Genetics
Question
This question has been answered by 70 people and has an average rating of 4.18 (based on 39 ratings)
A 13kb HindIII fragment is cut with EcoRI, BamHI and SalI in the combinations given below and Southern blotting is then used with three separate probes, Probe A, Probe B and Probe C, to determine the sizes of the obtained pieces. Following is the table of values obtained (sizes are in kb; a+b means that two fragments, of size a and b respectively, have been detected using the respective probe):
| Enzyme(s) | Probe A | Probe B | Probe C |
| HindIII | 13 | 13 | 13 |
| HindIII+EcoRI | 2+9 | 2 | 9 |
| HindIII+BamHI | 6+4 | 3 | 6+3 |
| HindIII+SauI | 9.5 | 3.5 | 9.5+3.5 |
| HindIII+EcoRI+BamHI+SauI | 2+3.5 | 2 | 1+4 |
To your best knowledge, what is the approximate size of the smallest fragment obtained when all restriction enzymes are placed together?
Alternatives
| Option | Alternative | First answers  |
Confirmed answers  |
|---|---|---|---|
| A |
0.5kb |
38 (54.29%) |
0 |
| B |
1.0kb |
24 (34.29%) |
0 |
| C |
1.5kb |
3 (4.29%) |
0 |
| D |
2.0kb |
4 (5.71%) |
0 |
| E |
2.5kb |
1 (1.43%) |
0 |
Explanation
The following explanation has been provided relating to this question:
Topics
The following topics have been indicated as being relevant to this question:
Comments
What you need to do here is build a map of the HindIII fragment and figure out where everything goes. The map you obtain should look something like this:
Note that everything is at scale.
If you're confused about building a fragment map, keep in mind the following:
1. Fragment maps are linear (unlike plasmid maps) and have a recognition site for the enzyme used to cut them from the plasmid of origin at each end, but never within the fragment (otherwise, you'd just get a smaller fragment).
2. The number of recognition sites for each probe depends on the total number of fragments it recognizes when you add all of the enzymes together and NOT on their sizes. If, for example, probe A recognizes two pieces when using all of the enzymes to cleave the original fragment, then probe A has AT LEAST two recognition sites on the respective fragment. You can use this approximation effectively when constructing a map based on the data you're given, but it might be useful to note that it may actually have even more recognition sites, but some of them are quite close together and find themselves on the same piece. To figure that out, you would just need to use as many restriction enzymes as possible, but that is beyond the scope of the question.
3. The fragment sizes recognised with each probe will give you the distances (in kb, in this case, or any other measurement unit you're using) between recognition sites of the enzymes used. However, they will not tell you which sites the fragment lies between. For example, when adding HindIII and EcoRI and using probe A, the 2kb fragment can have either one EcoRI and one HindIII, or two EcoRI sites bordering it (it cannot have two HindIII sites because it is clear that those are 13kb apart because of the fragment size obtained when using them on the plasmid in the first place).
4. Not every fragment given in the table on a row (that is, when using one individual enzyme in addition to the main one, here HindIII) will necessarily be a distinct fragment on the map: it may happen that some of the pieces have recognition sites for two probes rather than one (for example, the 9kb fragment obtained with both Probe A and Probe C is, actually, the same fragment; however, the two 2kb fragments obtained with Probe A and Probe B are distinct, and you can determine that by doing some simple maths). The best way to get around this when building your map is to first sum up all distinct values on that particular row (in the case of "HindIII+EcoRI", 9+2=11), then see which combinations of repeated values will help you reach the length of the initial fragment when added to that number: each repeated value doing that will mean there is one extra fragment of that size for every iteration (in this case, only the 2kb fragment does this, indicating there are two distinct 2kb fragments recognised by the two probes, because I didn't want to complicate stuff too much, but you may sometimes have to work with harder stuff). If no such combination is possible, it means that you are losing some fragments that don't have a recognition site for any of your probes at all and you need to investigate further to determine what their exact sizes are and how many you're losing.
5. Only building the map based on distances between restriction enzyme recognition sites will give you several possible maps. That shouldn't worry you, but you should know that only one is correct. To determine which one that is, you only need to start adding the probes on the map and see which stop making sense. You can eliminate any map that doesn't make sense immediately, you don't have to look at the remaining probes to see if they fit.
Also, this was a bit of a trick question. You may have gotten it correctly just by summing up the values on the last row and noticing they add up to 12.5 rather than 13 without building the map, but did you look at the repeating 2kb fragment obtained with both Probe A and Probe B? That might have been either a different fragment or the same, although, in this case, it was different. The same fragment would have left you with 2.5kb unaccounted for and a smallest fragment size of 1kb (recognised by Probe C), as far as your knowledge went.
This question might have been very easy, but I was aiming more at explaining how plasmid/plasmid fragment maps worked, as a few people I know seemed to have problems understanding them, so feedback on the clarity of the explanation would be much appreciated.