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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?


The contributor suggests A is the correct option
Option Alternative First


38 (54.29%)




24 (34.29%)




3 (4.29%)




4 (5.71%)




1 (1.43%)



The following explanation has been provided relating to this question:

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:


HindIII Fragment Map


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.


The following topics have been indicated as being relevant to this question:

Plasmids, practical 5, Gene mapping, Mapping, Maps


There are 8 comments for this question (8 top-level comments and 0 replies) 
Written: 4:56pm, 24 MarAuthor has: 2762 points

Really good explanation, well done.

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Written: 1:23pm, 30 AprAuthor has: 4252 points

FANTASTIC! really learnt a lot here. WELL DONE!

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Written: 7:59pm, 24 MarAuthor has: 1823 points

Thank you very much for your detailed explanation. Good job!!

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Written: 8:08am, 08 AprAuthor has: 1070 points

really good question structure and detail explaination

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Written: 5:35pm, 24 MarAuthor has: 3871 points

Good question, and great effort with the explanation, but there is a much simpler one.  consider the final row in the table - where all restriction enzymes are used.  The cumulative length of all detected fragments adds up to 12.5kb.  You mentioned that the original fragment was 13kb in length.  A quick calculation tells you that there is an additional fragment of length 0.5kb - the smallest possible answer, so this must be the correct answer.  Very easy to bypass the valuable working.

  Reply written by question author



Yes, I've mentioned that in the explanation and I also said what the glitch with using such a method would be. Finding fragments of the same size (and there are two occurrences of a 2kb fragment there) would force you to either reconsider the entire calculation in your head to determine by numbers whether it would be possible for the two different probes to recognise the same fragment or, much simpler, to draw the map and see for sure whether that can happen.


If I had added, for example, another Probe B recognition site on one of the fragments recognised by Probe A and would have boosted the sum to be over 13kb (which is also perfectly possible), that method wouldn't have been viable any more. I only decided not to do that so that I wouldn't complicate things too much with drawing the map, as I didn't see any use in making a painstakingly difficult map with little relevance to those not going into Genetics.


So, as originally stated in the explanation, yes, you can sometimes bypass drawing the map by doing the calculation, but you don't necessarily get the right answer that way. Even here, there could have been a 2.5kb fragment being lost rather than the 0.5kb one, depending on probe location (which is easiest to determine via a graphical representation).

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Written: 3:50pm, 25 MarAuthor has: 2545 points

I just add them all up to get 12.5, and obviously 0.5 is missing from it...


Good question though. Explanation very thorough!

  Reply written by question author

Again, read the clarifications above to see why, although that method may give you the correct answer, it is not guaranteed to.


Here is an example where the table wouldn't give you the right answer if you summed up the values on the last row, for a 10kb fragment (I just thought it up in two minutes and it was the first length to come to mind):


Enzyme(s) Probe A Probe B Probe C
HindIII 10 10 10
HindIII+EcoRI 1+9 9 9
HindIII+BamHI 4+3.5 4+2.5 4+3.5
HindIII+SauI 6+4 6+4 6+4
HindIII+EcoRI+BamHI+SauI 1+1.5 3+2.5



This one is pretty sloppy and many values are identical all throughout, but the main point is that the last row returns a sum of 12.5kb even though an entire 2kb fragment will be missing from it (between a BamHI recognition point and a SauI recognition point) because no probes happen to be on it. Additionally, also for a 10kb fragment cut the same way:


Enzyme(s) Probe A Probe B Probe C
HindIII 10 10 10
HindIII+EcoRI 1+9 9 9
HindIII+BamHI 4+3.5 4+3.5 3.5
HindIII+SauI 6 4 6
HindIII+EcoRI+BamHI+SauI 1+2 1.5



This table will give you a sum of 9.5kb on the last row, yet the smallest fragment actually cut by all of the restriction enzymes added together would actually be 1kb in length. Again, Probes A and C recognise areas on the same fragment, of length 2.0kb, whereas one 2.5kb fragment is not recognised by any of the probes. In this case, answering that the smallest cut fragment is 0.5kb would actually be wrong, as the smallest one would be the 1kb fragment recognised by Probe A.

Again when one of the values is repeated, simply adding things up does not necessarily provide a true answer unless you analyse the table thoroughly OR build a map. Of course, these two tables are very sloppy and it is quite apparent that some probes recognise similar fragments from the other cells, but I am only providing them as two examples of situations where calculations do NOT provide the correct answer, even when the sum on the final row is less than the total length of the fragment.


I did these examples by first creating the fragment map, then "backtracking" the values as they appear in the table, and I chose all fragments of different sizes for clarity. In reality, there is no guarantee that two identical values indicate the same fragment and not two fragments that happen to have the same size. The table in the question actually demonstrates that: there are TWO occurences of a 2kb value on the last row, but they happen to belong to DIFFERENT pieces of cut fragment, and that is why adding the values up HAPPENS to lead to the correct answer.

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Written: 3:43pm, 25 MarAuthor has: 3871 points

My bad.  Good job.

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Written: 1:28pm, 30 AprAuthor has: 4252 points

FANTASTIC! really learnt a lot here. WELL DONE!

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