on the restriction map. It does not matter which ones. We start by locating PvuII and EcoRV together on the map:
PvuII 7.0 4.0
PvuII + EcoRV 4.3 3.3 2.7 0.7
I think that it is easier to work with linear representations of the plasmid. Figure 1 below represent the plasmid more or less to scale with the fragments formed from PvuII and PvuII + EcoRV. I chose to start by locating the largest PvuII fragment (7.0) at 12 o'clock followed by the 4.0 fragment in the first line below.
We can rearrange the fragments once we recognize that the 7.0 PvuII fragment could be made from the 4.3 and 2.7 bands together. I changed the order of the bands accordingly (Figure 2).
We can now see that the order of the fragments in fits the observations so far. We do the same thing over again, but with EcoRV and PvuII:
EcoRV 6.0 5.0
PvuII + EcoRV 4.3 3.3 2.7 0.7
We recognize that the 6.0 EcoRV fragment could be made from the 3.3 and 2.7 bands together. I changed the order of the bands accordingly (Figure 4).
If we compare the order of the bands in Figure 4 with the one in Figure 2, we see that the 0.7 kb band is located between the 4.3 and the 3.3 bands in Figure 2, while in Figure 4 it is located between the 3.3 and the 4.3. We can simply switch the places of the 4.3 and 0.7 bands in Figure 4 to match the order (Figure 5).
The new fragment order (green) should satisfy all the observed fragments. We can superimpose one of the maps on a concatenation of the other to verify that they are in fact equivalent (Figure 6).
We continue by locating PvuII and BtrI together on the map. The size and order of the PvuII fragments are the same as in Figure 1.
PvuII 7.0 4.0
PvuII + BtrI 6.1 2.8 1.2 0.9
We rearrange the fragments once we recognize that the 7.0 PvuII fragment could be made from the 6.1 and 0.9 bands together. I changed the order of the bands accordingly (Figure 8).
We continue with BtrI and PvuII (Figure 9). We can see that the original order fits the digestion without reordering the fragments.
Comparing Figure 8 and Figure 9, we see that the 0.9 kb band is not in the same place. It is between the 6.1 and the 2.8 bands or between the 6.1 and 1.2 bands. We can move the fragment in Figure 8 like so (Figure 10).
We can verify that the two fragment orders are equivalent in the same way as in Figure 6.
The final task is to combine the two restriction maps. It is important to remember that they can be combined in two different directions relative to each other.
In Figure 12a, the final plasmid maps in Figure 2 and Figure 10 were superimposed. In B, the pink map was inverted and 7 kb PvuII fragment was matched on both maps. The overhang was folded on the other side to match the 4 kb PvuII fragment (C). I we look at the (A) map, we can see that it would produce fragments that are not observed in the list of fragments (3.4, 0.5 kb), so this orientation can not be correct. C on the other hand produces all the expected fragments (0.9, 1.8, 4.3, 0.7, 2.1 and 1.2). A circular map (D) was produced reflecting the fragment order in (C).
Question 1: Plasmid pRIT451 was cut with SmaI, BglII, and AvaI. From the data below, determine the map.
SmaI 5.9 4.3
BglII 8.2 2.0
AvaI 5.3 4.9
SmaI + BglII 5.4 2.8 1.5 0.5
SmaI + AvaI 3.3 2.6 2.3 2.0
AvaI + BglII 5.3 2.1 2.0 0.8
Question 2:
This is an individual question for each student. Follow this link that points to a Google Spreadsheet. You should find your name in the leftmost column. In cell F2, there is a link to a google document that has a table of fragment sizes immediately below your name.
Your task is to use your data and solve the restriction map. Formulate you restriction map with enzyme names and fragment sizes as for the first example student "Max Maximus".
It does not matter which enzyme you chose to begin your map.
You can solve one of more of the examples below if you feel that you need more practice.
Plasmid pRIT452 was cut with PstI, HindIII, and EcoRI. From the data below, determine the
map.
PstI 6.8 5.9
HindIII 6.5 6.2
EcoRI 9.2 3.5
PstI + HindIII 4.8 4.2 2.0 1.7
PstI + EcoRI 5.4 3.8 3.0 0.5
EcoRI + HindIII 6.2 3.5 1.8 1.2
Plasmid pRIT453 was cut with SmaI, HindIII, and EcoRI. From the data below, determine the
map.
EcoRI 7.7 1.6
HindIII 7.4 1.9
SmaI 6.6 2.7
EcoRI + HindIII 4.5 1.9 1.6 1.3
EcoRI + SmaI 5.7 2.0 0.9 0.7
SmaI + HindIII 2.7 2.5 2.2 1.9
Plasmid pRIT454 was cut with PstI, HindIII, and EcoRI. From the data below, determine the map.
PstI 6.0 5.3
HindIII 5.8 5.5
EcoRI 6.5 3.0 1.8
PstI + HindIII 4.0 3.8 2.0 1.5
PstI + EcoRI 3.5 3.0 2.5 1.8 0.5
EcoRI + HindIII 5.0 3.0 1.5 1.0 0.8
Plasmid pRIT455 was cut with BtrI, HindIII, and EcoRI. From the data below, determine the
map.
EcoRI 8.5 2.0 0.5
HindIII 5.6 5.4
BamHI 6.5 4.5
EcoRI + BamHI 4.5 2.2 2.0 1.8 0.5
EcoRI + HindIII 4.3 4.2 1.2 0.8 0.5
BamHI + HindIII 3.4 3.1 2.5 2.0
Plasmid pRIT456 was cut with PstI, HindIII, and EcoRI. From the data below, determine the
map.
EcoRI 5.8 3.3 2.9
PstI 7.4 4.6
HindIII 6.8 5.2
PstI + EcoRI 3.3 3.3 2.5 2.1 0.8
PstI + HindIII 6.3 4.1 1.1 0.5
EcoRI + HindIII 3.6 3.3 2.2 1.6 1.3
Plasmid pRIT457 was cut with BamHI, HindIII, and PstI. From the data below, determine the
map.
BamHI 5.1 4.5 3.8
HindIII 8.1 5.3
PstI 13.4
BamHI + HindIII 5.1 3.3 2.5 2.0 0.5
BamHI + PstI 5.1 4.5 2.5 1.3
HindIII + PstI 8.1 3.3 2.0