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Leena is a molecular biology student. She purifies two fragments of DNA, 800 and 300 base pairs long. These were obtained from a plasmid after digesting it with HindIII. Each of these fragments has a single EcoRI recognition site. Leena wants to join these two fragments to get a 1.1kb gene as shown in Figure 7.1. She suspects that this gene has a unique protein-coding sequence.

She, therefore, mixes the two fragments in the presence of excess DNA ligase in an appropriate buffer and incubates the mixture. She removes an aliquot (a small part of the reaction mixture) after 30 minutes and loads it on an agarose gel to check the results. She is surprised to find many bands along with the expected 1.1kb band in the gel.

fig 7.1

FIG 7.1

enter image description here

Leena is interested in the 1.1kb fragment shown in Figure 7.1. Hence, she elutes the 1.1kb fragment from the gel shown in Figure 7.2 and subjects part of this sample to HindIII digestion. She obtains the expected pattern with two bands, 800 and 300 base pairs long. To confirm its restriction map, she subjects the remaining sample to complete EcoRI digestion. Which pattern of bands would she obtain? enter image description here

The answer given is this and I understand why this is the answer and other options given in the question(not posted here) are definitely not the answer but I do not understand why there are no 0.4kb fragments, 1.1kb fragments here.

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@A. Kennard beat me to it. Here is my answer, assuming complete digestion. The two HindIII fragments can ligate to give four different products as shown below. I have marked one end of each HindIII fragment to help in seeing polarity, and I have left the ligated junction as a gap to help see what is going on. I don't understand the point about the 1 kb band.

enter image description here

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  • $\begingroup$ Nice graphics! I had a typo in my response, I meant to say that if you assume the digestion is incomplete, in addition to 0.4 kb and 1.1 kb bands, it should be possible by not cutting at one cut site to get an extra 1kb band not present in the complete digestion case. $\endgroup$
    – A. Kennard
    Nov 15, 2013 at 19:24
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You are being too smart for this question ;) If you assume that the enzyme is perfect, you've optimized the amount of template, and you leave the reaction for a really long time then there will be complete digestion. If you assume complete digestion, then there won't be any 1.1kb fragments. Furthermore, if you work out all the possible orientations of the 800bp and 300bp that were ligated together (each piece can be in the orientation shown and the opposite orientation) and you force an EcoRI cut at every available EcoRI site, then you won't have any 0.4 kb fragments from imcomplete digestion, either.

In reality, I've never been able to optimize my reaction conditions to get complete digestion of my template, and I always have some residual template around, so it's good your thinking that way for "real-world" preparation.

If you do not assume incomplete digestion, in addition to the 1.1kb and 0.4kb bands missing from this gel, I calculate that there should be a 1kb band as well.


Edit: I goofed up the last paragraph. It should read: "If you do not assume complete digestion, in addition to the 1.1kb and 0.4kb bands missing from the complete digestion gel, I calculate that there should be a 1kb band as well.

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  • $\begingroup$ I agree with @Alan Boyd. Even I do not understand why 1kb should be there ? $\endgroup$
    – biogirl
    Nov 15, 2013 at 17:45
  • $\begingroup$ I had a typo. I meant to say "if you do not assume complete digestion." Look at any of the possible restriction products that @Alan Boyd has displayed. If you get 800bp and 200bp products (or 700 and 300, or 900 and 100) with compete digestion, you would get a 1kb product with incomplete digestion. Sorry for the confusion! $\endgroup$
    – A. Kennard
    Nov 15, 2013 at 19:22

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