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    http://i1174.photobucket.com/albums/...658FC0BA30.jpg

    Picture is in the link, I'm at a complete loss at what to do!

    EDIT: if anyone can read it, this one as well... sorry, I'm hopeless at these restriction enzyme questions :/

    http://i1174.photobucket.com/albums/...66A45341E5.jpg
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    This what I think..

    Note, the question says the piece of DNA being cut is 20 kilobase pairs.

    Also, if you look at the values given for each restriction enzyme ( IE PstI 15 and 5). They all add up to 20.

    So we know that the answer has to equal 20 when you add up all the numbers together. That narrows it B and D.

    According to the values, if you use BamHI and PstI together then you automatically get 12,5 and 3.

    So we know after 2 enzymes are used that we are going to be left with 12,5 and 3.

    SO.....remember each restriction enzyme can only cut once. That means the answer has to contain two of 12,5 and 3.

    Answer D contains 5 and 3.

    So my answer is D!

    I hope that makes sense. Any line that doesn't. Please highlight!
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    A is the answer to the second question.

    Look at the requirements for the enzyme.

    It can only cut when

    A) Between CC
    B) Between CCGG

    That means for Allele 1. There is CCGG at the two ends so it can cut in 2 places. That gives 3 fragments W,Y and X. Now, Y and W are the two small fragments and X is the big fragment. The Y and W are created because the enzyme cut at both ends giving two small fragments.

    For Allele 2, there is no CCGG at one end. So it can only cut in 1 place at one end. That gives 2 fragments. One big and one small.

    Now, the Big one has to be bigger than X because it contains the end of the allele. Therefore, it will be further to the left on the graph. The remaining small one is the same size as Y because it is cut in the same place as before at CCGG.
 
 
 
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