Team:NUDT CHINA/Notebook

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In this way, we can cut the RBS and target gene out which means this route does not have line L (we link the 1, 2 (or 1, 3) to keep the circle structure of plasmid). Then we can test this route by the time of arriving at the ending node which reported by the report gene (we used the GFP).
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<p><h5>Second stage:</h5></p>
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<p>But we found that it was not easy get so many different good restrict enzyme. So we change the design slightly. (Fig. 3) </p>
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<p><center><img src="https://static.igem.org/mediawiki/2014/b/b7/NUDT_CHINA_Notes_idea_3part_fig3.png" width="302" height="113" /><br>Fig. 3 the position of two same restrict enzyme sites</center></p>
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<p>In this way, we can still finish the function of first stage but save my restrict enzymes.</p>
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<p><h5>Third stage:</h5></p>
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<p>The last problem is how to compare the length of different pathway. Firstly, we decide to build all the possible route and test them. That is to say, we need to cut out 1, 2, 3,…,device(s) so that we can test the entire route, which is named exhaustion, quite time-consuming. Later, we found a better way where we can only cut one device which means we only need to build <i>n</i> graph if the graph has <i>n</i> lines.</p>
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<p>Provided that we have 5 lines in a graph like Fig. 4.</p>
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<p><center><img src="https://static.igem.org/mediawiki/2014/2/2e/NUDT_CHINA_Notes_idea_3part_fig4.png" width="302" height="219" /><br>Fig. 4 a graph of 5 lines, where node 1 is the beginning node 4 is the ending</center></p>
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<p>According our method to find the shortest path, the first step is to build 6 different systems. Then we tested the “finished time” of each systems. Here is the systems and their “finish time”.</p>
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<p><table width="100%" boder=1 cellpadding=0 cellspacing=0>
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<tr><td width="20%"><strong>System</strong></td><td width="40%"><strong>Array</strong></td><td width="40%">Time (Theoretically)</td></tr><tr>
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<td>11111</td><td rowspan=6><img src="https://static.igem.org/mediawiki/2014/7/72/NUDT_CHINA_Notes_idea_3part_array.png" width="189" height="235" /></td><td>2 units</td></tr>
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<tr><td>01111</td><td>2 units</td></tr>
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<tr><td>10111</td><td>2 units</td></tr>
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<tr><td>11011</td><td>2 units</td></tr>
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<tr><td>11101</td><td>3 units</td></tr>
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<tr><td>11110</td><td>unreachable</td></tr>
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</table></p>
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<p>Noticed that without line L<font size="-3">5</font>, the ending is inaccessible which implicates the L5 is indispensable. Though it does not block the accessibility, the absence of L<font size="-3">4</font> slows down the arrival, which implies that L4 is on the shortest path. Thus, from the date of the chart above, we can pick out the shortest path is:</p>
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<p><img src="https://static.igem.org/mediawiki/2014/d/d8/NUDT_CHINA_Notes_idea_3part_lastfigure.png" width="189" height="49" />
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<h1 >WELCOME TO iGEM 2014! </h1>
 
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<p>Your team has been approved and you are ready to start the iGEM season!
 
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<br>On this page you can document your project, introduce your team members, document your progress <br> and share your iGEM experience with the rest of the world! </p>
 
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<p style="color:#E7E7E7"> <a href="https://2014.igem.org/wiki/index.php?title=Team:NUDT_CHINA/Notebook&action=edit"style="color:#FFFFFF"> Click here  to edit this page!</a> </p>
 
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<a href="https://2014.igem.org/wiki/index.php?title=Team:NUDT_CHINA/Notebook&action=edit"> Click here to edit this page!</a>
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<a href="https://2014.igem.org/Special:Upload">Click here to upload! </a>
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Revision as of 16:05, 17 October 2014


The Transformation of Our iGEM Ideas

Originally, we were attracted by the Shortest Path Problem, which aims to find the shortest path in a directed graph that can link the beginning node and ending node. Be inspired, we decided to design a cascade pathway in vivo which can help us read out the answer of Shortest Path Problem directly by the life stage of E. coli.

The main problem in the design are:

  • how to describe a route in E. coli;
  • how to build different routes in E. coli;
  • how to compare the length of different pathways.

To solve the first problem, we built devices that contains promoter, RBS, target gene (CDS) and terminator to simulate the nodes and lines in directed graph: the promoters are nodes and the target fragment are lines. When the promoter is activated, the target gene can coding protein which then activate the next promoter as regulatory element. This regulatory process can describe how a man the walking in a directed graph: he departed from the node P1, went through the line L and finally arrived at the spot P2. In this way, theoretically, we can describe any given directed graph in E. coli. (For further details about the coding of the graph, please refer to the project profile of our wiki) (Fig. 1)


Fig. 1 One promoter and one target gene can simulate one node and one line

After building one complete graph in E. coli, different route should be labelled and tested independently so that we can get the length indirectly by the time of whole travel from beginning to ending. Then we can selected the quickest one as the shortest route. However, we are stuck in the second problem: how to build different route.

First design:

In each regulatory unit, three different restrict enzyme sites are put between the promoter and RBS, target gene and terminator, terminator and next promoter, respectively. (Fig. 2)


Fig. 2 the position of three different restrict enzyme sites

In this way, we can cut the RBS and target gene out which means this route does not have line L (we link the 1, 2 (or 1, 3) to keep the circle structure of plasmid). Then we can test this route by the time of arriving at the ending node which reported by the report gene (we used the GFP).

Second stage:

But we found that it was not easy get so many different good restrict enzyme. So we change the design slightly. (Fig. 3)


Fig. 3 the position of two same restrict enzyme sites

In this way, we can still finish the function of first stage but save my restrict enzymes.

Third stage:

The last problem is how to compare the length of different pathway. Firstly, we decide to build all the possible route and test them. That is to say, we need to cut out 1, 2, 3,…,device(s) so that we can test the entire route, which is named exhaustion, quite time-consuming. Later, we found a better way where we can only cut one device which means we only need to build n graph if the graph has n lines.

Provided that we have 5 lines in a graph like Fig. 4.


Fig. 4 a graph of 5 lines, where node 1 is the beginning node 4 is the ending

According our method to find the shortest path, the first step is to build 6 different systems. Then we tested the “finished time” of each systems. Here is the systems and their “finish time”.

SystemArrayTime (Theoretically)
111112 units
011112 units
101112 units
110112 units
111013 units
11110unreachable

Noticed that without line L5, the ending is inaccessible which implicates the L5 is indispensable. Though it does not block the accessibility, the absence of L4 slows down the arrival, which implies that L4 is on the shortest path. Thus, from the date of the chart above, we can pick out the shortest path is:

Notebook

You should make use of the calendar feature on the wiki and start a lab notebook. This may be looked at by the judges to see how your work progressed throughout the summer. It is a very useful organizational tool as well.

Click here to edit this page! Click here to upload!